Defines test functions and distributions as continuous linear functionals on them, proves that every locally integrable function is a distribution while the delta is not, defines the distributional derivative and derives the jump formula for it, shows that a distribution with vanishing derivative is a constant, and ends with a worked derivative and the notion of a weak solution.
The support of a function f on Rn is the closure of {x:f(x)=0}. A test function is a function φ:Rn→C that is infinitely differentiable and has compact support. The set of all of them is written
D=D(Rn)=C0∞(Rn),
a complex vector space. A word on the scalars. We take test functions complex-valued, following Folland (p. 282), because the Fourier transform of Fourier transforms of distributions needs complex scalars. Stakgold and Holst (p. 94) and Strauss (p. 331) instead make C0∞(Rn) a real linear space and their functionals real-valued; everything in this unit holds verbatim in either convention, since each result quoted from them is real-linear and extends to complex φ by splitting into real and imaginary parts.
Non-trivial test functions exist: for the ball ∣x∣<1 the function
φ∗(x)=⎩⎨⎧exp(∣x∣2−11),0,∣x∣<1,∣x∣≥1,
is Stakgold and Holst's example (2.1.2). Note that the exponent is the reciprocal of ∣x∣2−1, which is negative on the ball: for ∣x∣<1 one may equally write φ∗(x)=exp(−1/(1−∣x∣2)), and in particular φ∗(0)=e−1 while φ∗(x)↓0 as ∣x∣↑1. It is smooth across the sphere ∣x∣=1 because every derivative of it is a rational function of ∣x∣, with poles only at ∣x∣=1, times the same exponential, and that exponential beats every power as ∣x∣↑1. No such function is the sum of a convergent Taylor series at ∣x∣=1: all its derivatives vanish there, so the series would be identically zero. Translating and rescaling, φ∗((x−x0)/ε) is a test function supported in the ball of radius ε about x0.
Two closure properties are used constantly: every partial derivative of a test function is a test function, and aφ is a test function whenever a∈C∞(Rn) and φ∈D — the product has compact support because φ does, and a need not decay at all.
Convergence in D is deliberately severe. Say φm→φin D if
all the φm vanish outside one fixed ball, and
for every multi-index k, maxx∣Dkφm(x)−Dkφ(x)∣→0 as m→∞.
So the functions and all their derivatives converge uniformly, inside a common bounded set. A sequence converging to 0 in this sense is what Stakgold and Holst call a null sequence; Folland phrases the same condition as convergence in C0∞(K) for one compact K. The second condition really bites: with φ the one-dimensional version of φ∗, the sequence m−1φ(mx) tends to 0 uniformly but its derivatives do not, so it is not a null sequence.
A distribution (or generalized function) on Rn is a linear functional f on D that is continuous, meaning that ⟨f,φm⟩→0 for every null sequence {φm} in D. The value of f at φ is written ⟨f,φ⟩. The pairing is linear in both slots — unlike the inner product of Hilbert space, there is no conjugation. The space of distributions is denoted D′(Rn); it is a vector space, with ⟨c1f1+c2f2,φ⟩=c1⟨f1,φ⟩+c2⟨f2,φ⟩, and f1=f2 means ⟨f1,φ⟩=⟨f2,φ⟩ for every test function.
Hassani's reading of the definition is the useful one for a physicist: a distribution is the thing that a formula such as ∫fφ computes, stripped of the pretence that f has point values.
Three standard examples.
The Dirac distribution at ξ∈Rn is ⟨δξ,φ⟩=φ(ξ). Linearity is immediate, and continuity is immediate too: a null sequence converges uniformly, so φm(ξ)→0. Strauss makes this rule the definition of the delta. We write δ for δ0 and, following the symbolic convention below, δ(x−ξ) for δξ.
φ↦Dkφ(x0) is a distribution for every multi-index k and every point x0 (Folland, p. 283).
A function f on Rn is locally integrable, written f∈Lloc1, if ∫Ω∣f∣ is finite for every bounded domain Ω (The Lebesgue integral). Behaviour at infinity is irrelevant; the condition restricts only the strength of singularities, and ∣x∣−a on Rn is locally integrable exactly when a<n. By the theorem below, each f∈Lloc1is a distribution, through
⟨f,φ⟩=∫Rnf(x)φ(x)dx,
and we do not distinguish the function from the functional. A distribution that can be written this way is called regular; all others are singular. For a singular distribution the integral sign is kept as pure notation: ⟨f,φ⟩ and ∫fφdx mean the same thing, which is why one writes ∫δ(x)φ(x)dx=φ(0) and speaks of "the delta function" (Stakgold and Holst's Eq. (2.1.5)).
Two more operations are needed. For a∈C∞ the product af is the distribution ⟨af,φ⟩=⟨f,aφ⟩, which is legitimate because aφ∈D; in particular a(x)δ(x)=a(0)δ(x). And a sequence of distributions converges weakly to f if ⟨fN,φ⟩→⟨f,φ⟩ for every test function (Strauss, Eq. (5) of §12.1); every delta family of The Dirac delta in calculations converges weakly to δ in this sense, and Delta-function identities makes that precise.
Definition(Functions are distributions; the delta is not a function)AdaptedAdapted from Stakgold & Holst (2011), §2.1, Theorem 2.1.1 and p. 97, and Example 3, p. 98
Let f∈Lloc1(Rn).
φ↦∫fφdx is a distribution.
Two locally integrable functions define the same distribution if and only if they are equal almost everywhere. In particular distinct continuous functions define distinct distributions, so nothing is lost by regarding a continuous function as a distribution.
There is no locally integrable f with ∫fφdx=φ(0) for every φ∈D(Rn). The Dirac distribution is singular.
Stakgold and Holst prove the "if" half of (2) and the continuous case; the converse implication is quoted there without proof, and is quoted here the same way.
Idea of the proofOriginal
Part (1) is one inequality: a test function from a null sequence is uniformly small on a fixed ball, and ∫Ω∣f∣ over that ball is a finite constant, so the pairing is small.
Part (3) is a competition between two scalings. Shrink a fixed bump to width a about the origin. Its height does not change, so the pairing φ↦φ(0) does not notice the shrinking at all. But a locally integrable f does notice: all of the mass of f that the bump can see is the mass inside a ball of radius a, and that tends to 0 as the ball closes up. A quantity that stays fixed cannot equal one that tends to zero.
ProofAdaptedAdapted from Stakgold & Holst (2011), §2.1, proof of Theorem 2.1.1, p. 96, and Example 3, p. 98
(1). Linearity of φ↦∫fφ is linearity of the integral. For continuity, let {φm} be a null sequence and let Ω be a bounded set outside which every φm vanishes — one such Ω exists by condition 1 of the definition of a null sequence. Then
The integral on the right is a finite number, by local integrability of f and boundedness of Ω; the maximum tends to 0 by condition 2 with k=0. Hence ⟨f,φm⟩→0. □
(2), the direction proved here. Suppose f1,f2∈Lloc1 are equal almost everywhere. Then ∫Ω∣f1−f2∣=0 for every bounded Ω, so ∣⟨f1−f2,φ⟩∣≤max∣φ∣∫suppφ∣f1−f2∣=0 for every test function, and the two distributions coincide.
Conversely, suppose f1,f2 are continuous and differ at some x0; without loss of generality f1(x0)>f2(x0). By continuity there is ε>0 with f1−f2>0 throughout the ball ∣x−x0∣<ε. Take φ(x)=φ∗((x−x0)/ε)≥0, which is positive exactly on that ball. The integrand (f1−f2)φ is then non-negative and strictly positive on an open set, so ⟨f1−f2,φ⟩>0 and the two distributions differ. (For merely locally integrable functions, equality of the distributions forces equality almost everywhere; we take that over from Stakgold and Holst, p. 97, who omit its proof.) □
(3). Suppose, for contradiction, that f∈Lloc1(Rn) satisfies ∫fφ=φ(0) for every test function. Apply this to the shrinking family
ψa(x)=φ∗(x/a),a>0,
each of which is a test function supported in ∣x∣≤a. Two facts about ψa matter, and both come straight from the formula for φ∗:
ψa(0)=φ∗(0)=e−1, the same number for every a;
0≤ψa≤e−1 everywhere, since on the support 0≤∣x/a∣<1, so −1≤∣x/a∣2−1<0 and therefore the exponent satisfies 1/(∣x/a∣2−1)≤−1.
The assumed identity gives ∫fψadx=ψa(0)=e−1 for every a. On the other hand ψa vanishes off the ball ∣x∣≤a, so
∫fψadx≤e−1∫∣x∣≤a∣f(x)∣dxa→0+0,
the limit being 0 because ∫∣x∣≤a∣f∣ is the integral of a fixed integrable function over sets shrinking to a point. So the constant e−1 would have to be 0. It is not, so no such f exists. The same argument with ψa(x)=φ∗((x−ξ)/a) shows δξ is singular for every ξ. □
Definition(The distributional derivative)AdaptedAdapted from Stakgold & Holst (2011), §2.1, Eqs. (2.1.12)–(2.1.14), p. 101; Strauss (2008), §12.1, Eqs. (11)–(12), p. 334
Let f∈C1(R) with f′∈Lloc1, and let φ∈D(R). Integrating by parts and using φ=0 outside a bounded interval, which kills the boundary term,
∫−∞∞f′(x)φ(x)dx=−∫−∞∞f(x)φ′(x)dx.
The right-hand side makes sense for any distribution. So for f∈D′(R) the derivativef′ is defined by
⟨f′,φ⟩=−⟨f,φ′⟩,
and in n variables, for a multi-index k of order ∣k∣,
⟨∂if,φ⟩=−⟨f,∂iφ⟩,⟨Dkf,φ⟩=(−1)∣k∣⟨f,Dkφ⟩.
These are distributions: −φ′ and Dkφ are again test functions, the maps φ↦−φ′ and φ↦Dkφ are linear, and they send null sequences to null sequences — condition 2 in the definition of a null sequence covers every derivative, which is exactly why it was imposed. Hence every distribution is infinitely differentiable, and for f∈C1 the new derivative agrees with the old one by the displayed integration by parts. Derivatives commute, ∂i∂jf=∂j∂if, because they do on test functions.
Weak limits may be differentiated freely: if fN→f weakly then fN′→f′ weakly, since ⟨fN′,φ⟩=−⟨fN,φ′⟩→−⟨f,φ′⟩. Term-by-term differentiation of a Fourier series, illegitimate pointwise, is therefore always legitimate in D′ (Strauss, §12.1, Examples 8–9).
Definition(The jump formula)AdaptedAdapted from Stakgold & Holst (2011), §2.1, Eq. (2.1.15) and Example 11, Eq. (2.1.17), pp. 102–103
Let f:R→C be locally integrable, smooth except at finitely many points a1<⋯<aN, and such that at each ai the one-sided limits of f and of all its derivatives exist. Write [f′] for the pointwise derivative computed on each smooth piece and ignoring the jumps, and
Δfi=f(ai+)−f(ai−)
for the jump of f at ai. Then, in D′(R),
f′=[f′]+i=1∑NΔfiδ(x−ai).
Each simple jump contributes a delta of weight equal to that jump. With f=H, which has the single jump Δf=1 at 0 and [H′]=0, this is H′=δ. With f(x)=∣x∣, which has no jump, it gives ∣x∣′=sgnx. Iterating, if f is smooth except at the single point 0 and Δf(m) denotes the jump of the m-th pointwise derivative there,
f′′=[f′′]+Δf(1)δ+Δf(0)δ′,f(m)=[f(m)]+j=0∑m−1Δf(m−1−j)δ(j).Proof(The jump formula)AdaptedAdapted from Stakgold & Holst (2011), §2.1, Example 11, Eqs. (2.1.16)–(2.1.17), pp. 102–103
The route is Stakgold and Holst's: subtract the steps to get a continuous function, derive the formula there, and put the steps back. Step 2, which supplies the integration by parts they assert without proof, is ours.
Step 1: reduce to a continuous function. Set
g=f−i=1∑NΔfiH(x−ai).
Each subtracted term jumps by exactly Δfi at ai and nowhere else, so g has no jumps: it is continuous, and smooth away from the ai, with a piecewise continuous pointwise derivative [g′]=[f′] (the step functions contribute nothing to the pointwise derivative away from their jumps).
Step 2: for such a g the two derivatives agree. Let φ∈D(R) have support in (−R,R) and put a0=−R, aN+1=R. On each closed subinterval [ai,ai+1] the function g is C1, so ordinary integration by parts applies:
The boundary terms telescope. At the two outer ends φ(±R)=0; at each interior ai the two contributions are g(ai−)φ(ai) and −g(ai+)φ(ai), which cancel because g is continuous, by Step 1. Hence ⟨g′,φ⟩=∫[g′]φdx, that is, g′=[g′]=[f′] as distributions.
Step 3: differentiate the steps. For a single step, with φ supported in (−R,R),
using the fundamental theorem of calculus on [a,R] and φ(R)=0. So H(⋅−a)′=δ(⋅−a).
Step 4: assemble. Differentiation is linear on D′, so from Step 1,
f′=g′+i∑ΔfiH(x−ai)′=[f′]+i∑Δfiδ(x−ai).
The higher-derivative formulas follow by applying the same result to [f(m)] repeatedly. □
Definition(Distributions with vanishing derivative)AdaptedAdapted from Stakgold & Holst (2011), §2.5, Lemmas 2.5.1–2.5.2 and pp. 164–166
Let u∈D′(R) satisfy u′=0, that is, ⟨u,φ′⟩=0 for every φ∈D(R). Then u is a constant: there is c∈C with ⟨u,φ⟩=c∫−∞∞φ(x)dx for all φ. Conversely every constant has zero derivative.
Consequently the equation u′=f has, for each f∈D′(R), a solution in D′(R) unique up to an additive constant: every distribution on the line has an antiderivative. Stakgold and Holst construct the particular solution explicitly.
Idea of the proofOriginal
A test function is the derivative of another test function exactly when its total integral vanishes — one direction is the fundamental theorem of calculus, the other is that an antiderivative of φ is only compactly supported if the net area is zero. So the test functions split: fix one φ0 with ∫φ0=1, and every φ is (∫φ)φ0 plus something of zero total integral, hence plus a derivative. The hypothesis u′=0 says u annihilates derivatives, so all it can see of φ is the one number ∫φ — and a functional of that form is a constant.
Proof(Distributions with vanishing derivative)AdaptedAdapted from Stakgold & Holst (2011), §2.5, Lemmas 2.5.1–2.5.2 and p. 165
Step 1: which test functions are derivatives. Let M={ψ′:ψ∈D(R)}. If φ=ψ′ with ψ∈D, then ∫−∞∞φ=ψ(∞)−ψ(−∞)=0. Conversely let φ∈D with ∫φ=0 and set χ(x)=∫−∞xφ(s)ds. Then χ is smooth with χ′=φ; it vanishes for x below the support of φ, and for x above the support it equals ∫−∞∞φ=0 — this is where the vanishing total integral is used. So χ has compact support, χ∈D and φ∈M. Hence
M={φ∈D(R):∫−∞∞φ=0}.
Step 2: split a test function. Fix once and for all a φ0∈D(R) with ∫φ0=1; for instance a rescaled φ∗. Given any φ∈D, put α=∫−∞∞φ(x)dx and ψ=φ−αφ0. Then ψ∈D and ∫ψ=α−α=0, so ψ∈M by Step 1, and
φ=αφ0+ψ,ψ∈M.
Step 3: evaluate u. Pair u with this decomposition. Since ψ∈M, say ψ=χ′, the hypothesis gives ⟨u,ψ⟩=⟨u,χ′⟩=0. Therefore
⟨u,φ⟩=α⟨u,φ0⟩=c∫−∞∞φ(x)dx,c:=⟨u,φ0⟩,
which is the statement that u is the constant c. Note that c does not depend on φ, only on the fixed φ0. Conversely, for a constant c, ⟨c,φ′⟩=c∫φ′=0, so constants do solve u′=0; and if u1′=u2′=f then (u1−u2)′=0, giving the uniqueness-up-to-a-constant claim. □
Remark(Where the hypotheses are used)Original
Compact support of test functions does two jobs. It kills the boundary term in the integration by parts that defines the derivative, and it makes ∫Ω∣f∣<∞ enough for continuity in part (1) of the theorem. Drop it and the derivative of a distribution is no longer defined by the same formula; the repair is to keep a decay condition instead, which is what Fourier transforms of distributions does.
Smoothness of test functions, to all orders, is what makes every distribution infinitely differentiable. If the test space were C1 only, φ↦−φ′ would leave it and f′′ would be undefined. It is also why continuity is imposed on all derivatives at once: without clause 2 of the null-sequence definition, f↦f′ need not map distributions to distributions.
Local integrability, not integrability, is the right hypothesis in part (1): f(x)=ex2 is a perfectly good distribution on R, because a test function cuts off the growth. What fails is a singularity that is too strong: 1/x on R is not locally integrable and is not a distribution by the formula ∫fφ; it has to be replaced by its principal value (Principal values and indented contours, and Fourier transforms of distributions for the distribution pv(1/x)).
Continuity of g in Step 2 of the jump proof is the whole content of that proof. Without Step 1 the boundary terms at the ai do not cancel and exactly reproduce the missing deltas. Discarding them is the standard error of writing H′=0 "because H is constant on each side": then ∫0xH′=0=H(x), and the fundamental theorem of calculus fails. With H′=δ it holds.
Products are not available.D′ is a vector space and C∞⋅D′⊂D′, but there is no product of two distributions: δ2 is meaningless, and so is δ(x)H(x). Every identity below multiplies a distribution by a smooth function only.
Finitely many jumps in the jump formula keeps the sum finite on the support of any test function. The formula survives for a locally finite set of jumps, since only finitely many lie in suppφ; it fails for a function with a dense set of jumps.
Method(Computing a distributional derivative)Original
To differentiate a function that is piecewise smooth but not differentiable:
Locate the bad points. List the points ai where f or a derivative of f fails to exist, and check that the one-sided limits exist there. Check also that f is locally integrable — a non-integrable singularity must be handled separately.
Differentiate pointwise, piece by piece, ignoring the bad points. This gives [f′].
Measure each jump as Δfi=f(ai+)−f(ai−), from the right minus from the left.
Add the deltas: f′=[f′]+∑iΔfiδ(x−ai).
Iterate for higher derivatives, applying steps 2–4 to [f′], and remembering that the jumps of [f′] generate deltas while the jumps of f itself now generate δ′ terms.
Check by pairing with a test function that equals 1 on a region containing all the ai: then ⟨f′,φ⟩=−∫fφ′ only picks up f where φ turns off, which is usually a one-line evaluation.
Example(A function with two jumps)Original
Let
f(x)=⎩⎨⎧0,cosx,2,x<0,0≤x<π/2,x≥π/2.
Step 1.f is bounded and smooth except at a1=0 and a2=π/2, with one-sided limits everywhere, and it is locally integrable.
Step 2. The piecewise derivative is [f′](x)=−sinx on (0,π/2) and 0 elsewhere.
Step 3. The jumps are
Δf1=cos0−0=1,Δf2=2−cos2π=2.
Step 4. Therefore
f′(x)=−sinx[H(x)−H(x−2π)]+δ(x)+2δ(x−2π).
Step 6: the check, with a number. Let φ be any test function with φ≡1 on [−1,2] and suppφ⊂(−2,3); such a φ exists — convolve the indicator function of [−1.5,2.5] with a normalised bump of width 21. Then φ′=0 on [−1,2], so
Both routes give 2. Numerical quadrature against the explicit φ just described reproduces 2.0000000 on both sides.
Differentiating once more: [f′′]=−cosx on (0,π/2), the jump of [f′] is −sin0−0=0 at x=0 and 0−(−sin2π)=1 at x=π/2, and the jumps of f now ride on δ′:
f′′(x)=−cosx[H(x)−H(x−2π)]+δ(x−2π)+δ′(x)+2δ′(x−2π).Application(Point sources and weak solutions)AdaptedAdapted from Strauss (2008), §12.1, Example 10 and the definition of a weak solution, pp. 336–337; Hassani (2013), §7.4, Example 7.4.3, Eq. (7.27), p. 234
Idealised sources. A point charge q at r0 has charge density ρ=qδ(r−r0), which as a distribution is ⟨ρ,φ⟩=qφ(r0) — Hassani's Eq. (7.27). This is not an approximation to a physical density but a different kind of object: ρ has no point values, only integrated ones, and that is exactly the information a measurement of total charge in a region returns. The same device gives surface layers and line charges, which are distributions supported on a surface or a curve rather than at a point.
Weak solutions. A differential equation can be imposed on a distribution by moving the derivatives onto the test function. For the wave equation utt=c2uxx, Strauss calls u∈D′(R2) a weak solution if
⟨u,φtt−c2φxx⟩=0for every φ∈D(R2),
the operator ∂t2−c2∂x2 being its own formal adjoint. This settles a real difficulty. The plucked string has a corner in its initial shape, so the classical solution u=21ϕ(x+ct)+21ϕ(x−ct) of The one-dimensional wave equation is not twice differentiable, and uxx contains deltas travelling along the characteristics; but utt−c2uxx=0 holds in D′, so the physics is intact. Even the jump u=H(x−ct) is a weak solution. Weak solutions are the entry point to the modern theory of partial differential equations (Weak derivatives and Sobolev spaces), where one first finds a distributional solution and then asks how smooth it is.
Green's functions. The defining property of a Green's function, LG=δ, is a statement in D′ and nothing else: G is typically singular where the source sits, and LG has no pointwise meaning there. The Dirac delta in calculations computes the basic instance ∇2(1/r)=−4πδ(3), and the whole of Green's function for a boundary-value problem rests on it.
Introduces the Schwartz space and the tempered distributions on it, defines the Fourier transform by duality and shows it is a bijection of that space obeying the usual operational rules, gives four routes for computing a transform, and derives the pairs for the delta, constants and powers, the Heaviside step and the principal value of 1/x, ending with a distributional solution of an ordinary differential equation.
The Fourier transform cannot act on D(Rn)=C0∞(Rn), because it destroys compact support: Folland notes (p. 293) that if φ∈C0∞ and φ^ vanishes on a non-empty open set, then φ=0. The repair is to trade compact support for rapid decay.
The Schwartz spaceS=S(Rn) is the set of infinitely differentiable φ:Rn→C for which every seminorm
∥φ∥α,β=x∈RnsupxαDβφ(x)
is finite, α and β running over all multi-indices. Equivalently, φ and all its derivatives decay faster than every inverse power of ∣x∣. Reed and Simon call these the functions of rapid decrease; Stakgold and Holst write C↓∞(R) in one variable and define convergence by maxx∣xkφm(l)(x)∣→0 for every k and l, which is convergence in all the seminorms. Every φ∈D lies in S, because a smooth function vanishing outside a bounded set has each xαDβφ continuous and compactly supported, hence bounded; the inclusion is strict, since e−x2 lies in S and not in D. Conversely D is dense in S (Folland, Proposition 9.9).
The tempered distributions are those functionals on S that are linear and continuous; write S′(Rn) for the collection of them. Continuity has a concrete form. Reed and Simon first bundle the seminorms into the directed family
∥φ∥k,m=∣α∣≤k,∣β∣≤mmax∥φ∥α,β,k,m=0,1,2,…,
and then record (p. 134, from their Theorem V.2) that T∈S′ if and only if there are a constant C and indices k,m with
∣⟨T,φ⟩∣≤C∥φ∥k,mfor all φ∈S.
Replacing the maximum by the sum over the same finite index set alters only the constant, so a bound by a finite sum of seminorms serves just as well, and that is how the criterion is verified below.
Because D⊂S with continuous inclusion and D is dense, restricting a tempered distribution to D gives a distribution, and the restriction determines it. So S′ is exactly the set of distributions of Test functions and distributions that extend continuously to S — a smaller space, since there are fewer functionals on a bigger test space. Stakgold and Holst call its elements distributions of slow growth.
Which ordinary functions qualify? A locally integrable f with
∣f(x)∣≤C∣x∣Nfor ∣x∣>R
for some constants is a function of slow growth (Stakgold and Holst, p. 153), and it is tempered: the proof below gives the required seminorm bound. So every polynomial, every bounded measurable function, every Lp function (Reed and Simon, Example 2) and δ itself are tempered. On the other hand ex on R is a distribution but not a tempered one, as Folland shows on p. 294 by testing it against e−xψ(x−j); rapid decay of the test function is not enough to tame exponential growth. The borderline is subtler than mere size: Folland's excosexis tempered, since differentiating sinex produces it, and sinex stays between −1 and 1, so one integration by parts moves the growth onto the test function, where it is harmless. Rapid oscillation can cancel large size.
Multiplication is restricted. φ↦aφ maps S to S, and hence aT makes sense for T∈S′, precisely when a∈C∞ and a together with all its derivatives grows at most polynomially (Folland, p. 294; Reed and Simon's class OM, p. 137). Such an a is called slowly increasing. Every polynomial is slowly increasing; ex2 is not.
Two tempered distributions recur below.
The principal value.pvx1 is defined on S(R) by
⟨pvx1,φ⟩=ε→0+lim∫∣x∣>εxφ(x)dx=∫0∞xφ(x)−φ(−x)dx.
The second form, which is Reed and Simon's Example 6, shows the limit exists: the integrand tends to 2φ′(0) as x→0 and is O(x−1∣φ∣) at infinity. The function 1/x itself is not locally integrable, so this is the only sense in which it is a distribution. It satisfies
x⋅pvx1=1
as distributions (Stakgold and Holst, p. 106), because xφ(x)/x=φ(x) leaves nothing to regularise. Stakgold and Holst write pf(1/x) for the same object.
δ and its derivatives.⟨δ,φ⟩=φ(0) and ⟨δ(n),φ⟩=(−1)nφ(n)(0) are bounded by ∥φ∥0,0 and ∥φ∥0,n, so they are tempered.
the asymmetric convention with the whole 2π on the inversion. The cited books differ, and the transforms of δ, 1 and H differ with them, so the convention must be checked before a formula is copied: Stakgold and Holst use f∧(u)=∫f(x)e+iuxdx with 1/2π on the inverse, so their u is our −k; Reed and Simon use the symmetric (2π)−n/2∫e−iλ⋅xf(x)dx; Folland uses ∫f(x)e−2πixξdx, which carries no 2π prefactor on the transform or on its inversion; 2π still appears inside the formulas, as in his (∂αF)∧=(2πiξ)αF^ and in H^(ξ)=21δ(ξ)−2πipv(1/ξ), which is his Exercise 19e (Folland (1999), §9.3, Exercise 19c–e, p. 299) rather than a displayed formula. The Fourier transform tabulates the conversions.
Translating to that convention is a rescaling of the transform variable, k=2πξ, not multiplication by a fixed factor, and how much a given entry moves depends on how homogeneous it is. A distribution homogeneous of degree −1 acquires a factor (2π)−1, since δ(2πξ)=(2π)−1δ(ξ) and pv2πξ1=(2π)−1pvξ1; one homogeneous of degree 0 acquires nothing, since sgn(2πξ)=sgnξ. So our 1^=2πδ(k) becomes δ(ξ) and loses its 2π, while our pv(1/x)=−iπsgnk becomes −iπsgnξ and loses none: no single factor rescales the whole table. Within one entry, however, the scaling can be uniform, and for H^ it is — both of its terms have degree −1, so both are divided by 2π, and πδ(k)−ipv(1/k) goes over into precisely the 21δ(ξ)−2πipv(1/ξ) quoted above. A complex conjugation exchanges the sign conventions.
Definition(The Fourier transform of a tempered distribution)AdaptedAdapted from Stakgold & Holst (2011), §2.4, Eq. (2.4.30), p. 157; Reed & Simon (1980), §IX.1, the definition of $\hat T$ and its consistency with the transform on $\mathcal{S}$, p. 322; Folland (1999), §9.2, p. 295
that is ⟨f^,φ⟩=⟨f,φ^⟩. Note that no conjugation appears: this is the bilinear pairing, not an inner product. The right-hand side needs only that φ^ be a legitimate test function, which it is, because the Fourier transform maps S into S (Fourier inversion). So for any F∈S′(Rn)define
⟨F^,φ⟩=⟨F,φ^⟩,φ∈S,
and likewise ⟨F∨,φ⟩=⟨F,φ∨⟩ for the inverse transform φ∨(x)=(2π)−1∫φ(k)eikxdk. This is the only possible definition if the transform is to extend the classical one and stay weakly continuous.
F^ is again a tempered distribution, and F↦F^ is a one-to-one linear map of S′(Rn) onto itself with inverse G↦G∨. It is weakly continuous, and no other weakly continuous map of S′ agrees with the classical transform on S (Reed and Simon, Theorem IX.2).
It agrees with the classical transform on L1: for f∈L1(Rn) the distributional transform of f is the distribution of the continuous function f^. It also agrees with the Plancherel transform on L2, but that is a separate fact and does not follow from the L1 case: L2⊂L1, and the L2 transform is itself defined by extension from L1∩L2 (Plancherel's theorem). Taking the two together, the distributional transform agrees with the classical one on L1+L2.
The operational rules of The Fourier transform hold unchanged. In one variable, for F∈S′(R) and a∈R, n∈N:
F(n)(k)=(ik)nF^(k),xnF(k)=indkndnF^(k),F(x−a)(k)=e−ikaF^(k),eiaxF(k)=F^(k−a),F(−x)(k)=F^(−k),F^^(x)=2πF(−x).
The first two involve multiplication of a tempered distribution by the slowly increasing function (ik)n or xn, which is legitimate; a product of two distributions never occurs.
Idea of the proofOriginal
Every construction here is the same move: an operation on distributions is defined by pushing it onto the test function, and it is the right definition when it reproduces the classical one on the functions where both make sense. Integration by parts does this for the derivative; the multiplication formula does it for the Fourier transform. The one thing to check each time is that the operation pushed onto the test function keeps it inside the test space — and that is exactly what rapid decay buys and compact support does not. Bijectivity is then free: F^^ on S′ inherits the inversion theorem from S, one line at a time.
The integral is finite: near the origin because f is locally integrable, and at infinity because the integrand is O(∣x∣N−2p) with N−2p<−1. The supremum is a finite combination of the seminorms ∥φ∥α,0 with α≤2p. That is the continuity criterion, so f∈S′. □
(1) F^ is a tempered distribution.φ↦φ^ is linear and maps S continuously into S, by the theorem of Fourier inversion (Stakgold and Holst prove it as a theorem on p. 156, by differentiating and integrating by parts under the integral sign; Reed and Simon as the lemma preceding their Theorem IX.1). So φ↦⟨F,φ^⟩ is a composition of a continuous linear map with a continuous linear functional, hence a continuous linear functional: F^∈S′.
For bijectivity, use the inversion theorem on S: (φ^)∨=φ and (φ∨)∧=φ. Then for every φ∈S,
⟨(F^)∨,φ⟩=⟨F^,φ∨⟩=⟨F,(φ∨)∧⟩=⟨F,φ⟩,
so (F^)∨=F, and symmetrically G∨=G. Hence the transform is a bijection of S′ with inverse ∨. Weak continuity is immediate: if ⟨Fj,ψ⟩→⟨F,ψ⟩ for all ψ∈S, then in particular for ψ=φ^, which says ⟨F^j,φ⟩→⟨F^,φ⟩. Uniqueness of the extension holds because S is weakly dense in S′.
(2) Consistency. If f∈L1 then f^ in the classical sense is a bounded continuous function, hence tempered, and the multiplication formula displayed in the definition says precisely that the classical f^ and the distributional f^ have the same action on every φ∈S. They are therefore equal.
(3) The rules. Each is a two-line calculation in which the corresponding rule for test functions is used once. For the derivative: differentiating φ^(x)=∫φ(k)e−ikxdk under the integral sign gives (φ^)′(x)=(−ikφ)(x), and therefore
where the definition of the distributional derivative was used at the second step and the definition of F^ at the first and fourth. For multiplication by x: integrating ∫φ(k)i∂k(e−ikx)dk by parts gives xφ^(x)=(−iφ′)(x), so
The shift and reflection rules come the same way from φ(x−a) and φ^(−x), and F^^(x)=2πF(−x) from φ^^(x)=2πφ(−x), which is the inversion theorem rearranged. Iterating gives the powers n. □
Method(Distributions killed by multiplication by k)AdaptedAdapted from Stakgold & Holst (2011), §2.4, p. 160, and the cited Exercise 2.5.11(c)
Let f∈S′(R) satisfy kf=0. Then f=Cδ for some constant C. Consequently the general solution of kf(k)=g(k), g given, is one particular solution plus Cδ.
Proof(Distributions killed by multiplication by k)Original
Step 1: f annihilates test functions that vanish at the origin. Let ψ∈S with ψ(0)=0 and set θ(k)=ψ(k)/k for k=0, θ(0)=ψ′(0). Then θ(k)=∫01ψ′(tk)dt, which is smooth in k, and ψ=kθ. For the seminorms, split the line at ∣k∣=1.
On ∣k∣≤1 nothing but continuity is needed: θ and each derivative Dβθ, obtained by differentiating ∫01ψ′(tk)dt under the integral sign, are continuous, hence bounded on that compact interval, and ∣kα∣≤1 there.
On ∣k∣≥1 use θ=ψ/k and the Leibniz rule: Dβθ=∑j=0βcβjk−1−jDβ−jψ with constants cβj depending only on β and j. So ∣kαDβθ∣ is a finite combination of the quantities ∣kα−1−jDβ−jψ∣ for 0≤j≤β, and each of those is at most ∣kαDβ−jψ∣≤∥ψ∥α,β−j, because α−1−j≤α and ∣k∣≥1.
Every seminorm of θ is therefore finite, so θ∈S, whence
⟨f,ψ⟩=⟨f,kθ⟩=⟨kf,θ⟩=0
by the hypothesis.
Step 2: split a general test function. Fix χ∈S with χ(0)=1, say χ(k)=e−k2. For any φ∈S write
φ=φ(0)χ+(φ−φ(0)χ),
and note the bracket vanishes at k=0 and lies in S. By Step 1 it is annihilated by f, so
⟨f,φ⟩=φ(0)⟨f,χ⟩=Cφ(0)=⟨Cδ,φ⟩,C:=⟨f,χ⟩.
Since φ was arbitrary, f=Cδ. The constant is independent of φ; it does depend on the choice of χ only apparently, because the left-hand side does not. □
Method(Computing the transform of a tempered distribution)AdaptedAdapted from Stakgold & Holst (2011), §2.4, Examples 5–8, pp. 159–161
Four routes, in increasing order of effort; Stakgold and Holst use routes 1, 3 and 4 on the Heaviside function alone.
Straight from the definition. Compute ⟨F,φ^⟩ and recognise the answer as the action of a known distribution on φ. This is immediate for δ, where ⟨δ,φ^⟩=φ^(0)=∫φ.
From a known pair, by the operational rules. If F is a derivative, a translate, a reflection or a polynomial multiple of something whose transform is known, read the answer off rule (3) of the statement. This is how the transforms of δ(n), δ(x−a) and xn are obtained from those of δ and 1.
Solve an algebraic equation. Transform a differential relation satisfied by F and solve the resulting equation in S′. The equation is algebraic, but its solution need not be unique, so the homogeneous solutions must be found and then fixed by a further condition on F — a symmetry, a known total integral, or a known value. The lemma just proved supplies the homogeneous solutions.
As a weak limit. Write F=limε→0+fε weakly with each fε an ordinary integrable function, transform classically, and take the weak limit of f^ε. This is legitimate because the transform is weakly continuous, by part (1) of the statement. It is the most useful route in practice, because the limit ε→0+ carries the physics: ε is an adiabatic switching rate, a damping, or a line width.
Method(The basic transform pairs)AdaptedAdapted from Stakgold & Holst (2011), §2.4, Eqs. (2.4.36)–(2.4.40a), Examples 5–8, pp. 159–161
In the convention f^(k)=∫f(x)e−ikxdx, with a∈R and n=0,1,2,…:
f
f^
δ(x)
1
δ(x−a)
e−ika
δ(n)(x)
(ik)n
1
2πδ(k)
xn
2πinδ(n)(k)
eiax
2πδ(k−a)
cosax
π[δ(k−a)+δ(k+a)]
sinax
−iπ[δ(k−a)−δ(k+a)]
pvx1
−iπsgnk
sgnx
−2ipvk1
H(x)
πδ(k)−ipvk1
Stakgold and Holst's Eqs. (2.4.36), (2.4.37) and (2.4.40a) are the entries for δ, for 1 and for H in their sign convention, which carries k↦−k on the right-hand column. The Heaviside line is worth checking explicitly, since it is the only entry where the substitution does visible work. Their (2.4.40a) reads H∧(u)=ipf(1/u)+πδ(u); putting u=−k and using that pf(1/⋅) is odd and δ even,
ipfu1u=−k=ipf−k1=−ipvk1,δ(u)u=−k=δ(k),
which is the table's πδ(k)−ipv(1/k). The pair δ↔1 is the precise content of the formal identity δ(x)=(2π)−1∫eikxdk used throughout physics (The Dirac delta in calculations): the integral diverges pointwise, but it is the inversion formula applied to δ^=1, and it is correct in S′.
Proof(The transform pairs)AdaptedAdapted from Stakgold & Holst (2011), §2.4, Examples 5–8, pp. 159–161
δ^=1 (route 1).⟨δ^,φ⟩=⟨δ,φ^⟩=φ^(0)=∫φ(x)dx=⟨1,φ⟩. The translate and the derivatives follow from the shift rule and δ(n)=(ik)nδ^=(ik)n (route 2).
1^=2πδ (route 2). Apply F^^(x)=2πF(−x) to F=δ: 1^=δ^^, and δ^^(k)=2πδ(−k)=2πδ(k), using that δ is even. Then xn=in(d/dk)n1^=2πinδ(n)(k), and eiax=1^(k−a)=2πδ(k−a) because multiplying by eiax shifts the transform; the cosine and sine lines are linear combinations of the last.
pv(1/x)=−iπsgnk (route 1, by hand). For φ∈S, the odd form of the principal value gives
The inner bracket is −2isinkx. The two integrations may not simply be exchanged. Over (0,∞)×R the double integral is not absolutely convergent: ∫∣φ(k)∣2∣sinkx∣dk does not tend to 0 as x→∞ — the oscillation of sinkx decorrelates from φ and the integral tends to (4/π)∥φ∥1 — and ∫∞dx/x of a quantity bounded away from 0 diverges. The singularity at x=0, which is the only thing a bound ∣sinkx∣/x≤∣k∣ settles, is not where the difficulty is.
So truncate first. Over the strip (0,R)×R the double integral is absolutely convergent, for every finite R: on 0<x≤1 the factor 2∣sinkx∣/x is at most 2∣k∣, so the inner integral is at most 2∫∣kφ(k)∣dk, finite because φ∈S; on 1≤x≤R it is at most 2/x≤2, so the inner integral is at most 2∥φ∥1. Fubini's theorem (Fubini and Tonelli theorems) applies there and gives
Now let R→∞. The left-hand side tends to the full integral, which converges absolutely because φ^∈S. On the right, substituting u=∣k∣x gives DR(k)=sgn(k)Si(∣k∣R) with Si(t)=∫0tu−1sinudu, and Si is non-negative on [0,∞) with its largest value at its first maximum, Si(π)=1.8519. So ∣φ(k)DR(k)∣≤Si(π)∣φ(k)∣, an integrable bound independent of R, while DR(k)→2πsgnk for every k by the Dirichlet integral ∫0∞x−1sinkxdx=2πsgnk (and trivially at k=0, where both sides vanish). Dominated convergence therefore gives
⟨pvx1,φ⟩=−2i⋅2π∫−∞∞φ(k)sgnkdk,
which is ⟨−iπsgnk,φ⟩. The conditional convergence of the Dirichlet integral is exactly what the truncation handles: DR converges pointwise but not absolutely, and it is the uniform bound, not absolute convergence, that licenses the passage to the limit.
sgn=−2ipv(1/k) (route 2). Transform the previous pair a second time and use F^^(x)=2πF(−x) with F=pv(1/x), which is odd:
−iπsgn(x)=(pvx1)(x)=2πpv−x1=−2πpvx1.
Dividing by the constant −iπ and relabelling the variable, sgn(k)=(2/i)pv(1/k)=−2ipv(1/k).
One solution is −ipv(1/k), because k⋅pv(1/k)=1. By the lemma above, every solution is
H^(k)=−ipvk1+Cδ(k).
To fix C, use H(x)+H(−x)=1 for x=0, hence as distributions. Transforming with the reflection rule, H^(k)+H^(−k)=2πδ(k). The principal value is odd, so its two copies cancel, and 2Cδ(k)=2πδ(k). Thus C=π and H^(k)=πδ(k)−ipv(1/k).
Route 4.H(x)e−εx→H(x) weakly as ε→0+, and each is in L1 with the classical transform
∫0∞e−εxe−ikxdx=ε+ik1=ε2+k2ε−iε2+k2k.
The first term is a Lorentzian whose total integral is the same for every ε, namely ∫−∞∞ε(ε2+k2)−1dk=π — not 1, since the unit-mass Lorentzian is π−1ε/(ε2+k2) — so it tends weakly to πδ(k) (Delta-function identities). For the second, k/(ε2+k2) is odd and tends to pv(1/k): pairing with φ and using oddness,
the convergence being dominated because ∣φ(k)−φ(−k)∣≤2ksup∣φ′∣ bounds the integrand uniformly in ε near k=0. So H^(k)=πδ(k)−ipv(1/k) again, now with C=π obtained rather than fitted. Since sgn=2H−1, this gives sgn=2πδ−2ipv(1/k)−2πδ=−2ipv(1/k), consistent with the entry above. □
Remark(Where the hypotheses are used)Original
Rapid decay, not compact support. The definition ⟨F^,φ⟩=⟨F,φ^⟩ requires φ^ to be in the test space. On D it is not, by Folland's observation quoted above, so the transform of a general distribution in D′ is not defined. This is the whole reason the Schwartz space exists.
Temperedness restricts what can be transformed.ex has no Fourier transform in S′. Taking φ^ of a rapidly decreasing φ cannot beat exponential growth, and no amount of regularisation fixes this inside S′; the usual response is a Laplace transform in a half-plane (The Laplace transform) rather than a Fourier transform.
Slow increase of the multiplier. Rules (3) multiply F^ by (ik)n and F by xn, both slowly increasing, so the products stay in S′. The analogous step with ek2 is illegal. The convolution theorem (The convolution theorem) is similarly restricted: F∗ψ is defined for F∈S′ and ψ∈S, and F∗ψ=ψ^F^ holds there, but two tempered distributions can be neither multiplied nor convolved in general.
Non-uniqueness in route 3. The lemma is needed exactly because kH^=−i does not determine H^. Forgetting the Cδ(k) loses the term that makes H^ fail to be odd, and with it the whole real part of H^ — in physical terms, the time-averaged value of a switched-on signal. Fixing C needs one extra piece of information about H, and the identity H(x)+H(−x)=1 is the cheapest.
1/x is not a distribution; pv(1/x) is. The limit defining the principal value exists because the odd part of φ supplies a factor x that cancels the singularity. For an even test function the symmetric-interval cancellation is doing all the work; for a one-sided cutoff the integral really does diverge. The same device at higher order needs Hadamard finite parts, which are not treated here.
Example(Solving u′′−u=δ in S′)Original
Find all u∈S′(R) with u′′−u=δ. This is the Green's function equation for the operator d2/dx2−1 on the line.
Step 1: transform. By rule (3) with n=2, u′′=(ik)2u^=−k2u^, and δ^=1, so
−k2u^(k)−u^(k)=1,that is−(1+k2)u^(k)=1.
Step 2: divide, legitimately. The function −(1+k2)−1 is smooth, bounded, and all its derivatives are bounded, so it is slowly increasing and may multiply a tempered distribution. Moreover 1+k2 never vanishes, so — unlike the Heaviside calculation — there is no homogeneous solution to add: if (1+k2)v=0 then v=(1+k2)−1(1+k2)v=0. Hence the solution is unique:
u^(k)=−1+k21.
Step 3: invert. From the standard pair F[e−a∣x∣](k)=2a/(a2+k2) for a>0 (The Fourier transform), taking a=1 gives F[e−∣x∣]=2/(1+k2), so
u(x)=−21e−∣x∣,u(0)=−21.
Step 4: check directly, by the jump formula.u is continuous with Δu=0 at the origin, and [u′](x)=21sgn(x)e−∣x∣, which jumps by 21−(−21)=1 there. So u′=[u′] and
u′′=[u′′]+1⋅δ=−21e−∣x∣+δ=u+δ,
since [u′′](x)=−21e−∣x∣ away from 0. That is u′′−u=δ. Pairing numerically with a smooth plateau function φ equal to 1 near the origin gives ⟨u,φ′′−φ⟩=1.0000000=φ(0), as it must.
The numbers to remember: the Green's function is −21e−∣x∣, its value at the source is −0.5, and its total integral is ∫u=u^(0)=−1.
Application(Switching, causality and dispersion relations)AdaptedAdapted from Reed & Simon (1980), §V.3, Eq. (V.4), p. 137
The iε prescription. Routes 3 and 4 above produce the same object in two guises, and comparing them gives the identity Reed and Simon record as their Eq. (V.4): in S′,
ε→0+limx±iε1=pvx1∓iπδ(x),
known in physics as the Sokhotski–Plemelj formula. Its derivation is the decomposition used in route 4: (x±iε)−1=x/(x2+ε2)∓iε/(x2+ε2), whose two terms converge weakly to pv(1/x) and ∓iπδ(x). This is why a physicist's ε is never a typographical decoration: the sign chosen in E−En±iε selects a boundary condition, and the delta term it generates is a real physical rate. In the Fermi golden rule, the δ is the energy-conserving factor and the principal value is the level shift.
Causal response functions. A linear system with impulse response G is causal when G(t)=H(t)g(t), so G^=(2π)−1H^∗g^ with H^=πδ−ipv(1/k) — the (2π)−1 is the one this unit's convention puts on the transform of a product, the dual of the factor-free F∗ψ=ψ^F^. Written out, G^(ω)=21g^(ω)−2πipv∫g^(ω′)/(ω−ω′)dω′. The principal-value half forces a relation between the real and imaginary parts of the response, which is the Kramers–Kronig relation of Dispersion relations: absorption and refraction in a medium are not independent, because the medium cannot respond before it is driven. The same calculation, read in the time domain, builds the retarded Green's functions of Causal Green's functions for initial-value problems.
Plane waves and lattices. The pair eiax↔2πδ(k−a) says a plane wave has a sharp wavenumber — not approximately, but as an identity between tempered distributions. Without S′ the statement is false, since eiax∈/L1∪L2 and has no classical transform. The corresponding statement for a periodic array of point sources, ∑nδ(x−nL), has transform (2π/L)∑mδ(k−2πm/L), which is the reciprocal lattice of a crystal and is Poisson summation formula in distributional form.
Collects the working identities for the Dirac delta — scaling, evenness, multiplication by a smooth function, composition with a function, the derivative rules, the transformation of the multidimensional delta under a change of coordinates, and the standard representations as limits — each with a proof in the distributional sense and a statement of the hypothesis it needs.
Note(What an identity between deltas means)Original
Every line below is an identity in D′(Rn): the two sides give the same number when paired with every test function. None of them is an identity between functions, and checking one pointwise is meaningless. The Dirac delta in calculations states most of these rules and derives them from delta sequences, which is how a physicist meets them; here each is proved from the definitions of Test functions and distributions, the hypotheses are made explicit, and the transformation of the multidimensional delta under a change of coordinates is derived.
Two conventions are fixed throughout. For a distribution f and real a=0, the translate and the rescaling are defined by
these being what the change of variable gives when f happens to be a function (Stakgold and Holst's Eqs. (2.1.7) and (2.1.8); the modulus and the reversal of the limits for a<0 are built in). And for a∈C∞, ⟨af,φ⟩=⟨f,aφ⟩.
Reference table(One-dimensional identities)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.11, Eqs. (1.158)–(1.161) and Exercise 1.11.6, pp. 78–81; Stakgold & Holst (2011), §2.1, Eqs. (2.1.8) and (2.1.11a), pp. 99–100
Let φ be a test function, b∈R, a∈R∖{0}, n∈N, and let u∈C∞(R).
Identity
Hypothesis
⟨δ(x−b),φ⟩=φ(b)
—
δ(−x)=δ(x)
—
δ(ax)=∣a∣1δ(x)
a=0
δ(a(x−b))=∣a∣1δ(x−b)
a=0
u(x)δ(x−b)=u(b)δ(x−b)
u∈C∞
xδ(x)=0
—
⟨δ(n)(x−b),φ⟩=(−1)nφ(n)(b)
—
δ(n)(−x)=(−1)nδ(n)(x)
—
δ(n)(ax)=∣a∣an1δ(n)(x)
a=0
xδ′(x)=−δ(x)
—
x2δ′′(x)=2δ(x)
—
u(x)δ′(x)=u(0)δ′(x)−u′(0)δ(x)
u∈C∞
δ(g(x))=i∑g′(ai)δ(x−ai)
g∈C1; the zeros ai of g in suppφ are finitely many and satisfy g′(ai)=0
The composition rule is Arfken's Eq. (1.160); the two displayed forms of it,
∫−∞∞φ(x)δ(g(x))dx=i∑∣g′(ai)∣φ(ai),
are what is used in practice. It contains the rescaling rule as the case g(x)=ax, and δ((x−x1)(x−x2))=[δ(x−x1)+δ(x−x2)]/∣x1−x2∣ for x1=x2 as another case.
Proof(One-dimensional identities)Original
Each is the definition of the operation, applied once.
Evenness and rescaling. By the convention above with a=−1, ⟨δ(−x),φ⟩=⟨δ,φ(−x)⟩=φ(−0)=φ(0), so δ(−x)=δ(x). For general a=0,
⟨δ(ax),φ⟩=∣a∣1⟨δ,φ(ax)⟩=∣a∣φ(0)=⟨∣a∣δ,φ⟩.
The shifted version follows by composing with a translation.
Multiplication.⟨uδ(⋅−b),φ⟩=⟨δ(⋅−b),uφ⟩=u(b)φ(b)=⟨u(b)δ(⋅−b),φ⟩. Taking u(x)=x, b=0 gives xδ(x)=0. Note that u must be smooth for uφ to be a test function; continuity at b is not enough to make uδ a distribution in general, though it is enough for this one identity.
Derivatives. Iterating ⟨f′,φ⟩=−⟨f,φ′⟩ gives ⟨δ(n)(⋅−b),φ⟩=(−1)nφ(n)(b). For the parity, ⟨δ(n)(−x),φ⟩=⟨δ(n),φ(−x)⟩=(−1)ndxndnφ(−x)x=0=(−1)n(−1)nφ(n)(0); comparing with ⟨δ(n),φ⟩=(−1)nφ(n)(0) gives the factor (−1)n. For the rescaling,
Products with powers.⟨xδ′,φ⟩=⟨δ′,xφ⟩=−(xφ)′(0)=−[φ(0)+0⋅φ′(0)]=−φ(0), so xδ′=−δ. Similarly ⟨x2δ′′,φ⟩=(x2φ)′′(0)=[2φ+4xφ′+x2φ′′]x=0=2φ(0), so x2δ′′=2δ. For a general smooth u, ⟨uδ′,φ⟩=−(uφ)′(0)=−u′(0)φ(0)−u(0)φ′(0)=⟨u(0)δ′−u′(0)δ,φ⟩. □
Reference table(Composition with a function)AdaptedAdapted from Arfken, Weber & Harris (2013), §1.11, Eq. (1.160), p. 79
Let ρ be continuous with support in [−1,1] and ∫−11ρ=1, and put ρε(t)=ε−1ρ(t/ε). Let g∈C1(R) and let φ be a test function such that the zeros of g in suppφ are a finite set a1,…,aN with g′(ai)=0 for each i. Then
and the limit does not depend on which ρ was used. This is the precise meaning of δ(g(x))=∑iδ(x−ai)/∣g′(ai)∣. Arfken obtains the same formula by replacing g near each zero by the leading term of its Taylor series.
Idea of the proofOriginal
A narrow spike in the variable t=g(x) becomes, in the variable x, a spike sitting at each solution of g(x)=0. Its height is unchanged but its width is divided by ∣g′∣, because that is the rate at which x must move to make t move; so each spike carries weight 1/∣g′(ai)∣ instead of 1. The proof is that sentence made into a substitution: near each simple zero, g is invertible, and the change of variable t=g(x) turns the spike back into the standard one, leaving behind the Jacobian factor 1/∣g′∣ evaluated at the zero.
Proof(Composition with a function)Original
Step 1: isolate the zeros. Since g′(ai)=0 and g′ is continuous, the inverse function theorem (Implicit and inverse function theorems) gives an open interval Ii∋ai on which g is a C1 bijection onto an open interval Ji∋0, with g′ of one sign on Ii. Shrink the Ii so that they are disjoint. The set K=suppφ∖⋃iIi is compact and contains no zero of g, so m=minK∣g∣>0.
Step 2: the remainder contributes nothing. For x∈K we have ∣g(x)/ε∣≥m/ε>1 once ε<m, and ρ vanishes outside [−1,1]. So ρε(g(x))=0 on K for all small ε: this is where the compact support of ρ is used, and it is the only place.
Step 3: change variables near each zero. On Ii substitute t=g(x), so x=gi−1(t) and dx=dt/∣g′(gi−1(t))∣ — the modulus absorbing the orientation when g′<0. Writing
Ψi(t)=g′(gi−1(t))φ(gi−1(t)),t∈Ji,
a continuous function on Ji with Ψi(0)=φ(ai)/∣g′(ai)∣, we get
∫Iiρε(g(x))φ(x)dx=∫Jiρε(t)Ψi(t)dt.
Step 4: let ε→0. For ε small enough that [−ε,ε]⊂Ji, substitute t=εs:
the limit by uniform convergence of Ψi(εs)→Ψi(0) for s∈[−1,1], which holds because Ψi is continuous at 0, and ∫ρ=1. Summing over i and adding the vanishing contribution of K from Step 2 gives the claim; the answer involves ρ only through ∫ρ=1, so it is independent of ρ. □
In Rn the delta at x0 is the functional φ↦φ(x0), and in Cartesian coordinates it factorises,
δ(n)(x−x0)=j=1∏nδ(xj−x0j),
which is Riley, Hobson and Bence's Eq. (13.21) in three dimensions. Let X:U→V be a C1 bijection between open subsets of Rn with Jacobian determinant
J(u)=det∂u∂X,J(u)=0 on U,
and let x0=X(u0) with u0∈U. Then, for every test function supported in V,
δ(n)(x−x0)=∣J(u0)∣1j=1∏nδ(uj−u0j),
meaning that the right-hand side, integrated against f(X(u))∣J(u)∣dnu, returns f(x0). Hassani writes the same thing with a weight w in place of ∣J∣. In the two standard systems of Cylindrical and spherical coordinates, where ∣J∣=r2sinθ and ∣J∣=ρ:
The first is the form behind Arfken's Eq. (1.162). For a delta at the origin paired with functions of r alone, the angular factors integrate to 4π and the correct expression is instead
δ(3)(r)=4πr2δ(r)on radial test functions,
since ∫0∞∫S2[δ(r)/4πr2]f(r)r2drdΩ=f(0). That last step uses the half-line convention∫0∞δ(r)f(r)dr=f(0), which assigns the whole unit mass to the endpoint r=0 rather than splitting it. This is exactly the boundary root that step 1 of the method below refuses to evaluate, so it is a convention and not a consequence of the identities above; it is forced here by requiring that δ(3) integrate to 1 over any ball about the origin, and it is the convention used throughout physics for radial deltas at the origin. Against a test function with angular dependence the expression is wrong, which is the restriction already stated.
Write φ~(u)=φ(X(u)), which is C1 on U with φ~(u0)=φ(x0). Pair the proposed right-hand side with φ, using the volume element ∣J(u)∣dnu as the statement prescribes:
the middle step by the sifting property of the product delta applied to the function u↦φ~(u)∣J(u)∣, which is continuous at u0. Since φ(x0) is also ⟨δ(n)(x−x0),φ⟩, the two distributions agree. The hypothesis J(u0)=0 is what allows the division; the hypothesis that X is a bijection is what makes u0 the unique preimage, so that no sum over branches appears — compare the composition rule above, where several zeros do contribute. □
Reference table(Representations as limits)AdaptedAdapted from Folland (1999), §9.1, Proposition 9.1, p. 283; Arfken, Weber & Harris (2013), §1.11, Eqs. (1.152)–(1.155) and Exercises 1.11.1–1.11.9, pp. 76–81; Hassani (2013), §7.3, Examples 7.3.1–7.3.3, pp. 229–232
Let ρ∈L1(Rn) with ∫Rnρ=A, and for ε>0 put
ρε(x)=ε−nρ(x/ε).
Then ρε→Aδ(n) weakly as ε→0+: for every bounded continuous φ on Rn, and in particular for every test function,
∫Rnρε(x)φ(x)dx⟶Aφ(0).
This is Folland's Proposition 9.1. No positivity, no monotonicity and no smoothness of ρ is required, only absolute integrability; with A=1 it covers every family in the table below except the last two rows.
Family in one dimension
∫
Source
επ1e−x2/ε2 (Gaussian)
1
Arfken (1.153), p. 76; Hassani, Example 7.3.1
2ε1 on ∣x∣<ε, else 0 (rectangular)
1
Arfken, Exercise 1.11.1
π1ε2+x2ε (Lorentzian)
1
Arfken (1.154), p. 77, and Exercise 1.11.2
2ε1sech2(x/ε)
1
Arfken, Exercise 1.11.9
the rectangular family again, as dxd of a piecewise-linear smoothing of the step
1
Hassani, Example 7.3.3
DT(x)=πxsinTx, T→∞ (Dirichlet)
—
Arfken (1.155); Hassani, Example 7.3.2
2πN1(sin(t/2)sin(Nt/2))2, N→∞ (Fejér)
1
Arfken, Exercise 1.11.3
The last two rows need separate arguments. DT is not in L1 — ∫∣DT∣ diverges — so the theorem does not apply to it; its convergence to δ is the Dirichlet-integral computation of Pointwise convergence of Fourier series, and the associated formal identity δ(x)=(2π)−1∫eikxdk (Arfken's Eq. (1.163), Hassani's Eq. (7.22)) is made rigorous in Fourier transforms of distributions as the inversion formula applied to δ^=1. The Fejér kernel is integrable but periodic rather than a rescaling of a fixed profile, so it converges to δ on an interval of length 2π rather than on the line; see Cesàro means and Fejér's theorem.
Proof(Representations as limits)Original
Let φ be bounded and continuous, with M=sup∣φ∣. Substituting x=εy, so that dnx=εndny and the factor ε−n cancels,
∫Rnρε(x)φ(x)dnx=∫Rnρ(y)φ(εy)dny.
The rescaling has been moved off the singular factor and onto the smooth one — that is the whole trick. Now let ε→0+ inside the integral. The integrands are dominated uniformly in ε,
ρ(y)φ(εy)≤M∣ρ(y)∣∈L1(Rn),
and converge pointwise to ρ(y)φ(0) by continuity of φ at 0. The dominated convergence theorem (Dominated convergence) applies and gives
∫ρ(y)φ(εy)dny⟶φ(0)∫ρ(y)dny=Aφ(0).□Remark(Where the hypotheses are used)Original
Simple zeros in the composition rule.g′(ai)=0 is used twice: to invert g near ai, and to divide by ∣g′(ai)∣. Without it the rule is false and the left side is usually undefined. The standard counterexample is g(x)=x2: ∫ρε(x2)φ(x)dx behaves like ε−1/2, so δ(x2) does not exist. Likewise δ(g) is undefined when g≡0 on an interval.
Finitely many zeros in the support. If g has infinitely many zeros accumulating inside suppφ — for example g(x)=xsin(1/x) near 0 — the sum need not converge and the proof's Step 1 fails, since the intervals Ii cannot be made disjoint.
Smoothness of the multiplier.uδ requires u∈C∞ so that uφ is a test function. The identity u(x)δ′(x)=u(0)δ′−u′(0)δ shows why: a derivative of u appears, so u merely continuous will not do. In particular δ(x)H(x), with H the step, is not defined, and neither is δ2. Any identity that appears to multiply two deltas has been written down carelessly.
The Jacobian must not vanish. The curvilinear formulas fail exactly at the coordinate singularities: at r0=0 or θ0∈{0,π} in spherical polars, and on the axis ρ0=0 in cylindrical polars, because ∣J∣=0 there and X is not injective. That is why the delta at the origin needs the separate expression δ(r)/4πr2, and why that expression is stated only against radial test functions: a general test function sees angular structure at the origin that δ(r)/4πr2 has averaged away.
Integrability in the limit theorem. Absolute integrability of ρ supplies the dominating function; the conclusion can fail without it, as DT shows — it does converge to δ, but the proof above cannot see this, and the convergence is genuinely more delicate, being conditional rather than absolute.
Continuity of φ at the origin only. Nothing in the proof needs φ smooth, or even continuous away from 0: a bounded φ continuous at the single point 0 suffices. This is the same hypothesis as in the sifting property of The Dirac delta in calculations.
Method(Evaluating an integral containing a delta)Original
To compute ∫φ(x)δ(g(x))dx over an interval or region:
Solve g(x)=0 and keep the roots inside the region of integration. A root on the boundary makes the integral ambiguous; a root outside contributes nothing.
Check each root is simple, g′(ai)=0. If one is not, stop: the integral is not defined by these rules.
Add up φ(ai)/∣g′(ai)∣ over the surviving roots. Take the modulus; a sign error here is the commonest mistake.
In several dimensions, first write the delta in the coordinates in which the integral is set up, dividing by ∣J∣ at the point where the delta sits, and then integrate with the volume element of those coordinates. The two Jacobian factors must cancel; if they do not, one of them has been put in twice.
Check dimensions.δ(g) carries the reciprocal of the units of g. A delta in energy has units of inverse energy, so the final answer's units are a quick and reliable check on step 3.
Example(A cubic argument, and a charged ring)Original
(a) One dimension. Evaluate
I=∫−∞∞(x2+1)δ(x3−4x)dx.
Step 1.g(x)=x3−4x=x(x−2)(x+2) vanishes at x=−2,0,2, all inside the range.
Step 2.g′(x)=3x2−4, so ∣g′(0)∣=4 and ∣g′(±2)∣=∣12−4∣=8. All three roots are simple.
Step 3. With φ(x)=x2+1, so φ(0)=1 and φ(±2)=5,
I=41+85+85=82+5+5=812=23=1.5.
A quick sanity check on the dominant term: the two outer roots are in a region where g is steep, so they are weighted less per unit of φ than the root at the origin — 1/8 against 1/4 — yet they still dominate because φ is five times larger there.
(b) Three dimensions. A total charge Q is spread uniformly on the circle r=R, θ=π/2 in spherical polars. Its density is a delta supported on a curve, and by the curvilinear rule the two angular-and-radial deltas must be divided by the Jacobian at the ring:
the factor 1/2π spreading the charge over the azimuth, and no δ(ϕ−ϕ0) appearing because the ring covers every ϕ. Checking the total charge with the spherical volume element r2sinθdrdθdϕ:
using sin(π/2)=1. Both the R2 and the 2π cancel exactly, which is the step-4 check: had the sinθ of the volume element been forgotten, the answer would have been Q/sin(π/2) — right here by luck, and wrong for a ring at any other latitude. For a ring at colatitude θ0 the density is Qδ(r−R)δ(θ−θ0)/(2πR2sinθ0) and the sinθ0 is essential.
Application(Energy conservation and phase space)Original
Fermi's golden rule and densities of states. A transition rate in quantum mechanics carries a factor δ(Ef−Ei), which is a delta in an energy and therefore has units of inverse energy; summing it over final states converts it into a density of states. Whenever the final states are labelled by a continuous variable k with E=E(k), the composition rule does the conversion:
∫δ(E(k)−E)dk=i∑∣E′(ki)∣1,
the sum running over the solutions of E(k)=E. The group velocity E′(k) in the denominator is why flat bands have large densities of states, and why a van Hove singularity occurs exactly where E′(ki)=0 — precisely the point at which the composition rule's hypothesis fails and the integral ceases to be finite.
Relativistic phase space. In a scattering calculation the on-shell condition is imposed by δ(p2−m2) with p2=(p0)2−∣p∣2. Regarded as a function of p0 at fixed p, this argument has simple zeros at p0=±Ep with Ep=(∣p∣2+m2)1/2, and derivative 2p0 there, so
δ(p2−m2)θ(p0)=2Epδ(p0−Ep),
the step function (The Heaviside step and related functions) discarding the negative-energy root. This single application of the composition rule is what makes d4pδ(p2−m2)θ(p0) the Lorentz-invariant measure d3p/2Ep on the mass shell, the normalisation that appears in front of every relativistic cross-section.
Idealised sources. The curvilinear forms are the standard way of writing the charge or mass density of an idealised ring, shell, wire or point, as in the worked example — Riley, Hobson and Bence's Eq. (13.20) for an impulse is the same idea in time. These densities are the right-hand sides of the Poisson and wave equations whose solutions are the Green's functions of Green's functions for Laplace and Poisson problems, and the Jacobian factors in them are what make a multipole expansion in Multipole expansions come out with the correct powers of the radius.
Defines the Heaviside step, the sign, ramp and rectangle functions and the identities relating them, proves the chain of distributional derivatives that ends at the delta and the reciprocal statement that the step is the integral of the delta, and gives a recipe with a worked trapezoidal pulse for transforming a switched-on signal by differentiating it into deltas.
which is Riley, Hobson and Bence's Eq. (13.22). Arfken writes u(t−k) for the step switched on at k and defines it the same way. As a distribution, H is the functional generated by the indicator function of (0,∞):
⟨H,φ⟩=∫0∞φ(x)dx,
Stakgold and Holst's Example 2. Three related functions are built from it:
the sign (or signum) function, the ramp function — Strauss's plus functionp(x)=x+, Reed and Simon's Example 8 — and the rectangle (or boxcar, or top-hat) function on (a,b). They are related by
each valid for every x=0 and hence as an identity of distributions. The rectangle relation is the one Arfken uses to write a square pulse of height A on [0,t0] as A[u(t)−u(t−t0)].
The value at the jump.H(0) is left undefined above. Riley, Hobson and Bence set H(0)=21, which is the value the Fourier and Dirichlet theories assign to a jump and the one Arfken's Eq. (20.146) produces, ∫0∞x−1sintxdx=2π[2u(t)−1], whose value at t=0 is 0. Other books take H(0)=1. The choice is immaterial for everything in this unit: functions differing at one point are equal almost everywhere and therefore define the same distribution (Test functions and distributions), so no formula below depends on it. The convention matters only when H is evaluated, not integrated.
The step as a limit.H is the pointwise limit of smooth switches, for instance Arfken's
un(x)=21(1+tanhnx)⟶H(x)(x=0),
whose derivative un′(x)=21nsech2nx is one of the delta families of Delta-function identities. Hassani's piecewise-linear smoothing gives the rectangular family the same way. Each such pair exhibits H′=δ before any distribution theory is invoked.
Conversely, H is the indefinite integral of δ: for every φ∈D(R),
H(x)=∫−∞xδ(t)dt
in the sense that x↦∫−∞xδ(t)dt, defined as the distribution whose derivative is δ and which vanishes for x<0, equals H. By the uniqueness statement of Test functions and distributions, H is the only such antiderivative up to an additive constant, and the two side conditions H=0 for x<0 and H=1 for x>0 fix the constant. Stakgold and Holst obtain this by integrating Eq. (2.1.15) formally; it is the statement that the fundamental theorem of calculus survives the jump.
In two variables, with x=(x1,x2),
∂x1∂x2∂2[H(x1)H(x2)]=δ(2)(x),
Stakgold and Holst's Eq. (2.1.16): the indicator function of the first quadrant is a two-dimensional ramp whose mixed second derivative is the point source at the corner.
Idea of the proofOriginal
All of these are the jump formula of Test functions and distributions read off a graph. Differentiating a piecewise-smooth function slides each corner down one order of singularity: a corner in the graph of x+ becomes a jump in H, the jump in H becomes a delta, the delta becomes a dipole. Nothing is lost at each stage, which is why integrating back recovers the original function — and that recovery is precisely what the naive answer H′=0 destroys.
Step 1: H′=δ. By the definition of the distributional derivative and the definition of H as a functional,
⟨H′,φ⟩=−⟨H,φ′⟩=−∫0∞φ′(x)dx=−[φ(x)]0R=φ(0),
by the fundamental theorem of calculus on [0,R], where φ is C1, together with φ(R)=0. The right-hand side is ⟨δ,φ⟩, so H′=δ. Replacing H by H(⋅−a) shifts the integral and gives H(⋅−a)′=δ(⋅−a).
Step 2: (x+)′=H. Here x+ is continuous, so no delta can appear; the computation checks it. Integrating by parts on [0,R], where x+=x is C1,
the boundary term vanishing at x=0 because of the factor x and at x=R because φ(R)=0. Combining with Step 1, (x+)′′=δ: Reed and Simon's observation that δ is the second derivative of a continuous function.
Step 3: the sign and the modulus.sgn=2H−1 and constants have zero derivative, so sgn′=2H′=2δ. And ∣x∣=2x+−x, so ∣x∣′=2H−1=sgnx and ∣x∣′′=2δ. (Note ∣x∣′=sgnx is the one case where the naive pointwise answer is also the right one — there is no jump in ∣x∣ to generate a delta.)
Step 4: the rectangle. Linearity and Step 1 give Πa,b′=H(⋅−a)′−H(⋅−b)′=δ(⋅−a)−δ(⋅−b): a positive spike where the pulse switches on and a negative one where it switches off.
Step 5: the step is the integral of the delta. Let u be a distribution with u′=δ and u=0 on (−∞,0). Then (u−H)′=δ−δ=0, so u−H is a constant c by the theorem on distributions with vanishing derivative in Test functions and distributions. Testing against a φ supported in (−∞,0), where both u and H vanish, gives c∫φ=0 for every such φ, hence c=0 and u=H. □
Remark(Where the hypotheses are used)Original
Compact support of φ supplies the vanishing boundary terms at x=R in Steps 1 and 2. Without it the same integrations by parts leave behind φ(∞), and H′=δ becomes false.
Why H′=0 is wrong and ∣x∣′=sgnx is right. Stakgold and Holst put the test crisply: ∫0xsgntdt=∣x∣, so discarding the single bad point in ∣x∣′ costs nothing, whereas ∫−∞x0dt=0=H(x), so discarding the jump in H′ loses the whole function. The information sits in a set of measure zero for the function and in a non-zero distribution for the derivative; only the second description survives integration.
The value H(0) is irrelevant because H enters every statement only through ∫Hφ. This is a consequence of part (2) of the theorem in Test functions and distributions, and it is why the dispute between H(0)=21 and H(0)=1 never shows up in a transform or a Green's function.
An alternative derivation, and what it needs. Stakgold and Holst define H′ instead as the weak limit of the difference quotients ε−1[H(x+ε)−H(x)], which is the rectangular family ε−1Π−ε,0; pairing with φ gives ε−1∫−ε0φ→φ(0), and the hypothesis used is continuity of φ at 0 and nothing more. The two definitions agree, which is their Exercise 2.1.3.
Products are still forbidden.x+=xH(x) is legitimate because x is smooth; H(x)δ(x) is not legitimate at all, and H2=H is an identity about functions that happens to survive, not a licence to multiply distributions.
How far does this go? Every one of these functions is a derivative of a continuous function, and Reed and Simon's regularity theorem (Theorem V.10, p. 139) says this is the general picture: every tempered distribution is Dαg for some continuous g of polynomial growth. So nothing more exotic than repeated differentiation of ordinary continuous functions is ever needed to produce a tempered distribution.
Method(Transforming a switched-on signal)AdaptedAdapted from Arfken, Weber & Harris (2013), §20.7, Eqs. (20.137)–(20.139), p. 1010, and §20.8, Eq. (20.166), p. 1023
A signal that is piecewise polynomial, or that switches on at a finite time, is transformed most cheaply by differentiating it until only deltas are left.
Write the signal with steps and ramps. Replace every "for t>a" clause by a factor H(t−a), and every straight segment by a ramp (t−a)+. A piecewise-linear signal is always a finite combination ∑jcj(t−aj)+, and a signal with jumps needs steps as well.
Differentiate until the deltas appear. For a piecewise-linear signal two derivatives suffice: by the chain above, dt2d2∑jcj(t−aj)+=∑jcjδ(t−aj), where cj is the change of slope at aj.
Transform the deltas. With f^(ω)=∫fe−iωtdt the deltas give pure phases, δ(⋅−a)=e−iωa; with the Laplace transform L{F}(s)=∫0∞F(t)e−stdt they give L{δ(t−a)}=e−as for a>0, which is Arfken's Eq. (20.139).
Undo the differentiations by dividing.f(n)=(iω)nf^, so f^(ω)=(iω)−nf(n)(ω); for the Laplace transform of a signal vanishing at t=0 together with its first n−1 derivatives, L{f}=s−nL{f(n)}.
Check the division at ω=0. Step 4 is only legitimate if f(n) vanishes to order n at the origin, that is, if the delta weights satisfy ∑jcj=0, ∑jcjaj=0, and so on up to order n−1. For a signal of compact support these hold automatically, and the limiting value f^(0)=∫f is a free check on the arithmetic.
For a signal switched on at t=k, use the shifting theorems directly instead: L{u(t−k)}=e−ks/s (Arfken's Eq. (20.138)) and L{F(t−k)u(t−k)}=e−ksL{F}(s), the Heaviside shifting theorem, Arfken's Eq. (20.166). The step function in the second formula is what restricts F to positive arguments; omitting it transforms the wrong function. The Fourier analogue of the simplest switched-on signal is H(t)e−at(ω)=(a+iω)−1 for a>0, whose a→0+ limit is the transform of H itself (Fourier transforms of distributions).
Example(A trapezoidal pulse)Original
A pulse rises linearly from 0 at t=0 to 1 at t=1, stays at 1 until t=3, and falls linearly back to 0 at t=4.
Step 1: write it with ramps. The slope is +1 on (0,1), 0 on (1,3), −1 on (3,4) and 0 after. The changes of slope are +1 at t=0, −1 at t=1, −1 at t=3 and +1 at t=4, so
Step 3: transform.T′′(ω)=1−e−iω−e−3iω+e−4iω, which factorises:
T′′(ω)=(1−e−iω)−e−3iω(1−e−iω)=(1−e−iω)(1−e−3iω).
Step 4: divide by (iω)2=−ω2.
T^(ω)=−ω2(1−e−iω)(1−e−3iω).
Step 5: the checks. The delta weights are 1,−1,−1,1 with positions 0,1,3,4, and
j∑cj=1−1−1+1=0,j∑cjaj=0−1−3+4=0,
so T′′ has a double zero at ω=0 and the division is legitimate. Expanding 1−e−iω=iω+O(ω2) and 1−e−3iω=3iω+O(ω2) gives a numerator ∼(iω)(3iω)=−3ω2, so
T^(0)=3,
which must be ∫Tdt — and indeed the trapezoid has area 2+21+21=3. At ω=π the exponentials are e−iπ=e−3iπ=−1, so
T^(π)=−π2(1+1)(1+1)=−π24=−0.405285,
real and negative. Direct numerical quadrature of ∫04T(t)e−iπtdt returns −0.4052847, and the agreement holds at other frequencies too: at ω=1 both routes give −0.796047−1.739393i.
The Laplace transform, for comparison. Here T(t)=0 for t<0, so by the Heaviside shifting theorem with L{t}=1/s2,
L{T}(s)=s21−e−s−e−3s+e−4s=s2(1−e−s)(1−e−3s),
the same function of s=iω, as it must be for a signal supported in t≥0. Letting s→0 again gives 3.
Switched sources. Every circuit that is switched on, every force applied at a definite instant, every beam that is gated, is modelled by multiplying the source by a step. Arfken's square pulse A[u(t)−u(t−t0)] is the elementary case, and its Laplace transform A(1−e−t0s)/s shows what the method above gives in general: the shape of the pulse enters only through a factor built from exponentials, so a train of pulses is a geometric series and can be summed in closed form. Riley, Hobson and Bence's complementary idealisation, the impulse j(t)=Jδ(t−t0), is legitimate precisely when the forcing lasts much less than the system's own response time; the step and the delta are the two ends of the same approximation, linked by H′=δ.
Step response and impulse response. Because H′=δ and a linear time-invariant system commutes with differentiation, the response to a step is the integral of the response to an impulse:
(step response)(t)=∫−∞t(impulse response)(τ)dτ.
This is why a single measurement suffices to characterise such a system, and why engineers measure the step response — easy to generate — and differentiate it to obtain the impulse response, which is the Green's function. The algebra is the derivative rule for transforms: dividing a transform by iω or by s integrates in time.
Causality. A causal response function is one supported in t≥0, that is, one of the form G(t)=H(t)g(t). The step is therefore not a cosmetic factor but the mathematical content of causality, and it has consequences: the transform of a product with H involves H^=πδ−ipv(1/ω) (Fourier transforms of distributions), whose principal-value part ties the real and imaginary parts of the response together into the dispersion relations of Dispersion relations. The retarded Green's functions of Causal Green's functions for initial-value problems are written with exactly this factor, and the jump conditions that determine them are the derivative identities of this unit read backwards.
Characteristics and shocks. A step is a weak solution of the wave equation and of first-order conservation laws, as Test functions and distributions notes: u=H(x−ct) satisfies utt=c2uxx in D′. Differentiating a piecewise-linear initial profile, as Strauss does for the plucked string, puts deltas on the characteristics and shows where the singularities of the solution travel — the starting point for Conservation laws and shocks.