University Mathematics — Year 2 · Bachelor Year 2

14Fourier Series

Can every periodic signal be rebuilt from pure sines and cosines? Fourier’s audacious “yes” created a century of analysis. This chapter proves the two pillars within MP* reach: Dirichlet’s theorem (pointwise reconstruction for piecewise $C^1$ functions, via the Dirichlet kernel) and Parseval’s identity (the energy of a signal is the sum of the energies of its harmonics), and harvests the classical numerical series — Basel’s $\sum 1/n^2 = \pi^2/6$ first among them.

Throughout, functions are $2\pi$ -periodic, piecewise continuous, complex-valued; $\mathcal{C}$ denotes the continuous ones.

14.1 Fourier coefficients

Definition 14.1

The Fourier coefficients of $f$ are

c_n(f) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)\,\eu^{-\iu n t}\,\dd t \qquad (n \in \Z),

and the real-form coefficients $a_n = c_n + c_{-n}$ , $b_n = \iu(c_n - c_{-n})$ , so that the Fourier partial sums are

S_N(f)(t) = \sum_{n=-N}^{N} c_n(f)\,\eu^{\iu nt} = \frac{a_0}{2} + \sum_{n=1}^{N}\bigl(a_n\cos nt + b_n \sin nt\bigr).

On $\mathcal{C}$ , define the Hermitian inner product $\langle f, g\rangle = \frac{1}{2\pi}\int_{-\pi}^{\pi}\conj f\,g$ : the exponentials $e_n(t) = \eu^{\iu nt}$ are orthonormal ( $\langle e_m, e_n\rangle = \delta_{mn}$ , direct computation), and $c_n(f) = \langle e_n, f\rangle$ : Fourier analysis is Hermitian geometry (Chapter 13) in infinite dimension.

Proposition 14.2 (Bessel’s inequality)

$S_N(f)$ is the orthogonal projection of $f$ onto the space $\mathcal{T}_N$ of trigonometric polynomials of degree $\leq N$ , and

\sum_{n=-N}^{N} \abs{c_n(f)}^2 \leq \norm f_2^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi} \abs f^2 :

the series $\sum \abs{c_n}^2$ converges, and $c_n(f) \to 0$ as $\abs n \to \infty$ (Riemann–Lebesgue for coefficients).

Proof. $f - S_N(f)$ is orthogonal to each $e_k$ , $\abs k \leq N$ ( $\langle e_k, f - S_N f\rangle = c_k - c_k = 0$ ): $S_Nf$ is the orthogonal projection onto $\mathcal{T}_N = \operatorname{Vect}(e_{-N}, \dots, e_N)$ (Year 1 volume’s projection theorem, verbatim in the Hermitian setting). Pythagoras: $\norm f_2^2 = \norm{S_Nf}_2^2 + \norm{f - S_Nf}_2^2 \geq \norm{S_Nf}_2^2 = \sum_{\abs n \leq N} \abs{c_n}^2$ ; let $N \to \infty$ . ∎

Example 14.3 (Best approximation, measured)

How well do low-degree trigonometric polynomials approximate the sawtooth $f(t) = t$ (on $\intoo{-\pi}{\pi}$ ) in the quadratic mean? By Proposition 14.2 the best degree- $N$ approximation is $S_N(f)$ , with squared error

\norm{f - S_Nf}_2^2 = \norm f_2^2 - \sum_{\abs n\leq N}\abs{c_n}^2 .

Here $\norm f_2^2 = \frac{1}{2\pi}\int_{-\pi}^\pi t^2\dd t = \frac{\pi^2}{3}$ , and from $b_n = \frac{2(-1)^{n+1}}{n}$ (Example 14.12): $\abs{c_n}^2 + \abs{c_{-n}}^2 = \frac{b_n^2}{2} = \frac{2}{n^2}$ . So

\norm{f - S_Nf}_2^2 = \frac{\pi^2}{3} - \sum_{n=1}^{N}\frac{2}{n^2} \qquad\text{: numerically } 1.29,\ 0.79,\ 0.57,\ 0.44

for $N = 1, 2, 3, 4$ — decreasing, but slowly: the tail $\sum_{n>N}\frac2{n^2} \sim \frac2N$ is governed by the slow $\frac1n$ decay of the coefficients, itself the signature of the jump (Exercise 14.6 read backwards). Closing insight: Parseval turns approximation quality into a tail of a numerical series — and predicts, before any picture, that jumps make Fourier series converge reluctantly.

Method 14.4 (Computing Fourier coefficients efficiently)

Before integrating anything:

Parity: even $f$ has $b_n = 0$ , odd $f$ has $a_n = 0$ , and the surviving integrals reduce to $\frac2\pi \int_0^\pi$ — half the work, twice the reliability.
Trigonometric polynomials are done already: linearize products ( $\cos^3$ , $\sin^2\cos$ , …) and read the coefficients off (Exercise 14.9); orthonormality makes any further integration redundant.
Complex exponentials for exponentials: for factors $\eu^{at}$ or damped oscillations, compute $c_n$ directly — one integral of $\eu^{(a - \iu n)t}$ beats two integrations by parts (Exercise 14.10).
Differentiate a known expansion: if $f'$ has known coefficients and $f$ is continuous, $c_n(f) = \frac{c_n(f')}{\iu n}$ ( $n \neq 0$ ) recovers all but $c_0$ , which is the mean — often the fastest route, and legitimate exactly under the hypotheses of Theorem 14.10 (1).

14.2 Dirichlet’s theorem

Lemma 14.5 (Dirichlet kernel)

$S_N(f)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x + u)\,D_N(u)\,\dd u$ , where

D_N(u) = \sum_{n=-N}^{N} \eu^{\iu nu} = \frac{\sin\bigl((N + \frac12)u\bigr)}{\sin\frac u2} \quad (u \notin 2\pi\Z), \qquad \frac{1}{2\pi}\int_{-\pi}^{\pi} D_N = 1 .

Proof. Insert the definition of $c_n$ into $S_N$ and swap sum and integral (legitimate: the sum is finite):

S_N(f)(x) = \sum_{n=-N}^{N}\Bigl(\frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)\,\eu^{-\iu nt}\dd t\Bigr)\eu^{\iu nx} = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t) \sum_{n=-N}^{N}\eu^{\iu n(x-t)}\,\dd t ;

substitute $u = t - x$ and slide the integration segment back to $\intcc{-\pi}{\pi}$ by $2\pi$ -periodicity of the integrand; the symmetric index range makes $\sum_n\eu^{-\iu nu} = D_N(u)$ . The closed form: geometric sum with ratio $\eu^{\iu u}$ ,

D_N(u) = \eu^{-\iu Nu}\,\frac{\eu^{\iu(2N+1)u} - 1}{\eu^{\iu u} - 1} = \frac{\eu^{\iu(N + \frac12)u} - \eu^{-\iu(N+\frac12)u}} {\eu^{\iu u/2} - \eu^{-\iu u/2}} ,

which is the sine quotient. Its mean is $1$ : only $n = 0$ contributes. ∎

Theorem 14.6 (Riemann–Lebesgue lemma)

For $g$ piecewise continuous on a segment, $\int_a^b g(t)\sin(\lambda t + \varphi)\,\dd t \to 0$ as $\lambda \to +\infty$ .

Proof. Approximate $g$ uniformly by step functions (Theorem 10.16 is not needed — the elementary step-function approximation of piecewise continuous functions suffices) and integrate each step explicitly: every piece contributes $O\bigl(\frac1\lambda\bigr)$ , and the approximation error contributes $\varepsilon(b - a)$ . This argument was carried out in full as the final exercise of the Year 1 volume’s integration chapter; for $C^1$ pieces one can instead integrate by parts and bound by $\frac C\lambda$ . ∎

Example 14.7 (How fast do coefficients die?)

Riemann–Lebesgue says the coefficients tend to $0$ ; their rate is a smoothness meter. Three specimens from this chapter and its exercises:

\text{square wave: } b_n = \frac{4}{\pi n}\ (n\ \text{odd}), \qquad \abs t : a_n = \frac{-4}{\pi n^2}\ (n\ \text{odd}), \qquad \abs{\sin t} : a_{2k} = \frac{-4}{\pi(4k^2-1)} .

A jump in $f$ (square wave, sawtooth) leaves coefficients of order $\frac1n$ : no normal convergence, Gibbs overshoot at the jumps. Continuity with a corner — a jump in $f'$ only — improves the order to $\frac{1}{n^2}$ : normal convergence, uniform reconstruction. In general $k$ derivatives buy $c_n = O(n^{-k})$ (Exercise 14.6), and conversely a spectrum decaying faster than every power forces $f$ to be $C^\infty$ (differentiate term by term, now legitimately). Closing insight: regularity of the signal and decay of the spectrum are the same information — an engineer reads one off the slope of the other without ever plotting the function.

Theorem 14.8 (Dirichlet)

Let $f$ be $2\pi$ -periodic and piecewise $C^1$ . Then for every $x$ ,

S_N(f)(x) \xrightarrow[N \to \infty]{} \frac{f(x^+) + f(x^-)}{2}

(the mean of one-sided limits) — in particular $S_N(f)(x) \to f(x)$ at every point of continuity.

Proof. By the kernel lemma and its unit mean, splitting the integral into $u > 0$ and $u < 0$ halves (each of mean $\frac12$ ):

\begin{align*} S_N(f)(x) - \frac{f(x^+) + f(x^-)}{2} &= \frac{1}{2\pi}\int_{0}^{\pi} \bigl(f(x+u) - f(x^+)\bigr)D_N(u)\,\dd u\\ &\quad+ \frac{1}{2\pi}\int_{-\pi}^{0}\bigl(f(x+u) - f(x^-)\bigr)D_N(u)\,\dd u . \end{align*}

Treat the first (the second is symmetric). Write

\bigl(f(x + u) - f(x^+)\bigr)\,D_N(u) = \underbrace{\frac{f(x+u) - f(x^+)}{\sin\frac u2}}_{g(u)}\, \sin\Bigl(\Bigl(N + \frac12\Bigr)u\Bigr) .

The function $g$ is piecewise continuous on $\intoc{0}{\pi}$ and has a finite limit at $0^+$ : writing

g(u) = \frac{f(x+u) - f(x^+)}{u}\cdot\frac{u}{\sin\frac u2} ,

the first factor tends to $f'(x^+)$ (one-sided differentiability, from piecewise $C^1$ ) and the second to $2$ (the standard limit $\frac{\sin v}{v} \to 1$ at $v = \frac u2$ ): $g(0^+) = 2f'(x^+)$ exists. So $g$ extends piecewise continuously to $\intcc{0}{\pi}$ , and Riemann–Lebesgue (Theorem 14.6) sends the integral to $0$ . This is the whole point of the hypothesis: without one-sided derivatives, the factor $\frac{1}{\sin(u/2)}$ blows up at $0$ faster than Riemann–Lebesgue can compensate, and pointwise convergence can genuinely fail for merely continuous $f$ — the gap that Fejér’s theorem (weekend problem) closes by averaging. ∎

Example 14.9 (Dirichlet at a jump)

For the sawtooth $f(t) = t$ on $\intoo{-\pi}{\pi}$ (Example 14.12 below), the periodic extension jumps at $t = \pi$ from $f(\pi^-) = \pi$ to $f(\pi^+) = -\pi$ . Dirichlet promises the value $\frac{\pi + (-\pi)}{2} = 0$ there, and indeed every term of $\sum \frac{2(-1)^{n+1}}{n}\sin nt$ vanishes at $t = \pi$ : the series politely converges to the midpoint, ignoring both one-sided values. Moving the evaluation point to $t = \frac\pi2$ instead (a continuity point) turns the same series into Leibniz’s $\frac\pi4$ . One series, two behaviors — exactly the two clauses of the theorem.

Theorem 14.10 (Normal convergence for $C^1$ ; Parseval)

If $f$ is continuous, $2\pi$ -periodic and piecewise $C^1$ , then $c_n(f') = \iu n\,c_n(f)$ , the Fourier series of $f$ converges normally on $\R$ , and its sum is $f$ .
(Parseval) For every piecewise continuous $2\pi$ -periodic $f$ :

$\frac{1}{2\pi}\int_{-\pi}^{\pi}\abs{f}^2 = \sum_{n=-\infty}^{\infty} \abs{c_n(f)}^2 = \frac{\abs{a_0}^2}{4} + \frac12\sum_{n\geq1} \bigl(\abs{a_n}^2 + \abs{b_n}^2\bigr).$

(Proved here for $f$ continuous piecewise $C^1$ ; admitted in general.)

Proof. (1) Integration by parts on each $C^1$ piece (boundary terms cancel by continuity and periodicity): $c_n(f') = \iu n c_n(f)$ . Then, by Cauchy–Schwarz on the two square-summable families (Proposition 14.2 for $f'$ ):

\sum_{n \neq 0} \abs{c_n(f)} = \sum_{n\neq0} \frac{\abs{c_n(f')}}{\abs n} \leq \Bigl(\sum \abs{c_n(f')}^2\Bigr)^{1/2} \Bigl(\sum_{n\neq0}\frac{1}{n^2}\Bigr)^{1/2} < \infty :

normal convergence of the Fourier series. Its sum is continuous and coincides with $f$ at every point by Dirichlet (Theorem 14.8: $f$ is continuous): the series converges to $f$ , uniformly.

(2) For such $f$ : $S_N f \to f$ uniformly, so $\norm{f - S_Nf}_2 \leq \norm{f - S_Nf}_\infty \to 0$ , and Pythagoras ( $\norm f_2^2 = \sum_{\abs n \leq N}\abs{c_n}^2 + \norm{f - S_Nf}_2^2$ ) passes to the limit. The real form is bookkeeping with $a_n, b_n$ . ∎

Example 14.11 (The $C^1$ tail bound, made quantitative)

The proof of Theorem 14.10 (1) hides a usable estimate. For continuous piecewise- $C^1$ $f$ , the same Cauchy–Schwarz applied to the tail only gives

\sum_{\abs n > N}\abs{c_n(f)} = \sum_{\abs n > N}\frac{\abs{c_n(f')}}{\abs n} \leq \Bigl(\sum_{\abs n > N}\abs{c_n(f')}^2\Bigr)^{\!1/2} \Bigl(\sum_{\abs n>N}\frac{1}{n^2}\Bigr)^{\!1/2} \leq \norm{f'}_2\,\sqrt{\frac{2}{N}} ,

using Bessel for $f'$ and $\sum_{n>N}n^{-2} \leq \frac1N$ . So the uniform error of the partial sums obeys

\norm{f - S_Nf}_\infty \leq \sum_{\abs n>N}\abs{c_n(f)} \leq \norm{f'}_2\,\sqrt{\frac2N} .

For $f(t) = \abs t$ : $\norm{f'}_2 = 1$ (the derivative is $\pm1$ ), so ten terms already reconstruct $\abs t$ uniformly within $\sqrt{0.2} \approx 0.45$ , and $N = 10^4$ within $0.015$ . Closing insight: one derivative buys the $\frac{1}{\sqrt N}$ uniform rate; comparing with the $\frac1n$ -coefficient world of the square wave (no uniform convergence at all), the dictionary of Example 14.7 acquires numbers.

Example 14.12 (Basel and friends)

Let $f(t) = t$ on $\intoo{-\pi}{\pi}$ , extended $2\pi$ -periodically (a sawtooth, piecewise $C^1$ ). Computing, $a_n = 0$ (oddness) and

b_n = \frac{1}{\pi}\int_{-\pi}^{\pi} t\sin nt\,\dd t = \frac{2(-1)^{n+1}}{n} .

Dirichlet at $t = \frac\pi2$ recovers Leibniz’s $\frac\pi4 = 1 - \frac13 + \frac15 - \dots$ ; Parseval gives

\frac{1}{2\pi}\int_{-\pi}^{\pi} t^2\,\dd t = \frac{\pi^2}{3} = \frac12\sum_{n\geq1}\frac{4}{n^2} \quad\Longrightarrow\quad \boxed{\;\sum_{n\geq1}\frac{1}{n^2} = \frac{\pi^2}{6}\;}

— Euler’s Basel sum, in two lines. The function $f(t) = t^2$ similarly yields $\sum \frac1{n^4} = \frac{\pi^4}{90}$ (Exercise 14.3).

Example 14.13 (A full expansion with built-in check: $\abs{\sin t}$ )

The function $f(t) = \abs{\sin t}$ is continuous, even, $\pi$ -periodic (hence $2\pi$ -periodic), piecewise $C^1$ . Evenness kills the $b_n$ ; $a_0 = \frac1\pi\int_{-\pi}^{\pi} \abs{\sin t}\dd t = \frac4\pi$ ; and for $n \geq 1$ , product-to-sum gives

a_n = \frac2\pi\int_0^\pi \sin t\cos nt\,\dd t = \frac{1}{\pi}\int_0^\pi\bigl(\sin(1+n)t + \sin(1-n)t\bigr)\dd t = \frac2\pi\cdot\frac{1 + \cos n\pi}{1 - n^2}

for $n \neq 1$ (and $a_1 = 0$ directly): zero for odd $n$ , and $a_{2k} = \frac{-4}{\pi(4k^2-1)}$ . By Theorem 14.10 (1) the convergence is normal, and

\abs{\sin t} = \frac{2}{\pi} - \frac{4}{\pi} \sum_{k\geq1}\frac{\cos(2kt)}{4k^2 - 1} \qquad (t \in \R) .

Built-in check at $t = 0$ : the identity demands $\sum_{k\geq1}\frac{1}{4k^2-1} = \frac12$ , which telescoping confirms:

\sum_{k\geq1}\frac{1}{4k^2-1} = \frac12\sum_{k\geq1}\Bigl(\frac{1}{2k-1} - \frac{1}{2k+1}\Bigr) = \frac12 . \checkmark

Closing insight: the spectrum of $\abs{\sin}$ lives on the even frequencies only — rectifying a sine doubles its frequency content, which is why full-wave rectifiers hum at $100$ or $120$ hertz, twice the mains frequency.

Example 14.14 (Parseval as a computing device)

Parseval turns expansions into numerical series wholesale. Apply it to $f(t) = \abs t$ (Exercise 14.2: $a_0 = \pi$ , $a_n = \frac{-4}{\pi n^2}$ for odd $n$ , the rest zero):

\frac{1}{2\pi}\int_{-\pi}^{\pi}t^2\,\dd t = \frac{\pi^2}{3} = \frac{a_0^2}{4} + \frac12\sum_{n \text{ odd}} a_n^2 = \frac{\pi^2}{4} + \frac{8}{\pi^2}\sum_{n\text{ odd}}\frac{1}{n^4} ,

whence

\sum_{n\text{ odd}}\frac{1}{n^4} = \frac{\pi^2}{8}\Bigl(\frac{\pi^2}{3} - \frac{\pi^2}{4}\Bigr) = \frac{\pi^4}{96} .

Cross-check against Exercise 14.3: splitting $\sum\frac1{n^4}$ into odd and even parts gives $\zeta$ -style bookkeeping $S = S_{\mathrm{odd}} + \frac{S}{16}$ , so $S = \frac{16}{15}\cdot\frac{\pi^4}{96} = \frac{\pi^4}{90}$ — exactly the value found there by a different function. Two expansions, one number: the consistency is Parseval’s isometry at work. Closing insight: each new Fourier expansion is a generating machine for series identities; the weekend problem’s Part II explains why the machine can never contradict itself.

The square wave (gray) and the Fourier partial sums S_1 (blue) and S_9 (red): convergence at every continuity point, but a persistent 9\% overshoot near the jumps — the Gibbs phenomenon. Uniform convergence fails exactly because the limit is discontinuous. — The square wave (gray) and the Fourier partial sums $S_1$ (blue) and $S_9$ (red): convergence at every continuity point, but a persistent $\sim 9\%$ overshoot near the jumps — the *Gibbs phenomenon*. Uniform convergence fails exactly because the limit is discontinuous.

Example 14.15 (Translation and modulation)

Two one-line rules generate many expansions from one. For $a \in \R$ , substituting $s = t - a$ :

c_n\bigl(f(\cdot - a)\bigr) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(t - a)\eu^{-\iu nt}\dd t = \eu^{-\iu na}\,c_n(f) \qquad\text{(translation modulates the spectrum)},

and directly from the definition,

c_n\bigl(\eu^{\iu kt}f\bigr) = c_{n-k}(f) \qquad\text{(modulation translates the spectrum)}.

Worked instance: the sawtooth shifted by $\pi$ , $g(t) = f(t - \pi)$ with $f(t) = t$ , has $b_n$ -coefficients $(-1)^n\cdot\frac{2(-1)^{n+1}}{n} = -\frac2n$ : the expansion $g \sim -2\sum\frac{\sin nt}{n}$ of the sawtooth that jumps at $0$ instead of at $\pi$ — no integral recomputed. Closing insight: time shifts only turn phases, never amplitudes ( $\abs{c_n}$ is shift-invariant), which is why energy (Parseval) and convergence class are properties of the signal shape, not of where the clock starts.

Remark 14.16 (Common pitfalls)

(i) Three convergences, three currencies: pointwise (Dirichlet: needs piecewise $C^1$ , pays the midpoint at each jump — never the one-sided value), uniform (needs a continuous limit; impossible across a jump, Gibbs is the visible symptom), and quadratic mean (Parseval: the most robust, blind to individual points). Always name which one you are claiming. (ii) No termwise differentiation by default: differentiating the sawtooth series of Example 14.12 term by term yields $\sum 2(-1)^{n+1}\cos nt$ , whose terms do not even tend to $0$ — the transfer theorems of the function-sequences chapter need uniform convergence of the derived series, which the jump destroys. Smoothness first, differentiation second (Exercise 14.6 is the dictionary). (iii) Symmetric partial sums: Dirichlet’s theorem concerns $S_N = \sum_{-N}^{N}$ ; rearranging or summing one side first can change divergence into convergence and back. (iv) Normalization drift: conventions differ across books ( $\frac{1}{2\pi}$ or $\frac1\pi$ in front, period $2\pi$ or $1$ ); the reliable invariants are the orthonormality relations — recompute $\langle e_m, e_n\rangle$ in the convention at hand before trusting any formula.

Remark 14.17 (Where this is used)

Parseval is the germ of the $L^2$ theory of Fourier series: the Year 3 volume completes the picture (the exponentials are a Hilbert basis of $L^2$ , and the map $f \mapsto (c_n)$ is a bijective isometry). Within this volume, the weekend problem proves Fejér’s theorem — the Cesàro means of the Fourier series converge uniformly for every continuous periodic $f$ — which upgrades the admitted general Parseval to a theorem, yields the trigonometric Weierstrass theorem, and pays two spectacular dividends: Weyl’s equidistribution theorem and the isoperimetric inequality. Applied mathematics reads this chapter daily: spectra of signals, harmonics of vibrating systems, and the fast Fourier transform (the finite avatar was Exercise 13.10).

Remark 14.18 (Perspectives within this volume)

Three chapters converse with this one. Backward: the Hermitian chapter supplied the geometry (orthonormal families, projections, Bessel), and the function-sequences chapter the analysis (uniform convergence, transfer theorems, approximate identities — Fejér’s kernel is to Fourier series what Bernstein’s polynomials were to Weierstrass). Sideways: the power-series chapter’s boundary theory returns through Abel summation, realized here by the Poisson kernel (Exercise 14.12) — the disk’s radius $r$ playing the role of the summation parameter. Forward: the differential-equations chapter decomposes periodic forcing into harmonics and feeds each to the oscillator’s frequency response; resonance happens when a Fourier mode of the input matches a natural frequency, which is why its weekend problem and this chapter’s are two halves of one story.

The Dirichlet kernel D_8 (blue) oscillates and takes negative values; the Fejér kernel F_8 (red) is nonnegative, concentrates at 0, and has mean 1: an approximate identity. Positivity is exactly what the Dirichlet kernel lacks, and what makes Fejér’s theorem in the weekend problem unconditional. — The Dirichlet kernel $D_8$ (blue) oscillates and takes negative values; the Fejér kernel $F_8$ (red) is nonnegative, concentrates at $0$ , and has mean $1$ : an *approximate identity*. Positivity is exactly what the Dirichlet kernel lacks, and what makes Fejér’s theorem in the weekend problem unconditional.

14.3 Exercises

Exercise 14.1 ★

Compute the Fourier coefficients of the square wave ( $f = -1$ on $\intoo{-\pi}{0}$ , $+1$ on $\intoo{0}{\pi}$ ), state Dirichlet’s conclusion at $t = \frac\pi2$ and at the jump $t = 0$ , and recover Leibniz’s series.

Solution

Solution of Exercise 14.1.

Oddness kills the $a_n$ . For $n \geq 1$ :

b_n = \frac{2}{\pi}\int_0^{\pi} \sin nt\,\dd t = \frac{2}{\pi}\cdot\frac{1 - (-1)^n}{n} = \begin{cases} \frac{4}{\pi n} & n \text{ odd},\\ 0 & n \text{ even}. \end{cases}

So $S(f)(t) = \frac{4}{\pi}\sum_{k\geq0} \frac{\sin\bigl((2k+1)t \bigr)}{2k+1}$ . Dirichlet at $t = \frac\pi2$ (a continuity point, value $1$ ): $\sin\bigl((2k+1)\frac\pi2\bigr) = (-1)^k$ , giving

1 = \frac4\pi \sum_{k\geq0}\frac{(-1)^k}{2k+1} \quad\Longrightarrow\quad \sum_{k\geq0}\frac{(-1)^k}{2k+1} = \frac{\pi}{4} \quad\text{(Leibniz)}.

At the jump $t = 0$ : the series sums to $0 = \frac{f(0^+) + f(0^-)}{2}$ , as Dirichlet prescribes (every term vanishes: consistent).

Exercise 14.2 ★

Expand $f(t) = \abs t$ ( $\abs t \leq \pi$ , $2\pi$ -periodic) in Fourier series; justify normal convergence; evaluate at $t = 0$ to obtain $\sum_{k\geq0}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8}$ , and re-derive Basel from it.

Solution

Solution of Exercise 14.2.

Evenness kills the $b_n$ ; $a_0 = \frac{1}{\pi}\int_{-\pi}^\pi\abs t\,\dd t = \pi$ , and for $n \geq 1$ :

a_n = \frac{2}{\pi}\int_0^\pi t\cos nt\,\dd t = \frac{2}{\pi}\cdot\frac{(-1)^n - 1}{n^2} = \begin{cases} -\frac{4}{\pi n^2} & n \text{ odd},\\ 0 & n \text{ even}, \end{cases}

(one integration by parts). Hence

\abs t = \frac{\pi}{2} - \frac{4}{\pi}\sum_{k\geq0} \frac{\cos\bigl((2k+1)t\bigr)}{(2k+1)^2} ,

with normal convergence ( $\sum (2k+1)^{-2} < \infty$ ) — as Theorem 14.10 (1) predicts for this continuous piecewise- $C^1$ function. At $t = 0$ :

0 = \frac\pi2 - \frac4\pi\sum_{k\geq0}\frac{1}{(2k+1)^2} \quad\Longrightarrow\quad \sum_{k\geq0}\frac{1}{(2k+1)^2} = \frac{\pi^2}{8} .

Splitting $\sum \frac{1}{n^2}$ into odd and even parts: $S = \frac{\pi^2}{8} + \frac S4$ , so $S = \frac{\pi^2}{6}$ : Basel again.

Exercise 14.3 ★

Expand $f(t) = t^2$ ( $\abs t \leq \pi$ ) and deduce

\sum_{n\geq1}\frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}, \qquad \sum_{n\geq1}\frac{1}{n^4} = \frac{\pi^4}{90} \quad\text{(Parseval)}.

Solution

Solution of Exercise 14.3.

Evenness: $b_n = 0$ ; $a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} t^2 = \frac{2\pi^2}{3}$ ; two integrations by parts give $a_n = \frac{4(-1)^n}{n^2}$ ( $n \geq 1$ ). Hence

t^2 = \frac{\pi^2}{3} + 4\sum_{n\geq1} \frac{(-1)^n}{n^2}\cos nt \qquad (\abs t \leq \pi),

normally convergent. At $t = 0$ : $0 = \frac{\pi^2}{3} + 4\sum\frac{(-1)^n}{n^2}$ , i.e. $\sum_{n\geq1} \frac{(-1)^{n+1}}{n^2} = \frac{\pi^2}{12}$ . Parseval:

\frac{1}{2\pi}\int_{-\pi}^{\pi} t^4\,\dd t = \frac{\pi^4}{5} = \frac{a_0^2}{4} + \frac12\sum_{n\geq1} a_n^2 = \frac{\pi^4}{9} + 8\sum_{n\geq1}\frac{1}{n^4} ,

so $\sum \frac{1}{n^4} = \frac18\bigl(\frac{\pi^4}{5} - \frac{\pi^4}{9}\bigr) = \frac{\pi^4}{90}$ .

Exercise 14.4 ★★

Let $f$ be continuous $2\pi$ -periodic with $c_n(f) = 0$ for all $n$ . Prove $f = 0$ (Parseval — for which class is it proved here? justify that continuity plus piecewise $C^1$ can be dropped by admitting general Parseval, or give the density argument in outline).

Solution

Solution of Exercise 14.4.

If moreover $f$ is piecewise $C^1$ : Parseval (proved) gives $\norm f_2^2 = \sum\abs{c_n}^2 = 0$ , and strict positivity of the integral of the continuous $\abs f^2$ forces $f = 0$ .

For merely continuous $f$ , admit general Parseval: same one-line proof. (Outline of the density route: Fejér/Weierstrass-type trigonometric approximation shows trigonometric polynomials are $\norm\cdot_2$ -dense among continuous periodic functions; since $f \perp$ all of them, $\norm f_2^2 = \langle f, f - P\rangle \leq \norm f_2\norm{f - P}_2$ for approximants $P$ , forcing $\norm f_2 = 0$ .)

Exercise 14.5 ★★

For $\alpha \notin \Z$ , expand $f(t) = \cos(\alpha t)$ ( $\abs t \leq \pi$ ) and deduce the partial-fraction expansion of the cotangent:

\pi\cot(\pi\alpha) = \frac{1}{\alpha} + \sum_{n\geq1} \frac{2\alpha}{\alpha^2 - n^2} .

Solution

Solution of Exercise 14.5.

Evenness: $b_n = 0$ ;

a_n = \frac{2}{\pi}\int_0^\pi \cos(\alpha t)\cos(nt)\,\dd t = \frac{2}{\pi}\cdot \frac{(-1)^n\,\alpha\sin(\pi\alpha)}{\alpha^2 - n^2}

(product-to-sum, then integrate; $a_0 = \frac{2\sin(\pi\alpha)}{\pi\alpha}$ ). Dirichlet at $t = \pi$ (a continuity point of the periodic extension, whose one-sided values agree by evenness):

\cos(\pi\alpha) = \frac{\sin(\pi\alpha)}{\pi\alpha} + \sum_{n\geq1} \frac{2\alpha\sin(\pi\alpha)}{\pi(\alpha^2 - n^2)}\,(-1)^n\cos(n\pi) = \frac{\sin(\pi\alpha)}{\pi}\Bigl(\frac{1}{\alpha} + \sum_{n\geq1}\frac{2\alpha}{\alpha^2 - n^2}\Bigr),

using $(-1)^n\cos n\pi = 1$ . Dividing by $\sin(\pi\alpha)/\pi$ gives the cotangent expansion.

Exercise 14.6 ★★

Prove that if $f$ is $2\pi$ -periodic and $C^k$ with $f^{(k)}$ piecewise continuous, then $c_n(f) = O\bigl(\abs n^{-k}\bigr)$ : smoothness of the signal $=$ decay of its spectrum.

Solution

Solution of Exercise 14.6.

Iterating $c_n(f') = \iu n\,c_n(f)$ ( $k$ times, integration by parts across the $C^{k}$ pieces with matching boundary values): $c_n(f^{(k)}) = (\iu n)^k c_n(f)$ . The coefficients of the piecewise continuous $f^{(k)}$ are bounded (indeed $\to 0$ , Bessel):

\abs{c_n(f)} = \frac{\abs{c_n(f^{(k)})}}{\abs n^k} = O\bigl(\abs n^{-k}\bigr) .

Exercise 14.7 ★★★

(Wirtinger’s inequality) Let $f$ be $C^1$ , $2\pi$ -periodic, with $\int_{-\pi}^{\pi} f = 0$ . Prove

\int_{-\pi}^{\pi} \abs{f}^2 \leq \int_{-\pi}^{\pi} \abs{f'}^2 ,

with equality iff $f(t) = a\cos t + b\sin t$ . (Parseval on both sides; compare $\abs{c_n}^2$ and $n^2\abs{c_n}^2$ .)

Solution

Solution of Exercise 14.7.

Parseval for $f$ and for $f'$ (both legitimate: $f$ is $C^1$ , $f'$ piecewise continuous — indeed continuous):

\frac{1}{2\pi}\int \abs f^2 = \sum_{n\neq0} \abs{c_n}^2 \quad (c_0 = 0 \text{ by the mean-zero hypothesis}), \qquad \frac{1}{2\pi}\int \abs{f'}^2 = \sum_{n\neq0} n^2\abs{c_n}^2 .

Termwise, $n^2\abs{c_n}^2 \geq \abs{c_n}^2$ for $\abs n \geq 1$ : the inequality follows. Equality forces $(n^2 - 1)\abs{c_n}^2 = 0$ for all $n$ , i.e. $c_n = 0$ for $\abs n \geq 2$ : $f(t) = c_1\eu^{\iu t} + c_{-1}\eu^{-\iu t} = a\cos t + b\sin t$ (real form); conversely such $f$ give equality.

Exercise 14.8 ★★★

(The Gibbs constant) For the square wave of Exercise 14.1, evaluate the partial sum at $x_N = \frac{\pi}{2N}$ : writing $u_k = \frac{(2k+1)\pi}{2N}$ and $\Delta u = \frac{\pi}{N}$ , show that

S_{2N-1}\Bigl(\frac{\pi}{2N}\Bigr) = \frac{4}{\pi}\sum_{k=0}^{N-1} \frac{\sin u_k}{2k+1} = \frac{2}{\pi}\sum_{k=0}^{N-1} \frac{\sin u_k}{u_k}\,\Delta u \xrightarrow[N\to\infty]{} \frac{2}{\pi}\int_0^{\pi}\frac{\sin u}{u}\,\dd u \approx 1.179 :

a Riemann sum of $\frac{2}{\pi}\cdot\frac{\sin u}{u}$ on $\intcc{0}{\pi}$ at midpoints. Conclude that the overshoot beyond the jump value $1$ does not vanish as $N \to \infty$ .

Solution

Solution of Exercise 14.8.

From Exercise 14.1, $S_{2N-1}(x) = \frac4\pi\sum_{k=0}^{N-1}\frac{\sin((2k+1)x)}{2k+1}$ . At $x_N = \frac{\pi}{2N}$ , with $u_k = (2k+1)x_N = \frac{(2k+1)\pi}{2N}$ :

S_{2N-1}(x_N) = \frac{4}{\pi}\sum_{k=0}^{N-1}\frac{\sin u_k}{2k+1} = \frac{4}{\pi}\sum_{k=0}^{N-1}\frac{\sin u_k}{u_k}\cdot \frac{u_k}{2k+1} = \frac{2}{\pi}\sum_{k=0}^{N-1}\frac{\sin u_k}{u_k}\cdot \frac{\pi}{N},

since $\frac{u_k}{2k+1} = \frac{\pi}{2N}$ . The points $u_k$ are the midpoints of the $N$ subintervals of $\intcc{0}{\pi}$ of length $\frac{\pi}{N}$ : the sum is a midpoint Riemann sum of the continuous $u \mapsto \frac{\sin u}{u}$ , hence converges to

\frac{2}{\pi}\int_0^\pi \frac{\sin u}{u}\,\dd u \approx \frac{2}{\pi}\times 1.8519 \approx 1.179 .

The partial sums near the jump overshoot the value $1$ by $\approx 18\%$ of the half-jump forever: Gibbs’s phenomenon, quantified.

Exercise 14.9 ★

Expand $\cos^3 t$ and $\sin^2 t\,\cos t$ in Fourier series (linearize; a trigonometric polynomial is its own Fourier series, by uniqueness of coefficients). What are $c_n$ , $a_n$ , $b_n$ for each?

Solution

Solution of Exercise 14.9.

From $\cos 3t = 4\cos^3t - 3\cos t$ :

\cos^3 t = \frac{3\cos t + \cos 3t}{4}, \qquad \sin^2t\,\cos t = \cos t - \cos^3 t = \frac{\cos t - \cos 3t}{4} .

Each is a trigonometric polynomial, hence equal to its own Fourier series (uniqueness of coefficients: two expansions would differ by a trigonometric polynomial with all coefficients zero). For $\cos^3t$ : $a_1 = \frac34$ , $a_3 = \frac14$ , all other $a_n$ and all $b_n$ zero; $c_{\pm1} = \frac38$ , $c_{\pm3} = \frac18$ . For $\sin^2t\cos t$ : $a_1 = \frac14$ , $a_3 = -\frac14$ ; $c_{\pm1} = \frac18$ , $c_{\pm3} = -\frac18$ .

Exercise 14.10 ★★

Let $a > 0$ and $f(t) = \eu^{at}$ on $\intoc{-\pi}{\pi}$ , extended $2\pi$ -periodically. Compute

c_n(f) = \frac{(-1)^n\sinh(a\pi)}{\pi(a - \iu n)} ,

apply Dirichlet’s theorem at the jump $t = \pi$ , and deduce the partial-fraction expansion of the hyperbolic cotangent:

\coth(\pi a) = \frac{1}{\pi a} + \sum_{n\geq1} \frac{2a}{\pi(a^2 + n^2)} .

Solution

Solution of Exercise 14.10.

Direct computation:

c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}\eu^{(a - \iu n)t}\dd t = \frac{\eu^{(a-\iu n)\pi} - \eu^{-(a - \iu n)\pi}} {2\pi(a - \iu n)} = \frac{(-1)^n\sinh(a\pi)}{\pi(a - \iu n)} ,

using $\eu^{\pm\iu n\pi} = (-1)^n$ . At $t = \pi$ the periodic extension jumps from $\eu^{a\pi}$ to $\eu^{-a\pi}$ ; Dirichlet (symmetric partial sums) gives

\cosh(a\pi) = \sum_{n\in\Z}(-1)^n c_n\, = \frac{\sinh(a\pi)}{\pi}\Bigl(\frac1a + \sum_{n\geq1}\Bigl(\frac{1}{a - \iu n} + \frac{1}{a + \iu n}\Bigr)\Bigr) = \frac{\sinh(a\pi)}{\pi}\Bigl(\frac1a + \sum_{n\geq1}\frac{2a}{a^2 + n^2}\Bigr) ,

the imaginary parts of the paired terms cancelling. Divide by $\sinh(a\pi)$ :

\coth(\pi a) = \frac{1}{\pi a} + \sum_{n\geq1}\frac{2a}{\pi(a^2 + n^2)} ,

the hyperbolic twin of Exercise 14.5.

Exercise 14.11 ★★

(Convolution) For $f, g$ continuous and $2\pi$ -periodic define

(f * g)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x - t)\,g(t)\, \dd t .

Show that $f * g = g * f$ , that $c_n(f*g) = c_n(f)\,c_n(g)$ (to swap the two integrals of a continuous integrand, compare the two functions of the upper limit: both vanish at the left endpoint and have the same derivative, by continuity and differentiation under the integral sign), and that $S_N(f) = f * D_N$ for the Dirichlet kernel. (The weekend problem’s Fejér means are likewise convolutions, $\sigma_N(f) = f * F_N$ .)

Solution

Solution of Exercise 14.11.

Commutativity: substitute $s = x - t$ and use periodicity of the integrand. For $c_n(f * g)$ , the integrand $(x, t) \mapsto f(x-t)g(t)\eu^{-\iu nx}$ is continuous; the two iterated integrals agree (both, as functions of the upper limit of the outer variable, vanish at the left endpoint and have the same derivative — continuity plus Theorem 9.10 justify differentiating the iterated integral). Hence

c_n(f*g) = \frac{1}{2\pi}\int_{-\pi}^{\pi} g(t)\,\eu^{-\iu nt} \Bigl(\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-t)\,\eu^{-\iu n(x-t)}\dd x\Bigr)\dd t = c_n(f)\,c_n(g),

the inner integral being $c_n(f)$ for every $t$ (substitution and periodicity). Finally Lemma 14.5 says $S_N(f)(x) = \frac{1}{2\pi}\int f(x+u)D_N(u)\dd u$ ; the substitution $u \mapsto -t$ and the evenness of $D_N$ turn this into $(f * D_N)(x)$ .

Exercise 14.12 ★★★

(Poisson kernel: Abel means of Fourier series) For $0 \leq r < 1$ set $P_r(t) = \sum_{n\in\Z} r^{\abs n}\eu^{\iu nt}$ .

Sum the two geometric series and show

$P_r(t) = \frac{1 - r^2}{1 - 2r\cos t + r^2} > 0, \qquad \frac{1}{2\pi}\int_{-\pi}^{\pi}P_r = 1 .$
Show that for $\delta \leq \abs t \leq \pi$ : $P_r(t) \leq \frac{1 - r^2}{1 - 2r\cos\delta + r^2} \to 0$ as $r \to 1^-$ , uniformly.
Deduce that for every continuous $2\pi$ -periodic $f$ , the Abel means $(f * P_r)(x) = \sum_n r^{\abs n}c_n(f)\,\eu^{\iu nx}$ converge to $f$ uniformly as $r \to 1^-$ — the continuous-parameter sibling of Fejér’s theorem, and the Fourier incarnation of Abel summation from the power-series chapter.

Solution

Solution of Exercise 14.12.

With $w = r\eu^{\iu t}$ :

$P_r(t) = 1 + 2\,\Re\frac{w}{1 - w} = \Re\frac{1 + w}{1 - w} = \frac{1 - \abs w^2}{\abs{1 - w}^2} = \frac{1 - r^2}{1 - 2r\cos t + r^2} > 0 .$

Mean $1$ : term-by-term integration of the normally convergent series keeps only $n = 0$ .
For $\delta \leq \abs t \leq \pi$ : $\cos t \leq \cos\delta$ , so $P_r(t) \leq \frac{1 - r^2}{1 - 2r\cos\delta + r^2}$ , whose denominator tends to $2 - 2\cos\delta > 0$ while the numerator tends to $0$ : uniform convergence to $0$ off any neighborhood of $0$ .
Term-by-term integration (normal convergence in $t$ ) gives $(f * P_r)(x) = \sum_n r^{\abs n}c_n(f)\eu^{\iu nx}$ . The approximate-identity argument: with mean $1$ and positivity,

$\abs{(f*P_r)(x) - f(x)} \leq \frac{1}{2\pi}\int_{-\pi}^{\pi} \abs{f(x-t) - f(x)}\,P_r(t)\,\dd t ,$

split at $\abs t = \delta$ : at most $\varepsilon$ (Heine) plus $2\norm f_\infty\sup_{\delta\leq\abs t\leq\pi}P_r \to \varepsilon$ : uniform convergence as $r \to 1^-$ . This is Abel summation of the Fourier series — the Fourier twin of the power-series chapter’s boundary theory.

14.4 Problem: Fejér’s theorem and its dividends

Problem 14.1

Dirichlet’s theorem needs $f$ piecewise $C^1$ ; for merely continuous $f$ the partial sums $S_N(f)$ can misbehave. Fejér’s discovery: their Cesàro means never do. The engine is the positivity of the Fejér kernel, and the harvest is immense: uniform trigonometric approximation (Weierstrass), uniqueness of Fourier coefficients, Parseval for every function of this chapter (removing the “admitted” in Theorem 14.10), Weyl’s equidistribution theorem, and — crowning a century of geometry — the isoperimetric inequality. Throughout, $f$ is $2\pi$ -periodic and piecewise continuous, and

\sigma_N(f) = \frac{S_0(f) + S_1(f) + \dots + S_{N-1}(f)}{N} .

Part I — The Fejér kernel.

Show that $\sigma_N(f)(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+u)\,F_N(u)\,\dd u$ with $F_N = \frac{D_0 + \dots + D_{N-1}}{N}$ , and that $\frac{1}{2\pi}\int_{-\pi}^{\pi}F_N = 1$ .
Prove the closed form, for $u \notin 2\pi\Z$ :

$F_N(u) = \frac{1}{N}\, \frac{\sin^2\bigl(\frac{Nu}{2}\bigr)} {\sin^2\bigl(\frac u2\bigr)} \;\geq\; 0$

(sum $\sin\bigl((n+\frac12)u\bigr)$ as the imaginary part of a geometric series).
Show the concentration estimate: for $0 < \delta \leq \abs u \leq \pi$ ,

$F_N(u) \leq \frac{1}{N\sin^2\frac\delta2} \xrightarrow[N\to\infty]{} 0 \quad\text{uniformly} :$

$(F_N)$ is a positive approximate identity.
(Fejér’s theorem) Prove: if $f$ is continuous and $2\pi$ -periodic, then $\sigma_N(f) \to f$ uniformly on $\R$ (split the integral of $\bigl(f(x+u) - f(x)\bigr)F_N(u)$ at $\abs u = \delta$ ; use Heine and questions 1–3).
For piecewise continuous $f$ , show the pointwise version $\sigma_N(f)(x) \to \frac{f(x^+) + f(x^-)}{2}$ at every $x$ , and the uniform bound $\norm{\sigma_N(f)}_\infty \leq \norm f_\infty$ (positivity!).

Part II — Weierstrass, uniqueness, Parseval.

(Trigonometric Weierstrass) Deduce: every continuous $2\pi$ -periodic function is a uniform limit of trigonometric polynomials.
(Uniqueness) Deduce: a continuous $f$ with $c_n(f) = 0$ for all $n$ is identically zero — two continuous periodic functions with the same Fourier coefficients coincide (Exercise 14.4, now with no admission).
(Parseval, continuous case) Using the projection property of $S_N$ (Proposition 14.2) and $\sigma_Nf \in \mathcal T_{N-1} \subseteq \mathcal T_N$ , prove

$\norm{f - S_Nf}_2 \leq \norm{f - \sigma_Nf}_2 \leq \norm{f - \sigma_Nf}_\infty \xrightarrow[N\to\infty]{} 0 ,$

and conclude Parseval’s identity for every continuous $2\pi$ -periodic $f$ .
(Parseval, piecewise continuous case) Given $f$ piecewise continuous and $\varepsilon > 0$ , construct a continuous periodic $g$ with $\norm{f - g}_2 \leq \varepsilon$ (replace $f$ by an affine interpolation on tiny intervals around the jumps), and deduce $\norm{f - S_Nf}_2 \to 0$ (use Bessel: $\norm{S_Nh}_2 \leq \norm h_2$ ): Parseval holds in the full generality stated in Theorem 14.10 — the “admitted” is gone.
(No Gibbs for Fejér) Contrast with Exercise 14.8: show that for the square wave $f$ , $\abs{\sigma_N(f)} \leq 1$ everywhere, for every $N$ — Cesàro averaging erases the overshoot that haunts $S_N$ . Explain in one sentence which property of $F_N$ is responsible.

Part III — Rates.

Prove the two kernel bounds, for $0 < \abs u \leq \pi$ :

$F_N(u) \leq N, \qquad F_N(u) \leq \frac{\pi^2}{N u^2}$

(for the first, $\abs{\sin N\theta} \leq N\abs{\sin\theta}$ by induction; for the second, $\sin\frac u2 \geq \frac{u}{\pi}$ on $\intcc{0}{\pi}$ ).
Deduce the first-moment estimate

$\frac{1}{2\pi}\int_{-\pi}^{\pi}\abs u\,F_N(u)\,\dd u \;\leq\; \frac{C\,\ln N}{N} \qquad (N \geq 2)$

for an explicit constant (split at $\abs u = \frac1N$ ).
Conclude: if $f$ is $L$ -Lipschitz and $2\pi$ -periodic, then

$\norm{\sigma_N f - f}_\infty \leq \frac{C\,L\ln N}{N} .$
(Saturation) Compute $\sigma_N(e_1)$ for $e_1(t) = \eu^{\iu t}$ and show $\norm{\sigma_Ne_1 - e_1}_\infty = \frac1N$ : even for the smoothest functions, Fejér converges no faster than $\frac1N$ — the exact analogue of the Bernstein saturation in the function-sequences chapter’s weekend problem.
(Localization) Show: if $f$ (piecewise continuous) vanishes on $\intoo{x - \delta}{x + \delta}$ , then $\sigma_N(f)(x) \to 0$ , however wild $f$ is elsewhere — convergence of the means at $x$ only sees $f$ near $x$ .

Part IV — Weyl’s equidistribution theorem. A sequence $(x_n)_{n\geq1}$ in $\intco{0}{1}$ is equidistributed when, for every interval $\intcc{a}{b} \subseteq \intcc{0}{1}$ ,

\frac{\#\{n \leq N : x_n \in \intcc ab\}}{N} \xrightarrow[N\to\infty]{} b - a .

Show that $(x_n)$ is equidistributed as soon as $\frac1N\sum_{n\leq N}f(x_n) \to \int_0^1 f$ for every continuous $1$ -periodic $f$ (squeeze the indicator of $\intcc ab$ between two continuous piecewise-affine functions whose integrals differ by $\varepsilon$ ).
(Weyl’s criterion, sufficiency) Suppose

$\frac{1}{N}\sum_{n=1}^{N}\eu^{2\iu\pi kx_n} \xrightarrow[N\to\infty]{} 0 \qquad\text{for every } k \in \Z\setminus\{0\} .$

Show $\frac1N\sum f(x_n) \to \int_0^1f$ first for trigonometric polynomials, then for all continuous $1$ -periodic $f$ by question 6 (transported to period $1$ ): with question 16, $(x_n)$ is equidistributed.
Let $\alpha$ be irrational and $x_n = \{n\alpha\}$ (fractional part). Bound the geometric sum

$\Bigl|\sum_{n=1}^{N}\eu^{2\iu\pi kn\alpha}\Bigr| \leq \frac{2}{\abs{1 - \eu^{2\iu\pi k\alpha}}} \qquad (k \neq 0),$

and conclude Weyl’s theorem: $(\{n\alpha\})$ is equidistributed in $\intco{0}{1}$ .
Deduce that $(\{n\alpha\})$ is dense in $\intcc{0}{1}$ for irrational $\alpha$ , and explain in one sentence why equidistribution is strictly stronger than density.
(Leading digits) Prove that the proportion of integers $n \leq N$ such that $2^n$ has leading (decimal) digit $1$ tends to $\log_{10}2 \approx 0.301$ (leading digit $1$ means $\{n\log_{10}2\} \in \intco{0}{\log_{10}2}$ ; show $\log_{10}2$ is irrational).

Part V — The isoperimetric inequality. Let $\Gamma$ be a closed simple $C^1$ curve of length $L$ enclosing a signed area $A$ , parametrized by scaled arc length: $z(t) = x(t) + \iu y(t)$ , $2\pi$ -periodic, with $\abs{z'(t)} = \frac{L}{2\pi}$ constant; the enclosed area is

A = \frac12\int_0^{2\pi}\bigl(x\,y' - y\,x'\bigr)\dd t = \frac{1}{2}\,\Im\int_0^{2\pi}\conj{z}\,z'\,\dd t

(taken here as the definition of the signed area; the chapter on multiple integrals proves it agrees with the intuitive one, via Green’s formula).

Expand $z(t) = \sum_{n\in\Z}c_n\eu^{\iu nt}$ (the series of a $C^1$ function, normally convergent) and prove, by Parseval applied to $z'$ :

$\frac{L^2}{2\pi} = \int_0^{2\pi}\abs{z'}^2\dd t = 2\pi\sum_{n\in\Z}n^2\abs{c_n}^2 .$
Prove likewise $A = \pi\sum_{n\in\Z} n\,\abs{c_n}^2$ (Parseval in its polarized form: $\frac{1}{2\pi}\int\conj f g = \sum \conj{c_n(f)}c_n(g)$ , applied to $f = z$ , $g = z'$ ).
(Hurwitz) Conclude:

$L^2 - 4\pi A = 4\pi^2\sum_{n\in\Z} (n^2 - n)\abs{c_n}^2 \;\geq\; 0 ,$

with equality if and only if $z(t) = c_0 + c_1\eu^{\iu t}$ — a circle. The isoperimetric inequality: among closed curves of length $L$ , only the circle encloses area $\frac{L^2}{4\pi}$ .
Sanity checks: verify equality for the circle of radius $R$ and the strict inequality for the square of side $a$ ; explain why $n^2 - n \geq 0$ for every integer $n$ , including the negative ones, and where the parametrization’s constant speed was used.
Synthesis. In one sentence each: (i) the single property of $F_N$ from which Parts I–III flow, and which $D_N$ lacks; (ii) how Cesàro summation here relates to the power-series chapter’s weekend problem (Frobenius); (iii) which dividend used only Weierstrass (question 6) and which needed full Parseval; (iv) one sentence on what the Year 3 volume adds ( $L^2$ completeness: Fourier series as a Hilbert basis).

Solution

Solution of Problem 14.1.

1. Averaging Lemma 14.5 over $n = 0, \dots, N-1$ (linearity of the integral) gives $\sigma_N(f)(x) = \frac{1}{2\pi}\int f(x+u)F_N(u)\dd u$ ; each $D_n$ has mean $1$ , so $F_N$ has mean $1$ .

2. With $\eu^{\iu u} - 1 = 2\iu\,\eu^{\iu u/2}\sin\frac u2$ :

\sum_{n=0}^{N-1}\sin\Bigl(\Bigl(n + \frac12\Bigr)u\Bigr) = \Im\Bigl[\eu^{\iu u/2}\,\frac{\eu^{\iu Nu} - 1}{\eu^{\iu u} - 1}\Bigr] = \Re\,\frac{1 - \eu^{\iu Nu}}{2\sin\frac u2} = \frac{1 - \cos Nu}{2\sin\frac u2} = \frac{\sin^2\frac{Nu}2}{\sin\frac u2} .

Dividing by $N\sin\frac u2$ :

F_N(u) = \frac1N\sum_{n=0}^{N-1} \frac{\sin\bigl((n+\frac12)u\bigr)}{\sin\frac u2} = \frac{1}{N}\, \frac{\sin^2\frac{Nu}{2}}{\sin^2\frac u2} \geq 0 .

3. On $\delta \leq \abs u \leq \pi$ : $\sin^2\frac u2 \geq \sin^2\frac\delta2$ and $\sin^2\frac{Nu}2 \leq 1$ : $F_N \leq \frac{1}{N\sin^2(\delta/2)} \to 0$ , uniformly there.

4. By the unit mean, $\sigma_Nf(x) - f(x) = \frac{1}{2\pi}\int\bigl(f(x+u) - f(x)\bigr)F_N(u)\dd u$ . Given $\varepsilon$ , Heine yields $\delta$ with $\abs{f(x+u) - f(x)} \leq \varepsilon$ for $\abs u \leq \delta$ , uniformly in $x$ . Then, using $F_N \geq 0$ and its unit mean,

\abs{\sigma_Nf(x) - f(x)} \leq \varepsilon + 2\norm f_\infty\cdot\frac{1}{2\pi} \int_{\delta\leq\abs u\leq\pi}F_N \leq \varepsilon + \frac{2\norm f_\infty}{N\sin^2\frac\delta2} \leq 2\varepsilon

for large $N$ , uniformly in $x$ : Fejér’s theorem.

5. $F_N$ is even with mean $1$ : each half $\intcc{0}{\pi}$ , $\intcc{-\pi}{0}$ carries mean $\frac12$ . Then

\sigma_Nf(x) - \frac{f(x^+)+f(x^-)}{2} = \frac{1}{2\pi}\int_0^\pi\bigl(f(x+u) - f(x^+)\bigr)F_N\,\dd u + \frac{1}{2\pi}\int_{-\pi}^0\bigl(f(x+u) - f(x^-)\bigr)F_N\,\dd u ;

on each half, split at $\abs u = \delta$ where the one-sided limit is $\varepsilon$ -close, and let question 3 kill the far part: both integrals tend to $0$ . The bound: $F_N \geq 0$ gives $\abs{\sigma_Nf(x)} \leq \frac{1}{2\pi}\int\abs{f(x+u)}F_N \leq \norm f_\infty$ .

6. Each $\sigma_N(f)$ is a trigonometric polynomial (an average of the $S_n(f)$ , $n < N$ ), and $\sigma_N(f) \to f$ uniformly: the trigonometric Weierstrass theorem.

7. $c_n(f) = 0$ for all $n$ makes every $S_n(f) = 0$ , hence every $\sigma_N(f) = 0$ ; by Fejér, $f = \lim\sigma_Nf = 0$ . Applying this to a difference: continuous periodic functions are determined by their Fourier coefficients.

8. $S_Nf$ is the orthogonal projection of $f$ onto $\mathcal T_N$ (Proposition 14.2), so it minimizes $\norm{f - P}_2$ over $P \in \mathcal T_N$ ; since $\sigma_Nf \in \mathcal T_{N-1} \subseteq \mathcal T_N$ :

\norm{f - S_Nf}_2 \leq \norm{f - \sigma_Nf}_2 \leq \norm{f - \sigma_Nf}_\infty \to 0

(the middle inequality because the mean of $\abs\cdot^2$ is at most the sup squared). Pythagoras $\norm f_2^2 = \sum_{\abs n\leq N}\abs{c_n}^2 + \norm{f - S_Nf}_2^2$ then passes to the limit: Parseval for every continuous periodic $f$ .

9. Let $t_1, \dots, t_p$ be the jumps of $f$ in a period, $M = \norm f_\infty$ . For small $\eta$ , define $g = f$ outside the intervals $\intoo{t_j - \eta}{t_j + \eta}$ and by the affine chord across each such interval: $g$ is continuous, periodic, $\norm g_\infty \leq M$ , and

\norm{f - g}_2^2 \leq \frac{1}{2\pi}\,p\cdot(2M)^2\cdot2\eta \leq \varepsilon^2

for $\eta$ small. Bessel makes $S_N$ a contraction for $\norm\cdot_2$ , so

\norm{f - S_Nf}_2 \leq \norm{f - g}_2 + \norm{g - S_Ng}_2 + \norm{S_N(g - f)}_2 \leq 2\varepsilon + \norm{g - S_Ng}_2 ,

and question 8 gives $\limsup_N\norm{f - S_Nf}_2 \leq 2\varepsilon$ for every $\varepsilon$ : $\norm{f - S_Nf}_2 \to 0$ , and Pythagoras yields Parseval for every piecewise continuous $f$ : the “admitted” in Theorem 14.10 is now a theorem.

10. The square wave has $\norm f_\infty = 1$ , so question 5 gives $\abs{\sigma_N(f)} \leq 1$ everywhere and for every $N$ — no overshoot, ever — while Exercise 14.8 shows $\sup_xS_{2N-1}(f)(x) \to \approx 1.179$ . The one responsible property: $F_N \geq 0$ , so $\sigma_Nf(x)$ is a weighted average of values of $f$ and can never leave $\intcc{\min f}{\max f}$ ; $D_N$ takes negative values, so $S_N$ can.

11. $\abs{\sin N\theta} \leq N\abs{\sin\theta}$ by induction ( $\abs{\sin(N{+}1)\theta} \leq \abs{\sin N\theta}\abs{\cos\theta} + \abs{\cos N\theta}\abs{\sin\theta} \leq (N+1)\abs{\sin\theta}$ ): with $\theta = \frac u2$ ,

F_N(u) = \frac{\sin^2\frac{Nu}2}{N\sin^2\frac u2} \leq \frac{N^2\sin^2\frac u2}{N\sin^2\frac u2} = N .

Concavity of $\sin$ on $\intcc{0}{\frac\pi2}$ gives $\sin\frac u2 \geq \frac{u}{\pi}$ for $0 \leq u \leq \pi$ , so $F_N(u) \leq \frac{1}{N(u/\pi)^2} = \frac{\pi^2}{Nu^2}$ .

12. By evenness and the split at $\frac1N$ :

\frac{1}{2\pi}\int_{-\pi}^{\pi}\abs uF_N = \frac1\pi\int_0^\pi uF_N \leq \frac1\pi\Bigl(\int_0^{1/N}uN\,\dd u + \int_{1/N}^{\pi}\frac{\pi^2}{Nu}\,\dd u\Bigr) = \frac{1}{2\pi N} + \frac{\pi\ln(\pi N)}{N} .

For $N \geq 2$ : $\ln(\pi N) \leq \bigl(1 + \frac{\ln\pi}{\ln2}\bigr)\ln N \leq 2.66\ln N$ and $\frac{1}{2\pi N} \leq \frac{\ln N}{N}$ , so the moment is $\leq \frac{9\ln N}{N}$ .

13. For $L$ -Lipschitz $f$ :

\abs{\sigma_Nf(x) - f(x)} \leq \frac{1}{2\pi}\int\abs{f(x+u) - f(x)}F_N(u)\dd u \leq L\cdot\frac{1}{2\pi}\int\abs uF_N \leq \frac{9L\ln N}{N},

uniformly in $x$ .

14. $c_0(e_1) = 0$ gives $S_0(e_1) = 0$ , while $S_n(e_1) = e_1$ for $n \geq 1$ : $\sigma_N(e_1) = \frac{N-1}{N}e_1$ , so $\norm{\sigma_Ne_1 - e_1}_\infty = \frac1N$ . Even for this entire, band-limited signal the rate is $\frac1N$ : Fejér saturates, exactly as Bernstein’s operator saturates at $\frac1n$ (Voronovskaya, in the function-sequences chapter’s weekend problem).

15. If $f$ vanishes on $\intoo{x-\delta}{x+\delta}$ , then $\sigma_Nf(x) = \frac{1}{2\pi}\int_{\delta \leq \abs u \leq \pi}f(x+u)F_N(u)\dd u$ , of absolute value at most $\frac{\norm f_\infty}{N\sin^2(\delta/2)} \to 0$ : the Cesàro means at $x$ only see $f$ near $x$ .

16. Given $\intcc ab$ and $\varepsilon$ , choose continuous $1$ -periodic piecewise-affine $\varphi^\pm$ with $\varphi^- \leq \mathbf 1_{\intcc ab} \leq \varphi^+$ and $\int_0^1(\varphi^+ - \varphi^-) \leq \varepsilon$ (trapezoids with slopes over intervals of total length $\varepsilon$ ). Then

\limsup_N\frac{\#\{n \leq N : x_n \in \intcc ab\}}{N} \leq \lim_N\frac1N\sum_{n\leq N}\varphi^+(x_n) = \int_0^1\varphi^+ \leq b - a + \varepsilon ,

and symmetrically $\liminf \geq b - a - \varepsilon$ : the proportion tends to $b - a$ .

17. For $f = \eu^{2\iu\pi k\cdot}$ with $k \neq 0$ the hypothesis gives the limit $0 = \int_0^1f$ ; for $k = 0$ both sides are $1$ ; linearity handles every trigonometric polynomial. For continuous $1$ -periodic $f$ and $\varepsilon > 0$ , question 6 (transported by $t = 2\pi x$ ) provides a trigonometric polynomial $P$ with $\norm{f - P}_\infty \leq \varepsilon$ :

\Bigl|\frac1N\sum_{n\leq N}f(x_n) - \int_0^1f\Bigr| \leq 2\varepsilon + \Bigl|\frac1N\sum_{n\leq N}P(x_n) - \int_0^1P\Bigr| \longrightarrow 2\varepsilon .

With question 16: $(x_n)$ is equidistributed.

18. For $k \neq 0$ and irrational $\alpha$ , $w = \eu^{2\iu\pi k\alpha} \neq 1$ :

\Bigl|\sum_{n=1}^{N}w^n\Bigr| = \Bigl|\frac{w(w^N - 1)}{w - 1}\Bigr| \leq \frac{2}{\abs{1 - w}} ,

a bound independent of $N$ ; dividing by $N$ gives Weyl’s criterion, and question 17 concludes: $(\{n\alpha\})$ is equidistributed.

19. Every subinterval receives asymptotic proportion equal to its length, in particular infinitely many points: $(\{n\alpha\})$ is dense. Equidistribution is stronger: a sequence can be dense while spending nearly all its time in one corner (density says where the sequence goes, equidistribution says how often).

20. $2^n$ has leading digit $1$ iff $10^m \leq 2^n < 2\cdot10^m$ for some $m$ , i.e. iff $\{n\log_{10}2\} \in \intco{0}{\log_{10}2}$ . Irrationality: $\log_{10}2 = \frac pq$ would give $2^q = 10^p = 2^p5^p$ , impossible for $p \geq 1$ by unique factorization. Weyl’s theorem (question 18 with $\alpha = \log_{10}2$ ; the half-open interval is squeezed between closed ones of nearby lengths) gives the proportion $\log_{10}2 \approx 0.301$ : the first digits of $2^n$ follow Benford’s law.

21. $z$ is $C^1$ , so its Fourier series converges normally with sum $z$ (Theorem 14.10 (1)), and $c_n(z') = \iu n\,c_n$ . Since $\abs{z'} = \frac{L}{2\pi}$ is constant,

\int_0^{2\pi}\abs{z'}^2\dd t = 2\pi\Bigl(\frac{L}{2\pi}\Bigr)^{\!2} = \frac{L^2}{2\pi} ,

and Parseval applied to the continuous $z'$ gives $\frac{1}{2\pi}\int_0^{2\pi}\abs{z'}^2 = \sum_n\abs{\iu nc_n}^2$ , i.e. $\frac{L^2}{2\pi} = 2\pi\sum_n n^2\abs{c_n}^2$ .

22. The polarized Parseval $\frac{1}{2\pi}\int\conj fg = \sum\conj{c_n(f)}c_n(g)$ follows from Parseval applied to $f + g$ and $f + \iu g$ (polarization identity), both continuous. With $f = z$ , $g = z'$ :

A = \frac12\,\Im\int_0^{2\pi}\conj z\,z' = \pi\,\Im\sum_n\conj{c_n}(\iu n c_n) = \pi\sum_n n\abs{c_n}^2 .

23. Combining questions 21–22:

L^2 - 4\pi A = 4\pi^2\sum_n n^2\abs{c_n}^2 - 4\pi^2\sum_n n\abs{c_n}^2 = 4\pi^2\sum_{n\in\Z}(n^2 - n)\abs{c_n}^2 \geq 0 ,

since $n^2 - n = n(n-1) \geq 0$ for every integer. Equality forces $c_n = 0$ for all $n \notin \{0, 1\}$ : $z(t) = c_0 + c_1\eu^{\iu t}$ , a circle of center $c_0$ and radius $\abs{c_1} = \frac{L}{2\pi}$ (constant speed). Hurwitz’s proof of the isoperimetric inequality: $A \leq \frac{L^2}{4\pi}$ , circle only.

24. Circle of radius $R$ : $L = 2\pi R$ , $A = \pi R^2$ : $L^2 = 4\pi^2R^2 = 4\pi A$ : equality. Square of side $a$ : $L^2 = 16a^2 > 4\pi a^2 = 4\pi A$ (as $16 > 4\pi \approx 12.57$ ). For negative $n$ , $n^2 - n = n(n - 1)$ is a product of two negative integers: positive — so backward-winding modes cost area twice. Constant speed entered in question 21, converting $\int\abs{z'}^2$ into $\frac{L^2}{2\pi}$ ; for a non-constant speed, Cauchy–Schwarz gives $\int\abs{z'}^2 \geq \frac{(\int\abs{z'})^2}{2\pi} = \frac{L^2}{2\pi}$ , so the inequality survives, with the circle still the only equality case.

25. (i) Everything flows from $F_N \geq 0$ (with unit mean and concentration); $D_N$ has unit mean and concentration of oscillation but not positivity, and Gibbs is the price. (ii) Cesàro summability of the Fourier series implies its Abel summability with the same sum (Frobenius, proved in the power-series chapter’s weekend problem) — the Poisson-kernel route of Exercise 14.12 is exactly Abel’s method. (iii) Weyl’s theorem needed only uniform approximation (question 6); the isoperimetric inequality needed Parseval itself (questions 8, 21–22). (iv) The Year 3 volume proves completeness: the exponentials form a Hilbert basis of $L^2$ , Parseval becomes an isometry of Hilbert spaces, and Fejér’s theorem becomes the statement that this isometry is computable by positive averages.