Operators

Bounded, compact and unbounded operators and their spectral theory.

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Bounded and compact operators
Unbounded operators and quantum mechanics

Bounded and compact operators

Definition

Bounded linear operators

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Defines the operator norm, proves that a linear map between normed spaces is continuous exactly when it is bounded, collects the algebra of the norm (submultiplicativity, completeness of $\mathcal{B}(X,Y)$ , extension from a dense domain), and computes the norm of the Volterra integral operator exactly.

Prerequisites: Normed and Banach spaces, Hilbert spaces, Orthonormal bases in Hilbert space, L^p spaces, Sturm-Liouville eigenvalue problems

Definition(Bounded operator and operator norm)AdaptedAdapted from Kreyszig (1978), §2.7, Def. 2.7-1 and (2), pp. 91–92; Evans (2010), Appendix D.3a, p. 676; Hassani (2013), Def. 17.2.2 and (17.3), pp. 513–514

Let $X$ and $Y$ be normed spaces over the same field ( $\mathbb{R}$ or $\mathbb{C}$ ) and let $T : \mathcal{D}(T) \to Y$ be linear on a subspace $\mathcal{D}(T) \subseteq X$ . Both norms are written $\|\cdot\|$ ; which one is meant is fixed by what stands inside.

$T$ is bounded if there is a real $c$ with

\|Tx\| \le c\,\|x\| \qquad \text{for every } x \in \mathcal{D}(T).

Equivalently, $T$ carries bounded subsets of $\mathcal{D}(T)$ to bounded subsets of $Y$ ; this is the reason for the word, and it is not the calculus sense, in which a bounded function has bounded range. A non-zero bounded operator never has bounded range, precisely because its domain is a subspace: if $Tx \neq 0$ then $\lambda x \in \mathcal{D}(T)$ for every scalar $\lambda$ and $\|T(\lambda x)\| = |\lambda|\,\|Tx\| \to \infty$ .

The smallest admissible $c$ is the operator norm

\|T\| \;=\; \sup_{\substack{x \in \mathcal{D}(T) \\ x \ne 0}} \frac{\|Tx\|}{\|x\|} \;=\; \sup_{\substack{x \in \mathcal{D}(T)\\ \|x\| = 1}} \|Tx\| \;=\; \sup_{\substack{x \in \mathcal{D}(T)\\ \|x\| \le 1}} \|Tx\| ,

with $\|T\| = 0$ when $\mathcal{D}(T) = \{0\}$ . The three expressions agree by homogeneity (part 1 of the statement below). By construction

\|Tx\| \le \|T\|\,\|x\| \qquad (x \in \mathcal{D}(T)),

and this inequality is the one actually used in computations.

When $\mathcal{D}(T) = X$ we write $T \in \mathcal{B}(X,Y)$ , where $\mathcal{B}(X,Y)$ collects every bounded linear $X \to Y$ ; and $\mathcal{B}(X) = \mathcal{B}(X,X)$ . Following Evans (Evans (2010), Appendix D.3a, p. 676), the null space is $N(T) = \{x : Tx = 0\}$ and the range is $R(T) = \{Tx : x \in X\}$ ; both are subspaces.

The supremum defining $\|T\|$ need not be attained: Stakgold and Holst (Stakgold & Holst (2011), §5.1, Example 4, p. 303) give multiplication by $x$ on $L^2(0,1)$ , for which $\|A\| = 1$ but $\|Au\| < \|u\|$ for every $u \ne 0$ . Accordingly, to prove that a number $m$ is $\|A\|$ one shows both that $\|Au\| \le m\|u\|$ for all $u$ and that some sequence of unit vectors $u_n$ has $\|Au_n\| \to m$ .

Definition(The operator norm and continuity)AdaptedAdapted from Kreyszig (1978), Lemma 2.7-2, pp. 92–93, Thm 2.7-8, p. 96, Thm 2.7-9, pp. 96–98, Cor. 2.7-10 and (7), p. 98, Thm 2.7-11, pp. 100–101; Hassani (2013), Prop. 17.2.8 and (17.5), pp. 515–516; Reed & Simon (1980), Thm III.2, p. 70

Let $X$ , $Y$ , $Z$ be normed spaces and $T : \mathcal{D}(T) \to Y$ linear, $\mathcal{D}(T) \subseteq X$ .

The three formulas agree, and $\|\cdot\|$ is a norm. The quotient, unit-sphere and unit-ball suprema above are equal, and $T \mapsto \|T\|$ is a norm on $\mathcal{B}(X,Y)$ , which is therefore a normed space.
Boundedness equals continuity. $T$ is bounded exactly when it is continuous; and continuity at one single point of $\mathcal{D}(T)$ already forces continuity everywhere on $\mathcal{D}(T)$ .
Consequences. If $T$ is bounded then $x_n \to x$ in $\mathcal{D}(T)$ implies $Tx_n \to Tx$ . If moreover $\mathcal{D}(T) = X$ , that is $T \in \mathcal{B}(X,Y)$ , the null space $N(T)$ is closed in $X$ ; on a proper domain it need not be, as the remark below shows. The range need not be closed even for $T \in \mathcal{B}(X,Y)$ (Reed and Simon (Reed & Simon (1980), §VI.1, p. 185)).
Finite dimension. If $\dim X < \infty$ , every linear operator on $X$ is bounded.
Submultiplicativity. For $S \in \mathcal{B}(Y,Z)$ and $T \in \mathcal{B}(X,Y)$ , $ST \in \mathcal{B}(X,Z)$ with $\|ST\| \le \|S\|\,\|T\|$ ; in particular $\|T^n\| \le \|T\|^n$ for $T \in \mathcal{B}(X)$ and $n \in \mathbb{N}$ .
Completeness. If $Y$ is a Banach space, so is $\mathcal{B}(X,Y)$ , for any normed $X$ . (Completeness of $X$ is not required.)
Extension from a dense domain. If $Y$ is a Banach space, a bounded linear $T : \mathcal{D}(T) \to Y$ has a unique bounded linear extension $\tilde T : \overline{\mathcal{D}(T)} \to Y$ , and $\|\tilde T\| = \|T\|$ . In particular an operator defined on a dense subspace of $X$ extends to all of $X$ with the same norm.

Note(Examples to keep in mind)AdaptedAdapted from Kreyszig (1978), Examples 2.7-3 to 2.7-7, pp. 93–95; Stakgold & Holst (2011), §5.1, Examples 4, 5, 6 and 9 with (5.1.9), pp. 303–306; Hassani (2013), Example 17.2.7, p. 515

Identity and zero. $\|I\| = 1$ on any $X \ne \{0\}$ , and $\|0\| = 0$ .
Matrices. A real $r \times n$ matrix acting on $\mathbb{R}^n$ with the Euclidean norm is bounded, and Kreyszig (Kreyszig (1978), Example 2.7-7, p. 95) gets $\|T\| \le \bigl(\sum_{j,k} a_{jk}^2\bigr)^{1/2}$ from the Cauchy-Schwarz inequality. The bound is usually not sharp; see matrix norms and condition numbers.
Integral operator, continuous kernel. On $C[0,1]$ with the supremum norm, $(Tu)(t) = \int_0^1 k(t,\tau)u(\tau)\,d\tau$ with $k$ continuous on the square satisfies $\|T\| \le \max|k|$ .
Integral operator, square-integrable kernel. On $L^2(a,b)$ , if $\int_a^b\!\!\int_a^b |k(x,y)|^2dx\,dy < \infty$ — a Hilbert-Schmidt kernel — then Cauchy-Schwarz applied in the $y$ variable gives $\|K\| \;\le\; \Bigl[\int_a^b\!\!\int_a^b |k(x,y)|^2\,dx\,dy\Bigr]^{1/2} ,$ which is Stakgold and Holst's (Stakgold & Holst (2011), §5.1, (5.1.9), p. 304) estimate (5.1.9). Their Exercise 5.1.3 notes that Holmgren kernels, satisfying $\max_z \int\!\!\int |k(x,y)k(x,z)|\,dx\,dy = M < \infty$ , are also bounded, with $\|K\| \le M^{1/2}$ , and that on $L^2(-\infty,\infty)$ a convolution kernel $k(x,y) = h(x-y)$ with $h$ integrable is Holmgren without being Hilbert-Schmidt. Compactness of these operators is a separate matter, taken up in compact operators and Hilbert-Schmidt theory.
Right shift. On a separable Hilbert space with orthonormal basis $(e_j)$ , $Ae_j = e_{j+1}$ gives $\|Au\| = \|u\|$ , so $\|A\| = 1$ ; the range is the closed subspace $\{u : (u,e_1) = 0\}$ , which is not dense.
Differentiation is unbounded. On the polynomials in $C[0,1]$ , $x_n(t) = t^n$ has $\|x_n\| = 1$ and $\|x_n'\| = n$ . On $L^2(a,b)$ , Hassani (Hassani (2013), Example 17.2.7, p. 515) uses $f(x) = \sqrt{x-a}$ , which lies in $L^2$ while $f' = \tfrac12 (x-a)^{-1/2}$ does not. Unbounded operators are the subject of unbounded and closed operators.
Unbounded on a dense domain. Multiplication by $1/x$ on $L^2(0,1)$ is defined on the dense subspace of $u$ with $u/x \in L^2$ , and the unit vectors supported on $(\varepsilon,1)$ show it is unbounded there. So part 7 above really does need boundedness: a merely densely defined linear operator has no continuous extension.

Idea of the proofOriginal

Everything here rests on one observation: a linear map is determined by what it does to the unit sphere, because $T(\lambda x) = \lambda\,Tx$ transports the behaviour at one scale to every other scale.

That is why the three formulas for $\|T\|$ agree, and it is why continuity at a single point spreads to the whole domain: a linear map commits the same error near every point that it commits near $0$ , since $Tx - Tx_0 = T(x-x_0)$ . Continuity at $0$ gives a $\delta$ -ball mapped into an $\varepsilon$ -ball; scaling that one statement by $\|y\|/\delta$ turns it into the inequality $\|Ty\| \le (\varepsilon/\delta)\|y\|$ valid for every $y$ . The converse direction is the displayed inequality $\|Tx\| \le \|T\|\,\|x\|$ read as a Lipschitz estimate with constant $\|T\|$ .

Completeness of $\mathcal{B}(X,Y)$ and the extension theorem are the same argument twice: a Cauchy sequence in the operator norm is Cauchy at each point $x$ uniformly in $\|x\| \le 1$ , so the pointwise limits exist in the complete space $Y$ and assemble into a linear map, and uniformity turns pointwise convergence into norm convergence.

Proof(Parts 1 to 5)AdaptedAdapted from Kreyszig (1978), Lemma 2.7-2, pp. 92–93, Thm 2.7-8, p. 96, Thm 2.7-9, pp. 96–98, Cor. 2.7-10, p. 98; Hassani (2013), Prop. 17.2.8, pp. 515–516

Part 1. For $x \ne 0$ put $y = x/\|x\|$ , so $\|y\| = 1$ and, by homogeneity of $T$ and of the norm, $\|Tx\|/\|x\| = \|Ty\|$ . As $x$ runs over the non-zero vectors of $\mathcal{D}(T)$ , $y$ runs over its unit vectors, so the quotient and unit-sphere suprema are equal. The unit-ball supremum is at least the unit-sphere one, since the sphere sits inside the ball; and it is at most the unit-sphere one, because for $0 < \|z\| \le 1$ the unit vector $w = z/\|z\|$ satisfies $\|Tz\| = \|z\|\,\|Tw\| \le \|Tw\|$ , while $z = 0$ contributes $\|T0\| = 0$ . For the norm axioms: $\|T\| \ge 0$ always; $\|T\| = 0$ forces $Tx = 0$ for all $x$ , that is $T = 0$ ; and taking suprema over the unit sphere in $\|(\alpha T)x\| = |\alpha|\,\|Tx\|$ and in $\|(S+T)x\| \le \|Sx\| + \|Tx\|$ gives homogeneity and the triangle inequality.

Part 2, bounded $\Rightarrow$ continuous. Let $x_0 \in \mathcal{D}(T)$ and $\varepsilon > 0$ . If $T = 0$ there is nothing to prove, so let $\|T\| > 0$ and set $\delta = \varepsilon/\|T\|$ . For $\|x - x_0\| < \delta$ , linearity and the displayed inequality give

\|Tx - Tx_0\| = \|T(x-x_0)\| \le \|T\|\,\|x - x_0\| < \|T\|\,\delta = \varepsilon .

Boundedness is what supplies the single Lipschitz constant valid at every $x_0$ at once.

Part 2, continuous at one point $\Rightarrow$ bounded. Suppose $T$ is continuous at some $x_0 \in \mathcal{D}(T)$ . Take $\varepsilon = 1$ and get $\delta > 0$ with $\|Tx - Tx_0\| \le 1$ whenever $x \in \mathcal{D}(T)$ and $\|x - x_0\| \le \delta$ . Given any $y \in \mathcal{D}(T)$ with $y \ne 0$ , the vector $x = x_0 + \delta y/\|y\|$ lies in $\mathcal{D}(T)$ (a subspace) and satisfies $\|x - x_0\| = \delta$ . Hence, using linearity twice,

1 \;\ge\; \|Tx - Tx_0\| \;=\; \Bigl\|T\Bigl(\frac{\delta}{\|y\|}y\Bigr)\Bigr\| \;=\; \frac{\delta}{\|y\|}\,\|Ty\| ,

so $\|Ty\| \le \delta^{-1}\|y\|$ , and $T$ is bounded with $\|T\| \le 1/\delta$ . Linearity is used twice here — once to say that $x$ is again in the domain, and once to pull the scalar out of $T$ — and the implication is false for non-linear maps. Combining the two halves also proves the second sentence of part 2: continuity at one point gives boundedness, which gives continuity everywhere.

Part 3. If $x_n \to x$ with all terms and the limit in $\mathcal{D}(T)$ , then $\|Tx_n - Tx\| \le \|T\|\,\|x_n - x\| \to 0$ . For the null space let $T \in \mathcal{B}(X,Y)$ , let $x \in \overline{N(T)}$ and choose $x_n \in N(T)$ with $x_n \to x$ . Here $\mathcal{D}(T) = X$ is what is used: it places the limit $x$ in the domain, so $Tx$ is defined and the first sentence applies, giving $Tx = \lim_n Tx_n = 0$ . Hence $x \in N(T)$ and $N(T)$ is closed.

Part 4. Let $\{e_1,\dots,e_n\}$ be a basis of $X$ and $x = \sum_j \xi_j e_j$ . Then $\|Tx\| \le \sum_j |\xi_j|\,\|Te_j\| \le \bigl(\max_j \|Te_j\|\bigr)\sum_j|\xi_j|$ . On a finite-dimensional space the map $x \mapsto \sum_j|\xi_j|$ is a norm, and all norms on a finite-dimensional space are equivalent (normed and Banach spaces), so $\sum_j |\xi_j| \le \gamma\|x\|$ for some $\gamma$ independent of $x$ . Hence $\|Tx\| \le \gamma\bigl(\max_j\|Te_j\|\bigr)\|x\|$ . Equivalence of norms is exactly the finite-dimensional input, and it is what fails in infinite dimensions.

Part 5. For $\|x\| \le 1$ , $\|STx\| \le \|S\|\,\|Tx\| \le \|S\|\,\|T\|$ ; take the supremum over the unit ball. Iterating gives $\|T^n\| \le \|T\|^n$ . $\square$

Proof(Parts 6 and 7: completeness and extension)AdaptedAdapted from Reed & Simon (1980), Thm III.2, p. 70; Kreyszig (1978), Thm 2.7-11, pp. 100–101

Part 6. Let $Y$ be complete and let $(T_n)$ be Cauchy in $\mathcal{B}(X,Y)$ .

Step 1: a candidate limit. For each fixed $x$ , $\|T_nx - T_mx\| \le \|T_n - T_m\|\,\|x\|$ , so $(T_nx)$ is Cauchy in $Y$ and converges, because $Y$ is complete; call the limit $Tx$ . Linearity of $T$ follows by passing to the limit in $T_n(\alpha x + \beta y) = \alpha T_nx + \beta T_ny$ , the algebraic operations being continuous.

Step 2: $T$ is bounded. A Cauchy sequence is bounded, say $\|T_n\| \le M$ for all $n$ . Then $\|Tx\| = \lim_n\|T_nx\| \le M\|x\|$ , using continuity of the norm.

Step 3: convergence in the operator norm, not merely pointwise. Let $\varepsilon > 0$ and pick $N$ with $\|T_n - T_m\| \le \varepsilon$ for $n,m \ge N$ . For $\|x\| \le 1$ and $n \ge N$ , $\|T_nx - T_mx\| \le \varepsilon$ for every $m \ge N$ ; letting $m \to \infty$ gives $\|T_nx - Tx\| \le \varepsilon$ . The bound is uniform in $x$ , so $\|T_n - T\| \le \varepsilon$ . The uniformity is the whole point: pointwise limits of bounded operators need not converge in norm.

Part 7. Let $T : \mathcal{D}(T) \to Y$ be bounded and $Y$ complete, and let $x \in \overline{\mathcal{D}(T)}$ .

Step 1: define $\tilde Tx$ . Choose $x_n \in \mathcal{D}(T)$ with $x_n \to x$ . Then $\|Tx_n - Tx_m\| \le \|T\|\,\|x_n - x_m\|$ , so $(Tx_n)$ is Cauchy and converges in $Y$ ; set $\tilde Tx = \lim_n Tx_n$ .

Step 2: the value does not depend on the sequence. If also $z_n \to x$ with $z_n \in \mathcal{D}(T)$ , interleave the two sequences into $x_1, z_1, x_2, z_2,\dots$ , which still converges to $x$ . By step 1 its image converges, and the images of the two subsequences, being subsequences of a convergent sequence, have that same limit.

Step 3: $\tilde T$ extends $T$ , is linear, and is bounded. Taking the constant sequence $x_n = x$ for $x \in \mathcal{D}(T)$ gives $\tilde Tx = Tx$ . Linearity passes to the limit as in part 6. From $\|Tx_n\| \le \|T\|\,\|x_n\|$ and continuity of both norms, $\|\tilde Tx\| \le \|T\|\,\|x\|$ , so $\|\tilde T\| \le \|T\|$ ; and $\|\tilde T\| \ge \|T\|$ because the supremum defining $\|\tilde T\|$ runs over a larger set. Hence $\|\tilde T\| = \|T\|$ .

Step 4: uniqueness. Two continuous maps agreeing on a dense subset of $\overline{\mathcal{D}(T)}$ agree everywhere on it. $\square$

Remark(Where the hypotheses are used)Original

Linearity is indispensable in part 2. For a non-linear map, continuity at a point says nothing about the rest of the domain, and "bounded" reverts to its calculus meaning. Take $f(t) = \sqrt{|t|}$ on $\mathbb{R}$ : it is continuous everywhere, yet no constant $c$ satisfies $|f(t)| \le c|t|$ for all $t$ , since $\sqrt{|t|}\,/\,|t| \to \infty$ as $t \to 0$ . Conversely $f(t) = t^2$ obeys no such inequality either, this time because of large $t$ . Only for linear maps do "continuous", "Lipschitz" and "bounded on the unit ball" collapse into one condition.
Completeness of $Y$ , not of $X$ , is what parts 6 and 7 need. In both proofs the only limit taken in a possibly incomplete space is a limit of images, and that is where $Y$ must be complete. This is why $\mathcal{B}(X,\mathbb{C})$ — the dual space of the Hahn-Banach theorem — is a Banach space for every normed $X$ whatever.
Density, in part 7, is what makes the extension unique. Without it there are many bounded extensions; producing one at all on a non-dense domain, with the same norm and over a general Banach space, is the content of the Hahn-Banach theorem, which is a genuinely harder statement.
Part 4 fails without finite dimension, and the failure is not marginal: differentiation is unbounded on every reasonable infinite-dimensional function space, and multiplication by $1/x$ is unbounded on its natural dense domain in $L^2(0,1)$ . The underlying reason is that the unit ball of an infinite-dimensional normed space is not compact.
The null space is closed but the range need not be. The null space is the preimage of the closed set $\{0\}$ under a continuous map; the range is a forward image, and forward images of closed sets under continuous maps are not closed in general. The operator $(Tu)_j = u_j/j$ on $\ell^2$ is bounded with dense, non-closed range: $(1/j)_j$ is not of the form $(u_j/j)_j$ with $u \in \ell^2$ , since that would need $u_j = 1$ for all $j$ . Closedness of the null space, on the other hand, needs the domain to be all of $X$ (or at least closed), because the proof evaluates $T$ at a limit of points of $N(T)$ . On the subspace $\mathcal{D}(T) \subset \ell^2$ of finitely supported sequences, the functional $Tu = \sum_j u_j/j$ is bounded, with $|Tu| \le (\pi/\sqrt6)\|u\|$ by Cauchy-Schwarz, and its null space is dense in a closed hyperplane of $\ell^2$ without being equal to it, so $N(T)$ is not closed in $\ell^2$ . (Kreyszig (Kreyszig (1978), Cor. 2.7-10(b), p. 98) states the closedness for an arbitrary domain $\mathcal{D}(T)$ ; we state it for $T \in \mathcal{B}(X,Y)$ , which is the case he uses everywhere afterwards.) The consequences of a closed range are drawn in the uniform boundedness and open mapping theorems.
Submultiplicativity is an inequality, not an identity. On $\mathbb{C}^2$ , the matrices $S = \begin{pmatrix}0&0\\0&1\end{pmatrix}$ and $T = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ have $\|S\| = \|T\| = 1$ while $ST = 0$ , so the inequality can be as slack as it gets. The order matters: the same two matrices give $TS = T$ , with $\|TS\| = 1 = \|T\|\,\|S\|$ , so equality occurs too. Equality in $\|ST\| \le \|S\|\,\|T\|$ requires the near-maximisers of $T$ to be near-maximisers of $S$ as well.

Method(Finding an operator norm)Original

To evaluate, or at least bracket, $\|T\|$ :

Get an upper bound by estimating $\|Tx\|$ with the inequality natural to the space: Cauchy-Schwarz for integral and matrix operators on $L^2$ and $\mathbb{R}^n$ , the triangle inequality and $\sup|k|$ for the supremum norm, Hölder's inequality on $L^p$ ( $L^p$ spaces). Any $c$ with $\|Tx\| \le c\|x\|$ for all $x$ gives $\|T\| \le c$ .
Get a matching lower bound by exhibiting one vector, or a sequence of unit vectors $u_n$ , with $\|Tu_n\| \to c$ . If $\|Tu\| = c\|u\|$ for a single $u \ne 0$ , the supremum is attained and $\|T\| = c$ exactly. This two-sided procedure is the one Stakgold and Holst (Stakgold & Holst (2011), §5.1, p. 300) prescribe.
If the two bounds do not meet, diagonalise instead. When $T$ maps an orthonormal basis $(e_n)$ to mutually orthogonal vectors, $Te_n = s_n f_n$ with $(f_n)$ orthonormal, then $\|T\| = \sup_n |s_n|$ : Parseval's identity (orthonormal bases) turns the supremum over the unit sphere into a supremum over a sequence of numbers. This is the route used in the example below, and it is the infinite-dimensional form of the singular value decomposition (SVD).
To prove unboundedness, exhibit $u_n \ne 0$ with $\|Tu_n\|/\|u_n\| \to \infty$ . Monomials of rising degree work for differentiation; indicator functions shrinking onto a singularity work for multiplication by an unbounded function.
To define an operator you only know on nice functions — polynomials, step functions, $C_c^\infty$ — check boundedness there, then invoke part 7 to extend. This is how the Fourier transform is defined on all of $L^2$ (Plancherel's theorem).

Example(The Volterra operator: Hilbert-Schmidt bound against the exact norm)AdaptedAdapted from Stakgold & Holst (2011), §5.1, Example 7, p. 305, and Exercise 5.1.4(a)–(c), pp. 306–307

On $H = L^2(0,1)$ define

(Au)(x) = \int_0^x u(y)\,dy .

This is the integral operator with kernel $k(x,y) = 1$ for $y < x$ and $0$ otherwise.

Step 1: the Hilbert-Schmidt bound. The kernel is square-integrable, with

\int_0^1\!\!\int_0^1 |k(x,y)|^2\,dy\,dx = \int_0^1 x\,dx = \tfrac12 ,

so by the estimate of the note above, $\|A\| \le 1/\sqrt{2} = 0.7071068$ .

Step 2: a basis adapted to $A$ . The family that diagonalises $A$ is the one Stakgold and Holst (Stakgold & Holst (2011), §5.1, Exercise 5.1.4(a), pp. 306–307) single out for this operator; they obtain it by reflecting functions on $(0,1)$ into $(0,2)$ , while we get it, together with its partner family, from a pair of Sturm-Liouville problems. Put $\omega_n = (2n-1)\pi/2$ for $n \in \mathbb{N}$ and

e_n(x) = \sqrt{2}\cos(\omega_n x), \qquad f_n(x) = \sqrt{2}\sin(\omega_n x) .

These are the eigenfunctions of $-v'' = \lambda v$ on $(0,1)$ with $v'(0) = 0,\ v(1) = 0$ and with $v(0) = 0,\ v'(1) = 0$ respectively, so each family is an orthonormal basis of $L^2(0,1)$ by Sturm-Liouville theory. (Orthonormality is also a two-line computation: $2\int_0^1\cos(\omega_n x)\cos(\omega_m x)dx = \delta_{nm}$ because $\omega_n \pm \omega_m$ is an integer multiple of $\pi$ for $n \ne m$ , and $\sin(2\omega_n) = \sin((2n-1)\pi) = 0$ .)

Step 3: $A$ is diagonal in this pair of bases. Integrating,

(Ae_n)(x) = \sqrt{2}\int_0^x \cos(\omega_n y)\,dy = \frac{\sqrt{2}}{\omega_n}\sin(\omega_n x) = \frac{1}{\omega_n}f_n(x) .

So $A$ sends the orthonormal basis $(e_n)$ to the mutually orthogonal vectors $f_n/\omega_n$ , with

s_n = \frac{1}{\omega_n} = \frac{2}{(2n-1)\pi} : \qquad s_1 = \frac{2}{\pi},\quad s_2 = \frac{2}{3\pi},\quad s_3 = \frac{2}{5\pi},\ \dots

Step 4: the exact norm. Write $u = \sum_n c_n e_n$ with $\sum_n |c_n|^2 = \|u\|^2$ (Parseval). By continuity of $A$ , $Au = \sum_n c_n s_n f_n$ , and since $(f_n)$ is orthonormal, Parseval again gives

\|Au\|^2 = \sum_{n=1}^{\infty} s_n^2 |c_n|^2 \;\le\; s_1^2 \sum_{n=1}^{\infty}|c_n|^2 = \frac{4}{\pi^2}\,\|u\|^2 ,

with equality for $u = e_1$ . Hence

\|A\| = \frac{2}{\pi} = 0.6366198 .

Step 5: check. Directly, $u = e_1 = \sqrt{2}\cos(\pi x/2)$ has $\|u\| = 1$ and $Au = (2/\pi)\sqrt{2}\sin(\pi x /2)$ , whose norm is $2/\pi$ ; so the supremum really is attained here, unlike for multiplication by $x$ . Discretising $A$ by the midpoint rule on a uniform grid of $N = 4000$ points, $h = 1/N$ — the matrix with entries $h$ strictly below the diagonal and $h/2$ on it — and computing its largest singular value numerically returns $0.63661976$ , against $2/\pi = 0.63661977$ . (The quadrature matters at this precision: the cruder left-endpoint matrix, with $h$ below the diagonal and $0$ on it, gives $0.6365402$ on the same grid, an $O(h)$ error rather than an $O(h^2)$ one.) The value $2/\pi$ agrees with the one Stakgold and Holst (Stakgold & Holst (2011), §5.1, Example 7, p. 305, and Exercise 5.1.4(c), p. 307) record for this operator.

What the example shows. The Hilbert-Schmidt bound overshoots by the factor $(1/\sqrt2)/(2/\pi) = \pi/(2\sqrt2) = 1.1107$ , about $11\%$ . That is typical: the bound adds the squares of all the numbers $s_n$ , since $\sum_n s_n^2 = (4/\pi^2)\sum_n (2n-1)^{-2} = (4/\pi^2)(\pi^2/8) = 1/2$ , exactly the kernel integral of step 1 — whereas the norm only sees the largest. The bound is sharp precisely when the operator has a single non-zero $s_n$ , that is when it has rank one.

Example(An operator norm that is a supremum and not a maximum)AdaptedAdapted from Stakgold & Holst (2011), §5.1, Example 4, p. 303

On $L^2(0,1)$ let $(Mu)(x) = x\,u(x)$ .

Upper bound. $\|Mu\|^2 = \int_0^1 x^2|u|^2dx \le \int_0^1 |u|^2dx = \|u\|^2$ , so $\|M\| \le 1$ .

Lower bound. For $0 < \varepsilon < 1$ let $u_\varepsilon = \varepsilon^{-1/2}\chi_{(1-\varepsilon,1)}$ , a unit vector. Then

\|Mu_\varepsilon\|^2 = \frac{1}{\varepsilon}\int_{1-\varepsilon}^1 x^2dx = \frac{1 - (1-\varepsilon)^3}{3\varepsilon} = 1 - \varepsilon + \frac{\varepsilon^2}{3} .

At $\varepsilon = 10^{-2}$ the squared norm is $0.99003\overline{3}$ and the norm itself is $\|Mu_\varepsilon\| = 0.9950042$ ; at $\varepsilon = 10^{-4}$ the squared norm is $0.9999000$ and the norm is $0.9999500$ . Letting $\varepsilon \to 0$ gives $\|M\| \ge 1$ , hence $\|M\| = 1$ .

Not attained. If $\|Mu\| = \|u\|$ for some $u$ , then $\int_0^1(1-x^2)|u(x)|^2dx = 0$ with a strictly positive integrand off $x = 1$ , forcing $u = 0$ almost everywhere. So no unit vector achieves the supremum — which is why the recipe of the method asks for a sequence. The operator, the bound and the approximating family supported on $(1-\varepsilon,1)$ are Stakgold and Holst's (Stakgold & Holst (2011), §5.1, Example 4, p. 303); what we add is the normalisation that makes $u_\varepsilon$ a unit vector, the arithmetic carried out explicitly, and a proof of the non-attainment that they state without one; the spectrum of this operator is computed in spectrum and resolvent.

ApplicationOriginal

Quantum mechanics. Observables are operators on $L^2$ , and the bounded ones — projections onto subspaces, spin matrices, bounded potentials acting by multiplication — are the ones for which no domain question arises. Position and momentum are not among them, and both live on $L^2(\mathbb{R})$ , not on a bounded interval; that distinction is the whole point, so it is worth saying exactly where the unboundedness comes from.

Position. The example above computes the multiplication operator $u \mapsto xu$ on $L^2(0,1)$ and finds $\|M\| = 1$ : on a bounded interval the multiplier is a bounded function and the operator is bounded, supremum not attained. The position observable $\hat x$ is the same formula on $L^2(\mathbb{R})$ , where the multiplier is unbounded, and there the operator is too: the unit vectors $u_n = \chi_{(n,\,n+1)}$ give $\|\hat x\,u_n\|^2 = \int_n^{n+1}x^2dx \ge n^2$ , so $\|\hat x\,u_n\| \ge n$ . The mechanism is the one in the last bullet of the note above, multiplication by $1/x$ on a dense domain: what decides a multiplication operator is whether the multiplier is a bounded function. On $(0,1)$ the multiplier $x$ is bounded by $1$ ; on $\mathbb{R}$ it is not.

Momentum. Differentiation is unbounded for a different reason — it amplifies oscillation without limit — and the two bullets above display that on $C[0,1]$ (where $t^n$ has derivative of norm $n$ ) and on $L^2(a,b)$ (Hassani's $\sqrt{x-a}$ , whose derivative leaves the space). On $L^2(\mathbb{R})$ , where $\hat p = -i\hbar\,d/dx$ lives, take $\phi$ smooth, compactly supported, of norm $1$ , and $u_n(x) = e^{inx}\phi(x)$ , again of norm $1$ ; then $\hat p\,u_n = \hbar\,e^{inx}(n\phi - i\phi')$ , so $\|\hat p\,u_n\| \ge \hbar\,(n - \|\phi'\|) \to \infty$ .

Both operators are therefore only densely defined, which is why quantum mechanics needs unbounded and closed operators. Time evolution $e^{-iHt/\hbar}$ is bounded with norm $1$ ; see adjoints, self-adjoint and unitary operators.

Integral equations. Whether $u - \lambda Ku = f$ can be solved by iterating $u_{n+1} = f + \lambda K u_n$ depends on $|\lambda|\,\|K\| < 1$ , so the operator norm is the convergence criterion of the Neumann series (Neumann series). The Hilbert-Schmidt bound is the usual way to check it in practice, and the Volterra example above shows how much margin that bound throws away.

Numerical analysis. The condition number $\|A\|\,\|A^{-1}\|$ of a linear solve, the stability constant of a finite-difference scheme, and the error amplification of a quadrature rule are all operator norms; see matrix norms and condition numbers and numerical linear systems. Stability of a time-stepping scheme is the statement that the norms of the powers $\|T^n\|$ stay bounded, and submultiplicativity is why $\|T\| \le 1$ suffices.

Partial differential equations. Solution operators, trace operators and embeddings between Sobolev spaces are bounded operators, and the constants in the energy estimates are their norms; see energy estimates and weak solutions and Sobolev spaces. The extension theorem (part 7) is used constantly: an operator is first constructed on smooth compactly supported functions, where integration by parts is legitimate, and then extended by density to the whole space.

Signal processing. A linear time-invariant filter is convolution by its impulse response, and its operator norm on $L^2$ is the supremum of the modulus of its transfer function, by Plancherel's theorem. "Gain" in engineering is the operator norm.

Definition

Adjoints, self-adjoint and unitary operators

Adapted

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Builds the Hilbert adjoint $T^*$ from the Riesz representation theorem, proves the algebraic rules including $\|T^*T\| = \|T\|^2$ , and characterises the three classes it defines — self-adjoint, unitary and orthogonal projection operators — together with the elementary consequences a physicist uses, namely real expectation values, norm preservation, and the correspondence between projections and closed subspaces.

Prerequisites: Bounded linear operators, The Riesz representation theorem, The projection theorem, Hilbert spaces, Inner products and norms

Definition(Hilbert adjoint, and the three classes it defines)AdaptedAdapted from Kreyszig (1978), Def. 3.9-1, p. 196, and Def. 3.10-1, p. 201; Reed & Simon (1980), §VI.2, pp. 186–188; Hassani (2013), §17.2.1, p. 517

Throughout, $H$ , $H_1$ , $H_2$ are complex Hilbert spaces and the inner product $(x,y)$ is linear in its first slot and conjugate-linear in its second, so that $(y,x) = \overline{(x,y)}$ .

Let $T \in \mathcal{B}(H_1,H_2)$ . The Hilbert adjoint of $T$ is the operator $T^* : H_2 \to H_1$ determined by

(Tx, y) = (x, T^*y) \qquad \text{for all } x \in H_1,\ y \in H_2 .

Part 1 of the statement below says that exactly one such operator exists. Physicists write $T^{\dagger}$ for $T^*$ and, with the inner product conjugate-linear in the first slot, state the same equation as $\langle Tx\mid y\rangle = \langle x\mid T^{\dagger}y\rangle$ (Hassani (Hassani (2013), §17.2.1, p. 517)); the operator defined is the same.

An operator $T \in \mathcal{B}(H)$ is

self-adjoint (or Hermitian) if $T^* = T$ , that is $(Tx,y) = (x,Ty)$ for all $x,y$ ;
unitary if $T$ is bijective and $T^* = T^{-1}$ ;
normal if $TT^* = T^*T$ ;
positive, written $T \ge 0$ , if $(Tx,x) \ge 0$ for all $x$ .

Self-adjoint and unitary operators are normal; a normal operator need not be either, as $T = 2iI$ shows ( $T^* = -2iI \ne T$ , and $T^{-1} = -\tfrac{i}{2}I \ne T^*$ ).

$P \in \mathcal{B}(H)$ is a projection if $P^2 = P$ , and an orthogonal projection if in addition $P = P^*$ . Following the usual practice, "projection" below means orthogonal projection.

For an unbounded operator one must distinguish symmetric ( $(Au,v) = (u,Av)$ for $u,v$ in the domain) from self-adjoint (the adjoint has the same domain as well); for bounded operators defined on all of $H$ the two coincide, as Stakgold and Holst (Stakgold & Holst (2011), §5.4, p. 317) note. The distinction is the subject of symmetric versus self-adjoint operators.

Definition(Existence and the algebra of the adjoint)AdaptedAdapted from Kreyszig (1978), Thm 3.9-2, pp. 196–197, and Thm 3.9-4, pp. 198–200; Reed & Simon (1980), Thm VI.3, pp. 186–187; Stakgold & Holst (2011), (5.4.1)–(5.4.2) and (5.4.10), pp. 316–317

Let $S, T \in \mathcal{B}(H_1,H_2)$ and $\alpha$ a scalar.

Existence. $T^*$ exists, is unique, is linear and bounded, and $\|T^*\| = \|T\|$ .
Rules. $(S+T)^* = S^* + T^*$ , $(\alpha T)^* = \bar\alpha\,T^*$ , $(T^*)^* = T$ , and $(ST)^* = T^*S^*$ whenever the product is defined. If $T$ has a bounded inverse then so does $T^*$ , with $(T^*)^{-1} = (T^{-1})^*$ . The map $T \mapsto T^*$ is thus a conjugate-linear isometry of $\mathcal{B}(H_1,H_2)$ onto $\mathcal{B}(H_2,H_1)$ .
The $C^*$ identity. $\|T^*T\| = \|TT^*\| = \|T\|^2$ ; in particular $T^*T = 0$ only if $T = 0$ .
Zero-operator lemma. For $Q \in \mathcal{B}(H_1,H_2)$ : $Q = 0$ if and only if $(Qx,y) = 0$ for all $x,y$ . If moreover $H_1 = H_2 = H$ is complex, then $(Qx,x) = 0$ for all $x$ already forces $Q = 0$ .

Definition(The three classes)AdaptedAdapted from Kreyszig (1978), Thm 3.10-3, p. 204, Thm 3.10-4, p. 204, Thm 3.10-5, p. 205, Thm 3.10-6, pp. 205–206, Problems 4 and 12, pp. 207–208, Thm 9.1-1, p. 461, and Thm 9.2-2, pp. 466–467; Reed & Simon (1980), §VI.2, pp. 187–188, and §VI.4, p. 195

Let $H$ be a complex Hilbert space and $T, S, U, V, P \in \mathcal{B}(H)$ .

Self-adjointness is the reality of $(Tx,x)$ . $T$ is self-adjoint if and only if $(Tx,x) \in \mathbb{R}$ for every $x \in H$ . Consequently every positive operator on a complex $H$ is self-adjoint.
Elementary consequences. If $T$ is self-adjoint, its eigenvalues are real and eigenvectors belonging to different eigenvalues are orthogonal; the norm is computable from the quadratic form, $\|T\| \;=\; \sup_{\|x\| = 1} \bigl|(Tx,x)\bigr| .$
Stability of the class. A real linear combination of self-adjoint operators is self-adjoint; $ST$ is self-adjoint if and only if $ST = TS$ ; and a norm limit of self-adjoint operators is self-adjoint. Every $T \in \mathcal{B}(H)$ splits uniquely as $T = T_1 + iT_2$ with $T_1 = \tfrac12(T+T^*)$ and $T_2 = \tfrac{1}{2i}(T - T^*)$ self-adjoint, and $T$ is normal exactly when $T_1T_2 = T_2T_1$ ; these last two are Problems 4 and 12 of Kreyszig (Kreyszig (1978), §3.10, pp. 207–208), proved below.
Unitary operators. If $U$ and $V$ are unitary then $U$ is isometric, $\|Ux\| = \|x\|$ for all $x$ ; $\|U\| = 1$ when $H \ne \{0\}$ ; $U^{-1} = U^*$ is unitary; $UV$ is unitary; and $U$ is normal. Conversely, a bounded linear $T$ on a complex $H$ is unitary if and only if it is both isometric and surjective. An isometry need not be unitary, the right shift on $\ell^2$ being an isometry whose range is a proper closed subspace.
Projections. For $P \in \mathcal{B}(H)$ with $P^2 = P = P^*$ , the range $R(P)$ is a closed subspace, $N(P) = R(P)^{\perp}$ , $P$ acts as the identity on $R(P)$ and as $0$ on $R(P)^{\perp}$ , and $\|P\| = 1$ unless $P = 0$ . Conversely each closed subspace $M \subseteq H$ carries exactly one orthogonal projection $P_M$ , namely the map sending $x$ to its component in $M$ under the decomposition $H = M \oplus M^{\perp}$ of the projection theorem. Closed subspaces and orthogonal projections therefore label one another, each determining the other uniquely.

Idea of the proofOriginal

The adjoint exists for one reason: in a Hilbert space every bounded linear functional is an inner product with a fixed vector. Hold $y$ fixed and look at $x \mapsto (Tx,y)$ . It is linear in $x$ and bounded by $\|T\|\,\|y\|\,\|x\|$ , so the Riesz representation theorem hands us a unique vector to pair $x$ with. Calling that vector $T^*y$ defines the adjoint, and every algebraic rule is then obtained by moving operators across the inner product one at a time and appealing to uniqueness.

The three classes are what one gets by asking how $T$ interacts with the geometry. Self-adjointness says $T$ is "real" with respect to the inner product, and the honest version of that statement is part 1 of the second statement block: $(Tx,x)$ , the quantity a physicist calls an expectation value, is a real number. Unitarity says $T$ preserves the geometry outright: lengths, hence angles, hence orthonormal bases. A projection says $T$ is a piece of the geometry — it names a closed subspace and forgets everything orthogonal to it.

Two proofs deserve advance warning. The $C^*$ identity $\|T^*T\| = \|T\|^2$ comes from reading $\|Tx\|^2 = (T^*Tx,x)$ in both directions, and it is much stronger than the submultiplicative bound. And the converse half of part 1 — $(Tx,x)$ real $\Rightarrow$ $T$ self-adjoint — needs complex scalars, because it is proved by evaluating a quadratic form at $x + y$ and at $x + iy$ and subtracting; a real space has no second evaluation to make.

Proof(Existence, the rules, and the

C^*

identity)AdaptedAdapted from Kreyszig (1978), Thm 3.9-2, pp. 196–197, Lemma 3.9-3, pp. 197–198, Thm 3.9-4, pp. 198–200; Stakgold & Holst (2011), (5.4.1)–(5.4.2), p. 316

Part 1: existence and uniqueness. Fix $y \in H_2$ and consider $\varphi_y : H_1 \to \mathbb{C}$ , $\varphi_y(x) = (Tx,y)$ . It is linear in $x$ because $T$ is linear and the inner product is linear in its first slot, and bounded because $|\varphi_y(x)| \le \|Tx\|\,\|y\| \le \|T\|\,\|y\|\,\|x\|$ by Cauchy-Schwarz. Here $H_1$ must be complete, since the Riesz representation theorem is what we invoke next: there is exactly one $z \in H_1$ with $\varphi_y(x) = (x,z)$ for all $x$ . Define $T^*y = z$ ; the defining identity holds by construction, and uniqueness of $z$ for each $y$ makes $T^*$ well defined. If $S$ also satisfies $(Tx,y) = (x,Sy)$ then $(x, (S - T^*)y) = 0$ for all $x$ ; taking $x = (S-T^*)y$ gives $S = T^*$ .

Linearity. For scalars $\alpha,\beta$ and $y_1,y_2 \in H_2$ and every $x$ ,

\bigl(x, T^*(\alpha y_1 + \beta y_2)\bigr) = (Tx, \alpha y_1 + \beta y_2) = \bar\alpha (Tx,y_1) + \bar\beta(Tx,y_2) = \bigl(x, \alpha T^*y_1 + \beta T^*y_2\bigr) ,

using conjugate-linearity in the second slot twice, so the two vectors agree.

The norm. Putting $x = T^*y$ in the defining identity gives $\|T^*y\|^2 = (T(T^*y), y) \le \|T\|\,\|T^*y\|\,\|y\|$ , whence $\|T^*y\| \le \|T\|\,\|y\|$ and $\|T^*\| \le \|T\|$ . Rule $(T^*)^* = T$ , proved next, then gives $\|T\| = \|(T^*)^*\| \le \|T^*\|$ , so the two norms are equal.

Part 4: the zero-operator lemma (proved here because the rules use it). If $Q = 0$ then certainly $(Qx,y) = 0$ for all $x,y$ . Conversely, taking $y = Qx$ gives $\|Qx\|^2 = 0$ for every $x$ . For the second sentence, let $H$ be complex and $(Qv,v) = 0$ for every $v \in H$ . Expand with $v = x + \alpha y$ :

0 = (Q(x+\alpha y), x + \alpha y) = (Qx,x) + \bar\alpha(Qx,y) + \alpha(Qy,x) + |\alpha|^2 (Qy,y) = \bar\alpha(Qx,y) + \alpha(Qy,x) .

Taking $\alpha = 1$ gives $(Qx,y) + (Qy,x) = 0$ . Taking $\alpha = i$ , so $\bar\alpha = -i$ , gives $-i(Qx,y) + i(Qy,x) = 0$ , that is $(Qx,y) - (Qy,x) = 0$ . Adding the two and halving gives $(Qx,y) = 0$ for all $x,y$ , so $Q = 0$ by the first sentence. The choice $\alpha = i$ is the step that needs complex scalars.

Part 2: the rules. Each is the same move. For additivity, $(x,(S+T)^*y) = ((S+T)x,y) = (Sx,y) + (Tx,y) = (x,S^*y) + (x,T^*y) = (x,(S^*+T^*)y)$ , and uniqueness finishes. For $(\alpha T)^*$ , $(x,(\alpha T)^*y) = (\alpha Tx, y) = \alpha (Tx,y) = \alpha(x,T^*y) = (x,\bar\alpha T^*y)$ — the conjugate appears because the scalar must cross the second slot. For $(T^*)^*$ , note first that the defining identity read backwards is $(T^*y,x) = \overline{(x,T^*y)} = \overline{(Tx,y)} = (y,Tx)$ ; comparing with the definition of $(T^*)^*$ applied to $T^*$ gives $(T^*)^* = T$ . For the composite, $(x,(ST)^*y) = (STx,y) = (Tx,S^*y) = (x,T^*S^*y)$ . For the inverse, apply the composite rule to $T^{-1}T = I = TT^{-1}$ and use $I^* = I$ : $T^*(T^{-1})^* = I = (T^{-1})^*T^*$ .

Part 3: the $C^*$ identity. By Cauchy-Schwarz and part 1,

\|Tx\|^2 = (Tx,Tx) = (T^*Tx, x) \le \|T^*Tx\|\,\|x\| \le \|T^*T\|\,\|x\|^2 ,

so taking the supremum over $\|x\| = 1$ gives $\|T\|^2 \le \|T^*T\|$ . In the other direction submultiplicativity (bounded linear operators) and $\|T^*\| = \|T\|$ give $\|T^*T\| \le \|T^*\|\,\|T\| = \|T\|^2$ . Hence $\|T^*T\| = \|T\|^2$ ; replacing $T$ by $T^*$ and using $(T^*)^* = T$ gives $\|TT^*\| = \|T^*\|^2 = \|T\|^2$ . Finally $T^*T = 0$ forces $\|T\|^2 = 0$ . $\square$

Proof(The three classes)AdaptedAdapted from Kreyszig (1978), Thm 3.10-3, p. 204, Thm 3.10-6, pp. 205–206, Thm 9.1-1, p. 461, Thm 9.2-2, pp. 466–467; Reed & Simon (1980), §VI.2, pp. 187–188, §VI.4, p. 195

Part 1: self-adjointness and the reality of $(Tx,x)$ . Put $Q = T - T^*$ . Using $(T^*x,x) = \overline{(x,T^*x)} = \overline{(Tx,x)}$ ,

(Qx,x) = (Tx,x) - \overline{(Tx,x)} = 2i\,\operatorname{Im}(Tx,x) .

So $(Tx,x)$ is real for every $x$ precisely when $(Qx,x) = 0$ for every $x$ , and on a complex $H$ that is equivalent to $Q = 0$ , that is $T = T^*$ , by part 4 above. If $T \ge 0$ then $(Tx,x)$ is real by assumption, so $T$ is self-adjoint.

Part 2: eigenvalues, eigenvectors and the norm. Let $T = T^*$ . If $Tx = \lambda x$ with $x \ne 0$ , then $\lambda\|x\|^2 = (Tx,x)$ is real by part 1 and $\|x\|^2 > 0$ , so $\lambda \in \mathbb{R}$ . If also $Ty = \mu y$ with $\mu \ne \lambda$ , then, $\mu$ being real,

\lambda (x,y) = (Tx,y) = (x,Ty) = (x,\mu y) = \bar\mu (x,y) = \mu(x,y) ,

so $(\lambda - \mu)(x,y) = 0$ and $(x,y) = 0$ .

For the norm, write $K = \sup_{\|x\|=1}|(Tx,x)|$ . Cauchy-Schwarz gives $|(Tx,x)| \le \|Tx\|\,\|x\| \le \|T\|$ on the unit sphere, so $K \le \|T\|$ . For the reverse, note first that $|(Ty,y)| \le K\|y\|^2$ for every $y$ , by scaling. Take a unit vector $z$ with $Tz \ne 0$ (if there is none, $T = 0$ and there is nothing to prove) and set

v = \|Tz\|^{1/2}z, \qquad w = \|Tz\|^{-1/2}Tz, \qquad y_1 = v+w,\quad y_2 = v-w ,

so $\|v\|^2 = \|w\|^2 = \|Tz\|$ and $(Tv,w) = (Tz,Tz) = \|Tz\|^2$ . Expanding and using self-adjointness, $(Tw,v) = (w,Tv) = \overline{(Tv,w)} = \|Tz\|^2$ , so the cross terms survive and the diagonal ones cancel:

(Ty_1,y_1) - (Ty_2,y_2) = 2\bigl[(Tv,w) + (Tw,v)\bigr] = 4\|Tz\|^2 .

On the other hand, by the triangle inequality, the bound $|(Ty,y)| \le K\|y\|^2$ and the parallelogram law $\|y_1\|^2 + \|y_2\|^2 = 2(\|v\|^2 + \|w\|^2)$ ,

\bigl|(Ty_1,y_1) - (Ty_2,y_2)\bigr| \le K\bigl(\|y_1\|^2 + \|y_2\|^2\bigr) = 2K\bigl(\|v\|^2 + \|w\|^2\bigr) = 4K\|Tz\| .

Comparing, $4\|Tz\|^2 \le 4K\|Tz\|$ , so $\|Tz\| \le K$ ; taking the supremum over unit $z$ gives $\|T\| \le K$ .

Part 3: stability. If $S^* = S$ , $T^* = T$ and $a,b \in \mathbb{R}$ , then $(aS+bT)^* = \bar a S^* + \bar b T^* = aS + bT$ . For the product, $(ST)^* = T^*S^* = TS$ , so $ST$ is self-adjoint exactly when $ST = TS$ . For a norm limit $T_n \to T$ with each $T_n$ self-adjoint, $\|T - T^*\| \le \|T - T_n\| + \|T_n - T_n^*\| + \|T_n^* - T^*\| = 2\|T-T_n\| \to 0$ , using $\|T_n^* - T^*\| = \|(T_n - T)^*\| = \|T_n - T\|$ from part 1 of the previous block; so $T = T^*$ . The splitting $T = T_1 + iT_2$ is checked by computing $T_1^*$ and $T_2^*$ from the rules, and uniqueness follows because $T_1 + iT_2 = S_1 + iS_2$ with all four self-adjoint gives $T_1 - S_1 = i(S_2 - T_2)$ , whose adjoint is $T_1 - S_1 = -i(S_2-T_2)$ ; adding, $T_1 = S_1$ . Finally, expanding with $T_1^* = T_1$ , $T_2^* = T_2$ ,

TT^* = (T_1+iT_2)(T_1-iT_2) = T_1^2 + T_2^2 + i(T_2T_1 - T_1T_2) ,

T^*T = (T_1-iT_2)(T_1+iT_2) = T_1^2 + T_2^2 + i(T_1T_2 - T_2T_1) ,

so $TT^* = T^*T$ if and only if $T_1T_2 = T_2T_1$ .

Part 4: unitary operators. Let $U$ be unitary. Then $\|Ux\|^2 = (Ux,Ux) = (x,U^*Ux) = (x,x) = \|x\|^2$ , so $U$ is isometric, and $\|U\| = 1$ once $H \ne \{0\}$ . Since $U$ is bijective, $(U^{-1})^* = (U^*)^* = U = (U^{-1})^{-1}$ , so $U^{-1}$ is unitary; and $(UV)^* = V^*U^* = V^{-1}U^{-1} = (UV)^{-1}$ , so $UV$ is unitary. Normality is $U^*U = UU^* = I$ .

Conversely let $T$ be isometric and surjective. Isometry gives injectivity, so $T$ is bijective. From $\|Tx\| = \|x\|$ ,

(T^*Tx,x) = (Tx,Tx) = (x,x) = (Ix,x) \quad\text{for all } x ,

so $((T^*T - I)x,x) = 0$ for all $x$ and, $H$ being complex, $T^*T = I$ by part 4 of the previous block. Multiplying on the right by $T^{-1}$ , which exists, gives $T^* = T^{-1}$ .

Part 5: projections. Let $P^2 = P = P^*$ and put $M = R(P)$ . If $x \in M$ , say $x = Pu$ , then $Px = P^2u = Pu = x$ : $P$ is the identity on $M$ . Hence $M = \{x : Px = x\} = N(I - P)$ , which is closed because $I - P$ is bounded (bounded linear operators). For any $x$ and any $u$ , $(x - Px, Pu) = (Px,u) - (P^*Px,u) = (Px,u) - (P^2x,u) = 0$ , so $x - Px \perp M$ ; thus $x = Px + (x - Px)$ is the decomposition of $x$ in $H = M \oplus M^{\perp}$ , which identifies $P$ as the orthogonal projection onto $M$ and gives $N(P) = M^{\perp}$ . By Pythagoras $\|x\|^2 = \|Px\|^2 + \|x - Px\|^2 \ge \|Px\|^2$ , so $\|P\| \le 1$ , with equality as soon as $M \ne \{0\}$ because $P$ fixes the unit vectors of $M$ .

Conversely, given a closed $M$ , the projection theorem writes each $x$ uniquely as $x = m + m^{\perp}$ ; set $P_Mx = m$ . Linearity and $P_M^2 = P_M$ follow from uniqueness of the decomposition, and for $x = m + m^{\perp}$ , $y = n + n^{\perp}$ ,

(P_Mx, y) = (m, n + n^{\perp}) = (m,n) = (m + m^{\perp}, n) = (x, P_My) ,

so $P_M^* = P_M$ . Distinct closed subspaces give distinct projections because $M = R(P_M)$ , so the correspondence is one-to-one. $\square$

Remark(Where the hypotheses are used)Original

Completeness of $H_1$ is used once, in the existence proof, through the Riesz representation theorem. On an inner-product space that is not complete, a bounded operator can fail to have an adjoint, because the functional $x \mapsto (Tx,y)$ may have no representing vector.
Complex scalars are needed in two places, and both are genuine. The implication " $(Tx,x)$ real $\Rightarrow$ $T$ self-adjoint" fails on $\mathbb{R}^2$ : Kreyszig's (Kreyszig (1978), §3.9, p. 198) counterexample is the rotation $Q$ through a right angle, for which $Qx \perp x$ and hence $(Qx,x) = 0$ for every $x$ , although $Q \ne 0$ . (On a real space $(Tx,x)$ is automatically real, so the criterion has no content there.) The same gap closes the converse half of part 4 of the second statement: the argument that an isometric surjection is unitary runs through the same lemma.
Surjectivity is what separates "unitary" from "isometric". The right shift $S$ on $\ell^2$ has $S^*S = I$ , so it preserves norms and inner products, yet $SS^* \ne I$ : $S^*$ is the left shift, and $SS^*$ is the projection killing the first coordinate. An isometry of a finite-dimensional space is automatically surjective, which is why the distinction is invisible for matrices.
Bounded plus everywhere-defined is what makes "symmetric" and "self-adjoint" the same. For an unbounded operator the condition $(Au,v) = (u,Av)$ on a domain $D_A$ says only that $A^*$ extends $A$ ; self-adjointness additionally pins down the domain of $A^*$ . Stakgold and Holst's (Stakgold & Holst (2011), §5.4, Example 1, pp. 318–319) $T_1u = u'$ with $u(0) = 0$ on $L^2(0,1)$ has $T_1^*v = -v'$ with $v(1) = 0$ : both the sign and the boundary condition move. See symmetric versus self-adjoint operators.
"Product of self-adjoint operators" needs commutativity, not merely self-adjointness. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.4, Example 5.4.3(c), p. 286) make the point in the physics setting: $AB$ is an observable only when $A$ and $B$ commute. For $A = \hat x$ and $B = \hat p$ the product is not self-adjoint, while the symmetrised combination $\tfrac12(\hat x\hat p + \hat p\hat x)$ is.
$P^2 = P$ alone is weaker than $P^2 = P = P^*$ . On $\mathbb{C}^2$ the matrix $\begin{pmatrix}1&1\\0&0\end{pmatrix}$ is idempotent, so it is a projection in the algebraic sense, but it is not self-adjoint; it projects along a direction that is not orthogonal to its range, and its norm is $\sqrt2 > 1$ . The identity $\|P\| = 1$ is therefore a statement about orthogonality, not about idempotence.
The $C^*$ identity is strictly stronger than submultiplicativity, and it is what makes $\|T\|$ recoverable from the self-adjoint operator $T^*T$ . It is also the reason the operator norm cannot be replaced by, say, the Hilbert-Schmidt norm in these arguments: that norm is submultiplicative but does not satisfy $\|T^*T\| = \|T\|^2$ .

Method(Computing an adjoint)Original

Write down $(Tu,v)$ , push $T$ across the inner product by whatever identity the space provides, and read off what acts on $v$ .

Matrices. In an orthonormal basis, the matrix of $T^*$ is the conjugate transpose of that of $T$ : $a^*_{ij} = \overline{a_{ji}}$ (Stakgold and Holst (Stakgold & Holst (2011), §5.4, (5.4.3), p. 316)). Self-adjoint means Hermitian, unitary means $U^{\dagger}U = I$ ; see Hermitian matrices and unitary and orthogonal matrices. In a non-orthonormal basis this is false — the Gram matrix intervenes.
Integral operators. For $(Ku)(x) = \int_a^b k(x,y)u(y)\,dy$ on $L^2(a,b)$ , interchanging the order of integration (legitimate by Fubini's theorem when $k$ is square-integrable) gives $k^*(x,y) = \overline{k(y,x)} ,$ so $K$ is self-adjoint exactly when $k(x,y) = \overline{k(y,x)}$ — for a real kernel, exactly when it is symmetric. Thus the Volterra kernel $k(x,y) = H(x-y)$ of bounded linear operators has adjoint kernel $H(y-x)$ , that is $(A^*z)(x) = \int_x^1 z(y)\,dy$ .
Shift and diagonal operators. Read off the adjoint on an orthonormal basis from $(T^*e_j, e_i) = \overline{(Te_i,e_j)}$ . The right shift's adjoint is the left shift; the adjoint of $u \mapsto (m_ju_j)_j$ on $\ell^2$ is $u \mapsto (\overline{m_j}u_j)_j$ , which is self-adjoint exactly for real $m_j$ .
Differential operators. Integrate by parts and keep the boundary terms: they decide the domain of the adjoint, and a formally self-adjoint expression may still fail to be self-adjoint. This is the route of adjoint and self-adjoint differential operators.
To check unitarity, verify $\|Tx\| = \|x\|$ (equivalently $T^*T = I$ ) and surjectivity. Checking only the first gives an isometry.
To recognise a projection, verify $P^2 = P$ and $P = P^*$ ; then $R(P)$ is the subspace it projects onto and $\|P\| = 1$ .

Example(A self-adjoint integral operator, diagonalised by hand)Original

On $H = L^2(0,1)$ let

(Ku)(x) = \int_0^1 (x+y)\,u(y)\,dy .

Step 1: $K$ is self-adjoint. The kernel $k(x,y) = x+y$ is real and symmetric, so $\overline{k(y,x)} = y + x = k(x,y)$ and rule 2 of the method gives $K^* = K$ . It is bounded, by the Hilbert-Schmidt estimate of bounded linear operators:

\|K\| \le \Bigl[\int_0^1\!\!\int_0^1 (x+y)^2dx\,dy\Bigr]^{1/2} = \Bigl[\tfrac13 + 2\cdot\tfrac14 + \tfrac13\Bigr]^{1/2} = \sqrt{7/6} = 1.0801234 .

Step 2: $K$ has rank two. Writing $a = \int_0^1 u$ and $b = \int_0^1 y\,u(y)\,dy$ ,

(Ku)(x) = a\,x + b ,

so the range lies in $\operatorname{span}\{1,x\}$ , and $N(K) = \{u : a = b = 0\} = \operatorname{span}\{1,x\}^{\perp}$ .

Step 3: the eigenvalue problem. Look for $u = \alpha + \beta x$ with $Ku = \lambda u$ . Then $a = \alpha + \beta/2$ and $b = \alpha/2 + \beta/3$ , and matching the constant and linear coefficients of $Ku = b + ax$ against $\lambda\alpha + \lambda\beta x$ gives

\lambda\begin{pmatrix}\alpha\\ \beta\end{pmatrix} = M\begin{pmatrix}\alpha\\ \beta\end{pmatrix}, \qquad M = \begin{pmatrix} 1/2 & 1/3 \\ 1 & 1/2 \end{pmatrix} .

(The matrix is not symmetric because $\{1,x\}$ is not orthonormal; self-adjointness of $K$ is a statement about the inner product, not about this array.) Its trace is $1$ and its determinant is $\tfrac14 - \tfrac13 = -\tfrac{1}{12}$ , so $\lambda^2 - \lambda - \tfrac1{12} = 0$ and

\lambda_{1,2} = \frac{1}{2} \pm \frac{1}{\sqrt3}: \qquad \lambda_1 = 1.0773503, \qquad \lambda_2 = -0.0773503 .

Both are real, as part 2 of the second statement requires. The eigenvectors come from $(\tfrac12 - \lambda)\alpha + \tfrac13\beta = 0$ , that is $\beta = \pm\sqrt3\,\alpha$ :

u_1(x) = 1 + \sqrt3\,x \ \ (\lambda_1), \qquad u_2(x) = 1 - \sqrt3\,x \ \ (\lambda_2) .

Step 4: orthogonality, as predicted. The eigenvalues differ, so the eigenvectors must be orthogonal, and they are:

(u_1,u_2) = \int_0^1 (1+\sqrt3 x)(1-\sqrt3 x)\,dx = \int_0^1 (1 - 3x^2)\,dx = 1 - 1 = 0 .

Their norms are $\|u_1\|^2 = \int_0^1(1+\sqrt3x)^2dx = 1 + \sqrt3 + 1 = 2+\sqrt3 = 3.7320508$ and, likewise, $\|u_2\|^2 = 2 - \sqrt3 = 0.2679492$ .

Step 5: the norm of $K$ . Put $\hat u_j = u_j/\|u_j\|$ . Then $\{\hat u_1,\hat u_2\}$ is an orthonormal basis of $\operatorname{span}\{1,x\}$ and $H = \operatorname{span}\{\hat u_1,\hat u_2\}\oplus N(K)$ . For $u = c_1\hat u_1 + c_2\hat u_2 + w$ with $w \in N(K)$ ,

\|Ku\|^2 = \lambda_1^2|c_1|^2 + \lambda_2^2|c_2|^2 \le \lambda_1^2\bigl(|c_1|^2 + |c_2|^2\bigr) \le \lambda_1^2\|u\|^2 ,

with equality at $u = \hat u_1$ . Hence

\|K\| = \lambda_1 = \frac12 + \frac{1}{\sqrt3} = 1.0773503 ,

which is $99.74\%$ of the Hilbert-Schmidt bound $\sqrt{7/6}$ of step 1 — a much tighter bound than for the Volterra operator, because here one eigenvalue dominates the other by a factor of nearly $14$ .

Step 6: checks. The quadratic-form formula of part 2 agrees: $(K\hat u_1,\hat u_1) = \lambda_1 = 1.0773503$ , so the supremum $\sup_{\|u\|=1}|(Ku,u)|$ is at least $\|K\|$ , and by the theorem it equals it. Directly, at $x = 0.4$ , $\int_0^1(0.4+y)(1+\sqrt3y)\,dy = 1.8237604$ and $\lambda_1 u_1(0.4) = 1.0773503 \times 1.6928203 = 1.8237604$ . Discretising the kernel on a $3000$ -point grid and diagonalising numerically returns eigenvalues $1.0773503$ and $-0.0773503$ with all others of size $10^{-15}$ , confirming both the values and the rank.

Example(A unitary operator and a projection on the same space)Original

Still on $L^2(0,1)$ , with $(e_n)_{n\ge1}$ the orthonormal basis $e_n(x) = \sqrt2\cos\bigl((2n-1)\pi x/2\bigr)$ used in bounded linear operators.

A unitary operator. Let $(Uu)(x) = e^{2\pi i x}u(x)$ . Then $|e^{2\pi ix}| = 1$ gives $\|Uu\| = \|u\|$ , so $U$ is isometric; and $U$ is surjective because $v \mapsto e^{-2\pi ix}v$ inverts it. By part 4 of the second statement $U$ is unitary, and indeed $(U^*v)(x) = \overline{e^{2\pi ix}}\,v(x) = e^{-2\pi ix}v(x)$ , which is $U^{-1}$ . Note $U$ is not self-adjoint: $(Uu,u) = \int_0^1 e^{2\pi ix}|u(x)|^2dx$ is not real in general — for $u \equiv 1$ it is $0$ , but for $u(x) = \sqrt{2}\,\chi_{(0,1/2)}(x)$ it is $2\int_0^{1/2}e^{2\pi ix}dx = 2i/\pi = 0.6366198\,i$ , which is purely imaginary and non-zero.

A projection. Let $P$ be the orthogonal projection onto $M = \operatorname{span}\{e_1,e_2\}$ , that is $Pu = (u,e_1)e_1 + (u,e_2)e_2$ . Then $P^2 = P = P^*$ by direct computation, $\|P\| = 1$ , and for the particular function $u \equiv 1$ ,

(u,e_n) = \sqrt2\int_0^1\cos\bigl(\omega_n x\bigr)dx = \frac{\sqrt2}{\omega_n}\sin\omega_n = \frac{\sqrt2\,(-1)^{n+1}}{\omega_n}, \qquad \omega_n = \frac{(2n-1)\pi}{2} ,

so $(u,e_1) = 2\sqrt2/\pi = 0.9003163$ and $(u,e_2) = -2\sqrt2/(3\pi) = -0.3001054$ . Hence $\|Pu\|^2 = 0.8105695 + 0.0900633 = 0.9006328$ and $\|u - Pu\| = \sqrt{1 - 0.9006328} = 0.3152257$ : two terms of the expansion already capture $90.06\%$ of the energy of the constant function. Parseval's identity (orthonormal bases) confirms the total, since $\sum_{n\ge1} 8/\bigl((2n-1)^2\pi^2\bigr) = (8/\pi^2)(\pi^2/8) = 1 = \|u\|^2$ .

ApplicationOriginal

Quantum mechanics. An observable is a self-adjoint operator, and part 1 of the second statement is the reason: the expectation value $\langle A\rangle = (A\psi,\psi)$ of an observable in a normalised state must be a real number, and self-adjointness is exactly the condition that makes it so for every state — this is Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), §5.4, p. 283) opening argument. Reality of eigenvalues says the possible measurement outcomes are real; orthogonality of eigenvectors for distinct eigenvalues says distinguishable outcomes correspond to orthogonal states. A measurement projects the state onto an eigenspace, which is an orthogonal projection, and $\|P\psi\|^2$ is the probability of the outcome. Time evolution $U(t) = e^{-iHt/\hbar}$ is unitary precisely because $H$ is self-adjoint, so probability is conserved; see Stone's theorem and unitary groups. The position and momentum operators are shown self-adjoint on $L^2(\mathbb{R})$ by Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.4, Example 5.4.1, pp. 284–285) — for $p = -i\,d/dx$ the integration by parts must have vanishing boundary terms, which is the first sign that domains will matter.

Changes of basis. A transformation between two orthonormal bases is a unitary operator with matrix $u_{\nu\mu} = (\varphi'_{\nu},\varphi_\mu)$ , and $U^{\dagger}U = I$ is exactly the statement that going to the new basis and back is the identity (Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.5, pp. 287–288)). The same fact is why the Fourier transform, being unitary on $L^2$ by Plancherel's theorem, preserves energy, and why discrete orthogonal transforms used in signal compression are numerically stable.

Least squares and the normal equations. For $A \in \mathcal{B}(H_1,H_2)$ , minimising $\|Ax - b\|$ leads to $A^*Ax = A^*b$ , because the residual must be orthogonal to the range of $A$ ; the operator $A^*A$ is self-adjoint and positive, and $\|A^*A\| = \|A\|^2$ is the statement that the condition number squares. See projections and least squares and the singular value decomposition.

Boundary value problems. A Sturm-Liouville operator is made self-adjoint by the right boundary conditions, and that self-adjointness is what delivers real eigenvalues, orthogonal eigenfunctions and a usable expansion; see Sturm-Liouville eigenvalue problems and eigenfunction expansions. The same mechanism in integral form, with a symmetric kernel, is Hilbert-Schmidt theory; the general statement is the spectral theorem for compact self-adjoint operators.

Statistics. The covariance operator of a square-integrable random field is self-adjoint and positive, so its eigenfunctions are orthogonal and real-eigenvalued; that is the principal-component decomposition, and in finite dimensions it is the spectral theorem for the covariance matrix.

Definition

Spectrum and resolvent

Adapted

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Splits the complex plane into the resolvent set and the point, continuous and residual spectrum of a bounded operator, shows the spectrum is a non-empty compact subset of the disc of radius $\|T\|$ , and proves that a bounded self-adjoint operator has real spectrum and no residual spectrum.

Prerequisites: Adjoints, self-adjoint and unitary operators, Bounded linear operators, Normed and Banach spaces, Hilbert spaces, The projection theorem, The Riesz representation theorem, The Hahn-Banach theorem, Liouville's theorem and the maximum modulus principle

Definition(Resolvent set and the three parts of the spectrum)AdaptedAdapted from Kreyszig (1978), §7.2, Def. 7.2-1 and the table, pp. 370–371

Let $X \ne \{0\}$ be a complex normed space and $T : \mathcal{D}(T) \to X$ linear, with $\mathcal{D}(T) \subseteq X$ . For $\lambda \in \mathbb{C}$ write

T_\lambda = T - \lambda I , \qquad R_\lambda(T) = T_\lambda^{-1} \ \text{ (when it exists)} ,

and call $R_\lambda(T)$ , or simply $R_\lambda$ , the resolvent of $T$ at $\lambda$ . Its domain is the range of $T_\lambda$ . We follow Kreyszig (Kreyszig (1978), §7.2, p. 370) in using $T - \lambda I$ ; some authors use $\lambda I - T$ , and the older integral-equation literature uses $(I - \mu T)^{-1}$ , so the sign and the role of $\lambda$ must be checked before comparing formulas between books.

$\lambda$ is a regular value of $T$ when all three hold:

(R1) $R_\lambda$ exists;
(R2) $R_\lambda$ is bounded;
(R3) $R_\lambda$ is defined on a dense subset of $X$ .

The resolvent set $\rho(T)$ is the set of regular values; the spectrum is $\sigma(T) = \mathbb{C}\setminus\rho(T)$ . It splits into three disjoint pieces:

	(R1)	(R2)	(R3)	$\lambda$ lies in
regular	yes	yes	yes	$\rho(T)$
point spectrum	no	—	—	$\sigma_p(T)$
continuous spectrum	yes	no	yes	$\sigma_c(T)$
residual spectrum	yes	—	no	$\sigma_r(T)$

Point spectrum. (R1) fails exactly when $T_\lambda x = 0$ for some $x \ne 0$ , that is when $\lambda$ is an eigenvalue with eigenvector $x$ . The eigenspace is $N(T_\lambda)$ and its dimension is the (geometric) multiplicity of $\lambda$ .
Continuous spectrum. $T_\lambda$ is injective with dense range, but the inverse is unbounded: the equation $T_\lambda x = y$ is solvable for a dense set of data, and the solution does not depend continuously on them.
Residual spectrum. $T_\lambda$ is injective but its range misses a whole closed subspace: there is a $y \ne 0$ with $y \perp R(T_\lambda)$ (in a Hilbert space), so a set of data of positive "thickness" admits no solution even approximately.

The four sets are disjoint and cover $\mathbb{C}$ . In finite dimensions $\sigma_c = \sigma_r = \varnothing$ and the spectrum is exactly the set of eigenvalues, because an injective linear map of a finite-dimensional space onto itself is surjective with bounded inverse; Hassani (Hassani (2013), §17.3, p. 518) emphasises that it is precisely this implication which fails in infinite dimensions. See eigenvalues and eigenvectors for the finite-dimensional theory.

Note(Three other conventions in the library)AdaptedAdapted from Reed & Simon (1980), §VI.3, p. 188, and the proposition on p. 194; Stakgold & Holst (2011), §5.6, pp. 326–327; Evans (2010), Appendix D.3a, p. 677; Hassani (2013), Def. 17.3.1, p. 517

The definition above is Kreyszig's (Kreyszig (1978), Def. 7.2-1, p. 371), and it is the one used throughout this unit. Three others appear in the books cited here; they are not in conflict about any concrete operator, but their words differ, so a result must be read with its author's definition in hand.

Bijection with bounded inverse. Reed and Simon (Reed & Simon (1980), §VI.3, p. 188) put $\lambda \in \rho(T)$ when $\lambda I - T$ is a bijection of $X$ with a bounded inverse, and note that for $T \in \mathcal{B}(X)$ with $X$ complete the boundedness is automatic, by the inverse mapping theorem (open mapping and bounded inverse). Hassani (Hassani (2013), Def. 17.3.1, p. 517) asks only that $(T - \lambda\mathbf 1)^{-1}$ exist and be bounded. His "exists" has to be read as exists as a two-sided inverse defined on all of $H$ , which is how he uses it: in Example 17.3.2, p. 518, he disqualifies a $\lambda$ from $\rho(T_r)$ precisely by showing that $T_r - \lambda\mathbf 1$ is not onto. Read literally — an inverse on the range alone — the right shift would be regular at $\lambda = 0$ , since it is injective with $\|Su\| = \|u\|$ , whereas the example below puts $0$ in the residual spectrum. (A related slip: that same Example 17.3.2 reads off $\sigma(T_r) = \{0 < |\lambda| \le 1\}$ , with $\lambda = 0$ omitted, which also contradicts his own Prop. 17.4.7, p. 521, that the spectrum is closed.) With that reading, all three definitions pick out the same set $\rho(T)$ for a bounded operator on a Banach space: by Lemma 7.2-3 of Kreyszig (Kreyszig (1978), §7.2, p. 373), (R1)–(R3) already force $R_\lambda$ to be defined on all of $X$ , and conversely a bounded two-sided inverse on all of $X$ satisfies (R1)–(R3) at once. They can differ for unbounded operators, which is why unbounded and closed operators restates the definition in that setting.
Compression spectrum instead of residual. Stakgold and Holst (Stakgold & Holst (2011), §5.6, p. 327) classify by the state of $T_\lambda$ and use three sets that are deliberately not disjoint: the point spectrum, the continuous spectrum (inverse exists but is unbounded) and the compression spectrum ( $R(\lambda) \ne H$ , the codimension of the range being the deficiency of $\lambda$ ). Point and continuous spectrum are disjoint, and their union is the approximate point spectrum, characterised by the existence of unit vectors $u_n$ with $\|Au_n - \lambda u_n\| \to 0$ . Their Theorem 5.6.1 says that $\lambda$ has deficiency $m$ in the compression spectrum exactly when $\bar\lambda$ is an eigenvalue of $A^*$ of multiplicity $m$ — their printed statement has $\lambda$ rather than $\bar\lambda$ there, while their own one-line proof, $R(A-\lambda I)^{\perp} = N\bigl((A-\lambda I)^*\bigr) = N(A^* - \bar\lambda I)$ , delivers the conjugate, which is the form we use and the form Reed and Simon (Reed & Simon (1980), §VI.3, p. 194) state; the two agree in the only case the book applies the theorem, $\lambda$ real; in Kreyszig's language, a $\lambda$ in the residual spectrum is a $\lambda$ in the compression spectrum but not the point spectrum. Reed and Simon (Reed & Simon (1980), §VI.3, p. 194) record the same duality: if $\lambda$ lies in the residual spectrum of $T$ then $\bar\lambda$ lies in the point spectrum of $T^*$ .
Real scalars. Evans (Evans (2010), Appendix D.3a, p. 677) works on a real Banach space and defines $\rho(A) = \{\eta \in \mathbb{R} : A - \eta I \text{ is one-to-one and onto}\}$ , $\sigma(A) = \mathbb{R}\setminus\rho(A)$ , invoking the closed graph theorem for boundedness of the inverse. This is the convention of partial differential equations, where the operators are real and self-adjoint and nothing is lost. It is not interchangeable with the complex definition: over $\mathbb{R}$ a rotation of the plane has empty spectrum, whereas over $\mathbb{C}$ no bounded operator does (part 4 of the statement below).

Definition(The geometry of the resolvent set)AdaptedAdapted from Kreyszig (1978), Thm 7.3-1, p. 375, Thm 7.3-2, p. 376, Thm 7.3-3, p. 377, Thm 7.3-4, pp. 377–378, Thm 7.5-4, pp. 390–391; Reed & Simon (1980), Thm VI.5 with (VI.1), pp. 190–191, and its corollary, p. 191; Stakgold & Holst (2011), Lemma 5.9.1, pp. 348–349

Let $X$ be a complex Banach space and $T \in \mathcal{B}(X)$ .

Neumann series. If $\|T\| < 1$ then $I - T$ is invertible, $(I-T)^{-1} \in \mathcal{B}(X)$ is given by the norm-convergent series $\sum_{j\ge0}T^j$ , and $\bigl\|(I-T)^{-1}\bigr\| \;\le\; \frac{1}{1 - \|T\|} .$
$\rho(T)$ is open and $R_\lambda$ is analytic on it. If $\lambda_0 \in \rho(T)$ then every $\lambda$ with $|\lambda - \lambda_0| < 1/\|R_{\lambda_0}\|$ lies in $\rho(T)$ and $R_\lambda = \sum_{j=0}^{\infty}(\lambda - \lambda_0)^{\,j}\,R_{\lambda_0}^{\,j+1} ,$ absolutely convergent in $\mathcal{B}(X)$ on that disc. Hence $\sigma(T)$ is closed. For $\lambda,\mu \in \rho(T)$ the resolvents commute and satisfy the first resolvent identity $R_\lambda - R_\mu = (\lambda - \mu)\,R_\lambda R_\mu .$ (Reed and Simon (Reed & Simon (1980), (VI.1), p. 190) write the right-hand side with the opposite sign, because their resolvent is $(\lambda I - T)^{-1}$ .)
$\sigma(T)$ is compact. Every $\lambda$ with $|\lambda| > \|T\|$ is regular, with $R_\lambda = -\lambda^{-1}\sum_{j\ge0}(T/\lambda)^j$ ; so $\sigma(T)$ lies in the closed disc of radius $\|T\|$ and, being closed and bounded, is compact.
$\sigma(T)$ is not empty, provided $X \ne \{0\}$ .

Definition(Bounded self-adjoint operators)AdaptedAdapted from Kreyszig (1978), Thm 9.1-2, pp. 461–463, Thm 9.1-3, pp. 463–464, Thm 9.2-4, p. 468; Reed & Simon (1980), Thm VI.8, p. 194; Stakgold & Holst (2011), Thm 5.6.2 and Thm 5.6.3, pp. 327–328

Let $H \ne \{0\}$ be a complex Hilbert space and $T \in \mathcal{B}(H)$ self-adjoint, $T = T^*$ .

A quantitative criterion for regularity. $\lambda \in \rho(T)$ if and only if there is a $c > 0$ with $\|T_\lambda x\| \ge c\,\|x\| \qquad \text{for every } x \in H .$
The spectrum is real. $\sigma(T) \subset \mathbb{R}$ . More precisely, if $\lambda = \alpha + i\beta$ with $\beta \ne 0$ then $\|T_\lambda x\| \ge |\beta|\,\|x\|$ , so $\lambda \in \rho(T)$ and $\|R_\lambda\| \le 1/|\beta|$ .
There is no residual spectrum. $\sigma_r(T) = \varnothing$ , so $\sigma(T) = \sigma_p(T)\,\dot\cup\,\sigma_c(T)$ , a disjoint union of real sets.
Eigenvalues of $T$ are real and eigenvectors belonging to distinct eigenvalues are orthogonal; this was proved in adjoints, self-adjoint and unitary operators and is used below.

A self-adjoint operator may have no eigenvalues at all — the multiplication operator of the first example has $\sigma_p = \varnothing$ — so part 4 is not a substitute for parts 2 and 3.

Note(Three further facts, used but not proved here)AdaptedAdapted from Kreyszig (1978), Def. 7.3-5 and (11), p. 378, Thm 7.5-5, p. 391, Thm 9.2-1, pp. 465–466, Thm 9.2-3, p. 467; Reed & Simon (1980), Thm VI.6, p. 192, Thm VI.7, p. 192

Spectral radius. $r_\sigma(T) = \sup\{|\lambda| : \lambda \in \sigma(T)\}$ satisfies $r_\sigma(T) \le \|T\|$ by part 3 above, and in fact Gelfand's formula $r_\sigma(T) = \lim_{n\to\infty}\|T^n\|^{1/n}$ holds (Kreyszig (Kreyszig (1978), Thm 7.5-5, p. 391); Reed and Simon (Reed & Simon (1980), Thm VI.6, p. 192)). Both obtain it by identifying $r_\sigma(T)^{-1}$ with the radius within which the expansion of part 3, read as a Laurent expansion at $\lambda = \infty$ , converges; that argument belongs to complex analysis rather than operator theory, and we do not reproduce it. For self-adjoint $A$ it gives $r_\sigma(A) = \|A\|$ : the $C^*$ identity of adjoints gives $\|A^2\| = \|A\|^2$ , hence $\|A^{2^k}\| = \|A\|^{2^k}$ , and the limit along that subsequence is $\|A\|$ . So for a self-adjoint operator the spectrum reaches the boundary of the disc of part 3, while for the nilpotent matrix $\bigl(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\bigr)$ , with $\|T\| = 1$ and $\sigma(T) = \{0\}$ , it does not.
Location between the extremes of the quadratic form. With $m = \inf_{\|x\|=1}(Tx,x)$ and $M = \sup_{\|x\|=1}(Tx,x)$ , Kreyszig (Kreyszig (1978), Thm 9.2-1, pp. 465–466, and Thm 9.2-3, p. 467) shows $\sigma(T) \subseteq [m,M]$ for self-adjoint $T$ , and that both endpoints belong to $\sigma(T)$ ; combined with $\|T\| = \sup_{\|x\|=1}|(Tx,x)|$ from adjoints, this is the same statement as $r_\sigma(T) = \|T\|$ with the sign resolved.
Adjoints. Reed and Simon's (Reed & Simon (1980), Thm VI.7, p. 192) theorem of Phillips gives $\sigma(T^*) = \{\bar\lambda : \lambda \in \sigma(T)\}$ and $R_\lambda(T^*) = R_{\bar\lambda}(T)^*$ in the Hilbert-space case. It is what makes the shift example below computable from its adjoint.

Idea of the proofOriginal

Three ideas carry the whole unit.

Invertibility is an open condition, because of the geometric series. If $\|T\| < 1$ then $\sum T^j$ converges and inverts $I - T$ , exactly as $(1-t)^{-1} = \sum t^j$ for a number. Writing $T - \lambda I = (T - \lambda_0 I)\bigl[I - (\lambda-\lambda_0)R_{\lambda_0}\bigr]$ turns "invertible at $\lambda_0$ " into "invertible near $\lambda_0$ ", which is why $\rho(T)$ is open; and factoring out $\lambda$ instead turns " $|\lambda|$ large" into "invertible", which is why $\sigma(T)$ is bounded.

Non-emptiness is Liouville's theorem. The same series shows $R_\lambda \to 0$ as $|\lambda| \to \infty$ . If the spectrum were empty, $R_\lambda$ would be an entire, bounded, operator-valued function, hence constant and hence $0$ — but $0$ is not invertible. The spectrum exists because an entire bounded function is constant.

Self-adjointness pushes the spectrum onto the real axis by a direct estimate. For self-adjoint $T$ the number $(Tx,x)$ is real, so the imaginary part of $((T-\lambda I)x,x)$ is $-\beta\|x\|^2$ , where $\beta = \operatorname{Im}\lambda$ . Cauchy-Schwarz then forces $\|T_\lambda x\| \ge |\beta|\,\|x\|$ . The work is in knowing that such a lower bound is enough for regularity — it gives injectivity and boundedness of the inverse at once, and self-adjointness is used a second time to show that the range is dense, and completeness a third time to show it is closed.

Proof(The geometry of the resolvent set)AdaptedAdapted from Kreyszig (1978), Thm 7.3-1, p. 375, Thm 7.3-2, p. 376, Thm 7.3-4, pp. 377–378, Thm 7.5-4, pp. 390–391; Reed & Simon (1980), Thm VI.5, pp. 190–191, and its corollary, p. 191; Stakgold & Holst (2011), Lemma 5.9.1, pp. 348–349

Part 1: Neumann series. Submultiplicativity gives $\|T^j\| \le \|T\|^j$ (bounded linear operators), so $\sum_j\|T^j\| \le \sum_j\|T\|^j = (1-\|T\|)^{-1} < \infty$ : the series is absolutely convergent. Because $X$ is complete, $\mathcal{B}(X)$ is complete, and in a Banach space absolute convergence implies convergence; call the sum $S$ . Telescoping the partial sums,

(I-T)\sum_{j=0}^n T^j = \Bigl(\sum_{j=0}^n T^j\Bigr)(I-T) = I - T^{n+1} ,

and $\|T^{n+1}\| \le \|T\|^{n+1} \to 0$ since $\|T\| < 1$ . Letting $n \to \infty$ gives $(I-T)S = S(I-T) = I$ , so $S = (I-T)^{-1}$ , and $\|S\| \le \sum_j\|T\|^j = (1-\|T\|)^{-1}$ , which is the quantitative form recorded by Stakgold and Holst (Stakgold & Holst (2011), Lemma 5.9.1, pp. 348–349).

Part 2: $\rho(T)$ open. Let $\lambda_0 \in \rho(T)$ . For a bounded $T$ on a Banach space, $R_{\lambda_0} \in \mathcal{B}(X)$ is defined on all of $X$ (Kreyszig (Kreyszig (1978), Lemma 7.2-3, p. 373)). Write

T - \lambda I = (T - \lambda_0 I) - (\lambda - \lambda_0)I = (T-\lambda_0 I)\bigl[I - (\lambda-\lambda_0)R_{\lambda_0}\bigr] ,

which is legitimate because $R_{\lambda_0}$ is a genuine two-sided inverse of $T - \lambda_0 I$ . If $|\lambda - \lambda_0|\,\|R_{\lambda_0}\| < 1$ , part 1 inverts the bracket as $\sum_j (\lambda-\lambda_0)^jR_{\lambda_0}^{\,j}$ , and a product of invertible operators is invertible with

R_\lambda = \bigl[I - (\lambda-\lambda_0)R_{\lambda_0}\bigr]^{-1}R_{\lambda_0} = \sum_{j=0}^{\infty}(\lambda-\lambda_0)^jR_{\lambda_0}^{\,j+1} \in \mathcal{B}(X) .

So the disc $|\lambda - \lambda_0| < 1/\|R_{\lambda_0}\|$ lies in $\rho(T)$ , which is therefore open, and $\sigma(T)$ is closed. The displayed power series shows $\lambda \mapsto R_\lambda$ is analytic on $\rho(T)$ . For the resolvent identity, start from $R_\lambda T_\mu R_\mu = R_\lambda$ and $R_\lambda T_\lambda R_\mu = R_\mu$ and subtract:

R_\lambda - R_\mu = R_\lambda\bigl(T_\mu - T_\lambda\bigr)R_\mu = (\lambda-\mu)\,R_\lambda R_\mu ,

since $T_\mu - T_\lambda = (\lambda-\mu)I$ with the convention $T_\lambda = T - \lambda I$ . Swapping the names $\lambda$ and $\mu$ negates the left side and also the factor $\lambda - \mu$ , so $R_\lambda R_\mu = R_\mu R_\lambda$ .

Part 3: compactness. Let $|\lambda| > \|T\|$ . Then $\|T/\lambda\| < 1$ and

T - \lambda I = -\lambda\Bigl(I - \frac{T}{\lambda}\Bigr) ,

so by part 1 it is invertible with $R_\lambda = -\lambda^{-1}\sum_{j\ge0}(T/\lambda)^j$ and $\|R_\lambda\| \le (|\lambda| - \|T\|)^{-1}$ . Hence $\sigma(T) \subseteq \{|\lambda| \le \|T\|\}$ is bounded, and closed by part 2, hence compact (compactness).

Part 4: non-emptiness. Suppose $\sigma(T) = \varnothing$ , so $\rho(T) = \mathbb{C}$ and $\lambda \mapsto R_\lambda$ is analytic on all of $\mathbb{C}$ . Fix $x \in X$ and a bounded linear functional $f$ on $X$ , and set $h(\lambda) = f(R_\lambda x)$ . Composing an analytic $\mathcal{B}(X)$ -valued function with the bounded linear maps $S \mapsto Sx$ and $f$ makes $h$ an entire scalar function. It is bounded: on the compact disc $|\lambda| \le 2\|T\|$ it is continuous, hence bounded; and for $|\lambda| > 2\|T\|$ the estimate of part 3 gives $\|R_\lambda\| \le (|\lambda|-\|T\|)^{-1} \le 1/\|T\|$ , so $|h(\lambda)| \le \|f\|\,\|x\|/\|T\|$ (if $T = 0$ then $\sigma(T) = \{0\} \ne \varnothing$ already). By Liouville's theorem, $h$ is constant; and since $h(\lambda) \to 0$ as $|\lambda| \to \infty$ , that constant is $0$ . As $f$ was an arbitrary bounded functional, and the bounded functionals separate the points of a normed space — the one consequence of the Hahn-Banach theorem used here, and the only result quoted in this unit that is proved later in this part rather than earlier — we get $R_\lambda x = 0$ for every $x$ and every $\lambda$ , that is $R_\lambda = 0$ . But $R_\lambda(T - \lambda I) = I \ne 0$ because $X \ne \{0\}$ — a contradiction. $\square$

Proof(Bounded self-adjoint operators)AdaptedAdapted from Kreyszig (1978), Thm 9.1-2, pp. 461–463, Thm 9.1-3, pp. 463–464, Thm 9.2-4, p. 468

Throughout, $T = T^*$ on a complex Hilbert space $H$ , and $T_\lambda = T - \lambda I$ .

Part 1, necessity. Let $\lambda \in \rho(T)$ , so $R_\lambda$ exists and is bounded with $\|R_\lambda\| = k$ ; and $k > 0$ , since $R_\lambda \ne 0$ . For any $x \in H$ the vector $y = T_\lambda x$ lies in the domain of $R_\lambda$ and $R_\lambda y = x$ , so $\|x\| \le k\,\|T_\lambda x\|$ . Thus the criterion holds with $c = 1/k$ .

Part 1, sufficiency. Suppose $\|T_\lambda x\| \ge c\|x\|$ for all $x$ , with $c > 0$ . We verify (R1), (R3), (R2) in turn.

(a) $T_\lambda$ is injective. If $T_\lambda x_1 = T_\lambda x_2$ then $c\|x_1 - x_2\| \le \|T_\lambda(x_1-x_2)\| = 0$ , so $x_1 = x_2$ . Hence $R_\lambda$ exists on $R(T_\lambda)$ .

(b) $R(T_\lambda)$ is dense. Let $x_0 \perp R(T_\lambda)$ . Then $(T_\lambda x, x_0) = 0$ for every $x$ , that is $(Tx,x_0) = \lambda(x,x_0)$ . Here self-adjointness enters: $(Tx,x_0) = (x,Tx_0)$ , while $\lambda(x,x_0) = (x,\bar\lambda x_0)$ , so $(x, Tx_0 - \bar\lambda x_0) = 0$ for all $x$ and therefore $Tx_0 = \bar\lambda x_0$ . If $x_0 \ne 0$ this makes $\bar\lambda$ an eigenvalue of $T$ , hence real by part 4 of the statement, hence $\bar\lambda = \lambda$ and $T_\lambda x_0 = 0$ ; but then the hypothesis gives $0 = \|T_\lambda x_0\| \ge c\|x_0\| > 0$ , a contradiction. So $x_0 = 0$ , and $R(T_\lambda)^{\perp} = \{0\}$ , which by the projection theorem means $R(T_\lambda)$ is dense.

(c) $R(T_\lambda)$ is closed. Let $y_n = T_\lambda x_n \to y$ . Then $c\|x_n - x_m\| \le \|y_n - y_m\|$ , so $(x_n)$ is Cauchy; here $H$ must be complete, and $x_n \to x$ for some $x$ . Since $T_\lambda$ is continuous, $y_n = T_\lambda x_n \to T_\lambda x$ , so $y = T_\lambda x \in R(T_\lambda)$ . With (b) this gives $R(T_\lambda) = H$ .

(d) $R_\lambda$ is bounded. For $y \in H$ write $y = T_\lambda x$ ; then $\|R_\lambda y\| = \|x\| \le c^{-1}\|T_\lambda x\| = c^{-1}\|y\|$ . So (R1), (R2), (R3) all hold and $\lambda \in \rho(T)$ , with $\|R_\lambda\| \le 1/c$ .

Part 2: the spectrum is real. Let $\lambda = \alpha + i\beta$ with $\alpha,\beta$ real and $\beta \ne 0$ , and let $x \ne 0$ . Since $T$ is self-adjoint, $(Tx,x)$ is real (adjoints), and $(x,x) = \|x\|^2$ is real, so

(T_\lambda x, x) = (Tx,x) - \lambda\|x\|^2 \quad\Longrightarrow\quad \operatorname{Im}(T_\lambda x, x) = -\beta\,\|x\|^2 .

The imaginary part of a complex number is at most its modulus, so by Cauchy-Schwarz

|\beta|\,\|x\|^2 = \bigl|\operatorname{Im}(T_\lambda x,x)\bigr| \le \bigl|(T_\lambda x,x)\bigr| \le \|T_\lambda x\|\,\|x\| .

Part 3: no residual spectrum. Suppose $\lambda \in \sigma_r(T)$ . By part 2, $\lambda$ is real. By the definition of $\sigma_r$ , $R_\lambda$ exists but its domain $R(T_\lambda)$ is not dense, so by the projection theorem there is $y \ne 0$ with $y \perp R(T_\lambda)$ :

(T_\lambda x, y) = 0 \qquad \text{for every } x \in H .

Since $T$ is self-adjoint and $\lambda$ is real, $T_\lambda$ is self-adjoint too, so $(x, T_\lambda y) = 0$ for every $x$ . Taking $x = T_\lambda y$ gives $T_\lambda y = 0$ with $y \ne 0$ , so $\lambda$ is an eigenvalue, that is $\lambda \in \sigma_p(T)$ . But $\sigma_p$ and $\sigma_r$ are disjoint by definition — a $\lambda$ in $\sigma_p$ fails (R1), a $\lambda$ in $\sigma_r$ satisfies it. The contradiction shows $\sigma_r(T) = \varnothing$ . $\square$

Remark(Where the hypotheses are used)Original

Complex scalars are used in every part. Over $\mathbb{R}$ the non-emptiness argument has no Liouville theorem to appeal to, and the conclusion is false: the rotation of $\mathbb{R}^2$ through a right angle has $A - \eta I$ invertible for every real $\eta$ , so its real spectrum is empty. This is the price of Evans's (Evans (2010), Appendix D.3a, p. 677) real-scalar convention, and it costs nothing in his setting because there the operators are self-adjoint and part 2 says the spectrum was going to be real anyway.
Completeness of $X$ is used three times: to sum the Neumann series (an absolutely convergent series needs a complete space to converge in), to make $R_{\lambda_0}$ defined on all of $X$ in the openness proof, and in step (c) of the self-adjoint proof, where the Cauchy sequence $(x_n)$ must have a limit. Without it the three parts of the spectrum do not even partition the plane in a stable way.
Boundedness of $T$ is what makes $\sigma(T)$ bounded, and nothing more. For an unbounded operator the spectrum can be all of $\mathbb{C}$ , or empty, or a half-plane; see unbounded and closed operators.
Self-adjointness is used twice in the real-spectrum proof, at different strengths. Part 2 needs only that $(Tx,x)$ is real, which is symmetry; step (b) of part 1 needs $(Tx,x_0) = (x,Tx_0)$ for all $x$ , which for a bounded everywhere-defined operator is the same thing. For unbounded operators the two separate, and the conclusion separates with them: a symmetric operator that is not self-adjoint can have spectrum equal to a half-plane or to all of $\mathbb{C}$ . This is the point of symmetric versus self-adjoint operators. Stakgold and Holst (Stakgold & Holst (2011), Thm 5.6.2 and Thm 5.6.3, pp. 327–328) state the split carefully: for a merely symmetric operator it is the approximate spectrum that is real, while full self-adjointness is what puts every non-real $\lambda$ in the resolvent set.
The three parts of the spectrum are defined by conditions on $R_\lambda$ , not on $T_\lambda$ alone, and the order in the table matters: a $\lambda$ failing both (R2) and (R3) is counted as residual, not continuous. Changing that choice is what produces the overlapping sets of Stakgold and Holst (Stakgold & Holst (2011), §5.6, p. 327).
A non-empty spectrum is not a non-empty point spectrum. The multiplication operator below is bounded and self-adjoint with $\sigma(T) = [0,1]$ and not a single eigenvector. Eigenvalues return in force only under compactness: see compact operators and the spectral theorem for compact self-adjoint operators.

Method(Locating the spectrum of a concrete operator)Original

Bound the region. Compute or estimate $\|T\|$ ; the spectrum lies in $|\lambda| \le \|T\|$ . If $T$ is self-adjoint, it lies in the real interval $[-\|T\|,\|T\|]$ , and in fact in $[m,M]$ with $m,M$ the extremes of the quadratic form.
Find the eigenvalues. Solve $Tx = \lambda x$ in the space concerned. Membership of the space, not the formal solvability of the equation, is what decides: the recursion for the right shift has solutions for every $\lambda$ , but only the zero one lies in $\ell^2$ .
Invert explicitly where you can. Solve $T_\lambda x = y$ for $x$ and inspect the formula. If it defines a bounded operator on all of $X$ , then $\lambda \in \rho(T)$ , and the formula usually also gives $\|R_\lambda\|$ .
For the remaining $\lambda$ , test (R2) and (R3) separately. (R2) fails exactly when there are unit vectors $u_n$ with $\|T_\lambda u_n\| \to 0$ — Stakgold and Holst's (Stakgold & Holst (2011), §5.6, p. 327) approximate point spectrum. (R3) fails exactly when some $y \ne 0$ is orthogonal to $R(T_\lambda)$ ; in a Hilbert space that happens exactly when $\bar\lambda$ is an eigenvalue of $T^*$ , which converts a question about ranges into a question about eigenvectors.
Use closedness to fill in the boundary. Having found the spectrum on an open set, remember that $\sigma(T)$ is closed: boundary points obtained as limits of spectral values are themselves spectral values, and they are usually the continuous spectrum.
Sanity checks. $\sigma(T) \ne \varnothing$ ; $\sigma(T^*) = \overline{\sigma(T)}$ ; for self-adjoint $T$ , $\sigma(T) \subset \mathbb{R}$ , $\sigma_r(T) = \varnothing$ and $\max|\sigma(T)| = \|T\|$ .

Example(A self-adjoint operator that is all continuous spectrum)Original

On $H = L^2(0,1)$ let $(Mu)(x) = x\,u(x)$ . It is self-adjoint, since the multiplier $x$ is real: $(Mu,v) = \int_0^1 x\,u\bar v = (u,Mv)$ . Its norm is $1$ (worked out in bounded linear operators), so $\sigma(M) \subseteq [-1,1]$ , and since $(Mu,u) = \int_0^1 x|u|^2 \in [0,\|u\|^2]$ , the sharper bound of the note gives $\sigma(M) \subseteq [0,1]$ .

Step 1: no eigenvalues. $Mu = \lambda u$ means $(x-\lambda)u(x) = 0$ almost everywhere. The factor $x - \lambda$ vanishes at a single point, a set of measure zero, so $u = 0$ almost everywhere: $u$ is the zero element of $L^2$ . Hence $\sigma_p(M) = \varnothing$ and $M_\lambda = M - \lambda I$ is injective for every $\lambda$ .

Step 2: $\lambda \notin [0,1]$ is regular. Let $d = \operatorname{dist}(\lambda,[0,1]) > 0$ . The function $x \mapsto (x-\lambda)^{-1}$ is continuous on $[0,1]$ with modulus at most $1/d$ , so $(R_\lambda f)(x) = f(x)/(x-\lambda)$ defines a bounded operator on all of $L^2(0,1)$ inverting $M_\lambda$ , and

\|R_\lambda\| = \sup_{x\in[0,1]}\frac{1}{|x-\lambda|} = \frac{1}{d} .

(The supremum is attained at the nearest point of $[0,1]$ and is a genuine operator norm because multiplication operators have norm equal to the essential supremum of the multiplier.) For instance at $\lambda = 1.25$ , $d = 0.25$ and $\|R_\lambda\| = 4$ ; the general bound $\|R_\lambda\| \le (|\lambda|-\|M\|)^{-1} = (1.25 - 1)^{-1} = 4$ of part 3 is attained here.

Step 3: $\lambda \in [0,1)$ lies in the continuous spectrum. Take $u_n = \sqrt{n}\,\chi_{(\lambda,\lambda+1/n)}$ for $n > (1-\lambda)^{-1}$ , a unit vector since $n\cdot(1/n) = 1$ . Then

\|M_\lambda u_n\|^2 = n\int_\lambda^{\lambda+1/n}(x-\lambda)^2dx = n\int_0^{1/n}t^2dt = \frac{1}{3n^2} , \qquad \|M_\lambda u_n\| = \frac{1}{n\sqrt3} .

At $n = 100$ this is $0.005773503$ while $\|u_n\| = 1$ , so no inequality $\|M_\lambda x\| \ge c\|x\|$ with $c > 0$ can hold: $R_\lambda$ is unbounded and (R2) fails. (By part 1 of the self-adjoint statement this already proves $\lambda \in \sigma(M)$ .) The range is nevertheless dense: every $f \in L^2$ vanishing on a neighbourhood of $\lambda$ is $M_\lambda u$ for $u = f/(x-\lambda) \in L^2$ , and such $f$ are dense, since truncating a general $f$ to zero on $(\lambda - \varepsilon, \lambda+\varepsilon)$ changes it by at most $\bigl(\int_{|x-\lambda|<\varepsilon}|f|^2\bigr)^{1/2} \to 0$ by absolute continuity of the integral. So (R3) holds and $\lambda \in \sigma_c(M)$ . The endpoint $\lambda = 1$ is handled by the same sequence placed to the left, $u_n = \sqrt n\,\chi_{(1-1/n,1)}$ , or simply by closedness of $\sigma(M)$ .

Conclusion.

\sigma(M) = \sigma_c(M) = [0,1], \qquad \sigma_p(M) = \sigma_r(M) = \varnothing .

This matches part 3 of the self-adjoint statement, which forbids a residual spectrum, and it is the example Stakgold and Holst (Stakgold & Holst (2011), §5.6, Example 5, p. 329) use for the same purpose. The lesson for applications is blunt: a perfectly well-behaved self-adjoint operator can have a spectrum with no eigenvectors anywhere in it, so "diagonalise it" must mean something other than "find its eigenvectors". What it means is the subject of the spectral theorem.

Example(The right shift: all three parts at once)Original

On $H = \ell^2$ with orthonormal basis $(e_j)_{j\ge1}$ let $S$ be the right shift, $S(\xi_1,\xi_2,\dots) = (0,\xi_1,\xi_2,\dots)$ . Then $\|Su\| = \|u\|$ , so $\|S\| = 1$ and $\sigma(S) \subseteq \{|\lambda| \le 1\}$ . Its adjoint is the left shift, $S^*(\xi_1,\xi_2,\dots) = (\xi_2,\xi_3,\dots)$ , since $(Su,v) = \sum_j \xi_j\bar\eta_{j+1} = (u,S^*v)$ .

Step 1: $S$ has no eigenvalues. $Su = \lambda u$ reads $0 = \lambda\xi_1$ and $\xi_j = \lambda\xi_{j+1}$ for $j \ge 1$ . If $\lambda = 0$ all $\xi_j$ vanish at once. If $\lambda \ne 0$ then $\xi_1 = 0$ , and the recursion propagates $\xi_j = 0$ upwards. So $\sigma_p(S) = \varnothing$ and $S_\lambda$ is injective for every $\lambda$ .

Step 2: the eigenvectors of $S^*$ . $S^*v = \mu v$ reads $\eta_{j+1} = \mu\eta_j$ , so $\eta_j = \mu^{\,j-1}\eta_1$ , and

\|v\|^2 = |\eta_1|^2\sum_{j\ge0}|\mu|^{2j} = \frac{|\eta_1|^2}{1-|\mu|^2} \quad \text{is finite exactly when } |\mu| < 1 .

So $\sigma_p(S^*) = \{|\mu| < 1\}$ , with eigenvector $v_\mu = (1,\mu,\mu^2,\dots)$ of norm $(1-|\mu|^2)^{-1/2}$ ; at $\mu = 1/2$ this is $2/\sqrt3 = 1.1547005$ .

Step 3: $|\lambda| < 1$ lies in the residual spectrum. Put $\mu = \bar\lambda$ , so $|\mu| < 1$ and $v_\mu \ne 0$ exists by step 2. For every $u$ ,

(S_\lambda u, v_\mu) = (u, (S^* - \bar\lambda I)v_\mu) = (u, (S^*-\mu I)v_\mu) = 0 ,

so $v_\mu \perp R(S_\lambda)$ and the range is not dense: (R3) fails while (R1) holds by step 1. Hence $\lambda \in \sigma_r(S)$ .

Step 4: $|\lambda| = 1$ lies in the continuous spectrum. The spectrum is closed by part 2 of the first statement, and it contains the open unit disc by step 3, so it contains the closed disc; in particular the circle. A $\lambda$ on the circle is not in $\sigma_p(S)$ by step 1. Nor is it in $\sigma_r(S)$ : a non-zero $v \perp R(S_\lambda)$ would satisfy $(S^* - \bar\lambda I)v = 0$ , making $\bar\lambda$ an eigenvalue of $S^*$ , which step 2 excludes for $|\lambda| = 1$ . So (R1) and (R3) hold, (R2) must fail, and $\lambda \in \sigma_c(S)$ .

The table.

	$\sigma$	$\sigma_p$	$\sigma_c$	$\sigma_r$
right shift $S$	$\lvert\lambda\rvert\le 1$	$\varnothing$	$\lvert\lambda\rvert = 1$	$\lvert\lambda\rvert<1$
left shift $S^*$	$\lvert\lambda\rvert\le 1$	$\lvert\lambda\rvert<1$	$\lvert\lambda\rvert = 1$	$\varnothing$

The second row follows from the first by Reed and Simon's (Reed & Simon (1980), Thm VI.7, p. 192) theorem of Phillips, $\sigma(S^*) = \overline{\sigma(S)}$ , together with steps 1 and 2. Reed and Simon (Reed & Simon (1980), §VI.3, pp. 192–194) work the same pair of shifts on a different space — the left shift on $\ell^1$ and its adjoint, the right shift, on $\ell^\infty$ — and their table on p. 194 agrees with our second row but not with our first: they put the whole closed disc $|\lambda| \le 1$ in the residual spectrum of the right shift, with no continuous spectrum, proving the circle case separately on pp. 193–194 by exhibiting a ball of radius $1/4$ about the vector $(\lambda^n)_n \in \ell^\infty$ that misses the range of $\lambda I - T'$ . Both tables are right; what differs is the space. On $\ell^2$ , failure of (R3) at $|\lambda| = 1$ would hand us a non-zero vector orthogonal to $R(S_\lambda)$ , hence an eigenvector of $S^*$ on the unit circle, which step 2 forbids. That argument needs the inner product twice over: the projection theorem turns a non-dense range into an orthogonal vector, and the Riesz representation theorem turns that vector back into an eigenvector of the Hilbert adjoint. On $\ell^\infty$ neither step is available — the functional that cuts the range off is produced by the Hahn-Banach theorem and need not be given by a sequence — and there the circle is residual after all. So when comparing with their table, compare the spaces first. Three checks: the spectral radius is $1 = \|S\|$ , as it must be for an isometry; $\sigma_r(S) \ne \varnothing$ shows $S$ is not self-adjoint, consistent with $SS^* \ne S^*S$ ; and $\sigma_p(S)$ is empty while $\sigma(S)$ is a full disc, so the spectrum of an infinite-dimensional operator carries far more than its eigenvalues. In finite dimensions a square matrix with no eigenvalues would have empty spectrum, which part 4 of the first statement forbids — there is no contradiction only because $\ell^2$ is infinite-dimensional.

ApplicationOriginal

Quantum mechanics. The spectrum of the Hamiltonian is the set of measurable energies. Its point spectrum gives the bound states, whose eigenfunctions are normalisable; its continuous spectrum gives the scattering states, which are not normalisable and are exactly the reason the multiplication-operator example above matters — the position operator is multiplication by $x$ , and it has no eigenvectors in $L^2$ , only the "eigenfunctions" $\delta(x-\lambda)$ of distributions. Self-adjointness is what guarantees real energies and the absence of a residual spectrum, so that the spectrum splits cleanly into the discrete and continuous parts a physicist draws on an energy diagram. See the mathematical framework of quantum mechanics.

Stability of equilibria and of numerical schemes. A linear evolution $\dot u = Au$ decays when the spectrum of $A$ sits in the left half-plane, and a time-stepping scheme $u^{n+1} = Bu^n$ can be stable only when $r_\sigma(B) \le 1$ — the von Neumann condition is necessary and not sufficient, as $B = \bigl(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\bigr)$ shows, with $r_\sigma(B) = 1$ and $B^n = \bigl(\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\bigr)$ growing linearly; what stability asks for is $\sup_n\|B^n\| < \infty$ ; the Neumann series of part 1 is the same computation that makes a fixed-point iteration converge. The gap between $r_\sigma(B)$ and $\|B\|$ is why a scheme with spectral radius below $1$ can still amplify errors transiently, and why practitioners check $\|B^n\|$ and not only the eigenvalues. See von Neumann stability analysis and matrix norms and condition numbers.

Integral equations. For $u - \lambda K u = f$ the solvable $\lambda$ are those with $1/\lambda \in \rho(K)$ , and the Neumann series of part 1 is exactly the iterative solution of the Neumann series method, convergent when $|\lambda|\,\|K\| < 1$ . Which $\lambda$ break this, and what replaces the series there, is the Fredholm alternative of Hilbert-Schmidt theory.

Resonances and the resolvent as a transfer function. The resolvent $R_\lambda$ is the Green's function of the operator in the spectral parameter, and the statement of Kreyszig's (Kreyszig (1978), Thm 7.5-3, p. 390) Theorem 7.5-3 — $\|R_\lambda\| \ge 1/\operatorname{dist}(\lambda,\sigma(T))$ — is the precise form of the engineer's rule that the response of a linear system blows up as the driving frequency approaches a resonance. Analyticity of $\lambda \mapsto R_\lambda$ on $\rho(T)$ is what licenses the contour-integral manipulations of residues in scattering theory and in semigroup theory (operator semigroups).

Boundary value problems. For a Sturm-Liouville problem the spectrum is the set of $\lambda$ for which the boundary value problem fails to be uniquely solvable, and the eigenvalues are the frequencies of the modes; the resolvent is the Green's function. See Sturm-Liouville eigenvalue problems and Green's functions for boundary value problems.

Definition

Compact operators

Adapted

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Defines the operators that carry bounded sets to sets with compact closure, shows that they form a closed two-sided ideal containing every bounded finite-rank operator, identifies them on a separable Hilbert space as exactly the norm limits of finite-rank operators, and gives the working criterion that an integral operator with a square-integrable kernel is compact.

Prerequisites: Bounded linear operators, Adjoints, self-adjoint and unitary operators, The Riesz representation theorem, Orthonormal bases in Hilbert space, Weak convergence, The Hahn-Banach theorem, Compactness, L^p spaces

Definition(Compact operator, finite rank, Hilbert–Schmidt class)AdaptedAdapted from Kreyszig (1978), §8.1, Def. 8.1-1 and Thm 8.1-3, pp. 405–407; Reed & Simon (1980), §VI.5, the definition on p. 199, and §VI.6, the definition of the Hilbert–Schmidt class, p. 210; Hassani (2013), §17.5, Defs 17.5.1, 17.5.4 and 17.5.8 and Thm 17.5.2, pp. 524–525; Stakgold & Holst (2011), §5.7, the definition, p. 336

Let $X$ and $Y$ be normed spaces. A linear operator $T\colon X\to Y$ is compact (the older name, still used by Kreyszig, Reed and Simon and Stakgold and Holst, is completely continuous) if it carries every bounded subset of $X$ to a subset of $Y$ whose closure is compact.

Because a subset of a metric space is compact exactly when every sequence in it has a subsequence converging inside it (Compactness), the definition can be restated for sequences, and that is the form used in practice:

$T$ is compact exactly when every bounded sequence $(x_{n})\subset X$ admits a subsequence along which the images converge in $Y$ .

Two subclasses matter.

$T$ has finite rank if its range $T(X)$ is finite-dimensional. Its rank is $\dim T(X)$ .
Let $H$ be a separable Hilbert space with orthonormal basis $\{e_{i}\}$ (Orthonormal bases in Hilbert space). A bounded operator $T$ on $H$ is Hilbert–Schmidt if $\operatorname{tr}(T^{*}T) \;=\; \sum_{i}\,(T^{*}Te_{i},e_{i}) \;=\; \sum_{i}\lVert Te_{i}\rVert^{2} \;<\;\infty ,$ the value being independent of the basis. We write $\lVert T\rVert_{2}=(\operatorname{tr}T^{*}T)^{1/2}$ for the Hilbert–Schmidt norm.

Throughout, $\lVert T\rVert$ is the operator norm (Bounded linear operators), and $T^{*}$ the Hilbert adjoint (Adjoints, self-adjoint and unitary operators). Reed and Simon use the notation $\mathcal{J}_{2}$ for the Hilbert–Schmidt class; Hassani writes the trace with a dagger, $\operatorname{tr}(T^{\dagger}T)$ .

Definition(What compactness gives)AdaptedAdapted from Kreyszig (1978), §8.1, Lemma 8.1-2, Thms 8.1-3, 8.1-4, 8.1-5 and 8.1-7, pp. 406–410; Reed & Simon (1980), §VI.5, Thms VI.11, VI.12 and VI.13, pp. 199–201, and §VI.6, Thms VI.22 and VI.23, pp. 210–211; Hassani (2013), §17.5, Example 17.5.3, Prop. 17.5.5, Thms 17.5.6, 17.5.7 and 17.5.9, pp. 524–525; Stakgold & Holst (2011), §5.7, Thms 5.7.1 and 5.7.2 with the corollary, pp. 336–337, and §6.1, the compactness theorem for Hilbert–Schmidt operators, pp. 357–358; Evans (2010), Appendix D.5, Thm 4, p. 679

Below $X$ , $Y$ and $Z$ denote normed spaces and $H$ a Hilbert space.

Compact implies bounded. A compact operator is automatically bounded, and so continuous. Boundedness is strictly weaker: for $\dim X=\infty$ the identity $I\colon X\to X$ fails to be compact.
Finite rank. A bounded operator of finite rank is compact. In particular every linear operator on a finite-dimensional space is compact.
An ideal. The compact operators from $X$ to $Y$ form a vector subspace of the bounded operators. If $T\colon X\to Y$ is compact and $S\colon Y\to Z$ is bounded, then $ST$ is compact; if $S\colon Z\to X$ is bounded, then $TS$ is compact. So on a single space the compact operators absorb multiplication by bounded operators from either side, and any polynomial in a compact operator with zero constant term is compact.
Norm closedness. Suppose $Y$ is complete, each $T_{n}\colon X\to Y$ is compact, and $\lVert T_{n}-T\rVert\to0$ . Then the limit $T$ is compact too. Convergence in the operator norm is essential here; pointwise ("strong") convergence $T_{n}x\to Tx$ is not enough.
Finite-rank approximation. On a separable $H$ , compactness of an operator is equivalent to its being approximable in operator norm, to any accuracy, by operators of finite rank. One direction is parts 2 and 4; the other is Reed and Simon's Theorem VI.13.
Adjoints. A bounded operator $T$ on a Hilbert space is compact if and only if $T^{*}$ is. (Reed and Simon state this for Banach spaces with the Banach-space adjoint $T'$ .)
Weak to strong. If $T\colon X\to Y$ is compact and $x_{n}\rightharpoonup x$ weakly in $X$ , then $Tx_{n}\to Tx$ in norm (Weak convergence). As a special case, if $\{e_{n}\}$ is an infinite orthonormal sequence in $H$ and $K$ is compact, then $Ke_{n}\to0$ .
Inverses are bad. Should a compact operator be invertible, with $\dim H=\infty$ , the inverse cannot be bounded. Put the other way: in infinite dimensions no compact operator has a bounded inverse.
Hilbert–Schmidt operators. Every Hilbert–Schmidt operator is compact, with $\lVert T\rVert\le\lVert T\rVert_{2}$ , and the Hilbert–Schmidt class is itself a two-sided $*$ -ideal in which the finite-rank operators are dense in $\lVert\cdot\rVert_{2}$ .
Square-integrable kernels. Let $(M,\mu)$ be a measure space and $H=L^{2}(M,\mathrm{d}\mu)$ . Membership of $A$ in the Hilbert–Schmidt class is equivalent to the existence of some $K\in L^{2}(M\times M,\mathrm{d}\mu\otimes\mathrm{d}\mu)$ representing it as $(Au)(x)=\int_{M}K(x,y)\,u(y)\,\mathrm{d}\mu(y), \qquad\text{and then}\qquad \lVert A\rVert_{2}^{2}=\iint_{M\times M}\lvert K(x,y)\rvert^{2}\,\mathrm{d}\mu(x)\,\mathrm{d}\mu(y) .$ Hence an integral operator whose kernel is square-integrable on the product is compact. This is the criterion that does nearly all the work in applications; the kernel side of it is set out in Classification of integral equations.

Idea of the proofOriginal

Compactness is the property that lets an infinite-dimensional operator behave like a matrix. A bounded operator may spread the unit ball out over infinitely many independent directions with no loss of size — the identity does exactly that — and then nothing finite-dimensional survives. A compact operator cannot: it squeezes the unit ball into a set small enough that sequences in the image must bunch up. Two consequences follow almost mechanically. First, the compact operators are closed under norm limits, because a diagonal subsequence can be chosen that works for every approximating operator at once, and an $\varepsilon/3$ estimate transfers the Cauchy property to the limit. Second, a finite-rank operator is compact because bounded sets in a finite-dimensional space are relatively compact, so anything that can be approximated in norm by finite-rank operators is compact — and on a separable Hilbert space nothing else is. The last part of the statement says when that approximation is available in practice: an $L^{2}$ kernel is a square-summable "infinite matrix" in any product orthonormal basis, and truncating the matrix truncates the operator.

Proof(Parts 1–4)AdaptedAdapted from Kreyszig (1978), §8.1, proofs of Lemma 8.1-2, Thms 8.1-3, 8.1-4 and 8.1-5, pp. 406–409; Stakgold & Holst (2011), §5.7, the boundedness argument and the proof of Thm 5.7.2, pp. 336–337; Hassani (2013), §17.5, Example 17.5.3, p. 524

Part 1. Let $T$ be compact and suppose $T$ were unbounded. Then there are unit vectors $x_{n}$ with $\lVert Tx_{n}\rVert\to\infty$ . The sequence $(x_{n})$ is bounded, so compactness gives a subsequence along which $(Tx_{n})$ converges, hence is bounded — contradicting $\lVert Tx_{n}\rVert\to\infty$ . (This is Stakgold and Holst's argument.) For the identity on an infinite-dimensional space, take an infinite orthonormal sequence, or more generally use Riesz's lemma to produce a bounded sequence whose terms stay a fixed distance apart; its image under $I$ is itself, which has no convergent subsequence. The hypothesis $\dim X=\infty$ is exactly what makes such a sequence exist.

Part 2. Let $T$ be bounded with $\dim T(X)=m<\infty$ , and let $(x_{n})$ be bounded, $\lVert x_{n}\rVert\le c$ . Then $\lVert Tx_{n}\rVert\le c\lVert T\rVert$ , so $(Tx_{n})$ is a bounded sequence in the $m$ -dimensional space $T(X)$ . Bounded sets in a finite-dimensional normed space are relatively compact (Bolzano–Weierstrass, coordinate by coordinate in any basis), so $(Tx_{n})$ has a convergent subsequence. If instead $\dim X<\infty$ , then $T$ is automatically bounded and $\dim T(X)\le\dim X<\infty$ , so the same applies.

Part 3. If $T_{1},T_{2}$ are compact and $(x_{n})$ is bounded, extract a subsequence along which $(T_{1}x_{n})$ converges, then a further subsequence along which $(T_{2}x_{n})$ converges too; along it $((\alpha T_{1}+\beta T_{2})x_{n})$ converges. For $ST$ with $S$ bounded: a convergent subsequence of $(Tx_{n})$ is carried to a convergent subsequence by the continuity of $S$ . For $TS$ with $S$ bounded: $(Sx_{n})$ is bounded because $S$ is, and $T$ then supplies the subsequence. These are the observations Hassani records in Example 17.5.3; the last one also makes a product of two compact operators compact, so $p(T)$ is compact for any polynomial $p$ with $p(0)=0$ .

Part 4. This is the one argument with real content, and we follow Kreyszig and Stakgold and Holst. Let $(x_{m})$ be bounded, $\lVert x_{m}\rVert\le c$ . Since $T_{1}$ is compact there is a subsequence $(x^{(1)}_{m})$ of $(x_{m})$ with $(T_{1}x^{(1)}_{m})$ convergent; since $T_{2}$ is compact there is a subsequence $(x^{(2)}_{m})$ of $(x^{(1)}_{m})$ with $(T_{2}x^{(2)}_{m})$ convergent; and so on. Put $y_{m}=x^{(m)}_{m}$ . For each fixed $k$ the tail of $(y_{m})$ from index $k$ on is a subsequence of $(x^{(k)}_{m})$ , so $(T_{k}y_{m})_{m}$ converges, and in particular is Cauchy. That is the diagonal trick.

Now let $\varepsilon>0$ . Choose $k$ with $\lVert T-T_{k}\rVert<\varepsilon/(3c)$ — this is where norm convergence is used — and then $N$ with $\lVert T_{k}y_{j}-T_{k}y_{l}\rVert<\varepsilon/3$ for $j,l>N$ . For such $j,l$ ,

\lVert Ty_{j}-Ty_{l}\rVert \le \lVert (T-T_{k})y_{j}\rVert + \lVert T_{k}y_{j}-T_{k}y_{l}\rVert + \lVert (T_{k}-T)y_{l}\rVert < \frac{\varepsilon c}{3c}+\frac{\varepsilon}{3}+\frac{\varepsilon c}{3c}=\varepsilon .

So $(Ty_{m})$ is Cauchy in $Y$ , and converges because $Y$ is complete — the second place a hypothesis is used. By the sequential criterion $T$ is compact. $\square$

Proof(Parts 5–8)AdaptedAdapted from Reed & Simon (1980), §VI.5, proofs of Thms VI.11 and VI.13, pp. 199–201; Evans (2010), Appendix D.5, proof of Thm 4, p. 679; Stakgold & Holst (2011), §5.7, proof of Thm 5.7.1 and its corollary, p. 337; Kreyszig (1978), §8.1, Thm 8.1-7, p. 410

Part 7 first, because the others use it. Let $x_{n}\rightharpoonup x$ and put $y_{n}=Tx_{n}$ , $y=Tx$ . A weakly convergent sequence is bounded (Weak convergence), so $(x_{n})$ is bounded. For a bounded functional $g$ on $Y$ , the composite $g\circ T$ is a bounded functional on $X$ , so $g(y_{n})=(g\circ T)(x_{n})\to(g\circ T)(x)=g(y)$ : that is, $y_{n}\rightharpoonup y$ weakly. Suppose $y_{n}\not\to y$ in norm. Then some subsequence satisfies $\lVert y_{n_{k}}-y\rVert\ge\varepsilon>0$ . The corresponding $x_{n_{k}}$ are bounded, so by compactness a further subsequence has $y_{n_{k_{j}}}\to\tilde y$ in norm with $\lVert \tilde y-y\rVert\ge\varepsilon$ , so $\tilde y\ne y$ . Norm convergence implies weak convergence, so that sub-subsequence converges weakly to $\tilde y$ ; but it also converges weakly to $y$ , so $g(\tilde y-y)=0$ for every bounded functional $g$ on $Y$ . In a general normed space it is the Hahn–Banach theorem (The Hahn-Banach theorem) that supplies, for each nonzero vector, a bounded functional not vanishing on it; so the functionals separate points, weak limits are unique, and $\tilde y=y$ . Contradiction. (When $Y$ is a Hilbert space, Riesz representation (The Riesz representation theorem) does the same job without Hahn–Banach.) For the orthonormal case, Bessel's inequality gives $\sum_{n}\lvert(u,e_{n})\rvert^{2}\le\lVert u\rVert^{2}<\infty$ for every $u$ , hence $(e_{n},u)\to0$ for every $u$ , which by the Riesz representation theorem (The Riesz representation theorem) is exactly $e_{n}\rightharpoonup0$ ; so $Ke_{n}\to0$ in norm. (Stakgold and Holst prove this last statement directly, by contradiction, from the continuity of the inner product.)

Part 8. Let $K$ be compact and injective with range all of $H$ , $\dim H=\infty$ . Take an infinite orthonormal sequence $(e_{n})$ , which exists because $H$ is infinite-dimensional. By part 7, $\lVert Ke_{n}\rVert\to0$ while $\lVert e_{n}\rVert=1$ , so there is no constant $c>0$ with $\lVert Ku\rVert\ge c\lVert u\rVert$ ; hence $K^{-1}$ is unbounded. (Alternatively: if $K^{-1}$ were bounded, $I=KK^{-1}$ would be compact by part 3, and it is not.)

Part 6. We give Evans's Hilbert-space proof. Let $K$ be compact and let $(u_{k})$ be bounded. Bounded sequences in a Hilbert space have weakly convergent subsequences, so pass to one with $u_{k}\rightharpoonup u$ . Moving the outer $K^{*}$ across the inner product by the defining property of the adjoint (Adjoints, self-adjoint and unitary operators), and using that the resulting quantity is real,

\lVert K^{*}u_{k}-K^{*}u\rVert^{2} =\bigl(K^{*}(u_{k}-u),\,K^{*}(u_{k}-u)\bigr) =\bigl(KK^{*}(u_{k}-u),\,u_{k}-u\bigr).

Since $K^{*}$ is bounded it is weakly continuous, so $K^{*}u_{k}\rightharpoonup K^{*}u$ ; since $K$ is compact, part 7 upgrades this to $KK^{*}u_{k}\to KK^{*}u$ in norm. The right-hand side is then an inner product of a null sequence with a bounded one, so it tends to $0$ . Thus $K^{*}u_{k}\to K^{*}u$ in norm, and $K^{*}$ is compact. Applying this to $K^{*}$ gives the converse.

Part 5. One direction is parts 2 and 4. For the other, let $K$ be compact on the separable Hilbert space $H$ with orthonormal basis $\{\varphi_{1},\varphi_{2},\dots\}$ , let $P_{n}$ be the orthogonal projection onto $\operatorname{span}\{\varphi_{1},\dots,\varphi_{n}\}$ , and set

\lambda_{n}=\sup\bigl\{\lVert K\psi\rVert \;:\; \psi\perp\varphi_{1},\dots,\varphi_{n},\ \lVert\psi\rVert=1\bigr\} = \lVert K(I-P_{n})\rVert .

Enlarging $n$ shrinks the set over which the supremum is taken, so $(\lambda_{n})$ is decreasing and non-negative, and converges to some $\lambda\ge0$ . Suppose $\lambda>0$ . Choose unit vectors $\psi_{n}\perp\varphi_{1},\dots,\varphi_{n}$ with $\lVert K\psi_{n}\rVert\ge\lambda/2$ . Each $\psi_{n}$ is orthogonal to the first $n$ basis vectors, so for every fixed $u\in H$ we have $(\psi_{n},u)=(\psi_{n},(I-P_{n})u)$ , whose modulus is at most $\lVert(I-P_{n})u\rVert\to0$ by completeness of the basis; thus $\psi_{n}\rightharpoonup0$ . By part 7, $\lVert K\psi_{n}\rVert\to0$ , contradicting $\lVert K\psi_{n}\rVert\ge\lambda/2>0$ . Hence $\lambda=0$ . Since $KP_{n}$ has rank at most $n$ and $\lVert K-KP_{n}\rVert=\lVert K(I-P_{n})\rVert=\lambda_{n}\to0$ , the operator $K$ is a norm limit of finite-rank operators. $\square$

Proof(Parts 9 and 10)AdaptedAdapted from Stakgold & Holst (2011), §6.1, the proof that a Hilbert–Schmidt operator is compact, pp. 357–358; Reed & Simon (1980), §VI.6, Thms VI.22 and VI.23, pp. 210–211; Hassani (2013), §17.5, Def. 17.5.8 and Example 17.5.10, pp. 525–526

Part 9, compactness. Let $T$ be Hilbert–Schmidt, $\{e_{i}\}$ an orthonormal basis, $P_{n}$ the projection onto the span of $e_{1},\dots,e_{n}$ . For a unit vector $u=\sum_{i}c_{i}e_{i}$ with $c_{i}=(u,e_{i})$ supported on $i>n$ , the triangle inequality and Cauchy–Schwarz give

\lVert Tu\rVert = \Bigl\lVert \sum_{i>n}c_{i}\,Te_{i}\Bigr\rVert \le \sum_{i>n}\lvert c_{i}\rvert\,\lVert Te_{i}\rVert \le \Bigl(\sum_{i>n}\lvert c_{i}\rvert^{2}\Bigr)^{1/2}\Bigl(\sum_{i>n}\lVert Te_{i}\rVert^{2}\Bigr)^{1/2} \le \Bigl(\sum_{i>n}\lVert Te_{i}\rVert^{2}\Bigr)^{1/2} .

Hence $\lVert T-TP_{n}\rVert = \lVert T(I-P_{n})\rVert\le\bigl(\sum_{i>n}\lVert Te_{i}\rVert^{2}\bigr)^{1/2}$ , which is the tail of the convergent series $\sum_{i}\lVert Te_{i}\rVert^{2}=\lVert T\rVert_{2}^{2}$ and so tends to $0$ . Each $TP_{n}$ has rank at most $n$ , so $T$ is compact by parts 2 and 4. Taking $n=0$ in the same estimate gives $\lVert T\rVert\le\lVert T\rVert_{2}$ . The remaining assertions of part 9 are Reed and Simon's Theorem VI.22, which we take from them.

Part 10. Reed and Simon's Theorem VI.23; we take the equivalence from them and record here the direction that is used in practice, with Stakgold and Holst's route. Let $K\in L^{2}(M\times M)$ and let $\{e_{i}\}$ be an orthonormal basis of $L^{2}(M,\mathrm{d}\mu)$ . Then $\{e_{i}(x)\overline{e_{j}(y)}\}_{i,j}$ is an orthonormal basis of $L^{2}(M\times M)$ , so

K(x,y)=\sum_{i,j}k_{ij}\,e_{i}(x)\overline{e_{j}(y)}, \qquad k_{ij}=\iint K(x,y)\,\overline{e_{i}(x)}\,e_{j}(y)\,\mathrm{d}\mu(x)\,\mathrm{d}\mu(y),

with $\sum_{i,j}\lvert k_{ij}\rvert^{2}=\iint\lvert K\rvert^{2}<\infty$ by Parseval's identity. Given $\varepsilon>0$ , choose $n$ so that the tail over every pair $(i,j)$ with $i>n$ or $j>n$ is small,

\sum_{\max(i,j)>n}\lvert k_{ij}\rvert^{2}<\varepsilon^{2},

which is possible because the full double sum converges; let $S_{n}$ be the operator with the truncated kernel $\sum_{i,j\le n}k_{ij}e_{i}(x)\overline{e_{j}(y)}$ . That kernel is a finite sum of products of a function of $x$ with a function of $y$ — a separable (degenerate) kernel in Stakgold and Holst's terminology (Separable kernels) — so $S_{n}$ has rank at most $n$ and is compact. The difference $A-S_{n}$ is the integral operator whose kernel is the discarded part $\sum_{\max(i,j)>n}k_{ij}e_{i}(x)\overline{e_{j}(y)}$ , of squared $L^{2}$ norm less than $\varepsilon^{2}$ by the choice of $n$ , and the elementary bound $\lVert A\rVert\le\bigl(\iint\lvert K\rvert^{2}\bigr)^{1/2}$ — itself just Cauchy–Schwarz applied inside the integral — gives $\lVert A-S_{n}\rVert\le\varepsilon$ . So $A$ is a norm limit of finite-rank operators, hence compact. Computing $\lVert A\rVert_{2}^{2}=\sum_{i}\lVert Ae_{i}\rVert^{2}=\sum_{i,j}\lvert(Ae_{i},e_{j})\rvert^{2}=\sum_{i,j}\lvert k_{ji}\rvert^{2}=\iint\lvert K\rvert^{2}$ identifies the two norms. $\square$

Remark(Where the hypotheses are used, and what fails without them)AdaptedAdapted from Kreyszig (1978), §8.1, Lemma 8.1-2(b) and the remark after Thm 8.1-5, pp. 406–409; Reed & Simon (1980), §VI.5, Thm VI.13, p. 200, and §VI.6, Thm VI.22(e), p. 210; Stakgold & Holst (2011), §5.7, p. 336, and §6.1, the examples after the compactness theorem, p. 358; Hassani (2013), §17.5, p. 525

Infinite dimension is where the concept lives. Part 2 says every operator on a finite-dimensional space is compact, so in finite dimensions "compact" is no condition at all. Hassani marks the point in the margin beside Proposition 17.5.5 (p. 524); every statement that distinguishes compact from bounded — parts 1, 5, 7, 8 — carries the hypothesis $\dim=\infty$ either openly or through the existence of an infinite orthonormal sequence.
Norm convergence in part 4, not strong convergence. The proof uses $\lVert T-T_{k}\rVert<\varepsilon/(3c)$ uniformly over the bounded sequence; a pointwise bound would depend on the vector and the $\varepsilon/3$ argument would collapse. Kreyszig's counterexample is the cleanest: on $\ell^{2}$ let $T_{n}x=(\xi_{1},\dots,\xi_{n},0,0,\dots)$ be truncation. Each $T_{n}$ has rank $n$ , so is compact, and $T_{n}x\to x$ for every $x$ because the tail $\sum_{j>n}\lvert\xi_{j}\rvert^{2}$ vanishes. The pointwise limit is the identity, which is not compact. In the operator norm the sequence does not converge at all: $\lVert I-T_{n}\rVert=1$ for every $n$ , as the basis vector $e_{n+1}$ shows.
Completeness of the target in part 4. The diagonal argument delivers only that $(Ty_{m})$ is Cauchy; it is completeness of $Y$ that turns this into convergence. Over an incomplete target the conclusion can fail.
Separability in part 5. Reed and Simon state the finite-rank approximation theorem for a separable Hilbert space, and their proof uses a countable orthonormal basis to build the projections $P_{n}$ . They note that the corresponding question on a general Banach space is a different matter and refer it to their notes; we do not state anything about it here. Part 5 is also what makes the compact operators the closure of the finite-rank operators — the smallest norm-closed ideal containing them.
Hilbert–Schmidt is strictly stronger than compact. Part 9 is one-way. Reed and Simon's Theorem VI.22(e) quantifies the gap: a compact operator is Hilbert–Schmidt exactly when its singular values are square-summable. A compact operator whose singular values are, say, $1/\log(n+1)$ is compact but not Hilbert–Schmidt. So failing the kernel test of part 10 does not prove an integral operator non-compact; it only means another argument is needed. Stakgold and Holst give exactly such a case: on the unit square the kernel $\lvert x-\xi\rvert^{-\alpha}$ is Hilbert–Schmidt for $0<\alpha<\tfrac12$ , and for $\tfrac12\le\alpha<1$ it is not, yet the operator remains compact.
The kernel must be square-integrable on the product, and the domain matters. Stakgold and Holst's standard warning is $K(x,\xi)=\mathrm{e}^{-\lvert x-\xi\rvert}$ on the whole line: bounded, symmetric, and bounded as an operator (it is a Holmgren kernel), but $\iint\lvert K\rvert^{2}$ diverges and the operator is not compact. Restricting the same kernel to a finite interval makes it Hilbert–Schmidt and compact. Conversely boundedness of the kernel is not needed: $\mathrm{e}^{-x^{2}-\xi^{2}}$ is Hilbert–Schmidt on the whole plane, and an unbounded kernel such as $\lvert x-\xi\rvert^{-1/4}$ on the unit square is too.
Part 7 does not reverse in general. Reed and Simon note that the converse of their Theorem VI.11 holds when $X$ is reflexive and leave the general case aside; we claim nothing beyond the stated direction.

Method(Showing that an operator is compact)Original

In order of how often each step settles the question:

Is the operator an integral operator with a square-integrable kernel? Compute $\iint\lvert K\rvert^{2}$ over the product of the domain with itself. If it is finite, part 10 gives compactness and the bound $\lVert A\rVert\le\lVert A\rVert_{2}$ at once. Watch for unbounded domains and for singularities of $K$ on the diagonal: both are places where the integral can diverge.
Is it a norm limit of finite-rank operators? Exhibit a sequence $T_{n}$ of finite-rank operators and estimate $\lVert T-T_{n}\rVert$ , not $\lVert Tx-T_{n}x\rVert$ for fixed $x$ . Truncating a diagonal matrix, truncating a double Fourier series of a kernel, and smoothing a singular kernel away from the diagonal are the three standard constructions.
Is it a product with a compact factor? By part 3, $STS'$ is compact as soon as one factor is and the others are bounded. This is how compactness is inherited: a bounded solution operator composed with a compact embedding is compact.
Is it an adjoint, or a sum, of operators already known to be compact? Parts 3 and 6.
Failing all of these, use the definition. Take a bounded sequence, extract a subsequence by a compactness theorem adapted to the space — Arzelà–Ascoli for continuous functions, a diagonal argument for sequence spaces — and show the image converges.
To prove an operator is not compact, produce a bounded sequence whose image has no convergent subsequence. An orthonormal sequence $(e_{n})$ is the standard probe: by part 7 a compact operator must send it to zero in norm, so if $\lVert Te_{n}\rVert\ge c>0$ for infinitely many $n$ , $T$ is not compact.

Example(The Volterra operator: compact, with every constant checked)Original

On $H=L^{2}(0,1)$ with $(u,v)=\int_{0}^{1}u\overline{v}$ , define

(Vu)(x)=\int_{0}^{x}u(t)\,\mathrm{d}t .

Step 1: $V$ is Hilbert–Schmidt, hence compact. Write $V$ as an integral operator with kernel

K(x,t)=\begin{cases}1,& 0\le t\le x,\\ 0,& x<t\le1,\end{cases}

the indicator of the lower triangle. Then

\iint_{[0,1]^{2}}\lvert K\rvert^{2}\,\mathrm{d}t\,\mathrm{d}x=\int_{0}^{1}\!\!\int_{0}^{x}1\,\mathrm{d}t\,\mathrm{d}x=\int_{0}^{1}x\,\mathrm{d}x=\frac12 ,

so $K\in L^{2}$ of the square and part 10 gives compactness, with $\lVert V\rVert_{2}=1/\sqrt2=0.70711$ and therefore $\lVert V\rVert\le0.70711$ .

Step 2: the singular values exactly. Since $(Vu,w)=\int_{0}^{1}\overline{w(x)}\int_{0}^{x}u=\int_{0}^{1}u(t)\overline{\int_{t}^{1}w(x)\,\mathrm{d}x}\,\mathrm{d}t$ (Fubini on the triangle, the integrand being absolutely integrable by Cauchy–Schwarz), the adjoint is $(V^{*}w)(t)=\int_{t}^{1}w$ . Put $g=V^{*}Vu$ . Differentiating twice,

g'(x)=-(Vu)(x)=-\int_{0}^{x}u, \qquad g''(x)=-u(x),

with $g'(0)=0$ and $g(1)=0$ . If $V^{*}Vu=\mu u$ with $\mu\neq0$ , then $g=\mu u$ and $g''=-u=-g/\mu$ , so $g(x)=\cos(x/\sqrt\mu)$ up to a constant, and $g(1)=0$ forces $\cos(1/\sqrt\mu)=0$ , that is

\mu_{n}=\frac{1}{\bigl(n-\tfrac12\bigr)^{2}\pi^{2}}, \qquad s_{n}=\sqrt{\mu_{n}}=\frac{2}{(2n-1)\pi}, \qquad n=1,2,\dots

(The operator $V^{*}V$ is compact and self-adjoint, so Spectral theorem for compact self-adjoint operators guarantees that these exhaust its nonzero spectrum.) Numerically $s_{1}=2/\pi=0.63662$ , $s_{2}=2/(3\pi)=0.21221$ , $s_{3}=0.12732$ .

Step 3: the two computations of the Hilbert–Schmidt norm agree. By part 9 and the singular-value description,

\sum_{n\ge1}s_{n}^{2}=\frac{1}{\pi^{2}}\sum_{n\ge1}\frac{1}{(n-\tfrac12)^{2}}=\frac{1}{\pi^{2}}\cdot\frac{\pi^{2}}{2}=\frac12 ,

using $\sum_{n\ge1}(n-\tfrac12)^{-2}=4\sum_{m\ \mathrm{odd}}m^{-2}=4\cdot\pi^{2}/8=\pi^{2}/2$ . This is the same $\tfrac12$ that step 1 got from $\iint\lvert K\rvert^{2}$ — the equality $\lVert A\rVert_{2}^{2}=\iint\lvert K\rvert^{2}$ of part 10, verified. The operator norm is $\lVert V\rVert=s_{1}=0.63662$ , comfortably below the Hilbert–Schmidt bound $0.70711$ of step 1, as part 9 requires.

Step 4: an explicit finite-rank approximation, with its error proved. Step 2 identified the eigenfunctions of $V^{*}V$ as multiples of $\cos\bigl((n-\tfrac12)\pi x\bigr)$ ; normalising them (the integral of $\cos^{2}$ over $(0,1)$ is $\tfrac12$ for each of these frequencies) gives

p_{n}(x)=\sqrt2\,\cos\bigl((n-\tfrac12)\pi x\bigr), \qquad V^{*}Vp_{n}=s_{n}^{2}\,p_{n}, \qquad n=1,2,\dots

Since $V^{*}Vu=0$ forces $\lVert Vu\rVert^{2}=(Vu,Vu)=(u,V^{*}Vu)=0$ and hence $u=0$ by step 5, the operator $V^{*}V$ has trivial kernel, so by Spectral theorem for compact self-adjoint operators the family $\{p_{n}\}$ is an orthonormal basis of $L^{2}(0,1)$ . Set

q_{n}=\frac{Vp_{n}}{s_{n}}, \qquad\text{which works out to}\qquad q_{n}(x)=\sqrt2\,\sin\bigl((n-\tfrac12)\pi x\bigr),

again an orthonormal family, since $(Vp_{n},Vp_{m})=(V^{*}Vp_{n},p_{m})=s_{n}^{2}\delta_{nm}$ and therefore $(q_{n},q_{m})=s_{n}^{2}\delta_{nm}/(s_{n}s_{m})=\delta_{nm}$ . Now define the rank- $N$ operator

V_{N}u=\sum_{n\le N}s_{n}\,(u,p_{n})\,q_{n} .

Expanding $u=\sum_{n}(u,p_{n})p_{n}$ in the basis and applying the bounded operator $V$ term by term gives $Vu=\sum_{n}s_{n}(u,p_{n})q_{n}$ , so by orthonormality of $\{q_{n}\}$ and the monotonicity of $(s_{n})$ ,

\lVert (V-V_{N})u\rVert^{2}=\sum_{n>N}s_{n}^{2}\,\lvert(u,p_{n})\rvert^{2}\le s_{N+1}^{2}\,\lVert u\rVert^{2},

with equality at $u=p_{N+1}$ . The error is therefore exactly the largest discarded singular value,

\lVert V-V_{N}\rVert=s_{N+1}=\frac{2}{(2N+1)\pi},

so $N=5$ already gives $\lVert V-V_{5}\rVert=2/(11\pi)=0.057875$ , and the error falls off like $1/N$ . A direct discretisation confirms the numbers: replacing $[0,1]$ by $2000$ equal cells and $V$ by the corresponding lower-triangular matrix times the cell width gives top singular values $0.63678$ , $0.21226$ , $0.12736$ against the exact $0.63662$ , $0.21221$ , $0.12732$ , and $\sum s_{n}^{2}=0.50025$ against $0.5$ .

Step 5: what $V$ is not. $V$ is compact but has no eigenvalues at all: $Vu=\lambda u$ with $\lambda\ne0$ forces $u=\lambda^{-1}\!\int_{0}^{x}u$ , hence $u$ is absolutely continuous with $u(0)=0$ and $\lambda u'=u$ , so $u=c\,\mathrm{e}^{x/\lambda}$ and $u(0)=0$ gives $c=0$ ; and $Vu=0$ gives $u=0$ by differentiating. By part 8 its inverse (differentiation, on the range) is unbounded, as it must be.

ApplicationOriginal

Why integral equations are solvable and differential equations are not, directly. The reason a boundary-value problem is recast as an integral equation (Classification of integral equations) is part 10. A differential operator is unbounded and has no finite-dimensional shadow; its inverse, the Green's operator, has a square-integrable kernel on a finite domain and is therefore compact. All the structure that makes the problem tractable — a discrete spectrum, finite multiplicities, the Fredholm alternative (The Fredholm alternative and modified Green's functions), an eigenfunction expansion (Spectral theorem for compact self-adjoint operators) — is a property of the compact inverse, inherited by the differential problem rather than possessed by it. Evans's Theorem 5 is the statement in its abstract form: for $K$ compact, the equation $u-Ku=f$ has a unique solution for every $f$ unless the homogeneous equation has a nonzero solution, and then the solution space is finite-dimensional.

Scattering theory and the Born series. In quantum scattering the Lippmann–Schwinger equation is $\psi=\psi_{0}+G_{0}V\psi$ , and whether the Born series $\sum (G_{0}V)^{n}\psi_{0}$ converges and whether bound states exist are questions about the operator $G_{0}V$ . For a potential decaying fast enough, $G_{0}V$ has a square-integrable kernel and part 10 applies; compactness then yields a discrete set of bound states of finite multiplicity, with the continuum lying elsewhere. The same structure appears in Neumann series and Scattering and the Born approximation.

Compact embeddings in partial differential equations. The modern use of compactness in analysis is not that a given operator is compact but that an inclusion is: the Rellich–Kondrachov theorem makes the embedding of a Sobolev space into $L^{2}$ on a bounded domain compact. Part 3 then promotes any bounded solution operator into that Sobolev space to a compact operator on $L^{2}$ , which is how existence proofs for elliptic problems obtain discrete spectra and how a minimising sequence is made to converge (Weak derivatives and Sobolev spaces, Variational characterization of eigenvalues). The hypothesis that the domain be bounded is exactly the hypothesis that fails for $\mathrm{e}^{-\lvert x-\xi\rvert}$ on the line in the remark above: it is the same loss of compactness, seen twice.

Numerics: why discretisation works, and when it does not. Part 5 says a compact operator is its finite-rank approximations, up to a norm error that goes to zero. This is the theoretical licence for solving an integral equation by collocation or by a Galerkin method on a finite basis (Separable kernels, Spectral methods): the discretised problem converges to the true one in operator norm, and with it the spectrum. Part 8 is the warning attached: a compact operator has an unbounded inverse, so the discretised matrices become arbitrarily ill-conditioned as the discretisation is refined (Matrix norms and condition numbers). Solving a first-kind integral equation numerically is therefore ill-posed by construction, and must be regularised; step 4 of the example shows the mechanism, with singular values marching down to zero at a known rate.

Theorem

Spectral theorem for compact self-adjoint operators

Adapted

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A compact self-adjoint operator has real eigenvalues of finite multiplicity that can only accumulate at zero, the largest in modulus equals the operator norm and is reached by maximising the Rayleigh quotient, and the eigenvectors form an orthonormal basis in which the operator is diagonal — the infinite-dimensional replacement for diagonalising a Hermitian matrix, and the reason a self-adjoint boundary-value problem has a complete set of normal modes.

Prerequisites: Compact operators, Adjoints, self-adjoint and unitary operators, Orthonormal bases in Hilbert space, Spectrum and resolvent, The projection theorem

Definition(The setting)AdaptedAdapted from Stakgold & Holst (2011), §6.3, p. 371; Reed & Simon (1980), §VI.5, Thm VI.16, p. 203; Hassani (2013), §17.6.1, p. 529

Let $H$ be a Hilbert space over $\mathbb{C}$ with inner product $(u,v)$ , linear in the first slot, and let

A\colon H\to H

be compact (Compact operators) and self-adjoint, $A^{*}=A$ , equivalently $(Au,v)=(u,Av)$ for all $u,v\in H$ (Adjoints, self-adjoint and unitary operators). Hassani and other physics texts say Hermitian and write $K^{\dagger}=K$ ; Evans, working over $\mathbb{R}$ , says symmetric. We assume $A\ne0$ , since the zero operator is a degenerate case of every statement below.

A scalar $\lambda$ is an eigenvalue with eigenvector $e\ne0$ if $Ae=\lambda e$ ; the eigenspace is $N_{\lambda}=\ker(A-\lambda I)$ and its dimension is the multiplicity of $\lambda$ . (The letter $M$ is reserved below for spans of eigenvectors: $M_{n}=\operatorname{span}\{e_{1},\dots,e_{n}\}$ , with $M_{n}^{\perp}$ its orthogonal complement.) The numerical range of $A$ is the set of values $(Au,u)$ for unit vectors $u$ ; because $A$ is self-adjoint these are all real, since $(Au,u)=(u,Au)=\overline{(Au,u)}$ . The quantity

R(u)=\frac{(Au,u)}{(u,u)},\qquad u\ne0,

is the Rayleigh quotient.

The spectrum $\sigma(A)$ , the resolvent set and the distinction between point, continuous and residual spectrum are those of Spectrum and resolvent.

Theorem(Spectral theorem for compact self-adjoint operators)AdaptedAdapted from Stakgold & Holst (2011), §6.3, Thms 6.3.1–6.3.5 and (6.3.6)–(6.3.12), pp. 371–374; Reed & Simon (1980), §VI.5, Thms VI.15 and VI.16, p. 203; Evans (2010), Appendix D.5, Thm 6, p. 682, and Appendix D.6, Thm 7, pp. 683–684; Hassani (2013), §17.5.1, Thm 17.5.11, p. 527, and §17.6.1, Lemma 17.6.4, Thm 17.6.5 and Cor. 17.6.6, pp. 529–531; Kreyszig (1978), §8.3, Thm 8.3-1, p. 421

Let $A\ne0$ be compact and self-adjoint on a Hilbert space $H$ .

An eigenvalue of maximal size exists, and the Rayleigh quotient finds it. The supremum $\sup_{\lVert v\rVert=1}\lvert(Av,v)\rvert \;=\; \lVert A\rVert$ is attained, at a unit vector $e_{1}$ which is an eigenvector: $Ae_{1}=\lambda_{1}e_{1}$ with $\lvert\lambda_{1}\rvert=\lVert A\rVert>0$ . No eigenvalue exceeds $\lvert\lambda_{1}\rvert$ in modulus, and $\lVert Av\rVert\le\lvert\lambda_{1}\rvert\lVert v\rVert$ for every $v$ .
Successive extremal principles. Having found orthonormal eigenvectors $e_{1},\dots,e_{n}$ with eigenvalues $\lambda_{1},\dots,\lambda_{n}$ , write $M_{n}=\operatorname{span}\{e_{1},\dots,e_{n}\}$ and let $M_{n}^{\perp}$ be its orthogonal complement, the set of vectors orthogonal to all of them. Then the supremum of $\lvert(Av,v)\rvert$ over unit vectors of $M_{n}^{\perp}$ is attained at a unit eigenvector $e_{n+1}\in M_{n}^{\perp}$ , with eigenvalue $\lambda_{n+1}$ satisfying $\lvert\lambda_{n+1}\rvert\le\lvert\lambda_{n}\rvert$ , and $\lVert Av\rVert\le\lvert\lambda_{n+1}\rvert\,\lVert v\rVert \qquad\text{for every }v\in M_{n}^{\perp}.$ Exactly one of two things happens. Either the supremum is $0$ at some finite stage $n+1$ , in which case $A$ has precisely $n$ nonzero eigenvalues counted with multiplicity and $A$ has finite rank; or the process never stops and produces an infinite sequence $\lvert\lambda_{1}\rvert\ge\lvert\lambda_{2}\rvert\ge\cdots>0,\qquad \lambda_{n}\to0 .$
Structure of the spectrum. Every eigenvalue is real. Eigenvectors belonging to different eigenvalues are orthogonal. Every nonzero eigenvalue has finite multiplicity, and the nonzero eigenvalues are countable in number with $0$ as their only possible accumulation point. Moreover $\sigma(A)\setminus\{0\}$ consists of eigenvalues only, and if $\dim H=\infty$ then $0\in\sigma(A)$ (though $0$ need not be an eigenvalue). Those last two sentences — finite multiplicity, countability and accumulation only at $0$ , and the identification of the nonzero spectrum with the eigenvalues — hold for every compact operator, self-adjoint or not: they are the Riesz–Schauder theorem. Self-adjointness is what adds the final clause: the list produced in part 2 contains every nonzero eigenvalue. That fails without it, as the Volterra operator of the remark below shows: it is compact and has no eigenvalues at all, so part 2 has nothing to produce.
Expansion theorem. With $\{e_{n}\}$ the orthonormal eigenvectors belonging to the nonzero eigenvalues, $Au \;=\; \sum_{n}\lambda_{n}\,(u,e_{n})\,e_{n} \qquad\text{for every }u\in H ,$ the series converging in $H$ . Hence $\{e_{n}\}$ is an orthonormal basis of $\overline{\operatorname{ran}A}$ , and every $u\in H$ splits in exactly one way as $u \;=\; h+\sum_{n}(u,e_{n})\,e_{n},\qquad Ah=0 ,$ with $h$ the orthogonal projection of $u$ on $\ker A$ . The family $\{e_{n}\}$ is an orthonormal basis of all of $H$ if and only if $\ker A=\{0\}$ .
Diagonalisation. Adjoining an orthonormal basis of $\ker A$ to $\{e_{n}\}$ gives an orthonormal basis of $H$ consisting entirely of eigenvectors of $A$ ; it is countable when $H$ is separable. In projection form, with $\lambda_{1},\lambda_{2},\dots$ now the distinct nonzero eigenvalues in decreasing order of modulus and $P_{j}$ the orthogonal projection onto the eigenspace of $\lambda_{j}$ , $A=\sum_{j}\lambda_{j}P_{j}, \qquad P_{i}P_{j}=0\ (i\ne j), \qquad \Bigl\lVert A-\sum_{j\le m}\lambda_{j}P_{j}\Bigr\rVert=\lvert\lambda_{m+1}\rvert \qquad (m=0,1,2,\dots),$ so the series converges in operator norm at a rate given exactly by the first discarded eigenvalue. Hassani states this last identity for the case of infinitely many distinct nonzero eigenvalues (Theorem 17.6.5(2), p. 531), where $\lambda_{m+1}$ always exists. If instead there are only $N$ distinct nonzero eigenvalues, the sum is finite and exact at $m=N$ , and the identity continues to hold for $m\ge N$ under the convention $\lambda_{m+1}=0$ .

Idea of the proofOriginal

A Hermitian matrix is diagonalised by repeatedly maximising the Rayleigh quotient: the maximum over all unit vectors is the largest eigenvalue, the maximum over the vectors orthogonal to the first eigenvector is the next, and so on. In infinite dimensions the step that can fail is the very first one — a supremum over the unit sphere need not be attained, because the unit sphere is not compact. Self-adjointness gives $\lVert A\rVert=\sup_{\lVert v\rVert=1}\lvert(Av,v)\rvert$ , so a maximising sequence of unit vectors is at hand; compactness then supplies a convergent subsequence of its images, which is just enough to produce an actual maximiser and hence an actual eigenvector. Everything else is bookkeeping: $A$ maps the orthogonal complement of an eigenvector into itself (this uses self-adjointness again), so the argument repeats there, and the resulting eigenvalues cannot fail to tend to zero, since otherwise the rescaled eigenvectors $e_{n}/\lambda_{n}$ — the $e_{n}$ themselves are the normalised ones — would be a bounded sequence whose images $e_{n}$ are mutually at distance $\sqrt2$ — impossible for a compact operator. The expansion theorem is then the statement that what is left after removing all the eigenvectors is annihilated, which the extremal bound $\lVert Av\rVert\le\lvert\lambda_{n+1}\rvert\lVert v\rVert$ delivers in the limit.

Proof(Part 1: the extremal principle)AdaptedAdapted from Hassani (2013), §17.6.1, Lemmas 17.6.3 and 17.6.4 with the proof of 17.6.4, pp. 529–530; Stakgold & Holst (2011), §6.3, Thm 6.3.1, p. 371; Evans (2010), Appendix D.6, the Lemma on bounds for the spectrum, p. 683

Write $M=\sup_{\lVert v\rVert=1}\lvert(Av,v)\rvert$ .

Step 1: $M=\lVert A\rVert$ . One inequality is immediate from Cauchy–Schwarz: $\lvert(Av,v)\rvert\le\lVert Av\rVert\lVert v\rVert\le\lVert A\rVert$ for unit $v$ , so $M\le\lVert A\rVert$ . For the reverse — this is Hassani's Lemma 17.6.3, whose proof he leaves as an exercise, so we give one — note first that $\lvert(Au,u)\rvert\le M\lVert u\rVert^{2}$ for every $u$ , by homogeneity. Self-adjointness makes the polarisation identity real:

(A(u+w),u+w)-(A(u-w),u-w)=2(Au,w)+2(Aw,u)=4\,\mathrm{Re}\,(Au,w) .

Hence, by the triangle inequality and the parallelogram law,

4\,\lvert\mathrm{Re}\,(Au,w)\rvert \le M\bigl(\lVert u+w\rVert^{2}+\lVert u-w\rVert^{2}\bigr)=2M\bigl(\lVert u\rVert^{2}+\lVert w\rVert^{2}\bigr).

Take $\lVert u\rVert=1$ and $w=Au/\lVert Au\rVert$ (assuming $Au\ne0$ ); then $\mathrm{Re}\,(Au,w)=\lVert Au\rVert$ and the right-hand side is $4M$ , so $\lVert Au\rVert\le M$ . Taking the supremum over unit $u$ gives $\lVert A\rVert\le M$ . Here self-adjointness is used in an essential way: without it the cross terms do not combine into a real quantity and the identity $\lVert A\rVert=M$ is false.

Step 2: the supremum is attained, at an eigenvector. We follow Hassani's proof of his Lemma 17.6.4. By step 1 there are unit vectors $x_{n}$ with $(Ax_{n},x_{n})\to\lambda$ , where $\lambda$ is real with $\lvert\lambda\rvert=\lVert A\rVert$ (pass to a subsequence so that the real numbers $(Ax_{n},x_{n})$ , which lie in $[-\lVert A\rVert,\lVert A\rVert]$ , converge to an endpoint of maximal modulus). Since $A$ is compact and $(x_{n})$ is bounded, we may pass to a further subsequence along which $Ax_{n}\to y$ in norm. Now expand

0\le\lVert Ax_{n}-\lambda x_{n}\rVert^{2} =\lVert Ax_{n}\rVert^{2}-2\lambda\,(Ax_{n},x_{n})+\lambda^{2} ,

using $\lambda$ real and $\lVert x_{n}\rVert=1$ . Letting $n\to\infty$ gives $0\le\lVert y\rVert^{2}-2\lambda^{2}+\lambda^{2}$ , that is $\lVert y\rVert^{2}\ge\lambda^{2}=\lVert A\rVert^{2}$ . On the other hand $\lVert y\rVert=\lim\lVert Ax_{n}\rVert\le\lVert A\rVert$ . So $\lVert y\rVert=\lvert\lambda\rvert=\lVert A\rVert\ne0$ , and the displayed expression tends to $0$ : $Ax_{n}-\lambda x_{n}\to0$ . Since $Ax_{n}\to y$ and $\lambda\ne0$ , this forces $x_{n}\to y/\lambda$ . Put $e_{1}=y/\lambda$ , a unit vector, and apply the continuous operator $A-\lambda I$ :

(A-\lambda I)e_{1}=\lim_{n}(A-\lambda I)x_{n}=0 .

So $\lambda_{1}=\lambda$ is an eigenvalue with unit eigenvector $e_{1}$ and $\lvert\lambda_{1}\rvert=\lVert A\rVert$ , and $(Ae_{1},e_{1})=\lambda_{1}$ shows the supremum is attained there. If $Ae=\mu e$ with $\lVert e\rVert=1$ then $\lvert\mu\rvert=\lVert Ae\rVert\le\lVert A\rVert=\lvert\lambda_{1}\rvert$ , so no eigenvalue is larger, and $\lVert Av\rVert\le\lVert A\rVert\lVert v\rVert=\lvert\lambda_{1}\rvert\lVert v\rVert$ is the definition of the operator norm. $\square$

Proof(Parts 2 and 3: iteration, and the shape of the spectrum)AdaptedAdapted from Stakgold & Holst (2011), §6.3, proofs of Thms 6.3.2 and 6.3.3, pp. 371–372, the two outcomes and the proof that $\\lambda_{n}\\to0$, pp. 372–373, and the argument on p. 374 that no eigenvalue is missed; Reed & Simon (1980), §VI.5, Thm VI.15, p. 203; Evans (2010), Appendix D.5, Thm 6 and its proof, p. 682; Kreyszig (1978), §8.3, Thm 8.3-1, p. 421

Part 2. Suppose $e_{1},\dots,e_{n}$ have been found. Let $M_{n}^{\perp}$ be the orthogonal complement of their span, a closed subspace and therefore a Hilbert space in its own right (The projection theorem). The key observation is that $A$ maps $M_{n}^{\perp}$ into itself: for $v\in M_{n}^{\perp}$ and each $k\le n$ ,

(Av,e_{k})=(v,Ae_{k})=(v,\lambda_{k}e_{k})=\overline{\lambda_{k}}\,(v,e_{k})=0 ,

where the first equality is self-adjointness and the last uses $v\perp e_{k}$ . (Reality of $\lambda_{k}$ , proved below, makes $\overline{\lambda_{k}}=\lambda_{k}$ ; either way the product vanishes.) The restriction $A|_{M_{n}^{\perp}}$ is again compact — a restriction of a compact operator to a closed invariant subspace is compact, since bounded sequences in the subspace are bounded in $H$ — and again self-adjoint. So part 1 applies on $M_{n}^{\perp}$ and delivers $e_{n+1}$ and $\lambda_{n+1}$ with $\lvert\lambda_{n+1}\rvert=\lVert A|_{M_{n}^{\perp}}\rVert$ , and the bound $\lVert Av\rVert\le\lvert\lambda_{n+1}\rvert\lVert v\rVert$ on $M_{n}^{\perp}$ . Because $M_{n}^{\perp}\subset M_{n-1}^{\perp}$ , the suprema cannot increase: $\lvert\lambda_{n+1}\rvert\le\lvert\lambda_{n}\rvert$ . This is Stakgold and Holst's chain of Theorems 6.3.1–6.3.3.

If at some stage the supremum is $0$ , then by step 1 of part 1 the restriction has zero norm, so $A$ vanishes on $M_{n}^{\perp}$ ; every vector there is an eigenvector for the eigenvalue $0$ , and $A$ has rank $n$ , with range the span of $e_{1},\dots,e_{n}$ . Otherwise the process continues for ever.

Part 2, $\lambda_{n}\to0$ . Suppose not; then $\lvert\lambda_{n}\rvert\ge c>0$ for all $n$ , and the vectors $u_{n}=e_{n}/\lambda_{n}$ satisfy $\lVert u_{n}\rVert\le1/c$ , a bounded sequence. Compactness would give a convergent subsequence of $(Au_{n})=(e_{n})$ . But the $e_{n}$ are orthonormal, so $\lVert e_{n}-e_{m}\rVert=\sqrt2$ for $n\ne m$ and no subsequence is Cauchy. Contradiction. (Stakgold and Holst's argument; equivalently, part 7 of Compact operators forces $Ae_{n}=\lambda_{n}e_{n}\to0$ , hence $\lambda_{n}\to0$ .)

Part 3, reality and orthogonality. If $Ae=\mu e$ with $e\ne0$ then $\mu\lVert e\rVert^{2}=(Ae,e)$ is real by self-adjointness, so $\mu$ is real. If $Ae=\mu e$ and $Af=\nu f$ with $\mu\ne\nu$ , both real, then

\mu(e,f)=(Ae,f)=(e,Af)=\nu(e,f),

so $(\mu-\nu)(e,f)=0$ and $(e,f)=0$ .

Part 3, finite multiplicity and countability. If some $\mu\ne0$ had an infinite-dimensional eigenspace, that eigenspace would contain an infinite orthonormal sequence $(f_{k})$ with $Af_{k}=\mu f_{k}$ , and $\lVert Af_{k}-Af_{l}\rVert=\lvert\mu\rvert\sqrt2$ for $k\ne l$ , so $(Af_{k})$ has no convergent subsequence although $(f_{k})$ is bounded — contradicting compactness. (This is Evans's step 1.) The same argument, run with eigenvectors for distinct eigenvalues of modulus at least $c>0$ , shows that only finitely many such eigenvalues exist; Kreyszig proves this for a general compact operator on a normed space using Riesz's lemma instead of orthogonality. Letting $c=1/m$ and taking the union over $m$ exhibits the nonzero eigenvalues as a countable set accumulating only at $0$ .

Part 3, no eigenvalue is missed. Let $Av=\mu v$ with $\mu\ne0$ and $v\ne0$ , and suppose $\mu$ is not among the $\lambda_{n}$ . By the orthogonality just proved, $v\perp e_{n}$ for every $n$ , so $v\in M_{n}^{\perp}$ for every $n$ and the extremal bound gives $\lVert Av\rVert\le\lvert\lambda_{n+1}\rvert\lVert v\rVert$ for all $n$ . Letting $n\to\infty$ gives $Av=0$ , so $\mu v=0$ and $v=0$ : a contradiction. (In the finite case the bound is $Av=0$ outright.) Hence the list is complete. $\square$

Note(The Riesz–Schauder clauses are quoted, not proved here)AdaptedAdapted from Reed & Simon (1980), §VI.5, Thm VI.15, p. 203; Hassani (2013), §17.5.1, Thm 17.5.11, p. 527; Evans (2010), Appendix D.5, Thm 6 and its proof, p. 682

Two clauses of part 3 are not proved above, because they need no self-adjointness and belong to the theory of a general compact operator: that every nonzero $\eta\in\sigma(A)$ is an eigenvalue, and that $0\in\sigma(A)$ when $\dim H=\infty$ . We take them from Reed and Simon's Theorem VI.15, Hassani's Theorem 17.5.11 and Evans's Theorem 6.

Evans's argument for the second is one line and worth recording: if $0\notin\sigma(A)$ then $A$ is boundedly invertible and $I=A\circ A^{-1}$ would be compact by the ideal property, which is impossible in infinite dimensions. His argument for the first runs through the Fredholm alternative for a compact operator — if $\ker(A-\eta I)=\{0\}$ with $\eta\ne0$ then the alternative gives $\operatorname{ran}(A-\eta I)=H$ , so $\eta$ is in the resolvent set after all — which is quoted, in Evans's form, in the application of Compact operators. Reed and Simon instead derive both from their analytic Fredholm theorem.

Proof(Parts 4 and 5: the expansion and the basis)AdaptedAdapted from Stakgold & Holst (2011), §6.3, proof of Thm 6.3.4 with (6.3.10)–(6.3.11) and the decomposition (6.3.12) with Thm 6.3.5, pp. 373–374; Reed & Simon (1980), §VI.5, proof of Thm VI.16, p. 203; Evans (2010), Appendix D.6, proof of Thm 7, pp. 683–684; Hassani (2013), §17.6.1, Thm 17.6.5 and Cor. 17.6.6, pp. 530–531

Part 4. Fix $u\in H$ and set $v_{n}=u-\sum_{k\le n}(u,e_{k})e_{k}$ , the component of $u$ orthogonal to $e_{1},\dots,e_{n}$ , so $v_{n}\in M_{n}^{\perp}$ and, by Bessel's inequality (Orthonormal bases in Hilbert space), $\lVert v_{n}\rVert\le\lVert u\rVert$ . Applying $A$ term by term to the finite sum and using $Ae_{k}=\lambda_{k}e_{k}$ ,

Av_{n}=Au-\sum_{k\le n}\lambda_{k}(u,e_{k})\,e_{k} .

The extremal bound of part 2 on $M_{n}^{\perp}$ now gives

\Bigl\lVert Au-\sum_{k\le n}\lambda_{k}(u,e_{k})e_{k}\Bigr\rVert=\lVert Av_{n}\rVert\le\lvert\lambda_{n+1}\rvert\,\lVert v_{n}\rVert\le\lvert\lambda_{n+1}\rvert\,\lVert u\rVert .

Since $\lambda_{n+1}\to0$ the right-hand side vanishes, which is the expansion. (If the list is finite the bound reads $Av_{n}=0$ and the finite sum is exact.) Consequently every element of $\operatorname{ran}A$ lies in the closed span of $\{e_{n}\}$ ; conversely each $e_{n}=A(e_{n}/\lambda_{n})$ lies in $\operatorname{ran}A$ . So $\{e_{n}\}$ , being orthonormal, is an orthonormal basis of $\overline{\operatorname{ran}A}$ .

For the splitting, set $h=u-\sum_{n}(u,e_{n})e_{n}$ ; the series converges by Bessel and the Riesz–Fischer theorem, and applying the expansion just proved gives $Ah=Au-\sum_{n}\lambda_{n}(u,e_{n})e_{n}=0$ . Uniqueness: if $Ah=0$ then for each $n$ , $\lambda_{n}(h,e_{n})=(h,Ae_{n})=(Ah,e_{n})=0$ and $\lambda_{n}\ne0$ , so $h\perp e_{n}$ ; thus $\ker A$ is exactly the orthogonal complement of the closed span of the $e_{n}$ , and the decomposition is the orthogonal one. In particular $\{e_{n}\}$ spans $H$ precisely when $\ker A=\{0\}$ . This is Stakgold and Holst's Theorem 6.3.5.

Part 5. Taking an orthonormal basis of the closed subspace $\ker A$ — every Hilbert space has one — and adjoining it to $\{e_{n}\}$ gives an orthonormal set whose closed span is $\ker A\oplus\overline{\operatorname{ran}A}=H$ by the splitting, so an orthonormal basis of $H$ , and every member of it is an eigenvector ( $0$ being the eigenvalue for the adjoined part). Countability when $H$ is separable is Evans's Theorem 7: a separable space has a countable orthonormal basis, so in particular $\ker A$ does.

Two other routes to the same conclusion are worth knowing. Reed and Simon let $\mathcal{M}$ be the closed span of all the eigenvectors; self-adjointness makes $\mathcal{M}^{\perp}$ invariant, the restriction of $A$ there is compact and self-adjoint with no nonzero eigenvalue, so by part 1 its norm is $0$ , so $A$ vanishes on $\mathcal{M}^{\perp}$ — but then every vector of $\mathcal{M}^{\perp}$ is an eigenvector with eigenvalue $0$ and therefore lies in $\mathcal{M}$ , forcing $\mathcal{M}^{\perp}=\{0\}$ . Evans's Theorem 7 is the same argument with the extremal principle replaced by his lemma $\sigma(S)\subset[m,M]$ with $m,M\in\sigma(S)$ , which gives $(\tilde Su,u)=0$ on the complement and hence $\tilde S=0$ by polarisation.

The projection form. Group the $e_{n}$ by distinct eigenvalue; $P_{j}$ is the (finite-rank) orthogonal projection onto the eigenspace of the $j$ th distinct nonzero eigenvalue, which part 5 calls $\lambda_{j}$ , and the expansion of part 4 reads $Au=\sum_{j}\lambda_{j}P_{j}u$ .

The norm estimate needs the index translation between part 2, which lists eigenvalues with multiplicity, and part 5, which lists them without. Fix $m$ and let

k=\sum_{j\le m}\dim P_{j}H

be the total multiplicity of the first $m$ distinct nonzero eigenvalues. Part 2 produces the $e_{n}$ grouped by eigenvalue in non-increasing order of modulus, so the span of those $m$ eigenspaces is exactly $M_{k}=\operatorname{span}\{e_{1},\dots,e_{k}\}$ , and the eigenvalue attached to the next vector $e_{k+1}$ is the next distinct one, of modulus $\lvert\lambda_{m+1}\rvert$ . Now $A-\sum_{j\le m}\lambda_{j}P_{j}$ vanishes on $M_{k}$ and on $\ker A$ and acts as $A$ on $M_{k}^{\perp}$ , where by part 2 its norm is $\lvert\lambda_{m+1}\rvert$ , attained at $e_{k+1}$ . This is Hassani's Theorem 17.6.5. If there are only $N$ distinct nonzero eigenvalues, then at $m=N$ the list is exhausted, $M_{k}^{\perp}=\ker A$ and the norm is $0$ — the convention $\lambda_{m+1}=0$ recorded in the statement. $\square$

Remark(Where the hypotheses are used, and what fails without them)AdaptedAdapted from Stakgold & Holst (2011), §6.3, the opening paragraph, p. 371, Thm 6.3.5, p. 374, and the definite and indefinite cases, pp. 375–376; Reed & Simon (1980), §VI.5, the multiplication operator example, p. 201, and Thm VI.17, pp. 203–204; Evans (2010), Appendix D.5, Thm 6, p. 682; Hassani (2013), §17.6.1, Cor. 17.6.6, p. 531

Self-adjointness is used four times, and dropping it destroys everything. It gives $\lVert A\rVert=\sup\lvert(Av,v)\rvert$ in step 1 of part 1, so that a maximising sequence for the Rayleigh quotient is a maximising sequence for the norm; it makes the eigenvalues real; it makes the orthogonal complement of an eigenvector invariant, which is what allows the iteration of part 2; and it makes the eigenvectors of distinct eigenvalues orthogonal, so that the $e_{n}$ assemble into an orthonormal family. The sharpest counterexample is the Volterra operator $(Vu)(x)=\int_{0}^{x}u$ on $L^{2}(0,1)$ , worked out in Compact operators: it is compact, it is not self-adjoint, and it has no eigenvalues at all, so $\sigma(V)=\{0\}$ and there is nothing to expand in. What survives for a general compact operator is the singular-value decomposition — Reed and Simon's canonical form (their Theorem VI.17, p. 204), which in the convention of this unit reads

Au=\sum_{n}\lambda_{n}\,(u,\psi_{n})\,\varphi_{n}

for two orthonormal families $\{\psi_{n}\}$ , $\{\varphi_{n}\}$ and positive $\lambda_{n}\to0$ , obtained by applying this theorem to the compact self-adjoint operator $A^{*}A$ (Singular value decomposition). Reed and Simon write the same operator as $\sum_{n}\lambda_{n}(\psi_{n},\cdot)\varphi_{n}$ because their inner product is conjugate-linear in the first slot; ours is linear in the first slot, so the functional $u\mapsto(u,\psi_{n})$ is the one that is linear here. The Volterra operator's singular values are computed in Compact operators and are perfectly well behaved; it is only the eigen-structure that is missing.

Compactness is used twice, and dropping it destroys part 1. It supplies the convergent subsequence of $(Ax_{n})$ that turns a maximising sequence into an actual maximiser, and it forces $\lambda_{n}\to0$ and finite multiplicities. Stakgold and Holst open §6.3 with the warning: a bounded self-adjoint operator may have no eigenvalues whatever. The simplest example is multiplication by the coordinate, $(A\varphi)(x)=x\varphi(x)$ , which Reed and Simon give on $L^{2}[0,2]$ (p. 201) to show that the Fredholm alternative fails for a non-compact operator; we take the same operator on $L^{2}[0,1]$ . There it is bounded with $\lVert A\rVert=1$ and self-adjoint, $(Au,v)=(u,Av)$ since $x$ is real, but $x\varphi(x)=\mu\varphi(x)$ forces $\varphi=0$ almost everywhere for every $\mu$ . Its spectrum is the whole interval $[0,1]$ and is entirely continuous. Recovering a spectral theorem in that generality needs spectral measures rather than eigenvectors, which is The spectral theorem.
$A\ne0$ is needed only so that part 1 produces a nonzero eigenvalue; for $A=0$ every statement holds vacuously with no $e_{n}$ at all and $\ker A=H$ .
$\lambda=0$ is the exceptional point and must be handled separately. It is in the spectrum whenever $\dim H=\infty$ , but it may or may not be an eigenvalue, and when it is, its multiplicity may be anything from $1$ to infinite. The theorem says nothing about the structure of $A$ on $\ker A$ because there is nothing to say: $A$ is zero there. This is also the reason the expansion theorem is about $\operatorname{ran}A$ , not about $H$ : a function with no component in $\overline{\operatorname{ran}A}$ has an eigenfunction expansion that converges to something other than itself.
Separability is needed only for countability. Hassani's Corollary 17.6.6 states that the eigenvectors form an orthonormal basis with no separability hypothesis, which is correct: the basis simply need not be a sequence, because $\ker A$ can be non-separable. The part that is always countable is the family belonging to the nonzero eigenvalues, by part 3. Evans states the theorem for a separable space, where everything is a sequence.
Definiteness sharpens part 2 into a clean maximum principle. Stakgold and Holst note that for a non-negative operator the absolute values can be dropped, $\lambda_{n+1}=\sup\{(Av,v):\lVert v\rVert=1,\ v\in M_{n}^{\perp}\}$ , and that for an indefinite operator the positive and negative eigenvalues each obey their own principle, with $\lambda_{1}^{-}\lVert v\rVert^{2}\le(Av,v)\le\lambda_{1}^{+}\lVert v\rVert^{2}$ . Without a sign condition the modulus is unavoidable, and the maximiser may belong to either end of the spectrum.

Method(Diagonalising a compact self-adjoint operator in practice)Original

Check the two hypotheses. Self-adjointness is usually a symmetry of a kernel, $K(x,t)=\overline{K(t,x)}$ , or of a matrix, $a_{ij}=\overline{a_{ji}}$ . Compactness is usually the square-integrability of the kernel or the decay of the diagonal (Compact operators). Both must be checked; neither implies the other.
Find the eigenvalues. If $A$ is the inverse of a differential operator, solve the differential eigenvalue problem instead: $Au=\lambda u$ with $\lambda\ne0$ is $Lu=\lambda^{-1}u$ with the boundary conditions built into $A$ (Sturm-Liouville eigenvalue problems). Otherwise look for a basis in which the kernel separates, or compute numerically.
Order and normalise. List the nonzero eigenvalues by decreasing modulus with multiplicity, apply Gram–Schmidt (Orthonormal bases and Gram-Schmidt) inside each eigenspace, and check orthogonality across eigenspaces as an arithmetic check — part 3 guarantees it.
Decide whether the eigenvectors span. By part 4 this is the single question: is $\ker A=\{0\}$ ? If $A$ is the inverse of an invertible operator, yes automatically.
Expand, and truncate with a known error. Write $Au=\sum\lambda_{n}(u,e_{n})e_{n}$ . Truncating after $m$ distinct eigenvalues costs exactly $\lvert\lambda_{m+1}\rvert$ in operator norm, by part 5 (and nothing at all once every distinct nonzero eigenvalue has been used, where $\lambda_{m+1}$ is read as $0$ ) — so the decay rate of the eigenvalues is the convergence rate of every computation built on the expansion.
Bound the largest eigenvalue without solving anything. Evaluate the Rayleigh quotient $R(v)=(Av,v)/(v,v)$ at any convenient trial vector $v$ . By part 1, $\lvert\lambda_{1}\rvert\ge\lvert R(v)\rvert$ , and the bound improves as $v$ is varied — this is the Rayleigh–Ritz method (The Rayleigh-Ritz method) read through the integral operator rather than the differential one.

Example(The Dirichlet Green's operator on the unit interval, with four numerical checks)Original

On $H=L^{2}(0,1)$ take the integral operator $A$ with kernel

K(x,t)=\min(x,t)-xt= \begin{cases} x(1-t), & x\le t,\\ t(1-x), & t<x,\end{cases}

the Green's function of $-\mathrm{d}^{2}/\mathrm{d}x^{2}$ with $u(0)=u(1)=0$ (Green's function for a boundary-value problem).

Hypotheses. $K$ is real, symmetric and continuous on the closed unit square, so $A$ is self-adjoint, and $\iint\lvert K\rvert^{2}<\infty$ makes it Hilbert–Schmidt, hence compact (Compact operators). Both hypotheses hold.

Eigenvalues and eigenvectors. If $Au=\lambda u$ with $\lambda\ne0$ then $w=Au=\lambda u$ satisfies $-w''=u=w/\lambda$ with $w(0)=w(1)=0$ , so $w''+w/\lambda=0$ and $w(x)=\sin(x/\sqrt\lambda)$ up to a constant, with $\sin(1/\sqrt{\lambda})=0$ . Hence

\lambda_{n}=\frac{1}{n^{2}\pi^{2}},\qquad e_{n}(x)=\sqrt2\,\sin(n\pi x),\qquad n=1,2,\dots,

with $\int_{0}^{1}2\sin(m\pi x)\sin(n\pi x)\,\mathrm{d}x=\delta_{mn}$ . All are simple and positive, decreasing to $0$ as part 2 requires, and $\ker A=\{0\}$ (if $Au=0$ then $u=-(Au)''=0$ ), so by part 4 the $\{e_{n}\}$ are an orthonormal basis of $L^{2}(0,1)$ — the familiar sine basis.

Check 1: the norm. Part 1 says $\lVert A\rVert=\lambda_{1}=1/\pi^{2}=0.1013212$ . The Hilbert–Schmidt bound of Compact operators gives $\lVert A\rVert\le\bigl(\iint\lvert K\rvert^{2}\bigr)^{1/2}$ , and

\iint_{[0,1]^{2}}\bigl(\min(x,t)-xt\bigr)^{2}\mathrm{d}t\,\mathrm{d}x=\frac{1}{90}=0.0111111,

so the bound is $1/\sqrt{90}=0.1054093$ — consistent, and tight to within $4\%$ here. Independently, $\sum_{n}\lambda_{n}^{2}=\pi^{-4}\sum n^{-4}=\pi^{-4}\cdot\pi^{4}/90=1/90$ : the two computations of the Hilbert–Schmidt norm agree exactly.

Check 2: the trace. Summing the eigenvalues, $\sum_{n}\lambda_{n}=\pi^{-2}\sum n^{-2}=\pi^{-2}\cdot\pi^{2}/6=1/6$ . Integrating the kernel along the diagonal, $\int_{0}^{1}K(x,x)\,\mathrm{d}x=\int_{0}^{1}(x-x^{2})\,\mathrm{d}x=\tfrac12-\tfrac13=1/6$ . Agreement.

Check 3: the expansion theorem on a concrete vector. Take $u\equiv1$ . Its coefficients are

(u,e_{n})=\sqrt2\int_{0}^{1}\sin(n\pi x)\,\mathrm{d}x=\frac{\sqrt2\,(1-\cos n\pi)}{n\pi} =\begin{cases}2\sqrt2/(n\pi),& n\ \text{odd},\\ 0,& n\ \text{even},\end{cases}

so part 4 predicts

(Au)(x)=\sum_{n\ \text{odd}}\frac{1}{n^{2}\pi^{2}}\cdot\frac{2\sqrt2}{n\pi}\cdot\sqrt2\sin(n\pi x) =\sum_{n\ \text{odd}}\frac{4\sin(n\pi x)}{n^{3}\pi^{3}} .

Computed directly, $(Au)(x)=\int_{0}^{1}\bigl(\min(x,t)-xt\bigr)\mathrm{d}t=\bigl(x-\tfrac{x^{2}}{2}\bigr)-\tfrac{x}{2}=\tfrac{x(1-x)}{2}$ , which at $x=\tfrac12$ equals $\tfrac18=0.125$ . The series at $x=\tfrac12$ is $\tfrac{4}{\pi^{3}}\bigl(1-\tfrac1{27}+\tfrac1{125}-\tfrac1{343}+\cdots\bigr)$ , whose partial sums are $0.129006$ , $0.124228$ , $0.125260$ , $0.124884$ , $0.125061$ , $0.124964$ — alternating about $0.125$ and squeezing it, as they must, with $\sum_{k\ge0}(-1)^{k}(2k+1)^{-3}=\pi^{3}/32$ giving the limit $4/\pi^{3}\cdot\pi^{3}/32=1/8$ exactly.

Check 4: the Rayleigh quotient bounds $\lambda_{1}$ from below. Take the trial function $v(x)=x(1-x)$ , which satisfies the boundary conditions but is not an eigenfunction. Then $Av=w$ solves $-w''=v$ , $w(0)=w(1)=0$ , giving $w(x)=\tfrac{x}{12}-\tfrac{x^{3}}{6}+\tfrac{x^{4}}{12}$ , and

(Av,v)=\int_{0}^{1}wv\,\mathrm{d}x=\frac{17}{5040}, \qquad (v,v)=\int_{0}^{1}x^{2}(1-x)^{2}\mathrm{d}x=\frac{1}{30}, \qquad R(v)=\frac{17}{168}=0.1011905 .

Part 1 requires $R(v)\le\lambda_{1}=0.1013212$ , and indeed it is smaller, by $0.13\%$ . Read back through $\lambda_{1}=1/\pi^{2}$ , this single polynomial trial function gives $\pi^{2}\le168/17=9.882353$ against $\pi^{2}=9.869604$ — the Rayleigh–Ritz bound for the fundamental frequency of a string, obtained without solving the eigenvalue problem.

ApplicationOriginal

Sturm–Liouville theory is a corollary. Let $L$ be a formally self-adjoint differential operator with real coefficients on $[a,b]$ , with boundary conditions making the problem self-adjoint and with $0$ not an eigenvalue (which costs nothing, since $L$ may be shifted by a constant). Its Green's function $g(x,\xi)$ is real and symmetric and generates a compact operator $\mathcal{G}$ on $L^{2}(a,b)$ , and the differential eigenvalue problem $Lu=\theta u$ becomes $u=\theta\,\mathcal{G}u$ . Parts 1–5 then deliver, in one stroke, the standard facts of Sturm-Liouville eigenvalue problems and Eigenfunction expansions: the eigenvalues are real and discrete with $\lvert\theta_{n}\rvert\to\infty$ , the eigenfunctions are orthogonal with finite-dimensional eigenspaces, and — because $\mathcal{G}$ has trivial null space, since $Lz=u$ with homogeneous data has the unique solution $z=\mathcal{G}u$ — they form an orthonormal basis. That is Stakgold and Holst's Theorem 6.3.6. The logical direction deserves emphasis: completeness of the normal modes of a vibrating string, a drumhead, a waveguide or a quantum well is not a fact about differential operators. It is a fact about compact self-adjoint operators, transferred to the differential problem through the Green's function. The kernel-level version of all this, including Mercer's theorem and the solution of the inhomogeneous integral equation, is Hilbert-Schmidt theory.

Normal modes and the variational characterisation of frequencies. Part 2 is the Rayleigh–Ritz principle in its exact form: the $n$ th eigenvalue is a constrained maximum of the Rayleigh quotient, so trial functions give one-sided bounds and a finite-dimensional trial space gives a convergent scheme (The Rayleigh-Ritz method, Variational characterization of eigenvalues, Generalized eigenproblems and normal modes). Check 4 of the example is the whole method in miniature. In structural engineering and in molecular vibration analysis, this is how a lowest resonance is estimated before any mode is computed; part 5's error identity $\lVert A-\sum_{j\le m}\lambda_{j}P_{j}\rVert=\lvert\lambda_{m+1}\rvert$ is the corresponding statement about how many modes must be kept.

Quantum mechanics: why bound states are discrete. A Hamiltonian is unbounded and this theorem does not apply to it directly. It applies to the resolvent $(H-z)^{-1}$ , which for a confining or sufficiently decaying potential is compact; the discreteness of the bound-state energies, the finite degeneracy of each level, and the orthogonality and completeness of the bound states in the relevant subspace are then parts 3 and 4 read back through $E\mapsto(E-z)^{-1}$ . The accumulation of eigenvalues at $0$ for the resolvent is the accumulation of energy levels at the ionisation threshold. The framework is set out in The mathematical framework of quantum mechanics, and the general tool for the non-compact case in The spectral theorem.

Data analysis: principal components and empirical orthogonal functions. A covariance operator $C$ defined by a square-integrable covariance kernel is self-adjoint, non-negative and compact. Part 1 says the direction of greatest variance exists and is an eigenvector; part 2 says the successive directions are obtained by the same maximisation restricted to the orthogonal complement; part 5 says that truncating after $m$ components leaves an error of exactly $\lambda_{m+1}$ in operator norm. That is the entire justification of principal component analysis and of empirical orthogonal functions in climate science (Principal components and empirical orthogonal functions), and the decay of $\lambda_{n}$ is what makes a handful of patterns describe a field with thousands of degrees of freedom. The matrix case is Spectral theorem for Hermitian matrices; this unit is the statement that nothing is lost when the index set becomes continuous, provided the operator is compact.

Theorem · advanced

Uniform boundedness and open mapping theorems

Adapted

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Two of the three pillars of linear functional analysis, both consequences of Baire's category theorem: a family of bounded operators on a Banach space that is bounded at every point is bounded in norm, and a bounded operator from one Banach space onto another maps open sets to open sets — from which the bounded inverse and closed graph theorems follow, and with them the fact that a symmetric operator defined on a whole Hilbert space cannot be unbounded.

Prerequisites: Metric spaces and completeness, Normed and Banach spaces, Bounded linear operators, Hilbert spaces

Definition(Category, open maps, graphs)AdaptedAdapted from Kreyszig (1978), §4.7, Def. 4.7-1, p. 247, §4.12, Def. 4.12-1, p. 286, and §4.13, Def. 4.13-1, p. 292; Folland (1999), §5.3, the terminology after Thm 5.9 and the definition of the graph, pp. 161–162; Evans (2010), Appendix D.3, the definition of a closed operator, p. 677

Let $X$ be a metric space. A subset $M\subset X$ is

nowhere dense (Kreyszig: rare) if its closure has empty interior — it contains no ball;
meager, or of the first category, if it is a countable union of nowhere dense sets;
nonmeager, or of the second category, otherwise. A set is called residual when what it omits is meager.

Let $X,Y$ be metric spaces. A map $f\colon X\to Y$ is open if $f(U)$ is open in $Y$ for every open $U\subset X$ . For a linear map between normed spaces this simplifies: since a linear map commutes with translations and with dilations, $f$ is open exactly when $f(B)$ contains a ball centred at $0$ in $Y$ , where $B$ is the open unit ball of $X$ . Openness is not continuity and does not follow from it; Kreyszig's reminder is $t\mapsto\sin t$ on $\mathbb{R}$ , continuous but carrying the open interval $(0,2\pi)$ to the closed set $[-1,1]$ .

Let $T\colon \mathcal{D}(T)\to Y$ be linear with domain $\mathcal{D}(T)\subset X$ . Its graph is

\Gamma(T)=\{(x,Tx)\;:\;x\in\mathcal{D}(T)\}\subset X\times Y ,

and $T$ is closed if $\Gamma(T)$ is a closed subset of $X\times Y$ with the norm $\lVert(x,y)\rVert=\lVert x\rVert+\lVert y\rVert$ . Unwinding the definition gives the form used in practice:

T\ \text{is closed}\iff \bigl[\,x_{n}\in\mathcal{D}(T),\ x_{n}\to x,\ Tx_{n}\to y\ \Rightarrow\ x\in\mathcal{D}(T)\ \text{and}\ Tx=y\,\bigr].

Compare continuity, which asserts that $x_{n}\to x$ by itself forces $Tx_{n}\to Tx$ . Closedness is weaker: it assumes the image sequence already converges and only identifies the limit.

Part 5 of the second statement below, and its proof, are the one place where a Hilbert space is needed. There $H$ is a complex Hilbert space (Hilbert spaces) with inner product $(x,y)$ , linear in the first slot and conjugate-linear in the second, so that $\lVert x\rVert^{2}=(x,x)$ . Two of its properties are used: the inner product is continuous in each slot separately, since $\lvert(x,y)-(x',y)\rvert\le\lVert x-x'\rVert\lVert y\rVert$ by Cauchy–Schwarz; and it is non-degenerate, so that $(z,v)=0$ for every $z$ forces $v=0$ (take $z=v$ ). Reed and Simon write the inner product conjugate-linear in the first slot instead; the symmetry hypothesis $(Ax,y)=(x,Ay)$ is the same statement in either convention.

Theorem(Baire's category theorem)AdaptedAdapted from Kreyszig (1978), §4.7, Thm 4.7-2, p. 247; Folland (1999), §5.3, Thm 5.9, p. 161; Reed & Simon (1980), §III.5, p. 81

A nonempty complete metric space is nonmeager in itself. Equivalently: if a complete metric space $X$ is written as a countable union $X=\bigcup_{k\ge1}A_{k}$ of closed sets, then at least one $A_{k}$ contains a ball. Equivalently again (Folland's form (a)): in a complete metric space the intersection of countably many dense open sets is dense.

The converse fails — an incomplete normed space can be nonmeager in itself — so completeness is sufficient, not necessary.

Theorem(Uniform boundedness, or the Banach–Steinhaus theorem)AdaptedAdapted from Kreyszig (1978), §4.7, Thm 4.7-3, p. 249; Folland (1999), §5.3, Thm 5.13, p. 163; Reed & Simon (1980), §III.5, Thm III.8, p. 81; Stakgold & Holst (2011), §5.9, Thm 5.9.4, p. 348

Let $X$ be a normed space, $Y$ a normed space, and $\mathcal{F}$ a family of bounded linear operators from $X$ to $Y$ .

(Folland's form.) If the set of $x$ at which $\sup_{T\in\mathcal{F}}\lVert Tx\rVert<\infty$ is nonmeager in $X$ , then $\sup_{T\in\mathcal{F}}\lVert T\rVert<\infty$ .
(The usual form.) If $X$ is a Banach space and for every $x\in X$ there is a constant $c_{x}$ with $\lVert Tx\rVert\le c_{x}\qquad\text{for all }T\in\mathcal{F},$ then there is a single constant $c$ with $\lVert T\rVert\le c$ for all $T\in\mathcal{F}$ .

Part 2 follows from part 1 because a Banach space is nonmeager in itself. Note what is not assumed: the family need not be countable, need not be a sequence, and nothing is assumed about convergence. The useful reading is often the contrapositive — if the norms $\lVert T\rVert$ are unbounded and $X$ is complete, then some single $x$ has $\sup_{T}\lVert Tx\rVert=\infty$ .

Theorem(Open mapping, bounded inverse, closed graph)AdaptedAdapted from Kreyszig (1978), §4.12, Thm 4.12-2 and Lemma 4.12-3, pp. 286–287, and §4.13, Thm 4.13-2, p. 292; Folland (1999), §5.3, Thm 5.10, Cor. 5.11 and Thm 5.12, pp. 162–163; Reed & Simon (1980), §III.5, Thms III.10, III.11 and III.12, pp. 82–83, and the Corollary on p. 86; Stakgold & Holst (2011), §5.9, Thms 5.9.1–5.9.3, p. 347; Evans (2010), Appendix D.3, Thm 1, p. 677

Let $X$ and $Y$ be Banach spaces.

Open mapping theorem (Banach–Schauder). A bounded linear operator $T\colon X\to Y$ that is onto $Y$ is an open mapping.
Bounded inverse theorem. If in addition $T$ is injective, then $T^{-1}\colon Y\to X$ is a bounded linear operator. So a bounded linear bijection between Banach spaces is an isomorphism of normed spaces.
Equivalence of complete norms. If $\lVert\cdot\rVert_{1}$ and $\lVert\cdot\rVert_{2}$ are two norms on one vector space $X$ , each making it complete, and $\lVert x\rVert_{2}\le C\lVert x\rVert_{1}$ for all $x$ , then the norms are equivalent: $a\lVert x\rVert_{1}\le\lVert x\rVert_{2}\le C\lVert x\rVert_{1}$ for some $a>0$ . (Apply part 2 to the identity map.)
Closed graph theorem. A closed linear operator $T\colon\mathcal{D}(T)\to Y$ whose domain $\mathcal{D}(T)$ is a closed subspace of $X$ — in particular one defined on all of $X$ — is bounded. Conversely, a bounded operator with closed domain is closed.
Hellinger–Toeplitz theorem. Let $H$ be a Hilbert space and $A\colon H\to H$ a linear operator defined on all of $H$ with $(Ax,y)=(x,Ay)$ for all $x,y\in H$ . Then $A$ is bounded.

Statements 1, 2 and 4 are equivalent to one another: Stakgold and Holst record this, and the proofs below derive 2 from 1 and 4 from 2.

Note(Names, and the company these theorems keep)AdaptedAdapted from Stakgold & Holst (2011), §5.9, the list of the three basic principles and the names attached to them, p. 347

Three theorems are traditionally singled out as the foundations of the linear theory, and Stakgold and Holst list them in that way. One is the extension theorem for functionals, due to Hahn and Banach (The Hahn-Banach theorem), which rests on Zorn's lemma (Zorn's lemma and the axiom of choice) rather than on completeness. The other two are the subject of this unit: Banach and Schauder's open mapping theorem, and the principle of uniform boundedness, which carries the names of Banach and Steinhaus. Both of the latter are consequences of Baire's theorem, and completeness is where their force comes from.

Idea of the proofOriginal

Both theorems convert a pointwise hypothesis into a uniform conclusion, and both do it the same way. Complete a metric space and you gain one thing: it cannot be assembled out of countably many negligible pieces. So if a space is covered by countably many closed sets defined by some bound — " $\lVert Tx\rVert\le n$ for every $T$ in the family", or " $y$ is within $n$ of the image of the unit ball" — one of those sets must be fat enough to contain a ball. A ball is all that linearity needs: translating it to the origin and rescaling turns a bound on a ball into a bound on the whole space. That is the uniform boundedness theorem in two sentences. The open mapping theorem needs one extra manoeuvre, because Baire's theorem produces a ball inside the closure of the image of the unit ball, and a closure is not good enough. The manoeuvre is a geometric series: hit the target to within half, correct the remainder to within a quarter using the rescaled statement, and so on; the corrections form an absolutely convergent series, which converges because $X$ is complete, and its sum is an exact preimage. The closed graph theorem is then free: the graph of a closed operator is itself a Banach space, and the projection from the graph onto $X$ is a bounded bijection, so its inverse $x\mapsto(x,Tx)$ is bounded — which says exactly that $T$ is.

Proof(Baire's category theorem)AdaptedAdapted from Kreyszig (1978), §4.7, proof of Thm 4.7-2, pp. 247–248; Folland (1999), §5.3, proof of Thm 5.9(a), p. 161

Let $X\ne\emptyset$ be complete and suppose, for contradiction, that $X=\bigcup_{k\ge1}M_{k}$ with every $M_{k}$ nowhere dense. We build a point lying in no $M_{k}$ .

Since $\overline{M_{1}}$ contains no ball and $X$ does, $\overline{M_{1}}\ne X$ , so its complement is nonempty and open; choose a closed ball $\overline{B}_{1}=\overline{B}(p_{1};r_{1})$ inside it with $r_{1}<\tfrac12$ . Inductively, suppose the closed ball $\overline{B}_{k}$ of radius $r_{k}<2^{-k}$ has been chosen with $\overline{B}_{k}\cap M_{k}=\emptyset$ . The set $\overline{M_{k+1}}$ contains no ball, so it does not contain the open ball $B_{k}$ ; hence $B_{k}\setminus\overline{M_{k+1}}$ is nonempty and open, and we may choose inside it a closed ball $\overline{B}_{k+1}$ of radius $r_{k+1}<2^{-(k+1)}$ . By construction $\overline{B}_{1}\supset\overline{B}_{2}\supset\cdots$ and $\overline{B}_{k}\cap M_{k}=\emptyset$ .

The centres satisfy $d(p_{m},p_{n})<2^{-m}$ for $n>m$ , since both lie in $\overline{B}_{m}$ ; so $(p_{k})$ is Cauchy and, by completeness, converges to some $p\in X$ . For each fixed $m$ the tail $(p_{k})_{k\ge m}$ lies in the closed set $\overline{B}_{m}$ , so $p\in\overline{B}_{m}$ , and therefore $p\notin M_{m}$ . As $m$ was arbitrary, $p$ lies in no $M_{m}$ , contradicting $X=\bigcup M_{k}$ .

For the second form, if $X=\bigcup A_{k}$ with each $A_{k}$ closed and none containing a ball, then each $A_{k}$ is nowhere dense and the above applies. $\square$

Proof(Uniform boundedness)AdaptedAdapted from Folland (1999), §5.3, proof of Thm 5.13, p. 163; Kreyszig (1978), §4.7, proof of Thm 4.7-3, pp. 249–250; Reed & Simon (1980), §III.5, proof of Thm III.8, p. 81

For $n\in\mathbb{N}$ put

E_{n}=\{x\in X:\ \lVert Tx\rVert\le n\ \text{for every }T\in\mathcal{F}\}=\bigcap_{T\in\mathcal{F}}\{x:\lVert Tx\rVert\le n\} .

Each set in the intersection is closed, because $x\mapsto\lVert Tx\rVert$ is continuous ( $T$ is bounded, the norm is continuous), and intersections of closed sets, however many, stay closed; so every $E_{n}$ is closed. The hypothesis of part 1 says that $\bigcup_{n}E_{n}$ contains a nonmeager set, hence is itself nonmeager, hence — being a countable union — cannot have all its terms nowhere dense. Some $E_{n}$ is therefore not nowhere dense; being closed, it contains a ball $B(x_{0};r)$ .

Now transport that ball to the origin. If $\lVert x\rVert<r$ then $x_{0}+x\in E_{n}$ and $x_{0}\in E_{n}$ , so for every $T\in\mathcal{F}$

\lVert Tx\rVert=\lVert T(x_{0}+x)-Tx_{0}\rVert\le\lVert T(x_{0}+x)\rVert+\lVert Tx_{0}\rVert\le 2n .

Taking the supremum over $\lVert x\rVert<r$ and rescaling by $r$ gives $\lVert T\rVert\le2n/r$ for every $T\in\mathcal{F}$ , a bound independent of $T$ . That proves part 1.

Part 2: if $X$ is a Banach space then by Baire's theorem $X$ is nonmeager in itself, and the hypothesis says the set where the family is pointwise bounded is all of $X$ ; apply part 1 with $c=2n/r$ . $\square$

Proof(Open mapping theorem)AdaptedAdapted from Folland (1999), §5.3, proof of Thm 5.10, p. 162; Kreyszig (1978), §4.12, proof of Lemma 4.12-3 and of Thm 4.12-2, pp. 287–289; Reed & Simon (1980), §III.5, the reductions and proof of Thm III.10, pp. 82–83

Write $B_{r}=\{x\in X:\lVert x\rVert<r\}$ . By the reduction noted in the definition, it suffices to prove that $T(B_{1})$ contains a ball about $0$ in $Y$ .

Stage 1: a ball inside the closure. Since every $x\in X$ lies in $B_{n}$ for $n$ large, $X=\bigcup_{n\ge1}B_{n}$ , and surjectivity gives $Y=\bigcup_{n\ge1}T(B_{n})\subset\bigcup_{n\ge1}\overline{T(B_{n})}$ . Each $\overline{T(B_{n})}$ is closed, and $Y$ is complete, so Baire's theorem says some $\overline{T(B_{n})}$ contains a ball. The dilation $y\mapsto y/n$ is a homeomorphism of $Y$ carrying $\overline{T(B_{n})}$ onto $\overline{T(B_{1})}$ , so $\overline{T(B_{1})}$ contains a ball $B(y_{0};4\rho)$ for some $y_{0}\in Y$ and $\rho>0$ .

Next remove the centre. Pick $\tilde y=T\tilde x\in T(B_{1})$ with $\lVert \tilde y-y_{0}\rVert<2\rho$ , so that $B(\tilde y;2\rho)\subset B(y_{0};4\rho)\subset\overline{T(B_{1})}$ . If $\lVert y\rVert<2\rho$ then $y+\tilde y\in\overline{T(B_{1})}$ , so

y=(y+\tilde y)-T\tilde x\in\overline{T(B_{1})}-T(B_{1})\subset\overline{T(B_{2})},

using linearity and $\lVert \tilde x\rVert<1$ . (The vectors $x_{1},x_{2},\dots$ of stage 2 below are unrelated to $\tilde x$ .) Halving, we conclude: there is $\rho>0$ with

\lVert y\rVert<\rho\ \Longrightarrow\ y\in\overline{T(B_{1})}, \qquad\text{hence, by dilation,}\qquad \lVert y\rVert<\rho\,2^{-n}\ \Longrightarrow\ y\in\overline{T(B_{2^{-n}})} . \tag{$\ast$}

Stage 2: removing the closure. Let $\lVert y\rVert<\rho/2$ . By $(\ast)$ with $n=1$ , $y\in\overline{T(B_{1/2})}$ , so there is $x_{1}\in B_{1/2}$ with $\lVert y-Tx_{1}\rVert<\rho/4$ . By $(\ast)$ with $n=2$ , the remainder $y-Tx_{1}$ lies in $\overline{T(B_{1/4})}$ , so there is $x_{2}\in B_{1/4}$ with $\lVert y-Tx_{1}-Tx_{2}\rVert<\rho/8$ . Continuing, we obtain $x_{k}\in B_{2^{-k}}$ with

\Bigl\lVert y-\sum_{k\le n}Tx_{k}\Bigr\rVert<\rho\,2^{-n-1}\qquad(n=1,2,\dots).

Since $\sum_{k}\lVert x_{k}\rVert<\sum_{k}2^{-k}=1$ , the series $\sum_{k}x_{k}$ is absolutely convergent and therefore converges in the complete space $X$ , to some $x$ with $\lVert x\rVert<1$ . By continuity of $T$ , $Tx=\lim_{n}\sum_{k\le n}Tx_{k}=y$ . So $y\in T(B_{1})$ .

Thus $T(B_{1})$ contains the ball of radius $\rho/2$ about $0$ in $Y$ , and $T$ is open.

Finally, for a general open $A\subset X$ and $y=Tx\in T(A)$ : $A$ contains a ball $B(x;r)$ , so $r^{-1}(A-x)$ contains $B_{1}$ , and by what was just proved $T(r^{-1}(A-x))=r^{-1}(T(A)-y)$ contains a ball about $0$ ; hence $T(A)$ contains a ball about $y$ . As $y$ was arbitrary, $T(A)$ is open. $\square$

Proof(Bounded inverse, closed graph, Hellinger–Toeplitz)AdaptedAdapted from Kreyszig (1978), §4.12, Thm 4.12-2, p. 289, and §4.13, proof of Thm 4.13-2, pp. 292–293; Folland (1999), §5.3, Cor. 5.11 and proof of Thm 5.12, pp. 162–163; Reed & Simon (1980), §III.5, Thms III.11 and III.12, p. 83, and the Corollary on p. 86

Bounded inverse (part 2). Let $T$ be a bounded bijection. For open $A\subset X$ , the preimage of $A$ under $T^{-1}$ is $T(A)$ , which is open by part 1; so $T^{-1}$ is continuous, and being linear (the inverse of a linear bijection is linear) it is bounded.

Equivalence of norms (part 3). Let $X_{1}=(X,\lVert\cdot\rVert_{1})$ and $X_{2}=(X,\lVert\cdot\rVert_{2})$ , both complete. The identity $\iota\colon X_{1}\to X_{2}$ is a linear bijection, and bounded because $\lVert x\rVert_{2}\le C\lVert x\rVert_{1}$ . By part 2 its inverse is bounded: $\lVert x\rVert_{1}\le C'\lVert x\rVert_{2}$ , which is the stated two-sided bound with $a=1/C'$ .

Closed graph (part 4). Suppose $\mathcal{D}(T)$ is closed in $X$ and $T$ is closed. First, $X\times Y$ with $\lVert(x,y)\rVert=\lVert x\rVert+\lVert y\rVert$ is complete: a Cauchy sequence $(x_{n},y_{n})$ has both coordinates Cauchy, hence convergent, hence converges in the product norm. Now $\Gamma(T)$ is a closed subspace of a Banach space, so is itself a Banach space, and $\mathcal{D}(T)$ is a closed subspace of $X$ , so is a Banach space too. Consider

P\colon \Gamma(T)\to\mathcal{D}(T),\qquad P(x,Tx)=x .

$P$ is linear, bijective (its inverse is $x\mapsto(x,Tx)$ ), and bounded, since $\lVert P(x,Tx)\rVert=\lVert x\rVert\le\lVert x\rVert+\lVert Tx\rVert=\lVert(x,Tx)\rVert$ . By part 2, $P^{-1}$ is bounded: there is $b$ with $\lVert x\rVert+\lVert Tx\rVert\le b\lVert x\rVert$ for all $x\in\mathcal{D}(T)$ . In particular $\lVert Tx\rVert\le b\lVert x\rVert$ , so $T$ is bounded. The converse is immediate: if $T$ is bounded with closed domain and $x_{n}\to x$ , $Tx_{n}\to y$ with $x_{n}\in\mathcal{D}(T)$ , then $x\in\mathcal{D}(T)$ by closedness of the domain and $Tx_{n}\to Tx$ by continuity, so $y=Tx$ .

Hellinger–Toeplitz (part 5). We follow Reed and Simon and show that $\Gamma(A)$ is closed; since $\mathcal{D}(A)=H$ is closed, part 4 then gives boundedness. Let $x_{n}\to x$ and $Ax_{n}\to y$ . For any $z\in H$ ,

(z,y)=\lim_{n}(z,Ax_{n})=\lim_{n}(Az,x_{n})=(Az,x)=(z,Ax),

where the first equality is continuity of the inner product in its second slot (applied to $Ax_{n}\to y$ ), the second is the symmetry hypothesis, the third is continuity of the inner product again (now applied to $x_{n}\to x$ ), and the fourth is symmetry once more. Hence $(z,y-Ax)=0$ for every $z$ , so $y=Ax$ and $(x,y)\in\Gamma(A)$ . $\square$

Remark(Where completeness is used, and what fails without it)AdaptedAdapted from Kreyszig (1978), §4.7, Application 4.7-4, pp. 250–251, §4.12, Problem 5, p. 290, and §4.13, Example 4.13-4 with the comment that follows it, pp. 294; Folland (1999), §5.3, the remark before Exercise 27 and Exercises 29–31, pp. 163–164; Reed & Simon (1980), §III.5, the emphasis after Thm III.12, p. 83

Uniform boundedness needs $X$ complete, not $Y$ . Completeness enters only once, to make $X$ nonmeager in itself. Kreyszig turns this into a test for incompleteness: on the space of polynomials with $\lVert x\rVert=\max_{i}\lvert a_{i}\rvert$ (the largest coefficient), let $f_{n}(x)=a_{0}+a_{1}+\dots+a_{n-1}$ be the sum of the first $n$ coefficients. Each $f_{n}$ is a bounded functional with $\lVert f_{n}\rVert\le n$ , and for a fixed polynomial of degree $N$ the values $\lvert f_{n}(x)\rvert\le(N+1)\max_{i}\lvert a_{i}\rvert$ are bounded in $n$ — so the family is pointwise bounded. But $x(t)=1+t+\dots+t^{n-1}$ has norm $1$ and $f_{n}(x)=n$ , so $\lVert f_{n}\rVert\ge n$ and the norms are unbounded. The only hypothesis available to fail is completeness, so this space is not a Banach space.
Folland's part 1 is genuinely stronger. It asks only that the pointwise bound hold on a nonmeager set, not on all of $X$ , and it does not mention completeness at all. The gain is real: a set can be nonmeager and still small in other senses, and in applications one often controls a family only on a dense or generic set.
The open mapping theorem needs both spaces complete, and surjectivity. Completeness of $Y$ licenses Baire in stage 1; completeness of $X$ makes the absolutely convergent series of stage 2 converge. Drop either and the theorem fails, and Folland supplies one counterexample for each side. For an incomplete target, his Exercise 29 takes $\mathcal{Y}=\ell^{1}$ and the incomplete dense subspace $X=\{f:\sum n\lvert f(n)\rvert<\infty\}$ with the $\ell^{1}$ norm; the map $S=T^{-1}$ , where $Tf(n)=nf(n)$ , is bounded from $\mathcal{Y}$ onto $X$ but not open. For an incomplete domain, his Exercise 31 takes Banach spaces $X,\mathcal{Y}$ and an unbounded linear $S\colon X\to\mathcal{Y}$ ; the graph $\Gamma(S)$ is then not complete, and the projection $\Gamma(S)\to X$ , $(x,Sx)\mapsto x$ , is bounded and onto the Banach space $X$ but not open. Surjectivity cannot be dropped either: the inclusion of $\ell^{1}$ into $\ell^{2}$ is bounded and injective between Banach spaces but not onto, and its inverse on the range is unbounded.
The closed graph theorem needs the domain closed, and that is exactly what fails in the examples one cares about. Kreyszig's Example 4.13-4: on $X=C[0,1]$ let $Tx=x'$ with $\mathcal{D}(T)$ the continuously differentiable functions. $T$ is closed — if $x_{n}\to x$ and $x_{n}'\to y$ uniformly, then integrating $x_{n}'$ and passing to the limit gives $x(t)=x(0)+\int_{0}^{t}y$ , so $x$ is differentiable with $x'=y$ — but $T$ is unbounded, as $x_{n}(t)=t^{n}$ shows, with $\lVert x_{n}\rVert=1$ and $\lVert x_{n}'\rVert=n$ . The theorem is not contradicted: $\mathcal{D}(T)$ is a dense but not closed subspace of $C[0,1]$ . Kreyszig draws the moral explicitly — were the domain closed, the operator would have to be bounded.
Closed and bounded are independent conditions. Closedness does not give boundedness (previous bullet) and boundedness does not give closedness unless the domain is closed (restrict a bounded operator to a non-closed dense subspace). Reed and Simon stress that in their closed graph theorem $T$ is defined on all of $X$ , precisely because the unbounded operators of Unbounded and closed operators are closed operators on proper dense domains.
Hellinger–Toeplitz is a theorem about domains, read backwards. Part 5 says a symmetric operator defined everywhere on a Hilbert space is bounded. Contrapositively, an unbounded symmetric operator cannot be defined on all of $H$ ; it must live on a proper dense domain. Reed and Simon call this the cause of much technical pain, since the energy, position and momentum operators of quantum mechanics are all unbounded and symmetric, and even defining $A+B$ or $AB$ requires care about $\mathcal{D}(A)\cap\mathcal{D}(B)$ — which can be $\{0\}$ even for two dense domains.

Method(Using the three theorems)Original

To upgrade pointwise bounds to a uniform bound, check that the domain space is complete and that each member of the family is individually bounded, then apply uniform boundedness. Typical uses: a sequence of approximation operators converges on a dense set and is pointwise bounded, so it converges everywhere; a weakly convergent sequence is bounded; a separately continuous bilinear form is jointly continuous (Reed and Simon's corollary).
To prove that something cannot work, run uniform boundedness backwards. Compute or bound $\lVert T_{n}\rVert$ from below, show the bound diverges, and conclude that some single input breaks the scheme. The input is not constructed — this is a pure existence argument — but the conclusion is definitive.
To prove an inverse is bounded, do not estimate it. Check that both spaces are complete and that the operator is a bounded bijection, and invoke the bounded inverse theorem. This is how one knows the resolvent $(A-\eta I)^{-1}$ of a bounded operator is bounded for $\eta$ in the resolvent set (Spectrum and resolvent). Evans draws that conclusion from the Closed Graph Theorem instead (Appendix D.3, p. 677), which by part 4 comes to the same thing; parts 1, 2 and 4 are equivalent.
To prove an everywhere-defined operator is bounded, use the closed graph theorem instead of an estimate. Folland's phrasing of the saving: to show $Tx_{n}\to Tx$ whenever $x_{n}\to x$ , you are allowed to assume that $Tx_{n}$ converges to something and need only identify the limit. That is usually much easier, because the limit can be identified by testing against a dense set or by an integral identity.
To compare two norms on the same space, if both are complete and one dominates the other, they are equivalent. There is no need to produce the reverse constant.

Example(Why high-order Newton–Cotes quadrature fails, with the constants computed)Original

A quadrature rule on $[0,1]$ using $n$ nodes is a linear functional

Q_{n}(f)=\sum_{k=0}^{n-1}w_{k}^{(n)}f(x_{k}^{(n)})

on the Banach space $C[0,1]$ with the supremum norm. Its norm is

\lVert Q_{n}\rVert=\sum_{k=0}^{n-1}\bigl\lvert w_{k}^{(n)}\bigr\rvert ,

since $\lvert Q_{n}(f)\rvert\le\lVert f\rVert\sum_{k}\lvert w_{k}\rvert$ always, and a continuous $f$ of norm $1$ can be chosen to equal $\operatorname{sign}w_{k}$ at each node (finitely many points, so interpolate linearly between them), giving equality.

The rule. Take the closed Newton–Cotes rule: equally spaced nodes $x_{k}=k/(n-1)$ , and weights obtained by integrating the Lagrange interpolation basis, $w_{k}=\int_{0}^{1}\prod_{j\ne k}\frac{x-x_{j}}{x_{k}-x_{j}}\mathrm{d}x$ . By construction $Q_{n}(p)=\int_{0}^{1}p$ for every polynomial $p$ of degree at most $n-1$ , and in particular $\sum_{k}w_{k}=Q_{n}(1)=1$ for every $n$ .

The weights eventually change sign. For $n\le8$ and for $n=10$ all the weights are positive, so $\lVert Q_{n}\rVert=\sum_{k}w_{k}=1$ . At $n=9$ a weight goes negative and $\lVert Q_{9}\rVert=6857/4725=1.451217$ . At $n=11$ the weights are

\frac{1}{598752}\bigl(16067,\;106300,\;-48525,\;272400,\;-260550,\;427368,\;-260550,\;272400,\;-48525,\;106300,\;16067\bigr),

which sum to $1$ as they must, while their absolute values sum to

\lVert Q_{11}\rVert=\frac{152921}{49896}=3.0647948 .

Continuing: $\lVert Q_{13}\rVert=7.5317$ , $\lVert Q_{15}\rVert=20.344$ , $\lVert Q_{21}\rVert=544.18$ , $\lVert Q_{31}\rVert=2.1196\times10^{5}$ . The growth is geometric, and $\sup_{n}\lVert Q_{n}\rVert=\infty$ .

The conclusion from uniform boundedness. $C[0,1]$ is a Banach space and each $Q_{n}$ is a bounded functional, so the contrapositive of part 2 of the uniform boundedness theorem applies: since $\sup_{n}\lVert Q_{n}\rVert=\infty$ , the family cannot be pointwise bounded. There exists a continuous function $f$ on $[0,1]$ with $\sup_{n}\lvert Q_{n}(f)\rvert=\infty$ — for which the Newton–Cotes approximations do not merely fail to converge to $\int_{0}^{1}f$ , they blow up. No such $f$ is exhibited; the theorem asserts existence only.

The positive half of the statement, for contrast. Suppose instead a family of rules has $\sup_{n}\lVert Q_{n}\rVert=M<\infty$ and $Q_{n}(p)\to\int_{0}^{1}p$ for every polynomial $p$ . Given $f\in C[0,1]$ and $\varepsilon>0$ , the Weierstrass approximation theorem — the polynomials are dense in $C[0,1]$ for the supremum norm, proved in Cesàro means and Fejér's theorem — supplies a polynomial $p$ with $\lVert f-p\rVert<\varepsilon$ , and then

\Bigl\lvert Q_{n}(f)-\int_{0}^{1}f\Bigr\rvert \le\lVert Q_{n}\rVert\,\lVert f-p\rVert+\Bigl\lvert Q_{n}(p)-\int_{0}^{1}p\Bigr\rvert+\Bigl\lvert\int_{0}^{1}(p-f)\Bigr\rvert \le (M+1)\varepsilon+\Bigl\lvert Q_{n}(p)-\int_{0}^{1}p\Bigr\rvert ,

and the last term tends to $0$ . So convergence on a dense set plus a uniform norm bound gives convergence everywhere. Gaussian quadrature has all weights positive, hence $\lVert Q_{n}\rVert=1$ for every $n$ , and therefore converges for every continuous integrand; that, and not the order of exactness, is the structural reason it is preferred (Numerical quadrature).

Checking the numbers. Two arithmetic checks are available at every $n$ , from exactness on $f\equiv1$ and on $f(x)=x$ : the weights must sum to $1$ , and $\sum_{k}w_{k}x_{k}$ must equal $\tfrac12$ . For $n=11$ the displayed numerators sum to $598752$ , giving $\sum_{k}w_{k}=1$ ; and weighting the $k$ th numerator by $k$ gives $2993760$ , so $\sum_{k}w_{k}x_{k}=2993760/(10\cdot598752)=\tfrac12$ exactly. A third check ties the norm to the sign pattern: since the weights sum to $1$ , the norm exceeds $1$ by twice the total size of the negative weights,

\lVert Q_{11}\rVert-1=\frac{2\,(48525+260550+260550+48525)}{598752}=\frac{1236300}{598752}=2.0647948 ,

which matches $152921/49896-1$ . It is this cancellation between large weights of opposite sign that destroys the rule: the exact integral of a polynomial survives it, but an arbitrary continuous function does not.

ApplicationOriginal

Divergent Fourier series of a continuous function. The oldest application of uniform boundedness is Kreyszig's 4.7-5. On the Banach space of continuous $2\pi$ -periodic functions, let $f_{n}$ be the functional sending $x$ to the value its Fourier partial sum of order $n$ takes at the origin. Writing the partial sum as an integral against the Dirichlet kernel shows that $\lVert f_{n}\rVert$ is the Lebesgue constant $\tfrac{1}{2\pi}\int_{0}^{2\pi}\lvert q_{n}\rvert$ , with $q_{n}(t)=\sin\bigl((n+\tfrac12)t\bigr)/\sin(t/2)$ , and bounding $\lvert\sin(t/2)\rvert\le t/2$ turns this into a multiple of a partial sum of the harmonic series, which diverges. Uniform boundedness therefore produces a periodic continuous function at whose origin the Fourier series fails to converge. The structure is identical to the quadrature example: a dense set of good inputs (trigonometric polynomials, there; polynomials, here) plus unbounded norms forces a bad input to exist. This is the negative half of the convergence theory in Pointwise convergence of Fourier series, and the reason the positive results there need a smoothness hypothesis, or a summation method as in Cesàro means and Fejér's theorem.

Numerical analysis: stability is a norm bound. The pattern "consistency on a dense set plus uniform stability implies convergence" is the Lax equivalence principle, and the uniform boundedness theorem is what makes the stability hypothesis unavoidable rather than merely convenient: if a family of discretisation operators is not uniformly bounded, some data must break it. The quadrature example is the simplest instance; the same argument governs interpolation at a fixed node family (Lebesgue constants again, Polynomial and spline interpolation) and finite-difference schemes (Stability, the CFL condition and von Neumann analysis).

Quantum mechanics: unbounded observables need domains. Hellinger–Toeplitz is the reason the mathematical framework of quantum mechanics is more delicate than the finite-dimensional analogy suggests. Position, momentum and energy are symmetric and unbounded, so part 5 forbids them from being defined on the whole Hilbert space; every one of them carries a dense domain, and the physics of self-adjointness — whether an operator has a unique self-adjoint extension, and hence a well-defined time evolution — is a question about those domains (Unbounded and closed operators, Symmetric versus self-adjoint operators, The mathematical framework of quantum mechanics). Part 5 also supplies half of the reason the canonical commutation relation $[x,p]=\mathrm{i}\hbar I$ cannot hold with both operators defined on all of $H$ . Hellinger–Toeplitz is the first half: $x$ and $p$ are symmetric, so everywhere-defined versions of them would be bounded. The second half is a purely algebraic fact about bounded operators, which Hellinger–Toeplitz does not give and which is worth separating out. If $A$ and $B$ are bounded with $[A,B]=cI$ and $c\ne0$ , rescale so that $c=1$ ; then induction on $n$ , using $[A,B^{n+1}]=[A,B^{n}]B+B^{n}[A,B]$ , gives $[A,B^{n}]=nB^{n-1}$ , whence

n\lVert B^{n-1}\rVert=\lVert AB^{n}-B^{n}A\rVert\le2\lVert A\rVert\,\lVert B\rVert\,\lVert B^{n-1}\rVert .

If no power of $B$ vanishes this forces $n\le2\lVert A\rVert\lVert B\rVert$ for every $n$ , which is false; and if $B^{m}=0$ for a smallest $m\ge1$ , then $mB^{m-1}=[A,B^{m}]=AB^{m}-B^{m}A=0$ gives $B^{m-1}=0$ , which contradicts the minimality of $m$ when $m\ge2$ and reads $I=B^{0}=0$ when $m=1$ — false either way. No bounded pair satisfies the relation. Putting the two halves together: $x$ and $p$ must live on proper dense domains, and $[x,p]=\mathrm{i}\hbar I$ can hold only on a subspace where both sides make sense. (The trace argument familiar from finite dimensions — take traces of $[x,p]=\mathrm{i}\hbar I$ and get $0=\mathrm{i}\hbar\dim H$ — is unavailable here, because in infinite dimensions $I$ is not trace class and neither side need have a trace.)

Weak boundedness and weak convergence. A weakly convergent sequence in a normed space is bounded, and a weakly bounded set is strongly bounded. Both follow by applying uniform boundedness to the family of functionals $\hat x(\Lambda)=\Lambda(x)$ on the Banach space $X^{*}$ : pointwise boundedness is weak boundedness, and the uniform conclusion is a norm bound. This is used without comment throughout Weak convergence and in the direct method for variational problems, where a minimising sequence is first shown to be bounded and then has a weakly convergent subsequence extracted.

Equivalence of complete norms, and why Sobolev norms are unambiguous. Part 3 says a vector space carries at most one complete norm topology up to equivalence once one norm dominates another. In practice this removes a whole class of worries: two natural definitions of the same function space, each complete, with an obvious inequality one way, are automatically equivalent, and no constant need be chased in the other direction. This is the standard argument behind the equivalence of the various norms on Sobolev spaces (Weak derivatives and Sobolev spaces) and of the graph norm with the original norm for a bounded operator.

Theorem · advanced

The Hahn-Banach theorem

Adapted

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Extends a linear functional from a subspace to the whole space without increasing the bound that controls it, and draws the consequences that make the dual space useful: a normed space always carries enough bounded functionals to reproduce its norm, to separate its points, to detect the distance to a closed subspace, and to embed the space isometrically in its bidual.

Prerequisites: Normed and Banach spaces, Vector spaces and bases, The Riesz representation theorem, L^p spaces

Definition(Sublinear functionals and the dual space)AdaptedAdapted from Kreyszig (1978), §4.2, eqs. (1)–(2), p. 214, and §4.3, eqs. (1)–(2), p. 219; Folland (1999), §5.2, p. 157; Stakgold & Holst (2011), §4.8, eq. (4.8.1), pp. 292–293

Let $X$ be a vector space over $\mathbb{K} = \mathbb{R}$ or $\mathbb{C}$ . A linear functional on $X$ is a linear map $f : X \to \mathbb{K}$ .

A sublinear functional on a real vector space $X$ is a map $p : X \to \mathbb{R}$ that is

subadditive: $p(x+y) \le p(x) + p(y)$ for all $x, y \in X$ , and
positively homogeneous: $p(\lambda x) = \lambda\, p(x)$ for all $x \in X$ and all real $\lambda > 0$ .

Nothing is assumed about $p(-x)$ , and $p$ may take negative values. A seminorm is a map $p : X \to \mathbb{R}$ that is subadditive and satisfies $p(\alpha x) = |\alpha|\,p(x)$ for every scalar $\alpha$ ; every norm is a seminorm, and every seminorm on a real space is a sublinear functional.

If $X$ is a normed space (see Normed and Banach spaces), its dual space is the set

X^{*} = \{\, f : X \to \mathbb{K} \ \text{linear with}\ \|f\| < \infty \,\}, \qquad \|f\| = \sup_{\|x\| \le 1} |f(x)| ,

with the operator norm. Since the scalar field is complete, $X^{*}$ is a Banach space whether or not $X$ is one; Stakgold and Holst (Stakgold & Holst (2011), Thm 4.8.4, p. 294) record this as one of the standard facts. Writers differ over the symbol: Kreyszig (Kreyszig (1978), §2.10, and §4.3, p. 223) and Stakgold and Holst write $X'$ , while Folland (Folland (1999), §5.2, p. 157) and Reed and Simon (Reed & Simon (1980), §III.2, p. 72) write $X^{*}$ ; we use $X^{*}$ .

On a complex space the real and complex functionals determine each other. If $f$ is complex linear and $u = \operatorname{Re} f$ , then $u$ is real linear and

f(x) = u(x) - i\,u(ix) ;

conversely this formula turns any real-linear $u$ into a complex-linear $f$ , and on a normed space $\|u\| = \|f\|$ . This is Folland's Prop. 5.5 (Folland (1999), Prop. 5.5, p. 157), and it is the device by which every complex statement below is reduced to a real one.

Theorem(Hahn-Banach: dominated extension)AdaptedAdapted from Kreyszig (1978), Thm 4.2-1, p. 214, Thm 4.3-1, p. 219, and Thm 4.3-2, pp. 221–222; Folland (1999), Thm 5.6, p. 157, and Thm 5.7, p. 158; Reed & Simon (1980), Thm III.5 and Corollary 1, pp. 75–77

(a) Real form. Let $X$ be a real vector space, $p$ a sublinear functional on $X$ , $M \subseteq X$ a subspace, and $f$ a linear functional on $M$ with

f(y) \le p(y) \qquad \text{for all } y \in M .

Then there is a linear functional $F$ on all of $X$ with $F|_M = f$ and $F(x) \le p(x)$ for every $x \in X$ .

(b) Seminorm form, real or complex. Let $X$ be a real or complex vector space, $p$ a seminorm on $X$ , $M \subseteq X$ a subspace, and $f$ a linear functional on $M$ with $|f(y)| \le p(y)$ for all $y \in M$ . Then there is a linear functional $F$ on $X$ with $F|_M = f$ and $|F(x)| \le p(x)$ for every $x \in X$ .

(c) Normed form. Let $X$ be a normed space, $M \subseteq X$ a subspace, and $f \in M^{*}$ . Then there is $F \in X^{*}$ with

F|_M = f , \qquad \|F\|_{X^{*}} = \|f\|_{M^{*}} .

No completeness, separability or closedness is assumed anywhere: $X$ need not be a Banach space and $M$ need not be closed.

Theorem(Consequences: the dual space is large)AdaptedAdapted from Folland (1999), Thm 5.8(a)–(d), p. 159; Kreyszig (1978), Thm 4.3-3 and Cor. 4.3-4, p. 223; Reed & Simon (1980), Corollaries 2 and 3, p. 77, and Thm III.4, p. 74; Stakgold & Holst (2011), Thms 4.8.2 and 4.8.3, p. 294

Let $X$ be a normed space.

Norming functional. For every $x \ne 0$ in $X$ there is $f \in X^{*}$ with $\|f\| = 1$ and $f(x) = \|x\|$ . Consequently $\|x\| = \max_{f \in X^{*},\, f \ne 0} \frac{|f(x)|}{\|f\|} ,$ the maximum being attained, and the dual norm determines the original norm.
Distance functional. If $M \subseteq X$ is a closed subspace and $x \in X \setminus M$ , put $\delta = \inf_{y \in M}\|x - y\| > 0$ . Then there is $f \in X^{*}$ with $\|f\| = 1 , \qquad f|_{M} = 0 , \qquad f(x) = \delta .$
Separation of points. If $x \ne y$ in $X$ there is $f \in X^{*}$ with $f(x) \ne f(y)$ . Equivalently, if $f(x) = 0$ for every $f \in X^{*}$ then $x = 0$ .
Embedding in the bidual. For $x \in X$ define $\hat{x} : X^{*} \to \mathbb{K}$ by $\hat{x}(f) = f(x)$ . Then $\hat{x} \in X^{**}$ and the map $x \mapsto \hat{x}$ is a linear isometry of $X$ into $X^{**}$ . A space for which this map is onto is called reflexive; every Hilbert space and every $L^{p}$ with $1 < p < \infty$ is reflexive, while $L^{1}$ and $c_{0}$ are not.

Part 1 is Kreyszig's Thm 4.3-3 (Kreyszig (1978), Thm 4.3-3, p. 223) and Folland's Thm 5.8(b) (Folland (1999), Thm 5.8(b), p. 159); part 2 is Folland's Thm 5.8(a) (Folland (1999), Thm 5.8(a), p. 159), which Reed and Simon (Reed & Simon (1980), Corollary 3, p. 77) state with the weaker normalisation $\|f\| \le 1$ ; part 4 is Reed and Simon's Thm III.4 (Reed & Simon (1980), Thm III.4, p. 74), whose list of reflexive and non-reflexive examples we quote in Examples 1-3 of their §III.2.

Note(The hypothesis on

p

differs between our sources)AdaptedAdapted from Reed & Simon (1980), Thm III.5, p. 75; Kreyszig (1978), Thm 4.2-1, p. 214, and Thm 4.3-1, p. 219; Folland (1999), Thm 5.6, p. 157; Stakgold & Holst (2011), Thm 4.8.1, p. 293

The four books in the library state the extension theorem with four different hypotheses on the dominating function, and it is worth knowing which is which before comparing statements.

Kreyszig (Kreyszig (1978), Thm 4.2-1, p. 214), Folland (Folland (1999), Thm 5.6, p. 157) and Stakgold and Holst (Stakgold & Holst (2011), Thm 4.8.1, p. 293) all take $p$ sublinear on a real space. That is the version stated in (a) above and proved below.
Reed and Simon (Reed & Simon (1980), Thm III.5, p. 75) assume only that $p$ is convex, $p(\alpha x + (1-\alpha)y) \le \alpha p(x) + (1-\alpha)p(y)$ for $\alpha \in [0,1]$ . A sublinear $p$ is convex, since for $0 < \alpha < 1$ subadditivity and positive homogeneity give $p(\alpha x + (1-\alpha)y) \le p(\alpha x) + p((1-\alpha)y) = \alpha p(x) + (1-\alpha)p(y)$ ; so their hypothesis is the weaker one and their theorem contains the other three. Their proof uses only convexity.
For the complex case Kreyszig (Kreyszig (1978), Thm 4.3-1, p. 219) assumes $p$ subadditive with $p(\alpha x) = |\alpha| p(x)$ , and Folland (Folland (1999), Thm 5.7, p. 158) assumes $p$ is a seminorm; these are the same condition.

Stakgold and Holst state the theorem for a real space only and refer the proof to books outside this library.

Idea of the proofOriginal

Everything rests on one observation about a single extra dimension. Suppose $f \le p$ on a subspace $M$ and we want to extend $f$ to $M + \mathbb{R}x$ for one new vector $x$ . The extension is determined by the single number $a = F(x)$ , and the requirement $F \le p$ on the larger space is a family of inequalities, one for each $y \in M$ and each sign of the coefficient of $x$ . Those inequalities say exactly that $a$ lies above a certain set of numbers built from $p$ and below another. Subadditivity of $p$ is precisely what makes the first set lie entirely below the second, so an admissible $a$ exists, and we have gained a dimension.

The rest is bookkeeping: the one-dimensional step can be repeated, and Zorn's lemma turns "can always be repeated" into "has a maximal element", whose domain must then be all of $X$ .

The complex case is not a separate argument. A complex functional is reconstructed from its real part, so one extends the real part by the real theorem and reassembles; the only point needing care is that the bound $|F| \le p$ is recovered from $\operatorname{Re} F \le p$ by rotating $x$ through the phase of $F(x)$ , which is where $p(\alpha x) = |\alpha| p(x)$ is used.

The corollaries all follow one pattern, which is worth naming because it is how the theorem is used: define the functional you want on the smallest subspace where its value is forced, check that the norm you want is correct there, then extend.

Proof(The real form, one dimension at a time)AdaptedAdapted from Kreyszig (1978), Thm 4.2-1, parts (a)–(c), pp. 214–217; Folland (1999), Thm 5.6, p. 158; Reed & Simon (1980), Thm III.5, pp. 75–76

Step 1: the inequality that creates room. Let $N$ be a subspace of $X$ , let $g$ be linear on $N$ with $g \le p$ there, and let $x \in X \setminus N$ . For any $y_1, y_2 \in N$ ,

g(y_1) + g(y_2) = g(y_1 + y_2) \le p(y_1 + y_2) = p\bigl((y_1 - x) + (y_2 + x)\bigr) \le p(y_1 - x) + p(y_2 + x) ,

using linearity of $g$ , the bound $g \le p$ on $N$ , and subadditivity of $p$ . Rearranging so that $y_1$ appears only on the left and $y_2$ only on the right,

g(y_1) - p(y_1 - x) \;\le\; p(y_2 + x) - g(y_2) .

Step 2: choose the value at $x$ . Since the left side does not involve $y_2$ and the right side does not involve $y_1$ , we may take the supremum over $y_1 \in N$ on the left and the infimum over $y_2 \in N$ on the right and keep the inequality. Hence

m_{0} = \sup_{y \in N}\bigl[g(y) - p(y - x)\bigr] \;\le\; \inf_{y \in N}\bigl[p(y + x) - g(y)\bigr] = m_{1} ,

and both are finite (take $y = 0$ to see $m_0 \ge -p(-x)$ and $m_1 \le p(x)$ ). Pick any real $a$ with $m_{0} \le a \le m_{1}$ and define, on $N + \mathbb{R}x$ ,

G(y + \lambda x) = g(y) + \lambda a \qquad (y \in N,\ \lambda \in \mathbb{R}) .

This is well defined because $x \notin N$ makes the decomposition $y + \lambda x$ unique, and it is linear and agrees with $g$ on $N$ .

Step 3: check $G \le p$ , separately for each sign of $\lambda$ . For $\lambda = 0$ the claim is $g \le p$ , which holds. For $\lambda > 0$ , use $a \le m_{1} \le p(y/\lambda + x) - g(y/\lambda)$ and multiply by $\lambda > 0$ , which is where positive homogeneity of $p$ enters:

G(y + \lambda x) = \lambda\Bigl[g(y/\lambda) + a\Bigr] \le \lambda\, p(y/\lambda + x) = p(y + \lambda x) .

For $\lambda = -\mu$ with $\mu > 0$ , use $a \ge m_{0} \ge g(y/\mu) - p(y/\mu - x)$ and multiply by $\mu > 0$ :

G(y - \mu x) = \mu\Bigl[g(y/\mu) - a\Bigr] \le \mu\, p(y/\mu - x) = p(y - \mu x) .

So $G \le p$ on $N + \mathbb{R}x$ , and we have extended $g$ by one dimension.

Step 4: Zorn's lemma. Let $\mathcal{E}$ be the set of all pairs $(N, g)$ with $M \subseteq N \subseteq X$ a subspace, $g$ linear on $N$ , $g|_{M} = f$ and $g \le p$ on $N$ . Order $\mathcal{E}$ by extension: $(N_1, g_1) \preceq (N_2, g_2)$ when $N_1 \subseteq N_2$ and $g_2|_{N_1} = g_1$ . The set is non-empty, since $(M, f) \in \mathcal{E}$ . If $\mathcal{C} \subseteq \mathcal{E}$ is a chain, the union of its domains is a subspace — any two elements of the union lie in a common member of the chain, because the chain is totally ordered — and the common value of the $g$ 's on it is unambiguous for the same reason; this gives an upper bound for $\mathcal{C}$ in $\mathcal{E}$ . By Zorn's lemma $\mathcal{E}$ has a maximal element $(N_{\max}, F)$ .

Step 5: the maximal domain is everything. If $N_{\max} \ne X$ , pick $x \in X \setminus N_{\max}$ ; Steps 1-3 produce a strictly larger member of $\mathcal{E}$ , contradicting maximality. Hence $N_{\max} = X$ , and $F$ is the required extension. $\square$

Proof(The seminorm form and the normed form)AdaptedAdapted from Kreyszig (1978), Thm 4.3-1, pp. 219–221, and Thm 4.3-2, pp. 221–222; Folland (1999), Thm 5.7, p. 158; Reed & Simon (1980), Corollary 1, p. 77

Real scalars. Let $p$ be a seminorm and $|f| \le p$ on $M$ . In particular $f \le p$ on $M$ , so part (a) gives a linear $F$ on $X$ with $F|_M = f$ and $F \le p$ . Then also $-F(x) = F(-x) \le p(-x) = p(x)$ , so $|F| \le p$ everywhere, using $p(-x) = p(x)$ .

Complex scalars. Write $u = \operatorname{Re} f$ on $M$ . Regard $X$ and $M$ as real vector spaces; $u$ is real linear and $u \le |f| \le p$ on $M$ . By part (a) there is a real-linear $U$ on $X$ with $U|_{M} = u$ and $U \le p$ , hence $|U| \le p$ as above. Define

F(x) = U(x) - i\,U(ix) .

By the correspondence recorded in the definition block, $F$ is complex linear; and on $M$ it agrees with $f$ , because $u = \operatorname{Re} f$ there and $f$ is determined by its real part.

It remains to bound $|F|$ . Fix $x$ with $F(x) \ne 0$ and write $F(x) = |F(x)|e^{i\theta}$ , so that $\alpha = e^{-i\theta}$ has modulus $1$ . Then

|F(x)| = \alpha F(x) = F(\alpha x) ,

which is real and therefore equals its own real part $U(\alpha x)$ . Hence

|F(x)| = U(\alpha x) \le p(\alpha x) = |\alpha|\,p(x) = p(x) ,

the last two steps using the homogeneity of the seminorm. (If $F(x) = 0$ the bound is immediate, since $p \ge 0$ : indeed $0 = p(0) \le p(x) + p(-x) = 2p(x)$ .)

Normed form. Let $f \in M^{*}$ and set $p(x) = \|f\|_{M^{*}}\,\|x\|$ , a seminorm on all of $X$ by the triangle inequality and absolute homogeneity of the norm. On $M$ we have $|f(y)| \le \|f\|_{M^{*}}\|y\| = p(y)$ , so the seminorm form gives a linear $F$ on $X$ with $F|_M = f$ and $|F(x)| \le \|f\|_{M^{*}}\|x\|$ for all $x$ ; thus $F \in X^{*}$ with $\|F\| \le \|f\|_{M^{*}}$ . The reverse inequality needs no theorem: the supremum defining $\|F\|$ runs over a larger set of vectors than the one defining $\|f\|_{M^{*}}$ , so $\|F\| \ge \|f\|_{M^{*}}$ . Hence the two norms are equal. $\square$

Proof(The consequences)AdaptedAdapted from Folland (1999), Thm 5.8, p. 159; Kreyszig (1978), Thm 4.3-3 and Cor. 4.3-4, p. 223; Reed & Simon (1980), Corollary 2, p. 77, and Thm III.4, p. 74

Part 2 first, since part 1 is its special case. Let $M$ be a closed subspace and $x \notin M$ , and let $\delta = \inf_{y \in M}\|x-y\|$ ; this is strictly positive because $M$ is closed, for $\delta = 0$ would exhibit $x$ as a limit of points of $M$ . On the subspace $M + \mathbb{K}x$ define

f_{0}(y + \lambda x) = \lambda\,\delta \qquad (y \in M,\ \lambda \in \mathbb{K}) .

The decomposition is unique because $x \notin M$ , so $f_{0}$ is a well-defined linear functional; it vanishes on $M$ and $f_{0}(x) = \delta$ .

Its norm on $M + \mathbb{K}x$ is exactly $1$ . For $\lambda \ne 0$ ,

|f_{0}(y + \lambda x)| = |\lambda|\,\delta \le |\lambda|\,\bigl\|x + \lambda^{-1}y\bigr\| = \|\lambda x + y\| ,

where the inequality is the definition of $\delta$ as an infimum over $M$ , applied to the element $-\lambda^{-1}y \in M$ ; for $\lambda = 0$ the left side is $0$ . Hence $\|f_{0}\| \le 1$ . Conversely, choose $y_{n} \in M$ with $\|x - y_{n}\| \to \delta$ ; then $f_{0}(x - y_{n}) = \delta$ while $\|x-y_{n}\| \to \delta$ , so $\|f_{0}\| \ge \delta/\|x-y_n\| \to 1$ . Thus $\|f_{0}\| = 1$ , and the normed form of the theorem extends $f_{0}$ to $f \in X^{*}$ with $\|f\| = 1$ , $f|_{M} = 0$ and $f(x) = \delta$ .

Part 1. Take $M = \{0\}$ , which is closed, and $x \ne 0$ : then $\delta = \|x\|$ and part 2 gives $f$ with $\|f\|=1$ and $f(x) = \|x\|$ . For the formula for $\|x\|$ : every $f \ne 0$ satisfies $|f(x)| \le \|f\|\,\|x\|$ , so the quotient $|f(x)|/\|f\|$ never exceeds $\|x\|$ , and the functional just constructed attains it.

Part 3. If $x \ne y$ , apply part 1 to $x - y \ne 0$ to get $f$ with $f(x-y) = \|x-y\| \ne 0$ , that is $f(x) \ne f(y)$ . Taking $y = 0$ gives the second formulation: if every $f \in X^{*}$ kills $x$ then $\|x\| = 0$ .

Part 4. Linearity of $x \mapsto \hat{x}$ and of each $\hat{x}$ is immediate from the definitions. The estimate $|\hat{x}(f)| = |f(x)| \le \|x\|\,\|f\|$ shows $\hat{x} \in X^{**}$ with $\|\hat{x}\| \le \|x\|$ . For the opposite inequality take the $f$ of part 1: $\|\hat{x}\| \ge |\hat{x}(f)| = |f(x)| = \|x\|$ since $\|f\|=1$ . So $\|\hat{x}\| = \|x\|$ and the map is an isometry. $\square$

Remark(Where the hypotheses are used, and what the theorem does not give)Original

Subadditivity of $p$ is used exactly once, in Step 1, to split $p(y_1+y_2)$ across the inserted pair $-x$ and $+x$ . It is what forces $m_{0} \le m_{1}$ ; without it the two sets could be in the wrong order and no admissible value $a$ would exist.
Positive homogeneity of $p$ is used only in Step 3, to rescale by $|\lambda|$ . Note that the two sign cases are genuinely different computations: the case $\lambda>0$ uses the upper bound $a \le m_1$ and the case $\lambda<0$ the lower bound $a \ge m_0$ . A one-sided choice of $a$ would not do.
Absolute homogeneity $p(\alpha x) = |\alpha| p(x)$ is needed only for the seminorm form; the real form (a) is false as stated if one tries to conclude $|F| \le p$ from $F \le p$ for a merely sublinear $p$ . For example, on $X = \mathbb{R}$ with $p(t) = t$ (sublinear but not a seminorm) the functional $F(t) = t$ satisfies $F \le p$ while $|F(-1)| = 1 > -1 = p(-1)$ .
Closedness of $M$ is used only in part 2 of the consequences, to guarantee $\delta > 0$ . If $M$ is a dense proper subspace then $\delta = 0$ for every $x$ , and the only $f \in X^{*}$ vanishing on $M$ is $f = 0$ by continuity.
Completeness is never used. The theorem and all four consequences hold in any normed space. What completeness buys is elsewhere: the other two basic principles of linear analysis, the uniform boundedness and open mapping theorems of Uniform boundedness and open mapping theorems, both need a Baire category argument and therefore a complete space.
No uniqueness. The theorem produces an extension, never the extension. In $\mathbb{R}^{2}$ with $\|y\|_{1} = |y_1| + |y_2|$ , let $Z = \{(t,0)\}$ and $f(t,0) = t$ , so $\|f\|_{Z^{*}} = 1$ . Every linear extension has the form $F_{c}(y) = y_{1} + c\,y_{2}$ , and the dual norm of $\|\cdot\|_1$ is $\|\cdot\|_\infty$ , so $\|F_{c}\| = \max(1, |c|)$ . Thus every $c \in [-1,1]$ gives a norm-preserving extension: a whole interval of them. In a Hilbert space this cannot happen: by The Riesz representation theorem a bounded functional on a subspace $Z$ is $(\,\cdot\,, z)$ for a unique $z \in \overline{Z}$ , and the same formula read on the whole space is the unique norm-preserving extension. Kreyszig (Kreyszig (1978), §4.3, p. 222) makes the same point: in Hilbert space the extension is immediate and no Zorn's lemma is needed.
The axiom of choice. Zorn's lemma enters only in Step 4. Kreyszig (Kreyszig (1978), §4.2, pp. 217–218) notes that if $X$ is exhausted by an increasing sequence of subspaces obtained by adjoining one vector at a time — in particular if $X$ is separable, where one can adjoin the elements of a countable dense set — then ordinary induction replaces it. The price of Zorn's lemma is that the proof is not a construction: it does not tell you what $F$ is.

Method(Building a bounded functional with prescribed behaviour)AdaptedAdapted from Reed & Simon (1980), §III.3, p. 75

Reed and Simon (Reed & Simon (1980), §III.3, p. 75) describe the standard two-step use of the theorem; written out as a recipe it is:

Identify the smallest subspace on which the functional is forced. If the requirements are " $f$ vanishes on $M$ " and " $f(x) = c$ ", that subspace is $M + \mathbb{K}x$ ; if the requirement is " $f(x) = \|x\|$ ", it is the line $\mathbb{K}x$ .
Write the functional there. On $M + \mathbb{K}x$ there is only one choice, $f_{0}(y + \lambda x) = \lambda c$ ; the decomposition is unique provided $x \notin M$ .
Compute $\|f_{0}\|$ on that subspace, and find the dominating function. For a norm bound take $p(\cdot) = \|f_{0}\|\,\|\cdot\|$ ; the point of the step is that on a one- or two-dimensional problem the norm is an explicit minimisation that can be done by hand.
Extend. Part (c) of the theorem delivers $F \in X^{*}$ with the same norm, still satisfying the prescriptions of step 1, since they were imposed on the subspace.
Check what you actually need. Only the properties imposed in step 1 and the norm survive; nothing else about $F$ is determined, so any further claim about it must be proved from those.

Example(A distance, and the functional that detects it)Original

Work in $X = \mathbb{R}^{3}$ with the norm $\|y\|_{1} = |y_{1}| + |y_{2}| + |y_{3}|$ . Let

u = (1, 2, -1), \qquad M = \operatorname{span}\{u\}, \qquad x = (3, 1, 0) .

$M$ is finite-dimensional, hence closed. We compute $\delta = \operatorname{dist}(x, M)$ and exhibit the functional of part 2.

Step 1: the distance. For $t \in \mathbb{R}$ ,

\varphi(t) = \|x - t u\|_{1} = |3 - t| + |1 - 2t| + |t| .

This is piecewise linear with corners at $t = 0, \tfrac12, 3$ . On $[0, \tfrac12]$ all three expressions inside the absolute values are non-negative in the order written, giving $\varphi(t) = (3-t) + (1-2t) + t = 4 - 2t$ , decreasing. On $[\tfrac12, 3]$ we get $\varphi(t) = (3-t) + (2t-1) + t = 2 + 2t$ , increasing. Outside $[0,3]$ the function only grows further. So the minimum is at the corner $t = \tfrac12$ :

\delta = \varphi(\tfrac12) = 4 - 1 = 3 .

Step 2: the dual norm. For $f(y) = a y_{1} + b y_{2} + c y_{3}$ we have $|f(y)| \le \max(|a|,|b|,|c|)\,\|y\|_{1}$ , with equality at the standard basis vector belonging to the largest of $|a|,|b|,|c|$ . Hence $\|f\| = \max(|a|,|b|,|c|)$ : the dual of the $\ell^{1}$ norm is the $\ell^{\infty}$ norm.

Step 3: the functional. Part 2 of the consequences asserts the existence of $f$ with $\|f\| = 1$ , $f(u) = 0$ and $f(x) = 3$ . Take

f(y) = y_{1} + y_{3}, \qquad\text{that is}\qquad (a,b,c) = (1,0,1) .

Then $\|f\| = \max(1,0,1) = 1$ ; $f(u) = 1 + (-1) = 0$ , so $f$ vanishes on all of $M$ ; and $f(x) = 3 + 0 = 3 = \delta$ . All three requirements hold, and the two computations agree:

\operatorname{dist}(x, M) = 3 = \max\{\, f(x) : \|f\|_{\infty} \le 1,\ f(u) = 0 \,\} .

Step 4: the maximum really is $3$ . The right-hand side is the linear program "maximise $3a + b$ subject to $a + 2b - c = 0$ and $|a|,|b|,|c| \le 1$ ". If $3a + b = 3$ with $b = 3 - 3a$ , the constraint gives $c = a + 2b = 6 - 5a$ , and $|c| \le 1$ forces $a \ge 1$ , hence $a = 1$ , $b = 0$ , $c = 1$ . So the value $3$ is attained, only by the $f$ above, and no larger value is possible. The duality is exact, not merely an inequality.

Note what this example shows about the recipe: the extension step was invisible because $M + \mathbb{R}x$ was already two-dimensional inside a three-dimensional space, so one extra dimension sufficed. In infinite dimensions the same two computations are made on the small subspace and the theorem supplies the rest.

Application(Weak formulations, and functionals that are not integrals)Original

Testing against functionals. The most-used consequence in applied analysis is the weakest-looking one, part 3: if $f(v) = 0$ for every $f \in X^{*}$ , then $v = 0$ . Stakgold and Holst (Stakgold & Holst (2011), Thm 4.8.3 and the remark following it, p. 294) single it out as the bridge between the weak form of an operator equation and its strong form, and use it repeatedly for boundary-value problems. The pattern is this. A weak formulation of $Au = g$ asks that

f(Au - g) = 0 \qquad \text{for every } f \text{ in a suitable dual space,}

which in a function space means testing against all $L^{q}$ functions, or all test functions, or all elements of $H^{1}_{0}$ . Part 3 says that this family of scalar equations is not weaker than the original one: if $Au - g$ lies in a space whose dual separates points, the weak equations force $Au = g$ exactly. Without the Hahn-Banach theorem one would not know that there are enough functionals for this, and the whole variational strategy of Weak derivatives and Sobolev spaces and The Lax-Milgram theorem and the Dirichlet problem would rest on nothing. The same remark explains why a weak limit is unique, which is the starting point of Weak convergence.

A state that is not a density. The theorem also produces objects that no formula can write down, and these are not pathologies: they are the reason the dual of $L^{\infty}$ is bigger than $L^{1}$ . On $X = L^{\infty}[0,1]$ consider the subspace $C[0,1]$ and the evaluation $f \mapsto f(0)$ , which is bounded with norm $1$ there because $|f(0)| \le \|f\|_{\infty}$ . Extend it to $\phi \in (L^{\infty})^{*}$ with $\|\phi\| = 1$ . Folland (Folland (1999), §6.4, p. 191) observes that $\phi$ cannot be integration against any $g \in L^{1}$ : the continuous functions $f_{n}(x) = \max(1 - nx, 0)$ have $\phi(f_{n}) = f_{n}(0) = 1$ for every $n$ , while $f_{n} \to 0$ pointwise on $(0,1]$ with $|f_n| \le 1$ , so dominated convergence (Dominated convergence) gives $\int f_{n} g \to 0$ for every $g \in L^{1}$ . Reed and Simon (Reed & Simon (1980), Problem 8(b), p. 84) set the same construction as an exercise on $L^{\infty}(\mathbb{R})$ .

In physics this is the standard way to make sense of an idealised measurement, or of an equilibrium state, that is specified only on a subalgebra of observables where its value is forced by symmetry or by a conservation law: one defines the state where it is determined, checks the normalisation there, and extends. The extension is not unique, and the non-uniqueness is physically meaningful — it is the freedom left over once the prescribed data are imposed.

Where the dual spaces themselves come from. The theorem says the dual is large, not what it is. Identifying it is separate work, done space by space: $(L^{p})^{*} = L^{q}$ for $1 \le p < \infty$ with $p^{-1} + q^{-1} = 1$ (see L^p spaces), $c_{0}^{*} = \ell^{1}$ and $(\ell^{1})^{*} = \ell^{\infty}$ , and a Hilbert space is its own dual by The Riesz representation theorem; these are Examples 1-3 of Reed and Simon (Reed & Simon (1980), §III.2, pp. 72–74). Their Thm III.7 adds a useful one-way test: if $X^{*}$ is separable then so is $X$ , which is how one shows that $\ell^{1}$ is not the dual of $\ell^{\infty}$ .

Unbounded operators and quantum mechanics

Definition · advanced

Unbounded and closed operators

Adapted

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Sets up the bookkeeping that unbounded operators force on us: an operator is a linear map on a dense domain, its graph decides whether it is closed or closable, its adjoint exists only because the domain is dense, and the adjoint determines the closure; the two running examples are multiplication by $x$ and $-i\,d/dx$ .

Prerequisites: Adjoints, self-adjoint and unitary operators, Bounded linear operators, Hilbert spaces, The projection theorem, The Riesz representation theorem, Uniform boundedness and open mapping theorems, L^p spaces

Definition(Operators in a Hilbert space: domain, graph, extension)AdaptedAdapted from Reed & Simon (1980), §VIII.1, pp. 249–250; Kreyszig (1978), §10.1, pp. 524–526, and Def. 10.3-1, pp. 535–536; Hassani (2013), Def. 19.1.1 and its footnotes, p. 563

Let $H$ be a complex Hilbert space. An operator in $H$ is a linear map

T : D(T) \longrightarrow H

defined on a linear subspace $D(T) \subseteq H$ , the domain of $T$ . Following Kreyszig (Kreyszig (1978), §10.1, p. 527) and Hassani (Hassani (2013), Def. 19.1.1 and footnote 2, p. 563) we say in $H$ when $D(T)$ may be a proper subspace and on $H$ when $D(T) = H$ . The domain is part of the data: two different domains carrying the same formula are two different operators, and almost every question below has a different answer for each of them.

Unless we say otherwise, $D(T)$ is assumed dense in $H$ ; the reason is given in the adjoint definition below.

$T$ is unbounded if there is no constant $c$ with $\|T\psi\| \le c\|\psi\|$ for all $\psi \in D(T)$ ; equivalently, if the quotients $\|T\psi\|/\|\psi\|$ are not bounded above on $D(T)\setminus\{0\}$ .

Extension. For operators $S, T$ in $H$ we write $S \subseteq T$ , and call $T$ an extension of $S$ , when $D(S) \subseteq D(T)$ and $T\psi = S\psi$ for every $\psi \in D(S)$ .

Graph. Von Neumann's device for handling all of this at once is the graph

\Gamma(T) = \bigl\{\, \langle \psi, T\psi\rangle : \psi \in D(T) \,\bigr\} \subseteq H \times H ,

where $H \times H$ is a Hilbert space under

\bigl(\langle \varphi_1,\psi_1\rangle, \langle \varphi_2,\psi_2\rangle\bigr) = (\varphi_1,\varphi_2) + (\psi_1,\psi_2) .

Then $S \subseteq T$ says exactly $\Gamma(S) \subseteq \Gamma(T)$ , and a subspace $\Gamma \subseteq H\times H$ is the graph of an operator precisely when $\langle 0, \psi\rangle \in \Gamma$ forces $\psi = 0$ , that is, when it is single-valued.

Sums and products. Following Kreyszig (Kreyszig (1978), §10.1, pp. 528–529), $S+T$ has domain $D(S)\cap D(T)$ and $TS$ has domain $\{\psi \in D(S) : S\psi \in D(T)\}$ , the largest sets on which the formulas make sense. These domains can be small, and in extreme cases trivial.

Definition(Closed, closable, closure)AdaptedAdapted from Reed & Simon (1980), §VIII.1, p. 250, and the definition of a core, §VIII.2, p. 256; Kreyszig (1978), Def. 10.3-1, Thm 10.3-2(a) and (c), and Def. 10.3-4, pp. 535–537; Stakgold & Holst (2011), §5.2, pp. 308–309, Remark 2, p. 309; Hassani (2013), Def. 19.1.3, p. 564

$T$ is closed if $\Gamma(T)$ is a closed subset of $H\times H$ . Spelled out in sequences, which is the form one checks in practice:

whenever $\psi_n \in D(T)$ , $\psi_n \to \psi$ and $T\psi_n \to \chi$ , it follows that $\psi \in D(T)$ and $T\psi = \chi$ .

Closedness is a different condition from continuity, not a weaker one. Continuity would make $T\psi_n$ converge whenever $\psi_n$ does, but it says nothing about where the limit $\psi$ lives; closedness says only that if both limits exist, then $\psi$ is back in the domain and the two limits are related correctly. Neither implies the other: the identity restricted to a dense proper subspace $D \subsetneq H$ is bounded, and $T\psi_n$ converges whenever $\psi_n$ does, yet it is not closed, because any $\psi \in H \setminus D$ is the limit of a sequence in $D$ and is not in the domain. The two conditions meet exactly when the domain is closed: Kreyszig's Thm 10.3-2(c) (Kreyszig (1978), Thm 10.3-2(c), p. 536) says that a bounded operator is closed exactly when its domain $D(T)$ is a closed subspace, and Stakgold and Holst (Stakgold & Holst (2011), §5.2, Remark 2, p. 309) record the half of this we use most: a bounded operator defined on a closed set is closed.

$T$ is closable if it has at least one closed extension. Whenever there is one there is a smallest one — that is part of the theorem proved below — and it is the closure $\bar{T}$ , characterised by

\Gamma(\bar{T}) = \overline{\Gamma(T)} .

The subtlety, which is why "closable" is not automatic, is that $\overline{\Gamma(T)}$ is always a closed subspace but need not be a graph: it may contain a pair $\langle 0, \chi\rangle$ with $\chi \ne 0$ . Stakgold and Holst (Stakgold & Holst (2011), §5.2, p. 308) phrase the condition in exactly those terms: $A$ is closable when, for every null sequence $u_n \to 0$ in $D(A)$ , the image sequence $Au_n$ either tends to $0$ or has no limit at all. Their construction of the closure is the direct one: adjoin to the domain every $u$ that is a limit of some $u_n \in D(A)$ for which $Au_n$ also converges, and let $\bar{A}u$ be that second limit.

If $T$ is closed, a subspace $D \subseteq D(T)$ is a core for $T$ when $\overline{T|_{D}} = T$ : the operator is already determined by its values on $D$ .

Definition(The adjoint of a densely defined operator)AdaptedAdapted from Kreyszig (1978), p. 527 and Def. 10.1-2, pp. 527–528; Reed & Simon (1980), §VIII.1, p. 252; Stakgold & Holst (2011), §5.4, p. 317

Let $T$ be densely defined in $H$ . Put

D(T^{*}) = \bigl\{\, \varphi \in H : \text{there is } \eta \in H \text{ with } (T\psi, \varphi) = (\psi, \eta) \text{ for all } \psi \in D(T) \,\bigr\} ,

and for such $\varphi$ set $T^{*}\varphi = \eta$ . The map $T^{*}$ is the adjoint of $T$ .

Two remarks make the definition work.

Density is exactly what makes $\eta$ unique. If $\eta_1$ and $\eta_2$ both serve, then $(\psi, \eta_1 - \eta_2) = 0$ for every $\psi \in D(T)$ , so $\eta_1 - \eta_2 \in D(T)^{\perp} = \{0\}$ by The projection theorem. Conversely, if $D(T)$ were not dense, any non-zero $\eta_0 \perp D(T)$ could be added to $\eta$ , and $T^{*}$ would not be a map at all. Kreyszig (Kreyszig (1978), §10.1, p. 527) gives precisely this two-sided argument.
A boundedness test for membership. $\varphi \in D(T^{*})$ if and only if $|(T\psi,\varphi)| \le C\|\psi\|$ for all $\psi \in D(T)$ , with $C$ depending on $\varphi$ . The bound makes $\psi \mapsto (T\psi,\varphi)$ a bounded linear functional on $D(T)$ ; being uniformly continuous there, it extends uniquely to a bounded linear functional on all of $H$ , because $D(T)$ is dense, and the Riesz representation (The Riesz representation theorem) applied to the extension produces $\eta$ with $(T\psi,\varphi) = (\psi,\eta)$ on $D(T)$ . Conversely the representation gives the bound back, with $C = \|\eta\|$ . This is Reed and Simon's (Reed & Simon (1980), §VIII.1, p. 252) formulation.

For bounded everywhere-defined operators this reduces to the Hilbert adjoint of Adjoints, self-adjoint and unitary operators, with $D(T^{*}) = H$ . In the unbounded case $D(T^{*})$ must be computed, and reversing an inclusion is the first thing it does:

S \subseteq T \quad\Longrightarrow\quad T^{*} \subseteq S^{*} ,

because the defining condition for $\varphi \in D(T^*)$ is a requirement over the larger set $D(T)$ . The algebra also picks up conjugates, as Stakgold and Holst (Stakgold & Holst (2011), eq. (5.4.10), p. 317) record: $(\alpha A)^{*} = \bar\alpha A^{*}$ and $(A + \alpha I)^{*} = A^{*} + \bar\alpha I$ , both on the domain $D(A^{*})$ .

Definition(Hellinger-Toeplitz: why unbounded operators need small domains)AdaptedAdapted from Kreyszig (1978), Thm 10.1-1, p. 525; Reed & Simon (1980), §VIII.1, p. 249; Hassani (2013), Thm 19.1.4, p. 564

Let $T$ be a linear operator defined on all of a complex Hilbert space $H$ with

(T\varphi, \psi) = (\varphi, T\psi) \qquad \text{for all } \varphi, \psi \in H .

Then $T$ is bounded.

Equivalently: an unbounded operator obeying that identity cannot be defined everywhere. Hassani (Hassani (2013), Thm 19.1.4, p. 564) records the companion statement, which he quotes from another source: a closed operator defined at every point of $H$ is bounded. The same conclusion is in Kreyszig's Thm 10.3-2(b) (Kreyszig (1978), Thm 10.3-2(b), p. 536) and Stakgold and Holst's Thm 5.3.3 (Stakgold & Holst (2011), Thm 5.3.3, p. 313) in the sharper form: a closed operator whose domain is a closed subspace is bounded.

Definition(The adjoint controls closability)AdaptedAdapted from Reed & Simon (1980), Thm VIII.1, pp. 252–253; Kreyszig (1978), Thm 10.3-3, p. 536, and §10.3, Problems 7–8, p. 541; Stakgold & Holst (2011), Properties of the Adjoint 1–2, p. 318

Let $T$ be densely defined in $H$ . Then:

(a) $T^{*}$ is closed, whether or not $T$ is.

(b) $T$ has a closed extension precisely when the domain of $T^{*}$ is dense, and in that case $\bar{T} = T^{**}$ .

(c) If $T$ is closable then $(\bar{T})^{*} = T^{*}$ .

Stakgold and Holst (Stakgold & Holst (2011), Properties of the Adjoint 1–2, p. 318) state (a) only for a closed densely defined $A$ — p. 317 announces a standing restriction to closed operators on dense domains — together with $A^{**} = A$ for such an $A$ , which is (b) specialised to a closed operator; their Property 2 is (c), stated for any closable $A$ . Our (a) assumes nothing beyond density of $D(T)$ , and in that generality it is Reed and Simon's Thm VIII.1(a) (Reed & Simon (1980), Thm VIII.1(a), p. 253) and Kreyszig's Thm 10.3-3 (Kreyszig (1978), Thm 10.3-3, p. 536), which Kreyszig states for a densely defined $T$ whether or not $T$ is closed. Stakgold and Holst draw the practical conclusion: to find the closure of a closable operator, compute $A^{**}$ .

Idea of the proofOriginal

Three facts drive everything in this unit, and all three are statements about graphs.

First, the adjoint is a graph construction in disguise. The condition $(T\psi,\varphi) = (\psi,\eta)$ says that $\langle\varphi,\eta\rangle$ is orthogonal in $H\times H$ to $\langle -T\psi, \psi\rangle$ . So if $V$ is the rotation $V\langle\varphi,\psi\rangle = \langle -\psi,\varphi\rangle$ of the product space, the graph of $T^{*}$ is $V[\Gamma(T)]^{\perp}$ . An orthogonal complement is always closed, which gives (a) at once and with no analysis.

Second, applying the construction twice brings you back to the closure. Because $V$ is unitary and $V^{2} = -I$ , two applications of "rotate and take the complement" amount to taking the closure of the original subspace. That is (b), once one knows that the resulting closed subspace is single-valued, which is precisely the condition that $D(T^{*})$ be dense.

Third, closedness is not continuity but is what survives of it. A bounded operator defined on a closed subspace is closed, and conversely a closed operator defined on a closed subspace is bounded — that is the closed graph theorem of Uniform boundedness and open mapping theorems. So for an unbounded operator, "closed" and "defined everywhere" are mutually exclusive, and since we want closedness (it is what makes limits behave), the domain must be a proper, non-closed, dense subspace. That is the whole reason for the bookkeeping.

Proof(Hellinger-Toeplitz)AdaptedAdapted from Kreyszig (1978), Thm 10.1-1, pp. 525–526

Suppose $T$ is defined on all of $H$ , satisfies $(T\varphi,\psi) = (\varphi,T\psi)$ , and is unbounded. Then there are unit vectors $\psi_{n}$ with $\|T\psi_{n}\| \to \infty$ .

Step 1: a family of bounded functionals. For each $n$ define $f_{n} : H \to \mathbb{C}$ by

f_{n}(\varphi) = (\varphi, T\psi_{n}) = (T\varphi, \psi_{n}) ,

the second expression by the hypothesis. Each $f_{n}$ is linear and defined on all of $H$ . From the first expression and Cauchy-Schwarz, $|f_{n}(\varphi)| \le \|T\psi_{n}\|\,\|\varphi\|$ , so each $f_{n}$ is bounded (with a constant that may be huge).

Step 2: pointwise boundedness. From the second expression and Cauchy-Schwarz, for each fixed $\varphi$ ,

|f_{n}(\varphi)| = |(T\varphi,\psi_{n})| \le \|T\varphi\| \,\|\psi_{n}\| = \|T\varphi\| ,

a bound independent of $n$ . This is the step that uses the symmetry hypothesis: without it there would be no way to move $T$ off $\psi_n$ .

Step 3: uniform boundedness. $H$ is complete, so by the uniform boundedness theorem (Uniform boundedness and open mapping theorems) the family $\{\|f_{n}\|\}$ is bounded, say $\|f_{n}\| \le k$ for all $n$ .

Step 4: the contradiction. Apply this to $\varphi = T\psi_{n}$ :

\|T\psi_{n}\|^{2} = (T\psi_{n}, T\psi_{n}) = f_{n}(T\psi_{n}) \le k\,\|T\psi_{n}\| ,

so $\|T\psi_{n}\| \le k$ for every $n$ , contradicting $\|T\psi_{n}\| \to \infty$ . Hence $T$ is bounded. $\square$

Proof(The adjoint controls closability)AdaptedAdapted from Reed & Simon (1980), Thm VIII.1 and its proof, pp. 252–253; Kreyszig (1978), §10.3, Problem 7, p. 541

Define $V : H\times H \to H \times H$ by $V\langle \varphi,\psi\rangle = \langle -\psi, \varphi\rangle$ . It is linear, preserves the product inner product, and is onto, hence unitary; and $V^{2} = -I$ , so $V[E]^{\perp} = V[E^{\perp}]$ for every subspace $E$ , and $V[E]$ is closed whenever $E$ is.

Step 1: identify the graph of the adjoint. For $\varphi,\eta \in H$ ,

\langle\varphi,\eta\rangle \perp V[\Gamma(T)] \iff \bigl(\langle\varphi,\eta\rangle, \langle -T\psi, \psi\rangle\bigr) = 0 \ \ \text{for all } \psi \in D(T) \iff -(\varphi, T\psi) + (\eta,\psi) = 0 \ \ \text{for all } \psi \in D(T) .

Taking complex conjugates, the last condition is $(T\psi,\varphi) = (\psi,\eta)$ for all $\psi \in D(T)$ , which is exactly the statement that $\varphi \in D(T^{*})$ with $T^{*}\varphi = \eta$ . Hence

\Gamma(T^{*}) = V[\Gamma(T)]^{\perp} .

Step 2: part (a). An orthogonal complement is a closed subspace, so $\Gamma(T^{*})$ is closed, that is, $T^{*}$ is a closed operator. (It is single-valued by the density argument in the definition.)

Step 3: a formula for the closure of the graph. Apply Step 1 to get $\Gamma(T^{*}) = V[\Gamma(T)]^{\perp}$ , then push $V$ through the complement and use $V^{2} = -I$ together with $-E = E$ for a subspace $E$ :

V[\Gamma(T^{*})] = V\bigl[\,V[\Gamma(T)]^{\perp}\,\bigr] = \bigl(V[V[\Gamma(T)]]\bigr)^{\perp} = \bigl(V^{2}[\Gamma(T)]\bigr)^{\perp} = \Gamma(T)^{\perp} .

Taking complements once more and using $E^{\perp\perp} = \overline{E}$ ,

V[\Gamma(T^{*})]^{\perp} = \Gamma(T)^{\perp\perp} = \overline{\Gamma(T)} .

This holds whether or not $D(T^{*})$ is dense: $\Gamma(T^{*})$ is a subspace of $H \times H$ in any case.

Step 4: part (b), the easy direction. Suppose $D(T^{*})$ is dense. Then $T^{**}$ is defined, and Step 1 applied to $T^{*}$ in place of $T$ gives $\Gamma(T^{**}) = V[\Gamma(T^{*})]^{\perp}$ , which by Step 3 is $\overline{\Gamma(T)}$ . So the closure of the graph is a graph, and $T^{**}$ is a closed extension of $T$ . Any closed extension $S \supseteq T$ has $\Gamma(S) \supseteq \overline{\Gamma(T)} = \Gamma(T^{**})$ , because $\Gamma(S)$ is closed and contains $\Gamma(T)$ ; hence $S \supseteq T^{**}$ . So $T^{**}$ is the smallest closed extension: $T$ is closable and $\bar{T} = T^{**}$ .

Step 5: part (b), the converse. Suppose $D(T^{*})$ is not dense, and pick $\chi \ne 0$ with $\chi \perp D(T^{*})$ . A typical element of $V[\Gamma(T^{*})]$ is $V\langle\varphi, T^{*}\varphi\rangle = \langle -T^{*}\varphi, \varphi\rangle$ with $\varphi \in D(T^{*})$ , and

\bigl(\langle 0,\chi\rangle,\ \langle -T^{*}\varphi,\varphi\rangle\bigr) = (0, -T^{*}\varphi) + (\chi,\varphi) = (\chi,\varphi) = 0

because $\chi \perp D(T^{*})$ . So $\langle 0,\chi\rangle \in V[\Gamma(T^{*})]^{\perp}$ , which by Step 3 is $\overline{\Gamma(T)}$ . Any closed extension of $T$ would have a graph containing $\overline{\Gamma(T)}$ and hence the pair $\langle 0,\chi\rangle$ with $\chi \ne 0$ , which no graph of a map can contain. So $T$ has no closed extension: it is not closable.

Step 6: part (c). Let $T$ be closable, so $D(T^{*})$ is dense and $\bar{T} = T^{**}$ by (b). Then

(\bar{T})^{*} = (T^{**})^{*} = (T^{*})^{**} .

Now apply (b) to the operator $T^{*}$ : it is densely defined, and its own adjoint $T^{**} = \bar{T}$ has domain containing the dense set $D(T)$ , so $T^{*}$ is closable with closure $(T^{*})^{**}$ . But $T^{*}$ is already closed by (a), so its closure is itself: $(T^{*})^{**} = T^{*}$ . Hence $(\bar{T})^{*} = T^{*}$ . $\square$

Remark(Where the hypotheses are used, and what fails without them)AdaptedAdapted from Reed & Simon (1980), Example 4, p. 252, and Example 5, pp. 254–255; Stakgold & Holst (2011), §5.3, Example 1(a)–(d), pp. 314–315; Kreyszig (1978), Thm 10.3-2(b), p. 536; Evans (2010), Thm 1 of §D.3, p. 677

Density of $D(T)$ is used first of all to define $T^{*}$ : drop it and $T^{*}$ is a relation, not an operator. It is then used once more, in Step 6 of the second proof, where part (b) is applied to $T^{*}$ and one needs $D(T^{**}) \supseteq D(T)$ to be dense.
Density of $D(T^{*})$ is the hypothesis in part (b), and it genuinely fails for some operators. Reed and Simon's Example 4 (Reed & Simon (1980), Example 4, p. 252) takes a bounded measurable $f \notin L^{2}(\mathbb{R})$ , fixes $\psi_{0} \in L^{2}$ , and defines a rank-one operator by pairing $\psi$ against $f$ and scaling $\psi_{0}$ by the result, on the dense domain $D(T) = \{\psi : f\psi \in L^{1}\}$ , which contains every $L^{2}$ function of compact support. Their inner product is linear in its second slot, ours in the first (see the style guide), so in our convention the operator is

T\psi = (\psi, f)\,\psi_{0}, \qquad \psi \in D(T) ,

and the domain is exactly the set on which $(\psi,f) = \int \psi\bar f$ converges absolutely. Then $(T\psi,\varphi) = (\psi,f)\,(\psi_{0},\varphi) = \bigl(\psi,\ \overline{(\psi_{0},\varphi)}\,f\bigr)$ , so

T^{*}\varphi = (\varphi, \psi_{0})\,f ,

and since $f \notin L^{2}$ this lies in $L^{2}$ only when the scalar $(\varphi,\psi_{0})$ vanishes. So $D(T^{*}) = \{\psi_{0}\}^{\perp}$ , and for a non-zero $\psi_{0}$ — which is what the conclusion needs, since $\psi_{0} = 0$ would make $T$ the zero operator and $\{\psi_0\}^\perp = H$ — this is a closed hyperplane and so not dense, $T^{*}$ is the zero operator on that domain, and $T$ is not closable. Reed and Simon remark that one can even arrange $D(T^{*}) = \{0\}$ .

Symmetry in Hellinger-Toeplitz is used only in Step 2 of its proof, and it is essential: an everywhere-defined unbounded linear operator on $H$ exists (build one on a Hamel basis), it simply cannot satisfy $(T\varphi,\psi) = (\varphi,T\psi)$ .
Closed is not the same as closed domain. A closed operator with a closed domain is bounded (Kreyszig Thm 10.3-2(b) (Kreyszig (1978), Thm 10.3-2(b), p. 536); the everywhere-defined case is the closed graph theorem as Evans (Evans (2010), Thm 1 of §D.3, p. 677) states it), so an unbounded closed operator must have a domain that is dense and not closed. This is why one never gets to work on a Banach space of nice functions and be done.
The domain decides the answer, not the formula. Two warnings from the library. Reed and Simon's Example 5 (Reed & Simon (1980), Example 5, pp. 254–255) puts $i\,d/dx$ on $L^{2}[0,1]$ with two domains, the absolutely continuous functions with $L^2$ derivative, and the same with $\varphi(0)=0$ ; both operators are closed and densely defined, the spectrum of the first is all of $\mathbb{C}$ and the spectrum of the second is empty. (For the vocabulary see Spectrum and resolvent.) And Stakgold and Holst (Stakgold & Holst (2011), §5.3, Example 1(a)–(d) and the principle stated after it, pp. 314–315) show that boundary conditions do not all survive closure: for a differential operator of order $n$ , conditions involving derivatives of order below $n$ survive, while conditions on derivatives of order $n$ or higher are lost, so that imposing $u'(0)=0$ on $d/dx$ gives an operator whose closure is the one with no boundary condition at all.
A closed operator need not be closable-to-anything-useful. Closability is about the existence of a closed extension, not about self-adjointness; which closed extensions are self-adjoint is a separate question, taken up in Symmetric versus self-adjoint operators.

Method(Computing the adjoint of a differential operator)AdaptedAdapted from Stakgold & Holst (2011), §5.4, the Heuristic Method and the Precise Method, pp. 318–319

Stakgold and Holst (Stakgold & Holst (2011), §5.4, pp. 318–319) give two versions of the same calculation, a quick one and a rigorous one. As a recipe:

Write the defining identity. You are looking for all pairs $(v, g)$ — Stakgold and Holst call them admissible pairs — with $(Tu, v) = (u, g) \qquad \text{for every } u \in D(T) .$ Every such $v$ lies in $D(T^{*})$ , with $T^{*}v = g$ .
Integrate by parts, keeping the boundary terms. For $T$ of order $n$ this produces $(Tu,v) = (u, T^{\dagger}v) + \bigl[\text{boundary form in } u,\dots,u^{(n-1)},\, \bar v,\dots,\bar v^{(n-1)}\bigr] ,$ where $T^{\dagger}$ is the formal adjoint of Adjoint and self-adjoint differential operators. At this stage one is assuming $v$ smooth enough; that is the heuristic step.
Kill the interior term first. Restrict to those $u \in D(T)$ that vanish near both endpoints. The boundary form vanishes for them, so $(u, g - T^{\dagger}v) = 0$ on a dense set, giving $g = T^{\dagger}v$ . The formal adjoint is therefore forced: the formula for $T^{*}$ is $T^{\dagger}$ , whatever the boundary conditions are.
Read the adjoint boundary conditions off the boundary form. Now let $u$ range over all of $D(T)$ . The surviving requirement is that the boundary form vanish for every such $u$ ; imposing the boundary conditions that define $D(T)$ and demanding the result vanish for all remaining free boundary values of $u$ yields the conditions defining $D(T^{*})$ .
Confirm with no smoothness assumed. The precise version avoids assuming $v$ differentiable: integrate by parts on the other side instead, writing $G(x) = -\int_{x}^{b} g$ , and deduce $v + G = 0$ by testing with a $u$ built from $v + G$ itself. Then $v$ is absolutely continuous because it equals an indefinite integral, and the regularity is a conclusion rather than an assumption.
Count the conditions. Imposing more conditions on $D(T)$ makes $D(T)$ smaller, hence $D(T^{*})$ larger: the number of adjoint boundary conditions goes down as the number of direct ones goes up. The operator is a candidate for self-adjointness only when the two counts balance.

Example(Multiplication by

x

, and

d/dx

with one boundary condition)AdaptedAdapted from Reed & Simon (1980), Example 1, pp. 249–250; Kreyszig (1978), Lemma 10.7-1 and §10.7, eq. (3), pp. 563–564; Stakgold & Holst (2011), §5.4, Example 1 and eq. (5.4.13), pp. 318–319

(i) The position operator is unbounded. On $H = L^{2}(\mathbb{R})$ let

D(X) = \Bigl\{\, \psi \in L^{2}(\mathbb{R}) : \int_{\mathbb{R}} x^{2}|\psi(x)|^{2}\,dx < \infty \,\Bigr\}, \qquad (X\psi)(x) = x\,\psi(x) ,

the largest domain on which the formula lands back in $L^{2}$ , as in Reed and Simon's Example 1 (Reed & Simon (1980), Example 1, p. 249). For $n \in \mathbb{N}$ take $\psi_{n} = \mathbf{1}_{[n,n+1]}$ . Then $\|\psi_{n}\|^{2} = 1$ and

\|X\psi_{n}\|^{2} = \int_{n}^{n+1} x^{2}\,dx = \frac{(n+1)^{3}-n^{3}}{3} = n^{2}+n+\frac{1}{3} .

For $n = 5$ this is $25+5+\tfrac13 = \tfrac{91}{3}$ , so $\|X\psi_{5}\|/\|\psi_{5}\| = \sqrt{91/3} \approx 5.5076$ ; as $n$ grows the ratio exceeds $n$ , so $X$ is unbounded. (Kreyszig's Lemma 10.7-1 (Kreyszig (1978), Lemma 10.7-1, p. 563) runs the same argument with the cruder bound $\|T x_n\| > n$ .) Note where the unboundedness comes from: the interval is infinite. On $L^{2}[a,b]$ multiplication by $x$ is bounded with norm at most $\max(|a|,|b|)$ and is defined on the whole space, as Kreyszig (Kreyszig (1978), §10.7, eq. (3), p. 564) points out.

(ii) A differentiation operator and its adjoint, computed by the recipe. On $H = L^{2}(0,1)$ put

D(T) = \{\, u \in AC[0,1] : u' \in L^{2}(0,1),\ u(0)=0 \,\}, \qquad Tu = u' .

Step 2 of the method gives, for $u \in D(T)$ and $v$ absolutely continuous,

(Tu, v) = \int_{0}^{1} u'\bar v \,dx = \bigl[u\bar v\bigr]_{0}^{1} - \int_{0}^{1} u\,\bar{v}'\,dx = u(1)\overline{v(1)} + (u, -v') ,

where $u(0)=0$ removed the lower endpoint. Step 3: for $u$ supported away from the endpoints the boundary term is absent, so $T^{*}v = -v'$ . Step 4: the remaining requirement is $u(1)\overline{v(1)} = 0$ for every $u \in D(T)$ , and $u(1)$ is unrestricted — take $u(x) = x$ — so $v(1) = 0$ . Hence

T^{*}v = -v', \qquad D(T^{*}) = \{\, v \in AC[0,1] : v' \in L^{2},\ v(1) = 0 \,\} ,

which is Stakgold and Holst's (Stakgold & Holst (2011), eq. (5.4.13), p. 318) result: the formal operator changes sign and the boundary condition moves to the other endpoint.

A numerical check of the identity. Take $u(x) = x^{2}$ , which satisfies $u(0)=0$ , and $v(x) = 1-x^{3}$ , which satisfies $v(1)=0$ . Both are real, so conjugation does nothing. Then

(Tu, v) = \int_{0}^{1} 2x\,(1-x^{3})\,dx = \Bigl[x^{2} - \tfrac{2}{5}x^{5}\Bigr]_{0}^{1} = 1 - \tfrac{2}{5} = \tfrac{3}{5} ,

(u, T^{*}v) = \int_{0}^{1} x^{2}\cdot\bigl(3x^{2}\bigr)\,dx = \Bigl[\tfrac{3}{5}x^{5}\Bigr]_{0}^{1} = \tfrac{3}{5} .

They agree: $\tfrac{3}{5} = 0.6$ .

A check that the boundary condition on $v$ is not optional. Take $v \equiv 1$ , so $-v' = 0$ . For $u(x) = x^{2}$ we get $(Tu, v) = \int_{0}^{1} 2x\,dx = 1$ , while $(u, 0) = 0$ . Since $1 \ne 0$ , no $\eta$ can represent the functional $u \mapsto (Tu, 1)$ in the required way for this $u$ ; indeed $|(Tu,1)| = |u(1)|$ is not bounded by any multiple of $\|u\|_{2}$ , as the functions $u_{k}(x) = x^{k}$ show: $u_k(1) = 1$ while $\|u_{k}\|_{2} = (2k+1)^{-1/2} \to 0$ . So $v \equiv 1 \notin D(T^{*})$ .

(iii) Unboundedness of $T$ , with a number. Parts (i) and (ii) follow set-ups from the books cited in them; this one is ours. Let $u_{n}(x) = \sqrt{2}\,\sin(n\pi x)$ for $n \in \mathbb{N}$ . Each lies in $D(T)$ : it is smooth and $u_{n}(0) = 0$ . Then $\|u_{n}\|^{2} = 2\int_{0}^{1}\sin^{2}(n\pi x)\,dx = 1$ and

\|Tu_{n}\|^{2} = 2n^{2}\pi^{2}\int_{0}^{1}\cos^{2}(n\pi x)\,dx = n^{2}\pi^{2} ,

so $\|Tu_{n}\|/\|u_{n}\| = n\pi$ . At $n = 3$ this is $3\pi \approx 9.42478$ , and it grows without bound: $T$ is unbounded, on a bounded interval, which is the contrast Kreyszig (Kreyszig (1978), §10.7, p. 569) draws with the multiplication operator.

Application(Why quantum mechanics cannot avoid this)Original

The observables of a one-particle quantum system are position and momentum,

(X\psi)(x) = x\,\psi(x), \qquad (P\psi)(x) = -i\hbar\,\psi'(x) ,

and the Hamiltonian built from them. Example (i) above shows $X$ is unbounded, and example (iii) shows that differentiation is already unbounded on the unit interval; $P$ inherits this, because the functions $u_{n}(x) = \sqrt{2}\,\sin(n\pi x)$ of example (iii), extended by zero outside $[0,1]$ , vanish at both endpoints and so are absolutely continuous on all of $\mathbb{R}$ with derivative in $L^{2}(\mathbb{R})$ , and satisfy $\|Pu_{n}\|/\|u_{n}\| = \hbar n\pi$ . By the Hellinger-Toeplitz theorem neither $X$ nor $P$ can be defined on all of $L^{2}$ while obeying $(T\varphi,\psi) = (\varphi,T\psi)$ . So a quantum observable is not specified by writing down a formula: it is specified by a formula together with a domain, and Reed and Simon (Reed & Simon (1980), §VIII.1, p. 254) warn that the choice of domain is usually dictated by the physics rather than by convenience, since the properties one cares about — the spectrum above all — change with it.

Three concrete consequences for practice.

Boundary conditions are the domain. A particle in a box, a particle on a ring, and a particle on a half-line all carry the same differential expression $-\hbar^{2}\psi''/2m$ and differ only in the domain. The energy levels differ because the operators differ. This is the same phenomenon as the dependence of eigenvalues on boundary conditions in Sturm-Liouville eigenvalue problems, now stated in the language that makes the spectral theorem applicable.
A core is enough. Because a closed operator is determined by its restriction to any core, one may do calculations on a convenient dense set — smooth compactly supported functions, or the Hermite functions of the harmonic oscillator used in Reed and Simon's Example 2 (Reed & Simon (1980), Example 2, p. 250), or the Fourier modes that Hassani (Hassani (2013), Example 19.1.2, p. 563) uses for $d/dx$ — and let the closure supply the rest. In practice this is how an operator is specified: give a formula, give a core, close.
Time evolution needs more than closedness. Solving $i\hbar\,\partial_{t}\psi = H\psi$ means exponentiating $H$ , and that requires the spectral theorem, which holds for self-adjoint operators and not for merely closed symmetric ones (The spectral theorem, Stone's theorem). The gap between the two conditions is the subject of Symmetric versus self-adjoint operators, and it is a real gap, not a technicality: it decides whether a given Hamiltonian determines a dynamics at all.

The same framework serves outside quantum mechanics. Parabolic and hyperbolic evolution equations are first-order systems $\dot{u} = Au$ with $A$ an unbounded closed operator on a function space, and the generation theory for the resulting semigroups (Evolution equations and operator semigroups) is stated entirely in terms of the domain, the closedness and the resolvent of $A$ . Stakgold and Holst (Stakgold & Holst (2011), §5.2, p. 307) list the alternatives honestly: work with the inverse operator, which is usually compact and bounded; or close the operator, as here; or change the norm on the domain, which is the Sobolev-space route of Weak derivatives and Sobolev spaces. All three are used, and they are not in competition.

Theorem · advanced

Symmetric versus self-adjoint operators

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Separates the identity $(T\varphi,\psi)=(\varphi,T\psi)$ from the equality of operators $T=T^{*}$ , proves the criterion that decides between them — a closed symmetric operator is self-adjoint exactly when $T^{*}\varphi = \pm i\varphi$ has only the zero solution — introduces the deficiency indices that count the failures, and settles the momentum operator on an interval, where a circle of self-adjoint extensions exists, and on a half-line, where none does.

Prerequisites: Unbounded and closed operators, Adjoints, self-adjoint and unitary operators, Orthonormal bases in Hilbert space, Spectrum and resolvent, L^p spaces

Definition(Symmetric, self-adjoint, essentially self-adjoint)AdaptedAdapted from Reed & Simon (1980), §VIII.2, pp. 255–256; Kreyszig (1978), Def. 10.2-3, Lemma 10.2-4 and Def. 10.2-5, pp. 533–534; Stakgold & Holst (2011), §5.4, eqs. (5.4.8)–(5.4.9), p. 317; Hassani (2013), Def. 19.1.5, p. 564

Let $T$ be densely defined in a complex Hilbert space $H$ , with adjoint $T^{*}$ as in Unbounded and closed operators.

$T$ is symmetric if $(T\varphi, \psi) = (\varphi, T\psi) \qquad \text{for all } \varphi, \psi \in D(T) ,$ equivalently if $T \subseteq T^{*}$ : the adjoint is an extension of $T$ . Kreyszig's Lemma 10.2-4 (Kreyszig (1978), Lemma 10.2-4, pp. 533–534) is exactly this equivalence, and he adds a third form: $T$ is symmetric if and only if $(T\varphi,\varphi)$ is real for every $\varphi \in D(T)$ .
$T$ is self-adjoint if $T = T^{*}$ , that is, $T$ is symmetric and $D(T^{*}) = D(T)$ .
$T$ is essentially self-adjoint if it is symmetric and its closure $\bar{T}$ is self-adjoint. A subspace $D \subseteq D(T)$ with $\overline{T|_{D}} = T$ is a core.

The whole content of this unit is in the two words "and $D(T^{*}) = D(T)$ ". For a symmetric $T$ , $T^{*}$ is always defined on at least $D(T)$ and may be defined on much more; self-adjointness is the statement that it is defined on no more.

The chain of inclusions. Let $T$ be symmetric. Then $D(T^{*}) \supseteq D(T)$ is dense, so by part (b) of the theorem in Unbounded and closed operators $T$ is closable, with $\bar{T} = T^{**}$ ; and applying $S \subseteq T \Rightarrow T^{*} \subseteq S^{*}$ to $T \subseteq T^{*}$ gives $T^{**} \subseteq T^{*}$ . Hence

T \ \subseteq\ \bar{T} = T^{**}\ \subseteq\ T^{*} ,

with $T = T^{**} \subseteq T^{*}$ when $T$ is closed and $T = T^{**} = T^{*}$ when $T$ is self-adjoint. Every symmetric operator therefore has a smallest closed symmetric extension, a fact Kreyszig (Kreyszig (1978), Thm 10.3-5, pp. 537–539) proves directly, by constructing the closure and checking that symmetry survives the limit.

For bounded everywhere-defined operators the distinction collapses: $D(T) = D(T^{*}) = H$ automatically, so symmetric and self-adjoint mean the same thing, which is why the question does not arise in Adjoints, self-adjoint and unitary operators. By the Hellinger-Toeplitz theorem, an unbounded symmetric operator can never be defined everywhere, so for unbounded operators the question always arises.

Definition(Deficiency subspaces and deficiency indices)AdaptedAdapted from Kreyszig (1978), §10.6, eq. (1), p. 556, and Problems 7 and 9, pp. 561–562

Let $T$ be a closed symmetric operator in $H$ . The deficiency subspaces of $T$ are

\mathcal{N}_{+} = \ker(T^{*} - iI) , \qquad \mathcal{N}_{-} = \ker(T^{*} + iI) ,

and the deficiency indices are the pair of Hilbert dimensions

n_{+} = \dim \mathcal{N}_{+} , \qquad n_{-} = \dim \mathcal{N}_{-} .

Equivalently, as the proof below records,

\mathcal{N}_{\pm} = \operatorname{Ran}(T \pm iI)^{\perp} ,

so $n_{\pm}$ measures how far $T \pm iI$ is from being onto: the indices are zero exactly when the two ranges are dense.

Kreyszig (Kreyszig (1978), §10.6, Problems 7 and 9, pp. 561–562) introduces them through the Cayley transform

U_{T} = (T - iI)(T + iI)^{-1} ,

defining the deficiency indices as the Hilbert dimensions of $\mathcal{D}(U_{T})^{\perp}$ and $\mathcal{R}(U_{T})^{\perp}$ , and states that both vanish exactly when $T$ is self-adjoint. Since $\mathcal{D}(U_{T}) = \operatorname{Ran}(T+iI)$ and $\mathcal{R}(U_{T}) = \operatorname{Ran}(T-iI)$ , that is the same pair of numbers as above. The motivation for the transform, which Kreyszig (Kreyszig (1978), §10.6, eq. (2), p. 556) gives, is that the Möbius map $t \mapsto (t-i)/(t+i)$ carries the real axis to the unit circle, so it should carry self-adjoint operators to unitary ones; Lemma 10.6-1 confirms this.

Conventions differ over which index carries which sign, so the pairing $\mathcal{N}_{+} = \ker(T^{*}-iI)$ fixed here should be checked against any other source before its results are quoted.

Theorem(Self-adjointness is decided at the two points

\pm i

)AdaptedAdapted from Reed & Simon (1980), Thm VIII.3 and its corollary, pp. 256–257

Let $H$ be a complex Hilbert space and let $T$ be symmetric in $H$ . The three conditions below are equivalent; the result is Reed and Simon's basic criterion (Reed & Simon (1980), Thm VIII.3, pp. 256–257).

(a) $T$ is self-adjoint.

(b) $T$ is closed and $\ker(T^{*} \pm iI) = \{0\}$ .

(c) $\operatorname{Ran}(T \pm iI) = H$ .

In the language of the previous block: a closed symmetric operator is self-adjoint exactly when $n_{+} = n_{-} = 0$ .

The corresponding statement for a symmetric operator that is not assumed closed is Reed and Simon's corollary (Reed & Simon (1980), Corollary to Thm VIII.3, p. 257): $T$ is essentially self-adjoint $\iff$ $\ker(T^{*} \pm iI) = \{0\}$ $\iff$ both ranges $\operatorname{Ran}(T \pm iI)$ are dense.

The number $i$ plays no special role beyond being non-real; what the proof uses is that $T \pm \lambda I$ is bounded below when $\operatorname{Im}\lambda \ne 0$ , which is also the content of Stakgold and Holst's Thm 5.6.2(b) (Stakgold & Holst (2011), Thm 5.6.2(b), p. 328): for a symmetric operator the approximate spectrum lies on the real axis.

Theorem(Equal deficiency indices are necessary for a self-adjoint extension)AdaptedAdapted from Kreyszig (1978), Lemma 10.6-1, pp. 556–557, and §10.6, Problems 5–9, pp. 561–562; Reed & Simon (1980), Thm VIII.3, pp. 256–257

Let $T$ be a closed symmetric operator in $H$ . If $T$ has a self-adjoint extension, then $n_{+} = n_{-}$ .

Consequently a closed symmetric operator with $n_{+} \ne n_{-}$ has no self-adjoint extension at all, and in particular is not essentially self-adjoint and generates no unitary dynamics.

The converse — that $n_{+} = n_{-} = n$ implies a family of self-adjoint extensions parametrised by the unitary maps $\mathcal{N}_{+} \to \mathcal{N}_{-}$ — is von Neumann's extension theorem. Our library does not contain a statement of it: Reed and Simon (Reed & Simon (1980), §VIII.2, p. 259) exhibit a circle of self-adjoint extensions for the operator $i\,d/dx$ on $[0,1]$ and refer the explanation forward to Section X.1 of their second volume, which is not held here. In the worked example below we therefore verify the extensions directly, by computing adjoints, rather than quoting the general theorem.

Note(Symmetric, hermitian, self-adjoint: what the books mean)AdaptedAdapted from Reed & Simon (1980), §VIII.2, p. 255; Hassani (2013), Def. 19.1.5 and footnote 3, p. 564; Stakgold & Holst (2011), §5.4, p. 317; Arfken, Weber & Harris (2013), §5.4 and Example 5.4.1, pp. 283–284

The vocabulary is not uniform across our sources, and misreading it is the commonest way to misread a statement about an unbounded operator.

Reed and Simon (Reed & Simon (1980), §VIII.2, p. 255) define symmetric and give Hermitian as a synonym for it, reserving self-adjoint for $T = T^{*}$ .
Hassani (Hassani (2013), Def. 19.1.5 and footnote 3, p. 564) uses hermitian for $T \subseteq T^{\dagger}$ and self-adjoint for $D(T)=D(T^{\dagger})$ , and then says plainly that the distinction is confined to that one discussion and that the two words are used interchangeably in the rest of his book.
Stakgold and Holst (Stakgold & Holst (2011), §5.4, p. 317) keep the two apart, and state the three facts to remember: every self-adjoint operator is symmetric; every bounded symmetric operator is self-adjoint; there are unbounded symmetric operators that are not self-adjoint.
Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.4 and Example 5.4.1, pp. 283–284) write self-adjoint (Hermitian) as one term, and check in Example 5.4.1 that multiplication by $x$ and $p = -i\,d/dx$ are self-adjoint on $L^{2}(-\infty,\infty)$ by comparing $\langle f | pg\rangle$ with $\langle pf | g\rangle$ , integrating by parts and discarding the integrated term at $\pm\infty$ . That calculation is the symmetry computation, performed on functions for which the boundary term vanishes; on the whole line its conclusion is also correct for self-adjointness, since Kreyszig's Thm 10.7-5 (Kreyszig (1978), Thm 10.7-5, p. 569) states that $i\,d/dx$ on its natural domain — the locally absolutely continuous $L^{2}$ functions with $L^{2}$ derivative — is self-adjoint. On an interval or a half-line the same computation no longer settles the question, and the example below shows what it misses.

In this unit symmetric always means $T \subseteq T^{*}$ and nothing more.

Idea of the proofOriginal

Why should the two particular numbers $\pm i$ decide everything?

Because for a symmetric operator they are the two points where $T - \lambda$ is as invertible as it can possibly be. Expanding the square and using that $(T\varphi,\varphi)$ is real,

\|(T \pm i)\varphi\|^{2} = \|T\varphi\|^{2} + \|\varphi\|^{2} ,

with no cross term. So $T \pm i$ is injective, bounded below, and — if $T$ is closed — has closed range. A closed subspace is the whole space exactly when its orthogonal complement vanishes, and that complement is a kernel of $T^{*} \mp i$ . The three conditions of the criterion are therefore three readings of one geometric fact: how much of $H$ the ranges of $T \pm i$ miss.

The proof that self-adjointness is equivalent to missing nothing then runs in a circle. If $T = T^{*}$ , an eigenvector of $T$ for the eigenvalue $i$ would make $(T\varphi,\varphi)$ non-real, so there is none. If there is none, the ranges are dense, and closedness upgrades dense to everything. If the ranges are everything, then any $\varphi \in D(T^{*})$ can be corrected by an element of $D(T)$ so that the difference is killed by $T^{*}-i$ , and injectivity of $T^{*}-i$ forces the difference to vanish, putting $\varphi$ in $D(T)$ .

Deficiency indices measure the failure, and the Cayley transform explains why they must be equal for an extension to exist. The transform turns $T$ into an isometry $U_{T}$ from $\operatorname{Ran}(T+i)$ onto $\operatorname{Ran}(T-i)$ , and self-adjoint extensions of $T$ correspond to unitary extensions of $U_{T}$ to all of $H$ . A unitary extension must carry the leftover space on one side isometrically onto the leftover space on the other; two leftover spaces of different dimension cannot be matched up, and no extension exists.

Proof(The chain and the reality of

(T\varphi,\varphi)

)AdaptedAdapted from Kreyszig (1978), Lemma 10.2-4, pp. 533–534, and §10.2, p. 534; Stakgold & Holst (2011), Thm 5.6.2(a), pp. 327–328

Symmetry as an inclusion. If $T \subseteq T^{*}$ then for $\varphi,\psi \in D(T) \subseteq D(T^{*})$ we have $(T\varphi,\psi) = (\varphi, T^{*}\psi) = (\varphi,T\psi)$ , since $T^{*}$ agrees with $T$ on $D(T)$ . Conversely, if $(T\varphi,\psi) = (\varphi,T\psi)$ for all $\varphi,\psi \in D(T)$ , then each $\psi \in D(T)$ satisfies the defining condition for membership of $D(T^{*})$ , with $T^{*}\psi = T\psi$ ; so $D(T) \subseteq D(T^{*})$ and $T \subseteq T^{*}$ .

Reality. If $T$ is symmetric and $\varphi \in D(T)$ then $(T\varphi,\varphi) = (\varphi,T\varphi) = \overline{(T\varphi,\varphi)}$ , the first equality by symmetry and the second by the conjugate symmetry of the inner product; a number equal to its own conjugate is real. The converse uses polarisation: if $(T\varphi,\varphi)$ is real for all $\varphi \in D(T)$ , then writing the sesquilinear form $B(\varphi,\psi) = (T\varphi,\psi) - (\varphi,T\psi)$ one checks that $B(\varphi,\varphi) = 0$ for all $\varphi$ and that $B$ is determined by its diagonal through the polarisation identity, so $B \equiv 0$ . This is the third characterisation quoted from Kreyszig (Kreyszig (1978), §10.2, p. 534). $\square$

Proof(The basic criterion)AdaptedAdapted from Reed & Simon (1980), Thm VIII.3, pp. 256–257

Throughout, $T$ is symmetric, so $(T\varphi,\varphi)$ is real for $\varphi \in D(T)$ .

Step 0: the identity with no cross term. For $\varphi \in D(T)$ ,

\|(T \pm i)\varphi\|^{2} = \|T\varphi\|^{2} \pm \overline{i}\,(T\varphi,\varphi) \pm i\,(\varphi,T\varphi) + \|\varphi\|^{2} = \|T\varphi\|^{2} + \|\varphi\|^{2} ,

because $(T\varphi,\varphi) = (\varphi,T\varphi)$ is real and $\bar i + i = 0$ . In particular $\|(T\pm i)\varphi\| \ge \|\varphi\|$ , so $T \pm i$ is injective.

Step 1: the orthogonal complement of a range is a kernel. For $\psi \in H$ ,

\psi \perp \operatorname{Ran}(T - i) \iff ((T-i)\varphi, \psi) = 0 \ \text{ for all } \varphi \in D(T) \iff (T\varphi,\psi) = i(\varphi,\psi) = (\varphi, -i\psi) \ \text{ for all } \varphi \in D(T) ,

and the last statement says exactly that $\psi \in D(T^{*})$ with $T^{*}\psi = -i\psi$ . So $\operatorname{Ran}(T-i)^{\perp} = \ker(T^{*}+i) = \mathcal{N}_{-}$ , and the same computation with the signs reversed gives $\operatorname{Ran}(T+i)^{\perp} = \ker(T^{*}-i) = \mathcal{N}_{+}$ .

Step 2: (a) implies (b). Let $T = T^{*}$ . Then $T$ is closed, because an adjoint is always closed. If $T^{*}\varphi = i\varphi$ then $\varphi \in D(T^{*}) = D(T)$ and $T\varphi = i\varphi$ , so

i\|\varphi\|^{2} = (i\varphi,\varphi) = (T\varphi,\varphi) = (\varphi,T\varphi) = (\varphi, i\varphi) = -i\|\varphi\|^{2} ,

forcing $\|\varphi\|^{2} = 0$ . The same with $-i$ in place of $i$ . Hence both kernels are trivial.

Step 3: (b) implies (c). By Step 1, $\ker(T^{*}+i) = \{0\}$ says $\operatorname{Ran}(T-i)^{\perp} = \{0\}$ , that is, $\operatorname{Ran}(T-i)$ is dense. It is also closed: if $(T-i)\varphi_{n} \to \chi$ , then by Step 0

\|\varphi_{n}-\varphi_{m}\|^{2} + \|T\varphi_{n}-T\varphi_{m}\|^{2} = \|(T-i)(\varphi_{n}-\varphi_{m})\|^{2} \to 0 ,

so both $(\varphi_{n})$ and $(T\varphi_{n})$ are Cauchy; let $\varphi_{n} \to \varphi$ and $T\varphi_{n}\to\eta$ . Here the hypothesis that $T$ is closed enters: it gives $\varphi \in D(T)$ and $T\varphi = \eta$ , whence $(T-i)\varphi = \chi$ and $\chi \in \operatorname{Ran}(T-i)$ . A dense closed subspace is the whole space, so $\operatorname{Ran}(T-i) = H$ ; likewise $\operatorname{Ran}(T+i) = H$ .

Step 4: (c) implies (a). Assume both ranges are all of $H$ . First, $\ker(T^{*}-i) = \operatorname{Ran}(T+i)^{\perp} = H^{\perp} = \{0\}$ by Step 1. Now take any $\varphi \in D(T^{*})$ . Since $\operatorname{Ran}(T-i) = H$ , there is $\eta \in D(T)$ with

(T - i)\eta = (T^{*}-i)\varphi .

Because $T$ is symmetric, $D(T) \subseteq D(T^{*})$ and $T^{*}\eta = T\eta$ ; so $\varphi - \eta \in D(T^{*})$ and $(T^{*}-i)(\varphi-\eta) = 0$ . By the triviality of $\ker(T^{*}-i)$ just established, $\varphi = \eta \in D(T)$ . Hence $D(T^{*}) \subseteq D(T)$ ; the reverse inclusion is symmetry, and the two operators agree where both are defined. So $T = T^{*}$ .

The corollary. Apply the theorem to $\bar{T} = T^{**}$ , which is closed and symmetric, and use $(\bar{T})^{*} = T^{*}$ from Unbounded and closed operators: then $\bar{T}$ is self-adjoint $\iff$ $\ker(T^{*}\pm i) = \{0\}$ $\iff$ $\operatorname{Ran}(\bar T \pm i) = H$ . Finally $\operatorname{Ran}(\bar{T}\pm i)$ is the closure of $\operatorname{Ran}(T \pm i)$ , since $\bar T$ is the closure of $T$ and Step 0 makes $T\pm i$ bounded below; so the last condition is that $\operatorname{Ran}(T\pm i)$ be dense. $\square$

Proof(Equal indices are necessary)AdaptedAdapted from Kreyszig (1978), Lemma 10.6-1, pp. 556–557, and §10.6, Problems 5–7, pp. 561–562; Reed & Simon (1980), Thm VIII.3, pp. 256–257

Let $T$ be closed and symmetric.

Step 1: the Cayley transform of $T$ is an isometry between two closed subspaces. By Step 0 of the previous proof, $T+i$ is injective, so

U_{T} = (T-i)(T+i)^{-1} : \operatorname{Ran}(T+i) \longrightarrow \operatorname{Ran}(T-i)

is a well-defined linear bijection, and Step 3 of that proof shows both ranges are closed (that step used only closedness of $T$ , not the triviality of the kernels). It is isometric: writing $\chi = (T+i)\varphi$ , Step 0 gives

\|U_{T}\chi\|^{2} = \|(T-i)\varphi\|^{2} = \|T\varphi\|^{2}+\|\varphi\|^{2} = \|(T+i)\varphi\|^{2} = \|\chi\|^{2} .

This is the computation Kreyszig (Kreyszig (1978), Lemma 10.6-1, p. 557) performs for a self-adjoint $T$ ; it uses only that $(T\varphi,\varphi)$ is real, hence only symmetry.

Step 2: a self-adjoint extension gives a unitary extension of $U_{T}$ . Let $S \supseteq T$ be self-adjoint. By the criterion, $\operatorname{Ran}(S \pm i) = H$ , so $U_{S} = (S-i)(S+i)^{-1}$ is defined on all of $H$ , maps onto $H$ , and is isometric by Step 1 applied to $S$ : it is unitary. And $U_{S}$ extends $U_{T}$ : if $\chi = (T+i)\varphi$ with $\varphi \in D(T) \subseteq D(S)$ , then $(S+i)\varphi = (T+i)\varphi = \chi$ , so $U_{S}\chi = (S-i)\varphi = (T-i)\varphi = U_{T}\chi$ .

Step 3: a unitary matches the leftovers. Write $M = \operatorname{Ran}(T+i)$ and $N = \operatorname{Ran}(T-i)$ , both closed. By Step 2, $U_{S}[M] = U_{T}[M] = N$ . A unitary operator is a surjective isometry, so it carries orthogonal complements to orthogonal complements: if $x \perp M$ then $(U_{S}x, U_{S}y) = (x,y) = 0$ for every $y \in M$ , so $U_{S}x \perp N$ ; and applying the same argument to $U_{S}^{-1}$ gives the reverse inclusion. Hence $U_{S}$ restricts to a unitary map $M^{\perp} \to N^{\perp}$ , and these two spaces have the same Hilbert dimension:

n_{+} = \dim \mathcal{N}_{+} = \dim \operatorname{Ran}(T+i)^{\perp} = \dim M^{\perp} = \dim N^{\perp} = \dim \operatorname{Ran}(T-i)^{\perp} = \dim\mathcal{N}_{-} = n_{-} ,

using Step 1 of the previous proof to identify the complements with the deficiency subspaces. $\square$

Remark(Where the hypotheses are used, and what fails without them)Original

Symmetry is used in Step 0 of the criterion, to kill the cross term; everything else follows from that identity. Without it, $T \pm i$ need not be bounded below and no part of the argument survives.
Closedness is used exactly once, in Step 3, to pass from dense range to full range. This is the whole difference between self-adjointness and essential self-adjointness: $-i\,d/dx$ on $C_{c}^{\infty}(\mathbb{R})$ has trivial deficiency subspaces and dense ranges, hence is essentially self-adjoint, but is not self-adjoint, because it is not closed. Its closure is.
Complex scalars are essential. On a real Hilbert space $\pm i$ do not exist, the criterion has no statement, and the reality of $(T\varphi,\varphi)$ carries no information. Every result here presupposes the complex field, as Kreyszig (Kreyszig (1978), Ch. 10, p. 524) states at the head of his chapter.
Density of $D(T)$ is needed before $T^{*}$ exists at all, and hence before "symmetric" can be phrased as $T \subseteq T^{*}$ .
Equal indices do not come for free. The half-line example below has $(n_{+},n_{-}) = (1,0)$ . For real operators — those commuting with a conjugation, such as a Schrödinger operator with real potential — the two indices are always equal and self-adjoint extensions always exist; that fact is not proved here and is not stated in our library.
Essential self-adjointness is a uniqueness statement. Reed and Simon (Reed & Simon (1980), §VIII.2, p. 256) note that an essentially self-adjoint $T$ has exactly one self-adjoint extension: if $S$ is self-adjoint with $S \supseteq T$ then $S$ is closed, so $S \supseteq T^{**}$ , hence $S = S^{*} \subseteq (T^{**})^{*} = T^{***} = T^{**}$ , giving $S = T^{**}$ . This is what makes it the useful property in practice: it says that specifying the operator on a convenient small domain already specifies the physics.
What is at stake. The spectral theorem (The spectral theorem) and the exponentiation $e^{-itT}$ that produces a unitary dynamics (Stone's theorem) hold for self-adjoint operators and fail for merely closed symmetric ones; Reed and Simon (Reed & Simon (1980), §VIII.2, p. 256) make this the reason the distinction matters. Note also that a closed symmetric $T$ falling short of self-adjointness has non-real points in its spectrum, even though $(T\varphi,\varphi)$ is real throughout its domain: Step 1 of the proof gives $\operatorname{Ran}(T+i)^{\perp} = \mathcal{N}_{+}$ and $\operatorname{Ran}(T-i)^{\perp} = \mathcal{N}_{-}$ , so $n_{+} \ne 0$ makes $T+i$ fail to be onto and puts $-i$ in the spectrum, while $n_{-} \ne 0$ puts $+i$ there (Spectrum and resolvent).

Method(Deciding whether a differential operator is self-adjoint)Original

For a formally self-adjoint differential expression $\tau$ on an interval $(a,b)$ , finite or infinite, and a proposed domain $D$ :

Check symmetry. Integrate by parts to isolate the boundary form, as in the method of Unbounded and closed operators. The operator is symmetric exactly when the boundary form vanishes for all pairs $u, v \in D$ .
Compute the adjoint. Repeat the integration by parts with $u \in D$ and $v$ unrestricted, and read off the boundary conditions on $v$ . The formula for $T^{*}$ is always $\tau$ ; only the domain is in question.
Compare the two domains. If they coincide, $T$ is self-adjoint and you are finished. If $D(T^{*})$ is strictly larger, continue.
Compute the deficiency indices. Solve $\tau v = \pm i v$ — an ordinary differential equation with constant or variable coefficients, solvable in closed form in the easy cases — and count how many of its solutions lie in $L^{2}(a,b)$ and in $D(T^{*})$ . Those counts are $n_{+}$ and $n_{-}$ . Solutions are discarded when they fail to be square-integrable, which is why the answer differs between a bounded interval, a half-line and the whole line.
Read off the verdict. If $n_{+} = n_{-} = 0$ and $T$ is closed, $T$ is self-adjoint; if $n_{+} = n_{-} = 0$ and $T$ is not closed, it is essentially self-adjoint and the closure is the operator you want. If $n_{+} \ne n_{-}$ , stop: there is no self-adjoint extension.
If $n_{+} = n_{-} = n > 0$ , hunt for the extensions by hand. Enlarge $D$ by imposing $n$ boundary conditions containing a parameter, recompute the adjoint domain as in step 2, and solve for the parameter values that make the two domains agree. For a first-order operator on a bounded interval this is a single equation in one unknown, done below.

Example(Momentum on an interval and on a half-line)Original

Throughout, $p = -i\,\dfrac{d}{dx}$ , the physicists' momentum with $\hbar = 1$ , and $(f,g) = \int f\bar g$ . For any $u,v$ absolutely continuous on $[c,d]$ with $L^{2}$ derivatives,

(pu, v) - (u, pv) = -i\int_{c}^{d}\bigl(u'\bar v + u \bar v'\bigr)\,dx = -i\Bigl[u\bar v\Bigr]_{c}^{d} , \tag{$\ast$}

since $\overline{-i v'} = i\bar v'$ . Everything below is read off from $(\ast)$ .

(A) Dirichlet domain on $[0,1]$ : symmetric, not self-adjoint. Let

D(p_{0}) = \{\, u \in AC[0,1] : u' \in L^{2},\ u(0) = u(1) = 0 \,\} , \qquad p_{0}u = -iu' .

By $(\ast)$ the boundary term vanishes for all $u,v \in D(p_{0})$ , so $p_{0}$ is symmetric. Its adjoint, computed as in Stakgold and Holst's Example 3 (Stakgold & Holst (2011), Example 3, p. 320), carries no boundary condition:

p_{0}^{*}v = -iv' , \qquad D(p_{0}^{*}) = \{\, v \in AC[0,1] : v' \in L^{2} \,\} .

Indeed $(\ast)$ shows the boundary term $-i[u\bar v]_{0}^{1}$ already vanishes for every $u \in D(p_{0})$ , whatever $v$ does. The domains are therefore different by two complex dimensions' worth of boundary values, and $p_{0}$ is not self-adjoint. Reed and Simon (Reed & Simon (1980), §VIII.2, pp. 257–258) obtain the same adjoint for $i\,d/dx$ on the same domain.

Deficiency indices. Solve $p_{0}^{*}v = \pm iv$ , that is $-iv' = \pm i v$ :

v' = -v \ \Rightarrow\ v = e^{-x} \quad (\text{for } +i), \qquad v' = +v \ \Rightarrow\ v = e^{x} \quad (\text{for } -i) .

Both are absolutely continuous on $[0,1]$ with $L^{2}$ derivative, so both lie in $D(p_{0}^{*})$ , and each solution space is one-dimensional:

n_{+} = \dim\ker(p_{0}^{*}-i) = 1 , \qquad n_{-} = \dim\ker(p_{0}^{*}+i) = 1 .

The indices are equal, so a self-adjoint extension is not excluded; we now produce them.

(B) The circle of self-adjoint extensions. For a complex number $\alpha \ne 0$ put

D(p_{\alpha}) = \{\, u \in AC[0,1] : u' \in L^{2},\ u(1) = \alpha\,u(0) \,\} , \qquad p_{\alpha}u = -iu' .

Each $p_{\alpha}$ extends $p_{0}$ , since $u(0)=u(1)=0$ satisfies the condition. By $(\ast)$ , for $u, v \in D(p_{\alpha})$ ,

(p_{\alpha}u,v) - (u,p_{\alpha}v) = -i\bigl[u(1)\overline{v(1)} - u(0)\overline{v(0)}\bigr] = -i\,u(0)\overline{v(0)}\bigl(|\alpha|^{2} - 1\bigr) ,

using $u(1) = \alpha u(0)$ and $\overline{v(1)} = \bar\alpha\,\overline{v(0)}$ . Since $u(0)$ and $v(0)$ are free, this vanishes identically exactly when $|\alpha| = 1$ . So $p_{\alpha}$ is symmetric precisely on the unit circle of parameters.

For $|\alpha|=1$ the operator is in fact self-adjoint. Let $v \in D(p_{\alpha}^{*})$ ; as in step 3 of the method the formula is forced to be $-iv'$ with $v \in AC[0,1]$ , and the remaining requirement from $(\ast)$ is

u(1)\overline{v(1)} - u(0)\overline{v(0)} = 0 \quad\text{for all } u \in D(p_{\alpha}) ,

that is $u(0)\bigl[\alpha\overline{v(1)} - \overline{v(0)}\bigr] = 0$ for all values of $u(0)$ , hence $\alpha \overline{v(1)} = \overline{v(0)}$ , hence $v(1) = \bar\alpha^{-1} v(0) = \alpha\,v(0)$ because $|\alpha|=1$ . The adjoint domain is $D(p_{\alpha})$ itself: $p_{\alpha}^{*} = p_{\alpha}$ .

This reproduces, with $\alpha = e^{i\theta}$ , the circle of distinct self-adjoint extensions that Reed and Simon (Reed & Simon (1980), §VIII.2, p. 259) exhibit for $i\,d/dx$ . Stakgold and Holst's Example 2 (Stakgold & Holst (2011), Example 2, p. 319) run the same computation but allow only real $a$ in the boundary condition, and therefore find exactly two self-adjoint members, $a = 1$ and $a = -1$ ; these are the points $\theta = 0$ and $\theta = \pi$ of the circle, and the two statements agree.

Spectrum of $p_{\alpha}$ . Write $\alpha = e^{i\theta}$ with $\theta \in [0,2\pi)$ . Solving $-iu' = \lambda u$ gives $u(x) = c\,e^{i\lambda x}$ , and the boundary condition $u(1) = \alpha u(0)$ reads $e^{i\lambda} = e^{i\theta}$ , so

\lambda_{n} = \theta + 2\pi n , \qquad n \in \mathbb{Z} ,

all real, as they must be for a self-adjoint operator. The normalised eigenfunctions are $e_{n}(x) = e^{i\lambda_{n}x}$ : they satisfy $\|e_{n}\|=1$ and

(e_{n},e_{m}) = \int_{0}^{1} e^{i(\lambda_{n}-\lambda_{m})x}\,dx = \int_{0}^{1} e^{2\pi i (n-m)x}\,dx = \delta_{nm} .

They are a complete orthonormal set, because $e_{n}(x) = e^{i\theta x}\,e^{2\pi i n x}$ and multiplication by the unimodular function $e^{i\theta x}$ is a unitary map of $L^{2}(0,1)$ onto itself, which carries the Fourier basis $\{e^{2\pi i n x}\}$ of Fourier series to $\{e_{n}\}$ (see Orthonormal bases in Hilbert space). So each $p_{\alpha}$ has a complete set of eigenfunctions and a purely real, purely discrete spectrum — and the spectra of different $\alpha$ are genuinely different sets, which is the sharpest possible statement that these are different operators.

A numerical check. Take $\alpha = i$ , that is $\theta = \pi/2$ , and

u(x) = 1 + (i-1)x ,

so $u(0) = 1$ and $u(1) = i = \alpha\,u(0)$ : thus $u \in D(p_{i})$ . Then $|u(x)|^{2} = (1-x)^{2} + x^{2}$ and

\|u\|^{2} = \int_{0}^{1}\bigl[(1-x)^{2}+x^{2}\bigr]dx = \tfrac13 + \tfrac13 = \tfrac23 .

Also $p_{i}u = -i(i-1) = 1+i$ , a constant, so

(p_{i}u, u) = (1+i)\int_{0}^{1}\overline{1+(i-1)x}\,dx = (1+i)\Bigl(1 + \frac{-1-i}{2}\Bigr) = (1+i)\cdot\frac{1-i}{2} = \frac{1+1}{2} = 1 .

This is real, as symmetry requires, and the expectation value is

\langle p \rangle = \frac{(p_{i}u,u)}{\|u\|^{2}} = \frac{1}{2/3} = \frac{3}{2} = 1.5 .

For comparison, the eigenvalues of $p_{i}$ are $\lambda_{n} = \tfrac{\pi}{2} + 2\pi n$ ; the two nearest zero are $\lambda_{-1} = -3\pi/2 \approx -4.7124$ and $\lambda_{0} = \pi/2 \approx 1.5708$ , and the value $1.5$ falls between them. Had the same calculation produced a complex number, or had it produced a real number for one state and a complex one for another, the operator would not have been symmetric.

Contrast with $|\alpha| \ne 1$ . Take $\alpha = 2$ and $u(x) = 1+x$ , so $u(0)=1$ , $u(1) = 2 = 2u(0)$ and $u \in D(p_{2})$ . Then $p_{2}u = -i$ and

(p_{2}u, u) = -i\int_{0}^{1}(1+x)\,dx = -\tfrac{3}{2}i ,

which is not real. By the reality criterion, $p_{2}$ is not symmetric — exactly as the boundary-form computation predicted, since $|2|^{2}-1 = 3 \ne 0$ .

(C) The half-line: no self-adjoint extension exists. On $H = L^{2}(0,\infty)$ let

D(p_{+}) = \{\, u : u \text{ locally absolutely continuous on } [0,\infty),\ u, u' \in L^{2}(0,\infty),\ u(0)=0 \,\} , \qquad p_{+}u = -iu' .

The domain contains $C_{c}^{\infty}(0,\infty)$ and is therefore dense.

A lemma: functions in the domain vanish at infinity. If $u$ and $u'$ are both in $L^{2}(0,\infty)$ and $u$ is locally absolutely continuous, then for $x > 0$

|u(x)|^{2} = |u(0)|^{2} + 2\operatorname{Re}\int_{0}^{x} u'(t)\overline{u(t)}\,dt ,

and $u'\bar u \in L^{1}(0,\infty)$ by Cauchy-Schwarz, so the right-hand side converges as $x \to \infty$ ; the limit of $|u(x)|^{2}$ must be $0$ , since a positive limit would contradict $u \in L^{2}$ .

Symmetry. Apply $(\ast)$ on $[0,R]$ and let $R \to \infty$ : the term at $R$ vanishes by the lemma and the term at $0$ vanishes because $u(0)=0$ . So $p_{+}$ is symmetric.

Deficiency indices. Let $v \in \ker(p_{+}^{*} - i)$ , so $(p_{+}u, v) = (u, iv)$ for every $u \in D(p_{+})$ . Testing with $u \in C_{c}^{\infty}(0,\infty)$ gives $-i\int u'\bar v = -i\int u \bar v$ , hence $\int u'\bar v = \int u\bar v$ for all test functions $u$ ; by the definition of the weak derivative (Weak derivatives and Sobolev spaces) this says $-\bar v' = \bar v$ , so $v(x) = c\,e^{-x}$ . And $e^{-x}$ really does belong: for $u \in D(p_{+})$ ,

(p_{+}u, e^{-x}) = -i\int_{0}^{\infty} u'e^{-x}dx = -i\Bigl(\bigl[ue^{-x}\bigr]_{0}^{\infty} + \int_{0}^{\infty}u e^{-x}dx\Bigr) = -i\int_{0}^{\infty}ue^{-x}dx = (u,\, i\,e^{-x}) ,

the boundary terms vanishing by $u(0)=0$ and the lemma. So $\mathcal{N}_{+} = \mathbb{C}\,e^{-x}$ and $n_{+} = 1$ ; for the record $\|e^{-x}\|^{2}_{L^{2}(0,\infty)} = \int_{0}^{\infty}e^{-2x}dx = \tfrac12$ .

The same computation with $-i$ gives $v(x) = c\,e^{x}$ , and

\int_{0}^{\infty} e^{2x}\,dx = \infty ,

so no non-zero multiple lies in $L^{2}(0,\infty)$ : $\mathcal{N}_{-} = \{0\}$ and $n_{-} = 0$ .

Verdict. $(n_{+}, n_{-}) = (1,0)$ . The indices are unequal, so by the theorem above the closure of $p_{+}$ has no self-adjoint extension whatever. The momentum of a particle confined to a half-line is not an observable — not because we have failed to find the right boundary condition, but because the count forbids one. Compare the three cases: on $(0,1)$ the indices are $(1,1)$ and there is a circle of answers; on $(0,\infty)$ they are $(1,0)$ and there is none; on $\mathbb{R}$ neither $e^{-x}$ nor $e^{x}$ is square-integrable, the indices are $(0,0)$ , and the operator is essentially self-adjoint on the smooth compactly supported functions, with the unique self-adjoint realisation that Kreyszig (Kreyszig (1978), Thm 10.7-5, p. 569) states for $i\,d/dx$ on the line.

Application(Boundary conditions as physics)Original

Why a physicist should care about a domain. An observable in quantum mechanics is a self-adjoint operator, not a symmetric one, because it is self-adjointness that delivers the spectral measure from which measurement probabilities are read (The spectral theorem, The mathematical framework of quantum mechanics) and the unitary group $e^{-iHt/\hbar}$ that moves states in time (Stone's theorem). Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.4, p. 283) state the requirement in its physical form: the expectation value of an observable must be real. Reality of $\langle A \rangle$ is symmetry, and the calculation above shows that on a bounded region symmetry alone leaves a whole circle of inequivalent operators, each with a different spectrum.

The particle in a box. For a particle confined to $0 \le x \le 1$ with the usual walls, the wavefunction satisfies $\psi(0) = \psi(1) = 0$ , and the momentum operator on that domain is case (A): symmetric, with deficiency indices $(1,1)$ , and not self-adjoint. There is no momentum observable for a particle in a box, and the familiar fact that the energy eigenfunctions $\sin(n\pi x)$ are not momentum eigenfunctions is the visible symptom. The Hamiltonian $-\tfrac{\hbar^{2}}{2m}\,d^{2}/dx^{2}$ on the same domain is self-adjoint, which is why the energies are well defined even though the momentum is not; a formally self-adjoint expression can thus be self-adjoint at one order and not at another on the same boundary conditions.

The particle on a ring. Identify $x=0$ with $x=1$ and the natural domains are the $p_{\alpha}$ of case (B). The parameter $\alpha = e^{i\theta}$ is a physical quantity: it is the phase a wavefunction picks up on going once round, and the spectrum $\theta + 2\pi n$ shifts rigidly with it. A magnetic flux threading the ring produces exactly this, because the gauge transformation $\psi \mapsto e^{i\theta x}\psi$ carries the periodic domain $\theta = 0$ to the twisted domain of parameter $\theta$ while preserving the inner product; the flux is not a term in the operator but a choice of self-adjoint extension. The same structure appears as the Bloch phase of a band electron in a periodic crystal, where $\theta$ is the crystal momentum.

Half-lines and radial problems. Case (C) is the reason a radial coordinate has to be handled with care: the radial momentum $-i\partial_{r}$ on $(0,\infty)$ has unequal deficiency indices and is not an observable, although the radial kinetic energy is. More generally, the deficiency-index count is the rigorous content of the physicist's question "which boundary conditions are allowed?": for a singular Sturm-Liouville problem (Sturm-Liouville eigenvalue problems) it is the number of solutions of $\tau v = \pm i v$ that survive in $L^{2}$ near each singular endpoint that decides how many conditions must be imposed, and whether the problem is determined at all. The limit-point and limit-circle vocabulary of that theory is the same count under another name.

And in applied mathematics generally. The pattern recurs wherever an evolution equation is posed on a bounded domain. A diffusion or wave problem is well posed only when the generator with its boundary conditions is a genuine self-adjoint (or generating) operator; the family of admissible boundary conditions is the family of extensions, and each one encodes a different physical boundary — reflecting, absorbing, periodic. The operator-semigroup framework of Evolution equations and operator semigroups takes this as its starting hypothesis, and the work of verifying it is precisely the work done in this unit.

Theorem · advanced

The spectral theorem

Adapted

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States and proves the spectral theorem for a possibly unbounded self-adjoint operator in its three equivalent forms - unitary equivalence with a multiplication operator, a Borel functional calculus, and a projection-valued measure - and shows on the momentum operator how the three are read off from one another.

Prerequisites: Symmetric versus self-adjoint operators, Unbounded and closed operators, Adjoints, self-adjoint and unitary operators, Bounded linear operators, Spectrum and resolvent, Hilbert spaces, Orthonormal bases in Hilbert space, Uniform boundedness and open mapping theorems, Zorn's lemma and the axiom of choice, The Lebesgue integral, Sigma-algebras and measures, L^p spaces, Monotone convergence and Fatou's lemma, Dominated convergence, The Radon-Nikodym theorem, Plancherel's theorem

Note(What the theorem does)Original

A Hermitian matrix can be diagonalised: there is an orthonormal basis of eigenvectors, and in that basis the matrix acts by multiplying each coordinate by a real number (Spectral theorem for Hermitian matrices). A compact self-adjoint operator behaves the same way, with a sequence of eigenvalues tending to zero (Spectral theorem for compact self-adjoint operators). But most of the operators that matter in physics have no eigenvectors at all. Multiplication by the coordinate on $L^2(\mathbb{R})$ has none, and neither does $-i\,d/dx$ ; their spectra are the whole real line, filled with "eigenvalues" whose "eigenfunctions" are not square-integrable.

The spectral theorem says that nothing essential is lost. Every self-adjoint operator, bounded or not, still is a multiplication operator - not on a sequence space indexed by eigenvalues, but on an $L^2$ space over a measure space that the operator itself determines. Sums over an eigenbasis become integrals against a measure. Once that is known, one can form $h(A)$ for any bounded Borel function $h$ , and in particular the spectral projections $P_\Omega$ that answer "how much of $\psi$ lives in the part of the spectrum lying in $\Omega$ ?" - the question quantum mechanics asks of every measurement.

Definition(Multiplication operators, spectral representations, spectral measures)AdaptedAdapted from Reed & Simon (1980), §VII.2, the definitions of cyclic vector and of spectral measure, pp. 226-228, the essential range, p. 229, and §VIII.3, Proposition 1 and the p.v.m. axioms, pp. 259, 263

Throughout, $\mathcal{H}$ is a complex separable Hilbert space, and a non-zero one, $\mathcal{H}\ne\{0\}$ - the exclusion is needed only in Step 1, which uses the non-emptiness of the spectrum - and the inner product $(\varphi,\psi)$ is linear in the first slot, as elsewhere in this handbook (Inner products and norms). An operator $A$ is self-adjoint when $A = A^{*}$ as operators with their domains, so that $\mathcal{D}(A) = \mathcal{D}(A^{*})$ ; symmetry alone is not enough (Symmetric versus self-adjoint operators).

Multiplication operator. Let $(M,\mu)$ be a measure space and let $f : M \to \mathbb{C}$ be measurable and finite $\mu$ -almost everywhere. The multiplication operator $T_f$ on $L^2(M,d\mu)$ is

(T_f \varphi)(m) \;=\; f(m)\,\varphi(m) , \qquad \mathcal{D}(T_f) \;=\; \bigl\{ \varphi \in L^2(M,d\mu) \;:\; f\varphi \in L^2(M,d\mu) \bigr\} .

The domain is forced on us: if $f$ is unbounded, $f\varphi$ leaves $L^2$ for some $\varphi$ , and there is nothing to be done about it. Complex-valued $f$ is allowed here because the proof below passes through $T_g$ with $g$ bounded and complex (Step 10) and through $T_{h\circ f}$ with $h$ a bounded complex Borel function (Step 13). On a finite measure space $T_f$ is self-adjoint exactly when $f$ is real almost everywhere - one direction is Step 8, the other the third part of Step 11 - and part A of the theorem asserts a real $f$ .

Essential range. A number $\lambda \in \mathbb{C}$ lies in the essential range of $f$ if

\mu\bigl( \{ m : \lvert f(m) - \lambda\rvert < \varepsilon \} \bigr) \;>\; 0 \qquad \text{for every } \varepsilon > 0 .

This is the set of values $f$ takes "on a set of positive measure, up to an arbitrarily small error"; changing $f$ on a null set does not change it. Reed and Simon (Reed & Simon (1980), §VII.2, p. 229) define the essential range only for a real-valued $f$ and a real $\lambda$ , with the condition written as $\mu(\{m : \lambda - \varepsilon < f(m) < \lambda + \varepsilon\}) > 0$ ; for real $f$ and $\lambda$ that is the same condition, and the form with a modulus, which makes sense for complex $f$ and complex $\lambda$ , is ours. The widening is used in one place only: it is what lets part B compare $\sigma(A)$ , a subset of $\mathbb{C}$ a priori, with the essential range and so conclude $\sigma(A) \subseteq \mathbb{R}$ . The $f$ it is applied to is the real one of part A.

Spectral representation. A spectral representation of a self-adjoint operator $A$ is a triple $\bigl((M,\mu),\,U,\,f\bigr)$ with $U : \mathcal{H} \to L^2(M,d\mu)$ unitary and $U A U^{-1} = T_f$ , domains included.

Cyclic vector. A vector $\psi \in \mathcal{H}$ is cyclic for a bounded self-adjoint $A$ if the finite linear combinations of $\psi, A\psi, A^2\psi, \dots$ are dense in $\mathcal{H}$ . For unbounded $A$ , Reed and Simon (Reed & Simon (1980), §VIII.3, p. 262) use instead the condition that $\{ g(A)\psi : g \in C_\infty(\mathbb{R})\}$ be dense, $C_\infty$ being the continuous functions vanishing at infinity.

Projection-valued measure. A family $\{P_\Omega\}$ of operators indexed by the Borel sets $\Omega \subseteq \mathbb{R}$ is a projection-valued measure (p.v.m.) if

every $P_\Omega$ is an orthogonal projection;
$P_{\varnothing} = 0$ and $P_{\mathbb{R}} = I$ ;
if $\Omega = \bigcup_{n\ge 1}\Omega_n$ with the $\Omega_n$ pairwise disjoint, then $P_\Omega = \lim_{N\to\infty}\sum_{n=1}^{N} P_{\Omega_n}$ in the strong sense, that is, applied to each fixed vector;
$P_{\Omega_1}P_{\Omega_2} = P_{\Omega_1 \cap \Omega_2}$ .

Spectral measure of a vector. Given a p.v.m. and $\psi \in \mathcal{H}$ , put

\mu_\psi(\Omega) \;=\; (P_\Omega \psi, \psi) \;=\; \lVert P_\Omega \psi\rVert^2 ,

the second equality because $P_\Omega$ is an orthogonal projection, so $P_\Omega = P_\Omega^{*} = P_\Omega^2$ . By axioms 1-3, $\mu_\psi$ is a finite positive Borel measure on $\mathbb{R}$ of total mass $\lVert\psi\rVert^2$ ; it is the spectral measure of $\psi$ . Because $\mu_\psi(\Omega)$ is a squared norm it is real, so it does not matter which slot of the inner product is taken linear - a point worth recording, since the sources differ (see the note below).

Theorem(The spectral theorem)AdaptedAdapted from Reed & Simon (1980), §VIII.3, Proposition 1, p. 259, Thm VIII.4, pp. 260-261, Thm VIII.5, p. 262, Thm VIII.6 and Proposition 3, pp. 263 and 264; Reed & Simon (1980), §VII.2, Thm VII.3 and the essential-range proposition, pp. 227-229

Let $A$ be a self-adjoint operator on a complex separable Hilbert space $\mathcal{H}$ , with domain $\mathcal{D}(A)$ .

A. Multiplication-operator form. There are a measure space $(M,\mu)$ with $\mu$ a finite measure, a unitary $U : \mathcal{H} \to L^2(M,d\mu)$ , and a real-valued measurable $f$ on $M$ , finite $\mu$ -a.e., such that

$\psi \in \mathcal{D}(A)$ if and only if $f\cdot(U\psi) \in L^2(M,d\mu)$ , and
$\bigl(UAU^{-1}\varphi\bigr)(m) = f(m)\varphi(m)$ for every $\varphi \in U[\mathcal{D}(A)]$ .

Conversely every $T_f$ of the definition above, on a finite measure space and with $f$ real and a.e. finite, is self-adjoint; so "self-adjoint operator" and "multiplication by a real function" name the same class of objects up to unitary equivalence. The measure space and $f$ may in addition be chosen so that $f \in L^p(M,d\mu)$ for every $p$ with $1 \le p < \infty$ . That last clause is no part of Theorem VIII.4, which says nothing about the integrability of $f$ ; it is the separate Proposition 3 of Reed and Simon (Reed & Simon (1980), §VIII.3, Proposition 3, p. 264), and we record it with the rest because their Proposition 2 (p. 260), which identifies the cores of $T_f$ , assumes it. Step 11 gives their proof.

B. The spectrum. $\sigma(A)$ is the essential range of $f$ . In particular $\sigma(A) \subseteq \mathbb{R}$ , as it must be for a self-adjoint operator (Spectrum and resolvent).

C. Borel functional calculus. There is exactly one map $h \mapsto h(A)$ from the bounded Borel functions on $\mathbb{R}$ into the bounded operators on $\mathcal{H}$ with these properties:

it is an algebraic $*$ -homomorphism: $(h_1h_2)(A) = h_1(A)h_2(A)$ , $(\alpha h)(A) = \alpha\,h(A)$ , $1(A) = I$ and $\bar h(A) = h(A)^{*}$ ;
it is norm-continuous, $\lVert h(A)\rVert \le \lVert h\rVert_\infty$ ;
if $h_n$ are bounded Borel functions with $h_n(x) \to x$ for every $x$ and $\lvert h_n(x)\rvert \le \lvert x\rvert$ for all $x$ and $n$ , then $h_n(A)\psi \to A\psi$ for every $\psi \in \mathcal{D}(A)$ ;
if $h_n \to h$ pointwise and $\sup_n \lVert h_n\rVert_\infty < \infty$ , then $h_n(A) \to h(A)$ strongly.

It automatically satisfies two more: if $A\psi = \lambda\psi$ then $h(A)\psi = h(\lambda)\psi$ ; and $h \ge 0$ implies $h(A) \ge 0$ . Concretely $h(A) = U^{-1} T_{h \circ f} U$ .

D. Projection-valued-measure form. Setting $P_\Omega = \chi_\Omega(A)$ for Borel $\Omega \subseteq \mathbb{R}$ gives a projection-valued measure, and

A \;=\; \int_{\mathbb{R}} \lambda \, dP_\lambda ,

meaning that $\mathcal{D}(A) = \bigl\{\psi : \int \lambda^2 \, d\mu_\psi(\lambda) < \infty\bigr\}$ and $(A\psi,\varphi) = \int \lambda \, d(P_\lambda\psi,\varphi)$ for $\psi \in \mathcal{D}(A)$ and $\varphi \in \mathcal{H}$ . Reed and Simon (Reed & Simon (1980), Thm VIII.6, p. 263) state this for $\varphi \in \mathcal{D}(A)$ , which is what polarisation gives directly, since polarising $(A\cdot,\cdot)$ needs $\psi + i^{k}\varphi \in \mathcal{D}(A)$ . The extension to every $\varphi \in \mathcal{H}$ is ours; it rests on a Cauchy-Schwarz estimate made inside the multiplication representation, which produces a positive finite measure dominating the complex measure $\Omega \mapsto (P_\Omega\psi,\varphi)$ and integrating $\lvert\lambda\rvert$ to at most $\lVert A\psi\rVert\,\lVert\varphi\rVert$ , so that the integral converges absolutely for every $\varphi\in\mathcal{H}$ ; it is proved in Step 14. This correspondence is one-to-one: every projection-valued measure on $\mathbb{R}$ is the one belonging to exactly one self-adjoint operator. More generally, for a real Borel $g$ (not necessarily bounded), $g(A) = \int g(\lambda)\,dP_\lambda$ is self-adjoint on $\mathcal{D}_g = \bigl\{\psi : \int \lvert g(\lambda)\rvert^2 d\mu_\psi(\lambda) < \infty \bigr\}$ . The construction below builds the calculus for bounded Borel functions and then treats the single unbounded function $g(\lambda)=\lambda$ in Step 14; the unbounded calculus in this generality is quoted from Reed and Simon (Reed & Simon (1980), Thm VIII.5 and the discussion of unbounded functions of a self-adjoint operator, pp. 262-263) rather than proved here, and it is listed with the other quoted results in Step 10. For bounded $h$ it agrees with part C:

\bigl( h(A)\psi, \psi \bigr) \;=\; \int_{\mathbb{R}} h(\lambda)\, d\mu_\psi(\lambda) .

E. Direct-sum form. (Proved below for bounded $A$ ; for unbounded $A$ it is quoted from Reed and Simon (Reed & Simon (1980), §VIII.3, the cyclic-vector paragraph following Thm VIII.5, p. 262, which realises the measure space of Thm VIII.4, pp. 260-261, as a union of copies of $\mathbb{R}$), who assert it there without proof.) One may take $(M,\mu)$ to be a disjoint union of at most countably many copies of $\mathbb{R}$ , so that $U$ becomes a unitary from $\mathcal{H}$ onto $\bigoplus_{n=1}^{N} L^2(\mathbb{R}, d\mu_n)$ , with $N \in \{1,2,\dots\} \cup \{\infty\}$ , carrying $A$ to multiplication by the variable $\lambda$ in each summand. The measures $\mu_n$ are the spectral measures of a family of cyclic vectors, and $N = 1$ is possible exactly when $A$ has a cyclic vector. One direction is Steps 5-7; the other is read off any representation with $N = 1$ , since in $L^2(\mathbb{R},d\mu_1)$ with $A$ multiplication by $\lambda$ the constant function $1$ is cyclic - it lies in $L^2$ because $\mu_1$ is finite, and $\{g\cdot 1 : g\in C_\infty(\mathbb{R})\}$ contains $C_c(\mathbb{R})$ , which is dense in $L^2(\mathbb{R},d\mu_1)$ ; when $\mu_1$ has compact support the polynomials in $\lambda$ serve as well. "Cyclic" is meant in the sense of the definition above that fits the case at hand: for bounded $A$ the polynomial sense, which is what Steps 5-7 use, and for unbounded $A$ the $C_\infty$ sense that Reed and Simon use on p. 262, the polynomial sense being unavailable there because $A^{n}\psi$ need not be defined for every $\psi$ once $n \ge 1$ , so the span of $\{A^{n}\psi\}$ is not available for an arbitrary $\psi$ . It is not that $C^{\infty}$ vectors are scarce — $\bigcap_{n}\mathcal{D}(A^{n})$ is in fact dense for every self-adjoint $A$ : in the representation of part A the vector $\psi_k = P_{[-k,k]}\psi$ has $U\psi_k = \chi_{[-k,k]}(f)\,U\psi$ , so $\lvert f^{m}\,U\psi_k\rvert \le k^{m}\lvert U\psi\rvert \in L^2$ for every $m$ , and conditions 1 and 2 of part A, applied $n$ times over, then put $\psi_k$ in $\mathcal{D}(A^{n})$ for every $n$ , while $\psi_k \to \psi$ by axiom 3 — but that the cyclicity has to be asked of a class of vectors on which the whole calculus acts, which is why the definition uses $\{g(A)\psi : g \in C_\infty(\mathbb{R})\}$ instead.

Note(The spectral-family form, and the two sources)AdaptedAdapted from Kreyszig (1978), §9.7, Definition 9.7-1 with (7)-(9), pp. 494-495; Kreyszig (1978), §9.9, Thm 9.9-1 with (1) and (1*), pp. 505-506; Kreyszig (1978), §10.6, the Cayley transform (1) and Thm 10.6-3 with (8), pp. 556-561

There is a second standard way to package part D, used by Kreyszig (Kreyszig (1978), §9.7 and §10.6, pp. 494-495, 558-561) and by much of the older literature. A spectral family (or decomposition of unity) is a one-parameter family $(E_\lambda)_{\lambda\in\mathbb{R}}$ of projections with

E_\lambda E_\nu = E_{\min(\lambda,\nu)} , \qquad E_\lambda \to 0 \ \ (\lambda\to-\infty) , \qquad E_\lambda \to I \ \ (\lambda\to+\infty) , \qquad E_{\nu} \to E_\lambda \ \ (\nu \to \lambda{+}0) ,

the last three limits taken strongly. A spectral family and a projection-valued measure are the same data: $E_\lambda = P_{(-\infty,\lambda]}$ and, conversely, $P_{(a,b]} = E_b - E_a$ . Kreyszig's Theorem 10.6-3 then states, for a densely defined self-adjoint $T$ ,

(Tx,x) \;=\; \int_{-\infty}^{\infty} \lambda \, dv(\lambda) , \qquad v(\lambda) = (F_\lambda x, x) , \qquad x \in \mathcal{D}(T),

an ordinary Riemann-Stieltjes integral, where $F_\lambda = E_{2\arctan\lambda}$ and $(E_\theta)$ is the spectral family in the spectral representation of $-U$ , with $U = (T-iI)(T+iI)^{-1}$ the Cayley transform of $T$ . The minus sign is Kreyszig's and is load-bearing. His Lemma 10.6-2 (p. 557) shows that $1$ is not an eigenvalue of $U$ , hence that $-1$ is not an eigenvalue of $-U$ ; that is what lets him prove $(E_\theta)$ continuous at $\theta = \pm\pi$ , and the continuity at the endpoints is what makes the substitution $\theta = 2\arctan\lambda$ , which sweeps $(-\pi,\pi)$ as $\lambda$ sweeps $\mathbb{R}$ , lose nothing.

Two differences between the sources should be recorded, because they look like disagreements and are not.

Which slot is linear. Reed and Simon (Reed & Simon (1980), §II.1, p. 37) take the inner product conjugate-linear in the first slot and write the functional calculus as $(\psi, h(A)\psi) = \int h \, d\mu_\psi$ . Kreyszig, and this handbook, take it linear in the first slot, giving $(h(A)\psi,\psi)$ . These two expressions denote the same number, because the conjugation sits in the first slot for them and in the second for us. What is not true, and would be an easy slip, is that the two orders agree inside one convention: in ours $(\psi, h(A)\psi) = \overline{(h(A)\psi,\psi)}$ , and for complex-valued $h$ those differ. So the order has to be written as each book writes it; we write $(h(A)\psi,\psi) = \int h\,d\mu_\psi$ throughout. For real $h$ both numbers are real and equal, which is why $\mu_\psi$ itself is free of the convention.
Which route to the proof. Reed and Simon reduce the unbounded case to the bounded one through the resolvent $(A+i)^{-1}$ ; Kreyszig reduces it through the Cayley transform $(A-i)(A+i)^{-1}$ , which is unitary rather than merely bounded and normal. The two are visibly relatives of one another, and they deliver the same object. We follow Reed and Simon, because their route produces the multiplication-operator form directly, and the multiplication-operator form is the one that makes the functional calculus a triviality.

Idea of the proofOriginal

First make sense of $h(A)$ for polynomials, then for everything else. For a bounded self-adjoint $A$ there is one identity that does all the work: for a polynomial $P$ ,

\lVert P(A)\rVert \;=\; \sup_{\lambda \in \sigma(A)} \lvert P(\lambda)\rvert .

The left-hand side is an operator norm, the right-hand side a supremum of numbers; the identity says that the map $P \mapsto P(A)$ is an isometry for the supremum norm on $\sigma(A)$ . Polynomials are dense in $C(\sigma(A))$ by Stone-Weierstrass, so the map extends to all continuous functions by continuity, and we have a continuous functional calculus.

Numbers out of operators: the spectral measures. For fixed $\psi$ , the assignment $h \mapsto (h(A)\psi,\psi)$ is a positive linear functional on $C(\sigma(A))$ . The Riesz-Markov theorem turns a positive linear functional on continuous functions into integration against a measure, and that measure $\mu_\psi$ is the whole content of $A$ as seen from $\psi$ .

From measures back to a multiplication operator. If a single $\psi$ is cyclic, the map $h(A)\psi \mapsto h$ sends $\mathcal{H}$ onto $L^2(\sigma(A), d\mu_\psi)$ and is isometric by the previous paragraph; under it $A$ becomes multiplication by $\lambda$ . A general $A$ splits $\mathcal{H}$ into countably many subspaces on each of which some vector is cyclic, and the pieces are reassembled as a direct sum.

The unbounded case costs one trick. If $A$ is self-adjoint and unbounded, $A \pm i$ are bijections of $\mathcal{D}(A)$ onto $\mathcal{H}$ with bounded inverses, and $R = (A+i)^{-1}$ is a bounded normal operator with trivial kernel. Diagonalise $R$ - which takes the bounded spectral theorem one step past the self-adjoint case, to bounded normal operators, a step we quote from Reed and Simon rather than prove: $R$ becomes multiplication by some bounded $g \ne 0$ a.e. Then $A$ must be multiplication by $f = g^{-1} - i$ , which is finite a.e. but need not be bounded - exactly as it should be. The domain condition $f\cdot(U\psi) \in L^2$ is nothing but a restatement of $\psi \in \mathrm{Ran}\,R$ .

Proof(Part A for bounded operators)AdaptedAdapted from Reed & Simon (1980), §VII.1, Thm VII.1 with Lemmas 1 and 2, pp. 222-224; Reed & Simon (1980), §VII.2, Lemma 1 with its proof, Lemma 2 and Thm VII.3 with its corollary, pp. 226-228; Reed & Simon (1980), §VI.3, Thm VI.6, p. 192; Folland (1999), Thm 4.51, p. 141, Thm 7.2, p. 212, and Thm 7.8 with Prop. 7.9, p. 217

Let $A$ be bounded and self-adjoint, so $\sigma(A)$ is a compact subset of $\mathbb{R}$ (Spectrum and resolvent).

Step 1: the spectrum of a polynomial in $A$ . For any bounded operator $B$ and polynomial $P$ one has $\sigma(P(B)) = \{P(\lambda) : \lambda \in \sigma(B)\}$ . For " $\supseteq$ ": $P(x) - P(\lambda)$ vanishes at $x=\lambda$ , so $P(x)-P(\lambda) = (x-\lambda)Q(x)$ for a polynomial $Q$ , whence $P(B)-P(\lambda) = (B-\lambda)Q(B)$ with the two factors commuting; if this product had an inverse, $B-\lambda$ would have one too, so $\lambda \in \sigma(B)$ forces $P(\lambda)\in\sigma(P(B))$ . For " $\subseteq$ ": a constant $P \equiv c$ gives $\sigma(cI) = \{c\}$ and $\{P(\lambda)\} = \{c\}$ , since $\sigma(B) \ne \varnothing$ - this is the one place the standing hypothesis $\mathcal{H}\ne\{0\}$ is used, the spectrum of a bounded operator being non-empty only on a non-zero space (Spectrum and resolvent) - so assume $P$ non-constant and factor $P(x) - \nu = c\prod_{j}(x-\lambda_j)$ with $c \ne 0$ ; if no $\lambda_j$ were in $\sigma(B)$ the product of the inverses would invert $P(B)-\nu$ , so some $\lambda_j \in \sigma(B)$ and $\nu = P(\lambda_j)$ .

Step 2: the isometry. Let $P$ be a polynomial. Then $P(A)^{*}P(A) = (\bar P P)(A)$ , where $\bar P$ is the polynomial with conjugated coefficients, and $\bar P P$ has real coefficients - the coefficient of $x^{k}$ in it is $\sum_{i+j=k}\bar a_i a_j$ , which is its own conjugate - so $(\bar PP)(A)^{*} = \bigl(\overline{\bar PP}\bigr)(A) = (\bar PP)(A)$ and the operator $(\bar PP)(A)$ is self-adjoint. (That $\bar PP$ takes real values on $\sigma(A)\subseteq\mathbb{R}$ would not do at this stage: only the polynomial calculus exists so far, and $\sigma(A)$ may be a finite set, so the values on it do not determine the operator.) Two named facts now do the work: the $C^{*}$ identity $\lVert T\rVert^2 = \lVert T^{*}T\rVert$ , valid for every bounded operator on a Hilbert space (Adjoints, self-adjoint and unitary operators), and the equality of norm and spectral radius for a self-adjoint operator (Reed and Simon (Reed & Simon (1980), Thm VI.6, p. 192)). They give, in that order,

\lVert P(A)\rVert^2 = \bigl\lVert P(A)^{*}P(A)\bigr\rVert = \bigl\lVert (\bar P P)(A)\bigr\rVert = \sup\bigl\{ \lvert\nu\rvert : \nu \in \sigma\bigl((\bar PP)(A)\bigr)\bigr\} = \sup_{\lambda\in\sigma(A)} \lvert P(\lambda)\rvert^2 ,

the last equality by Step 1. This is the identity the idea promised, and it is where self-adjointness of $A$ enters: for a general bounded operator the norm can exceed the spectral radius, as it does for the nilpotent $2\times 2$ matrix with a single $1$ above the diagonal, whose spectral radius is $0$ and whose norm is $1$ . (Self-adjointness is sufficient, not necessary: a unitary has norm and spectral radius both $1$ .)

Step 3: continuous functions. The polynomials with complex coefficients form a subalgebra $\mathcal{A}$ of $C(\sigma(A))$ . It contains the constants; it separates the points of $\sigma(A)$ , already through $P(x)=x$ ; and it contains $\bar P$ along with $P$ , the conjugate of a polynomial restricted to $\sigma(A)\subseteq\mathbb{R}$ being the polynomial with conjugated coefficients. The uniform closure of $\mathcal{A}$ inherits all three. So the complex Stone-Weierstrass theorem (Folland (Folland (1999), Thm 4.51, p. 141)), which we use here and do not reprove, leaves it two possibilities: all of $C(\sigma(A))$ , or the functions vanishing at one point of $\sigma(A)$ . The constants exclude the second. So the polynomials are dense. (The complex form is quoted because it applies directly to the complex algebra $\mathcal{A}$ ; the real form, Folland's Thm 4.45, p. 139, would also serve, at the cost of running it on real polynomials in $C(\sigma(A),\mathbb{R})$ and then splitting a complex $h$ into real and imaginary parts. His remark on p. 141 that the conclusion fails for complex algebras not closed under conjugation does not bite here, since $\mathcal{A}$ was just shown to be conjugation-closed.) By Step 2 the map $P \mapsto P(A)$ is isometric for $\lVert\cdot\rVert_\infty$ on $\sigma(A)$ ; a densely defined isometry into a complete space extends uniquely to the closure. This gives a map $h \mapsto h(A)$ on $C(\sigma(A))$ , isometric and multiplicative, with $x \mapsto x$ sent to $A$ .

Step 4: the spectral measures. Fix $\psi \in \mathcal{H}$ . The functional $h \mapsto (h(A)\psi,\psi)$ is linear on $C(\sigma(A))$ and positive: if $h \ge 0$ then $h = g^2$ with $g$ real and continuous, so $(h(A)\psi,\psi) = (g(A)^{*}g(A)\psi,\psi) = \lVert g(A)\psi\rVert^2 \ge 0$ , using that $g(A)$ is self-adjoint because $g$ is real. By the Riesz-Markov theorem - the Riesz representation theorem for positive linear functionals on continuous functions, Folland (Folland (1999), Thm 7.2, p. 212) - there is a unique positive Radon measure $\mu_\psi$ on the compact set $\sigma(A)$ with

\bigl( h(A)\psi,\psi \bigr) \;=\; \int_{\sigma(A)} h\, d\mu_\psi \qquad \text{for all } h \in C(\sigma(A)) .

It is finite, of total mass $\mu_\psi(\sigma(A)) = (1(A)\psi,\psi) = \lVert\psi\rVert^2$ . Two words about what "unique" covers. Folland's theorem is stated for $C_c(X)$ , which is $C(\sigma(A))$ here because $\sigma(A)$ is compact; and it gives uniqueness only among Radon measures. That is uniqueness among all finite Borel measures in this case, because $\sigma(A)$ is a compact metric space, hence second countable and locally compact Hausdorff, and on such a space every Borel measure finite on compact sets is Radon (Folland (Folland (1999), Thm 7.8, p. 217)).

Step 5: the cyclic case. Suppose some $\psi$ is cyclic. Define $V$ on the dense set $\{h(A)\psi : h \in C(\sigma(A))\}$ by $V\bigl(h(A)\psi\bigr) = h$ . It is isometric, because by Step 4

\bigl\lVert h(A)\psi \bigr\rVert^2 = \bigl( (\bar h h)(A)\psi,\psi\bigr) = \int \lvert h\rvert^2 d\mu_\psi = \lVert h\rVert^2_{L^2(d\mu_\psi)} ,

and the same computation shows $V$ is well defined. What well-definedness asks is that one vector of the dense set receive one function, that is, that $h(A)\psi = h'(A)\psi$ force $h = h'$ in $L^2(d\mu_\psi)$ ; and applying the display to $h - h'$ turns the first equality into $\lVert h-h'\rVert_{L^2(d\mu_\psi)} = 0$ , which is the second. (Read the other way, the same display gives the converse implication, $h = h'$ in $L^2(d\mu_\psi) \Rightarrow h(A)\psi = h'(A)\psi$ ; that is the injectivity of $V$ , not its well-definedness.) An isometry defined on a dense subspace extends to an isometry of $\mathcal{H}$ , and its range contains $C(\sigma(A))$ , which is dense in $L^2(\sigma(A), d\mu_\psi)$ because $\mu_\psi$ is a Radon measure on the compact space $\sigma(A)$ (Folland (Folland (1999), Prop. 7.9, p. 217), quoted and not reproved); so the extension is unitary onto $L^2(\sigma(A),d\mu_\psi)$ . Finally, for continuous $h$ ,

\bigl(VAV^{-1}h\bigr)(\lambda) = \bigl(V A h(A)\psi\bigr)(\lambda) = \bigl( V (x h)(A)\psi \bigr)(\lambda) = \lambda\, h(\lambda) ,

and by continuity this extends from $C(\sigma(A))$ to all of $L^2$ . So $A$ is multiplication by $\lambda$ .

Step 6: removing the cyclicity assumption. Separability gives a decomposition $\mathcal{H} = \bigoplus_{n=1}^{N}\mathcal{H}_n$ , $N \le \infty$ , into closed subspaces each invariant under $A$ and each carrying a vector $\varphi_n$ cyclic for the restriction of $A$ to it: take a maximal family of mutually orthogonal cyclic subspaces, which exists by Zorn's lemma, and suppose the closed span $\mathcal{W}$ of the family were not all of $\mathcal{H}$ . Then $\mathcal{W}^{\perp} \ne \{0\}$ , and $\mathcal{W}^{\perp}$ is $A$ -invariant: $\mathcal{W}$ is $A$ -invariant, being spanned by $A$ -invariant subspaces, so for $\varphi\in\mathcal{W}^{\perp}$ and $w\in\mathcal{W}$ we get $(A\varphi,w) = (\varphi,Aw) = 0$ , using that $A$ is bounded and self-adjoint. Any unit vector of $\mathcal{W}^{\perp}$ therefore generates, inside $\mathcal{W}^{\perp}$ , one more cyclic subspace orthogonal to every member of the family, contradicting maximality. That the family is at most countable - which is what the index $N \le \infty$ above asserts, and what the summability in Step 7 and the countable direct sum of part E need - is where separability is used: the members are non-zero and mutually orthogonal, so picking a unit vector in each produces an orthonormal set with one element per member, and in a separable Hilbert space every orthonormal set is at most countable (Orthonormal bases in Hilbert space). Label the members $\mathcal{H}_1,\mathcal{H}_2,\dots$ accordingly. Applying Step 5 in each summand and writing $\mu_n = \mu_{\varphi_n}$ gives a unitary $U : \mathcal{H} \to \bigoplus_{n} L^2(\mathbb{R},d\mu_n)$ carrying $A$ to multiplication by $\lambda$ in each slot. This is part E for bounded $A$ ; for unbounded $A$ part E is quoted rather than proved, as its statement says, and Step 10 lists it among the imports.

Step 7: one finite measure space. A direct sum of $L^2$ spaces is itself an $L^2$ space, over the disjoint union of the underlying measure spaces. So label $N$ disjoint copies of $\mathbb{R}$ by $n$ , give the $n$ th the measure $\mu_n$ , and call the resulting space $(M,\mu)$ : the unitary of Step 6 becomes a unitary onto $L^2(M,d\mu)$ , and the operator that acted as multiplication by $\lambda$ in every slot becomes $T_f$ for the single function $f$ whose value at the point $\lambda$ of the $n$ th copy is $\lambda$ itself. Only the finiteness of $\mu$ still has to be arranged, and the cyclic vectors are free to be rescaled: a non-zero multiple $c\varphi_n$ generates the same subspace and has $\mu_{c\varphi_n} = \lvert c\rvert^2\mu_{\varphi_n}$ , since the functional of Step 4 scales by $\lvert c\rvert^2$ . Normalise each $\varphi_n$ to length $2^{-n}$ . Then $\mu_n(\mathbb{R}) = \lVert\varphi_n\rVert^2 = 4^{-n}$ , whence $\mu(M) = \sum_{n} 4^{-n} \le \tfrac13$ , a finite total mass; any summable choice of lengths would do as well. This is part A for bounded $A$ . $\square$

Proof(Part A in general, and parts B, C, D)AdaptedAdapted from Reed & Simon (1980), §VIII.3, Proposition 1 with its proof and Thm VIII.4 with its proof, pp. 259-261, Thm VIII.5 and the construction of the p.v.m., pp. 262-263, Thm VIII.6, p. 263, and Proposition 3, p. 264; Reed & Simon (1980), Ch. VII, Problems 3-5, p. 246

Step 8: multiplication operators are self-adjoint. Let $\mu$ be finite and $f$ real and a.e. finite. $T_f$ is densely defined, since the sets $\{\lvert f\rvert \le n\}$ exhaust $M$ up to a null set and every $\varphi\in L^2$ is the $L^2$ -limit of its restrictions to them; and $T_f$ is symmetric because $f$ is real. Let $\chi_n$ be the indicator of $\{\lvert f\rvert \le n\}$ and let $\psi \in \mathcal{D}(T_f^{*})$ . For every $\varphi$ with $\lVert\varphi\rVert \le 1$ we have $\chi_n\varphi \in \mathcal{D}(T_f)$ , so

\bigl\lvert (\chi_n f \psi, \varphi) \bigr\rvert = \bigl\lvert (\psi, T_f \chi_n\varphi) \bigr\rvert = \bigl\lvert (T_f^{*}\psi, \chi_n \varphi)\bigr\rvert \le \lVert T_f^{*}\psi\rVert .

Taking the supremum over such $\varphi$ gives $\lVert \chi_n f\psi\rVert \le \lVert T_f^{*}\psi\rVert$ for every $n$ , and $\lvert \chi_n f \psi\rvert$ increases pointwise a.e. to $\lvert f\psi\rvert$ ; by the monotone convergence theorem (Monotone convergence and Fatou's lemma) $f\psi \in L^2$ , that is $\psi \in \mathcal{D}(T_f)$ . Hence $\mathcal{D}(T_f^{*}) \subseteq \mathcal{D}(T_f)$ , which with symmetry gives $T_f = T_f^{*}$ .

Step 9: the resolvent at $\pm i$ . Let $A$ be self-adjoint. By the basic criterion for self-adjointness (Symmetric versus self-adjoint operators), $A \pm i$ map $\mathcal{D}(A)$ onto $\mathcal{H}$ , and they are injective because $\lVert (A\pm i)\psi\rVert^2 = \lVert A\psi\rVert^2 + \lVert\psi\rVert^2 \ge \lVert\psi\rVert^2$ for $\psi \in \mathcal{D}(A)$ , the cross terms cancelling since $A$ is symmetric. So $R = (A+i)^{-1}$ is defined on all of $\mathcal{H}$ ; it is closed because $A+i$ is, and therefore bounded by the closed graph theorem (Uniform boundedness and open mapping theorems). The identity

\bigl( (A-i)^{-1}\psi, (A+i)\varphi \bigr) = \bigl( (A-i)(A-i)^{-1}\psi, \varphi \bigr) = (\psi,\varphi) \qquad \bigl(\psi\in\mathcal{H},\ \varphi\in\mathcal{D}(A)\bigr)

identifies $R^{*} = (A-i)^{-1}$ , since $\mathrm{Ran}(A+i) = \mathcal{H}$ . The first equality moves $A+i$ out of the second slot as its adjoint $(A+i)^{*} = A-i$ , which is legitimate because $(A-i)^{-1}\psi$ lies in $\mathcal{D}(A)$ ; the second is $(A-i)(A-i)^{-1} = I$ on $\mathcal{H}$ . (Reed and Simon (Reed & Simon (1980), p. 261) print the two factors in the opposite order at this point; the order above is the one the manipulation produces, and only this order is a composition of maps that can be applied.) The two resolvents commute. Reed and Simon (Reed & Simon (1980), §VIII.3, p. 261) take this from their Theorem VIII.2; it is also immediate from the algebra. On $\mathcal{D}(A^{2}) = \{\psi\in\mathcal{D}(A) : A\psi\in\mathcal{D}(A)\}$ the two products $(A-i)(A+i)$ and $(A+i)(A-i)$ are both $A^{2}+1$ , and each is a bijection of $\mathcal{D}(A^{2})$ onto $\mathcal{H}$ , the two factors being bijections between different pairs of spaces: $A+i$ carries $\mathcal{D}(A^{2})$ onto $\mathcal{D}(A)$ - if $\chi\in\mathcal{D}(A)$ and $\psi = (A+i)^{-1}\chi$ , then $A\psi = \chi - i\psi$ lies in $\mathcal{D}(A)$ , so $\psi\in\mathcal{D}(A^{2})$ - and $A-i$ carries $\mathcal{D}(A)$ onto $\mathcal{H}$ , and the same with the two factors exchanged; inverting the equality $(A-i)(A+i) = (A+i)(A-i)$ therefore gives

(A+i)^{-1}(A-i)^{-1} \;=\; (A^{2}+1)^{-1} \;=\; (A-i)^{-1}(A+i)^{-1}

on all of $\mathcal{H}$ . Hence $R^{*}R = RR^{*}$ : $R$ is a bounded normal operator. Finally $\ker R = \{0\}$ , since $R$ is a bijection onto $\mathcal{D}(A)$ .

Step 10: diagonalising $R$ . Steps 1-7 give the multiplication-operator form for a bounded self-adjoint operator, and $R$ is only normal. What is needed is one step further out, and we quote it rather than prove it:

The bounded normal case. Let $N$ be a bounded normal operator on a separable Hilbert space $\mathcal{H}$ . Then there are a finite measure space $(M,\mu)$ , a unitary $U : \mathcal{H}\to L^2(M,d\mu)$ and a bounded, in general complex-valued, Borel function $g$ on $M$ with $UNU^{-1} = T_g$ .

This is Problem 5(b) of Chapter VII of Reed and Simon (Reed & Simon (1980), Ch. VII, Problems 3-5, p. 246), who set it as an exercise rather than proving it in the text; it is not proved here either. It is the only step of the spectral-theorem construction itself that is taken from an exercise, but it is in no way the only thing the unit takes on quotation. The substantive imports - results quoted outright - are: the equality of norm and spectral radius for a self-adjoint operator in Step 2; the complex Stone-Weierstrass theorem in Steps 3 and 13a; the Riesz-Markov theorem and Folland's Thm 7.8 on Radon measures in Step 4, and his Prop. 7.9 on the density of $C(\sigma(A))$ in $L^2(\sigma(A),d\mu_\psi)$ in Step 5; the basic criterion for self-adjointness, which is a prerequisite unit, in Step 9; the bounded normal case here; the surjectivity half of Reed and Simon's Thm VIII.6 in Step 14; the direct-sum form of part E for unbounded $A$ , which Reed and Simon assert on p. 262 without proof; and the functional calculus for unbounded real Borel functions named at the end of part D - that $g(A) = \int g\,dP_\lambda$ is self-adjoint on $\mathcal{D}_g$ - which is Reed and Simon's Thm VIII.5 with the discussion of unbounded functions following it, pp. 262-263, and of which the construction below proves the single case $g(\lambda) = \lambda$ in Step 14 and nothing more. Their Problems 3-5 lay out the route they intend, which is not a two-line variant of Steps 1-7: write $N = B + iC$ with $B = \tfrac12(N+N^{*})$ and $C = \tfrac1{2i}(N-N^{*})$ , both self-adjoint and, because $N$ is normal, commuting; show that the spectral projections of $B$ commute with those of $C$ ; set $\Phi(\chi_{\Omega_1\times\Omega_2}) = P_{\Omega_1}(B)\,P_{\Omega_2}(C)$ on the rectangles of $\mathbb{R}^2$ and extend by linearity; pass to continuous functions on a product of intervals by the bounded-linear-transformation extension theorem, and then to Borel functions; and read the multiplication form off the two-variable calculus so obtained. Alongside these the proof draws without comment on the standard apparatus of the prerequisite units: Zorn's lemma in Step 6 (Zorn's lemma and the axiom of choice), the closed graph theorem in Step 9 (Uniform boundedness and open mapping theorems), the monotone and dominated convergence theorems in Steps 8, 13 and 14 (Monotone convergence and Fatou's lemma, Dominated convergence), the extension of a densely defined isometry in Steps 3 and 5, the operator norm and its submultiplicativity throughout (Bounded linear operators), the $C^{*}$ -identity $\lVert B^{*}B\rVert = \lVert B\rVert^2$ in Step 2 (Adjoints, self-adjoint and unitary operators), the non-emptiness of the spectrum of a bounded operator in Step 1 and the compactness of $\sigma(A)$ and its reality for a bounded self-adjoint $A$ , used in the opening line of the bounded proof and again in Steps 3 and 4 (Spectrum and resolvent), and the substitution rule for an image measure in Step 14, which is not quoted but proved on the spot there, in a line, from the definition of an image measure and monotone convergence.

Applied to $N = R$ , which is normal by Step 9, this produces a finite measure space $(M,\mu)$ , a unitary $U : \mathcal{H}\to L^2(M,d\mu)$ and a bounded measurable complex $g$ with $URU^{-1} = T_g$ .

Step 11: back to $A$ . Put $f := g^{-1} - i$ , which is defined and finite a.e. because $\ker R = \{0\}$ makes $\{g = 0\}$ a $\mu$ -null set. By Step 9, $\varphi \mapsto \psi = R\varphi$ is a bijection of $\mathcal{H}$ onto $\mathcal{D}(A)$ , and it carries two identities with it: $U\psi = g\,U\varphi$ , because $URU^{-1} = T_g$ , and $A\psi = \varphi - i\psi$ , because $(A+i)\psi = \varphi$ . Four things remain.

The action, condition 2. Apply $U$ to $A\psi = \varphi - i\psi$ and substitute $U\varphi = g^{-1}U\psi$ :

(UA\psi)(m) = (U\varphi)(m) - i (U\psi)(m) = \bigl( g(m)^{-1} - i \bigr)(U\psi)(m) = f(m)\,(U\psi)(m) .

The domain, condition 1. For $\psi \in \mathcal{D}(A)$ the display just obtained reads $f\,U\psi = U(A\psi)$ , which is in $L^2$ . Conversely let $\psi \in \mathcal{H}$ with $f\,U\psi \in L^2$ . Then $g^{-1}U\psi = (f+i)U\psi \in L^2$ , so it equals $U\varphi$ for exactly one $\varphi \in \mathcal{H}$ ; multiplying by $g$ gives $U(R\varphi) = g\,U\varphi = U\psi$ , so $\psi = R\varphi \in \mathcal{D}(A)$ .

$f$ is real. By the two conditions $T_f = UAU^{-1}$ is self-adjoint. Suppose $\mathrm{Im}\,f > 0$ on a set of positive measure. That set is the countable union of the sets $S_{k,n} = \{\mathrm{Im}\,f \ge 1/k\} \cap \{\lvert f\rvert \le n\}$ , so some $S = S_{k,n}$ has $\mu(S) > 0$ ; on it $f$ is bounded, so $\chi_S \in \mathcal{D}(T_f)$ , and

\mathrm{Im}\,(T_f \chi_S, \chi_S) = \int_S \mathrm{Im}\,f \, d\mu \;\ge\; \mu(S)/k \;>\; 0 ,

which no self-adjoint operator permits. Applying this to $-f$ rules out $\mathrm{Im}\,f < 0$ .

The integrability clause. This is Proposition 3 of Reed and Simon (Reed & Simon (1980), §VIII.3, Proposition 3, p. 264), stated by them separately from Theorem VIII.4, and what follows is their reweighting proof. Keep $f$ and replace $\mu$ by $d\tilde\mu = e^{-f^2}d\mu$ and $U$ by $VU$ , where $(V\varphi)(m) = e^{f(m)^2/2}\varphi(m)$ . Then $\int \lvert V\varphi\rvert^2 d\tilde\mu = \int\lvert\varphi\rvert^2 d\mu$ , so $V$ is unitary from $L^2(d\mu)$ onto $L^2(d\tilde\mu)$ ; it commutes with multiplication by $f$ , so conditions 1 and 2 survive verbatim, and $\tilde\mu(M) \le \mu(M) < \infty$ . Finally $c_p := \sup_{t\ge0} t^p e^{-t^2}$ is finite for every $p \in [1,\infty)$ , so $\int \lvert f\rvert^p d\tilde\mu = \int \lvert f\rvert^p e^{-f^2} d\mu \le c_p\,\mu(M) < \infty$ , that is, $f \in L^p(M,d\tilde\mu)$ for every such $p$ .

Step 12: the spectrum. By part A it suffices to compute $\sigma(T_f)$ . If $\lambda$ is not in the essential range of $f$ , then $\lvert f-\lambda\rvert \ge \varepsilon$ a.e. for some $\varepsilon>0$ , so $(f-\lambda)^{-1}$ is bounded by $1/\varepsilon$ and multiplication by it maps $L^2$ into $\mathcal{D}(T_f)$ , since $f\,(f-\lambda)^{-1} = 1 + \lambda(f-\lambda)^{-1}$ is bounded by $1+\lvert\lambda\rvert/\varepsilon$ ; it is therefore a two-sided inverse of $T_f - \lambda$ , and $\lambda \in \rho(T_f)$ . If $\lambda$ is in the essential range, put $S_n = \{\lvert f - \lambda\rvert < 1/n\}$ , which has positive measure, and $\varphi_n = \chi_{S_n}/\lVert\chi_{S_n}\rVert$ . Then $\lVert\varphi_n\rVert = 1$ and $\lVert (T_f-\lambda)\varphi_n\rVert \le 1/n$ , so $T_f-\lambda$ has no bounded inverse and $\lambda\in\sigma(T_f)$ . (This is Weyl's criterion in a concrete case.)

Step 13: the functional calculus. Define $h(A) = U^{-1}T_{h\circ f}U$ for bounded Borel $h$ . Properties 1 and 2 of part C are immediate from the corresponding statements for multiplication operators, $\lVert T_{h\circ f}\rVert \le \lVert h\rVert_\infty$ because $h\circ f$ is bounded by $\lVert h\rVert_\infty$ . For property 4, if $h_n \to h$ pointwise with $\sup_n\lVert h_n\rVert_\infty = C < \infty$ , then for fixed $\varphi \in L^2$ the functions $\lvert (h_n\circ f) \varphi - (h\circ f)\varphi\rvert^2$ tend to $0$ a.e. and are dominated by $4C^2\lvert\varphi\rvert^2 \in L^1$ , so the dominated convergence theorem (Dominated convergence) gives $T_{h_n\circ f}\varphi \to T_{h\circ f}\varphi$ . Property 3 is the same argument with the domination $\lvert h_n(f)\varphi\rvert \le \lvert f \varphi\rvert$ , which is in $L^2$ precisely when $\psi = U^{-1}\varphi \in \mathcal{D}(A)$ - property 3 is the only one of the four that touches the domain condition of part A, and the condition itself is called on again in Step 14, where it describes $\mathcal{D}(A)$ through the spectral measures.

The two further properties listed in part C, which are (e) and (f) of Reed and Simon (Reed & Simon (1980), §VIII.3, Thm VIII.5, p. 262), come off the same formula in a line each. Eigenvectors. Let $A\psi = \lambda\psi$ and put $\varphi = U\psi$ . By condition 2 of part A, $f\varphi = \lambda\varphi$ a.e., so $\varphi$ vanishes a.e. on $\{f \ne \lambda\}$ ; hence $(h\circ f)\varphi = h(\lambda)\varphi$ a.e., which is $h(A)\psi = h(\lambda)\psi$ . Positivity. If $h \ge 0$ then for every $\psi$

\bigl(h(A)\psi,\psi\bigr) = \int_M h\bigl(f(m)\bigr)\,\lvert (U\psi)(m)\rvert^2\,d\mu(m) \;\ge\; 0 ,

the integrand being a product of two non-negative functions; that is $h(A) \ge 0$ .

Step 13a: the calculus is unique. Polynomials are not available here: they are not bounded on $\mathbb{R}$ , and properties 1, 2 and 4 alone mention $A$ nowhere. Property 3 is the only link to $A$ , and the argument has to start from it. Let $\Phi$ and $\Phi'$ be two maps with properties 1-4.

They agree on one resolvent function. Let $r(x) = (x-i)^{-1}$ , which is bounded and continuous on $\mathbb{R}$ , and let $h_n(x) = x\chi_{[-n,n]}(x)$ , which are bounded Borel with $h_n(x)\to x$ for every $x$ and $\lvert h_n\rvert \le \lvert x\rvert$ - exactly the hypotheses of property 3. Fix $\psi \in \mathcal{D}(A)$ . Property 3 gives $\Phi(h_n)\psi \to A\psi$ , and $\Phi(r)$ is bounded, so

\Phi(r)(A-i)\psi \;=\; \lim_{n\to\infty} \Phi(r)\bigl(\Phi(h_n)\psi - i\psi\bigr) \;=\; \lim_{n\to\infty}\Phi(r\,h_n)\psi \;-\; i\,\Phi(r)\psi ,

the second equality by property 1. Now $\lvert r h_n\rvert \le \lvert x\rvert/\sqrt{x^2+1} \le 1$ and $r h_n \to x/(x-i)$ pointwise, so property 4 turns the limit into $\Phi\bigl(x/(x-i)\bigr)\psi$ . Since $x/(x-i) - i/(x-i) = 1$ and $\Phi(1) = I$ ,

\Phi(r)(A-i)\psi = \psi \qquad \text{for all } \psi \in \mathcal{D}(A) .

By Step 9, $A-i$ maps $\mathcal{D}(A)$ onto $\mathcal{H}$ , so $\Phi(r) = (A-i)^{-1}$ : a bounded operator fixed by $A$ alone. The same holds for $\Phi'$ , and by property 1, $\Phi(\bar r) = \Phi(r)^{*} = \Phi'(\bar r)$ .

They agree on continuous functions. Read $r$ as a continuous function on the one-point compactification $\mathbb{R}\cup\{\infty\}$ with $r(\infty) = 0$ . The unital $*$ -algebra generated by $r$ contains the constants, is closed under conjugation, and separates the points of $\mathbb{R}\cup\{\infty\}$ , because $x \mapsto (x-i)^{-1}$ is injective there. Its uniform closure inherits those properties and is closed, so the complex Stone-Weierstrass theorem (Folland (Folland (1999), Thm 4.51, p. 141)), applied exactly as in Step 3, makes that closure either all of $C(\mathbb{R}\cup\{\infty\})$ or the functions vanishing at one point - and the constants rule the second out. So the algebra is dense in $C(\mathbb{R}\cup\{\infty\})$ for the supremum norm. Property 1 makes $\Phi$ and $\Phi'$ agree on that algebra, and property 2 makes both contractions for $\lVert\cdot\rVert_\infty$ , so they agree on its uniform closure: on every continuous $h$ with a limit at infinity.

They agree everywhere. Let $\mathcal{C}$ be the set of bounded Borel $h : \mathbb{R}\to\mathbb{C}$ with $\Phi(h) = \Phi'(h)$ . Property 1 makes $\mathcal{C}$ an algebra containing the constants: if $h_1,h_2\in\mathcal{C}$ then $\Phi(h_1h_2) = \Phi(h_1)\Phi(h_2) = \Phi'(h_1)\Phi'(h_2) = \Phi'(h_1h_2)$ , and likewise for linear combinations. Property 4 closes $\mathcal{C}$ under pointwise limits of uniformly bounded sequences. The previous paragraph puts every continuous function with a limit at infinity into $\mathcal{C}$ . Those three facts suffice, in three moves, and no result outside this handbook is needed for them.

(i) Indicators of compact sets. Let $K\subseteq\mathbb{R}$ be compact and non-empty, and put $g_n(x) = \max\bigl(0,\,1 - n\,\mathrm{dist}(x,K)\bigr)$ . Each $g_n$ is continuous with values in $[0,1]$ and vanishes outside a bounded set, so it has limit $0$ at infinity and lies in $\mathcal{C}$ ; and $g_n(x) \to \chi_K(x)$ at every $x$ , because $\mathrm{dist}(x,K) = 0$ exactly on $K$ , $K$ being closed. The sequence is bounded by $1$ , so $\chi_K\in\mathcal{C}$ ; and $\chi_\varnothing = 0\in\mathcal{C}$ trivially, so $\chi_K\in\mathcal{C}$ for every compact $K$ .

(ii) Indicators of all Borel sets. Let $\mathcal{D} = \{E\subseteq\mathbb{R}\text{ Borel} : \chi_E\in\mathcal{C}\}$ . It contains $\mathbb{R}$ , since $\chi_{\mathbb{R}} = 1$ ; it is closed under complements, since $\chi_{E^{c}} = 1 - \chi_E$ and $\mathcal{C}$ is a vector space containing $1$ ; it is closed under finite intersections, since $\chi_{E\cap F} = \chi_E\chi_F$ and $\mathcal{C}$ is an algebra; and it is closed under countable increasing unions, since the indicators of $E_1\cup\cdots\cup E_N$ are bounded by $1$ and converge pointwise to the indicator of $\bigcup_n E_n$ . An algebra of sets closed under countable increasing unions is a $\sigma$ -algebra, because an arbitrary countable union is the increasing union of the finite ones. By (i) $\mathcal{D}$ contains every compact set, while each open $V\subseteq\mathbb{R}$ is the union of the countably many compact sets $\{x\in V : \lvert x\rvert\le n,\ \mathrm{dist}(x,V^{c})\ge 1/n\}$ ; so $\mathcal{D}$ contains the open sets and hence all of the Borel $\sigma$ -algebra they generate (Sigma-algebras and measures).

(iii) All bounded Borel functions. By (ii) and linearity $\mathcal{C}$ contains every simple Borel function. Any bounded Borel $h$ is a uniform limit of such: given $\varepsilon>0$ , cut the disc $\{\lvert z\rvert \le \lVert h\rVert_\infty\}$ into finitely many disjoint Borel pieces of diameter less than $\varepsilon$ , choose a point in each, and let the approximant take that value on the preimage of that piece. A uniform limit is in particular a uniformly bounded pointwise limit, so $h\in\mathcal{C}$ . Hence $\Phi = \Phi'$ .

Step 14: the projection-valued measure. Put $P_\Omega = \chi_\Omega(A)$ . Axiom 1 holds since $\chi_\Omega^2 = \bar\chi_\Omega = \chi_\Omega$ , so $P_\Omega$ is idempotent and self-adjoint; axiom 2 and axiom 4 are the identities $\chi_{\varnothing}=0$ , $\chi_{\mathbb{R}}=1$ , $\chi_{\Omega_1}\chi_{\Omega_2} = \chi_{\Omega_1\cap\Omega_2}$ transported through the homomorphism; axiom 3 is property 4 applied to the partial sums $\sum_{n\le N}\chi_{\Omega_n}$ , which are uniformly bounded by $1$ and converge pointwise to $\chi_\Omega$ . For $\psi\in\mathcal{H}$ the set function $\mu_\psi(\Omega) = (P_\Omega\psi,\psi)$ is therefore a finite positive Borel measure, and by construction $\mu_\psi$ is the image under $f$ of the measure $\lvert U\psi\rvert^2 d\mu$ ; hence for bounded Borel $h$ ,

\bigl(h(A)\psi,\psi\bigr) = \int_M h(f(m))\,\lvert (U\psi)(m)\rvert^2 d\mu(m) = \int_{\mathbb{R}} h(\lambda)\, d\mu_\psi(\lambda) ,

by the substitution rule for an image measure, which needs nothing from outside: if $\kappa$ is a finite positive measure on $M$ and $\rho(\Omega) = \kappa\bigl(f^{-1}(\Omega)\bigr)$ is its image under $f$ , then $\int_{\mathbb{R}} g\,d\rho = \int_M g\bigl(f(m)\bigr)\,d\kappa(m)$ for every Borel $g\ge0$ - for $g=\chi_\Omega$ this is the definition of $\rho$ , linear combinations give the simple $g$ , and monotone convergence along simple $g_k\uparrow g$ (Monotone convergence and Fatou's lemma) gives the rest - and hence for every Borel $g$ with $\int\lvert g\rvert\,d\rho<\infty$ , by splitting the real and imaginary parts of $g$ into their positive and negative parts. Here $\kappa = \lvert U\psi\rvert^2 d\mu$ and $g = h$ ; the rule is used again in the two-vector paragraph below, with $\kappa$ running over $\lvert U\psi\rvert\lvert U\varphi\rvert\,d\mu$ and the four non-negative pieces of $(U\psi)\overline{(U\varphi)}\,d\mu$ . The same rule with $g(\lambda) = \lambda^2$ gives $\int\lambda^2 d\mu_\psi = \int_M \lvert f\,U\psi\rvert^2 d\mu$ , finite exactly when $\psi\in\mathcal{D}(A)$ by condition 1 of part A and equal then to $\lVert U(A\psi)\rVert^2 = \lVert A\psi\rVert^2$ ; so $\mathcal{D}(A) = \{\psi : \int\lambda^2 d\mu_\psi<\infty\}$ . Take now $h_n(\lambda) = \lambda\chi_{[-n,n]}(\lambda)$ and a $\psi\in\mathcal{D}(A)$ . On the left $(h_n(A)\psi,\psi)\to(A\psi,\psi)$ , by property 3 of part C. On the right the limit may be taken inside the integral, but not for nothing: $\mu_\psi$ is finite and $\lambda\in L^2(\mu_\psi)$ , so Cauchy-Schwarz gives $\int\lvert\lambda\rvert\,d\mu_\psi \le \bigl(\int\lambda^2 d\mu_\psi\bigr)^{1/2}\mu_\psi(\mathbb{R})^{1/2} = \lVert A\psi\rVert\,\lVert\psi\rVert<\infty$ , and $\lvert h_n\rvert \le \lvert\lambda\rvert \in L^1(\mu_\psi)$ with $h_n\to\lambda$ pointwise, so dominated convergence (Dominated convergence) applies. Hence $(A\psi,\psi) = \int\lambda\,d\mu_\psi$ for every $\psi\in\mathcal{D}(A)$ .

The version with two vectors. For $\psi,\varphi\in\mathcal{H}$ write $F = U\psi$ and $G = U\varphi$ , and put $\nu_{\psi\varphi}(\Omega) = (P_\Omega\psi,\varphi)$ . Since $h(A) = U^{-1}T_{h\circ f}U$ and $U$ is unitary,

\bigl(h(A)\psi,\varphi\bigr) \;=\; \bigl( (h\circ f)\,F,\,G\bigr)_{L^2(M,d\mu)} \;=\; \int_M h\bigl(f(m)\bigr)\,F(m)\,\overline{G(m)}\,d\mu(m)

for every bounded Borel $h$ - the computation that opened this step, with $G$ in the second slot in place of $F$ . Read with $h = \chi_\Omega$ , it says that $\nu_{\psi\varphi}$ is the image under $f$ of $F\bar G\,d\mu$ , and $F\bar G \in L^1(M,d\mu)$ by Cauchy-Schwarz. Splitting $F\bar G$ into the positive and negative parts of its real and imaginary parts, and taking the image of each under $f$ , therefore writes

\nu_{\psi\varphi} \;=\; \rho_1 - \rho_2 + i(\rho_3 - \rho_4) , \qquad \rho_1,\dots,\rho_4 \ \text{finite positive Borel measures on } \mathbb{R} .

This is at once the countable additivity of $\nu_{\psi\varphi}$ and the meaning of $\int g\,d\nu_{\psi\varphi}$ : the same combination of four ordinary integrals, defined whenever $g$ is integrable against each $\rho_j$ - automatic for bounded $g$ , the $\rho_j$ being finite - and equal then, by the substitution rule, to $\int_M g(f)\,F\bar G\,d\mu$ . In particular

\bigl(h(A)\psi,\varphi\bigr) \;=\; \int_{\mathbb{R}} h\,d\nu_{\psi\varphi} \qquad (\psi,\varphi\in\mathcal{H},\ h \text{ bounded Borel}) ,

an identity polarisation would also give, both sides being bounded sesquilinear forms which agree on the diagonal by the first display of this step.

What polarisation does not give is the passage from bounded $h$ to $h(\lambda) = \lambda$ , and that passage is the whole of part D's claim for $\varphi\notin\mathcal{D}(A)$ . Let $\rho_{\psi\varphi}$ be the image under $f$ of the positive measure $\lvert F\rvert\,\lvert G\rvert\,d\mu$ , which is finite of total mass $\int_M\lvert F\rvert\lvert G\rvert\,d\mu \le \lVert\psi\rVert\lVert\varphi\rVert$ by Cauchy-Schwarz in $L^2(M,d\mu)$ . Each of the four densities above is at most $\lvert F\bar G\rvert = \lvert F\rvert\lvert G\rvert$ pointwise, so $\rho_j \le \rho_{\psi\varphi}$ for every $j$ : the positive finite measure $\rho_{\psi\varphi}$ dominates $\nu_{\psi\varphi}$ , and running the same Cauchy-Schwarz over $f^{-1}(\Omega)$ instead of over $M$ ,

\lvert \nu_{\psi\varphi}(\Omega)\rvert \;\le\; \rho_{\psi\varphi}(\Omega) \;=\; \int_{f^{-1}(\Omega)} \lvert F\rvert\,\lvert G\rvert\,d\mu \;\le\; \mu_\psi(\Omega)^{1/2}\,\mu_\varphi(\Omega)^{1/2} ,

so $\nu_{\psi\varphi}$ is controlled by the two spectral measures and vanishes on every set null for either of them. Now let $\psi\in\mathcal{D}(A)$ , so that $fF = U(A\psi) \in L^2(M,d\mu)$ . One more change of variables and one more Cauchy-Schwarz give

\int_{\mathbb{R}} \lvert\lambda\rvert\,d\rho_{\psi\varphi}(\lambda) \;=\; \int_M \lvert f\rvert\,\lvert F\rvert\,\lvert G\rvert\,d\mu \;\le\; \lVert fF\rVert_{L^2}\,\lVert G\rVert_{L^2} \;=\; \lVert A\psi\rVert\,\lVert\varphi\rVert \;<\;\infty ,

and $\rho_j \le \rho_{\psi\varphi}$ makes $\lambda$ integrable against each of the four. That is the absolute convergence promised in part D, for every $\varphi\in\mathcal{H}$ and not only for $\varphi\in\mathcal{D}(A)$ , and it is had without ever forming the total variation of $\nu_{\psi\varphi}$ : the dominating measure is a positive one, built by hand out of $\lvert U\psi\rvert$ and $\lvert U\varphi\rvert$ . The value of the integral is then read off by the substitution rule, this time with $g(\lambda) = \lambda$ :

\int_{\mathbb{R}} \lambda\,d\nu_{\psi\varphi}(\lambda) \;=\; \int_M f\,F\,\bar G\,d\mu \;=\; \bigl( T_f F, G\bigr)_{L^2(M,d\mu)} \;=\; \bigl( U(A\psi), U\varphi\bigr) \;=\; (A\psi,\varphi) .

Hence $(A\psi,\varphi) = \int\lambda\,d(P_\lambda\psi,\varphi)$ for $\psi\in\mathcal{D}(A)$ and every $\varphi\in\mathcal{H}$ . Reed and Simon (Reed & Simon (1980), §VIII.3, Thm VIII.6, p. 263) state the identity for $\varphi\in\mathcal{D}(A)$ , where polarising $(A\cdot,\cdot)$ itself is already legitimate; the passage to all of $\mathcal{H}$ is the estimate just made.

Two things are left in the one-to-one correspondence of part D, and they are not of the same kind. Injectivity is immediate from what has just been proved: if self-adjoint $A$ and $A'$ have the same p.v.m. $\{P_\Omega\}$ , then each $\psi$ has one and the same spectral measure $\mu_\psi$ for both, so the domain description gives $\mathcal{D}(A) = \{\psi : \int\lambda^2 d\mu_\psi<\infty\} = \mathcal{D}(A')$ , and on that common domain the two-vector identity just proved gives $(A\psi,\varphi) = \int\lambda\,d(P_\lambda\psi,\varphi) = (A'\psi,\varphi)$ , whence $A\psi = A'\psi$ . Injectivity does not in fact need the extension to all of $\mathcal{H}$ : $\mathcal{D}(A) = \mathcal{D}(A')$ is dense, so the identity for $\varphi$ in that dense set already pins $A\psi$ down. The extension earns its place in part D, where it strengthens what Reed and Simon state, not here. Surjectivity - that every p.v.m. on $\mathbb{R}$ is the p.v.m. of some self-adjoint operator, namely of $\int\lambda\,dP_\lambda$ on $\{\psi : \int\lambda^2 d\mu_\psi<\infty\}$ - is the other half of Theorem VIII.6 of Reed and Simon (Reed & Simon (1980), §VIII.3, Thm VIII.6, p. 263), and we quote it rather than prove it: it starts from a p.v.m. with no operator attached, so Steps 4-7 would have to be run again with $\{P_\Omega\}$ in place of the continuous functional calculus, and that is a second construction rather than a corollary of this one. $\square$

Remark(Where the hypotheses are used)Original

Self-adjointness, not symmetry. Step 9 is the whole of it: the proof needs $A \pm i$ to be onto, which is precisely the basic criterion for self-adjointness (Symmetric versus self-adjoint operators), and nothing weaker will do. The failure is not technical. Reed and Simon (Reed & Simon (1980), §VIII.2, pp. 257-259) exhibit $T = i\,d/dx$ on the absolutely continuous functions on $[0,1]$ vanishing at both ends: it is closed and symmetric, and it has a whole circle of distinct self-adjoint extensions $T_\alpha$ , indexed by $\lvert\alpha\rvert=1$ , with different spectra. There is no single spectral decomposition of $T$ to be had, because there is no single operator.
Separability. Used in Step 6, to get an at most countable family of cyclic subspaces, and again inside the bounded normal case quoted in Step 10, whose proof needs it for the same reason - Reed and Simon's Problem 5 says only "let $A$ be a normal operator", but it builds on their Problem 4, which does assume a separable space, so the hypothesis is one we carry rather than one they print; so everything downstream of Step 10 carries the hypothesis too. Dropping it costs more than the countability of the direct sum in part E: it costs the finiteness of $\mu$ that part A asserts. On the non-separable $\mathcal{H} = \ell^{2}([0,1])$ take the bounded self-adjoint $A e_t = t\,e_t$ . Every $t \in [0,1]$ is then an eigenvalue with a one-dimensional eigenspace, and the $\lambda$ -eigenspace of a multiplication operator $T_f$ is $L^{2}(f^{-1}(\lambda),\mu)$ , so each $f^{-1}(t)$ would have to have positive measure - and, the eigenspace being one-dimensional, to be an atom. Positive measure is already enough to settle the matter: uncountably many disjoint sets of positive measure do not fit inside a finite measure space, since for some $k$ uncountably many of them have measure $> 1/k$ . So $\mu(M) = \infty$ , and no finite measure space represents this $A$ - nor does a $\sigma$ -finite one, since in a countable exhaustion of $M$ by sets of finite measure some single piece would meet uncountably many of the $f^{-1}(t)$ in positive measure, and the same count rules that out. What the construction still delivers for bounded $A$ is the multiplication form over a disjoint union of finite measure spaces - one copy of $\mathbb{R}$ carrying $\mu_n$ for each member of the maximal family of Step 6, exactly as in Step 7, with only the countability of the index set lost - and parts B, C and D are read off it as before, Step 12's approximate eigenvectors now being formed inside a single piece, where the measure is finite. For unbounded $A$ what stops is our route, not the theorem: the bounded normal case quoted in Step 10 carries separability with it, so the reduction through the resolvent of Step 9 is unavailable, and nothing said here shows that an unbounded self-adjoint operator on a non-separable space is without a multiplication form.
Finiteness of $\mu$ . This is part of the statement as Reed and Simon give it. Where it is arranged depends on the case: for bounded $A$ , in Step 7, by scaling the cyclic vectors so that the masses $\mu_n(\mathbb{R}) = \lVert\varphi_n\rVert^2$ are summable; for unbounded $A$ , not in our hands at all, since the finite measure space arrives with the bounded normal theorem quoted in Step 10. Where it is used is Steps 11 and 12: both need $\chi_S \in L^2(M,d\mu)$ for a set $S$ of positive measure on which $f$ is bounded - in Step 11 to contradict self-adjointness when $\operatorname{Im} f \ne 0$ , in Step 12 to build the approximate eigenvectors - and $\chi_S \in L^2$ asks for $\mu(S) < \infty$ . It is a convenience, not a constraint: the example below realises the momentum operator on $L^2(\mathbb{R}, dk)$ , with Lebesgue measure, which is only $\sigma$ -finite, and the reweighting at the end of Step 11 converts that to a finite measure without changing the operator.
Boundedness of $h$ in part C. Needed for $h(A)$ to be everywhere defined and bounded. For unbounded $h$ the operator $h(A) = \int h\,dP$ - the clause of part D that is quoted from Reed and Simon rather than built here, as Step 10 records - exists but only on $\mathcal{D}_h$ , and $\mathcal{D}_h = \mathcal{H}$ if and only if $h$ is essentially bounded on the spectrum - bounded off a set null for every spectral measure $\mu_\psi$ , the same essential-versus-literal distinction drawn in part B and in the definition of the essential range. Literal boundedness is not necessary: for multiplication by $x$ on $L^2(\mathbb{R},dx)$ every $\mu_\psi$ is absolutely continuous, so the Borel function equal to $1$ off $\mathbb{Q}$ and to $q$ at the rational $p/q$ written in lowest terms with $q \ge 1$ is unbounded on $\sigma(A)=\mathbb{R}$ while $\int\lvert h\rvert^2\,d\mu_\psi = \lVert\psi\rVert^2$ for every $\psi$ , giving $\mathcal{D}_h = \mathcal{H}$ . Part A is the case $h(\lambda)=\lambda$ .
The two conditions in part A are not interchangeable. Condition 1 fixes the domain and condition 2 the action. Giving only condition 2 would leave the operator undetermined: multiplication by $f$ restricted to a smaller dense domain is a symmetric operator with the same formula and a different, usually much larger, spectrum.

Method(Finding a spectral representation)Original

Check self-adjointness on the stated domain. Symmetric is not enough. In practice one exhibits a core and applies the basic criterion, or quotes a known result (Symmetric versus self-adjoint operators).
Look for a unitary that already diagonalises. A translation-invariant operator on $\mathbb{R}^n$ is diagonalised by the Fourier transform; a Sturm-Liouville operator on a finite interval by its eigenfunction expansion (Sturm-Liouville eigenvalue problems); a compact self-adjoint operator by its eigenbasis (Spectral theorem for compact self-adjoint operators). In each case read off $M$ , $\mu$ and $f$ .
Read the spectrum. $\sigma(A)$ is the essential range of $f$ - the set of values $f$ takes on sets of positive measure. A point $\lambda$ is an eigenvalue exactly when $\mu\bigl(f^{-1}\{\lambda\}\bigr) > 0$ , since $\chi_{f^{-1}\{\lambda\}}$ is then a non-zero element of $L^2(M,d\mu)$ with $f\,\chi_{f^{-1}\{\lambda\}} = \lambda\,\chi_{f^{-1}\{\lambda\}}$ , while conversely an eigenvector of $T_f$ for $\lambda$ must vanish a.e. off $f^{-1}\{\lambda\}$ (Step 13); in the direct-sum normal form of part E, where $f$ is the coordinate, that says some $\mu_n$ has an atom at $\lambda$ . Where the measure has no such level set the spectrum is continuous, as in Step 2 of the example below.
Compute the spectral measure of the state you care about. $\mu_\psi$ is the image under $f$ of $\lvert U\psi\rvert^2 d\mu$ . Everything about $A$ that $\psi$ can see is in this one finite measure.
Apply functions. For bounded Borel $h$ , $h(A) = U^{-1}T_{h\circ f}U$ and $(h(A)\psi,\psi) = \int h\,d\mu_\psi$ . For unbounded $h$ , first check $\int \lvert h\rvert^2 d\mu_\psi < \infty$ .

Example(The momentum operator and its spectral data)Original

Take $\mathcal{H} = L^2(\mathbb{R})$ and

A = -i\frac{d}{dx} , \qquad \mathcal{D}(A) = \{ \psi \in L^2(\mathbb{R}) : \psi \text{ is absolutely continuous and } \psi' \in L^2(\mathbb{R}) \} ,

which is self-adjoint (Symmetric versus self-adjoint operators).

Step 1: the spectral representation. Let $U$ be the Fourier-Plancherel transform in the unitary convention

(U\psi)(k) = \hat\psi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x)\,e^{-ikx}\,dx ,

a unitary of $L^2(\mathbb{R},dx)$ onto $L^2(\mathbb{R},dk)$ (Plancherel's theorem). Differentiation becomes multiplication: $\widehat{\psi'}(k) = ik\hat\psi(k)$ , so $\widehat{A\psi}(k) = k\hat\psi(k)$ . So $M = \mathbb{R}$ , $f(k) = k$ , and condition 1 of part A reads: $\psi \in \mathcal{D}(A)$ exactly when $k\hat\psi(k) \in L^2$ - which is the Plancherel statement that $\psi' \in L^2$ . The measure $dk$ is $\sigma$ -finite but not finite; to match the statement literally, replace it by $d\tilde\mu(k) = \pi^{-1/2}e^{-k^2}dk$ , a probability measure, and conjugate by the unitary $V : L^2(dk)\to L^2(d\tilde\mu)$ , $(V\varphi)(k) = \pi^{1/4}e^{k^2/2}\varphi(k)$ . Since $V$ commutes with multiplication by $k$ , the operator is unchanged.

Step 2: the spectrum. The essential range of $f(k)=k$ is all of $\mathbb{R}$ : every interval around every $\lambda$ has positive Lebesgue measure. So $\sigma(A) = \mathbb{R}$ , and there are no eigenvalues, since $\{k=\lambda\}$ is Lebesgue-null and $k\hat\psi = \lambda\hat\psi$ forces $\hat\psi = 0$ a.e.

Step 3: a state and its spectral measure. Take the normalised Gaussian

\psi(x) = \pi^{-1/4} e^{-x^2/2} , \qquad \int_{-\infty}^{\infty}\lvert\psi\rvert^2 dx = 1 .

Its transform is $\hat\psi(k) = \pi^{-1/4}e^{-k^2/2}$ (the Gaussian is a fixed point of this convention), so

d\mu_\psi(k) = \lvert\hat\psi(k)\rvert^2 dk = \frac{1}{\sqrt\pi}\,e^{-k^2}\,dk ,

a Gaussian probability measure of mean $0$ and variance $\tfrac12$ . Three checks follow directly from it.

Total mass. $\mu_\psi(\mathbb{R}) = \pi^{-1/2}\int e^{-k^2}dk = 1 = \lVert\psi\rVert^2$ .
A spectral projection. With $\Omega = [-1,1]$ ,

(P_{[-1,1]}\psi,\psi) = \frac{1}{\sqrt\pi}\int_{-1}^{1} e^{-k^2}dk = \operatorname{erf}(1) = 0.842700792949\ldots

A second moment. $\int k^2 d\mu_\psi = \tfrac12$ , and indeed $\lVert A\psi\rVert^2 = \int\lvert\psi'\rvert^2 dx = \pi^{-1/2}\int x^2 e^{-x^2}dx = \tfrac12$ .

Step 4: a bounded function of $A$ , checked two ways. Take $h(\lambda) = e^{i\lambda}$ , which is bounded with $\lVert h\rVert_\infty = 1$ . By the recipe, $h(A) = U^{-1}T_{e^{ik}}U$ , and

\bigl(h(A)\psi,\psi\bigr) = \int_{\mathbb{R}} e^{ik}\,d\mu_\psi(k) = \frac{1}{\sqrt\pi}\int_{-\infty}^{\infty} e^{ik}e^{-k^2}dk = e^{-1/4} .

Independently, multiplying $\hat\psi$ by $e^{ik}$ is translation: $\bigl(h(A)\psi\bigr)(x) = \psi(x+1)$ , so

\bigl(h(A)\psi,\psi\bigr) = \int_{-\infty}^{\infty}\psi(x+1)\psi(x)\,dx = \frac{1}{\sqrt\pi}\int_{-\infty}^{\infty} e^{-(x+1)^2/2}e^{-x^2/2}dx = \frac{1}{\sqrt\pi}\,e^{-1/4}\!\int e^{-(x+\frac12)^2}dx = e^{-1/4} .

Both routes give

\bigl(e^{iA}\psi,\psi\bigr) = e^{-1/4} = 0.778800783071\ldots ,

which is the functional calculus delivering a number that can be checked without it. (That $e^{iA}$ should be the translation by one unit is no accident; it is Stone's theorem.)

Application(Why unbounded domains give integrals and not series)Original

Separation of variables on a bounded domain produces a series: a Fourier series, a Bessel series, a sum over normal modes. On an unbounded domain it produces an integral - a Fourier transform, a Hankel transform, a Weyl-Titchmarsh expansion. Students usually meet this as a change of technique. The spectral theorem says it is a change in the operator's spectral measure, and nothing else.

A Sturm-Liouville operator on a finite interval with sensible boundary conditions has compact resolvent, hence a purely atomic spectral measure; part E then has $\mu_n$ a sum of point masses and the $L^2$ space is a sequence space, which is the eigenfunction series. On $\mathbb{R}$ the atoms are gone: $-d^2/dx^2$ there, and the first-order $-i\,d/dx$ of the example above, both have spectral measures with a density, so $L^2(\mathbb{R},d\mu)$ is a genuine function space and the expansion is an integral. Reed and Simon (Reed & Simon (1980), §VII.2, Example 4, p. 229, and the Dirac-notation remark, pp. 227-228) put the point bluntly by observing that the Fourier transform is a spectral representation of $-i\,d/dx$ , and that the multiplication-operator form is a rigorous version of the physicists' continuum of improper eigenstates.

This also supplies the standard vocabulary for scattering. Decomposing each $\mu_\psi$ into its pure point, absolutely continuous and continuous singular parts (The Radon-Nikodym theorem) splits the Hilbert space as

\mathcal{H} = \mathcal{H}_{\mathrm{pp}} \oplus \mathcal{H}_{\mathrm{ac}} \oplus \mathcal{H}_{\mathrm{sing}} ,

each summand invariant under $A$ ; on the first, $A$ has a complete set of eigenvectors. For a Hamiltonian, $\mathcal{H}_{\mathrm{pp}}$ holds the bound states - the discrete energy levels a spectroscope measures - and $\mathcal{H}_{\mathrm{ac}}$ holds the states that escape to infinity. The singular part is the pathological case one works to rule out. None of this vocabulary is available without the theorem, because without it a Hamiltonian with no eigenvectors has no decomposition at all.

Theorem · advanced

Stone's theorem

Adapted

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Identifies the strongly continuous one-parameter unitary groups on a Hilbert space with the self-adjoint operators: every self-adjoint $A$ exponentiates to a group $U(t)=e^{itA}$ whose strong derivative at the origin recovers $A$ on exactly its own domain, and conversely every such group is of that form for one and only one self-adjoint operator, which is the mathematical content of the statement that a quantum Hamiltonian generates the time evolution.

Prerequisites: The spectral theorem, Symmetric versus self-adjoint operators, Unbounded and closed operators, Adjoints, self-adjoint and unitary operators, Bounded linear operators, Hilbert spaces, Orthonormal bases in Hilbert space, Dominated convergence, The Lebesgue integral

Definition(One-parameter unitary groups and their generators)AdaptedAdapted from Reed & Simon (1980), §VIII.4, the definition following Thm VIII.7, p. 265, and the definition of the infinitesimal generator, p. 268

Throughout, $\mathcal{H}$ is a complex Hilbert space, the inner product $(\varphi,\psi)$ is taken linear in its first argument, and $\mathcal{D}(A)$ denotes the domain of an operator $A$ (Unbounded and closed operators).

A map $t \mapsto U(t)$ from $\mathbb{R}$ into the operators on $\mathcal{H}$ is a strongly continuous one-parameter unitary group if

every $U(t)$ is unitary and $U(t+s) = U(t)U(s)$ for all real $s,t$ ; and
$U(t)\varphi \to U(t_0)\varphi$ in the norm of $\mathcal{H}$ whenever $t \to t_0$ , for each fixed $\varphi \in \mathcal{H}$ .

Condition 1 forces $U(0) = I$ and $U(-t) = U(t)^{-1} = U(t)^{*}$ . Condition 2 is continuity of each orbit $t \mapsto U(t)\varphi$ , not continuity of $t \mapsto U(t)$ in the operator norm; the distinction matters, because norm continuity is a much stronger demand that the interesting groups do not meet (see the remark below).

Given a self-adjoint $A$ , the operator

e^{itA} \;=\; h_t(A) , \qquad h_t(\lambda) = e^{it\lambda} ,

is defined by the Borel functional calculus of The spectral theorem; $h_t$ is bounded with $\lVert h_t\rVert_\infty = 1$ , so $e^{itA}$ is a bounded operator defined on all of $\mathcal{H}$ , whatever the domain of $A$ . When $A$ is bounded this agrees with the norm-convergent power series $\sum_{n\ge 0}(it)^nA^n/n!$ , but for unbounded $A$ the series is unavailable and the functional calculus is the definition.

If $U(t)$ is a strongly continuous one-parameter unitary group and $A$ is the self-adjoint operator with $U(t) = e^{itA}$ , then $A$ is the infinitesimal generator of $U$ . Part C of the statement below says that there is only one such $A$ , so the definite article is justified.

Theorem(Stone's theorem)AdaptedAdapted from Reed & Simon (1980), §VIII.4, Thm VIII.7, p. 265, Thm VIII.8, p. 266, and Thms VIII.10 and VIII.11, p. 269

A. Exponentiating a self-adjoint operator. Let $A$ be self-adjoint on $\mathcal{H}$ and set $U(t) = e^{itA}$ . Then

each $U(t)$ is unitary and $U(t+s) = U(t)U(s)$ for all $s,t \in \mathbb{R}$ ;
$U(t)\varphi \to U(t_0)\varphi$ for every $\varphi \in \mathcal{H}$ as $t \to t_0$ ;
for every $\psi \in \mathcal{D}(A)$ ,

\frac{U(t)\psi - \psi}{t} \;\longrightarrow\; iA\psi \qquad (t \to 0) ;

conversely, if $\psi \in \mathcal{H}$ is such that $\lim_{t\to 0} t^{-1}\bigl(U(t)\psi - \psi\bigr)$ exists in $\mathcal{H}$ , then $\psi \in \mathcal{D}(A)$ .

So $U$ is a strongly continuous one-parameter unitary group, and parts 3 and 4 together say that the strong derivative of $U$ at the origin exists on $\mathcal{D}(A)$ and nowhere else.

B. The converse (Stone). Conversely, suppose $t \mapsto U(t)$ is a unitary group in the sense of the definition above, strong continuity included. Then some self-adjoint operator $A$ satisfies

U(t) \;=\; e^{itA} \qquad \text{for every real } t .

C. Uniqueness. The operator $A$ of part B is unique. Indeed, if $e^{itA} = e^{itB} = U(t)$ with $A$ and $B$ self-adjoint, then by part A.4 applied to each of them, $\mathcal{D}(A)$ and $\mathcal{D}(B)$ are both equal to the set of $\psi$ for which $t^{-1}(U(t)\psi-\psi)$ converges; and on that common domain part A.3 gives $iA\psi = iB\psi$ . Hence $A = B$ as operators, domains included.

D. Cores. Two statements let one compute the generator on a convenient dense set rather than on all of $\mathcal{D}(A)$ .

Suppose $D \subseteq \mathcal{H}$ is dense, every $U(t)$ carries $D$ into itself, and $t \mapsto U(t)\varphi$ has a strong derivative for each $\varphi \in D$ . Then $i^{-1}$ times that derivative at $t=0$ , taken with domain $D$ , is an essentially self-adjoint operator whose closure generates $U$ .
Let $A$ be self-adjoint and let $D$ be a dense linear subset of $\mathcal{D}(A)$ . If $e^{itA}D \subseteq D$ for every $t$ , then $D$ is a core for $A$ , that is, $A$ is the closure of its restriction to $D$ .

Note(How little continuity is needed, and the

n

-parameter version)AdaptedAdapted from Reed & Simon (1980), §VIII.4, the computation preceding Thm VIII.9 and the statement of Thm VIII.9 itself, both on p. 268, whose proof runs onto p. 269, and Thm VIII.12, p. 270

Hypothesis 2 of the definition looks like the substantial one. It is not.

Weak continuity suffices. Expanding $\lVert U(t)\varphi - \varphi\rVert^2$ and using $\lVert U(t)\varphi\rVert = \lVert\varphi\rVert$ gives

\lVert U(t)\varphi - \varphi\rVert^2 \;=\; 2\lVert\varphi\rVert^2 - (U(t)\varphi,\varphi) - (\varphi, U(t)\varphi) ,

so if each $t \mapsto (U(t)\varphi,\psi)$ is continuous, the right side tends to $0$ as $t \to 0$ and the group is strongly continuous. Unitarity is what makes this work: it keeps the first term constant.

Measurability suffices. Reed and Simon (Reed & Simon (1980), Thm VIII.9, p. 268) record von Neumann's theorem: if $\mathcal{H}$ is separable, $U$ is a one-parameter group of unitaries, and $(U(t)\psi,\varphi)$ is a measurable function of $t$ for every $\psi,\varphi$ , then $U$ is strongly continuous. They note that this is useful in practice because one can often exhibit $(U(t)\psi,\varphi)$ as a pointwise limit of continuous functions, which settles measurability at once.
Several parameters. For a strongly continuous unitary representation of $\mathbb{R}^n$ on a separable $\mathcal{H}$ , Theorem VIII.12 (Reed & Simon (1980), p. 270) gives $n$ commuting self-adjoint generators $A_1,\dots,A_n$ , all essentially self-adjoint on the common domain of smoothed vectors used in the proof below, and a single projection-valued measure $P$ on $\mathbb{R}^n$ with $(U(t)\varphi,\psi) = \int_{\mathbb{R}^n} e^{it\cdot\lambda}\,d(P_\lambda\varphi,\psi)$ .

Note(The physicists' version)AdaptedAdapted from Arfken, Weber & Harris (2013), §17.7, Eqs. (17.35)-(17.37) and (17.40) with Example 17.7.1, pp. 846-847

In the physics literature the same correspondence is usually met in finite dimensions and derived in one line. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §17.7, pp. 846-847) start from an infinitesimal group element $U(\delta\varphi) = 1 + i\,\delta\varphi\,S$ - the factor $i$ kept outside $S$ by convention - compose $N$ copies at parameter $\varphi/N$ , and pass to the limit

U(\varphi) \;=\; \lim_{N\to\infty}\Bigl(1 + \frac{i\varphi S}{N}\Bigr)^{\!N} \;=\; \exp(i\varphi S) ,

recovering the generator by $S = -i\,dU/d\varphi$ at the identity. Unitarity of $U(\varphi)$ gives $\exp(-i\varphi S) = \exp(-i\varphi S^{\dagger})$ , hence $S = S^{\dagger}$ . Their Example 17.7.1 runs the argument on $SO(2)$ , where $S$ is the Pauli matrix $\sigma_2$ and $U(\varphi) = 1_2\cos\varphi + i\sigma_2\sin\varphi$ is the familiar rotation matrix.

Every step of that derivation is sound in finite dimensions, where $S$ is a matrix, "Hermitian" and "self-adjoint" mean the same thing (Symmetric versus self-adjoint operators) and the exponential series converges. In infinite dimensions each step needs replacing: the series by the functional calculus, "Hermitian" by self-adjoint rather than merely symmetric, and the derivative by a strong derivative that exists only on a domain. Stone's theorem is exactly the price and the reward of doing so.

Idea of the proofOriginal

One direction is the functional calculus and nothing else. Once $A$ is unitarily a multiplication operator by a real function, $e^{itA}$ is multiplication by $e^{it\lambda}$ , which has modulus one; so it is unitary, the group law is $e^{i(t+s)\lambda} = e^{it\lambda}e^{is\lambda}$ , and every claim about $U$ becomes a claim about the scalar function $e^{it\lambda}$ integrated against the spectral measure $\mu_\psi$ . The two analytic facts needed are $\lvert e^{it\lambda}-1\rvert \le 2$ and $\lvert e^{it\lambda}-1\rvert \le \lvert t\lambda\rvert$ : the first gives a dominating function on $\mathcal{H}$ , the second a dominating function on $\mathcal{D}(A)$ , and dominated convergence does the rest.

The converse must manufacture a domain out of nothing. (The route we take is the one Reed and Simon (Reed & Simon (1980), the preamble to the proof of Thm VIII.8, p. 266) set out before their proof; the estimates and the commentary below are ours.) Going the other way we are handed only the group. Differentiating it at $t=0$ is the obvious move, but a priori the derivative might exist at no vector except $0$ . The fix is averaging: replace $\varphi$ by

\varphi_f \;=\; \int_{-\infty}^{\infty} f(t)\,U(t)\varphi\,dt ,

with $f$ smooth and compactly supported. Such vectors are smooth along the group by construction, because translating the group parameter now only translates $f$ , and differentiation can be moved off $U$ and onto $f$ , where it is harmless. Taking $f$ to run through an approximate identity recovers $\varphi$ in the limit, so these vectors are dense.

Then the hard part is self-adjointness, not symmetry. On that dense set the derivative defines a symmetric operator at once. To upgrade symmetry to essential self-adjointness we use the criterion of Symmetric versus self-adjoint operators: show that $A^{*}u = \pm iu$ has only the zero solution. The group makes this a one-line ordinary differential equation. If $A^{*}u = iu$ , then $g(t) = (U(t)\varphi,u)$ satisfies $g' = g$ , so $g$ grows like $e^{t}$ ; but $\lvert g(t)\rvert \le \lVert\varphi\rVert\,\lVert u\rVert$ because $U(t)$ is unitary, and a bounded solution of $g'=g$ is identically zero. Boundedness of the group kills the deficiency.

Finally the two groups are compared by an energy argument. With $A$ in hand, $w(t) = U(t)\varphi - e^{itA}\varphi$ is differentiable and satisfies $w' = iAw$ , whence $\frac{d}{dt}\lVert w\rVert^2 = 0$ . Since $w(0)=0$ , $w$ vanishes for all $t$ .

Proof(Part A: the exponential of a self-adjoint operator)AdaptedAdapted from Reed & Simon (1980), §VIII.4, Thm VIII.7 with its proof, p. 265

Write $P$ for the projection-valued measure of $A$ and $\mu_\psi(\Omega) = (P_\Omega\psi,\psi)$ for the spectral measure of a vector $\psi$ , as in The spectral theorem. Recall from that unit that $h \mapsto h(A)$ is a $*$ -homomorphism, that $\lVert h(A)\rVert \le \lVert h\rVert_\infty$ , and that $\bigl(h(A)\psi,\psi\bigr) = \int h\,d\mu_\psi$ for bounded Borel $h$ .

Step 1 (part 1): unitarity and the group law. For each real $t$ the function $h_t(\lambda) = e^{it\lambda}$ satisfies $\overline{h_t}h_t = 1$ and $h_th_s = h_{t+s}$ pointwise on $\mathbb{R}$ . Applying the $*$ -homomorphism property:

e^{itA}\bigl(e^{itA}\bigr)^{*} = (h_t\overline{h_t})(A) = 1(A) = I , \qquad e^{itA}e^{isA} = (h_th_s)(A) = e^{i(t+s)A} .

The first identity, together with the same computation in the other order, says $e^{itA}$ is unitary; the second is the group law.

Step 2 (part 2): strong continuity at the origin. Fix $\varphi \in \mathcal{H}$ . Since $\lvert h_t - 1\rvert^2$ is a bounded Borel function,

\lVert U(t)\varphi - \varphi\rVert^2 = \bigl( (h_t-1)(A)\varphi, (h_t-1)(A)\varphi \bigr) = \int_{\mathbb{R}} \bigl\lvert e^{it\lambda} - 1 \bigr\rvert^2 \, d\mu_\varphi(\lambda) ,

the last equality because $(h_t-1)^{*}(h_t-1) = \lvert h_t-1\rvert^2$ under the functional calculus. For each fixed $\lambda$ the integrand tends to $0$ as $t \to 0$ , and it is bounded above by the constant $4$ , which is $\mu_\varphi$ -integrable because $\mu_\varphi$ is a finite measure of total mass $\lVert\varphi\rVert^2$ . By dominated convergence (Dominated convergence) the integral tends to $0$ . So $U(t)\varphi \to \varphi$ .

Step 3 (part 2 in general). For arbitrary $t_0$ , the group law gives $U(t)\varphi - U(t_0)\varphi = U(t_0)\bigl(U(t-t_0)\varphi - \varphi\bigr)$ , and $U(t_0)$ preserves norms, so continuity at $0$ propagates to every $t_0$ . This is the only place the group law is used in part 2, and it is indispensable: continuity at one point of a group is continuity everywhere, but without the group law nothing connects the two.

Step 4 (part 3): the derivative on $\mathcal{D}(A)$ . Let $\psi \in \mathcal{D}(A)$ , which by the spectral theorem means $\int\lambda^2\,d\mu_\psi(\lambda) < \infty$ . The same computation as in Step 2 gives

\Bigl\lVert \frac{U(t)\psi-\psi}{t} - iA\psi \Bigr\rVert^2 = \int_{\mathbb{R}} \Bigl\lvert \frac{e^{it\lambda}-1}{t} - i\lambda \Bigr\rvert^2 d\mu_\psi(\lambda) .

The integrand tends to $0$ pointwise, since $\frac{d}{dt}e^{it\lambda}\big|_{t=0} = i\lambda$ . For the dominating function use $\lvert e^{i\theta}-1\rvert \le \lvert\theta\rvert$ for real $\theta$ , which follows from $e^{i\theta}-1 = i\int_0^{\theta}e^{iu}\,du$ ; it gives $\lvert (e^{it\lambda}-1)/t\rvert \le \lvert\lambda\rvert$ , hence the integrand is at most $(2\lvert\lambda\rvert)^2 = 4\lambda^2$ . That bound is $\mu_\psi$ -integrable precisely because $\psi \in \mathcal{D}(A)$ . Dominated convergence again gives the limit $0$ .

Step 5 (part 4): nothing outside $\mathcal{D}(A)$ is differentiable. Let $\mathcal{D}(B)$ be the set of $\psi$ for which $t^{-1}(U(t)\psi-\psi)$ converges as $t \to 0$ , and define $iB\psi$ to be that limit. $\mathcal{D}(B)$ is a linear subspace and $B$ is linear. $B$ is symmetric: for $\psi,\chi \in \mathcal{D}(B)$ ,

(B\psi,\chi) = \lim_{t\to 0}\Bigl(\frac{U(t)\psi-\psi}{it},\chi\Bigr) = \lim_{t\to 0}\Bigl(\psi, \frac{U(-t)\chi-\chi}{-it}\Bigr) = (\psi,B\chi) ,

where the middle equality moves $U(t)$ across the inner product as $U(t)^{*} = U(-t)$ and uses that $\overline{1/(it)} = -1/(it)$ . By Step 4, $\mathcal{D}(A) \subseteq \mathcal{D}(B)$ and $B$ agrees with $A$ there, that is, $B$ extends $A$ . But a self-adjoint operator has no proper symmetric extension: if $A \subseteq B$ with $B$ symmetric, then taking adjoints reverses the inclusion, $B^{*} \subseteq A^{*} = A \subseteq B \subseteq B^{*}$ , forcing $B = A$ (Symmetric versus self-adjoint operators). Hence $\mathcal{D}(B) = \mathcal{D}(A)$ , which is part 4. $\square$

Proof(Part B: every group is an exponential)AdaptedAdapted from Reed & Simon (1980), §VIII.4, Thm VIII.8 with its proof, pp. 266-267, the route Reed and Simon attribute in their Notes, p. 306, to Gårding and Wightmann

Let $U$ be as in part B.

Step 1: a supply of smooth vectors. For $f \in C_0^{\infty}(\mathbb{R})$ and $\varphi \in \mathcal{H}$ put

\varphi_f \;=\; \int_{-\infty}^{\infty} f(t)\,U(t)\varphi\,dt .

The integrand is a continuous $\mathcal{H}$ -valued function of $t$ (strong continuity) supported in the compact support of $f$ , so the integral exists as a Riemann integral of a Banach-space-valued function, and it obeys $\bigl\lVert\int h\bigr\rVert \le \int\lVert h\rVert$ , proved from Riemann sums exactly as for scalar integrals. Write $D$ for the linear span of all these $\varphi_f$ , as $\varphi$ and $f$ vary.

Step 2: $D$ is dense. Let $j_\varepsilon \ge 0$ be smooth, supported in $[-\varepsilon,\varepsilon]$ , with $\int j_\varepsilon = 1$ . Then

\lVert \varphi_{j_\varepsilon} - \varphi\rVert = \Bigl\lVert \int j_\varepsilon(t)\bigl(U(t)\varphi - \varphi\bigr)\,dt \Bigr\rVert \;\le\; \sup_{\lvert t\rvert \le \varepsilon} \lVert U(t)\varphi - \varphi\rVert ,

using $\int j_\varepsilon = 1$ to subtract $\varphi$ inside the integral and then the norm inequality of Step 1. Strong continuity at $0$ makes the right side tend to $0$ with $\varepsilon$ . So every $\varphi$ is a limit of elements of $D$ .

Step 3: differentiating along the group moves onto $f$ . For $\varphi_f \in D$ and $s \ne 0$ , apply the group law and substitute $\tau = s+t$ :

\frac{U(s)-I}{s}\,\varphi_f = \int f(t)\,\frac{U(s+t)-U(t)}{s}\,\varphi\,dt = \int \frac{f(\tau-s) - f(\tau)}{s}\,U(\tau)\varphi\,d\tau .

As $s \to 0$ the difference quotient $\bigl(f(\tau-s)-f(\tau)\bigr)/s$ converges to $-f'(\tau)$ uniformly in $\tau$ , with supports in one fixed compact set; by the norm inequality the integral therefore converges to $\varphi_{-f'}$ . So the strong derivative exists on $D$ , and we may define

A\varphi_f \;=\; i^{-1}\varphi_{-f'} , \qquad \mathcal{D}(A) = D ,

extended linearly. The right-hand side involves $f$ , while a single vector of $D$ may be written as $\varphi_f$ for several pairs $(\varphi,f)$ ; the formula is nevertheless well defined, because the left-hand side has just been identified as $-i$ times the strong derivative of $s \mapsto U(s)$ at that vector, a quantity depending on the vector alone and not on how it was presented. Three consequences are immediate from the formula: $A$ maps $D$ into $D$ , because $-f' \in C_0^{\infty}(\mathbb{R})$ ; $U(t)$ maps $D$ into $D$ , because $U(t)\varphi_f = \varphi_{f(\cdot - t)}$ by the same substitution; and $U(t)A\varphi_f = AU(t)\varphi_f$ , since $U(t)$ commutes with the integral.

Step 4: $A$ is symmetric on $D$ . For $\varphi_f, \varphi_g \in D$ , Step 3 gives

(A\varphi_f,\varphi_g) = \lim_{s\to 0}\Bigl( \frac{U(s)-I}{is}\varphi_f , \varphi_g \Bigr) = \lim_{s\to 0}\Bigl( \varphi_f , \frac{U(-s)-I}{i(-s)}\varphi_g \Bigr) = (\varphi_f, A\varphi_g) ,

moving $U(s)$ across the inner product as $U(s)^{*} = U(-s)$ , and then noting that a scalar leaves the first argument as itself and enters the second as its conjugate, so that $1/(is)$ becomes $\overline{1/(is)} = 1/(i(-s))$ . Writing $s' = -s$ , the last limit is the difference quotient of Step 3 at $s'$ , which converges to $A\varphi_g$ .

Step 5: $A$ is essentially self-adjoint on $D$ . By the basic criterion (Symmetric versus self-adjoint operators) it is enough to show that $A^{*}u = iu$ and $A^{*}u = -iu$ have only the solution $u=0$ . Suppose $A^{*}u = iu$ and let $\varphi \in D$ . Put $g(t) = (U(t)\varphi, u)$ . Since $U(t)\varphi \in D = \mathcal{D}(A)$ and $\frac{d}{dt}U(t)\varphi = iAU(t)\varphi$ by Step 3 and the group law,

g'(t) = \bigl(iAU(t)\varphi,\, u\bigr) = i\bigl(AU(t)\varphi,\, u\bigr) = i\bigl(U(t)\varphi,\, A^{*}u\bigr) = i\bigl(U(t)\varphi,\, iu\bigr) = i\cdot(-i)\,\bigl(U(t)\varphi,\, u\bigr) = g(t) ,

the third equality being the definition of the adjoint, and the scalar $i$ leaving the first argument unchanged while leaving the second as its conjugate $-i$ , so that the two cancel to $+1$ . Hence $g(t) = g(0)e^{t}$ . But $\lvert g(t)\rvert \le \lVert U(t)\varphi\rVert\,\lVert u\rVert = \lVert\varphi\rVert\,\lVert u\rVert$ for every $t$ , because $U(t)$ is unitary; a multiple of $e^{t}$ is bounded on $\mathbb{R}$ only if it vanishes, so $g(0) = (\varphi,u) = 0$ . As $\varphi$ ranged over the dense set $D$ , $u = 0$ . The case $A^{*}u = -iu$ runs identically, with $i\cdot(+i) = -1$ in place of $i\cdot(-i)$ , so $g' = -g$ and $g(t) = g(0)e^{-t}$ , which is again bounded only if $g(0)=0$ . So $A$ is essentially self-adjoint on $D$ ; write $A$ again for its closure, which is self-adjoint.

Step 6: the two groups agree. Let $V(t) = e^{itA}$ , which is a strongly continuous unitary group by part A, and fix $\varphi \in D \subseteq \mathcal{D}(A)$ . By part A.3 and the group law, $V(t)\varphi \in \mathcal{D}(A)$ and $\frac{d}{dt}V(t)\varphi = iAV(t)\varphi$ ; by Step 3, $U(t)\varphi \in D$ and $\frac{d}{dt}U(t)\varphi = iAU(t)\varphi$ . So $w(t) = U(t)\varphi - V(t)\varphi$ is a strongly differentiable curve in $\mathcal{D}(A)$ with $w'(t) = iAw(t)$ , and

\frac{d}{dt}\lVert w(t)\rVert^2 = \bigl(w'(t),w(t)\bigr) + \bigl(w(t),w'(t)\bigr) = \bigl(iAw,w\bigr) + \bigl(w,iAw\bigr) = i\bigl[(Aw,w) - \overline{(Aw,w)}\,\bigr] = 0 ,

because $(Aw,w)$ is real for self-adjoint $A$ . Since $w(0) = 0$ we get $w \equiv 0$ , so $U(t)\varphi = V(t)\varphi$ for all $\varphi \in D$ ; both operators are bounded and $D$ is dense, so $U(t) = V(t) = e^{itA}$ . $\square$

Remark(Where the hypotheses are used, and what fails without them)Original

Self-adjointness, not symmetry (part A). The functional calculus that defines $e^{itA}$ exists only for self-adjoint $A$ . A symmetric operator that is not essentially self-adjoint has no spectral measure, hence no $e^{itA}$ , and the deficiency can be seen directly. On $L^2(0,\infty)$ let $B = -i\,d/dx$ on the absolutely continuous $\psi$ with $\psi$ and $\psi'$ both in $L^2(0,\infty)$ and $\psi(0)=0$ - the integrability of $\psi'$ is what makes the deficiency computation available - which is symmetric with deficiency indices $(1,0)$ (Symmetric versus self-adjoint operators) and so has no self-adjoint extension at all. The family that would have to play the part of $e^{-itB}$ is the shift $\bigl(S(t)\psi\bigr)(x) = \psi(x-t)$ for $x \ge t$ and $0$ for $x < t$ , whose strong derivative at $t=0$ is $-\psi' = -iB\psi$ , as it should be. For $t \ge 0$ each $S(t)$ is isometric and $S(t)S(s) = S(t+s)$ , but the range of $S(t)$ consists of the functions vanishing on $[0,t)$ , so $S(t)$ is not onto and therefore not unitary, and there is no way to define $S(t)$ for $t<0$ . One gets a semigroup, never a group.
Unitarity (part B, Steps 4 and 5). Unitarity enters the converse twice, and in two different strengths. Step 4 uses the identity $U(s)^{*} = U(-s)$ to move the group across the inner product; that is unitarity itself, and without it $A$ need not be symmetric on $D$ . Step 5 needs no more than the norm bound $\lVert U(t)\rVert = 1$ , whose only role is to bound $g(t) = (U(t)\varphi,u)$ and so defeat the exponentially growing solution of $g'=g$ . Drop unitarity and keep only a strongly continuous group of bounded operators and the conclusion fails: $\mathcal{H} = \mathbb{C}$ and $U(t) = e^{t}$ is such a group whose generator $A = -i$ is not self-adjoint. What survives in that generality is the Hille-Yosida theory of semigroups, which replaces self-adjointness by resolvent bounds.
Strong continuity (part B, Steps 1 and 2). It is used twice: to make $t \mapsto U(t)\varphi$ Riemann integrable, and to make $\varphi_{j_\varepsilon} \to \varphi$ , which is what makes $D$ dense. Without a dense $D$ there is no operator to construct. By the note above, the hypothesis can be weakened to measurability of the matrix elements on a separable space, but it cannot be dropped: the construction has no starting point.
Compact support and smoothness of $f$ (part B, Step 3). Smoothness gives the uniform convergence of the difference quotients of $f$ ; compact support keeps all the integrands inside one fixed compact set, so uniform convergence of the integrands implies convergence of the integrals. With $f$ merely integrable neither step is available.
Strong, not norm, continuity (the definition). Requiring $\lVert U(t)-I\rVert \to 0$ would make the theorem nearly empty: norm continuity of a unitary group forces its generator to be bounded, so every unbounded Hamiltonian would be excluded. In the example below $U(t)$ moves the basis vector $e_n$ to $e^{int}e_n$ , so $U(t)-I$ is diagonal in that basis and

\lVert U(t)-I\rVert \;=\; \sup_{n\in\mathbb{Z}}\bigl\lvert e^{int}-1\bigr\rvert \;=\; \sup_{n\in\mathbb{Z}} 2\bigl\lvert \sin(nt/2)\bigr\rvert .

This is not always $2$ . If $t/2\pi$ is irrational the points $nt$ are dense modulo $2\pi$ and the supremum is indeed $2$ . If instead $t = 2\pi p/q$ in lowest terms with $q \ge 2$ , the points $nt$ run through the multiples of $2\pi/q$ and the supremum is $2\sin\bigl(\pi\lfloor q/2\rfloor/q\bigr)$ : that is $2$ for even $q$ , but $\sqrt3 = 1.7320508\ldots$ at $q=3$ and $1.9021130\ldots$ at $q=5$ , and it increases with $q$ thereafter. So the sharp statement is

\lVert U(t)-I\rVert \;\ge\; \sqrt3 \qquad \text{for every } t \notin 2\pi\mathbb{Z} ,

with equality exactly at $t \equiv \pm 2\pi/3 \pmod{2\pi}$ . The distance from the identity therefore never shrinks as $t \to 0$ - it stays bounded below by $\sqrt3$ - so $U$ is not norm continuous at the origin, though it is strongly continuous.

Part D.1 needs invariance. If $D$ is not invariant under the group, the derivative computed on $D$ need not have anything to do with a self-adjoint operator: the restriction of $-i\,d/dx$ to functions vanishing at both ends of an interval is symmetric but not essentially self-adjoint, and the reason is exactly that the translations do not leave that domain, so part D.1 does not apply to it.

Method(Identifying the generator of a given group)Original

Given a concrete group $U(t)$ , the following produces its generator together with a proof that it is the generator, without ever computing a spectral measure.

Check the two axioms. Verify $U(t+s) = U(t)U(s)$ and that each $U(t)$ is unitary (an isometry which is onto). Then verify $U(t)\varphi \to \varphi$ as $t \to 0$ for $\varphi$ in some dense set only; boundedness of the $U(t)$ by $1$ and an $\varepsilon/3$ argument extend it to all of $\mathcal{H}$ , and the group law extends it to all $t_0$ .
Guess the generator by differentiating. Pick a convenient dense set $D$ - test functions, finite linear combinations of an orthonormal basis, or the smoothed vectors $\varphi_f$ of the proof - and compute $A_0\varphi = -i\frac{d}{dt}U(t)\varphi\big|_{t=0}$ on it.
Check that $D$ is invariant, that is, $U(t)D \subseteq D$ for every $t$ .
Conclude by part D.1. Steps 2 and 3 say $A_0$ is essentially self-adjoint on $D$ and that $\overline{A_0}$ is the generator. There is nothing left to prove: in particular you need not verify self-adjointness of $\overline{A_0}$ separately.
Or, if a self-adjoint $A$ is already in hand, check instead that $D \subseteq \mathcal{D}(A)$ is dense and $e^{itA}D \subseteq D$ ; part D.2 then says $D$ is a core, so an identity verified on $D$ extends to $\mathcal{D}(A)$ by closure.
Read off the spectrum when the group is periodic. If $U(T) = I$ for some $T>0$ , then $e^{iT\lambda} = 1$ for $\mu_\psi$ -almost every $\lambda$ and every $\psi$ , so $\sigma(A) \subseteq (2\pi/T)\mathbb{Z}$ : periodicity in $t$ quantises the generator.

Example(The rotation group on a circle, and the quantisation of angular momentum)Original

Take $\mathcal{H} = L^2(0,2\pi)$ with functions regarded as $2\pi$ -periodic, and let

\bigl(U(t)\psi\bigr)(\theta) \;=\; \psi(\theta + t) .

Step 1: it is a strongly continuous unitary group. Each $U(t)$ preserves the integral of $\lvert\psi\rvert^2$ over a period and is inverted by $U(-t)$ , so it is unitary; $U(t+s) = U(t)U(s)$ is the associativity of adding angles. For $\psi$ continuous and periodic, $U(t)\psi \to \psi$ uniformly, hence in $L^2$ ; such $\psi$ are dense, so by Step 1 of the method the group is strongly continuous.

Step 2: the generator. Let $e_n(\theta) = (2\pi)^{-1/2}e^{in\theta}$ , $n \in \mathbb{Z}$ , the orthonormal basis of Orthonormal bases in Hilbert space, and let $D$ be the finite linear combinations of the $e_n$ . Then $U(t)e_n = e^{int}e_n$ , so $D$ is invariant, and

-i\frac{d}{dt}\,U(t)e_n\Big|_{t=0} \;=\; -i(in)e_n \;=\; n\,e_n .

So $A_0 e_n = n e_n$ on $D$ , that is, $A_0 = -i\,d/d\theta$ . By part D.1 its closure $A$ is the generator, with $Ae_n = ne_n$ and

\mathcal{D}(A) = \Bigl\{ \psi = \sum_n c_n e_n \;:\; \sum_n n^2\lvert c_n\rvert^2 < \infty \Bigr\} .

Step 3: the spectrum is the integers. Here $U(2\pi) = I$ , so by Step 6 of the method $\sigma(A) \subseteq \mathbb{Z}$ ; and every integer is an eigenvalue, witnessed by $e_n$ . This is the whole of the quantisation of angular momentum: the eigenvalues of $-i\,d/d\theta$ are integers because a rotation by $2\pi$ is the identity, not because of any extra assumption.

Step 4: a state, and three numbers. Take

\psi(\theta) \;=\; \frac{1 + \cos\theta}{\sqrt{3\pi}} .

It is normalised: $\int_0^{2\pi}(1+\cos\theta)^2 d\theta = \int_0^{2\pi}\bigl(1 + 2\cos\theta + \tfrac12 + \tfrac12\cos 2\theta\bigr)d\theta = 2\pi + 0 + \pi + 0 = 3\pi$ . Writing $1+\cos\theta = 1 + \tfrac12 e^{i\theta} + \tfrac12 e^{-i\theta}$ and multiplying by $\sqrt{2\pi}/\sqrt{3\pi}$ gives the coefficients

c_0 = \sqrt{\tfrac23} , \qquad c_{1} = c_{-1} = \tfrac12\sqrt{\tfrac23} , \qquad \lvert c_0\rvert^2 + 2\lvert c_1\rvert^2 = \tfrac23 + 2\cdot\tfrac14\cdot\tfrac23 = 1 .

The spectral measure of $\psi$ is therefore the three-point measure putting mass $\tfrac23$ at $\lambda = 0$ and $\tfrac16$ at $\lambda = \pm 1$ .

A matrix element of the group. By the functional calculus, $\bigl(U(t)\psi,\psi\bigr) = \sum_n e^{int}\lvert c_n\rvert^2 = \tfrac23 + \tfrac13\cos t$ . Checking this directly at $t = 0.7$ : the integral $\frac{1}{3\pi}\int_0^{2\pi}(1+\cos(\theta+0.7))(1+\cos\theta)\,d\theta$ evaluates to $0.921614062428\ldots$ , and $\tfrac23 + \tfrac13\cos(0.7) = 0.921614062428\ldots$ as well.
How far the state moves. $\lVert U(t)\psi - \psi\rVert^2 = 2 - 2\operatorname{Re}\bigl(U(t)\psi,\psi\bigr) = \tfrac23(1-\cos t)$ . At $t = \pi/3$ this is $\tfrac23\cdot\tfrac12 = \tfrac13 = 0.3333\ldots$ , so the rotation by $60^{\circ}$ displaces this state by $\lVert U(\pi/3)\psi-\psi\rVert = 1/\sqrt3 = 0.57735\ldots$ in $L^2$ .
The derivative, verified numerically. Part A.3 predicts $t^{-1}(U(t)\psi-\psi) \to iA\psi$ . In the basis the error has squared norm $\sum_n \lvert c_n\rvert^2\bigl\lvert (e^{int}-1)/t - in\bigr\rvert^2$ ; the $n=0$ term vanishes and the two others are equal, so the error equals

\Bigl\lVert \frac{U(t)\psi-\psi}{t} - iA\psi \Bigr\rVert = \frac{1}{\sqrt3}\Bigl\lvert \frac{e^{it}-1}{t} - i \Bigr\rvert = \frac{1}{\sqrt3}\sqrt{\Bigl(\frac{\cos t - 1}{t}\Bigr)^2 + \Bigl(\frac{\sin t}{t}-1\Bigr)^2} .

At $t = 0.1$ this is $0.028859495597\ldots$ , about $t/(2\sqrt3) = 0.02887$ , the first-order behaviour one expects. For comparison, $\lVert A\psi\rVert^2 = \sum n^2\lvert c_n\rvert^2 = \tfrac13$ , matching the direct computation $\int_0^{2\pi}\lvert\psi'\rvert^2 d\theta = \frac{1}{3\pi}\int_0^{2\pi}\sin^2\theta\,d\theta = \tfrac13$ .

Step 5: why the convergence in Step 4 is not uniform in $\psi$ . Applied to $e_n$ alone the error is $\lvert (e^{int}-1)/t - in\rvert$ , which at fixed $t$ grows with $n$ . This is the failure of norm continuity recorded in the remark, seen from the other side: the derivative exists at each vector of $\mathcal{D}(A)$ but at no uniform rate, which is exactly why $A$ is unbounded.

Application(Dynamics, symmetry and conservation)Original

Time evolution. In the Hilbert-space model of quantum mechanics (The mathematical framework of quantum mechanics) the states of a system are unit vectors and the evolution over a time $t$ is a map $\varphi \mapsto U(t)\varphi$ . Three physical demands pin down $U$ : probabilities must be preserved ( $U(t)$ unitary), evolving for $t$ and then for $s$ must equal evolving for $t+s$ (the group law), and the state must not jump ( $U(t)\varphi$ continuous in $t$ ). Stone's theorem converts these three demands into a single object, a self-adjoint $H$ with $U(t) = e^{-itH}$ , and part A.3 then says that for $\varphi \in \mathcal{D}(H)$ ,

\frac{d}{dt}\,U(t)\varphi \;=\; -iH\,U(t)\varphi ,

the Schrödinger equation (The Schrodinger equation; we write $e^{-itH}$ rather than $e^{itH}$ because physics fixes the sign of the Hamiltonian, and units are chosen with $\hbar = 1$ ). The logical direction is worth stating plainly: one does not postulate the Schrödinger equation and hope a solution exists. One postulates unitary, continuous, composable dynamics, and the equation together with its generator is handed back, with a domain attached saying precisely which states are differentiable in time.

This also explains why quantisation insists on self-adjoint rather than merely symmetric operators, and why the deficiency-index analysis of Symmetric versus self-adjoint operators is physics and not pedantry. A symmetric Hamiltonian with deficiency indices $(1,1)$ - a particle on a half-line, say - generates no dynamics until a boundary condition is chosen; the circle of self-adjoint extensions is a circle of physically different theories, and the theorem says nothing can decide between them except more physics.

Symmetry and conservation. A symmetry is a one-parameter group $V(s) = e^{isG}$ commuting with the dynamics, $V(s)U(t) = U(t)V(s)$ for all $s,t$ . Differentiating $U(-t)V(s)U(t) = V(s)$ in $s$ at $s = 0$ on a common core shows that $G$ is constant along the motion, so its distribution in the state $U(t)\varphi$ does not depend on $t$ : the generator of a symmetry is a conserved quantity. This is Noether's theorem in its quantum form, and Stone's theorem is what makes the phrase "the generator of the symmetry" refer to an operator rather than to a formal expression. Translations give momentum, rotations give angular momentum - quantised to the integers by the periodicity computation in the example above - and time translation gives the energy, which is why $H$ is both the generator of the dynamics and the observable called energy.

The same statement in geometry. On a smooth manifold the analogous correspondence runs between global flows and vector fields: Lee (Lee (2013), Ch. 9, pp. 209-210) defines a global flow as a continuous action of the additive group $\mathbb{R}$ , so that $\theta_t\circ\theta_s = \theta_{t+s}$ and $\theta_0$ is the identity, and its infinitesimal generator as the vector field obtained by differentiating each orbit at time $0$ ; Proposition 9.7 there shows this vector field is smooth and that the orbits are its integral curves (Flows and integral curves). The dictionary is exact - group of transformations on the left, generator on the right - and so is the difference: on a compact manifold every smooth vector field generates a global flow, whereas on a Hilbert space the generator must be self-adjoint and not merely symmetric, and a merely symmetric generator is the infinite-dimensional analogue of a vector field whose integral curves escape to infinity in finite time.

Application

The mathematical framework of quantum mechanics

Adapted

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Assembles the Hilbert-space model of a quantum system - unit vectors up to a phase as states, densely defined self-adjoint operators as observables, the projection-valued measure of an observable as the probability distribution of measuring it, and a strongly continuous unitary group as the dynamics - and derives from it the expectation and variance of a measurement, the commutator inequality, and the Heisenberg uncertainty relation, with the statistics of two concrete states computed in full.

Prerequisites: The spectral theorem, Stone's theorem, Symmetric versus self-adjoint operators, Unbounded and closed operators, Adjoints, self-adjoint and unitary operators, Hilbert spaces, Orthonormal bases in Hilbert space, Sigma-algebras and measures, L^p spaces, Expectation, variance and covariance, Plancherel's theorem

Definition(The model)AdaptedAdapted from Reed & Simon (1980), §VIII.11, pp. 302-303; Kreyszig (1978), §11.1, eqs. (1)-(3) and (9)-(11), pp. 573-576

A quantum system is described by four pieces of data in a complex separable Hilbert space $\mathcal{H}$ , with the inner product $(\varphi,\psi)$ linear in the first slot as elsewhere in this handbook.

States. Every vector of norm one in $\mathcal{H}$ determines a physical state, and two unit vectors determine the same state exactly when they differ by a complex factor of modulus one. A state is therefore an equivalence class $\{\beta\varphi : \lvert\beta\rvert = 1\}$ , equivalently the one-dimensional subspace that $\varphi$ spans. In the one-particle model on $\mathcal{H} = L^2(\mathbb{R})$ the vector itself is the wave function $\psi(q)$ , and $\lvert\psi(q)\rvert^2$ is a probability density on the line: the probability of finding the particle in a set $J$ is $\int_J \lvert\psi\rvert^2\,dq$ , which is why unit norm is demanded and why the unimodular factor is invisible.

Observables. Every observable is a self-adjoint operator $A$ on $\mathcal{H}$ , densely defined, with $\mathcal{D}(A) = \mathcal{D}(A^{*})$ (Symmetric versus self-adjoint operators). Symmetry is not enough, and this is not a technicality: the whole of what follows is read off the projection-valued measure of $A$ , which exists only for self-adjoint operators (The spectral theorem).

Measurement. Let $\{P_\Omega\}$ be the projection-valued measure of $A$ . When the state is $\varphi$ and we measure the observable $A$ , the value obtained is a random variable whose distribution is the spectral measure of $\varphi$ :

\operatorname{Prob}\bigl(\text{value} \in \Omega\bigr) \;=\; \mu_\varphi(\Omega) \;=\; (P_\Omega\varphi,\varphi) , \qquad \Omega \subseteq \mathbb{R} \text{ Borel} ;

in particular the probability of a value in $[a,b]$ is $(P_{[a,b]}\varphi,\varphi)$ .

Dynamics. Time evolution is a strongly continuous one-parameter unitary group $U(t)$ : a system in the state $\varphi$ at time $0$ is in the state $U(t_0)\varphi$ at time $t_0$ . By Stone's theorem there is exactly one self-adjoint operator generating it. That operator is the Hamiltonian $H$ , the observable corresponding to the classical energy, and for $\varphi \in \mathcal{D}(H)$ the orbit satisfies the Schrödinger equation; Reed and Simon (Reed & Simon (1980), §VIII.11, p. 303) write it as

\frac{d}{dt}\bigl[U(t)\varphi\bigr] \;=\; iH\bigl[U(t)\varphi\bigr] ,

while on p. 305 of the same section they write the group of the $n$ -electron atom as $U(t) = e^{-itH}$ . This unit takes $U(t) = e^{-itH}$ , the form used on p. 305 and in the physics literature; the note below explains why both appear and why nothing measurable depends on the choice.

Following Kreyszig (Kreyszig (1978), §11.1, eqs. (9)-(11), pp. 575-576), write for an observable $T$ and a state $\psi$

\mu_\psi(T) = (T\psi,\psi) , \qquad \operatorname{var}_\psi(T) = \bigl( (T - \mu_\psi(T)I)^2\psi, \psi \bigr) , \qquad \operatorname{sd}_\psi(T) = \sqrt{\operatorname{var}_\psi(T)} ,

for the mean value, variance and standard deviation of the observable in that state. The commutator of two observables is $C = ST - TS$ with domain $\mathcal{D}(C) = \mathcal{D}(ST)\cap\mathcal{D}(TS)$ .

Application(What the model yields)AdaptedAdapted from Reed & Simon (1980), §VIII.11, pp. 302-303; Kreyszig (1978), §11.1, eqs. (6)-(11), pp. 575-576, and §11.2, Thm 11.2-1 and Thm 11.2-2 with eq. (8), pp. 579-581

Let $A$ , $S$ , $T$ be observables on $\mathcal{H}$ and $\psi$ a state.

A. The measurement rule is consistent. $\mu_\psi^{A}(\Omega) = (P_\Omega\psi,\psi) = \lVert P_\Omega\psi\rVert^2$ defines a probability measure on the Borel sets of $\mathbb{R}$ , supported in $\sigma(A)$ . So a measurement of $A$ never returns a value outside the spectrum, and every neighbourhood of every spectral point has positive probability for some state.

B. Mean and variance are matrix elements. If $\psi \in \mathcal{D}(A)$ then the distribution $\mu_\psi^{A}$ has a finite first moment and

\int_{\mathbb{R}} \lambda \, d\mu_\psi^{A}(\lambda) \;=\; (A\psi,\psi) \;=\; \mu_\psi(A) ,

which is real; and $\operatorname{var}_\psi(A) = \int (\lambda - \mu_\psi(A))^2 d\mu_\psi^{A}(\lambda) = \lVert A\psi\rVert^2 - \mu_\psi(A)^2$ , finite for the same reason. So the probabilistic mean and variance of the measurement agree with the operator expressions of the definition.

C. Sharp values are eigenvalues. $\operatorname{var}_\psi(A) = 0$ if and only if $A\psi = \mu_\psi(A)\,\psi$ , that is, exactly when $\psi$ is an eigenvector. A state in which $A$ has a certain, dispersion-free value is an eigenvector of $A$ , and the value is the eigenvalue; if $A$ has no point spectrum no state gives a sharp value.

D. Stationary states. If $H\psi = E\psi$ then $U(t)\psi = e^{-itE}\psi$ in the sign convention adopted above (and $e^{+itE}\psi$ in the other one), which in either case is the same state for every $t$ . Reed and Simon (Reed & Simon (1980), §VIII.11, p. 303) note that the point spectrum of $H$ therefore holds the stationary states, and that a system stimulated from outside typically responds by moving between two of them and radiating at a frequency proportional to the gap between the corresponding eigenvalues.

E. The commutator inequality. Theorem 11.2-1 of Kreyszig (Kreyszig (1978), §11.2, p. 579) states: let $S$ and $T$ be self-adjoint operators with domain and range in $L^2(-\infty,+\infty)$ , and $C = ST - TS$ . Then

\bigl\lvert \mu_\psi(C) \bigr\rvert \;\le\; 2\,\operatorname{sd}_\psi(S)\,\operatorname{sd}_\psi(T)

for every $\psi$ in the domain of $C$ . Two observables whose commutator has a non-zero mean in a state cannot both be sharp in that state.

F. The Heisenberg relation. On $L^2(-\infty,+\infty)$ let $Q$ be multiplication by $q$ and let $D\psi = \frac{h}{2\pi i}\frac{d\psi}{dq}$ on the absolutely continuous functions with $D\psi \in L^2$ . A differentiation gives the commutation relation

DQ - QD \;=\; \frac{h}{2\pi i}\,\tilde I

on $\mathcal{D}(DQ)\cap\mathcal{D}(QD)$ , where $\tilde I$ is the identity on that domain, and Theorem 11.2-2 (Kreyszig (1978), §11.2, p. 580) then gives

\operatorname{sd}_\psi(D)\,\operatorname{sd}_\psi(Q) \;\ge\; \frac{h}{4\pi} .

In the units of the examples below, where $\hbar = h/2\pi = 1$ and $P = -i\,d/dq$ , this reads $\operatorname{sd}_\psi(P)\operatorname{sd}_\psi(Q) \ge \tfrac12$ .

Note(Three conventions, and one sign that is not a disagreement)AdaptedAdapted from Reed & Simon (1980), §VIII.11, pp. 303 and 305, and §II.1, p. 37; Kreyszig (1978), §11.2, eq. (2) and footnote 2, p. 576; Arfken, Weber & Harris (2013), §5.3, eq. (5.43), p. 276

Which slot is linear. Reed and Simon (Reed & Simon (1980), §II.1, p. 37) make the inner product conjugate-linear in the first slot and write the measurement probability as $(\varphi, P_{[a,b]}\varphi)$ ; Kreyszig and this handbook take the first slot linear and write $(P_{[a,b]}\varphi,\varphi)$ . The number is the same, since it equals $\lVert P_{[a,b]}\varphi\rVert^2$ , and no statement above changes.
$h$ or $\hbar$ . Kreyszig keeps Planck's constant $h$ and defines $D = \frac{h}{2\pi i}\frac{d}{dq}$ , writing $P$ for projections instead; Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.3, p. 276) and Reed and Simon work in units with $\hbar = h/2\pi = 1$ and $p = -i\,d/dx$ , in which $[x,p] = i$ . Setting $\hbar = 1$ turns Kreyszig's $h/4\pi$ into $\tfrac12$ and his $DQ-QD = \frac{h}{2\pi i}\tilde I$ into $PQ - QP = -i\tilde I$ , which is Arfken's (5.43) with the two entries exchanged. The examples below use $\hbar = 1$ and $P = -i\,d/dq$ throughout.
The sign in the exponent. On p. 303 Reed and Simon write the Schrödinger equation as $\frac{d}{dt}[U(t)\varphi] = iH[U(t)\varphi]$ , which by Stone's theorem means $U(t) = e^{itH}$ ; two pages later, for the $n$ -electron atom, they write $U(t) = e^{-itH}$ . The two differ by replacing $H$ with $-H$ , and the second is the one the physics literature uses, since $i\hbar\,\partial_t\psi = H\psi$ gives $\psi(t) = e^{-itH/\hbar}\psi(0)$ . This unit follows the $p.\,305$ form, $U(t) = e^{-itH}$ with $\frac{d}{dt}U(t)\varphi = -iHU(t)\varphi$ , and states part D with that sign; the only quantity affected is the sign of the phase $e^{-itE}$ of a stationary state, which is unobservable because states are defined up to a unimodular factor. Nothing else in this unit, and no expectation value or probability anywhere, depends on the choice.

Idea of the proofOriginal

The spectral theorem is what turns an operator into a random variable. A classical observable is a function on phase space, and a state is a probability measure there; the observable's distribution is the push-forward of the state along the function. Quantum mechanics has no phase space, but the spectral theorem supplies the missing push-forward: the projection-valued measure of $A$ turns each unit vector $\psi$ into an honest probability measure $\mu_\psi^A$ on the real line, and everything statistical - probabilities, means, variances, sharp values - is read off that one measure. The model is not "operators instead of numbers"; it is "one probability measure per observable per state, manufactured by the spectral theorem".

Why variances cannot both vanish. Centre $S$ and $T$ at their means. Centring changes neither the variances nor the commutator, because constants commute with everything. The commutator of the centred operators is then sandwiched between two inner products that the Schwarz inequality bounds by the product of the two centred norms - which are exactly the two standard deviations. The commutator is thus a lower bound on the product of the spreads, and when it is a non-zero constant, as for position and momentum, neither spread can be made small without the other growing.

Where the difficulty really sits. Nothing in the model is hard once the observables are known to be self-adjoint. The hard part comes before: physics hands over a formal differential expression with no domain attached, and whether it has one self-adjoint realisation, many, or none is a genuine mathematical question whose answer is physics (Symmetric versus self-adjoint operators).

Proof(Parts A to D)AdaptedAdapted from Reed & Simon (1980), §VIII.11, pp. 302-303, read against the projection-valued-measure form of the spectral theorem, §VIII.3, pp. 262-263

Throughout, $P$ is the projection-valued measure of $A$ and $\mu_\psi(\Omega) = (P_\Omega\psi,\psi)$ .

Part A. The axioms of a projection-valued measure (The spectral theorem) give exactly the axioms of a measure. Each $P_\Omega$ is an orthogonal projection, so $(P_\Omega\psi,\psi) = (P_\Omega^2\psi,\psi) = (P_\Omega\psi,P_\Omega\psi) = \lVert P_\Omega\psi\rVert^2 \ge 0$ , which is both non-negativity and the stated identity. $P_{\varnothing} = 0$ gives $\mu_\psi(\varnothing) = 0$ , and $P_{\mathbb{R}} = I$ gives total mass $\lVert\psi\rVert^2 = 1$ . For countable additivity, let $\Omega = \bigcup_n \Omega_n$ with the $\Omega_n$ disjoint; strong $\sigma$ -additivity of $P$ gives $P_\Omega\psi = \lim_N \sum_{n\le N}P_{\Omega_n}\psi$ , and taking the inner product with $\psi$ , which is continuous, gives $\mu_\psi(\Omega) = \sum_n \mu_\psi(\Omega_n)$ . Finally, if $\Omega$ misses $\sigma(A)$ then $P_\Omega = 0$ , so the measure is supported in the spectrum.

Part B. By the spectral theorem, $\psi \in \mathcal{D}(A)$ is equivalent to $\int\lambda^2\,d\mu_\psi(\lambda) < \infty$ , and then $\int\lvert\lambda\rvert\,d\mu_\psi \le \bigl(\int\lambda^2 d\mu_\psi\bigr)^{1/2} < \infty$ by Cauchy-Schwarz against the finite measure $\mu_\psi$ , so the first moment exists. The spectral theorem also gives $(A\psi,\psi) = \int\lambda\,d\mu_\psi(\lambda)$ and $\lVert A\psi\rVert^2 = \int\lambda^2 d\mu_\psi(\lambda)$ . The first integral is real because $\mu_\psi$ is a positive measure on $\mathbb{R}$ ; expanding the square in the definition of the variance,

\int(\lambda-m)^2 d\mu_\psi = \int\lambda^2 d\mu_\psi - 2m\int\lambda\,d\mu_\psi + m^2 = \lVert A\psi\rVert^2 - m^2 , \qquad m = \mu_\psi(A) ,

using $\mu_\psi(\mathbb{R}) = 1$ in the last term. The same expansion performed on operators gives $\bigl((A-mI)^2\psi,\psi\bigr) = \lVert (A-mI)\psi\rVert^2$ , which is the operator form of the same number.

Part C. By part B, $\operatorname{var}_\psi(A) = \lVert (A - mI)\psi\rVert^2$ with $m = \mu_\psi(A)$ . A squared norm vanishes exactly when the vector does, that is, exactly when $A\psi = m\psi$ . Conversely an eigenvector has $\mu_\psi(A) = (A\psi,\psi) = m\lVert\psi\rVert^2 = m$ , so the sharp value is the eigenvalue.

Part D. If $H\psi = E\psi$ then $h(H)\psi = h(E)\psi$ for every bounded Borel $h$ , a property of the functional calculus recorded in The spectral theorem; applying it to $h(\lambda) = e^{-it\lambda}$ gives $U(t)\psi = e^{-itE}\psi$ . Since $\lvert e^{-itE}\rvert = 1$ and states are unit vectors up to a unimodular factor, $U(t)\psi$ is the same state as $\psi$ for every $t$ . $\square$

Proof(Parts E and F)AdaptedAdapted from Kreyszig (1978), §11.2, Thm 11.2-1 with its proof and eq. (8), pp. 579-580, and Thm 11.2-2, p. 580

Part E. Let $\psi \in \mathcal{D}(C)$ , so $\psi \in \mathcal{D}(ST)\cap\mathcal{D}(TS)$ ; since $\mathcal{D}(ST)\subseteq\mathcal{D}(T)$ and $\mathcal{D}(TS)\subseteq\mathcal{D}(S)$ , every quantity below is defined.

Step 1: centre the operators. Put $m_1 = \mu_\psi(S)$ , $m_2 = \mu_\psi(T)$ and

A = S - m_1 I , \qquad B = T - m_2 I .

Both $m_1$ and $m_2$ are real, being mean values of self-adjoint operators (part B), so $A$ and $B$ are self-adjoint. Expanding and cancelling the terms in which a scalar meets an operator,

AB - BA = ST - m_2 S - m_1 T + m_1m_2 I - \bigl(TS - m_1T - m_2S + m_1m_2I\bigr) = ST - TS = C ,

so centring changes nothing on the left-hand side of the inequality.

Step 2: write the mean of $C$ as a difference of two inner products. Using self-adjointness of $A$ and of $B$ to move each across the inner product,

\mu_\psi(C) = \bigl((AB - BA)\psi,\psi\bigr) = (AB\psi,\psi) - (BA\psi,\psi) = (B\psi, A\psi) - (A\psi, B\psi) .

Step 3: bound it. The two terms are complex conjugates of one another, so they have equal modulus; by the triangle inequality and then the Schwarz inequality,

\lvert\mu_\psi(C)\rvert \;\le\; \lvert (B\psi,A\psi)\rvert + \lvert (A\psi,B\psi)\rvert \;\le\; 2\,\lVert A\psi\rVert\,\lVert B\psi\rVert .

Step 4: identify the two norms as standard deviations. By the computation in part B, $\lVert B\psi\rVert^2 = \bigl((T-m_2I)^2\psi,\psi\bigr) = \operatorname{var}_\psi(T)$ , so $\lVert B\psi\rVert = \operatorname{sd}_\psi(T)$ , and likewise $\lVert A\psi\rVert = \operatorname{sd}_\psi(S)$ . Substituting into Step 3 gives the inequality.

Part F. For $\psi$ in the common domain, the product rule gives

DQ\psi(q) = \frac{h}{2\pi i}\frac{d}{dq}\bigl(q\psi(q)\bigr) = \frac{h}{2\pi i}\psi(q) + q\,\frac{h}{2\pi i}\psi'(q) = \frac{h}{2\pi i}\psi(q) + QD\psi(q) ,

which is the commutation relation. Hence $C = DQ - QD$ acts as multiplication by the constant $h/2\pi i$ , so $\mu_\psi(C) = \frac{h}{2\pi i}\lVert\psi\rVert^2$ and $\lvert\mu_\psi(C)\rvert = h/2\pi$ for every state in the domain. Part E with $S = D$ and $T = Q$ gives $h/2\pi \le 2\operatorname{sd}_\psi(D)\operatorname{sd}_\psi(Q)$ , which is the Heisenberg relation. $\square$

Remark(Where the hypotheses are used, and what the relation does not say)Original

Self-adjointness, in every part. Part A needs the projection-valued measure, which exists only for self-adjoint operators; part B needs $\mu_\psi(A)$ real, which is reality of the spectrum; parts E and F need $A$ and $B$ self-adjoint in order to move them across the inner product in Step 2. A merely symmetric operator fails all three. The point is not academic: the momentum of a particle confined to a box is symmetric with deficiency indices $(1,1)$ and the momentum on a half-line has indices $(1,0)$ , so neither is an observable in this model (Symmetric versus self-adjoint operators), while the Hamiltonian on the same domains is self-adjoint and is one.
The domain of the commutator, in part E. The inequality is asserted only for $\psi \in \mathcal{D}(C)$ . For $\psi$ outside it, one or more of the four quantities in the statement may be undefined, and the inequality is not merely unproved but meaningless. Kreyszig (Kreyszig (1978), §11.2, p. 579) records without proof that $\mathcal{D}(DQ)\cap\mathcal{D}(QD)$ is dense in $L^2(-\infty,+\infty)$ , so there are plenty of such $\psi$ ; he observes that it contains the Hermite functions (Hermite polynomials), which are total in the space.
Finite variance, in parts B, E and F. If $\psi \notin \mathcal{D}(A)$ the distribution $\mu_\psi^A$ may have an infinite second moment - and even an infinite first moment - while remaining a perfectly good probability measure. Part A survives, parts B onwards do not. A state can therefore have well-defined measurement probabilities and no expected value, which is the Hilbert-space shadow of a heavy-tailed distribution (Expectation, variance and covariance).
What the uncertainty relation is about. It is a statement about two distributions determined by one state, not about the disturbance caused by an apparatus. Kreyszig (Kreyszig (1978), §11.2, pp. 580-581) is careful on this: the standard deviations measure the precision attainable in principle, and the limitation is not an imperfection of instruments. The derivation above uses no model of a measuring device at all; it uses the Schwarz inequality.
Non-commuting is not the same as never simultaneously sharp. Part E bounds $\operatorname{sd}_\psi(S)\operatorname{sd}_\psi(T)$ below by $\lvert\mu_\psi(C)\rvert/2$ , which is useless in states where $\mu_\psi(C) = 0$ . For position and momentum, $\mu_\psi(C)$ is a non-zero constant and the bound bites in every state; for a pair such as two components of angular momentum, $\mu_\psi(C)$ does vanish in some states, and in those the inequality permits both spreads to be small.

Method(Computing the statistics of a measurement)Original

Given a state $\psi$ and an observable $A$ , the following produces every number the model predicts.

Check that $A$ is self-adjoint on a stated domain, not merely symmetric. If it is a differential expression on an interval or a half-line, count deficiency indices first (Symmetric versus self-adjoint operators). If this step fails there is nothing to compute.
Find a spectral representation of $A$ : an orthonormal basis of eigenvectors if the spectrum is a point spectrum, a unitary carrying $A$ to multiplication by a real function otherwise (The spectral theorem).
Write down $\mu_\psi^{A}$ . With eigenvectors $\varphi_n$ and $\psi = \sum_n c_n\varphi_n$ , it is the point measure giving mass $\lvert c_n\rvert^2$ to the eigenvalue $\lambda_n$ ; with $U\psi = \tilde\psi$ and $UAU^{-1}$ multiplication by $f$ , it is the push-forward of $\lvert\tilde\psi\rvert^2 d\mu$ along $f$ .
Read off probabilities as $\mu_\psi^{A}(\Omega)$ , summing $\lvert c_n\rvert^2$ over the eigenvalues in $\Omega$ or integrating $\lvert\tilde\psi\rvert^2$ over $f^{-1}(\Omega)$ .
Read off the mean and variance as $\int\lambda\,d\mu_\psi^A$ and $\int\lambda^2 d\mu_\psi^A - (\int\lambda\,d\mu_\psi^A)^2$ , and check them independently against $(A\psi,\psi)$ and $\lVert A\psi\rVert^2 - (A\psi,\psi)^2$ . The two routes must agree, which is a strong arithmetic check on the whole computation.
Evolve the state by multiplying each coefficient by $e^{-it\lambda_n}$ , or $\tilde\psi$ by $e^{-itf}$ , when $A = H$ . The distribution of the energy is then unchanged in time, while the distribution of any observable not commuting with $H$ generally moves.

Example(A particle in a box: the energy and the position)Original

Take $\mathcal{H} = L^2(0,1)$ , units with $\hbar = 1$ and mass $m = 1$ , and the Hamiltonian

H = -\tfrac12\,\frac{d^2}{dq^2} , \qquad \mathcal{D}(H) = \{\varphi : \varphi,\varphi' \text{ absolutely continuous}, \ \varphi'' \in L^2,\ \varphi(0)=\varphi(1)=0\} ,

which is self-adjoint on that domain (Symmetric versus self-adjoint operators), with

\varphi_n(q) = \sqrt2\,\sin(n\pi q) , \qquad E_n = \tfrac12 n^2\pi^2 , \qquad n = 1,2,\dots,

an orthonormal basis of eigenvectors (Sturm-Liouville eigenvalue problems). Take the state

\psi = \frac{3\varphi_1 + 2\varphi_2 + \varphi_3}{\sqrt{14}} ,

which has norm one because $9 + 4 + 1 = 14$ .

Step 1: the spectral measure of $\psi$ for $H$ . By step 3 of the method it is the three-point measure with masses

\tfrac{9}{14} \text{ at } E_1, \qquad \tfrac{4}{14} \text{ at } E_2 , \qquad \tfrac{1}{14} \text{ at } E_3 ,

and numerically $E_1 = 4.934802200545\ldots$ , $E_2 = 19.739208802179\ldots$ , $E_3 = 44.413219804902\ldots$ .

Step 2: probabilities. The probability of measuring the ground-state energy is $9/14 = 0.642857\ldots$ . The probability of a result below $20$ is $\mu_\psi^H\bigl((-\infty,20)\bigr) = (9+4)/14 = 13/14 = 0.928571\ldots$ , since $E_1$ and $E_2$ lie below $20$ and $E_3$ does not.

Step 3: mean and variance of the energy, two ways. From the measure,

\mu_\psi(H) = \frac{9E_1 + 4E_2 + E_3}{14} = \frac{\pi^2}{28}\bigl(9\cdot 1 + 4\cdot 4 + 1\cdot 9\bigr) = \frac{34\pi^2}{28} = \frac{17\pi^2}{14} = 11.984519629894\ldots

Independently, each $\varphi_n$ is multiplied by $n^2\pi^2/2$ , so $H\psi = \frac{\pi^2}{2\sqrt{14}}(3\varphi_1 + 8\varphi_2 + 9\varphi_3)$ and $(H\psi,\psi) = \frac{\pi^2}{2\cdot 14}(3\cdot 3 + 8\cdot 2 + 9\cdot 1) = \frac{34\pi^2}{28}$ , the same number. For the variance,

\int\lambda^2\,d\mu_\psi^H = \frac{\pi^4}{4}\cdot\frac{9\cdot 1 + 4\cdot 16 + 1\cdot 81}{14} = \frac{154\pi^4}{56} = \frac{11\pi^4}{4} ,

\operatorname{var}_\psi(H) = \frac{11\pi^4}{4} - \frac{289\pi^4}{196} = \frac{539 - 289}{196}\pi^4 = \frac{125\pi^4}{98} = 124.246289584187\ldots , \qquad \operatorname{sd}_\psi(H) = \frac{5\pi^2\sqrt{10}}{14} = 11.146581968666\ldots

The spread is comparable to the mean, as it must be for a state spread over three widely separated levels.

Step 4: the position in the same state. Position and energy are different observables of the same state, so they are read from different spectral measures. $Q\psi(q) = q\psi(q)$ has purely continuous spectrum $[0,1]$ , and $\mu_\psi^{Q}$ has density $\lvert\psi(q)\rvert^2$ on $[0,1]$ . Its mean needs the integrals

\int_0^1 2q\,\sin(m\pi q)\sin(n\pi q)\,dq = \begin{cases} \tfrac12 , & m = n , \\[2pt] -\dfrac{2}{\pi^2}\Bigl(\dfrac{1}{(m-n)^2} - \dfrac{1}{(m+n)^2}\Bigr), & m+n \text{ odd} , \\[6pt] 0 , & m \ne n,\ m+n \text{ even} , \end{cases}

which follow from $2\sin a\sin b = \cos(a-b) - \cos(a+b)$ and $\int_0^1 q\cos(k\pi q)\,dq = \bigl((-1)^k-1\bigr)/(k^2\pi^2)$ for $k \ne 0$ . Only the pairs $(1,2)$ and $(2,3)$ have $m+n$ odd, contributing $-16/(9\pi^2)$ and $-48/(25\pi^2)$ , each weighted by twice the product of the coefficients, $2\cdot\frac{3\cdot 2}{14} = \frac67$ and $2\cdot\frac{2\cdot 1}{14} = \frac27$ . The diagonal contributes $\tfrac12\sum_n \lvert c_n\rvert^2 = \tfrac12$ . Hence

\mu_\psi(Q) = \frac12 - \frac67\cdot\frac{16}{9\pi^2} - \frac27\cdot\frac{48}{25\pi^2} = \frac12 - \frac{1088}{525\pi^2} = 0.290023908947\ldots

The particle sits left of centre: the interference between $\varphi_1$ and $\varphi_2$ piles probability into the left half, and nothing in the energy distribution of Step 1 shows this, because the energy distribution is blind to the relative phases that the position distribution sees.

Step 5: what is not computed, and why. The momentum $-i\,d/dq$ on this domain is symmetric with deficiency indices $(1,1)$ and is not self-adjoint (Symmetric versus self-adjoint operators). By step 1 of the method the computation stops there: the model assigns this system no momentum distribution at all, and the familiar observation that $\sin(n\pi q)$ is not a momentum eigenfunction is the visible symptom.

Example(A Gaussian on the line: equality in the uncertainty relation)Original

Take $\mathcal{H} = L^2(\mathbb{R})$ , $\hbar = 1$ , $Q\psi(q) = q\psi(q)$ and $P = -i\,d/dq$ , both self-adjoint on their natural domains (Symmetric versus self-adjoint operators), and the state

\psi_\sigma(q) = \bigl(2\pi\sigma^2\bigr)^{-1/4} e^{-q^2/(4\sigma^2)} , \qquad \sigma > 0 .

Step 1: the position distribution. $\lvert\psi_\sigma(q)\rvert^2 = (2\pi\sigma^2)^{-1/2}e^{-q^2/(2\sigma^2)}$ , the normal density of mean $0$ and variance $\sigma^2$ ; in particular $\lVert\psi_\sigma\rVert = 1$ , $\mu_\psi(Q) = 0$ and $\operatorname{var}_\psi(Q) = \sigma^2$ , so $\operatorname{sd}_\psi(Q) = \sigma$ .

Step 2: the momentum distribution. The Fourier-Plancherel transform in the unitary convention $\hat\psi(p) = (2\pi)^{-1/2}\int\psi(q)e^{-ipq}dq$ carries $P$ to multiplication by $p$ (Plancherel's theorem), so it is the spectral representation of $P$ and $\mu_\psi^{P}$ has density $\lvert\hat\psi_\sigma(p)\rvert^2$ . Using $\int e^{-aq^2}e^{-ipq}dq = \sqrt{\pi/a}\,e^{-p^2/(4a)}$ with $a = 1/(4\sigma^2)$ ,

\hat\psi_\sigma(p) = (2\pi\sigma^2)^{-1/4}\,\sigma\sqrt2\;e^{-\sigma^2p^2} , \qquad \lvert\hat\psi_\sigma(p)\rvert^2 = \sigma\sqrt{\tfrac{2}{\pi}}\;e^{-2\sigma^2p^2} ,

which is the normal density of mean $0$ and variance $1/(4\sigma^2)$ . Hence $\operatorname{sd}_\psi(P) = 1/(2\sigma)$ .

Step 3: the uncertainty product. $\operatorname{sd}_\psi(P)\operatorname{sd}_\psi(Q) = \frac{1}{2\sigma}\cdot\sigma = \tfrac12$ for every $\sigma$ , which is equality in part F. Squeezing the position distribution by shrinking $\sigma$ widens the momentum distribution by exactly the reciprocal factor.

Step 4: the commutator inequality is also an equality here. $C = PQ - QP$ acts as $(PQ-QP)\psi = -i(q\psi)' + iq\psi' = -i\psi$ , so $C = -i\tilde I$ and $\lvert\mu_\psi(C)\rvert = 1$ , while $2\operatorname{sd}_\psi(P)\operatorname{sd}_\psi(Q) = 2\cdot\tfrac12 = 1$ . Part E is saturated, and one can see both of its inequalities become equalities. Here $\mu_\psi(P) = \mu_\psi(Q) = 0$ , so $A = P$ and $B = Q$ in the proof. Differentiating gives

P\psi_\sigma = -i\psi_\sigma' = \frac{i}{2\sigma^2}\,q\,\psi_\sigma = \frac{i}{2\sigma^2}\,Q\psi_\sigma ,

so $P\psi_\sigma$ and $Q\psi_\sigma$ are parallel and the Schwarz inequality is sharp; and

(Q\psi_\sigma, P\psi_\sigma) = \int q\psi_\sigma \overline{\Bigl(\frac{iq}{2\sigma^2}\psi_\sigma\Bigr)}\,dq = -\frac{i}{2\sigma^2}\int q^2\lvert\psi_\sigma\rvert^2 dq = -\frac{i}{2\sigma^2}\cdot\sigma^2 = -\frac{i}{2}

is purely imaginary, so in Step 2 of that proof the two conjugate terms add rather than cancel. Gaussians are exactly the states of minimum uncertainty, which is the operator form of the equality case in The uncertainty principle.

Step 5: numbers, with $\sigma = 2$ . Then $\operatorname{sd}_\psi(Q) = 2$ and $\operatorname{sd}_\psi(P) = 1/4$ , and $\psi_2(q) = (8\pi)^{-1/4}e^{-q^2/16}$ .

The probability that a position measurement returns a value within one standard deviation of the mean is

\int_{-2}^{2}\lvert\psi_2(q)\rvert^2 dq = \operatorname{erf}\bigl(1/\sqrt2\bigr) = 0.682689492137\ldots ,

and the probability that a momentum measurement returns a value in $[-\tfrac14,\tfrac14]$ is the same number, both distributions being normal and both intervals being one standard deviation wide.

For a free particle of unit mass, $H = \tfrac12 P^2$ , so by part B and Step 2

\mu_\psi(H) = \tfrac12\int_{\mathbb{R}} p^2\,d\mu_\psi^{P}(p) = \tfrac12\operatorname{var}_\psi(P) = \frac{1}{8\sigma^2} = \frac{1}{32} = 0.03125 .

Checking this on the other side of the Fourier transform: $\psi_2'(q) = -\frac{q}{2\sigma^2}\psi_2(q) = -\frac{q}{8}\psi_2(q)$ , so

\tfrac12\lVert P\psi_2\rVert^2 = \tfrac12\int_{\mathbb{R}}\lvert\psi_2'\rvert^2 dq = \frac{1}{2}\cdot\frac{1}{64}\int_{\mathbb{R}} q^2\lvert\psi_2\rvert^2 dq = \frac{1}{128}\cdot\sigma^2 = \frac{4}{128} = \frac{1}{32} ,

the same $0.03125$ . The agreement is Plancherel's theorem doing the arithmetic twice.

Application(The three problems, and why the model is not the physics)AdaptedAdapted from Reed & Simon (1980), §VIII.11, the three problems and the n-electron atom example, pp. 303-305

Reed and Simon (Reed & Simon (1980), §VIII.11, pp. 303-304) isolate three mathematical problems that any quantum model raises, and the division still organises the subject.

Self-adjointness. Physical reasoning supplies a formal expression - a differential operator - with no domain attached. The first task is to prove essential self-adjointness on some natural domain; failing that, to survey the self-adjoint extensions and decide which of them deserves the name of the observable. Everything in this unit is unavailable until that is settled.
Spectral analysis. The second task is to locate the spectra of the observables, that of $H$ before all others, and to pin down where the eigenvalues sit and how degenerate each one is. That is where the energy levels of a spectroscope come from.
Scattering theory. The third is to describe the behaviour of $U(t)\varphi$ for large $t$ .

Their worked case is the $n$ -electron atom. The Hilbert space is $L^2(\mathbb{R}^{3n})$ ; each classical momentum component is replaced by $-i\partial$ in the corresponding variable and each coordinate by multiplication, in units with $\hbar = 1$ ; and the classical energy, the sum of the kinetic terms, the attraction $-ne^2/\lvert r_k\rvert$ of each electron to the nucleus and the repulsion $e^2/\lvert r_k - r_l\rvert$ between pairs, becomes

H = -\sum_{k=1}^{n}\frac{1}{2m}\Delta_k + V , \qquad \text{dynamics } U(t) = e^{-itH} .

All of these operators are essentially self-adjoint on the Schwartz space, although Reed and Simon note that for $H$ this is far from immediate and defer the proof. The probability that the $x$ -coordinate of the $k$ -th electron is found in $(a,b)$ at time $t_0$ is then the integral of $\lvert U(t_0)\varphi\rvert^2$ over that slab, exactly as the measurement rule requires.

The honest closing observation is theirs: this model ignores electron spin and the Pauli exclusion principle, treats the nucleus as fixed, and is non-relativistic. The Hilbert-space framework is therefore a frame rather than a physical theory. It says what a state, an observable and a measurement are, and guarantees that the predictions are consistent probabilities; which Hilbert space, which operators and which Hamiltonian describe a given piece of the world is a separate question that no amount of functional analysis will answer.