Fourier series

Definition and convergence theory of Fourier series.

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Fourier series and their convergence

Fourier series and their convergence

Definition

Fourier series

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Defines the Fourier coefficients and Fourier series of an integrable function on an interval of any length, in both the complex-exponential and the real sine-cosine form, shows that the trigonometric system is orthogonal and that the coefficients of a uniformly convergent trigonometric series are forced to be these integrals, and sets up the even and odd extensions that turn a function given on half an interval into a cosine or sine half-range series.

Prerequisites: Inner products and norms

Definition(Fourier coefficients and Fourier series)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, §1.1, pp. 34–35; Stakgold & Holst (2011), §2.3, eqs. (2.3.1)–(2.3.4) and (2.3.7)–(2.3.8), pp. 127–129, and Change of Scale, p. 137; Arfken, Weber & Harris (2013), §19.1, eqs. (19.1)–(19.6), pp. 935–936, and eqs. (19.10)–(19.12), pp. 939–940; Riley, Hobson & Bence (2006), §12.2, eqs. (12.4)–(12.6), pp. 417–418, and §12.7, eqs. (12.9)–(12.11), pp. 424–425

Fix a period $L>0$ and a base point $a\in\mathbb{R}$ , and let $f$ be integrable on $I=[a,a+L]$ , meaning $\int_{I}\lvert f\rvert<\infty$ . The $n$ th Fourier coefficient of $f$ is the number

\hat f(n) \;=\; \frac{1}{L}\int_{a}^{a+L} f(x)\,\mathrm{e}^{-2\pi \mathrm{i} n x/L}\,\mathrm{d}x , \qquad n\in\mathbb{Z},

and the Fourier series of $f$ is the expression

\sum_{n=-\infty}^{\infty}\hat f(n)\,\mathrm{e}^{2\pi \mathrm{i} n x/L}.

Writing $f\sim\sum_n \hat f(n)\mathrm{e}^{2\pi\mathrm{i}nx/L}$ is shorthand for "the displayed series is the one built from $f$ by the formula above"; the symbol $\sim$ carries no claim about convergence, and none is made here. Following Stein and Shakarchi we always read convergence as the limit of the symmetric partial sums

S_N(f)(x) \;=\; \sum_{n=-N}^{N}\hat f(n)\,\mathrm{e}^{2\pi \mathrm{i} n x/L}, \qquad N\in\mathbb{N},

each of which is a trigonometric polynomial of degree $N$ . A general trigonometric series is any $\sum_n c_n \mathrm{e}^{2\pi\mathrm{i}nx/L}$ with $c_n\in\mathbb{C}$ ; its coefficients need not be the Fourier coefficients of anything.

Since every $\mathrm{e}^{2\pi\mathrm{i}nx/L}$ has period $L$ , so does the series. Hence the natural home of the definition is a function of period $L$ on the whole line, or equivalently a function on a circle of circumference $L$ ; a function given only on $I$ is extended to the line by $f(x+L)=f(x)$ , which is what the series will represent. The coefficients do not depend on the base point $a$ : see part 1 of the statement below.

The real form. It is often convenient to write the period as $L=2l$ and use the interval $(-l,l)$ . Then the series above can be rearranged into the real sine-cosine form

\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n\cos\frac{n\pi x}{l}+b_n\sin\frac{n\pi x}{l}\right), \qquad \begin{aligned} a_n &= \frac{1}{l}\int_{-l}^{l} f(x)\cos\frac{n\pi x}{l}\,\mathrm{d}x, &&n=0,1,2,\dots\\ b_n &= \frac{1}{l}\int_{-l}^{l} f(x)\sin\frac{n\pi x}{l}\,\mathrm{d}x, &&n=1,2,\dots \end{aligned}

The constant term is written $a_0/2$ rather than $a_0$ precisely so that one formula gives every $a_n$ , including $n=0$ ; this is the normalisation of Arfken, Weber and Harris and of Riley, Hobson and Bence. The two sets of coefficients are related by

a_n = \hat f(n)+\hat f(-n), \qquad b_n = \mathrm{i}\bigl(\hat f(n)-\hat f(-n)\bigr), \qquad \hat f(\pm n) = \tfrac12\bigl(a_n \mp \mathrm{i}b_n\bigr) \quad (n\ge1),

with $a_0=2\hat f(0)$ . In particular $\hat f(0)=a_0/2$ is the mean value of $f$ over one period. If $f$ is real-valued then $\hat f(-n)=\overline{\hat f(n)}$ for every $n$ , equivalently all $a_n$ and $b_n$ are real. The reverse implication is weaker than it looks and must be stated carefully: if $\hat f(-n)=\overline{\hat f(n)}$ for every $n$ , then the function $f-\overline f$ has every Fourier coefficient zero, and by Stein and Shakarchi's Theorem 2.1 (quoted in the remark below) it therefore vanishes at every point where it is continuous. So all one may conclude is that $f$ is real at each of its points of continuity; it may be non-real on a set of measure zero, and no statement about the coefficients can exclude that.

Conventions differ, and the differences are only bookkeeping. Stein and Shakarchi put the $2\pi$ in the exponent and work with $\mathrm{e}^{2\pi\mathrm{i}nx/L}$ on an interval of length $L$ ; Stakgold and Holst, Arfken, Weber and Harris, and most physics texts take $L=2\pi$ and write $\mathrm{e}^{\mathrm{i}nx}$ , rescaling at the end; Riley, Hobson and Bence keep a general period $L$ in the real form, so their cosine is $\cos(2\pi n x/L)$ where ours, with $L=2l$ , is $\cos(n\pi x/l)$ — the same function. Strauss writes the full series on $(-l,l)$ exactly as above. We use $L$ for the period throughout and set $L=2l$ whenever the real form is in play, so that $l$ is the half-period, and we flag any departure.

Definition(Even and odd extensions; half-range series)AdaptedAdapted from Strauss (2008), §5.2, eqs. (6)–(7) and the identification of the sine and cosine series with the odd and even extensions, p. 115, and §5.1, eqs. (1), (4), (6), (7), pp. 104–106; Riley, Hobson & Bence (2006), §12.5, Figure 12.4 and the discussion of possible continuations, p. 422; Arfken, Weber & Harris (2013), §19.1, eqs. (19.13)–(19.14) and the naming of the sine and cosine series, p. 941

Let $\varphi$ be integrable on $(0,l)$ . Its even extension to $(-l,l)$ is

\varphi_{\mathrm{even}}(x) = \begin{cases}\varphi(x), & 0<x<l,\\ \varphi(-x), & -l<x<0,\end{cases}

and its odd extension is

\varphi_{\mathrm{odd}}(x) = \begin{cases}\varphi(x), & 0<x<l,\\ -\varphi(-x), & -l<x<0,\\ 0,& x=0.\end{cases}

Strauss sets the odd extension to $0$ at the origin, as we do; the even extension need not be defined there. Each is then extended to the line with period $2l$ . These are the only even and odd extensions there are, so the construction involves no choice.

The half-range sine series of $\varphi$ on $(0,l)$ is

\sum_{n=1}^{\infty}B_n\sin\frac{n\pi x}{l}, \qquad B_n = \frac{2}{l}\int_{0}^{l}\varphi(x)\sin\frac{n\pi x}{l}\,\mathrm{d}x ,

and the half-range cosine series is

\frac{A_0}{2}+\sum_{n=1}^{\infty}A_n\cos\frac{n\pi x}{l}, \qquad A_n = \frac{2}{l}\int_{0}^{l}\varphi(x)\cos\frac{n\pi x}{l}\,\mathrm{d}x , \quad n\ge 0 .

The name records that the coefficient integral runs over half the period. Arfken, Weber and Harris call these the Fourier sine and Fourier cosine series. Part 4 of the statement identifies them with the full series of $\varphi_{\mathrm{odd}}$ and $\varphi_{\mathrm{even}}$ , which is why the factor is $2/l$ here and $1/l$ in the full series: the integrand is even in both cases, so half the interval carries half the integral.

Definition(Orthogonality, and that the coefficients are forced)AdaptedAdapted from Riley, Hobson & Bence (2006), §12.2, eqs. (12.1)–(12.3), p. 417, and §12.7, the exponential orthogonality relation, p. 425; Stakgold & Holst (2011), §2.3, eqs. (2.3.5)–(2.3.8), pp. 128–129; Strauss (2008), §5.1, eq. (2), p. 104, and §5.2, pp. 115–116; Arfken, Weber & Harris (2013), §19.1, the orthonormal systems, p. 936, and eqs. (19.13)–(19.14), p. 941; Stein & Shakarchi (2003), Ch. 2, §1.1, p. 35

Let $L=2l>0$ and write $e_n(x)=\mathrm{e}^{2\pi\mathrm{i}nx/L}$ . Equip the space of square-integrable functions on an interval $I$ of length $L$ with the inner product $(u,v)=\int_I u\overline{v}$ , in the convention of Inner products and norms.

Orthogonality. For all $m,n\in\mathbb{Z}$ and every base point $a$ , $\frac{1}{L}\int_{a}^{a+L} e_n\overline{e_m} = \delta_{nm},$ so $\{e_n\}_{n\in\mathbb{Z}}$ is orthogonal with $\lVert e_n\rVert^2=L$ , and $\{L^{-1/2}e_n\}$ is orthonormal. Equivalently, in real form, for integers $r,p\ge0$ , $\int_{a}^{a+L}\!\!\sin\frac{r\pi x}{l}\cos\frac{p\pi x}{l}\,\mathrm{d}x = 0 \ \text{ for all } r,p,$ $\int_{a}^{a+L}\!\!\cos\frac{r\pi x}{l}\cos\frac{p\pi x}{l}\,\mathrm{d}x =\begin{cases}L,& r=p=0,\\ \tfrac12 L,& r=p>0,\\ 0,& r\neq p,\end{cases} \qquad \int_{a}^{a+L}\!\!\sin\frac{r\pi x}{l}\sin\frac{p\pi x}{l}\,\mathrm{d}x =\begin{cases}\tfrac12 L,& r=p>0,\\ 0,& \text{otherwise.}\end{cases}$ Separately, and not as a consequence of these relations, $\hat f(n)$ is unchanged if the interval $[a,a+L]$ is replaced by any other interval of length $L$ , provided $f$ has period $L$ ; this is proved on its own below.
The coefficients are forced. Suppose a trigonometric series $\sum_{n}c_n e_n$ converges uniformly on an interval of length $L$ , with sum $f$ . Then $f$ is continuous, and it extends to a continuous function of period $L$ on the whole line — the series converges uniformly on every interval of length $L$ , since each $e_n$ has period $L$ — and, with $\hat f(n)$ computed from that extension, $c_n = \hat f(n) \quad\text{for every } n\in\mathbb{Z}.$ In the real form: if $\tfrac12\alpha_0+\sum_{n\ge1}(\alpha_n\cos(n\pi x/l)+\beta_n\sin(n\pi x/l))$ converges uniformly to $f$ , then $\alpha_n=a_n$ and $\beta_n=b_n$ . In particular a function has at most one uniformly convergent trigonometric expansion.
Symmetry. Let $f$ be integrable on $(-l,l)$ . If $f$ is even then $b_n=0$ for all $n$ ; if $f$ is odd then $a_n=0$ for all $n$ , including $a_0$ . In either case the surviving coefficients may be computed on $(0,l)$ alone, with the factor $2/l$ .
Half-range series. Let $\varphi$ be integrable on $(0,l)$ . The half-range sine series of $\varphi$ is exactly the full Fourier series of $\varphi_{\mathrm{odd}}$ on $(-l,l)$ , and the half-range cosine series of $\varphi$ is exactly the full Fourier series of $\varphi_{\mathrm{even}}$ : that is, $B_n=b_n[\varphi_{\mathrm{odd}}]$ and $A_n=a_n[\varphi_{\mathrm{even}}]$ , with all other coefficients zero. Both series therefore have period $2l$ . Nothing here asserts that either series converges, let alone to $\varphi$ : the claim is only that the two coefficient families coincide, so that whatever one series does on $(0,l)$ the other does to the reflected data on $(-l,0)$ .

Idea of the proofOriginal

Everything here is one move from finite-dimensional linear algebra, done in a space of functions. The exponentials $e_n$ are mutually orthogonal for the integral inner product over one period, and each has the same length $\sqrt L$ . In a finite-dimensional space with an orthogonal basis, the coefficient of a basis vector in an expansion is recovered by taking one inner product and dividing by the squared length; the formula for $\hat f(n)$ is that recipe, nothing more. What distinguishes the infinite-dimensional case is that the expansion is now an infinite sum, so recovering a coefficient requires interchanging an integral with a limit. Uniform convergence is exactly the hypothesis that licenses the interchange, which is why part 2 asks for it and asks for nothing else. The symmetry and half-range statements then follow from a single observation: the coefficient integral of an odd function over a symmetric interval vanishes, and $\sin$ is odd while $\cos$ is even.

ProofAdaptedAdapted from Stakgold & Holst (2011), §2.3, the derivation of (2.3.7)–(2.3.8), p. 128; Strauss (2008), §5.1, the proof of (2) and the termwise integration leading to (4), pp. 104–105, and §5.2, p. 116; Riley, Hobson & Bence (2006), §12.2, the derivation of (12.5)–(12.6) from (12.7), pp. 417–418

Part 1, the exponential relations. Put $k=n-m$ . If $k=0$ the integrand $e_n\overline{e_m}$ is identically $1$ and the integral is $L$ . If $k\neq0$ then $\mathrm{e}^{2\pi\mathrm{i}kx/L}$ has the antiderivative $(L/2\pi\mathrm{i}k)\,\mathrm{e}^{2\pi\mathrm{i}kx/L}$ , so

\int_{a}^{a+L}\mathrm{e}^{2\pi\mathrm{i}kx/L}\,\mathrm{d}x =\frac{L}{2\pi\mathrm{i}k}\Bigl(\mathrm{e}^{2\pi\mathrm{i}k(a+L)/L}-\mathrm{e}^{2\pi\mathrm{i}ka/L}\Bigr) =\frac{L}{2\pi\mathrm{i}k}\,\mathrm{e}^{2\pi\mathrm{i}ka/L}\bigl(\mathrm{e}^{2\pi\mathrm{i}k}-1\bigr)=0,

because $k$ is an integer. Dividing by $L$ gives $\delta_{nm}$ , and $\lVert e_n\rVert^2=\int_I\lvert e_n\rvert^2=L$ since $\lvert e_n\rvert=1$ . Note that the base point $a$ dropped out of the calculation. All that was used is that each $e_n$ has period $L$ , which is where $\mathrm{e}^{2\pi\mathrm{i}k}=1$ came from; the periodicity of $f$ plays no part here, and is a separate hypothesis used only in the base-point paragraph two paragraphs below.

Part 1, the real relations. Write $\theta_r=r\pi x/l=2\pi r x/L$ , so that $\mathrm{e}^{\mathrm{i}\theta_r}=e_r(x)$ , and replace each cosine and sine by $\cos\theta_r=\tfrac12(\mathrm{e}^{\mathrm{i}\theta_r}+\mathrm{e}^{-\mathrm{i}\theta_r})$ and $\sin\theta_r=\tfrac1{2\mathrm{i}}(\mathrm{e}^{\mathrm{i}\theta_r}-\mathrm{e}^{-\mathrm{i}\theta_r})$ . Each product becomes a sum of four exponentials, at the frequencies $r+p$ , $r-p$ , $-(r-p)$ and $-(r+p)$ (the frequency of $\mathrm{e}^{2\pi\mathrm{i}kx/L}$ being the integer $k$ ). By the exponential relations, integrating over $[a,a+L]$ gives each such term $L$ times its coefficient if its frequency is zero, and nothing otherwise, so only the zero-frequency terms have to be counted.

For the sine-cosine product, $\sin\theta_r\cos\theta_p=\tfrac1{4\mathrm{i}}\bigl(\mathrm{e}^{\mathrm{i}(\theta_r+\theta_p)}+\mathrm{e}^{\mathrm{i}(\theta_r-\theta_p)}-\mathrm{e}^{-\mathrm{i}(\theta_r-\theta_p)}-\mathrm{e}^{-\mathrm{i}(\theta_r+\theta_p)}\bigr)$ . If $r\neq p$ and $r+p>0$ , no frequency is zero and the integral is $0$ . If $r=p>0$ the two middle terms both have frequency zero and their coefficients are $+\tfrac1{4\mathrm{i}}$ and $-\tfrac1{4\mathrm{i}}$ , so they cancel; if $r=p=0$ the integrand is identically $0$ . So the integral vanishes for all $r,p\ge0$ .

For the two like products, $\cos\theta_r\cos\theta_p=\tfrac14\bigl(\mathrm{e}^{\mathrm{i}(\theta_r+\theta_p)}+\mathrm{e}^{\mathrm{i}(\theta_r-\theta_p)}+\mathrm{e}^{-\mathrm{i}(\theta_r-\theta_p)}+\mathrm{e}^{-\mathrm{i}(\theta_r+\theta_p)}\bigr)$ and $\sin\theta_r\sin\theta_p=-\tfrac14\bigl(\mathrm{e}^{\mathrm{i}(\theta_r+\theta_p)}-\mathrm{e}^{\mathrm{i}(\theta_r-\theta_p)}-\mathrm{e}^{-\mathrm{i}(\theta_r-\theta_p)}+\mathrm{e}^{-\mathrm{i}(\theta_r+\theta_p)}\bigr)$ . With $r,p\ge0$ and $r\neq p$ every frequency is nonzero and both integrals vanish. For $r=p>0$ it is the two middle terms that survive, the pair at frequency $r-p=0$ ; each carries coefficient $+\tfrac14$ in both products, contributing $L/4$ apiece, so each integral is $L/4+L/4=L/2$ . For $r=p=0$ the cosine integrand is $1$ and the integral is $L$ — here all four exponentials have frequency zero, with coefficients summing to $1$ — while the sine integrand is identically $0$ .

Part 1, independence of the base point. Let $f$ have period $L$ and let $b$ be any other base point. Then $\int_b^{b+L}f e_{-n} - \int_a^{a+L}f e_{-n}$ is a difference of integrals over two intervals of equal length $L$ ; since the integrand $fe_{-n}$ has period $L$ , the part of $[b,b+L]$ outside $[a,a+L]$ is a translate by $\pm L$ of the part of $[a,a+L]$ outside $[b,b+L]$ , and the two contributions are equal. Hence the difference is zero.

Part 2. Let $s_N=\sum_{\lvert n\rvert\le N}c_n e_n$ and suppose $s_N\to f$ uniformly on $I=[a,a+L]$ . Each $s_N$ is continuous, and continuity survives passage to a uniform limit (Uniform convergence). That argument is a single line, so we give it here rather than lean on another unit for it: for $x,y\in I$ ,

\lvert f(x)-f(y)\rvert\le 2\max_I\lvert f-s_N\rvert+\lvert s_N(x)-s_N(y)\rvert ,

and choosing $N$ to make the first term small and then $y$ close to $x$ to make the second small shows $f$ is continuous at $x$ . Each $e_n$ has period $L$ , hence so does each $s_N$ ; the same series therefore converges uniformly on every translate of $I$ by a multiple of $L$ , and the function it defines on the whole line is continuous of period $L$ and agrees with $f$ on $I$ . That extension is the $f$ meant in the statement. Fix $m$ and multiply by the bounded function $\overline{e_m}$ , which does not disturb uniform convergence since $\lvert e_m\rvert=1$ : $s_N\overline{e_m}\to f\overline{e_m}$ uniformly on $I$ . A uniformly convergent sequence on an interval of finite length may be integrated term by term, because

\left\lvert\int_I s_N\overline{e_m}-\int_I f\overline{e_m}\right\rvert \le \int_I\lvert s_N-f\rvert \le L\max_I\lvert s_N-f\rvert\longrightarrow 0 .

For each fixed $N$ the integral of the finite sum is a finite sum of integrals, and by part 1 only the term $n=m$ survives, so $\int_I s_N\overline{e_m}=Lc_m$ as soon as $N\ge\lvert m\rvert$ . Letting $N\to\infty$ gives $Lc_m=\int_I f\overline{e_m}$ , which is $c_m=\hat f(m)$ . The real statement follows by applying this to the exponential form obtained from the dictionary in the definition, or by repeating the argument with $\cos(m\pi x/l)$ and $\sin(m\pi x/l)$ in place of $\overline{e_m}$ and using the real orthogonality relations, which is the route Riley, Hobson and Bence take. Uniqueness is immediate: both expansions have the same coefficients, namely those of their common sum.

Part 3. Take the interval $(-l,l)$ , which is symmetric about $0$ . If $f$ is even then $f(x)\sin(n\pi x/l)$ is a product of an even and an odd function, hence odd, and the integral of an odd function over a symmetric interval vanishes: $b_n=0$ . If $f$ is odd then $f(x)\cos(n\pi x/l)$ is odd for every $n\ge0$ — the case $n=0$ included, since $\cos 0=1$ — so every $a_n$ vanishes, $a_0$ among them. In the surviving cases the integrand is even, so $\int_{-l}^{l}=2\int_0^l$ , which converts the factor $1/l$ into $2/l$ .

Part 4. Apply part 3 to $f=\varphi_{\mathrm{odd}}$ . It is odd by construction, so all its $a_n$ vanish, and

b_n[\varphi_{\mathrm{odd}}]=\frac1l\int_{-l}^{l}\varphi_{\mathrm{odd}}(x)\sin\frac{n\pi x}{l}\,\mathrm{d}x =\frac2l\int_{0}^{l}\varphi(x)\sin\frac{n\pi x}{l}\,\mathrm{d}x = B_n ,

using that $\varphi_{\mathrm{odd}}=\varphi$ on $(0,l)$ and that the single point $x=0$ contributes nothing to an integral. So the full series of $\varphi_{\mathrm{odd}}$ is $\sum_n B_n\sin(n\pi x/l)$ , the half-range sine series. The cosine case is the same computation with $f=\varphi_{\mathrm{even}}$ , which is even, so its $b_n$ vanish and $a_n[\varphi_{\mathrm{even}}]=A_n$ . Both extensions were given period $2l$ , so both series do. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, §2, Thm 2.1 and Cor. 2.2, pp. 39–41; Riley, Hobson & Bence (2006), §12.5, the warning about the continuation at the end-points, pp. 422 and 424; Arfken, Weber & Harris (2013), §19.1, Example 19.1.2 and Figure 19.3, pp. 941–942; Strauss (2008), §5.2, eqs. (8)–(10), p. 115

Integrability is all that is needed for $\hat f(n)$ to exist: $\lvert f e_{-n}\rvert=\lvert f\rvert$ , so the defining integral converges absolutely whenever $\int_I\lvert f\rvert<\infty$ . Nothing weaker will do, and nothing stronger is required to write down the series.

Uniform convergence in part 2 is used at exactly one step, the interchange of the integral with the limit, and the displayed estimate shows how: the error is bounded by the period times the uniform error. Two consequences of insisting on it are worth separating. First, it forces the sum $f$ to be continuous, so part 2 says nothing whatever about a function with a jump — for those, the series is defined by the coefficient integrals and convergence has to be proved, which is the business of Pointwise convergence of Fourier series and Cesàro means and Fejér's theorem. Second, part 2 is a statement about a series that is given as convergent; it does not say that the Fourier series of a function converges to it. Weakening the hypothesis to pointwise convergence invalidates the proof, since the estimate above then has no uniform bound to work with.

Why the coefficients cannot determine $f$ everywhere. Changing $f$ at finitely many points leaves every $\hat f(n)$ unchanged, so no theorem can conclude $f=g$ pointwise from $\hat f=\hat g$ . Stein and Shakarchi's Theorem 2.1 gives the sharp positive statement in this direction: if $f$ is integrable on the circle with $\hat f(n)=0$ for every $n$ , then $f(\theta_0)=0$ at every point $\theta_0$ where $f$ is continuous. Their Corollary 2.2 is the clean case: a continuous $f$ with all coefficients zero is identically zero. Our part 2 is consistent with this and weaker, since it starts from a convergent series.

Periodicity is what makes the base point irrelevant, and only that. For an $f$ merely given on $[a,a+L]$ with $f(a)\neq f(a+L)$ , the periodic extension has a jump at the end-points, and the coefficients computed from a shifted interval are the coefficients of a different periodic function. This is the trap Riley, Hobson and Bence flag for extensions: a continuation that is discontinuous at an end-point of the interval of interest will not converge to the wanted value there. Their own example is the odd extension of $x^2$ from $(0,2]$ , whose series converges at $x=2$ to $0$ rather than to $4$ , because the odd extension jumps from $4$ to $-4$ across $x=2$ . The cure is to choose the extension, when one is free to, so that no jump is created: $\varphi_{\mathrm{even}}$ is continuous at $0$ and at $\pm l$ whenever $\varphi$ is continuous on $[0,l]$ , whereas $\varphi_{\mathrm{odd}}$ is continuous at $0$ only if $\varphi(0+)=0$ and at $l$ only if $\varphi(l-)=0$ .

Which extension to use is dictated by the problem, not by taste. Strauss's correspondence is the operational version: homogeneous Dirichlet conditions $u(0)=u(l)=0$ go with the odd extension and hence a sine series, because every $\sin(n\pi x/l)$ vanishes at both ends; homogeneous Neumann conditions $u'(0)=u'(l)=0$ go with the even extension and a cosine series, because every $\cos(n\pi x/l)$ has vanishing derivative at both ends; genuinely periodic conditions go with the plain periodic extension and the full series. Arfken, Weber and Harris make the complementary point with $f(x)=x$ on $[0,\pi]$ : its sine series, its cosine series and its power series all represent it on $[0,\pi]$ and disagree sharply outside, because the three carry different assumptions about symmetry and periodicity.

A hypothesis we did not impose. Riley, Hobson and Bence state four Dirichlet conditions on $f$ — periodicity, single-valuedness with at most finitely many finite discontinuities, finitely many maxima and minima in a period, and convergence of $\int\lvert f\rvert$ over a period — as the conditions under which a Fourier expansion is valid; Arfken, Weber and Harris give the same name to the last three and call such an $f$ piecewise regular, remarking that they are sufficient but not necessary. None of them are needed for anything in this unit: they are convergence hypotheses, and they belong with the convergence theorems.

Method(Computing a Fourier series)AdaptedAdapted from Strauss (2008), §5.1, eqs. (4), (7), (9)–(10), pp. 104–107, and §5.2, eqs. (6)–(10), p. 115; Riley, Hobson & Bence (2006), §12.3, p. 419, and §12.5, p. 422; Arfken, Weber & Harris (2013), §19.1, Summing Fourier Series, Example 19.1.3, p. 944, and eqs. (19.11)–(19.12), p. 939

Fix the period and the extension. Decide what the function is outside the interval you were given. If it is already periodic, use its period. If it is given on $(0,l)$ and you are free to choose, choose the extension the application forces: odd (sine series) for data that must vanish at both ends, even (cosine series) for data whose derivative must vanish at both ends, plain periodic otherwise. Record the resulting period: $2l$ for either half-range series.
Use symmetry before integrating. By part 3, an even function has no sine terms and an odd function no cosine terms, which halves the work. For a half-range series only one family appears by construction.
Choose the form. The real form is usually easier for polynomial and piecewise-constant data; the complex form is usually easier when the integrand contains $\mathrm{e}^{\alpha x}$ , $\cosh$ , or a product of trigonometric functions, since then everything is an exponential and the integral is elementary. Convert at the end with the dictionary $\hat f(\pm n)=\tfrac12(a_n\mp\mathrm{i}b_n)$ .
Evaluate the integrals. For a polynomial integrand, integrate by parts repeatedly, each step lowering the degree by one and producing a boundary term; for data defined piecewise, split the integral at the break-points. Arfken, Weber and Harris's complementary device is worth remembering in the other direction: to identify the sum of a given trigonometric series, replace the trigonometric functions by exponentials and look for a Maclaurin series you recognise — their example sums $\sum_{n\ge1}n^{-1}\cos nx$ to $-\ln\lvert 2\sin(x/2)\rvert$ by recognising two copies of $-\ln(1-z)$ .
Check before you use it. Four cheap tests catch most errors. (i) $a_0/2=\hat f(0)$ must equal the mean of $f$ over a period; in particular it vanishes for an odd function. (ii) The parity predicted by step 2 must actually hold in the answer. (iii) Evaluate the series at a point where you know $f$ and where the trigonometric functions are simple ( $x=0$ , $x=l/2$ , $x=l$ ) and compare; this both tests the coefficients and, as a by-product, evaluates a numerical series. (iv) Check the decay rate of the coefficients against the smoothness of the periodic extension you built in step 1, using Smoothness and decay of Fourier coefficients: coefficients falling like $1/n$ signal a jump, like $1/n^2$ a corner. If step 1 produced a jump you did not intend, this is where it shows up.

Example(The two half-range series of a parabolic profile)Original

Take $\varphi(x)=x(L-x)$ on $(0,L)$ , with $L>0$ , so the half-period is $l=L$ and both half-range series have period $2L$ . (Note the clash of notation with the definition block, where $L$ was the period: in this example and in the application below, $L$ is the length of the interval the function is given on, so the period of either half-range series is $2L$ .) The function is non-negative, vanishes at both ends, and peaks at $L^2/4$ in the middle: it is the shape of a string pulled aside with a smooth parabolic profile.

Step 1: the sine series. With $\varphi(0)=\varphi(L)=0$ the odd extension is continuous, so the sine series is the natural choice. Integrating by parts twice, with $s=\sin(n\pi x/L)$ and $c=\cos(n\pi x/L)$ ,

\int_0^L x(L-x)\,s\,\mathrm{d}x =\Bigl[-\frac{L}{n\pi}x(L-x)\,c\Bigr]_0^L+\frac{L}{n\pi}\int_0^L (L-2x)\,c\,\mathrm{d}x .

The boundary term vanishes at both ends because $x(L-x)$ does. In the remaining integral, integrate by parts once more:

\frac{L}{n\pi}\int_0^L(L-2x)\,c\,\mathrm{d}x =\frac{L}{n\pi}\Bigl(\Bigl[\frac{L}{n\pi}(L-2x)\,s\Bigr]_0^L+\frac{2L}{n\pi}\int_0^L s\,\mathrm{d}x\Bigr) =\frac{2L^2}{n^2\pi^2}\cdot\frac{L}{n\pi}\bigl(1-(-1)^n\bigr),

where the $s$ boundary term vanished because $\sin 0=\sin n\pi=0$ , and $\int_0^L s\,\mathrm{d}x=(L/n\pi)(1-\cos n\pi)$ . Multiplying by $2/L$ ,

B_n=\frac{4L^2\bigl(1-(-1)^n\bigr)}{n^3\pi^3} =\begin{cases}\dfrac{8L^2}{n^3\pi^3},& n \text{ odd},\\[4pt] 0,& n \text{ even},\end{cases} \qquad\text{so}\qquad x(L-x)\;\sim\;\frac{8L^2}{\pi^3}\sum_{n\ \mathrm{odd}}\frac{1}{n^3}\sin\frac{n\pi x}{L}.

The even coefficients vanish for a reason visible in the picture: $\varphi$ is symmetric about $x=L/2$ , and $\sin(n\pi x/L)$ is antisymmetric about that point for even $n$ .

Step 2: a checkable number. Put $x=L/2$ , where $\varphi=L^2/4$ and $\sin(n\pi/2)=(-1)^{(n-1)/2}$ for odd $n$ . The series does converge to $\varphi$ there, by the following two steps. First, the series converges uniformly on $\mathbb{R}$ , by the Weierstrass M-test (Uniform convergence); the reason is one line, so we give it rather than lean on it: $\lvert B_n\sin(n\pi x/L)\rvert\le B_n$ for every $x$ , so the tail beyond $N$ is bounded by $\sum_{n>N,\ n\ \mathrm{odd}}8L^2/(n^3\pi^3)$ , a bound independent of $x$ which tends to $0$ because $\sum n^{-3}$ converges. Write $s$ for the sum. By part 2, $s$ is continuous, has period $2L$ , and has Fourier coefficients exactly $B_n$ in the sine slots and $0$ elsewhere. Second, $\varphi_{\mathrm{odd}}$ has those same coefficients, by part 4, and it is continuous on the whole line: it vanishes at $0$ and at $\pm L$ , which are the only places the pieces of the $2L$ -periodic odd extension are joined. So $s-\varphi_{\mathrm{odd}}$ is continuous with every Fourier coefficient zero, and Stein and Shakarchi's Corollary 2.2 (quoted in the remark above) forces $s=\varphi_{\mathrm{odd}}$ identically. Part 2 alone could not have done this — it fixes the coefficients of $s$ and nothing more — which is exactly the point made in the remark. Evaluating at $x=L/2$ we get $L^2/4=(8L^2/\pi^3)\sum_{k\ge0}(-1)^k(2k+1)^{-3}$ , that is

\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32}=0.968946146\ldots

Numerically the left side is $0.9689461462593694$ to sixteen figures and $\pi^3/32=0.9689461462593694$ ; the partial sums of the Fourier series at $x=L/2$ , in units of $L^2$ , run $0.2580$ , $0.2485$ , $0.2505$ , $0.24993$ , $0.25001$ for $1,2,3,6,11$ nonzero terms, oscillating in towards $0.25$ as the alternating series requires.

Step 3: the cosine series of the same function. Now extend evenly instead. The mean is

\frac{A_0}{2}=\frac1L\int_0^L x(L-x)\,\mathrm{d}x=\frac1L\Bigl(\frac{L^3}{2}-\frac{L^3}{3}\Bigr)=\frac{L^2}{6},

and two integrations by parts in the same pattern — this time the first boundary term vanishes because $\sin$ does at both ends, and the second survives because $\cos n\pi=(-1)^n$ — give

A_n=\frac{2}{L}\int_0^L x(L-x)\cos\frac{n\pi x}{L}\,\mathrm{d}x =\frac{2L^2\bigl((-1)^{n+1}-1\bigr)}{n^2\pi^2} =\begin{cases}0,& n\text{ odd},\\[2pt] -\dfrac{4L^2}{n^2\pi^2},& n\text{ even},\end{cases}

so, writing $n=2k$ ,

x(L-x)\;\sim\;\frac{L^2}{6}-\frac{L^2}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}\cos\frac{2k\pi x}{L}.

Step 4: the second checkable number. The same two steps apply here: $\sum_k k^{-2}<\infty$ makes the cosine series uniformly convergent, with a continuous sum whose coefficients are the $A_n$ ; $\varphi_{\mathrm{even}}$ is continuous on the line, since the even extension of $x(L-x)$ takes the value $0$ at both $x=0$ and $x=\pm L$ , where its pieces join; so the two continuous functions have the same coefficients and Corollary 2.2 identifies them. Setting $x=0$ , where $\varphi_{\mathrm{even}}=0$ and every cosine equals $1$ , the two displays therefore force $0=L^2/6-(L^2/\pi^2)\sum_{k\ge1}k^{-2}$ , that is $\sum_{k\ge1}k^{-2}=\pi^2/6$ — the correct value, which is the test passing rather than a new derivation.

What the two series show. The same $\varphi$ on $(0,L)$ has been written in two different ways, and both are right on $(0,L)$ . They differ in their periods' content outside it ( $\varphi_{\mathrm{odd}}$ against $\varphi_{\mathrm{even}}$ ) and in their decay: $n^{-3}$ for the sine series against $n^{-2}$ for the cosine series. The reason is step 1 of the method. The odd extension of $x(L-x)$ is $x(L+x)$ on $(-L,0)$ , and it is continuous with a continuous first derivative at $x=0$ and at $x=\pm L$ : its derivative is $L-2\lvert x\rvert$ on $(-L,L)$ , an even function, which matches itself across the join at $0$ and takes the common value $-L$ at both ends of the period. The even extension is $-x(L+x)$ on $(-L,0)$ , whose derivative $-(L+2x)$ tends to $-L$ at $0-$ against $+L$ at $0+$ : a corner at each join. One extension is smoother than the other, and the coefficients report it. Every coefficient and every numerical value above was checked symbolically with sympy and numerically to twelve figures with mpmath.

ApplicationOriginal

A plucked string, mode by mode. A string of length $L$ fixed at both ends, released from rest with displacement $\varphi(x)=x(L-x)$ scaled to a peak of $h$ — so the initial profile is $4h\,x(L-x)/L^2$ — vibrates as $u(x,t)=\sum_n C_n\sin(n\pi x/L)\cos(n\pi ct/L)$ . The boundary conditions are Dirichlet at both ends, so step 1 of the method sends us to the sine series, and the example delivers the amplitudes outright:

C_n=\frac{4h}{L^2}B_n=\frac{32h}{n^3\pi^3}\ \ (n\text{ odd}),\qquad C_n=0\ \ (n\text{ even}).

The fundamental carries amplitude $32h/\pi^3\approx1.032\,h$ , the third harmonic $1/27$ of that, the fifth $1/125$ . Energy goes as amplitude squared times $n^2$ , so the third harmonic holds about $(1/27)^2\times 9\approx1.2\%$ of the fundamental's energy: a smoothly plucked string is very nearly a pure tone, and the even harmonics are missing entirely because the pluck is symmetric about the midpoint. Pluck it off-centre and the symmetry argument fails, the even harmonics appear, and the tone brightens — which is why guitarists pick near the bridge for a harder sound. The same coefficients, with $\cos(n\pi ct/L)$ replaced by $\mathrm{e}^{-n^2\pi^2\kappa t/L^2}$ , solve the heat equation on the same interval with the same initial profile and ends held at zero; there the high harmonics die fastest, and after a short time the profile is the fundamental alone.

Insulated ends need the other series. Replace "ends held at zero temperature" by "ends perfectly insulated" and the boundary conditions become $u_x=0$ at both ends. Now the admissible modes are $\cos(n\pi x/L)$ , the extension to use is even, and the right expansion of the same initial profile is the cosine series of step 3 — including its constant term, which for the scaled profile $4h\,x(L-x)/L^2$ is $(4h/L^2)(L^2/6)=2h/3$ — the mean of the initial profile, and exactly the final steady state, since every other term decays. (The bare $L^2/6$ of step 3 is the constant term of the unscaled $x(L-x)$ .) Using the sine series here would impose the wrong boundary condition and predict total cooling instead of equilibration at the mean. The choice of extension is a physical statement.

Harmonic content of a periodic drive. When a linear system is driven by a periodic force of period $T$ , writing the force as a Fourier series in $t$ replaces one hard problem by a family of easy ones: each term $\hat F(n)\mathrm{e}^{2\pi\mathrm{i}nt/T}$ drives the system at a single frequency $n/T$ , the response to it is the transfer function at that frequency times the coefficient, and the total response is the sum. The numbers $\lvert\hat F(n)\rvert^2$ are the power spectrum of the drive, and the usefulness of the decomposition depends entirely on linearity: for a nonlinear system the responses do not add, and the harmonics of the drive are not the harmonics of the response. It is also the reason a sharp-edged drive is hard on a resonant system: a jump in the forcing makes the coefficients decay only like $1/n$ , so the drive has appreciable amplitude at arbitrarily high harmonics, and any one of them that coincides with a resonance is excited.

Theorem

Pointwise convergence of Fourier series

Adapted

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Writes the symmetric partial sum as an average of the function against the Dirichlet kernel and proves that it converges at any point where the one-sided limits and the one-sided derivatives of the function exist, the limit being the average of the two one-sided values, so that a piecewise-smooth function is recovered everywhere except that each jump is split down the middle.

Prerequisites: Fourier series, Dominated convergence

Definition(The Dirichlet kernel)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, §1.1, Example 4, p. 37; Strauss (2008), §5.5, eqs. (4)–(5), p. 137; Stakgold & Holst (2011), §2.3, eqs. (2.3.15)–(2.3.16), p. 133; Folland (1999), §8.5, eq. (8.40), p. 264

Throughout this unit the period is $2\pi$ , so a function on the circle is a $2\pi$ -periodic function on $\mathbb{R}$ , the Fourier coefficients of Fourier series read

\hat f(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)\,\mathrm{e}^{-\mathrm{i}ny}\,\mathrm{d}y ,

and the symmetric partial sum is $S_N f(x)=\sum_{\lvert n\rvert\le N}\hat f(n)\mathrm{e}^{\mathrm{i}nx}$ . Nothing is lost by this choice: a period $T$ is brought to $2\pi$ by the substitution $x\mapsto 2\pi x/T$ , which is Stakgold and Holst's change of scale, and every statement below transfers with $N+\tfrac12$ rescaled accordingly.

The $N$ th Dirichlet kernel is the trigonometric polynomial

D_N(x)=\sum_{n=-N}^{N}\mathrm{e}^{\mathrm{i}nx} =\frac{\sin\bigl[(N+\tfrac12)x\bigr]}{\sin(x/2)} ,

the second expression being understood by its limit, $D_N(0)=2N+1$ , at $x=0$ and at the other multiples of $2\pi$ . It is real, even, $2\pi$ -periodic, and normalised so that

\frac{1}{2\pi}\int_{-\pi}^{\pi}D_N(x)\,\mathrm{d}x = 1 .

Conventions. We keep the $1/2\pi$ outside the kernel. Stakgold and Holst absorb it, writing $D_n(x)=\sin[(n+\tfrac12)x]/[2\pi\sin(x/2)]$ with $\int_{-\pi}^{\pi}D_n=1$ ; Strauss writes the same function as ours but calls it $K_N$ ; Folland works on an interval of length $1$ with characters $\mathrm{e}^{2\pi\mathrm{i}kx}$ , where the closed form becomes $D_m(x)=\sin[(2m+1)\pi x]/\sin(\pi x)$ . These are the same object in different coordinates. The letter $D_N$ and the name are Stein and Shakarchi's and are standard.

Theorem(Convergence at a point with one-sided derivatives)AdaptedAdapted from Stakgold & Holst (2011), §2.3, Thm 2.3.3 with Remarks 1 and 2, pp. 132–133, the class $D(S)$, pp. 130–131, and the theorem on $C^1(S)$, p. 134; Strauss (2008), §5.4, Thm 4 and Thm 4$^\\infty$, p. 129

Let $f$ be $2\pi$ -periodic and integrable over one period. Let $x_0\in\mathbb{R}$ be a point at which the four one-sided data

f(x_0\pm)=\lim_{h\to0\pm}f(x_0+h), \qquad f'(x_0\pm)=\lim_{h\to0\pm}\frac{f(x_0+h)-f(x_0\pm)}{h}

all exist and are finite. Then

\lim_{N\to\infty}S_N f(x_0)=\frac{f(x_0+)+f(x_0-)}{2}.

Three readings of this, in increasing order of convenience and decreasing order of generality.

At a point of differentiability. If $f$ is differentiable at $x_0$ , then both one-sided limits equal $f(x_0)$ and both one-sided derivatives equal $f'(x_0)$ , so $S_Nf(x_0)\to f(x_0)$ . Only differentiability at the one point is used; the behaviour of $f$ elsewhere enters only through the requirement that it be integrable.
For piecewise-smooth data. Say that $f$ satisfies the Dirichlet conditions, written $f\in D$ , if $f$ is piecewise continuous and the circle can be cut into finitely many arcs on each of which $f$ has a continuous derivative, with one-sided derivatives existing at the ends of each arc. This is Stakgold and Holst's class $D(S)$ . Every $f\in D$ meets the hypotheses at every point, so the Fourier series of such an $f$ converges everywhere, to $f$ at each point of continuity and to the midpoint of the jump at each of the finitely many points of discontinuity. This is Strauss's Theorem 4 $^\infty$ , stated there as: $f$ and $f'$ piecewise continuous.
With continuity added. If $f\in D$ and $f$ is also continuous, the series converges to $f$ at every point. If moreover $f$ is continuously differentiable on the whole circle, the convergence is uniform, by a separate argument given in Smoothness and decay of Fourier coefficients.

The half-sum is forced, not chosen: $S_Nf$ cannot distinguish $f$ from the function obtained by resetting $f$ at its jumps to the half-sum there, since changing a function at finitely many points changes no Fourier coefficient. Stakgold and Holst make this the convention, redefining $f$ at a jump to be $\tfrac12[f(x_0+)+f(x_0-)]$ before stating the theorem; with that convention the conclusion reads $S_Nf(x_0)\to f(x_0)$ in all cases.

Idea of the proofOriginal

Summing the Fourier series to order $N$ is the same as averaging $f$ against the Dirichlet kernel, because substituting the coefficient integrals into $S_Nf$ and exchanging a finite sum with an integral produces exactly that average. If $D_N$ were a bump of unit mass, non-negative and concentrating at the origin, the average would be a weighted mean of nearby values of $f$ and convergence would be routine. It is not: $D_N$ oscillates and changes sign, and the total mass of $\lvert D_N\rvert$ grows without bound. So cancellation, not concentration, has to do the work.

The way to exploit the cancellation is to use the one thing $D_N$ does have, unit mass, to subtract off the answer before estimating. What is left is the integral of $\sin[(N+\tfrac12)\theta]$ against the quotient $[f(x_0+\theta)-f(x_0\pm)]/\sin(\theta/2)$ . That quotient looks singular at $\theta=0$ , where the denominator vanishes — and it is precisely the existence of the one-sided derivatives that makes the numerator vanish at the same rate, so the quotient stays bounded. Once the quotient is an honest integrable function, the oscillation of $\sin[(N+\tfrac12)\theta]$ at ever-higher frequency makes the integral die, because the functions $\sin[(N+\tfrac12)\theta]$ are mutually orthogonal and no fixed function can have large components along infinitely many orthogonal directions.

ProofAdaptedAdapted from Strauss (2008), §5.5, the five steps for $C^1$ data, eqs. (3)–(11), pp. 137–139, and the modification for jumps, eqs. (12)–(14), p. 140; Stakgold & Holst (2011), §2.3, the proof of Thm 2.3.3 and eqs. (2.3.15)–(2.3.18), pp. 133–134, and Remark 1 on p. 130; Folland (1999), §8.5, Lemma 8.42 (the half-masses of $D_m$), p. 265

Step 1: the closed form of $D_N$ . Multiply the defining sum by $\mathrm{e}^{\mathrm{i}x/2}-\mathrm{e}^{-\mathrm{i}x/2}$ and look at what each term contributes:

\bigl(\mathrm{e}^{\mathrm{i}x/2}-\mathrm{e}^{-\mathrm{i}x/2}\bigr)D_N(x) =\sum_{n=-N}^{N}\Bigl(\mathrm{e}^{\mathrm{i}(n+\frac12)x}-\mathrm{e}^{\mathrm{i}(n-\frac12)x}\Bigr).

Consecutive terms cancel: the first part of the $n$ th summand is the second part of the $(n+1)$ st. Only the two ends survive, leaving $\mathrm{e}^{\mathrm{i}(N+\frac12)x}-\mathrm{e}^{-\mathrm{i}(N+\frac12)x}=2\mathrm{i}\sin[(N+\tfrac12)x]$ . Since the prefactor is $2\mathrm{i}\sin(x/2)$ , dividing gives the closed form wherever $\sin(x/2)\neq0$ . At $x=0$ the original sum has $2N+1$ terms each equal to $1$ , and the quotient has the same limit by l'Hôpital, so the two expressions agree there too.

Step 2: unit mass, and unit mass on each half. Integrating the defining sum term by term over $(-\pi,\pi)$ , every $n\neq0$ contributes $0$ by the orthogonality relations of Fourier series, and $n=0$ contributes $2\pi$ . Hence $(2\pi)^{-1}\int_{-\pi}^{\pi}D_N=1$ . Because $D_N$ is even — the closed form is a ratio of two odd functions — each half carries half:

\frac{1}{2\pi}\int_{0}^{\pi}D_N(\theta)\,\mathrm{d}\theta=\frac{1}{2\pi}\int_{-\pi}^{0}D_N(\theta)\,\mathrm{d}\theta=\frac12 .

This is the second half of Folland's Lemma 8.42, and it is what lets the two one-sided limits be subtracted separately in Step 3.

Step 3: the partial sum is an average against $D_N$ , and the error is one integral. Substituting the coefficient integrals into $S_Nf(x)$ and exchanging the finite sum with the integral,

S_Nf(x)=\sum_{\lvert n\rvert\le N}\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)\mathrm{e}^{\mathrm{i}n(x-y)}\,\mathrm{d}y =\frac{1}{2\pi}\int_{-\pi}^{\pi}f(y)\,D_N(x-y)\,\mathrm{d}y .

Put $y=x+\theta$ . The interval of integration becomes $[-\pi-x,\pi-x]$ , but the integrand $f(x+\theta)D_N(\theta)$ has period $2\pi$ in $\theta$ , so by part 1 of the statement in Fourier series the integral over any interval of length $2\pi$ is the same, and $D_N(-\theta)=D_N(\theta)$ lets us drop the sign:

S_Nf(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+\theta)\,D_N(\theta)\,\mathrm{d}\theta .

Now fix $x=x_0$ and subtract, using Step 2 on each half separately — this is where having the two half-masses, rather than only the total, pays:

S_Nf(x_0)-\frac{f(x_0+)+f(x_0-)}{2} =\frac{1}{2\pi}\int_{0}^{\pi}\!\!\bigl[f(x_0+\theta)-f(x_0+)\bigr]D_N(\theta)\,\mathrm{d}\theta +\frac{1}{2\pi}\int_{-\pi}^{0}\!\!\bigl[f(x_0+\theta)-f(x_0-)\bigr]D_N(\theta)\,\mathrm{d}\theta .

Insert the closed form of $D_N$ and define

g_\pm(\theta)=\frac{f(x_0+\theta)-f(x_0\pm)}{\sin(\theta/2)} ,

so that the error is

E_N:=S_Nf(x_0)-\frac{f(x_0+)+f(x_0-)}{2} =\frac{1}{2\pi}\int_{0}^{\pi}g_+(\theta)\,\phi_N(\theta)\,\mathrm{d}\theta +\frac{1}{2\pi}\int_{-\pi}^{0}g_-(\theta)\,\phi_N(\theta)\,\mathrm{d}\theta , \qquad \phi_N(\theta)=\sin\bigl[(N+\tfrac12)\theta\bigr].

Everything now rests on showing that both integrals vanish as $N\to\infty$ .

Step 4: the oscillating factors are an orthogonal family. On $(0,\pi)$ , for integers $N,M\ge0$ , the product-to-sum identity gives

\int_0^\pi \phi_N\phi_M =\frac12\int_0^\pi\Bigl(\cos\bigl[(N-M)\theta\bigr]-\cos\bigl[(N+M+1)\theta\bigr]\Bigr)\mathrm{d}\theta .

Both $N-M$ and $N+M+1$ are integers, and $\int_0^\pi\cos(k\theta)\,\mathrm{d}\theta=\sin(k\pi)/k=0$ for every nonzero integer $k$ , while $N+M+1\ge1$ is never zero. So the second term always vanishes, and the first vanishes unless $N=M$ , in which case it contributes $\pi/2$ . Thus $\{\phi_N\}_{N\ge0}$ is orthogonal on $(0,\pi)$ with $\lVert\phi_N\rVert^2=\pi/2$ . Each $\phi_N$ is odd, so each product $\phi_N\phi_M$ is even; the identical computation on $(-\pi,0)$ therefore gives the same conclusion there.

Step 5: a square-integrable function has vanishing components along an orthogonal family. Let $\{\psi_N\}$ be an orthogonal family of nonzero vectors in an inner product space and let $g$ be square-integrable, so that $\lVert g\rVert^{2}=(g,g)<\infty$ . Put $c_N=(g,\psi_N)/\lVert\psi_N\rVert^2$ . Expanding the square of the norm of the residual, and using orthogonality to kill the cross terms,

0\le\Bigl\lVert g-\sum_{N\le M}c_N\psi_N\Bigr\rVert^2 =\lVert g\rVert^2-\sum_{N\le M}\frac{\lvert(g,\psi_N)\rvert^2}{\lVert\psi_N\rVert^2} ,

so the partial sums of the non-negative series $\sum_N\lvert(g,\psi_N)\rvert^2/\lVert\psi_N\rVert^2$ are bounded above by $\lVert g\rVert^2$ . A non-negative series with bounded partial sums converges, and the terms of a convergent series tend to zero; hence $(g,\psi_N)\to0$ . (This inequality is Bessel's, and it is proved in its own right in Mean-square convergence and Parseval's identity; the two lines above are given here so that this unit does not depend on that one.)

Step 6: $g_\pm$ is bounded near $0$ , by the one-sided derivative hypothesis. Factor the quotient as

g_+(\theta)=\underbrace{\frac{f(x_0+\theta)-f(x_0+)}{\theta}}_{\to\,f'(x_0+)}\cdot\underbrace{\frac{\theta}{\sin(\theta/2)}}_{\to\,2} \qquad(\theta\to0+).

The second factor is harmless on all of $(0,\pi]$ : Jordan's inequality $\sin u\ge 2u/\pi$ on $[0,\pi/2]$ , applied with $u=\theta/2$ , gives $\sin(\theta/2)\ge\theta/\pi$ and hence $0<\theta/\sin(\theta/2)\le\pi$ . The first factor converges to $f'(x_0+)$ by hypothesis, so it is bounded on some interval $(0,\delta]$ . Therefore $g_+$ is bounded on $(0,\delta]$ , say by $C$ . On the remainder $[\delta,\pi]$ the denominator is bounded below by $\sin(\delta/2)>0$ , so

\lvert g_+(\theta)\rvert\le\frac{\lvert f(x_0+\theta)\rvert+\lvert f(x_0+)\rvert}{\sin(\delta/2)} ,

which is integrable because $f$ is. So $g_+$ is integrable on $(0,\pi)$ , and bounded near the only point that threatened it. The same holds for $g_-$ on $(-\pi,0)$ .

Step 7: from integrable to square-integrable, and the conclusion. If $f$ is bounded — which covers every piecewise-continuous $f$ , hence the whole of class $D$ — then $g_+$ is bounded on $(0,\pi)$ by Step 6 and so square-integrable there, and Steps 4 and 5 give $\int_0^\pi g_+\phi_N\to0$ at once. For a general integrable $f$ one extra line is needed, and it is Stakgold and Holst's device of splitting off a bounded part. Given $\varepsilon>0$ , let $u$ be the truncation of $g_+$ at height $M$ , that is $u=g_+$ where $\lvert g_+\rvert\le M$ and $u=0$ elsewhere. Then $\int_0^\pi\lvert g_+-u\rvert=\int_{\{\lvert g_+\rvert>M\}}\lvert g_+\rvert$ , which tends to $0$ as $M\to\infty$ because $g_+$ is integrable (Dominated convergence); fix $M$ so large that this is below $2\pi\varepsilon$ . Since $\lvert\phi_N\rvert\le1$ ,

\left\lvert\frac{1}{2\pi}\int_0^\pi g_+\phi_N\right\rvert \le\left\lvert\frac{1}{2\pi}\int_0^\pi u\,\phi_N\right\rvert+\frac{1}{2\pi}\int_0^\pi\lvert g_+-u\rvert <\left\lvert\frac{1}{2\pi}\int_0^\pi u\,\phi_N\right\rvert+\varepsilon ,

and the first term tends to $0$ by Steps 4 and 5, since $u$ is bounded and hence square-integrable. So $\limsup_N\lvert(2\pi)^{-1}\int_0^\pi g_+\phi_N\rvert\le\varepsilon$ for every $\varepsilon>0$ , which makes the limit zero. The same argument applies to $g_-$ on $(-\pi,0)$ . Both terms of $E_N$ vanish in the limit, so $E_N\to0$ , which is the assertion. $\square$

The three readings. Reading 1 is the case $f(x_0+)=f(x_0-)=f(x_0)$ . Reading 2 holds because on each arc of the subdivision $f$ has a continuous derivative extending to the endpoints, so at an interior point of an arc $f$ is differentiable and at an endpoint both one-sided derivatives exist by assumption; a piecewise-continuous $f$ is bounded, so Step 7 needs only its first sentence. Reading 3 adds continuity, under which the half-sum is $f(x_0)$ .

Remark(Where the hypotheses are used)AdaptedAdapted from Folland (1999), §8.5, the attribution to Dirichlet in 1829, p. 264, Thm 8.43, p. 266, Thm 8.44 and Cor. 8.45, pp. 267–268, and the discussion of $\\lVert D_m\\rVert_1$ with Exercises 34–35, pp. 264 and 269; Strauss (2008), §5.4, p. 125; Stakgold & Holst (2011), §2.3, Remarks 1 and 2 after Thm 2.3.3, p. 133, and the description of the class $D$, p. 131; Arfken, Weber & Harris (2013), §19.1, the footnote to the Dirichlet conditions, p. 935; Stein & Shakarchi (2003), Ch. 2, §1.1, p. 39; Riley, Hobson & Bence (2006), §12.1, pp. 415–416

The one-sided derivatives are used once, in Step 6, and for one purpose: to cancel the zero of $\sin(\theta/2)$ in the denominator of $g_\pm$ . Nothing else in the proof refers to them. This is why the hypothesis is local — it is imposed at $x_0$ and nowhere else — which is Stakgold and Holst's first remark on the theorem: the convergence at a point is decided by the behaviour of $f$ there, even though every coefficient is a global integral. Folland's Theorem 8.44 is the sharp form of that observation: if $f$ and $g$ agree on an open interval $I$ , then $S_Nf-S_Ng\to0$ uniformly on compact subsets of $I$ , whatever the two functions do elsewhere.

Integrability of $f$ over a period is used twice: to have coefficients at all, and in Step 6 to control $g_\pm$ away from the origin. Step 7 is the only place where the difference between integrable and bounded matters, and it is a technicality.

Some local hypothesis is genuinely needed; continuity alone is not enough. The reason is visible in Step 3: $D_N$ has unit mass but $\lVert D_N\rVert_{1}=(2\pi)^{-1}\int_{-\pi}^{\pi}\lvert D_N\rvert$ is unbounded in $N$ , so the maps $f\mapsto S_Nf(x_0)$ have unbounded norms on the continuous functions. Folland draws the conclusion (his Exercises 34 and 35, by the uniform boundedness principle): there are continuous functions whose Fourier series diverge at a point, indeed at a dense set of points. Strauss records a sharper classical fact still: some integrable functions have a Fourier series that converges nowhere at all. In the other direction, Stein and Shakarchi note Carleson's theorem: for square-integrable $f$ the series converges at almost every point. So the theorem above is not a technical convenience hiding a stronger truth; without a hypothesis of this kind, pointwise convergence fails.

The hypothesis here is sufficient, never necessary, and does not contain the alternatives. Riley, Hobson and Bence and Arfken, Weber and Harris both quote a list called the Dirichlet conditions — periodicity, single-valuedness with finitely many finite jumps, finitely many maxima and minima per period, and convergence of $\int\lvert f\rvert$ over a period — and Arfken, Weber and Harris add explicitly that these are sufficient but not necessary. Folland proves a different sufficient condition: if $f$ is of bounded variation then $S_Nf(x)\to\tfrac12[f(x+)+f(x-)]$ at every $x$ , a result he attributes to Dirichlet in 1829 for piecewise continuous, piecewise monotone $f$ and to later refinements for bounded variation. Neither that hypothesis nor ours implies the other:

Bounded variation without one-sided derivatives. Let $f$ be the odd $2\pi$ -periodic extension of $\sqrt{x}$ on $(0,\pi)$ . It is increasing on $(-\pi,\pi)$ and hence of bounded variation, so Folland's theorem applies at $x_0=0$ and gives $S_Nf(0)\to0$ . Our hypothesis fails there: the difference quotient $\sqrt{h}/h=h^{-1/2}$ has no finite limit, so $f'(0+)$ does not exist.
One-sided derivatives without bounded variation. Let $f(x)=x^{2}\sin(x^{-3})$ for $x\neq0$ and $f(0)=0$ , extended $2\pi$ -periodically. Then $f(h)/h=h\sin(h^{-3})\to0$ , so $f'(0\pm)=0$ exist and our theorem applies at $x_0=0$ . But $f'(x)$ contains the term $-3x^{-2}\cos(x^{-3})$ , whose absolute integral over $(0,\delta)$ diverges, so $f$ is not of bounded variation near $0$ .

So the two theorems are genuinely different statements, and a function may satisfy either, both or neither. Stakgold and Holst's class $D$ is also incomparable with continuity in the same spirit: $D$ admits jumps, which continuity forbids, and excludes cusps such as $\sqrt{\lvert x\rvert}$ , which continuity allows.

What fails without the half-sum. The conclusion cannot be improved to " $S_Nf(x_0)\to f(x_0)$ " at a jump, for the reason given under the statement: $\hat f$ does not see the value of $f$ at a single point, so no theorem about $\hat f$ can predict it. Nor is the convergence uniform near a jump, since a uniform limit of continuous functions is continuous (Uniform convergence); what actually happens there is the subject of The Gibbs phenomenon. If uniform convergence is what is wanted from data with jumps, one changes the summation method rather than the hypotheses: see Cesàro means and Fejér's theorem.

Method(Deciding where a Fourier series converges, and to what)Original

Write down the function the series actually represents. That is the periodic extension fixed in step 1 of the method of Fourier series: the plain periodic extension for a full series, the odd one for a sine series, the even one for a cosine series. Every question below is about that function on the whole line, not about the formula on the interval you started with.
List the candidate bad points in one period. They are: the break-points of a piecewise definition; the ends of the period interval, where the extension may jump; the origin and the ends of the half-interval, for a half-range series; and any point where $f$ has a vertical tangent or a cusp.
At every other point the data this method is written for — piecewise-smooth $f$ , the class $D$ of the statement — is differentiable, so the series converges to $f(x)$ there and no computation is needed. (For a general integrable $f$ no such blanket statement is available: the theorem has to be checked point by point, and the remark below records that for merely continuous $f$ it can fail.)
At each candidate point, check the four one-sided data. If $f(x_0\pm)$ and $f'(x_0\pm)$ all exist and are finite, the series converges to $\tfrac12[f(x_0+)+f(x_0-)]$ — equal to $f(x_0)$ if $f$ is continuous there, and the midpoint of the jump otherwise. For data built out of polynomials, exponentials and trigonometric functions on finitely many pieces, this check always passes, and it is usually enough to say so.
If a one-sided derivative is infinite, the theorem is silent, and the fallback is the bounded-variation test. Folland's Theorem 8.43 asks for bounded variation over a whole period, not merely near $x_0$ : if the periodic extension is of bounded variation on $[-\pi,\pi]$ — which holds, for instance, when it is piecewise monotone with finitely many pieces, as $\sqrt{x}$ , $x^{1/3}$ and the like are — his theorem delivers the half-sum at every point, $x_0$ among them. If the function is well behaved only near $x_0$ , buy the global hypothesis with the localization theorem: choose a $\tilde f$ of bounded variation on the period that agrees with $f$ on an open interval $I\ni x_0$ (cutting $f$ off outside $I$ will usually do), read the half-sum off Theorem 8.43 for $\tilde f$ , and transfer it to $f$ by Theorem 8.44, which gives $S_Nf-S_N\tilde f\to0$ on compact subsets of $I$ . If neither test applies, retreat to Cesàro means (Cesàro means and Fejér's theorem) or to mean-square convergence (Mean-square convergence and Parseval's identity), both of which need far less.
Read off the end-point values without computing anything. A sine series on $(0,l)$ converges to $0$ at $x=0$ and at $x=l$ , whatever $\varphi$ was, because its odd $2l$ -periodic extension has the half-sum $0$ there; so a sine series represents $\varphi$ at the ends only if $\varphi(0+)=\varphi(l-)=0$ . A cosine series converges to $\varphi(0+)$ and $\varphi(l-)$ at those points, because the even extension has no jump there. This single observation explains most of the surprises in half-range expansions.
Expect slow convergence wherever the extension jumps. If step 2 found a genuine jump, the coefficients decay only like $1/n$ (Smoothness and decay of Fourier coefficients) and the partial sums converge like $1/N$ at the jump and oscillate near it; budget terms accordingly, and see The Gibbs phenomenon before trusting values close to the discontinuity.

Example(The periodic extension of

\mathrm{e}^{x}

, and the value of the series at the jump)Original

Let $f$ be the $2\pi$ -periodic function with $f(x)=\mathrm{e}^{x}$ for $-\pi<x<\pi$ . On the open interval it is as smooth as a function can be; the whole of its bad behaviour sits at the odd multiples of $\pi$ , where the extension drops from $\mathrm{e}^{\pi}$ back to $\mathrm{e}^{-\pi}$ .

Step 1: the coefficients, by the complex route. For every $n\in\mathbb{Z}$ ,

\hat f(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\mathrm{e}^{x}\mathrm{e}^{-\mathrm{i}nx}\,\mathrm{d}x =\frac{1}{2\pi}\left[\frac{\mathrm{e}^{(1-\mathrm{i}n)x}}{1-\mathrm{i}n}\right]_{-\pi}^{\pi} =\frac{\mathrm{e}^{-\mathrm{i}n\pi}\bigl(\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}\bigr)}{2\pi(1-\mathrm{i}n)} =\frac{(-1)^{n}\sinh\pi}{\pi(1-\mathrm{i}n)} ,

using $\mathrm{e}^{\pm\mathrm{i}n\pi}=(-1)^{n}$ and $\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}=2\sinh\pi$ . Rationalising, $\hat f(n)=(-1)^{n}\sinh\pi\,(1+\mathrm{i}n)/[\pi(1+n^{2})]$ , so by the dictionary of Fourier series,

a_n=2\operatorname{Re}\hat f(n)=\frac{2(-1)^{n}\sinh\pi}{\pi(1+n^{2})}, \qquad b_n=-2\operatorname{Im}\hat f(n)=\frac{-2(-1)^{n}n\sinh\pi}{\pi(1+n^{2})} .

The first few: $a_0/2=\sinh\pi/\pi=3.676078$ , $a_1=-\sinh\pi/\pi=-3.676078$ , $b_1=+\sinh\pi/\pi=3.676078$ . The mean check of step 5 of the method in Fourier series passes: $(2\pi)^{-1}\int_{-\pi}^{\pi}\mathrm{e}^{x}\mathrm{d}x=\sinh\pi/\pi$ , as it must.

Step 2: apply the theorem at the jump. Take $x_0=\pi$ . The four data exist:

f(\pi-)=\mathrm{e}^{\pi}=23.140693,\qquad f(\pi+)=f(-\pi+)=\mathrm{e}^{-\pi}=0.043214,

and the one-sided derivatives are $f'(\pi-)=\mathrm{e}^{\pi}$ and $f'(\pi+)=\mathrm{e}^{-\pi}$ , since $f$ coincides with a smooth exponential on each side. So the theorem applies and predicts

\lim_{N\to\infty}S_Nf(\pi)=\frac{\mathrm{e}^{\pi}+\mathrm{e}^{-\pi}}{2}=\cosh\pi=11.5919533 .

The jump itself is $\mathrm{e}^{\pi}-\mathrm{e}^{-\pi}=2\sinh\pi=23.097479$ , and the predicted limit sits exactly halfway up it — well away from either one-sided value.

Step 3: check the prediction against the series, and harvest an identity. At $x=\pi$ every $\sin n\pi$ vanishes and $\cos n\pi=(-1)^n$ , so

S_Nf(\pi)=\frac{\sinh\pi}{\pi}\left(1+2\sum_{n=1}^{N}\frac{1}{1+n^{2}}\right).

Setting the limit equal to $\cosh\pi$ and dividing by $\sinh\pi/\pi$ turns the theorem into the statement $1+2\sum_{n\ge1}(1+n^{2})^{-1}=\pi\coth\pi$ , that is

\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}=\frac{\pi\coth\pi-1}{2}=1.0766740474686 .

Summing the left-hand side numerically and independently gives $1.0766740474686$ : the prediction is confirmed to thirteen figures.

Step 4: the same check at an interior point. At $x=0$ , $f$ is differentiable, so reading 1 applies and the series must converge to $f(0)=1$ . There $\sin 0=0$ and $\cos 0=1$ , so the series reads $(\sinh\pi/\pi)\bigl(1+2\sum_{n\ge1}(-1)^{n}(1+n^{2})^{-1}\bigr)=1$ , which rearranges to

\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}+1}=\frac{1}{2}\left(\frac{\pi}{\sinh\pi}-1\right)=-0.3639854725089 ,

and the numerical sum is $-0.3639854725089$ . Both identities are consequences of the theorem, not inputs to it.

Step 5: how fast. Convergence at the jump is slow, exactly as step 7 of the method warns, because the coefficients decay only like $1/n$ :

$N$	$S_Nf(\pi)$	error	$N\times$ error
$5$	$10.273058$	$-1.318896$	$-6.59$
$20$	$11.233666$	$-0.358287$	$-7.17$
$100$	$11.518801$	$-0.073153$	$-7.32$
$1000$	$11.584605$	$-0.007348$	$-7.35$

The last column settles down, so the error behaves like $c/N$ ; the numbers identify $c\approx-7.35$ , and $2\sinh\pi/\pi=7.3521$ is the obvious candidate, the jump divided by $\pi$ . Getting three correct decimal places at the jump therefore costs of order $10^{4}$ terms. All numbers in this example were computed to twenty digits with mpmath and the coefficients checked symbolically.

ApplicationOriginal

What a reconstruction returns at a switching instant. A square-wave clock, a rectified sinusoid, a sampled bit-stream: all are modelled by piecewise-smooth periodic functions, all are in class $D$ , and all therefore have everywhere-convergent Fourier series. But at each switching instant the series returns the mean of the two levels, not either level. A simulation that synthesises a $0/1$ logic waveform from harmonics and then samples it exactly at an edge reads $0.5$ , which is neither value and is not a numerical error; it is the theorem. The practical consequence is to sample strictly between edges, and the diagnostic consequence is that a reconstruction which returns something other than the midpoint at a known edge has a bug in its coefficients.

Initial data with a step, in separation of variables. Solving the heat equation $u_t=\kappa u_{xx}$ on $(0,l)$ with ends held at zero and a step initial temperature — hot on the left half, cold on the right — produces $u(x,t)=\sum_n B_n\mathrm{e}^{-\kappa n^{2}\pi^{2}t/l^{2}}\sin(n\pi x/l)$ with $B_n$ the sine coefficients of the step. For every $t>0$ the exponential factors make the series converge absolutely and the solution is smooth, so nothing in this unit is needed. It is needed at exactly one instant, $t=0$ : the theorem is what licenses the claim that the series represents the initial data, and it says precisely in what sense — everywhere except at the step, where it gives the mean of the two temperatures. A statement that the series equals the initial data everywhere would be false, and the failure is at one point, so it does not affect the solution for $t>0$ .

Where the convergence rate comes from, and what it costs. The table in the example is the general picture: a jump in the periodic extension forces $1/n$ coefficients and $1/N$ convergence, so three-figure accuracy near a discontinuity costs thousands of harmonics. This is the quantitative reason spectral methods for partial differential equations (Spectral methods) demand smooth, periodic data and lose their accuracy advantage the moment a shock forms, and the reason a filter designer who truncates a frequency response at a sharp cutoff gets a slowly converging and badly behaved impulse response. The check in step 2 of the method — is the periodic extension continuous? — is thus not pedantry but the first estimate of cost.

A template for other eigenfunction expansions. The architecture of the proof survives far beyond trigonometric series: express the partial sum of an expansion in the eigenfunctions of a Sturm-Liouville problem as an integral against a kernel built from those eigenfunctions, show the kernel has unit mass, subtract the candidate limit, and read off what local regularity makes the remaining quotient integrable. The resulting theorems for Legendre, Bessel and Hermite expansions have the same shape, including the half-sum at a jump; and so, as Arfken, Weber and Harris note, does the overshoot near one.

Example

The Gibbs phenomenon

Adapted

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Computes exactly what a truncated Fourier series does next to a jump: the partial sum of the square wave reaches 2Si(pi)/pi = 1.17898 times the half-jump at a point just under half a wavelength of the highest retained harmonic away from the discontinuity, so it overshoots each one-sided value by 8.95 percent of the jump, and the spike narrows without shrinking as more terms are taken.

Prerequisites: Pointwise convergence of Fourier series

Definition(Overshoot, and the three ways it is quoted)AdaptedAdapted from Strauss (2008), §5.5, eq. (18) and the surrounding paragraph, p. 142; Folland (1999), §8.5, pp. 268–269; Stakgold & Holst (2011), §2.3, p. 136; Hassani (2013), §9.1.1, eq. (9.14) and Box 9.1.5, pp. 274–275; Arfken, Weber & Harris (2013), §19.3, p. 961

Let $f$ be $2\pi$ -periodic and in the class $D$ of Pointwise convergence of Fourier series, and let $f$ have a jump at $x=a$ :

J = f(a+)-f(a-)\neq0, \qquad m=\frac{f(a+)+f(a-)}{2} .

By Pointwise convergence of Fourier series the partial sums $S_Nf$ converge at every point: to $f(x)$ at each point of continuity, and to the midpoint of the jump at each of the finitely many points where $f$ jumps, so to $m$ at $a$ . Pointwise, then, nothing is amiss. The Gibbs phenomenon is what happens to the graphs: for every large $N$ the graph of $S_Nf$ carries a spike beside $x=a$ that overruns both one-sided values — rising above the higher of $f(a\pm)$ and dipping below the lower — by amounts that do not shrink as $N$ grows, although the spike itself narrows. Equivalently, $S_Nf\to f$ is not uniform on any interval containing $a$ — which it cannot be, since a uniform limit of continuous functions would be continuous — and the failure is of a definite, computable size.

The size is always quoted through the same number,

\frac{2}{\pi}\operatorname{Si}(\pi)=1.1789797444722\ldots, \qquad \operatorname{Si}(u)=\int_0^u\frac{\sin t}{t}\,\mathrm{d}t ,

but divided by three different things, which is why the literature reports $9\%$ and $18\%$ for the same effect. Writing $P=\tfrac2\pi\operatorname{Si}(\pi)$ :

Per side, relative to the jump. Beside $a$ the extreme values of $S_Nf$ tend to $m\pm\tfrac12\lvert J\rvert P$ , so the upper one exceeds the higher of $f(a\pm)$ by $\tfrac12\lvert J\rvert(P-1)=0.0894899\,\lvert J\rvert$ : about $9\%$ of the jump, which is how Stakgold and Holst and Strauss state it. There is an equal dip below the lower of the two. (Nothing here depends on the sign of $J$ . The absolute value is carried rather than assumed away because for a downward jump, $J<0$ , it is $f(a-)$ that is the higher level and $f(a+)$ the lower, so the two roles are exchanged while the sizes are not.)
Per side, relative to the half-jump. The same excess is $(P-1)=17.898\%$ of the half-jump $\tfrac12\lvert J\rvert$ , the distance from the midpoint to either level. Folland's $0.5895$ for a unit jump is "about $18\%$ greater than $\phi(0+)=\tfrac12$ " in this sense, and Arfken, Weber and Harris's "some $18\%$ " is the factor $P$ applied to their amplitude $h/2$ .
Across the jump. The spike spans $\lvert J\rvert P$ in total instead of $\lvert J\rvert$ , so the jump of the partial sum is $1.179$ times the jump of the function. This is Hassani's eq. (9.14) and his Box 9.1.5.

All three are the one statement that the extremes of $S_Nf$ beside $a$ tend to $m\pm\tfrac12\lvert J\rvert P$ . We quote the first: the overshoot beyond each one-sided value is $0.0895\,\lvert J\rvert$ , about nine per cent of the jump, and we say so explicitly whenever a percentage appears.

Example(The limiting shape of the spike for the square wave, and for any jump)AdaptedAdapted from Arfken, Weber & Harris (2013), §19.3, eqs. (19.32)–(19.38) and (19.43), pp. 958–960; Stakgold & Holst (2011), §2.3, the sine series of $\\operatorname{sgn} x$ and the reduction to $\\mathrm{Si}$, pp. 135–136; Folland (1999), §8.5, Exercise 37, p. 270, and the subtraction argument, pp. 268–269; Hassani (2013), §9.1.1, eq. (9.14), p. 275

Part 1: the model jump. Let $f$ be the $2\pi$ -periodic square wave with $f(x)=\operatorname{sgn}x$ on $(-\pi,\pi)$ , so $f(0\pm)=\pm1$ , $J=2$ and $m=0$ . Its Fourier series is the sine series

f(x)\sim\frac{4}{\pi}\sum_{k=0}^{\infty}\frac{\sin\bigl[(2k+1)x\bigr]}{2k+1},

and we write $S_n$ for the partial sum retaining the terms $k=0,\dots,n$ , whose highest harmonic is therefore $2n+1$ . Put $M=2n+2$ . Then:

$\displaystyle S_n(x)=\frac{2}{\pi}\int_0^x\frac{\sin(Mt)}{\sin t}\,\mathrm{d}t$ for all $x$ .
On $(0,\pi)$ the critical points of $S_n$ are exactly $x_j=j\pi/M$ for $j=1,\dots,M-1$ ; they alternate between maxima (odd $j$ ) and minima (even $j$ ), and $S_n$ attains its global maximum on $[0,\pi]$ at $x_1=\pi/M$ . (By the symmetry of $S_n$ about $\pi/2$ established in Step 4 of the proof, the same value is attained at $x_{M-1}=\pi-\pi/M$ and at no other point. For $n\ge1$ that is a second, distinct point, and when we speak of the maximum below we mean the first of the two; for $n=0$ the two coincide at $\pi/2$ .)
For each fixed $j\ge1$ , $\lim_{n\to\infty}S_n\!\left(\frac{j\pi}{M}\right)=\frac{2}{\pi}\operatorname{Si}(j\pi).$ In particular $\lim_{n\to\infty}\ \max_{[0,\pi]}S_n=\frac{2}{\pi}\operatorname{Si}(\pi)=1.1789797444722\ldots>1=f(0+),$ while the position $\pi/M$ of that maximum tends to $0$ . The spike therefore narrows like $1/n$ without losing height, and by oddness of $f$ the mirror statement holds below $-1$ .
The rescaled graphs converge. Because $f$ has no even harmonics, the ordinary symmetric partial sum $S_Nf$ of Fourier series coincides with $S_n$ for $n=\lfloor(N-1)/2\rfloor$ , for every $N\ge1$ . With that reading, for every fixed real $c\neq0$ , $\lim_{N\to\infty}S_Nf\!\left(\frac{c}{N}\right)=\frac{2}{\pi}\operatorname{Si}(c) ,$ so in the stretched coordinate $c=Nx$ the graphs converge to the single curve $c\mapsto\tfrac2\pi\operatorname{Si}(c)$ , which rises to $P$ at $c=\pi$ before settling to $1$ .

Part 2: an arbitrary jump. Let $g$ be $2\pi$ -periodic and in class $D$ , let $a$ be a point at which $g$ jumps, and put

J=g(a+)-g(a-)\neq0,\qquad m=\frac{g(a+)+g(a-)}{2} .

$S_Ng\to g$ uniformly on every closed interval containing no discontinuity of $g$ .
For every fixed real $c\neq0$ , $\lim_{N\to\infty}S_Ng\!\left(a+\frac{c}{N}\right)=m+\frac{J}{2}\cdot\frac{2}{\pi}\operatorname{Si}(c) .$ In the stretched coordinate $c=N(x-a)$ the graphs of the partial sums therefore converge to one fixed curve, the same curve for every $g$ once $m$ and $J$ are fixed.
That curve attains its extreme values at $c=\pm\pi$ , where it reaches $m\pm\tfrac12 JP$ . Since $P>1$ , those two values lie beyond $g(a+)=m+\tfrac12 J$ and $g(a-)=m-\tfrac12 J$ respectively, whichever sign $J$ has. The partial sums therefore overrun each one-sided value, in the limit by $\frac{\lvert J\rvert}{2}\left(\frac{2}{\pi}\operatorname{Si}(\pi)-1\right)=0.0894899\,\lvert J\rvert$ — about nine per cent of the jump, independently of everything else about $g$ — at points a distance $\approx\pi/N$ either side of $a$ , essentially half a wavelength of the highest harmonic retained.

The overshoot is thus a property of the jump alone, not of the function carrying it.

Idea of the proofOriginal

Everything follows from differentiating the partial sum instead of the function. The derivative of $S_n$ is a sum of cosines, and sums of cosines collapse: $S_n'$ is a single quotient whose numerator is $\sin(Mt)$ , so the critical points of $S_n$ are simply the zeros of one sine, spaced $\pi/M$ apart. Integrating back, $S_n$ is the running integral of an oscillation of period $2\pi/M$ divided by $\sin t$ . Its value at the $j$ th zero is a sum of $j$ alternating arches. On $(0,\pi/2]$ the arches get smaller as $t$ grows, because $1/\sin t$ decreases there; that is the whole range one needs, since $S_n$ turns out to be symmetric about $\pi/2$ (Step 4), so the arches on $[\pi/2,\pi)$ are reflected copies of the earlier ones rather than a continuation of the decrease. So the running integral overshoots on the first arch and then oscillates in: the maximum is at the very first zero, and it is above the final value because the first arch is bigger than the whole alternating tail that follows.

Rescaling $t$ by $M$ is what makes the limit exist. The integrand becomes $\sin u / [M\sin(u/M)]$ , and since $M\sin(u/M)\to u$ , the integral tends to $\int_0^{j\pi}u^{-1}\sin u\,\mathrm{d}u$ — a fixed number, independent of $n$ . The rescaling is also the reason the effect never goes away: $n$ has been absorbed entirely into the width of the picture, so increasing $n$ shrinks the spike horizontally and does nothing to it vertically.

The step from one jump to all jumps is pure bookkeeping. Subtract a multiple of the square wave to cancel the jump; what is left is continuous near that point and has a bounded derivative off finitely many others, hence is absolutely continuous there, and absolute continuity is exactly the hypothesis under which partial sums are known to converge uniformly on compact sets. So the remainder contributes no overshoot. Only the square-wave pieces can, and part 1 says exactly how much each one does.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §19.3, eqs. (19.32)–(19.43), pp. 958–960; Folland (1999), §8.5, Exercise 37(a)–(c), p. 270; Stakgold & Holst (2011), §2.3, the reduction of $S_{2n+1}$ to the sine integral and the decomposition of a function in $D$, pp. 135–136; Strauss (2008), §5.5, the change of variables and the bounds taken in the limit, pp. 143–144

Step 1: the series of the square wave. The function is odd, so by part 3 of the statement in Fourier series only sine terms appear, and

b_n=\frac{2}{\pi}\int_0^{\pi}\sin(nx)\,\mathrm{d}x=\frac{2}{\pi}\cdot\frac{1-\cos n\pi}{n}=\frac{2\bigl(1-(-1)^n\bigr)}{\pi n},

which is $4/(\pi n)$ for odd $n$ and $0$ for even $n$ . That is the displayed series.

Step 2: the derivative collapses (part 1, item 1). Differentiating the finite sum term by term,

S_n'(x)=\frac{4}{\pi}\sum_{k=0}^{n}\cos\bigl[(2k+1)x\bigr] =\frac{4}{\pi}\operatorname{Re}\sum_{k=0}^{n}\mathrm{e}^{\mathrm{i}(2k+1)x} =\frac{4}{\pi}\operatorname{Re}\frac{\mathrm{e}^{\mathrm{i}x}\bigl(\mathrm{e}^{2\mathrm{i}(n+1)x}-1\bigr)}{\mathrm{e}^{2\mathrm{i}x}-1},

the last step being the sum of a geometric progression with ratio $\mathrm{e}^{2\mathrm{i}x}$ , valid when $\mathrm{e}^{2\mathrm{i}x}\neq1$ . Cancel $\mathrm{e}^{\mathrm{i}x}$ against the denominator:

\frac{\mathrm{e}^{\mathrm{i}x}}{\mathrm{e}^{2\mathrm{i}x}-1}=\frac{1}{\mathrm{e}^{\mathrm{i}x}-\mathrm{e}^{-\mathrm{i}x}}=\frac{1}{2\mathrm{i}\sin x},

so, taking the real part of $z/(2\mathrm{i}\sin x)$ , which is $\operatorname{Im}(z)/(2\sin x)$ , with $z=\mathrm{e}^{2\mathrm{i}(n+1)x}-1$ ,

S_n'(x)=\frac{4}{\pi}\cdot\frac{\sin\bigl[2(n+1)x\bigr]}{2\sin x}=\frac{2}{\pi}\cdot\frac{\sin(Mx)}{\sin x}, \qquad M=2n+2 .

Since $S_n(0)=0$ — every term of the series vanishes at the origin — integrating from $0$ gives item 1. The integrand extends continuously to $t=0$ with value $M$ , so the integral is a perfectly ordinary one.

Step 3: the critical points (part 1, item 2, first half). On $(0,\pi)$ we have $\sin t>0$ , so $S_n'(x)=0$ exactly when $\sin(Mx)=0$ , that is at $x_j=j\pi/M$ with $1\le j\le M-1$ . On each interval $(x_{j},x_{j+1})$ the factor $\sin(Mx)$ keeps one sign and alternates from interval to interval, starting positive on $(0,x_1)$ ; so $S_n$ increases up to $x_1$ , decreases to $x_2$ , and so on. The odd $x_j$ are local maxima, the even ones local minima. (Folland's Exercise 37(a) is the same observation made about the error $S_m\phi-\phi$ , whose derivative is the Dirichlet kernel itself.)

Step 4: $x_1$ is the global maximum (part 1, item 2, second half). First, $S_n$ is symmetric about $\pi/2$ : replacing $x$ by $\pi-x$ sends $\sin[(2k+1)x]$ to $\sin[(2k+1)\pi-(2k+1)x]=\sin[(2k+1)x]$ , since $2k+1$ is odd. So it is enough to compare the local maxima lying in $(0,\pi/2]$ . For odd $j$ with $x_{j+2}\le\pi/2$ ,

S_n(x_{j+2})-S_n(x_j)=\frac{2}{\pi}\left(\int_{x_j}^{x_{j+1}}\frac{\sin(Mt)}{\sin t}\,\mathrm{d}t+\int_{x_{j+1}}^{x_{j+2}}\frac{\sin(Mt)}{\sin t}\,\mathrm{d}t\right),

in which the first integrand is negative and the second positive. Substituting $t\mapsto t+\pi/M$ carries the first interval onto the second and $\lvert\sin(Mt)\rvert$ onto itself, while $1/\sin t$ strictly decreases on $(0,\pi/2]$ . Hence the positive second integral is strictly smaller in magnitude than the negative first one, the bracket is negative, and $S_n(x_{j+2})<S_n(x_j)$ . The local maxima in $(0,\pi/2]$ therefore decrease, so the largest is the first, $x_1$ ; by the symmetry just established the maxima in $[\pi/2,\pi)$ are copies of these, and $S_n(0)=S_n(\pi)=0$ . So $\max_{[0,\pi]}S_n=S_n(\pi/M)$ .

Step 5: the limit (part 1, item 3). Substitute $t=u/M$ in item 1 at $x=j\pi/M$ :

S_n\!\left(\frac{j\pi}{M}\right)=\frac{2}{\pi}\int_0^{j\pi}\frac{\sin u}{M\sin(u/M)}\,\mathrm{d}u .

We compare the integrand with $\sin u/u$ uniformly on $[0,j\pi]$ , which is the estimate Arfken, Weber and Harris and Hassani both take for granted when they replace $\sin(u/M)$ by $u/M$ . Two elementary bounds do it, for $M\ge 2j$ — that is, for $n\ge j-1$ — so that $u/M\le\pi/2$ throughout:

$\bigl\lvert M\sin(u/M)-u\bigr\rvert\le u^3/(6M^2)$ , from the alternating Taylor series of $\sin$ ;
$M\sin(u/M)\ge 2u/\pi$ , from Jordan's inequality $\sin v\ge 2v/\pi$ on $[0,\pi/2]$ .

Therefore, for $0<u\le j\pi$ ,

\left\lvert\frac{\sin u}{M\sin(u/M)}-\frac{\sin u}{u}\right\rvert =\lvert\sin u\rvert\cdot\frac{\bigl\lvert u-M\sin(u/M)\bigr\rvert}{u\cdot M\sin(u/M)} \le\frac{u^{3}/(6M^{2})}{u\cdot(2u/\pi)} =\frac{\pi u}{12M^{2}} \le\frac{\pi^{2}j}{12M^{2}} .

The bound is independent of $u$ and tends to $0$ , so

\left\lvert S_n\!\left(\frac{j\pi}{M}\right)-\frac{2}{\pi}\operatorname{Si}(j\pi)\right\rvert \le\frac{2}{\pi}\cdot j\pi\cdot\frac{\pi^{2}j}{12M^{2}} =\frac{\pi^{2}j^{2}}{6M^{2}}\longrightarrow0 ,

valid for every $n\ge j-1$ . Since $j$ is fixed and $n\to\infty$ , all but finitely many $n$ satisfy that, which is item 3, with the explicit rate $O(n^{-2})$ .

Step 5b: the rescaled limit (part 1, item 4). Let $N\ge1$ , $n=\lfloor(N-1)/2\rfloor$ and $M=2n+2$ , so that $M=N+1$ for odd $N$ and $M=N$ for even $N$ ; in either case $M/N\to1$ . Item 1 and the substitution $u=Mt$ give, for $c>0$ ,

S_Nf\!\left(\frac{c}{N}\right)=\frac{2}{\pi}\int_0^{Mc/N}\frac{\sin u}{M\sin(u/M)}\,\mathrm{d}u .

Two differences separate this from $\tfrac2\pi\operatorname{Si}(c)$ . The integrand differs from $\sin u/u$ by at most $\pi u/(12M^{2})\le\pi c'/(6M^{2})$ on $[0,c']$ with $c'=2c$ , by the estimate of Step 5, which contributes $O(M^{-2})$ after integration. And the upper limit $Mc/N$ differs from $c$ by $O(1/N)$ , over which range the integrand is bounded: Jordan's inequality gives $M\sin(u/M)\ge2u/\pi$ , so $\lvert\sin u\rvert/[M\sin(u/M)]\le\pi/2$ . Both differences vanish as $N\to\infty$ , leaving $\tfrac2\pi\int_0^c u^{-1}\sin u\,\mathrm{d}u$ . For $c<0$ apply this to $-c$ and use that $f$ is odd, hence so is every $S_Nf$ , and so is $\operatorname{Si}$ .

Step 6: the limiting maximum exceeds $1$ . By Steps 4 and 5 the limiting maximum is $\tfrac2\pi\operatorname{Si}(\pi)$ , and $f(0+)=1=\tfrac2\pi\cdot\tfrac\pi2$ , so the claim is $\operatorname{Si}(\pi)>\pi/2$ . Since $\operatorname{Si}(\infty)=\pi/2$ ,

\operatorname{Si}(\pi)-\frac{\pi}{2}=-\int_\pi^{\infty}\frac{\sin u}{u}\,\mathrm{d}u =-\sum_{i=1}^{\infty}\int_{(2i-1)\pi}^{(2i+1)\pi}\frac{\sin u}{u}\,\mathrm{d}u ,

and each of those two-arch integrals is negative, by the same pairing argument as in Step 4 applied with $1/u$ decreasing. So the difference is positive. This is Arfken, Weber and Harris's eq. (19.42). Its value follows from the power series of $\operatorname{Si}$ , obtained by integrating that of $\sin u/u$ term by term (legitimate because the series converges uniformly on $[0,\pi]$ , by the M-test):

\operatorname{Si}(\pi)=\sum_{i=0}^{\infty}\frac{(-1)^{i}\pi^{2i+1}}{(2i+1)\,(2i+1)!} =\pi-\frac{\pi^{3}}{18}+\frac{\pi^{5}}{600}-\frac{\pi^{7}}{35280}+\cdots

The successive partial sums are $3.141593$ , $1.419022$ , $1.929055$ , $1.843445$ , $1.852573$ , $1.851903$ , $1.851938$ , $1.851937$ , converging to $\operatorname{Si}(\pi)=1.8519370519825$ , whence $\tfrac2\pi\operatorname{Si}(\pi)=1.1789797444722$ .

Step 7: part 2, from the model jump to any. For item 1, let $I$ be a closed interval containing no discontinuity of $g$ . Enlarge it slightly to an open $I'\supset I$ still free of discontinuities. Shrink $I'$ if necessary so that it meets only finitely many arcs of the class- $D$ subdivision of $g$ . On each of those closed arcs $g'$ extends continuously to the endpoints, by the definition of class $D$ , hence is bounded there; so $g'$ is bounded on $I'$ , and $g$ , being continuous on $I'$ with a bounded derivative off finitely many points, is the integral of that derivative, that is absolutely continuous on $I'$ , with $g'\in L^{p}(I')$ for every $p>1$ ; Folland's Corollary 8.45(b) then gives $S_Ng\to g$ uniformly on compact subsets of $I'$ , and $I$ is one of them.

For item 2, set

u=g-\frac{J}{2}\,f(\,\cdot-a),

with $f$ the square wave of part 1. Both terms are in class $D$ , so $u$ is. At $a$ their jumps cancel: $f$ jumps from $-1$ to $+1$ there, so $u(a+)=g(a+)-\tfrac12 J=m$ and $u(a-)=g(a-)+\tfrac12 J=m$ , so the two one-sided limits of $u$ at $a$ agree. After redefining $u(a)=m$ — which Folland does explicitly at the corresponding step, and which changes no coefficient and so nothing downstream — $u$ is continuous at $a$ . Choose $\delta>0$ small enough that $(a-\delta,a+\delta)$ contains no other discontinuity of $g$ and neither of the two discontinuities of $f(\,\cdot-a)$ other than $a$ itself — the nearest of those sits at $a\pm\pi$ . Then $u$ has no discontinuity in $(a-\delta,a+\delta)$ , so item 1 applied to $u$ gives $S_Nu\to u$ uniformly on $[a-\delta/2,a+\delta/2]$ .

(This is where the square wave has to be handled locally rather than globally. Folland subtracts his one-jump-per-period sawtooth $\phi$ and gets a remainder that is continuous everywhere; Stakgold and Holst subtract Heaviside steps on an interval. Our model $f$ has two jumps per period, at $a$ and at $a+\pi$ , so $u$ need not be continuous on the whole circle — only near $a$ , which is all the limit below requires.)

Now fix $c\neq0$ and take $N$ so large that $\lvert c\rvert/N<\delta/2$ . Splitting the partial sum along the definition of $u$ ,

S_Ng\!\left(a+\frac{c}{N}\right)=\frac{J}{2}\,S_Nf\!\left(\frac{c}{N}\right)+S_Nu\!\left(a+\frac{c}{N}\right).

The first term tends to $\tfrac J2\cdot\tfrac2\pi\operatorname{Si}(c)$ by part 1, item 4. For the second, write

\left\lvert S_Nu\!\left(a+\frac cN\right)-m\right\rvert \le\max_{[a-\delta/2,\,a+\delta/2]}\lvert S_Nu-u\rvert+\left\lvert u\!\left(a+\frac cN\right)-u(a)\right\rvert ,

in which the first term vanishes by the uniform convergence just established and the second by the continuity of $u$ at $a$ , since $a+c/N\to a$ . So the second term tends to $m$ , and item 2 follows.

Item 3 is the maximisation of $\operatorname{Si}$ . Its derivative $\operatorname{Si}'(c)=\sin c/c$ is positive on $(0,\pi)$ and changes sign at every multiple of $\pi$ , so the critical points are $c=j\pi$ ; the pairing argument of Step 6, with $1/u$ decreasing, shows $\operatorname{Si}(\pi)>\operatorname{Si}(3\pi)>\cdots$ and that every even- $j$ value is smaller still. Hence the maximum over $c>0$ is at $c=\pi$ , with value $\operatorname{Si}(\pi)$ ; by oddness the minimum over $c<0$ is at $c=-\pi$ . Multiplying by $J/2$ and adding $m$ gives the stated extreme values and, subtracting $g(a\pm)=m\pm\tfrac12 J$ , the stated overshoot. The location $c=\pm\pi$ means $x-a=\pm\pi/N$ . $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Stakgold & Holst (2011), §2.3, p. 136, the statement that the entire overshoot comes from the step terms, the filter reading and the Cesaro remark, and p. 137, the $13\\%$ figure for piecewise-linear fits; Arfken, Weber & Harris (2013), §19.3, the approximation step before (19.38), p. 960, and the closing paragraph on other eigenfunction expansions, p. 961; Folland (1999), §8.5, the remark on $\\lVert D_m\\rVert_1$, p. 264, and Exercise 37(c), p. 270; Strauss (2008), §5.5, eq. (18), p. 142

A genuine jump is needed, and nothing else about $f$ is. If $J=0$ the square-wave term is absent and no overshoot occurs; this is the whole content of the statement that the overshoot is proportional to the jump. Conversely the proportionality constant $0.0894899$ does not depend on $f$ at all, which makes it a usable diagnostic: a reconstruction that overshoots by appreciably more or less than nine per cent of a known jump has something else wrong with it.

Class $D$ is used only in Step 7, and there twice: to make the remainder $u$ continuous and piecewise smooth near $a$ , and to let Folland's Corollary 8.45(b) convert that into uniform convergence of $S_Nu$ near $a$ . It cannot be weakened to continuity alone: the Fourier series of a merely continuous function need not converge uniformly, or even pointwise (Pointwise convergence of Fourier series), so the split into "a locally uniformly convergent part plus a square wave" would collapse and the nine-per-cent figure would not be justified by this argument. Note also that the subtraction is local by necessity, not by preference: the square wave carries a second jump half a period away, and only a model function with a single jump per period — Folland's sawtooth $\phi$ — gives a remainder that is continuous on the whole circle.

Symmetric partial sums are assumed throughout, and this matters more than it looks. The overshoot is a property of the summation method, not of the function: it is caused by the sign changes of the Dirichlet kernel, which is also what makes $\lVert D_N\rVert_1$ unbounded. Replace the partial sums by Cesàro means and the kernel becomes the Fejér kernel, which is non-negative; a non-negative kernel of unit mass produces a weighted average of values of $f$ , and an average can never leave the range of $f$ , so no overshoot is possible. Stakgold and Holst record exactly this, that Cesàro sums avoid the overshoot, and it is proved in Cesàro means and Fejér's theorem.

Nothing here contradicts pointwise convergence. The limit in part 1, item 3 is taken along a moving point $j\pi/M$ , which runs into the discontinuity. At any point $x$ held fixed away from the jump, Pointwise convergence of Fourier series still gives $S_Nf(x)\to f(x)$ ; and at the jump itself $S_Nf(a)\to m$ . Strauss's way of recording the tension is that $\max\lvert S_N-f\rvert$ does not tend to zero although $S_N(x)-f(x)$ does for each $x$ away from the jump. Nor does Gibbs obstruct mean-square convergence: the spike has fixed height but width $O(1/N)$ , so its contribution to $\int\lvert S_Nf-f\rvert^{2}$ is $O(1/N)$ and vanishes — see Mean-square convergence and Parseval's identity.

Two places where we supplied what a source assumed. Arfken, Weber and Harris pass from their eq. (19.37) to (19.38) by replacing $\sin(\xi/2p)$ with $\xi/2p$ and calling it a good approximation; Hassani does the same with $\sin(\phi/2)$ . Neither estimates the error. Step 5 supplies the estimate, and it is better than needed: the error is $O(n^{-2})$ , which is also why the numerical table in the example converges so fast. Separately, Arfken, Weber and Harris locate the maximum at $x=\pi/p$ with $p=r+\tfrac12$ , and Hassani at $\pi/(N+\tfrac12)$ , whereas our exact statement for this square wave is $x=\pi/(2n+2)$ ; the three agree to leading order in the number of retained harmonics and differ in which index is being counted.

Notation for the sine integral varies, and the numbers look different as a result. We use the standard $\operatorname{Si}(u)=\int_0^u t^{-1}\sin t\,\mathrm{d}t$ , with $\operatorname{Si}(\infty)=\pi/2$ and $\operatorname{Si}(\pi)=1.8519$ . Stakgold and Holst normalise so that their $\mathrm{Si}(\infty)=1$ and their $\mathrm{Si}(\pi)\approx1.18$ ; theirs is $\tfrac2\pi$ times ours. Arfken, Weber and Harris use the complementary function $\operatorname{si}(u)=-\int_u^{\infty}t^{-1}\sin t\,\mathrm{d}t=\operatorname{Si}(u)-\pi/2$ , so their statement " $\operatorname{si}(\pi)>0$ indicates an overshoot" is our Step 6, and $\operatorname{si}(\pi)=\operatorname{Si}(\pi)-\pi/2=0.28114$ .

The phenomenon is not special to Fourier series. Arfken, Weber and Harris note that it occurs in other eigenfunction expansions, and Stakgold and Holst go further: it appears whenever the approximation is a least-squares fit, their example being the best piecewise-linear fit to a square wave on equally spaced nodes, which overshoots by about $13\%$ of the jump as the spacing is refined. The common cause is that a least-squares fit from a space of continuous functions must trade an error near the jump against an error further away, and the trade is settled by the geometry of the space, not by the data.

Method(Working next to a jump)AdaptedAdapted from Stakgold & Holst (2011), §2.3, the low-pass filter reading and the Cesaro remark, p. 136; Arfken, Weber & Harris (2013), §19.3, the closing paragraph and Additional Readings (the Lanczos convergence technique), pp. 961–962; Folland (1999), §8.5, Cor. 8.45, p. 268

Find out whether there is a jump at all. Run step 2 of the method of Pointwise convergence of Fourier series on the periodic extension, not on the formula you were given. Most Gibbs surprises are jumps accidentally created at the ends of the period interval by a careless extension.
Quarantine a neighbourhood of each jump. Within a distance of a few times $\pi/N$ of a discontinuity, a truncated series carries an $O(1)$ error, not an $O(1/N)$ one. Do not read values off there, do not difference them, and do not feed them to a nonlinear step such as a limiter, a logarithm or a square root.
Predict the overshoot and check it. Beyond each one-sided value, expect $0.0895$ times the jump, with the first extremum about $\pi/N$ away. If your output disagrees badly, suspect the coefficients rather than the theorem.
Choose a cure deliberately. In rough order of cost:
- Change the summation. Cesàro or Fejér means remove the overshoot entirely, because their kernel is non-negative (Cesàro means and Fejér's theorem); the price is slower convergence where $f$ is smooth.
- Taper the coefficients. Multiply $\hat f(n)$ by a window $w(n/N)$ falling smoothly to zero at $n=N$ instead of cutting off abruptly — a raised cosine, or the Lanczos smoothing that Arfken, Weber and Harris point to. This is the same device as the previous one with a different taper, and it is what signal-processing practice means by windowing.
- Subtract the jump analytically. Use the decomposition of Step 7: subtract $\tfrac12 J\,f(\cdot-a)$ with $f$ the square wave, expand the continuous remainder — which now converges uniformly — and add the closed-form square-wave series back at the end. This keeps full accuracy away from the jump and is exact, but it requires knowing where the jumps are.
- Change the representation. Put the discontinuity on a cell boundary and use a finite-volume or discontinuous-Galerkin discretisation, or expand in a basis that already contains the jump.
Do not take more terms and hope. Item 3 of part 1 is precisely the statement that this does not work: more terms move the spike closer to the jump and leave its height alone.
Check which norm you actually need. If the requirement is mean-square accuracy, or an integral of the reconstruction, Gibbs is harmless and no cure is called for; the spike's area is $O(1/N)$ . Only pointwise or uniform requirements near the jump are affected.

Example(The square wave counted out, and the same nine per cent on a sawtooth)Original

Part A: the square wave, exactly. Take $f=\operatorname{sgn}$ on $(-\pi,\pi)$ , extended with period $2\pi$ . By Step 1 the partial sum with $n+1$ nonzero terms, highest harmonic $2n+1$ , is

S_n(x)=\frac{4}{\pi}\sum_{k=0}^{n}\frac{\sin\bigl[(2k+1)x\bigr]}{2k+1} .

Take $n=9$ : ten terms, harmonics $1,3,\dots,19$ . By item 2 the maximum sits at $x_1=\pi/M=\pi/20=0.15707963$ . Evaluating there, with $\sin[(2k+1)\pi/20]$ for $k=0,\dots,9$ :

S_9\!\left(\frac{\pi}{20}\right)=\frac{4}{\pi}\left(\frac{\sin 9^{\circ}}{1}+\frac{\sin 27^{\circ}}{3}+\frac{\sin 45^{\circ}}{5}+\cdots+\frac{\sin 171^{\circ}}{19}\right)=1.17981402 .

The function's value just to the right of the jump is $1$ , so ten terms overshoot by $0.17981$ , which is $8.991\%$ of the jump $J=2$ . The predicted limit is $0.0894899\times2=0.1789797$ . Running $n$ up:

terms $n+1$	$x_1=\pi/(2n+2)$	$S_n(x_1)$	overshoot	as % of $J=2$
$3$	$0.5235988$	$1.18835691$	$0.18835691$	$9.418\%$
$5$	$0.3141593$	$1.18232821$	$0.18232821$	$9.116\%$
$10$	$0.1570796$	$1.17981402$	$0.17981402$	$8.991\%$
$20$	$0.0785398$	$1.17918814$	$0.17918814$	$8.959\%$
$50$	$0.0314159$	$1.17901308$	$0.17901308$	$8.951\%$
$100$	$0.0157080$	$1.17898808$	$0.17898808$	$8.949\%$
$500$	$0.0031416$	$1.17898008$	$0.17898008$	$8.949\%$
$\infty$	$0$	$1.17897974$	$0.17897974$	$8.94899\%$

Two things to read off. The height column is converging to $\tfrac2\pi\operatorname{Si}(\pi)$ , and the convergence is quadratic in $1/n$ exactly as Step 5 predicts: the errors against the limit are $3.35\times10^{-3}$ , $8.34\times10^{-4}$ and $2.08\times10^{-4}$ for $5$ , $10$ and $20$ terms, falling by a factor of four each time the term count doubles. The position column is converging to $0$ like $1/n$ . Height fixed, width shrinking: that is the whole phenomenon in two columns.

Part B: the ladder of critical values. Item 3 predicts $S_n(j\pi/M)\to\tfrac2\pi\operatorname{Si}(j\pi)$ for every $j$ , alternately above and below $1$ . With $n=199$ (two hundred terms, $M=400$ ):

$j$	$S_{199}(j\pi/400)$	$\tfrac{2}{\pi}\operatorname{Si}(j\pi)$
$1$	$1.1789818$	$1.1789797$
$2$	$0.9028192$	$0.9028233$
$3$	$1.0661927$	$1.0661865$
$4$	$0.9499310$	$0.9499393$

Five or six figures agree in each row, and the maxima $1.17898>1.06619>\cdots$ decrease while the minima $0.90282<0.94994<\cdots$ increase, confirming both the alternation of Step 3 and the claim of Step 4 that $j=1$ gives the global maximum.

Part C: a different jump, to test the proportionality. Now take the sawtooth: $g(x)=x$ on $(-\pi,\pi)$ , extended with period $2\pi$ , whose series is $g\sim2\sum_{n\ge1}(-1)^{n+1}n^{-1}\sin nx$ . At $x=\pi$ it jumps from $g(\pi-)=\pi$ down to $g(\pi+)=-\pi$ , so $J=-2\pi$ and $\lvert J\rvert=6.2831853$ . Part 2 predicts an overshoot above $g(\pi-)=\pi$ of $0.0894899\times6.2831853=0.5622816$ ; the exact value of that overshoot is $\bigl(\tfrac2\pi\operatorname{Si}(\pi)-1\bigr)\pi=0.5622815$ , the last figure differing only because the constant was rounded to seven places before multiplying. The limiting maximum is therefore

\pi+\Bigl(\frac{2}{\pi}\operatorname{Si}(\pi)-1\Bigr)\pi=2\operatorname{Si}(\pi)=3.7038741,

which is the square-wave result rescaled by $\lvert J\rvert/2=\pi$ , as it must be. Maximising $S_N g$ numerically near $x=\pi$ :

$N$	$\max S_Ng$	distance from $\pi$	$(N+1)\times$ distance
$20$	$3.5530870$	$0.14959965$	$3.1415927$
$100$	$3.6727179$	$0.03110488$	$3.1415927$
$500$	$3.6976014$	$0.00627064$	$3.1415927$
$2000$	$3.7023040$	$0.00157001$	$3.1415927$
$\infty$	$3.7038741$	$0$	$\pi$

The maxima climb to $2\operatorname{Si}(\pi)=3.7038741$ , confirming the nine-per-cent figure for a jump $\pi$ times larger than the square wave's. The last column is $\pi$ in every row, and exactly so: the critical points here solve $D_N(s)=1$ rather than $\sin(Ms)=0$ , and $s=\pi/(N+1)$ is a root because $(N+\tfrac12)s=\pi-\tfrac12 s$ there, which makes the numerator and denominator of the Dirichlet kernel equal. The first extremum therefore sits at distance $\pi/(N+1)$ from the discontinuity — the same $\pi/(H+1)$ rule as the square wave, with $H$ the highest harmonic kept, and to the eye half a wavelength of that harmonic. Every number in this example was computed to twenty significant figures with mpmath and the two Fourier series were checked symbolically; the entries above are the correctly rounded values.

ApplicationOriginal

Ringing in a filtered signal. An ideal "brick-wall" low-pass filter passes frequencies below a cutoff and removes the rest, which is precisely truncating a Fourier series; Stakgold and Holst make exactly this identification, noting that the output for a square-wave input is a partial sum. So a logic edge, or any fast transition, emerges from such a filter with a $9\%$ overshoot of the step height followed by decaying oscillations at the cutoff frequency, spaced by half a period of that frequency. The engineering response is step 4 of the method: no practical filter uses a brick wall, and the Hann, Hamming and Kaiser windows of filter design are coefficient tapers in the sense of that step. The trade is explicit and unavoidable — a gentler taper reduces the overshoot and widens the transition.

Truncation artefacts in imaging. Magnetic resonance imaging samples a finite region of $k$ -space, which is a two-dimensional Fourier truncation; computed tomography reconstructions truncate a filtered projection. Both therefore show Gibbs ringing at sharp boundaries — bright and dark fringes paralleling a tissue interface, whose first bright line is $9\%$ of the intensity step above the brighter tissue and lies about one pixel-scale from the edge. Radiologists call it truncation or ringing artefact, and the standard mitigation, apodisation of the $k$ -space data, is again the taper of step 4. Knowing the constant matters clinically: a $9\%$ bright rim is an artefact to be read past, whereas a $50\%$ one is not.

Shocks in spectral methods. A Fourier spectral discretisation of a conservation law (Spectral methods, Conservation laws and shocks) is accurate beyond any fixed order while the solution stays smooth, and loses that advantage the instant a shock forms: the numerical solution then oscillates about the shock with amplitude set by the jump, not by the resolution, and refining the grid only narrows the oscillation. Worse, the oscillations can push a physically positive quantity — a density, a pressure, a mass fraction — negative on the undershooting side, at which point an equation of state or a square root fails and the computation stops. This is step 2 of the method as an operational rule, and it is why production spectral codes carry a filter or a spectral-viscosity term, and why shock-capturing schemes are built on finite volumes instead.

Block-transform image compression. Quantising or discarding the high-frequency coefficients of an $8\times8$ cosine transform block is a truncation of the same kind, and near a sharp edge it produces the same overshoot, visible as the shimmering halo that image coders call mosquito noise. The size of the artefact scales with the contrast across the edge, exactly as part 2 says, which is why high-contrast text on a flat background is the worst case for such a coder and why text is usually coded by other means.

Theorem

Cesàro means and Fejér's theorem

Adapted

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Averaging the Fourier partial sums replaces the sign-changing Dirichlet kernel by the non-negative Fejér kernel, and the averages converge to the function at every point of continuity, uniformly when the function is continuous; this makes the trigonometric polynomials dense in the continuous periodic functions.

Prerequisites: Fourier series, Pointwise convergence of Fourier series, Uniform convergence

Definition(Cesàro means, good kernels, the Fejér kernel)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, §4, p. 48 and §5.1–5.2, pp. 51–53; Folland (1999), §8.4, p. 262

Throughout, integrable on the circle means Riemann integrable on $[-\pi,\pi]$ and extended $2\pi$ -periodically to $\mathbb{R}$ , and

\hat f(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(\theta)\,e^{-in\theta}\,d\theta, \qquad S_N(f)(\theta) = \sum_{|n| \le N} \hat f(n)\,e^{in\theta}

are the Fourier coefficients and the symmetric partial sums of the Fourier series. Convolution on the circle is $(f * g)(\theta) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(\theta - y)\,g(y)\,dy$ , and $S_N(f) = f * D_N$ with $D_N(y) = \sum_{|n| \le N} e^{iny} = \sin\bigl((N+\tfrac12)y\bigr)/\sin(y/2)$ the Dirichlet kernel of pointwise convergence.

Cesàro mean. For a numerical series $\sum_{k \ge 0} c_k$ with partial sums $s_n = \sum_{k \le n} c_k$ , the $N$ th Cesàro mean is the average $\sigma_N = (s_0 + s_1 + \dots + s_{N-1})/N$ of the first $N$ partial sums. If $\sigma_N \to \sigma$ , the series is Cesàro summable to $\sigma$ . Applied to the Fourier series of $f$ this gives the Fejér means

\sigma_N(f) = \frac{1}{N}\bigl(S_0(f) + S_1(f) + \dots + S_{N-1}(f)\bigr), \qquad N \in \mathbb{N},

each a trigonometric polynomial of degree at most $N-1$ .

Good kernel. A family $\{K_n\}_{n \ge 1}$ of integrable functions on the circle is a family of good kernels when

(a) $\frac{1}{2\pi}\int_{-\pi}^{\pi} K_n = 1$ for every $n$ : unit mass.
(b) there is one constant $M$ , the same for every $n$ , with $\int_{-\pi}^{\pi} \lvert K_n \rvert \le M$ : the total absolute mass stays bounded.
(c) for each fixed $\delta \in (0,\pi)$ , $\int_{\delta \le \lvert y \rvert \le \pi} \lvert K_n \rvert \to 0$ as $n \to \infty$ : the mass concentrates at the origin.

When $K_n \ge 0$ , (b) follows from (a) with $M = 2\pi$ . Such a family is also called an approximate identity, or, in the language of Stakgold and Holst (Stakgold & Holst (2011), Exercise 2.3.6, p. 144), a delta sequence.

Fejér kernel. $F_N = \dfrac{1}{N}\bigl(D_0 + D_1 + \dots + D_{N-1}\bigr)$ , so that $\sigma_N(f) = f * F_N$ by linearity of the convolution in its second slot.

Note(Conventions differ)AdaptedAdapted from Folland (1999), §8.4, p. 262 and Exercise 33, p. 269; Stakgold & Holst (2011), Exercise 2.3.6, p. 144

Three normalisations are in use and the closed forms look different because of it.

Ours, following Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 2, Lemma 5.1, p. 53): period $2\pi$ , the factor $1/2\pi$ in the convolution, $N$ partial sums $S_0,\dots,S_{N-1}$ , and $F_N(y) = \sin^2(Ny/2)/\bigl(N\sin^2(y/2)\bigr)$ .
Stakgold and Holst (Stakgold & Holst (2011), Exercise 2.3.6, p. 144) put no factor in front of the integral, so their kernel absorbs it: $F_n(x) = \sin^2(nx/2)/\bigl(2\pi n \sin^2(x/2)\bigr)$ .
Folland (Folland (1999), §8.4, p. 262 and Exercise 33b, p. 269) works on the circle of circumference one with characters $e^{2\pi i kx}$ and averages the $m+1$ sums $S_0,\dots,S_m$ , giving $F_m(x) = \sin^2\bigl((m+1)\pi x\bigr)/\bigl((m+1)\sin^2 \pi x\bigr)$ .

All three are the same object after the change of variable and the change of index; only the labelling of "the $N$ th mean" shifts by one. Mixing them silently is the usual source of a stray factor of $2\pi$ .

Theorem(Approximation by a good kernel)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, Thm 4.1, p. 49

Take $f$ integrable on the circle and $\{K_n\}$ any family of good kernels. Convolving against the family recovers $f$ wherever $f$ is unambiguous:

$(f * K_n)(x) \to f(x)$ as $n \to \infty$ , for each single $x$ at which $f$ is continuous;
when $f$ is continuous at every point of the circle, $\sup_x \lvert (f*K_n)(x) - f(x)\rvert \to 0$ .

For this reason $\{K_n\}$ is also called an approximate identity: the operation $f \mapsto f * K_n$ approaches the identity operator.

Theorem(Fejér's theorem)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, Lemma 5.1 and Thm 5.2, pp. 53–54

Let $f$ be integrable on the circle. Then:

The kernel. $\{F_N\}_{N \ge 1}$ is a family of good kernels, with $F_N(y) \;=\; \sum_{\lvert n \rvert < N}\Bigl(1 - \frac{\lvert n \rvert}{N}\Bigr) e^{iny} \;=\; \frac{1}{N}\,\frac{\sin^2(Ny/2)}{\sin^2(y/2)} \;\ge\; 0 \quad (y \notin 2\pi\mathbb{Z}), \qquad F_N(0) = N .$
At a point. $\sigma_N(f)(x) \to f(x)$ for each $x$ at which $f$ is continuous. Equivalently: the Fourier series of $f$ is Cesàro summable to the value $f(x)$ there, whether or not it converges there.
Everywhere at once. If $f$ has no discontinuity on the circle, then $\sigma_N(f) \to f$ uniformly on $\mathbb{R}$ .

Theorem(Three corollaries)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, Cor. 5.3 and Cor. 5.4, p. 54; Kreyszig (1978), Thm 4.11-5, pp. 280–281

Uniqueness. An integrable function on the circle whose Fourier coefficients all vanish is zero wherever it is continuous. Equivalently, the Fourier coefficients determine an integrable function at each of its points of continuity: if $\hat f = \hat g$ coefficientwise, then $f = g$ at every point at which both are continuous.
Trigonometric density (the periodic Weierstrass theorem). The trigonometric polynomials are dense in $C(S)$ , the continuous functions on the circle, for the uniform norm: every continuous function on the circle — that is, every $f$ continuous on $[-\pi,\pi]$ with matching endpoint values $f(-\pi) = f(\pi)$ — is the uniform limit on $[-\pi,\pi]$ of a sequence of trigonometric polynomials, and the sequence $\sigma_N(f)$ will do.
Weierstrass approximation theorem for polynomials. For every real continuous $g$ on a finite interval $[a,b]$ and every $\varepsilon > 0$ there is a real polynomial $p$ with $\max_{t \in [a,b]} \lvert g(t) - p(t) \rvert < \varepsilon$ ; that is, the polynomials are dense in the real space $C[a,b]$ .

Idea of the proofOriginal

A partial sum is already an average of $f$ against a weight, the Dirichlet kernel, but that weight takes both signs and its total absolute mass grows without bound, so cancellation rather than concentration decides the limit. Averaging the partial sums averages the weights, and the average of the Dirichlet kernels turns out to be a perfect square divided by $N$ : non-negative, of unit mass, and squeezed towards the origin as $N$ grows. Once the weight is a genuine probability density piling up at $0$ , the conclusion is a one-line idea — an average of numbers that all lie within $\varepsilon$ of $f(x)$ lies within $\varepsilon$ of $f(x)$ — and uniform continuity makes the one $\varepsilon$ work at every $x$ at once. The corollaries follow because each Fejér mean is a trigonometric polynomial.

Proof(Approximation by a good kernel)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, Thm 4.1, pp. 49–50

Fix $x$ with $f$ continuous at $x$ , and let $\varepsilon > 0$ . Because $f$ is integrable on the circle it is bounded, say $\lvert f \rvert \le B$ .

Step 1: write the error as one integral. Property (a) lets us insert $f(x)$ under the integral sign:

(f * K_n)(x) - f(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} \bigl[f(x - y) - f(x)\bigr] K_n(y)\,dy .

This is the only place where (a) is used, and it is what makes the whole error an integral of a difference of values of $f$ .

Step 2: choose the window. Continuity of $f$ at $x$ gives a $\delta \in (0,\pi)$ with $\lvert f(x-y) - f(x) \rvert < \varepsilon$ whenever $\lvert y \rvert < \delta$ .

Step 3: split at the window and bound each part. Taking absolute values in step 1 and splitting the range of integration,

\lvert (f * K_n)(x) - f(x) \rvert \;\le\; \underbrace{\frac{1}{2\pi}\int_{\lvert y \rvert < \delta} \lvert f(x-y) - f(x)\rvert\, \lvert K_n(y)\rvert \,dy}_{\text{near part}} \;+\; \underbrace{\frac{1}{2\pi}\int_{\delta \le \lvert y \rvert \le \pi} \lvert f(x-y) - f(x)\rvert\, \lvert K_n(y)\rvert \,dy}_{\text{far part}} .

On the near part the first factor is below $\varepsilon$ by step 2, and the remaining integral is at most $M$ by property (b); the near part is therefore at most $\varepsilon M / 2\pi$ — a bound independent of $n$ . On the far part the first factor is at most $2B$ , so the far part is at most $\frac{B}{\pi}\int_{\delta \le \lvert y\rvert \le \pi}\lvert K_n \rvert$ , which tends to $0$ as $n \to \infty$ by property (c), with $\delta$ now held fixed.

Step 4: let $n \to \infty$ , then $\varepsilon \to 0$ . The far part vanishes in the limit, so

\limsup_{n \to \infty} \ \lvert (f * K_n)(x) - f(x) \rvert \;\le\; \frac{M\varepsilon}{2\pi} .

The left-hand side does not depend on $\varepsilon$ and $\varepsilon > 0$ was arbitrary, so the $\limsup$ is $0$ , which is the first claim.

Step 5: uniformity. If $f$ is continuous at every point of the circle, then it is continuous on the compact interval $[-2\pi, 2\pi]$ and hence uniformly continuous there; periodicity extends uniform continuity to $\mathbb{R}$ . So in step 2 the same $\delta$ serves every $x$ , and the bounds in step 3 then involve no $x$ at all. Taking the supremum over $x$ before letting $n \to \infty$ gives $\sup_x \lvert (f*K_n)(x) - f(x)\rvert \to 0$ . $\square$

Proof(Fejér's theorem)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, Lemma 5.1, p. 53

Step 1: the coefficients of $F_N$ . By definition $D_n = \sum_{|m| \le n} e^{imy}$ , so in the average $F_N = \frac1N\sum_{n=0}^{N-1} D_n$ the exponential $e^{imy}$ appears once for each $n$ with $\lvert m \rvert \le n \le N-1$ , that is $N - \lvert m \rvert$ times when $\lvert m \rvert < N$ and never otherwise. Dividing by $N$ ,

F_N(y) = \sum_{\lvert m \rvert < N}\Bigl(1 - \frac{\lvert m \rvert}{N}\Bigr) e^{imy} .

In particular $F_N$ is a trigonometric polynomial. At $y = 0$ every exponential equals $1$ , so $F_N(0) = \frac1N\sum_{\lvert m\rvert < N}(N - \lvert m\rvert) = \frac1N\Bigl(N + 2\sum_{m=1}^{N-1}(N-m)\Bigr) = \frac1N\bigl(N + N(N-1)\bigr) = N .$

Step 2: the closed form. Sum the Dirichlet kernels directly instead. For $y \notin 2\pi\mathbb{Z}$ each $D_n(y) = \sin\bigl((n+\frac12)y\bigr)/\sin(y/2)$ , so

N F_N(y)\,\sin\tfrac{y}{2} = \sum_{n=0}^{N-1} \sin\Bigl(\bigl(n+\tfrac12\bigr)y\Bigr) = \operatorname{Im}\Bigl( e^{iy/2}\sum_{n=0}^{N-1} e^{iny}\Bigr) = \operatorname{Im}\Bigl( e^{iy/2}\,\frac{e^{iNy}-1}{e^{iy}-1}\Bigr) ,

the geometric sum being legitimate because $e^{iy} \ne 1$ . Factoring $e^{iNy} - 1 = e^{iNy/2}\bigl(e^{iNy/2} - e^{-iNy/2}\bigr) = 2i\,e^{iNy/2}\sin(Ny/2)$ and likewise $e^{iy} - 1 = 2i\,e^{iy/2}\sin(y/2)$ , the quotient is $e^{i(N-1)y/2}\sin(Ny/2)/\sin(y/2)$ , so

N F_N(y)\,\sin\tfrac{y}{2} = \operatorname{Im}\bigl(e^{iNy/2}\bigr)\frac{\sin(Ny/2)}{\sin(y/2)} = \frac{\sin^2(Ny/2)}{\sin(y/2)} ,

and dividing by $N\sin(y/2)$ gives $F_N(y) = \sin^2(Ny/2)/\bigl(N\sin^2(y/2)\bigr)$ . The right-hand side is a square over a square, hence $\ge 0$ .

Step 3: the three properties. (a) In the expansion of step 1 every term with $m \ne 0$ integrates to $0$ over a full period and the term $m = 0$ has coefficient $1$ , so $\frac{1}{2\pi}\int_{-\pi}^{\pi}F_N = 1$ . (b) is automatic from (a) and $F_N \ge 0$ , with $M = 2\pi$ . (c) Fix $\delta \in (0,\pi)$ . For $\delta \le \lvert y \rvert \le \pi$ we have $\lvert y \rvert/2 \in [\delta/2, \pi/2]$ , where $\sin$ is increasing and positive, so $\sin^2(y/2) \ge \sin^2(\delta/2) > 0$ ; with $\sin^2(Ny/2) \le 1$ this gives the pointwise bound $F_N(y) \le 1/\bigl(N \sin^2(\delta/2)\bigr)$ and hence

\int_{\delta \le \lvert y \rvert \le \pi} F_N \;\le\; \frac{2\pi}{N \sin^2(\delta/2)} \;\longrightarrow\; 0 \qquad (N \to \infty) .

So $\{F_N\}$ is a family of good kernels, and since $\sigma_N(f) = f * F_N$ , the approximation theorem above delivers both conclusions. $\square$

Proof(The corollaries)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 2, Cor. 5.3 and Cor. 5.4, p. 54; Kreyszig (1978), Thm 4.11-5, pp. 280–281

1. Uniqueness. If every $\hat f(n)$ vanishes then every $S_N(f)$ is the zero function, hence so is every average $\sigma_N(f)$ . By Fejér's theorem $\sigma_N(f)(x) \to f(x)$ at each point of continuity, and a constant sequence of zeros has limit $0$ ; so $f(x) = 0$ there. For the second form apply this to $f - g$ , which is integrable and continuous at every point where both $f$ and $g$ are, and whose coefficients are $\hat f(n) - \hat g(n) = 0$ .

2. Trigonometric density. The hypothesis $f(-\pi) = f(\pi)$ is exactly what makes the $2\pi$ -periodic extension continuous on the circle, so Fejér's theorem gives $\sup_x\lvert f(x) - \sigma_N(f)(x)\rvert \to 0$ ; and each $\sigma_N(f)$ is a trigonometric polynomial by step 1 of the previous proof. So the sequence $\sigma_N(f)$ of trigonometric polynomials converges to $f$ uniformly, which is density in the uniform norm.

3. Polynomials on $[a,b]$ . We follow the route of Kreyszig (Kreyszig (1978), Thm 4.11-5, pp. 280–281), replacing his piecewise-linear intermediate approximation by part 2 above.

Reduce to equal endpoint values. Given $g \in C[a,b]$ , set $\gamma = \bigl(g(b)-g(a)\bigr)/(b-a)$ and $u(t) = g(t) - \gamma\,(t-a)$ , so that $u(a) = u(b)$ . If $q$ is a polynomial with $\max\lvert u - q\rvert < \varepsilon$ then $p(t) = q(t) + \gamma (t - a)$ satisfies $g - p = u - q$ , hence $\max\lvert g-p\rvert<\varepsilon$ . So we may assume $g(a) = g(b)$ .
Rescale to the circle. Put $\varphi(x) = g\bigl(a + (b-a)x/2\pi\bigr)$ for $x \in [0,2\pi]$ . Then $\varphi$ is continuous with $\varphi(0) = \varphi(2\pi)$ , so its $2\pi$ -periodic extension is continuous on the circle. Part 2 is stated on $[-\pi,\pi]$ , but it is a statement about continuous functions on the circle, and $[0,2\pi]$ is another full period: applying part 2 to the shifted function $x \mapsto \varphi(x+\pi)$ and then shifting the approximating polynomial back by $-\pi$ — which keeps it a trigonometric polynomial, since $e^{in(x-\pi)} = (-1)^n e^{inx}$ — gives a trigonometric polynomial $P(x) = \sum_{\lvert n \rvert \le M} c_n e^{inx}$ with $\max_{x \in [0,2\pi]} \lvert \varphi(x) - P(x)\rvert < \varepsilon/2$ . Taking real parts if necessary, $P$ may be taken real, a finite combination of $\cos nx$ and $\sin nx$ .
Replace the trigonometric functions by polynomials. Each derivative of $x \mapsto \cos nx$ is $\pm n^{j}$ times $\cos nx$ or $\sin nx$ , so the $j$ th derivative is bounded by $n^{j}$ in absolute value, and the same holds for $\sin nx$ — the chain rule contributes one factor $n$ per differentiation. Taylor's theorem with Lagrange remainder therefore bounds the error of the degree- $k$ Taylor polynomial of $\cos(nx)$ or $\sin(nx)$ about $x = 0$ by $n^{k+1}(2\pi)^{k+1}/(k+1)! = (2\pi n)^{k+1}/(k+1)!$ on $[0,2\pi]$ , and this tends to $0$ as $k \to \infty$ for each fixed $n$ , the factorial outgrowing any fixed geometric factor. Truncating each of the finitely many ( $\lvert n\rvert \le M$ ) Taylor series far enough out produces a polynomial $r$ in $x$ with $\max_{x \in [0,2\pi]}\lvert P(x) - r(x)\rvert < \varepsilon/2$ .
Undo the rescaling. $x \mapsto 2\pi (t-a)/(b-a)$ is an affine map, so $q(t) = r\bigl(2\pi(t-a)/(b-a)\bigr)$ is a polynomial in $t$ , and by the triangle inequality $\max_{t}\lvert g(t) - q(t)\rvert < \varepsilon$ . $\square$

Remark(Where the hypotheses are used)Original

Property (a) is used once, in step 1 of the first proof, to turn the error into an integral of $f(x-y) - f(x)$ . Without unit mass the limit would be a multiple of $f(x)$ rather than $f(x)$ .
Property (b) is what the Dirichlet kernel lacks, and it is used on the near part, where the only control on the kernel is its total absolute mass. The Dirichlet kernels satisfy (a) but not (b): $\int_{-\pi}^{\pi}\lvert D_N\rvert \ge c\log N$ as $N \to \infty$ (Stein & Shakarchi (2003), Ch. 2, §4, p. 50 and Problem 2), and Folland (Folland (1999), §8.5, p. 264 and Exercise 34, p. 269) records the same fact as $\lVert D_m\rVert_1 \to \infty$ . Numerically $\frac{1}{2\pi}\int_{-\pi}^{\pi}\lvert D_n\rvert$ equals $2.2234$ , $3.1388$ and $4.0702$ at $n = 10$ , $100$ and $1000$ , consistent with the growth $\tfrac{4}{\pi^2}\log n + O(1)$ . Because (b) fails, the argument above gives nothing for $S_N(f)$ — and indeed the Fourier series of a continuous function can diverge at a point (Folland (1999), §8.5, p. 264 and Exercise 35, p. 269).
Property (c) is used on the far part, after $\delta$ has been fixed. The order matters: $\delta$ is chosen from $\varepsilon$ before $n$ is sent to infinity. Reversing the order proves nothing, which is why the proof ends with a $\limsup$ in $n$ and only then lets $\varepsilon \to 0$ .
Continuity at $x$ is used only in step 2. At a jump the general theorem simply says nothing, and it cannot be patched at the level of (a), (b), (c): a family of good kernels is free to favour one side. Take $K_n = 2\pi n\,\mathbf{1}_{[0,1/n]}$ on the circle. It satisfies (a), since $\frac{1}{2\pi}\int K_n = n\cdot\frac1n = 1$ ; (b) with $M = 2\pi$ , since $K_n \ge 0$ ; and (c), since $K_n$ vanishes on $\delta \le \lvert y\rvert \le \pi$ as soon as $n > 1/\delta$ . Yet $(f * K_n)(x) = n\int_0^{1/n} f(x-y)\,dy \to f(x^-)$ , the left-hand limit alone, not the mean. What rescues the Fejér case is a property the definition of a good kernel does not require: $F_N$ is even, so by (a) each half carries mass $\frac{1}{2\pi}\int_0^{\pi}F_N = \frac{1}{2\pi}\int_{-\pi}^{0}F_N = \frac12$ . Suppose then that the one-sided limits $f(x^\pm)$ exist. Splitting step 1 at $y = 0$ and subtracting $f(x^-)$ on $y > 0$ and $f(x^+)$ on $y < 0$ — which is legitimate precisely because each half has mass $\frac12$ — and running steps 2–4 on each half separately gives $\sigma_N(f)(x) \to \tfrac12\bigl(f(x^-) + f(x^+)\bigr)$ . So at a jump the conclusion changes rather than fails, but only for an even kernel.
Uniform continuity, not mere continuity, is what makes step 5 work. It comes from compactness of the circle, so it is free here; on $\mathbb{R}$ it would be an extra hypothesis.
Nothing is lost by averaging. Cesàro summation is a regular summability method in the sense of Kreyszig (Kreyszig (1978), §4.10, p. 270 and Thm 4.10-1, pp. 270–271): its matrix $a_{Nk} = 1/N$ for $k \le N$ and $0$ otherwise satisfies the three conditions of the Toeplitz limit theorem, so every convergent sequence is summable with the same limit. Hence Fejér's theorem does not contradict the Dirichlet convergence theorem; where the partial sums already converge, the means converge to the same value.
Positivity has a consequence the partial sums do not share. Since $F_N \ge 0$ has unit mass, $\sigma_N(f)$ is an honest weighted average of values of $f$ , so $\inf f \le \sigma_N(f) \le \sup f$ for real $f$ and every $N$ . No overshoot is possible, in contrast with the Gibbs phenomenon of the plain partial sums; Stakgold and Holst (Stakgold & Holst (2011), §2.3, p. 136) make the same remark.

Method(Approximating a continuous periodic function to a prescribed tolerance)Original

Check that the function is continuous on the circle: continuity on $[-\pi,\pi]$ is not enough, the endpoint values must agree. If they do not, subtract a linear function first, as in step 1 of the proof of corollary 3.
Compute the coefficients $\hat f(n)$ for $\lvert n \rvert < N$ once. The Fejér mean needs no new integrals: $\sigma_N(f)(\theta) = \sum_{\lvert n \rvert < N}\Bigl(1 - \frac{\lvert n \rvert}{N}\Bigr)\hat f(n)\,e^{in\theta} ,$ i.e. taper the coefficients of $S_{N-1}(f)$ by the triangular factor $1 - \lvert n\rvert/N$ .
To certify a tolerance $\varepsilon$ without computing the error, run the proof's two bounds with explicit constants: find $\delta$ from the modulus of continuity of $f$ so that $\lvert y \rvert < \delta$ forces $\lvert f(\theta - y) - f(\theta)\rvert < \varepsilon/2$ for all $\theta$ , then take $N > \frac{4\,\sup\lvert f\rvert}{\varepsilon\,\sin^{2}(\delta/2)} ,$ which makes the far part at most $\varepsilon/2$ as well. This bound is pessimistic but needs nothing beyond $\sup\lvert f \rvert$ and $\delta$ .
For a function with jumps, do not look for uniform convergence: use the means to get convergence at each point of continuity, and use the bound $\inf f \le \sigma_N(f) \le \sup f$ to guarantee no overshoot.
If polynomials rather than trigonometric polynomials are wanted on $[a,b]$ , follow corollary 3: subtract the chord, rescale to $[0,2\pi]$ , taper, then truncate the Taylor series of the finitely many sines and cosines.

Example(The square wave: the means stay inside the data, the partial sums do not)Original

Let $f$ be the $2\pi$ -periodic extension of $f(\theta) = \operatorname{sgn}\theta$ on $(-\pi,\pi)$ , an integrable function with a jump of size $2$ at $\theta = 0$ . It is odd, so only sine terms survive, and a direct integration gives the classical sine series

f(\theta) \sim \frac{4}{\pi}\sum_{j \ge 0} \frac{\sin\bigl((2j+1)\theta\bigr)}{2j+1} , \qquad\text{so}\qquad S_{N}(f)(\theta) = \frac{4}{\pi}\sum_{\substack{k \le N \\ k \ \mathrm{odd}}} \frac{\sin k\theta}{k} .

The Fejér mean. Tapering by $1 - k/N$ as in the method gives, for $N = 10$ ,

\sigma_{10}(f)(\theta) = \frac{4}{\pi}\sum_{k \in \{1,3,5,7,9\}}\Bigl(1 - \frac{k}{10}\Bigr)\frac{\sin k\theta}{k} = \frac{2}{315\pi}\bigl(567\sin\theta + 147\sin 3\theta + 63 \sin 5\theta + 27\sin 7\theta + 7 \sin 9\theta\bigr).

Only odd harmonics appear, and $\sin\bigl(k(\pi - \theta)\bigr) = \sin k\theta$ for odd $k$ , so both $S_9(f)$ and $\sigma_{10}(f)$ are symmetric about $\theta = \pi/2$ . Evaluating there, where every $\sin(k\pi/2) = \pm 1$ :

\sigma_{10}(f)\bigl(\tfrac{\pi}{2}\bigr) = \frac{2(567 - 147 + 63 - 27 + 7)}{315\pi} = \frac{926}{315\pi} = 0.935730\ldots , \qquad S_{9}(f)\bigl(\tfrac{\pi}{2}\bigr) = \frac{4}{\pi}\Bigl(1 - \tfrac13 + \tfrac15 - \tfrac17 + \tfrac19\Bigr) = \frac{1052}{315\pi} = 1.063054\ldots

The comparison. A numerical search over $\theta \in (0,\pi)$ gives $\max_\theta S_9(f) = 1.18233$ , attained at $\theta = \pi/10$ , while $\max_\theta \sigma_{10}(f) = 926/(315\pi) = 0.93573$ , attained at $\pi/2$ . Refining:

$N$	$\max_\theta S_{N-1}(f)$	$\max_\theta \sigma_N(f)$
$10$	$1.18233$	$0.93573$
$20$	$1.17981$	$0.96825$
$50$	$1.17911$	$0.98726$

The partial sums overshoot the value $1$ by about $18\%$ of the half-jump and keep doing so as $N$ grows; the means stay below $1$ and climb to it. The bound $\inf f \le \sigma_N(f) \le \sup f$ of the remark is exactly the reason, since here $\inf f = -1$ and $\sup f = 1$ . At the jump itself, $\sigma_N(f)(0) = 0$ for every $N$ because every term is odd — the mean of the one-sided limits $\pm 1$ , as the remark predicts.

ApplicationOriginal

Tapered reconstructions in signal processing. Replacing $\hat f(n)$ by $(1 - \lvert n\rvert/N)\hat f(n)$ is the triangular, or Bartlett, window. An ideal low-pass filter returns a plain partial sum and therefore rings at every edge in the input; a Bartlett-tapered filter returns a Fejér mean, and because the Fejér kernel is a non-negative weight of unit mass the output of a filter applied to a signal confined to $[m, M]$ stays confined to $[m, M]$ . For a square-wave input this removes the overshoot of the previous example, which is the observation Stakgold and Holst (Stakgold & Holst (2011), §2.3, p. 136) make about low-pass filters and Cesàro sums. The price is a wider transition region: the kernel's main lobe has width $\sim 2\pi/N$ instead of the partial sum's $\sim \pi/N$ .

Why the trigonometric system is complete. Corollary 2 is the step that upgrades Bessel's inequality to Parseval's identity: a continuous function can be approximated uniformly, hence in mean square, by trigonometric polynomials, and the best mean-square approximation of each degree is the partial sum. In quantum mechanics the same corollary is what justifies treating $\{e^{in\theta}/\sqrt{2\pi}\}$ as a complete basis for states on a ring, so that no state escapes the expansion.

Summability where the series itself diverges. Fejér's theorem recovers the function from its Fourier coefficients under the weakest hypothesis available — mere continuity at the point — and that is how a Fourier coefficient sequence is inverted in practice when nothing is known about smoothness, for example in reconstructing a crystal potential from measured structure factors. The same averaging idea, applied with a different good kernel, gives Abel summation and the Poisson kernel, which solves the Dirichlet problem in the disc.

Uniform polynomial approximation. Corollary 3 is the theoretical licence for polynomial models of calibration curves and equations of state: on any finite interval, a continuous law can be matched to any prescribed tolerance by a polynomial. It says nothing about the degree needed, which is why practical fitting uses Chebyshev or spline bases rather than this construction.

Theorem

Mean-square convergence and Parseval's identity

Adapted

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The partial sums of a Fourier series are the best mean-square approximations by trigonometric polynomials, they always converge to the function in mean square, and the energy of the function equals the sum of the squared moduli of its coefficients, from which the Riemann-Lebesgue lemma follows at once.

Prerequisites: Fourier series, Cesàro means and Fejér's theorem, Inner products and norms, Orthogonal projection and least squares

Definition(The mean-square setting)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 3, §1.1 Example 2 and §1.2, pp. 75–77

Let $\mathcal{R}$ be the complex vector space of functions that are Riemann integrable on $[-\pi,\pi]$ , regarded as $2\pi$ -periodic. On $\mathcal{R}$ put

(f,g) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(\theta)\,\overline{g(\theta)}\,d\theta , \qquad \lVert f \rVert = (f,f)^{1/2} = \Bigl(\frac{1}{2\pi}\int_{-\pi}^{\pi} \lvert f(\theta)\rvert^2 d\theta\Bigr)^{1/2} ,

linear in the first slot, as in inner-product spaces. Write $e_n(\theta) = e^{in\theta}$ . Then

(e_n, e_m) = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{i(n-m)\theta}\,d\theta = \delta_{nm} , \qquad \hat f(n) = (f, e_n) , \qquad S_N(f) = \sum_{\lvert n \rvert \le N} \hat f(n)\,e_n ,

so $\{e_n\}_{n \in \mathbb{Z}}$ is an orthonormal family and the Fourier coefficients of the Fourier series are the coordinates of $f$ along it. Convergence $\lVert f - g_N\rVert \to 0$ is called mean-square convergence, or convergence in the mean; Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.1, p. 262) stress that it permits the approximant to deviate from $f$ at isolated points, which is why it is the natural mode of convergence for Fourier series.

Two cautions about $\mathcal{R}$ , both from Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 3, §1.1, p. 76): $\lVert f\rVert = 0$ forces $f$ to vanish only at its points of continuity, and $\mathcal{R}$ is not complete. Neither affects the statements below; see the remark.

Theorem(Pythagoras, best approximation, Bessel)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 3, §1.2, eq. (3) and Lemma 1.2, pp. 77–78, and Remark 1, p. 80; Reed & Simon (1980), Thm II.1 and its corollary, pp. 37–38

Let $f \in \mathcal{R}$ and let $N \in \mathbb{N}$ .

Exact splitting. $\displaystyle \lVert f \rVert^2 = \lVert f - S_N(f)\rVert^2 + \sum_{\lvert n \rvert \le N} \lvert \hat f(n)\rvert^2 .$
Best approximation. For every choice of complex numbers $c_n$ , $\bigl\lVert f - S_N(f) \bigr\rVert \;\le\; \Bigl\lVert f - \sum_{\lvert n \rvert \le N} c_n e_n \Bigr\rVert ,$ with equality exactly when $c_n = \hat f(n)$ for every $\lvert n \rvert \le N$ . The partial sum is the orthogonal projection of $f$ onto the span of $e_{-N},\dots,e_N$ .
Bessel's inequality. $\displaystyle \sum_{n \in \mathbb{Z}} \lvert \hat f(n)\rvert^2 \le \lVert f \rVert^2 ,$ so the coefficient sequence is square-summable. More generally this holds for the coefficients $(f, u_n)$ along any orthonormal family $\{u_n\}$ in $\mathcal{R}$ , complete or not.

Theorem(Mean-square convergence and Parseval's identity)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 3, Thm 1.3 and Lemma 1.5, pp. 80–81; Stakgold & Holst (2011), §2.3, (2.3.20), p. 135

Let $f, g \in \mathcal{R}$ . Then

Mean-square convergence. $\displaystyle \lVert f - S_N(f)\rVert^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}\bigl\lvert f(\theta) - S_N(f)(\theta)\bigr\rvert^2 d\theta \longrightarrow 0$ as $N \to \infty$ .
Parseval's identity. Bessel's inequality is an equality: $\sum_{n \in \mathbb{Z}} \lvert \hat f(n)\rvert^2 \;=\; \lVert f \rVert^2 \;=\; \frac{1}{2\pi}\int_{-\pi}^{\pi} \lvert f(\theta)\rvert^2 d\theta .$
Two functions. $\displaystyle (f,g) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f\,\overline{g} \;=\; \sum_{n \in \mathbb{Z}} \hat f(n)\,\overline{\hat g(n)} ,$ the series converging absolutely.

The same identities hold for every $f$ in $L^2(-\pi,\pi)$ , a statement Stakgold and Holst (Stakgold & Holst (2011), §2.3, (2.3.20), p. 135) record here and prove in their Chapter 4; we prove only the Riemann-integrable case.

Theorem(Riemann-Lebesgue lemma)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 3, Thm 1.4, pp. 80–81; Stakgold & Holst (2011), §2.3, (2.3.10) and Remark 1 with (2.3.13), pp. 129–130

If $f$ is integrable on the circle then $\hat f(n) \to 0$ as $\lvert n \rvert \to \infty$ ; equivalently

\int_{-\pi}^{\pi} f(\theta)\cos(N\theta)\,d\theta \to 0 \qquad\text{and}\qquad \int_{-\pi}^{\pi} f(\theta)\sin(N\theta)\,d\theta \to 0 .

Stakgold and Holst (Stakgold & Holst (2011), §2.3, Remark 1 and (2.3.13), p. 130) record two extensions, both beyond what the proof below gives: the conclusion holds for every $f \in L_1(-\pi,\pi)$ , and for $f \in L_1(a,b)$ with $(a,b)$ finite or infinite the integrals $\int_a^b f(x)e^{-ikx}dx$ , $\int_a^b f(x)\cos kx\,dx$ and $\int_a^b f(x)\sin kx\,dx$ all tend to $0$ as $\lvert k \rvert \to \infty$ , with $k$ no longer restricted to integers.

Note(Other normalisations of Parseval)AdaptedAdapted from Riley, Hobson & Bence (2006), §12.8, (12.13), p. 426; Arfken, Weber & Harris (2013), Exercise 19.1.13(a), p. 948

For period $L$ and $f(x) \sim \sum_r c_r e^{2\pi i r x/L} = \tfrac12 a_0 + \sum_{r \ge 1}\bigl(a_r \cos\frac{2\pi r x}{L} + b_r\sin\frac{2\pi r x}{L}\bigr)$ , Riley, Hobson and Bence (Riley, Hobson & Bence (2006), §12.8, (12.13), p. 426) write the identity as

\frac{1}{L}\int_{x_0}^{x_0+L}\lvert f(x)\rvert^2 dx = \sum_{r=-\infty}^{\infty}\lvert c_r\rvert^2 = \bigl(\tfrac12 a_0\bigr)^2 + \tfrac12 \sum_{r=1}^{\infty}\bigl(a_r^2 + b_r^2\bigr) ,

the average of $\lvert f\rvert^2$ over one period. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Exercise 19.1.13(a), p. 948) state the same thing for period $2\pi$ as $\frac{1}{\pi}\int_{-\pi}^{\pi}\lvert f\rvert^2 = \frac{a_0^2}{2} + \sum_{n\ge1}(a_n^2+b_n^2)$ , which is twice the previous line; the discrepancy is only the factor $1/L$ against $1/\pi = 2/L$ . Our $\lVert f\rVert^2$ is the period average, so it matches the first form with $L = 2\pi$ . Arfken calls the identity a completeness relation, Riley, Hobson and Bence call it a conservation law; both names describe the same equation.

Idea of the proofOriginal

Everything follows from one picture in finitely many dimensions. The exponentials are orthonormal, so the partial sum $S_N(f)$ is the orthogonal projection of $f$ onto a finite-dimensional subspace, and Pythagoras splits $\lVert f\rVert^2$ into the piece captured by the projection and the piece missed. Two things come free: the projection is the closest point of the subspace (best approximation), and the captured piece never exceeds the whole (Bessel). The only genuinely infinite-dimensional input is that nothing is missed in the limit, and that is exactly the density of trigonometric polynomials supplied by Fejér's theorem: if some trigonometric polynomial of degree $M$ comes within $\varepsilon$ , then by best approximation $S_N(f)$ does too for every $N \ge M$ . Parseval is then the splitting identity with the error term sent to zero, and the Riemann-Lebesgue lemma is the remark that the terms of a convergent series tend to zero.

Proof(Pythagoras, best approximation, Bessel)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 3, §1.2, pp. 77–78; Riley, Hobson & Bence (2006), §8.1.3(iii), pp. 246–247

Step 1: the error is orthogonal to the subspace. Fix $N$ and let $V_N$ be the span of $e_{-N},\dots,e_N$ . For $\lvert m \rvert \le N$ ,

\bigl(f - S_N(f),\, e_m\bigr) = (f, e_m) - \sum_{\lvert n \rvert \le N}\hat f(n)\,(e_n, e_m) = \hat f(m) - \hat f(m) = 0 ,

using orthonormality of $\{e_n\}$ at the second step and the definition $\hat f(m) = (f,e_m)$ at the third. By linearity, $f - S_N(f) \perp v$ for every $v \in V_N$ . This is the only computation in the proof.

Step 2: splitting. Write $f = \bigl(f - S_N(f)\bigr) + S_N(f)$ . The two pieces are orthogonal by step 1, since $S_N(f) \in V_N$ , so expanding $\lVert f\rVert^2 = (f,f)$ kills the cross terms:

\lVert f\rVert^2 = \lVert f - S_N(f)\rVert^2 + \lVert S_N(f)\rVert^2 .

Orthonormality again gives $\lVert S_N(f)\rVert^2 = \sum_{\lvert n\rvert \le N}\lvert \hat f(n)\rvert^2$ , which is part 1.

Step 3: best approximation. Let $P = \sum_{\lvert n\rvert\le N} c_n e_n \in V_N$ and set $b_n = \hat f(n) - c_n$ . Then $f - P = \bigl(f - S_N(f)\bigr) + \sum_{\lvert n\rvert\le N} b_n e_n$ , again an orthogonal decomposition by step 1, so

\lVert f - P\rVert^2 = \lVert f - S_N(f)\rVert^2 + \sum_{\lvert n\rvert \le N}\lvert b_n\rvert^2 \;\ge\; \lVert f - S_N(f)\rVert^2 ,

with equality iff every $b_n = 0$ , that is iff $c_n = \hat f(n)$ . This is part 2, and it identifies $S_N(f)$ as the orthogonal projection onto $V_N$ , the Fourier case of least squares by projection.

Step 4: Bessel. Dropping the non-negative term $\lVert f - S_N(f)\rVert^2$ from part 1 gives $\sum_{\lvert n\rvert\le N}\lvert \hat f(n)\rvert^2 \le \lVert f\rVert^2$ for every $N$ . The left-hand side increases with $N$ and is bounded above by a constant, so it converges, and its limit obeys the same bound. Nothing in steps 1–3 used the particular family $\{e_n\}$ beyond orthonormality, so the same argument gives Bessel's inequality for any orthonormal family — this is how Riley, Hobson and Bence (Riley, Hobson & Bence (2006), §8.1.3(iii), pp. 246–247) and Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.1, (5.22), p. 262) obtain it, by expanding the squared error directly. $\square$

Proof(Mean-square convergence, Parseval, Riemann-Lebesgue)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 3, §1.2, pp. 79–81, using Ch. 2, Lemma 3.2, p. 47

Let $\varepsilon > 0$ .

Step 1: the sup norm dominates the mean-square norm. For any $h \in \mathcal{R}$ ,

\lVert h\rVert^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}\lvert h\rvert^2 \le \sup_\theta \lvert h(\theta)\rvert^2 ,

because the average of a function over the circle does not exceed its supremum. So uniform approximation implies mean-square approximation.

Step 2: continuous $f$ . Suppose first that $f$ is continuous on the circle. By Fejér's theorem there is a trigonometric polynomial $P$ , of some degree $M$ , with $\sup_\theta\lvert f(\theta) - P(\theta)\rvert < \varepsilon$ ; by step 1, $\lVert f - P\rVert < \varepsilon$ . For any $N \ge M$ the polynomial $P$ lies in $V_N$ , so best approximation gives $\lVert f - S_N(f)\rVert \le \lVert f - P\rVert < \varepsilon$ . As $\varepsilon$ was arbitrary, $\lVert f - S_N(f)\rVert \to 0$ .

Step 3: integrable $f$ . A Riemann integrable $f$ is bounded, say $\lvert f \rvert \le B$ . It need not be approximable uniformly by continuous functions, but it is approximable in the sense of Stein and Shakarchi's (Stein & Shakarchi (2003), Ch. 2, Lemma 3.2, p. 47) approximation lemma: there are continuous $g$ on the circle with $\sup\lvert g\rvert \le B$ and $\int_{-\pi}^{\pi}\lvert f - g\rvert$ as small as we please. Choose such a $g$ with $\int_{-\pi}^{\pi}\lvert f-g\rvert < \varepsilon^2$ . Then $\lvert f - g\rvert \le 2B$ pointwise, so one factor of $\lvert f-g\rvert$ can be traded for the constant $2B$ :

\lVert f - g\rVert^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}\lvert f-g\rvert\cdot\lvert f-g\rvert \;\le\; \frac{2B}{2\pi}\int_{-\pi}^{\pi}\lvert f-g\rvert \;<\; \frac{B}{\pi}\,\varepsilon^2 ,

that is $\lVert f - g\rVert < \varepsilon\sqrt{B/\pi}$ . This is the step that needs boundedness, and it is why the squared norm is estimated rather than the norm.

Now apply step 2 to the continuous function $g$ : there is a trigonometric polynomial $P$ of degree $M$ with $\lVert g - P\rVert < \varepsilon$ . The triangle inequality gives $\lVert f - P\rVert < \varepsilon\bigl(1 + \sqrt{B/\pi}\bigr)$ , and best approximation gives the same bound for $\lVert f - S_N(f)\rVert$ once $N \ge M$ . The constant $1 + \sqrt{B/\pi}$ does not depend on $\varepsilon$ , so part 1 of the theorem follows.

Step 4: Parseval. Rearranging the splitting identity,

\lVert f\rVert^2 - \sum_{\lvert n\rvert \le N}\lvert\hat f(n)\rvert^2 = \lVert f - S_N(f)\rVert^2 .

By step 3 the right-hand side tends to $0$ , so the partial sums on the left converge to $\lVert f\rVert^2$ . That is part 2.

Step 5: two functions. In any complex inner-product space the polarisation identity

(F,G) = \tfrac14\Bigl(\lVert F+G\rVert^2 - \lVert F-G\rVert^2 + i\lVert F+iG\rVert^2 - i\lVert F-iG\rVert^2\Bigr)

holds; it is verified by expanding each norm as $(F \pm \lambda G, F \pm \lambda G)$ and collecting terms. Apply Parseval to each of the four functions $f \pm g$ , $f \pm i g$ , whose $n$ th coefficients are $\hat f(n) \pm \hat g(n)$ and $\hat f(n) \pm i\hat g(n)$ by linearity of $f \mapsto \hat f(n)$ . Polarisation in $\ell^2(\mathbb{Z})$ then reassembles the four sums into $\sum_n \hat f(n)\overline{\hat g(n)}$ , giving part 3. Absolute convergence of that series holds because $2\lvert \hat f(n)\overline{\hat g(n)}\rvert \le \lvert \hat f(n)\rvert^2 + \lvert\hat g(n)\rvert^2$ and both of those series converge by Bessel.

Step 6: Riemann-Lebesgue. Parseval (or already Bessel) makes $\sum_n \lvert \hat f(n)\rvert^2$ a convergent series of non-negative terms; the terms of a convergent series tend to $0$ , so $\lvert \hat f(n)\rvert^2 \to 0$ and hence $\hat f(n) \to 0$ as $\lvert n\rvert \to \infty$ . For the real form, $\int_{-\pi}^{\pi} f\cos N\theta\,d\theta = \pi\bigl(\hat f(N) + \hat f(-N)\bigr)$ and $\int_{-\pi}^{\pi} f \sin N\theta\,d\theta = i\pi\bigl(\hat f(N) - \hat f(-N)\bigr)$ , both combinations of quantities tending to $0$ . $\square$

Remark(Where the hypotheses are used)Original

Orthonormality alone gives parts 1–3 of the first statement. Completeness of the family is never used there, which is why Bessel's inequality holds for any orthonormal family and why it is, as Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §5.1, p. 262) put it, a test for completeness rather than a consequence of it: equality for every $f$ is exactly completeness. Drop $e_0$ from the family and take $f \equiv 1$ : then every coefficient is $0$ while $\lVert f\rVert^2 = 1$ , so the inequality is strict.
Completeness enters exactly once, at step 2, through the density of trigonometric polynomials. That density is a theorem, not a definition, and it is the one place where Fejér's theorem is needed; everything else is finite-dimensional geometry.
Boundedness is used at step 3 to trade one factor of $\lvert f-g\rvert$ for $2B$ . Without it an $L^1$ function can fail to be square-integrable, and then neither Bessel nor Parseval need hold — the warning Stakgold and Holst (Stakgold & Holst (2011), §2.3, Remark 2, p. 130) give explicitly. Riemann integrability already implies boundedness, so the hypothesis is not visible in the statement.
Mean-square convergence says nothing pointwise. $\lVert f - S_N(f)\rVert \to 0$ allows the partial sums to misbehave on a small set at every $N$ ; see pointwise convergence and the Gibbs phenomenon, where the overshoot has fixed height but shrinking width and so does not obstruct mean-square convergence.
Parseval does not make $\mathcal{R}$ complete. The map $f \mapsto \{\hat f(n)\}$ is isometric into $\ell^2(\mathbb{Z})$ but not onto: Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 3, Remark 2, p. 80) note that there are square-summable sequences that are the coefficient sequence of no Riemann integrable function. Surjectivity is the Riesz-Fischer theorem and needs the Lebesgue space $L^2$ ; see orthonormal bases and Reed and Simon (Reed & Simon (1980), Thm II.6, p. 45), who state the two-sided version in a Hilbert space.
The Riemann-Lebesgue lemma proved here is weaker than the usual one. Our route goes through square-summability, so it needs $f$ bounded. The general $L_1$ statement is obtained instead by splitting $f = g + h$ with $g$ bounded and $\int\lvert h\rvert$ small, as Stakgold and Holst (Stakgold & Holst (2011), §2.3, Remark 1, p. 130) do.

Method(Summing a numerical series with Parseval)Original

Choose a $2\pi$ -periodic $f$ whose Fourier coefficients are, up to constants, the square roots of the terms you want to sum. A term $1/(n^2+a^2)$ asks for $\hat f(n) \propto 1/(n - ia)$ or $1/(n+ia)$ ; a term $1/n^4$ asks for $\hat f(n) \propto 1/n^2$ , hence for an $f$ that is continuous with a jump in its first derivative.
Compute the coefficients from the definition, and simplify using $e^{\pm i n\pi} = (-1)^n$ .
Compute $\lVert f\rVert^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}\lvert f\rvert^2$ in closed form. This integral is usually elementary even when the series is not.
Equate the two sides of Parseval and solve for the sum. Remember that the sum runs over all of $\mathbb{Z}$ : separate the $n = 0$ term and halve what is left if the series you want runs over $n \ge 1$ .
Check numerically. A partial sum to $\lvert n\rvert \le 100$ should match to the accuracy the decay rate allows; coefficients of size $1/n$ give only two or three digits at that point.
For a cross term $\sum \hat f(n)\overline{\hat g(n)}$ , use part 3 with two different functions rather than squaring one. Riley, Hobson and Bence (Riley, Hobson & Bence (2006), §12.8, pp. 426–427) caution that choosing the right $f$ for a given series is not a systematic procedure; the useful pairings are found, not derived.

Example(Parseval applied to

e^{\theta}

)Original

Let $f$ be the $2\pi$ -periodic function with $f(\theta) = e^{\theta}$ for $-\pi < \theta < \pi$ . It is bounded and continuous except for a jump at the odd multiples of $\pi$ , so it lies in $\mathcal{R}$ .

Coefficients. For every $n \in \mathbb{Z}$ ,

\hat f(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{(1-in)\theta}\,d\theta = \frac{1}{2\pi}\,\frac{e^{(1-in)\pi} - e^{-(1-in)\pi}}{1-in} = \frac{(-1)^n}{2\pi}\,\frac{e^{\pi}-e^{-\pi}}{1-in} = \frac{(-1)^n \sinh \pi}{\pi\,(1-in)} ,

where $e^{\mp i n \pi} = (-1)^n$ was used. Hence

\lvert \hat f(n)\rvert^2 = \frac{\sinh^2\pi}{\pi^2\,\lvert 1-in\rvert^2} = \frac{\sinh^2 \pi}{\pi^2\,(1+n^2)} .

Energy. $\displaystyle \lVert f\rVert^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi} e^{2\theta}\,d\theta = \frac{e^{2\pi}-e^{-2\pi}}{4\pi} = \frac{\sinh 2\pi}{2\pi} = 42.61292\ldots$

Parseval. Equating,

\frac{\sinh^2\pi}{\pi^2}\sum_{n \in \mathbb{Z}} \frac{1}{1+n^2} = \frac{\sinh 2\pi}{2\pi} \quad\Longrightarrow\quad \sum_{n \in \mathbb{Z}} \frac{1}{1+n^2} = \frac{\pi \sinh 2\pi}{2 \sinh^2\pi} = \frac{\pi \cdot 2\sinh\pi\cosh\pi}{2\sinh^2\pi} = \pi\coth\pi ,

using $\sinh 2\pi = 2\sinh\pi\cosh\pi$ . Numerically $\pi\coth\pi = 3.1533480949\ldots$ , and separating $n = 0$ gives the one-sided form

\sum_{n=1}^{\infty}\frac{1}{1+n^2} = \frac{\pi\coth\pi - 1}{2} = 1.0766740474\ldots

Checking, and the size of the error. Bessel's inequality with $N = 3$ uses $\sum_{\lvert n\rvert \le 3}(1+n^2)^{-1} = 1 + 2(\tfrac12+\tfrac15+\tfrac1{10}) = \tfrac{13}{5}$ , so the splitting identity gives the exact mean-square error of the best trigonometric approximation of degree $3$ :

\lVert f - S_3(f)\rVert^2 = \frac{\sinh 2\pi}{2\pi} - \frac{13\,\sinh^2\pi}{5\,\pi^2} = 42.61292 - 35.13523 = 7.47770 , \qquad \lVert f - S_3(f)\rVert = 2.73454 .

That is still $42\%$ of $\lVert f\rVert = 6.52786$ , because $\hat f(n)$ only decays like $1/\lvert n\rvert$ : the jump of size $e^{\pi}-e^{-\pi}$ at $\theta = \pi$ is expensive, exactly as the smoothness-decay correspondence predicts. At $N = 100$ the error has fallen only to $\lVert f - S_{100}(f)\rVert = 0.51857$ .

ApplicationOriginal

Energy and power. For a periodic current $I(t)$ through a resistance $R$ , the average dissipated power is $R\,\langle I^2\rangle$ , and Parseval says this is $R\sum_r \lvert c_r\rvert^2$ : the total power is the sum of the powers carried by the individual harmonics, with no interference terms, because distinct harmonics are orthogonal over a period. This is the sense in which Riley, Hobson and Bence (Riley, Hobson & Bence (2006), §12.8, p. 426) call the identity a conservation law. The same statement for the root-mean-square value of a waveform is what justifies quoting a signal's spectrum in power per harmonic.

Normalisation in quantum mechanics. A state $\psi$ on a ring expands as $\psi = \sum_n c_n e_n$ with $e_n$ the momentum eigenfunctions; Parseval is the statement $\sum_n \lvert c_n\rvert^2 = \lVert\psi\rVert^2 = 1$ , that the probabilities of the possible momentum measurements sum to one. Bessel's inequality applied to a truncated basis is the statement that an incomplete set of states cannot account for all of the probability, which is why completeness of the eigenfunction set is not a technicality.

Truncation error in spectral methods. Part 1 of the first statement is an exact error formula: the mean-square error of the $N$ -term Fourier approximation is the tail $\sum_{\lvert n\rvert > N}\lvert\hat f(n)\rvert^2$ . Because of best approximation, no other trigonometric polynomial of the same degree does better, so this tail is the true benchmark against which a numerical scheme is judged, and the convergence rate of a spectral method is read off from the decay of the coefficients.

Energy inequalities for differential equations. Applied to $f'$ rather than $f$ , Parseval turns $\int\lvert f'\rvert^2$ into $\sum n^2\lvert\hat f(n)\rvert^2$ , and comparing the two sums gives the Poincaré-type inequality $\int\lvert f\rvert^2 \le \int\lvert f'\rvert^2$ for mean-zero periodic $f$ — a one-line proof of a bound that controls the decay of solutions of the heat equation on a ring.

Theorem

Smoothness and decay of Fourier coefficients

Adapted

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Differentiating a periodic function multiplies its Fourier coefficients by $in$ , so each extra continuous derivative buys one more power of decay; this makes term-by-term integration of a Fourier series always legitimate, while term-by-term differentiation is legitimate only when the differentiated series still converges uniformly.

Prerequisites: Fourier series, Pointwise convergence of Fourier series, Mean-square convergence and Parseval's identity, Integrating and differentiating series term by term, Uniform convergence

Definition(Smoothness classes on the circle, and

O

notation)AdaptedAdapted from Stakgold & Holst (2011), §2.3, pp. 130–131; Stein & Shakarchi (2003), Ch. 2, §2, pp. 42 and 44

Write $S$ for the unit circle and identify a function on $S$ with a $2\pi$ -periodic function on $\mathbb{R}$ . Following Stakgold and Holst (Stakgold & Holst (2011), §2.3, pp. 130–131):

$f \in C^k(S)$ means $f$ is continuous and has continuous derivatives of every order up to and including $k$ on the circle — equivalently, the periodic extension of $f$ has $k$ continuous derivatives on all of $\mathbb{R}$ , so that $f^{(j)}(-\pi) = f^{(j)}(\pi)$ for $0 \le j \le k$ . In particular $C^0(S) = C(S)$ is the class of continuous $2\pi$ -periodic functions. Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 2, §2, p. 44) write $C^k$ for the same class.
$f \in D(S)$ ( $f$ obeys Dirichlet conditions) means $f$ is piecewise continuous and the circle admits a finite subdivision on each of whose subintervals $f$ has a continuous derivative, with one-sided derivatives existing at the subdivision points. Stakgold and Holst point out that $D$ and $C$ are not nested: $D$ admits simple jumps, which $C$ does not, while $D$ excludes continuous functions with a cusp such as $\sqrt{\lvert x\rvert}$ .

Integrable on the circle means Riemann integrable on $[-\pi,\pi]$ and extended $2\pi$ -periodically; such a function is in particular bounded. Fourier coefficients are $c_m = \hat f(m) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(\theta)e^{-im\theta}d\theta$ as in the Fourier series; in real form $f \sim \frac{a_0}{2} + \sum_{m \ge 1}(a_m\cos m\theta + b_m \sin m\theta)$ with $a_m = \hat f(m) + \hat f(-m)$ and $b_m = i\bigl(\hat f(m) - \hat f(-m)\bigr)$ .

Finally, $g(m) = O\bigl(h(m)\bigr)$ as $\lvert m\rvert \to \infty$ means that $\lvert g(m)\rvert \le C \lvert h(m)\rvert$ for some constant $C$ and all large $\lvert m\rvert$ ; $g(m) = o\bigl(h(m)\bigr)$ means $g(m)/h(m) \to 0$ , which is the stronger statement.

Theorem(Differentiation multiplies the coefficient by

im

)AdaptedAdapted from Stakgold & Holst (2011), §2.3, p. 134; Strauss (2008), §5.5, eqs. (15)–(16), p. 141

Let $f \in C^1(S)$ and let $c_m$ and $d_m$ be the Fourier coefficients of $f$ and of $f'$ . Then

d_m = i\,m\,c_m \qquad (m \in \mathbb{Z}), \qquad\text{equivalently}\qquad c_m = \frac{d_m}{i m}\ \ (m \ne 0) .

In the real coefficients of Strauss (Strauss (2008), §5.5, eqs. (15)–(16), p. 141), writing $A_m, B_m$ for those of $f$ and $A_m', B_m'$ for those of $f'$ : $A_m = -B_m'/m$ and $B_m = A_m'/m$ . The Fourier series of $f'$ is therefore the term-by-term derivative of the Fourier series of $f$ — as a statement about coefficients; whether the differentiated series converges to $f'$ is a separate question, settled below.

Theorem(Smoothness forces decay)AdaptedAdapted from Stakgold & Holst (2011), §2.3, Thm 2.3.1, p. 131 and the theorem on p. 134; Stein & Shakarchi (2003), Ch. 2, Cor. 2.4, pp. 42–43

Let $p \ge 1$ be an integer.

Decay. If $f \in C^p(S)$ then $\displaystyle \lim_{\lvert m\rvert \to \infty} m^p c_m = 0$ , i.e. $c_m = o\bigl(\lvert m\rvert^{-p}\bigr)$ .
One derivative already gives absolute convergence. If $f \in C^1(S)$ then $\sum_{m \in \mathbb{Z}} \lvert c_m\rvert < \infty$ , and the Fourier series of $f$ converges to $f$ absolutely and uniformly on $\mathbb{R}$ . Strauss (Strauss (2008), §5.5, p. 142) notes that the same proof works when $f$ is continuous and $f'$ is merely piecewise continuous, the example being $f(x) = \lvert x\rvert$ .
The $C^2$ form. If $f$ is twice continuously differentiable on the circle then $\hat f(m) = O\bigl(1/\lvert m\rvert^{2}\bigr)$ , and the Fourier series of $f$ converges absolutely and uniformly to $f$ .

Parts 1 and 3 are two readings of the same computation: with $p = 2$ , part 1 gives the formally stronger $o(\lvert m\rvert^{-2})$ , while the route of Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 2, Cor. 2.4, pp. 42–43) gives the explicit constant $2\pi \lvert m\rvert^2 \lvert \hat f(m)\rvert \le \int_{-\pi}^{\pi}\lvert f''\rvert$ (they write the integral over $[0,2\pi]$ , which is the same by periodicity).

Theorem(Term-by-term integration and term-by-term differentiation)AdaptedAdapted from Strauss (2008), §5.4, p. 130 and Exercise 5.4.11; Arfken, Weber & Harris (2013), Exercise 19.1.12, p. 947

Let $f$ be integrable on the circle, with coefficients $c_m$ .

Integration is always legitimate. For every $x \in [-\pi,\pi]$ , $\int_{-\pi}^{x} f(t)\,dt \;=\; c_0\,(x+\pi) \;+\; \sum_{m \ne 0} \frac{c_m}{i m}\Bigl(e^{imx} - (-1)^m\Bigr) ,$ and the series on the right converges absolutely and uniformly in $x$ . The right-hand side is exactly what term-by-term integration of $\sum_m c_m e^{imt}$ from $-\pi$ to $x$ produces — even when that series itself diverges. It is the precise form of the remark of Strauss (Strauss (2008), §5.4, p. 130 and Exercise 5.4.11), who calls term-by-term integration "not delicate" and usually valid, in contrast with differentiation, which he calls delicate.
Differentiation is not. Term-by-term differentiation of a convergent Fourier series may produce a divergent series. Strauss's example: the Fourier sine series of the constant $f \equiv 1$ on $(0,\pi)$ is $\sum_{n\ \mathrm{odd}}\frac{4}{n\pi}\sin nx$ , which converges pointwise to $1$ there; differentiating term by term gives $\frac{4}{\pi}\sum_{n\ \mathrm{odd}}\cos nx$ , whose terms do not tend to $0$ , so the $n$ th-term test denies convergence — this is how Strauss (Strauss (2008), §5.4, p. 130) dismisses it — while $f' \equiv 0$ . The single exception on $(0,\pi)$ is $x = \pi/2$ : there $\cos nx = 0$ for every odd $n$ , so every term vanishes and the series converges, to $0$ . Everywhere else in $(0,\pi)$ the terms fail to tend to $0$ and the series diverges, so the differentiated series does not represent $f'$ on the interval.
When it is. If $f \in C^2(S)$ then the differentiated series $\sum_m i m\, c_m e^{im\theta}$ converges absolutely and uniformly, and its sum is $f'$ .

Theorem(The converse: decay forces smoothness)Original

Let $k \ge 0$ be an integer and let $(c_m)_{m \in \mathbb{Z}}$ be complex numbers with

\sum_{m \in \mathbb{Z}} \lvert m \rvert^{k}\,\lvert c_m\rvert < \infty .

Then $g(\theta) = \sum_{m} c_m e^{im\theta}$ converges absolutely and uniformly, $g \in C^k(S)$ , the coefficients of $g$ are exactly the $c_m$ , and for every $0 \le j \le k$

g^{(j)}(\theta) = \sum_{m \in \mathbb{Z}} (im)^j c_m e^{im\theta} ,

each of these series converging absolutely and uniformly. Together with part 1 of the decay theorem this gives a two-sided dictionary: coefficients decaying faster than every power correspond to $f \in C^\infty(S)$ .

Idea of the proofOriginal

One integration by parts is the whole story. Moving a derivative off $f$ and onto $e^{-im\theta}$ produces a factor $im$ , and the boundary term it leaves behind is killed by periodicity — not by smallness, but by the fact that the two endpoints of a circle are the same point. So $p$ derivatives convert $c_m$ into $\widehat{f^{(p)}}(m)/(im)^p$ , and the Riemann-Lebesgue lemma says the numerator still tends to zero. For the converse one reads the same identity backwards: enough decay makes the differentiated series converge uniformly, and uniform convergence of a differentiated series is precisely the hypothesis under which differentiation may be taken inside a sum. Integration behaves better than differentiation because it divides by $m$ rather than multiplying, and dividing by $m$ turns the square-summability that Bessel's inequality gives for free into absolute summability, by the Cauchy-Schwarz inequality against $\sum m^{-2} < \infty$ .

Proof(Differentiation multiplies the coefficient by

im

, and smoothness forces decay)AdaptedAdapted from Stakgold & Holst (2011), §2.3, Thm 2.3.1, p. 131 and the theorem on p. 134; Stein & Shakarchi (2003), Ch. 2, Cor. 2.4, pp. 42–43

Step 1: one integration by parts. Let $f \in C^1(S)$ and $m \in \mathbb{Z}$ . Then $f$ and $e^{-im\theta}$ are both continuously differentiable on $[-\pi,\pi]$ , so

d_m = \frac{1}{2\pi}\int_{-\pi}^{\pi} f'(\theta)\,e^{-im\theta}\,d\theta = \frac{1}{2\pi}\Bigl[f(\theta)e^{-im\theta}\Bigr]_{-\pi}^{\pi} + \frac{im}{2\pi}\int_{-\pi}^{\pi} f(\theta)e^{-im\theta}\,d\theta .

The bracket is $f(\pi)e^{-im\pi} - f(-\pi)e^{im\pi} = (-1)^m\bigl(f(\pi)-f(-\pi)\bigr) = 0$ , because $f$ is continuous on the circle. Hence $d_m = i m\,c_m$ . This is the step, and the only step, at which periodicity is used. The real forms follow by taking $a_m = c_m + c_{-m}$ and $b_m = i(c_m - c_{-m})$ .

Step 2: part 1, by iterating. Let $f \in C^p(S)$ . Then $f, f', \dots, f^{(p-1)}$ all belong to $C^1(S)$ , so step 1 applies $p$ times in succession and gives

\widehat{f^{(p)}}(m) = (im)^p\,c_m , \qquad\text{that is}\qquad m^p c_m = \frac{\widehat{f^{(p)}}(m)}{i^{\,p}} .

Now $f^{(p)}$ is continuous on the circle, hence integrable, so the Riemann-Lebesgue lemma gives $\widehat{f^{(p)}}(m) \to 0$ as $\lvert m\rvert \to \infty$ . Taking absolute values, $\lvert m^p c_m\rvert = \lvert\widehat{f^{(p)}}(m)\rvert \to 0$ .

Step 3: part 2, by Cauchy-Schwarz. Let $f \in C^1(S)$ and let $d_m = i m c_m$ be the coefficients of $f'$ . Splitting off $m = 0$ and inserting $1 = \frac{1}{\lvert m\rvert}\cdot\lvert m\rvert$ ,

\sum_{m \in \mathbb{Z}} \lvert c_m\rvert = \lvert c_0\rvert + \sum_{m \ne 0}\frac{\lvert d_m\rvert}{\lvert m\rvert} \;\le\; \lvert c_0 \rvert + \Bigl(\sum_{m \ne 0}\frac{1}{m^2}\Bigr)^{1/2}\Bigl(\sum_{m \ne 0}\lvert d_m\rvert^2\Bigr)^{1/2} ,

by the Cauchy-Schwarz inequality for sequences. The first factor is $\sqrt{\pi^2/3}$ , finite; the second is finite because $f'$ is continuous, hence square integrable, so Bessel's inequality bounds $\sum_m \lvert d_m\rvert^2$ by $\lVert f'\rVert^2$ . Therefore $\sum_m \lvert c_m\rvert < \infty$ .

Consequently $\lvert c_m e^{im\theta}\rvert = \lvert c_m\rvert$ is a summable majorant independent of $\theta$ , so the Fourier series converges absolutely and uniformly by the Weierstrass $M$ -test; see uniform convergence. Its sum is continuous, and it agrees with $f$ at every point by the pointwise convergence theorem, whose hypotheses a $C^1(S)$ function certainly meets.

Step 4: part 3, with the explicit constant. Let $f \in C^2(S)$ and $m \ne 0$ . Applying step 1 twice,

2\pi\,\lvert m\rvert^2\,\lvert \hat f(m)\rvert = \Bigl\lvert \int_{-\pi}^{\pi} f''(\theta)e^{-im\theta}\,d\theta\Bigr\rvert \le \int_{-\pi}^{\pi}\lvert f''(\theta)\rvert\,d\theta = C ,

a constant independent of $m$ because $f''$ is continuous on a compact set. Hence $\hat f(m) = O(1/\lvert m\rvert^2)$ , and since $\sum_{m\ne0} \lvert m\rvert^{-2} < \infty$ the series converges absolutely and uniformly, again with sum $f$ . $\square$

Proof(Term-by-term integration, and when differentiation is allowed)AdaptedAdapted from Strauss (2008), §5.4, p. 130 and Exercise 5.4.11

Let $f$ be integrable on the circle, with coefficients $c_m$ , and define

F(x) = \int_{-\pi}^{x}\bigl(f(t) - c_0\bigr)\,dt , \qquad -\pi \le x \le \pi .

Step 1: $F$ is continuous on the circle. As an indefinite integral of an integrable function, $F$ is continuous (indeed Lipschitz, since $f$ is bounded). At the endpoints, $F(-\pi) = 0$ and

F(\pi) = \int_{-\pi}^{\pi} f - 2\pi c_0 = 2\pi c_0 - 2\pi c_0 = 0 ,

by the definition of $c_0$ . Subtracting $c_0$ is exactly what makes $F$ periodic; without it $F$ would increase by $2\pi c_0$ over one period.

Step 2: the coefficients of $F$ . Fix $m \ne 0$ and write $g = f - c_0$ , so $F(x) = \int_{-\pi}^{x} g$ . We want to integrate by parts, moving the derivative off $F$ ; but $F$ need not be differentiable everywhere, since $g$ is only Riemann integrable. What licenses the step is not a null-set argument about where $F' = g$ — it is the fact that $F$ is an indefinite Riemann integral, which lets us exchange the order of integration over the triangle $\{-\pi \le t \le x \le \pi\}$ :

\int_{-\pi}^{\pi} F(x)\,e^{-imx}\,dx = \int_{-\pi}^{\pi}\!\!\int_{-\pi}^{x} g(t)\,e^{-imx}\,dt\,dx = \int_{-\pi}^{\pi} g(t)\Bigl(\int_{t}^{\pi} e^{-imx}\,dx\Bigr) dt = \int_{-\pi}^{\pi} g(t)\,\frac{e^{-imt} - e^{-im\pi}}{im}\,dt .

(The exchange is Fubini's theorem for the Riemann integral of the bounded function $g(t)\,e^{-imx}\,\mathbf{1}_{t \le x}$ on the square; it is the integrated form of integration by parts, and it needs nothing of $F$ beyond its definition.) The term $e^{-im\pi}\int_{-\pi}^{\pi}g$ vanishes because $\int_{-\pi}^{\pi} g = F(\pi) = 0$ by step 1 — this is where periodicity of $F$ enters. Dividing by $2\pi$ and noting that the constant $c_0$ inside $g$ contributes nothing for $m \ne 0$ ,

\hat F(m) = \frac{1}{im}\cdot\frac{1}{2\pi}\int_{-\pi}^{\pi}\bigl(f(t)-c_0\bigr)e^{-imt}\,dt = \frac{c_m}{im} .

Step 3: the series for $F$ converges absolutely. As in step 3 of the previous proof, $f$ is bounded, hence square integrable, so Bessel's inequality gives $\sum_m \lvert c_m\rvert^2 < \infty$ , and Cauchy-Schwarz gives

\sum_{m \ne 0}\lvert \hat F(m)\rvert = \sum_{m\ne0}\frac{\lvert c_m\rvert}{\lvert m\rvert} \le \Bigl(\sum_{m\ne0}\frac{1}{m^2}\Bigr)^{1/2}\Bigl(\sum_{m}\lvert c_m\rvert^2\Bigr)^{1/2} < \infty .

So $\sum_m \hat F(m)e^{imx}$ converges absolutely and uniformly, to a continuous function. Its coefficients are the $\hat F(m)$ (uniform convergence permits integrating term by term against $e^{-imx}$ , by term-by-term integration of a uniformly convergent series), which are also the coefficients of the continuous function $F$ . By the uniqueness corollary of Fejér's theorem the two continuous functions coincide:

F(x) = \hat F(0) + \sum_{m \ne 0}\frac{c_m}{im}e^{imx} \qquad\text{for every } x .

Step 4: fix the constant and restore $c_0$ . Put $x = -\pi$ , where $F(-\pi) = 0$ and $e^{-im\pi} = (-1)^m$ ; this gives $\hat F(0) = -\sum_{m\ne0}\frac{c_m}{im}(-1)^m$ . Substituting back and adding $\int_{-\pi}^{x}c_0\,dt = c_0(x+\pi)$ to both sides gives the stated formula. Term-by-term integration of the original series produces $\int_{-\pi}^{x}c_m e^{imt}dt = \frac{c_m}{im}\bigl(e^{imx} - (-1)^m\bigr)$ for $m \ne 0$ and $c_0(x+\pi)$ for $m = 0$ , which is the same expression.

Step 5: part 3, when differentiation is allowed. Let $f \in C^2(S)$ . Then $f' \in C^1(S)$ , and by step 1 of the previous proof the coefficients of $f'$ are $d_m = i m\,c_m$ . Applying part 2 of the decay theorem to $f'$ in place of $f$ gives $\sum_m \lvert d_m\rvert = \sum_m \lvert m\rvert\lvert c_m\rvert < \infty$ , so the differentiated series $\sum_m i m\,c_m e^{im\theta}$ is the Fourier series of $f'$ and converges to $f'$ absolutely and uniformly. Note which hypothesis did the work: $f \in C^1(S)$ alone would give the coefficient identity but no convergence, which is exactly the gap Strauss's example exploits. $\square$

Proof(The converse)Original

Assume $\sum_m \lvert m\rvert^k \lvert c_m\rvert < \infty$ .

Step 1: every series up to order $k$ converges uniformly. For $m \ne 0$ and $0 \le j \le k$ we have $\lvert m\rvert^{j} \le \lvert m\rvert^{k}$ when $\lvert m \rvert \ge 1$ , so

\sum_{m}\bigl\lvert (im)^j c_m e^{im\theta}\bigr\rvert = \sum_m \lvert m\rvert^j \lvert c_m\rvert \le \lvert c_0\rvert + \sum_{m \ne 0}\lvert m\rvert^{k}\lvert c_m\rvert < \infty ,

a bound independent of $\theta$ . By the Weierstrass $M$ -test each of the $k+1$ series $G_j(\theta) = \sum_m (im)^j c_m e^{im\theta}$ , $0 \le j \le k$ , converges absolutely and uniformly on $\mathbb{R}$ , and each $G_j$ is continuous and $2\pi$ -periodic as a uniform limit of such functions.

Step 2: the sums differentiate into one another. Fix $0 \le j < k$ . The series for $G_j$ converges uniformly, its terms are differentiable, and the series of their derivatives is the series for $G_{j+1}$ , which also converges uniformly by step 1. Those are exactly the hypotheses of term-by-term differentiation, so $G_j$ is differentiable with $G_j' = G_{j+1}$ .

Step 3: conclusion. Writing $g = G_0$ and applying step 2 for $j = 0, 1, \dots, k-1$ in turn gives $g^{(j)} = G_j$ for $0 \le j \le k$ . Each $G_j$ is continuous and $2\pi$ -periodic, so $g \in C^k(S)$ and the displayed formula for $g^{(j)}$ holds.

Step 4: the coefficients are the $c_m$ . The series for $g$ converges uniformly, so it may be integrated against the bounded function $e^{-in\theta}$ term by term (term-by-term integration); orthonormality leaves only the term $m = n$ , giving $\hat g(n) = c_n$ . $\square$

Remark(Where the hypotheses are used)Original

Periodicity, not smoothness, kills the boundary term. Step 1 of the first proof needs $f(\pi) = f(-\pi)$ , nothing more. A function that is as smooth as one likes on the closed interval but does not match at the ends gets no benefit: $f(\theta) = \theta$ on $(-\pi,\pi)$ is real-analytic there, yet its periodic extension jumps, and $\hat f(m) = i(-1)^m/m$ decays only like $1/\lvert m\rvert$ . This is why the statements are about $C^k(S)$ , the circle, and not about $C^k[-\pi,\pi]$ .
One jump costs exactly one power. A function in $D(S)$ with a simple jump has coefficients of order $1/\lvert m\rvert$ and no better — this is Stakgold and Holst (Stakgold & Holst (2011), §2.3, Thm 2.3.2 and its proof, p. 131), whose proof isolates the jump term $e^{-im\xi}\bigl(f(\xi^+)-f(\xi^-)\bigr)$ and observes that it has modulus independent of $m$ , so $m c_m$ stays bounded while $m^{1+\varepsilon}c_m$ does not. Hence $\sum\lvert c_m\rvert$ diverges and the convergence cannot be uniform near the jump; that is the mechanism behind the Gibbs phenomenon. Stakgold and Holst (Stakgold & Holst (2011), §2.3, Example 2 and Example 3, p. 132) tabulate the next two cases: $\lvert x\rvert$ , continuous with a derivative in $D$ , has $a_m$ of order $1/m^2$ ; and the periodised $\cos a x$ with $a \notin \mathbb{Z}$ , continuous with a derivative jumping at odd multiples of $\pi$ , has $m^2 a_m$ bounded.
Decay alone, without periodicity of the derivatives, proves nothing about differentiability. The converse statement needs the full hypothesis $\sum \lvert m\rvert^k\lvert c_m\rvert < \infty$ ; square-summability of $m^k c_m$ is not enough, because the $M$ -test in step 1 needs an $\ell^1$ bound and uniform convergence of the differentiated series is what term-by-term differentiation requires.
The gap between the two directions is one power. Part 1 gives $f \in C^p(S) \Rightarrow c_m = o(\lvert m\rvert^{-p})$ , while the converse needs $\sum\lvert m\rvert^{k}\lvert c_m\rvert < \infty$ , which for a power law means $c_m = O(\lvert m\rvert^{-k-1-\varepsilon})$ . So the dictionary is not exact at any finite order; it becomes exact only in the limit, where both sides read "faster than every power" versus $C^\infty(S)$ .
Term-by-term differentiation fails for the reason the $M$ -test fails. In Strauss's example the function $1$ on $(0,\pi)$ is extended oddly, which creates jumps at $0$ and at $\pi$ ; the coefficients are of order $1/n$ , so multiplying by $n$ leaves coefficients that do not tend to zero. Nothing is wrong with the coefficient identity $d_m = imc_m$ — it is the hypothesis $f \in C^1(S)$ that fails, so there is no $f'$ on the circle whose coefficients those would be.
Term-by-term integration never fails because dividing by $m$ can only help, and because the square-summability needed for the Cauchy-Schwarz step is free from Bessel's inequality for any bounded $f$ . Note what the conclusion is not: the integrated series is not in general a Fourier series, since the term $c_0(x+\pi)$ is not periodic. It becomes one exactly when $c_0 = 0$ .

Method(Reading and using the decay rate)Original

Classify the function on the circle, not on the interval. Check the endpoint matching $f^{(j)}(-\pi) = f^{(j)}(\pi)$ for $j = 0, 1, 2, \dots$ and stop at the first $j$ that fails. If matching holds up to $j = p$ but not $j = p+1$ , expect $c_m$ of order $\lvert m \rvert^{-(p+2)}$ in the generic case with only $o(\lvert m\rvert^{-p})$ guaranteed by the theorem.
Predict the convergence mode. $\sum\lvert c_m\rvert < \infty$ (guaranteed from $f \in C^1(S)$ ) means absolute and uniform convergence; a jump means $c_m \sim 1/m$ , no absolute convergence, and overshoot near the jump.
To differentiate a Fourier series term by term, first multiply the coefficients by $im$ and test the new series for uniform convergence (the $M$ -test on $\sum\lvert m c_m\rvert$ is usually decisive). If it converges uniformly, the differentiated series is $f'$ . If it does not, do not differentiate: either work with the integrated form instead, or move to the distributional derivative of distributions, where differentiation is always permitted.
To integrate a Fourier series term by term, just do it; no hypothesis beyond integrability of $f$ is needed. Remember to treat $c_0$ separately: it contributes a linear term, not a periodic one. This is the trick Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Exercise 19.1.12, p. 947) use to turn the series for $f(x) = x$ into a value for $\sum_{n\ge1}(-1)^{n+1}/n^2$ .
To estimate a truncation error, combine the decay rate with Parseval's identity: if $\lvert c_m\rvert \le A\lvert m\rvert^{-q}$ then for $q > 1/2$ the mean-square error of the $N$ -term partial sum obeys $\lVert f - S_N(f)\rVert^2 = \sum_{\lvert m\rvert > N}\lvert c_m\rvert^2 \le 2A^2\int_N^{\infty} t^{-2q}dt$ , that is $\lVert f - S_N(f)\rVert \le \bigl(2A^2/(2q-1)\bigr)^{1/2} N^{-q+1/2}$ , so one extra derivative of smoothness buys one extra power of $N$ in the convergence rate.

Example(A function that is

C^1

but not

C^2

on the circle)Original

Let $h$ be the $2\pi$ -periodic extension of

h(\theta) = \theta\,(\pi^2 - \theta^2) , \qquad -\pi \le \theta \le \pi .

Which class is it in? $h(\pm\pi) = \pm\pi(\pi^2-\pi^2) = 0$ , so $h$ is continuous on the circle. Next $h'(\theta) = \pi^2 - 3\theta^2$ , and $h'(\pi) = h'(-\pi) = -2\pi^2$ : the derivative matches too, so $h \in C^1(S)$ . But $h''(\theta) = -6\theta$ gives $h''(\pi) = -6\pi$ and $h''(-\pi) = +6\pi$ , which differ; so $h \notin C^2(S)$ . Step 1 of the method predicts matching up to $j=1$ , hence decay somewhere between $o(m^{-1})$ (guaranteed) and $m^{-3}$ (generic).

The coefficients. $h$ is odd, so only sine terms survive and $b_m = \frac{2}{\pi}\int_0^{\pi}(\pi^2\theta - \theta^3)\sin m\theta\,d\theta$ . Two standard integrations by parts give

\int_0^\pi \theta \sin m\theta\,d\theta = \frac{(-1)^{m+1}\pi}{m} , \qquad \int_0^\pi \theta^3\sin m\theta\,d\theta = \frac{(-1)^{m+1}\pi^3}{m} + \frac{6(-1)^{m}\pi}{m^3} ,

so the $\pi^3/m$ parts cancel and

b_m = \frac{2}{\pi}\Bigl(-\frac{6(-1)^m \pi}{m^3}\Bigr) = \frac{12\,(-1)^{m+1}}{m^3} , \qquad h(\theta) = 12\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m^3}\,\sin m\theta .

The decay is $m^{-3}$ , the generic rate, and the theorem's guarantee $m\,b_m \to 0$ is comfortably met with room to spare.

Check the sum. At $\theta = \pi/2$ the sines are $1, 0, -1, 0, 1, \dots$ , so the series collapses to $12\sum_{j\ge0}(-1)^j(2j+1)^{-3} = 12\beta(3) = 12\cdot\pi^3/32$ , while $h(\pi/2) = \frac{\pi}{2}\bigl(\pi^2 - \frac{\pi^2}{4}\bigr) = \frac{3\pi^3}{8}$ . These agree: $12\pi^3/32 = 3\pi^3/8 = 11.627354$ . Absolute convergence is visible too, since $\sum_m\lvert b_m\rvert = 12\zeta(3) = 14.424683 < \infty$ .

Differentiate once — legitimate. The differentiated series is $12\sum_m (-1)^{m+1}m^{-2}\cos m\theta$ , with $\sum 12\,m^{-2} = 12\zeta(2) = 2\pi^2 = 19.739209 < \infty$ : uniformly convergent by the $M$ -test, so by step 3 of the method its sum is $h'(\theta) = \pi^2 - 3\theta^2$ . Testing at $\theta = 0$ :

12\sum_{m=1}^{\infty}\frac{(-1)^{m+1}}{m^2} = 12\cdot\frac{\pi^2}{12} = \pi^2 = h'(0) ,

and at $\theta = \pi$ , $12\sum_m (-1)^{m+1}(-1)^m m^{-2} = -12\zeta(2) = -2\pi^2 = h'(\pi)$ , as it must be.

Differentiate twice — not legitimate. The next formal derivative is $-12\sum_m (-1)^{m+1}m^{-1}\sin m\theta$ . Its coefficients are of order $1/m$ , so $\sum\lvert{\cdot}\rvert$ diverges and the $M$ -test gives nothing. It does converge pointwise on $(-\pi,\pi)$ , to $-12\cdot\frac{\theta}{2} = -6\theta = h''(\theta)$ ; but at $\theta = \pi$ every term is zero, so the series sums to $0$ there while $h''(\pi^-) = -6\pi$ . The convergence is therefore not uniform, exactly as $h \notin C^2(S)$ requires. A third formal differentiation gives $-12\sum(-1)^{m+1}\cos m\theta$ , whose terms do not tend to $0$ : divergent everywhere.

Integrate once — always legitimate. Integrating the series for $h$ from $0$ to $x$ term by term, and using $\int_0^x \sin m\theta\,d\theta = (1-\cos mx)/m$ ,

\int_0^x h(\theta)\,d\theta = \frac{\pi^2x^2}{2} - \frac{x^4}{4} = 12\sum_{m=1}^{\infty}\frac{(-1)^{m+1}\bigl(1 - \cos mx\bigr)}{m^4} .

At $x = \pi$ the factor $1-\cos m\pi$ is $2$ for odd $m$ and $0$ for even $m$ , so the right-hand side is $24\sum_{j\ge0}(2j+1)^{-4} = 24\cdot\frac{\pi^4}{96} = \frac{\pi^4}{4}$ , and the left-hand side is $\frac{\pi^4}{2} - \frac{\pi^4}{4} = \frac{\pi^4}{4} = 24.352273$ . They agree, and the integrated coefficients decay like $m^{-4}$ — one power better than before, as integration always gives.

ApplicationOriginal

Convergence rates of spectral methods. A Fourier spectral solver for a periodic PDE represents the solution by $N$ coefficients; by part 1 of the decay theorem the truncation error is $O(N^{-p})$ when the solution is in $C^p$ , and faster than every power — "spectral accuracy" — when it is $C^\infty$ . This is why a shock or a corner in the data destroys the advantage of a spectral method over a finite-difference one, and why practitioners inspect the measured decay rate of $\lvert c_m\rvert$ on a log-log plot as a diagnostic: a slope of $-1$ announces a jump, $-2$ a kink, and no straight line at all announces smoothness.

Smoothing by diffusion. Solving the heat equation on a ring multiplies the $m$ th coefficient of the initial data by $e^{-m^2 t}$ . The result decays faster than every power of $m$ for every $t > 0$ , so by the converse statement the solution is $C^\infty$ on the circle however rough the initial data were. Stakgold and Holst (Stakgold & Holst (2011), §2.3, Example 3, p. 132) make exactly this observation about the temperature in an infinite rod. Run backwards, the same factor amplifies high frequencies, which is the precise sense in which backward heat conduction is ill-posed.

Numerical differentiation amplifies noise. Differentiating a measured periodic signal multiplies the $m$ th coefficient by $im$ . Measurement noise is broadband, so its high-frequency coefficients do not decay; multiplying them by $m$ makes the differentiated signal worse the finer the sampling. The standard remedy — low-pass filtering before differentiating — is the statement that one must restore the decay the derivative destroyed.

Reading structure off a spectrum. Because a jump contributes a $1/m$ tail and a kink a $1/m^2$ tail, the asymptotic envelope of a measured Fourier spectrum identifies the strongest singularity of the underlying periodic signal without locating it. In vibration monitoring, the appearance of a $1/m$ tail where a $1/m^3$ tail was expected signals that something in the system has begun to strike rather than to flex.