---
title: "Fourier Series"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 14
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/14-fourier-series
---

# Chapter 14 — Fourier 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 reach at this level: *Dirichlet’s theorem* ([pointwise](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) reconstruction for piecewise $C^1$ functions, via the [Dirichlet kernel](#lem-b2-fourier-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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), complex-valued; $\mathcal{C}$ denotes the [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](https://one-course.com/books/math/4/en/chapter/13-hermitian-forms#def-b2-hermitian-def) $\langle f,
g\rangle = \frac{1}{2\pi}\int_{-\pi}^{\pi}\conj f\,g$: 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](https://one-course.com/books/math/4/en/chapter/13-hermitian-forms#def-b2-hermitian-adjoint) geometry ([Chapter 13](https://one-course.com/books/math/4/en/chapter/13-hermitian-forms#ch-b2-hermitian)) 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](https://one-course.com/books/math/4/en/chapter/13-hermitian-forms#def-b2-hermitian-adjoint) 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](#prop-b2-fourier-bessel) 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](#ex-b2-fourier-basel)): $\abs{c_n}^2 + \abs{c_{-n}}^2 =
\frac{b_n^2}{2} = \frac{2}{n^2}$. 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](#exo-b2-fourier-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:

1. *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.
2. *Trigonometric polynomials are done already:* linearize products ( $\cos^3$ , $\sin^2\cos$ , …) and read the coefficients off ( [Exercise 14.9](#exo-b2-fourier-9) ); orthonormality makes any further integration redundant.
3. *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](#exo-b2-fourier-10) ).
4. *Differentiate a known expansion:* if $f'$ has known coefficients and $f$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) , $c_n(f) =  \frac{c_n(f')}{\iu n}$ ( $n \neq 0$ ) recovers all but $c_0$ , which is the mean — often the fastest route, and legitimate exactly under the hypotheses of [Theorem 14.10](#thm-b2-fourier-parseval) (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](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on a segment, $\int_a^b g(t)\sin(\lambda t + \varphi)\,\dd t \to 0$ as $\lambda
\to +\infty$.

**Proof.** Approximate $g$ [uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) by step functions ([Theorem 10.16](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#thm-b2-funcseq-weierstrass) is not needed — the elementary step-function approximation of piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series), Gibbs overshoot at the jumps. [Continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) with a corner — a jump in $f'$ only — improves the order to $\frac{1}{n^2}$: [normal convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series), uniform reconstruction. In general $k$ derivatives buy $c_n =
O(n^{-k})$ ([Exercise 14.6](#exo-b2-fourier-6)), and conversely a [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) 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](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-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](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint)). 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) to $\intcc{0}{\pi}$, and Riemann–Lebesgue ([Theorem 14.6](#thm-b2-fourier-riemannlebesgue)) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) can genuinely fail for merely [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f$ — 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](#ex-b2-fourier-basel) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-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 C1C^1C1; Parseval).**

1. If $f$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) , $2\pi$ -periodic and piecewise $C^1$ , then $c_n(f') = \iu n\,c_n(f)$ , the Fourier series of $f$ converges *[normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series)* on $\R$ , and its sum is $f$ .
2. (Parseval) For every piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $2\pi$-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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) piecewise $C^1$; admitted in general.)*

**Proof.** (1) Integration by parts on each $C^1$ piece (boundary terms cancel by [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and periodicity): $c_n(f') = \iu n c_n(f)$. Then, by Cauchy–Schwarz on the two [square-summable](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable) families ([Proposition 14.2](#prop-b2-fourier-bessel) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) of the Fourier series. Its sum is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and coincides with $f$ at every point by Dirichlet ([Theorem 14.8](#thm-b2-fourier-dirichlet): $f$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)): the series converges to $f$, [uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def).

(2) For such $f$: $S_N f \to f$ [uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def), 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 C1C^1C1 tail bound, made quantitative).**

The proof of [Theorem 14.10](#thm-b2-fourier-parseval) (1) hides a usable estimate. For [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) at all), the dictionary of [Example 14.7](#ex-b2-fourier-decaytable) 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](#exo-b2-fourier-3)).

**Example 14.13 (A full expansion with built-in check: ∣sin⁡t∣\abs{\sin t}∣sint∣).**

The function $f(t) = \abs{\sin t}$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), 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](#thm-b2-fourier-parseval) (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](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) 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](#exo-b2-fourier-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](#exo-b2-fourier-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.](https://one-course.com/images/onecourse/chapters/math-4/b2-fourier/fig-6ce52b3434ba.svg)

*The square wave (gray) and the [Fourier partial sums](#def-b2-fourier-coefficients) $S_1$ (blue) and $S_9$ (red): convergence at every [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) point, but a persistent $\sim 9\%$ overshoot near the jumps — the *Gibbs phenomenon*. [Uniform convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) (Dirichlet: needs piecewise $C^1$, pays the *midpoint* at each jump — never the one-sided value), uniform (needs a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](#ex-b2-fourier-basel) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) of the *derived* series, which the jump destroys. Smoothness first, differentiation second ([Exercise 14.6](#exo-b2-fourier-6) is the dictionary). *(iii) [Symmetric](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) for every [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](https://one-course.com/books/math/4/en/chapter/13-hermitian-forms#exo-b2-hermitian-10)).

**Remark 14.18 (Perspectives within this volume).**

Three chapters converse with this one. Backward: the [Hermitian](https://one-course.com/books/math/4/en/chapter/13-hermitian-forms#def-b2-hermitian-adjoint) chapter supplied the geometry (orthonormal families, projections, Bessel), and the function-sequences chapter the analysis ([uniform convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def), 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](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-abel), realized here by the Poisson kernel ([Exercise 14.12](#exo-b2-fourier-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.](https://one-course.com/images/onecourse/chapters/math-4/b2-fourier/fig-01aa640fe8e7.svg)

*The [Dirichlet kernel](#lem-b2-fourier-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](#lem-b2-fourier-kernel) lacks, and what makes Fejér’s theorem in the weekend problem unconditional.*

## 14.3 Exercises

**Exercise 14.1 ★.**

Compute the [Fourier coefficients](#def-b2-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 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-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 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](#thm-b2-fourier-parseval) (1) predicts for this [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $2\pi$-periodic with $c_n(f) = 0$ for all $n$. Prove $f = 0$ *(Parseval — for which class is it proved here? justify that [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) plus piecewise $C^1$ can be dropped by admitting general Parseval, or give the density argument in outline)*.

**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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $\abs f^2$ forces $f = 0$.

For merely [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f$, 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), then $c_n(f) = O\bigl(\abs n^{-k}\bigr)$: smoothness of the signal $=$ decay of its [spectrum](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen).

**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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f^{(k)}$ 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 of Exercise 14.7.**

Parseval for $f$ and for $f'$ (both legitimate: $f$ is $C^1$, $f'$ piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) — indeed [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)):

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

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](#exo-b2-fourier-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 of Exercise 14.8.**

From [Exercise 14.1](#exo-b2-fourier-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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $u \mapsto \frac{\sin u}{u}$, 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 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 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](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-adjoint) 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](#exo-b2-fourier-5).

**Exercise 14.11 ★★.**

(Convolution) For $f, g$ [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) integrand, compare the two functions of the upper limit: both vanish at the left endpoint and have the same derivative, by [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and differentiation under the integral sign)*, and that $S_N(f) =
f * D_N$ for the [Dirichlet kernel](#lem-b2-fourier-kernel). (The weekend problem’s Fejér means are likewise convolutions, $\sigma_N(f) = f *
F_N$.)

**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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity); 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) plus [Theorem 9.10](https://one-course.com/books/math/4/en/chapter/9-integration#thm-b2-integration-continuity) 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](#lem-b2-fourier-kernel) 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 $P_r(t) = \sum_{n\in\Z} r^{\abs n}\eu^{\iu nt}$.

1. Sum the two geometric series and show $$P_r(t) = \frac{1 - r^2}{1 - 2r\cos t + r^2} > 0,  \qquad  \frac{1}{2\pi}\int_{-\pi}^{\pi}P_r = 1 .$$
2. 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) .
3. Deduce that for every [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $2\pi$ -periodic $f$ , the *Abel means* $(f * P_r)(x) = \sum_n  r^{\abs n}c_n(f)\,\eu^{\iu nx}$ converge to $f$ [uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) as $r \to 1^-$ — the continuous-parameter sibling of Fejér’s theorem, and the Fourier incarnation of [Abel summation](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-abel) from the power-series chapter.

**Solution of Exercise 14.12.**

1. 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) convergent series keeps only $n = 0$.
2. 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) to $0$ off any neighborhood of $0$ .
3. Term-by-term integration ([normal convergence](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) as $r \to 1^-$. This is [Abel summation](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#thm-b2-series-abel) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f$ 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](#def-b2-fourier-coefficients), Parseval for every function of this chapter (removing the “admitted” in [Theorem 14.10](#thm-b2-fourier-parseval)), Weyl’s equidistribution theorem, and — crowning a century of geometry — the isoperimetric inequality. Throughout, $f$ is $2\pi$-periodic and piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), and

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

**Part I — The Fejér kernel.**

1. 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 $\frac{1}{2\pi}\int_{-\pi}^{\pi}F_N = 1$ .
2. 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)*.
3. 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.
4. (Fejér’s theorem) Prove: if $f$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and $2\pi$ -periodic, then $\sigma_N(f) \to f$ *[uniformly](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def)* 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)* .
5. For piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f$ , show the [pointwise](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) 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.**

6. (Trigonometric Weierstrass) Deduce: every [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $2\pi$ -periodic function is a uniform limit of trigonometric polynomials.
7. (Uniqueness) Deduce: a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f$ with $c_n(f) = 0$ for all $n$ is identically zero — two [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) periodic functions with the same [Fourier coefficients](#def-b2-fourier-coefficients) coincide ( [Exercise 14.4](#exo-b2-fourier-4) , now with no admission).
8. (Parseval, [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) case) Using the projection property of $S_N$ ([Proposition 14.2](#prop-b2-fourier-bessel)) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)* $2\pi$-periodic $f$.
9. (Parseval, piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) case) Given $f$ piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) and $\varepsilon > 0$ , construct a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](#thm-b2-fourier-parseval) — the “admitted” is gone.
10. (No Gibbs for Fejér) Contrast with [Exercise 14.8](#exo-b2-fourier-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.**

11. 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 $\intcc{0}{\pi}$)*.
12. 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 $\abs u =  \frac1N$)*.
13. Conclude: if $f$ is $L$-Lipschitz and $2\pi$-periodic, then $$\norm{\sigma_N f - f}_\infty \leq \frac{C\,L\ln N}{N} .$$
14. (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.
15. (Localization) Show: if $f$ (piecewise [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ) 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 $\intcc{a}{b} \subseteq \intcc{0}{1}$,

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

16. 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)* $1$ -periodic $f$ *(squeeze the indicator of $\intcc ab$ between two [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) piecewise-affine functions whose integrals differ by $\varepsilon$)* .
17. (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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $1$-periodic $f$ by question 6 (transported to period $1$): with question 16, $(x_n)$ is equidistributed.
18. 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 $\intco{0}{1}$.
19. 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.
20. (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).

21. Expand $z(t) = \sum_{n\in\Z}c_n\eu^{\iu nt}$ (the series of a $C^1$ function, [normally](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) convergent) and prove, by Parseval applied to $z'$: $$\frac{L^2}{2\pi} = \int_0^{2\pi}\abs{z'}^2\dd t  = 2\pi\sum_{n\in\Z}n^2\abs{c_n}^2 .$$
22. 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 $f = z$, $g = z'$)* .
23. (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 $\frac{L^2}{4\pi}$.
24. 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.
25. 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) : Fourier series as a Hilbert basis).

**Solution of Problem 14.1.**

**1.** Averaging [Lemma 14.5](#lem-b2-fourier-kernel) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-def): 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) periodic functions are determined by their [Fourier coefficients](#def-b2-fourier-coefficients).

**8.** $S_Nf$ is the orthogonal projection of $f$ onto $\mathcal T_N$ ([Proposition 14.2](#prop-b2-fourier-bessel)), 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), 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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f$: the “admitted” in [Theorem 14.10](#thm-b2-fourier-parseval) 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](#exo-b2-fourier-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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $1$-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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $1$-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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) 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](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#def-b2-funcseq-series) with sum $z$ ([Theorem 14.10](#thm-b2-fourier-parseval) (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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $z'$ 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](https://one-course.com/books/math/4/en/chapter/12-quadratic-forms#def-b2-quadratic-def)), both [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity). 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](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable) of the Fourier series implies its Abel [summability](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#def-b2-series-summable) with the same sum (Frobenius, proved in the power-series chapter’s weekend problem) — the Poisson-kernel route of [Exercise 14.12](#exo-b2-fourier-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](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete): 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.