---
title: "The Fourier Transform"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 14
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform
---

# Chapter 14 — The Fourier Transform

Fourier series decompose periodic signals into discrete harmonics; the Fourier *transform* does the same for signals on the whole line, with a continuum of frequencies. It converts differentiation into multiplication, convolution into products, and Gaussians into Gaussians — the reasons it solves differential equations, drives signal processing, and will prove the central limit theorem in [Chapter 23](https://one-course.com/books/math/5/en/chapter/23-characteristic-functions-and-the-central-limit-theorem#ch-b3-clt). This chapter develops the $L^1$ theory (Riemann–Lebesgue, inversion, injectivity), the [Schwartz class](#def-b3-fouriertransform-schwartz) where the transform is a [perfect](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#prop-b3-galois-perfect) bijection, and the $L^2$ theory (Plancherel: the transform is, up to a constant, a unitary operator), with two showpiece applications: the heat equation, solved end to end in the weekend problem, and the Poisson summation formula. Convention:

$$
\hat f(\xi) = \int_\R f(x)\,\eu^{-\iu\xi x}\,\dd x .
$$

## 14.1 The transform on $L^1$

**Proposition 14.1.**

For $f \in L^1(\R)$: $\hat f$ is well defined, bounded ($\norm{\hat f}_\infty \leq \norm f_1$), [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and:

1. $\widehat{\tau_af}(\xi) = \eu^{-\iu a\xi}\hat f(\xi)$ and $\widehat{\eu^{\iu ax}f}(\xi) = \hat f(\xi - a)$ ;
2. $\widehat{f(\cdot/\lambda)}(\xi) = \lambda\hat  f(\lambda\xi)$ for $\lambda > 0$ ;
3. if $xf \in L^1$ , then $\hat f$ is $\mathcal C^1$ with $(\hat f)'(\xi) = \widehat{(-\iu x)f}(\xi)$ ;
4. if $f \in \mathcal C^1$ with $f' \in L^1$ (and $f \to  0$ at $\pm\infty$ , automatic here), then $\widehat{f'}(\xi) = \iu\xi\hat f(\xi)$ ;
5. $\widehat{f * g} = \hat f\,\hat g$ for $f, g \in L^1$ .

**Proof.** Boundedness: $\abs{\hat f} \leq \int\abs f$. [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): DCT with dominator $\abs f$ ([Theorem 10.14](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#thm-b3-lebesgue-paramcont)). (1), (2): substitutions ([Theorem 11.10](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-linearchange)). (3): differentiation under the integral, dominator $\abs{xf}$ ([Theorem 10.15](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#thm-b3-lebesgue-paramdiff)). (4): first, $f(x) = f(0)
+ \int_0^xf'$ has a limit at $\pm\infty$ ($f' \in L^1$), which must be $0$ ($f \in L^1$); then integrate by parts on $[-A,
A]$ and let $A \to \infty$. (5): Fubini, legitimate since $(x,y)\mapsto f(x - y)g(y)\eu^{-\iu\xi x}$ is absolutely [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) ([Theorem 11.9](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-convolution)):

$$
\widehat{f*g}(\xi) = \iint f(x - y)g(y)\eu^{-\iu\xi(x - y)}
\eu^{-\iu\xi y}\dd x\,\dd y = \hat f(\xi)\,\hat g(\xi).
$$

∎

**Example 14.2.**

The Gaussian: for $a > 0$,

$$
\widehat{\eu^{-ax^2}}(\xi) =
\sqrt{\frac\pi a}\;\eu^{-\xi^2/4a} :
$$

by [Exercise 10.7](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#exo-b3-lebesgue-7) (the ODE $F' = -\frac\xi{2}F$ trick, rescaled), or by (3): $g = \widehat{\eu^{-ax^2}}$ satisfies $g'(\xi) = -\frac{\xi}{2a}g(\xi)$ (integrate by parts), $g(0) = \sqrt{\pi/a}$. Gaussians are fixed points of the transform up to scaling — the deep reason they rule the central limit theorem.

**Theorem 14.3 (Riemann–Lebesgue).**

For $f \in L^1(\R)$: $\hat f(\xi) \to 0$ as $\abs\xi \to
\infty$. Thus $\widehat{\phantom f} \colon L^1 \to \mathcal
C_0(\R)$ ([continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions vanishing at infinity).

**Proof.** For an indicator of an interval, $\widehat{\mathbf 1_{\intcc
ab}}(\xi) = \frac{\eu^{-\iu a\xi} - \eu^{-\iu b\xi}}{\iu\xi}
\to 0$; hence for step functions. Step functions are dense in $L^1$ ([Theorem 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-density)(1) plus approximation of [finite-measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure) sets by finite unions of intervals, [Exercise 9.7](https://one-course.com/books/math/5/en/chapter/9-measure-theory#exo-b3-measure-7)), and the transform is $\norm\cdot_\infty$-$\norm\cdot_1$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): for $\norm{f -
s}_1 < \varepsilon$, $\limsup_{\abs\xi\to\infty}\abs{\hat
f(\xi)} \leq \varepsilon$. ∎

## 14.2 Inversion and injectivity

**Lemma 14.4 (Multiplication formula).**

For $f, g \in L^1(\R)$: $\displaystyle\int \hat f\,g =
\int f\,\hat g$.

**Proof.** Both sides equal $\iint f(x)g(\xi)\eu^{-\iu x\xi}\dd
x\,\dd\xi$ (Tonelli–Fubini: the double integral of the absolute value is $\norm f_1\norm g_1$). ∎

**Theorem 14.5 (Inversion).**

Let $f \in L^1(\R)$.

1. (Gaussian summability) For every $x$, $$(f * g_\varepsilon)(x) =  \frac1{2\pi}\int_\R \hat f(\xi)\,  \eu^{-\varepsilon\xi^2}\,\eu^{\iu x\xi}\,\dd\xi,  \qquad\text{where } g_\varepsilon(y) =  \frac{1}{2\sqrt{\pi\varepsilon}}\,  \eu^{-y^2/4\varepsilon},$$ and $f * g_\varepsilon \to f$ in $L^1$ as $\varepsilon  \to 0$.
2. If moreover $\hat f \in L^1$, then for almost every $x$ $$f(x) = \frac{1}{2\pi}\int_\R \hat  f(\xi)\,\eu^{\iu x\xi}\,\dd\xi ,$$ and $f$ has a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) representative.
3. (Injectivity) If $\hat f = 0$ then $f = 0$ a.e.

**Proof.** (1) Fix $x$ and apply [Lemma 14.4](#lem-b3-fouriertransform-multiplication) to $f$ and $g(\xi) = \frac1{2\pi}\eu^{-\varepsilon\xi^2}\eu^{\iu x\xi}$: by [Example 14.2](#ex-b3-fouriertransform-gaussian) (with the modulation rule),

$$
\hat g(y) = \frac1{2\pi}\sqrt{\frac\pi\varepsilon}\,
\eu^{-(y - x)^2/4\varepsilon} = g_\varepsilon(x - y),
$$

so $\frac1{2\pi}\int\hat f(\xi)\eu^{-\varepsilon\xi^2}
\eu^{\iu x\xi}\dd\xi = \int f(y)g_\varepsilon(x - y)\dd y =
(f*g_\varepsilon)(x)$. The $g_\varepsilon$ are an approximate identity: $g_\varepsilon \geq 0$, $\int g_\varepsilon = 1$ (Gaussian integral), concentrating at $0$; the proof of [Theorem 12.9](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-regularization)(2) applies verbatim (only $\int g_\varepsilon = 1$ and concentration were used: for the tail, $\int_{\abs y > \delta}g_\varepsilon \to
0$): $\norm{f * g_\varepsilon - f}_1 \to 0$.

(2) If $\hat f \in L^1$: the right side of (1) converges, by DCT (dominator $\abs{\hat f}$), to $\frac1{2\pi}\int\hat
f(\xi)\eu^{\iu x\xi}\dd\xi$ for *every* $x$, and this limit function is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (DCT again). On the other hand $f
* g_\varepsilon \to f$ in $L^1$, so along a subsequence a.e. ([Theorem 12.4](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-complete)): the two limits agree a.e.

(3) $\hat f = 0$ makes the right side of (1) vanish: $f *
g_\varepsilon = 0$ for all $\varepsilon$, and $f *
g_\varepsilon \to f$ in $L^1$: $f = 0$ a.e. ∎

## 14.3 The Schwartz class

**Definition 14.6.**

The *Schwartz class* $\mathcal
S(\R)$ consists of the $\mathcal C^\infty$ functions $f$ with $\sup_x\abs{x^m f^{(n)}(x)} < \infty$ for all $m, n \geq 0$ (all derivatives decay faster than any power). Examples: $\eu^{-ax^2}$, $\mathcal C_c^\infty$. Clearly $\mathcal S
\subseteq L^p$ for every $p$ (bound by $C(1 + x^2)^{-1}$), and $\mathcal S$ is stable under derivatives, multiplication by polynomials, and products.

**Theorem 14.7.**

The Fourier transform maps $\mathcal S(\R)$ bijectively onto itself, with inverse $\check g(x) = \frac1{2\pi}\int
g(\xi)\eu^{\iu x\xi}\dd\xi$.

**Proof.** Let $f \in \mathcal S$. Iterating [Proposition 14.1](#prop-b3-fouriertransform-basic)(3), $(\hat f)^{(n)} =
\widehat{(-\iu x)^nf}$ (each $x^kf \in L^1$); iterating (4) with $h = (-\iu x)^nf \in \mathcal S$ (all of whose derivatives are [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1)), $(\iu\xi)^m\hat h =
\widehat{h^{(m)}}$. Combining,

$$
\abs{\xi^m\,(\hat f)^{(n)}(\xi)}
= \bigl|\widehat{\,h^{(m)}}(\xi)\bigr|
\leq \bigl\|\bigl((-\iu x)^nf\bigr)^{(m)}\bigr\|_1 < \infty
$$

uniformly in $\xi$: $\hat f \in \mathcal S$. Since $\hat f \in L^1$, inversion ([Theorem 14.5](#thm-b3-fouriertransform-inversion)(2)) holds everywhere (both sides [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)): $\check{\hat f} = f$, and symmetrically $\widehat{\check g} = g$ (the check transform is $g \mapsto \frac1{2\pi}\hat g(-\cdot)$, again preserving $\mathcal S$): bijection. ∎

## 14.4 Plancherel and $L^2$

**Theorem 14.8 (Plancherel).**

For $f \in L^1 \cap L^2(\R)$:

$$
\norm{\hat f}_2^2 = 2\pi\,\norm f_2^2 .
$$

Consequently $\widehat{\phantom f}$ extends uniquely to a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) linear map $\mathcal F \colon L^2(\R) \to L^2(\R)$ with $\norm{\mathcal Ff}_2 = \sqrt{2\pi}\norm f_2$; $\mathcal
F$ is bijective, with $\mathcal F^{-1} =
\frac1{2\pi}\,\mathcal F\circ\sigma$ where $\sigma f =
f(-\cdot)$, and it preserves [inner products](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner) up to the factor $2\pi$.

**Proof.** Let $f \in L^1\cap L^2$ and $h = f * \tilde f$ with $\tilde f(x) = \overline{f(-x)}$. Then $h \in L^1$ ([Theorem 11.9](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#thm-b3-product-convolution)), $h$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and bounded ([Exercise 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#exo-b3-lp-6): $f, \tilde f \in L^2$), $h(0) =
\int f\bar f = \norm f_2^2$, and $\hat h = \hat
f\,\widehat{\tilde f} = \hat f\,\overline{\hat f} =
\abs{\hat f}^2 \geq 0$ (compute $\widehat{\tilde f} =
\overline{\hat f}$). Apply [Theorem 14.5](#thm-b3-fouriertransform-inversion)(1) to $h$ at $x = 0$:

$$
(h * g_\varepsilon)(0) = \frac1{2\pi}\int \hat
h(\xi)\,\eu^{-\varepsilon\xi^2}\dd\xi .
$$

As $\varepsilon \to 0$: the left side tends to $h(0)$ ($h$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) bounded: $(h*g_\varepsilon)(0) - h(0) = \int(h(-y)
- h(0))g_\varepsilon(y)\dd y \to 0$ by splitting small/large $y$); the right side increases to $\frac1{2\pi}\int\hat h$ by MCT ($\hat h \geq 0$). Hence $\frac1{2\pi}\int\abs{\hat f}^2 =
\norm f_2^2$, finite or not *a priori* — and finite, proving both membership and the identity.

Extension: $L^1\cap L^2 \supseteq \mathcal C_c$ is dense in $L^2$ ([Theorem 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-density)); the transform is $\sqrt{2\pi}$-isometric there, so extends uniquely to an isometry-up-to-constant $\mathcal F$ on $L^2$ ([Theorem 7.2](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-extension)). Inversion for $\mathcal S$ ([Theorem 14.7](#thm-b3-fouriertransform-schwartz)) transfers by the same [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) (both sides $L^2$-continuous): $\mathcal
F\bigl(\frac1{2\pi}\mathcal F(\sigma f)\bigr) = f$ on $\mathcal S$, hence on $L^2$: bijectivity. [Inner products](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-inner): polarization from the norm identity. ∎

**Theorem 14.9 (Poisson summation).**

Let $f \in \mathcal S(\R)$ ([continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$ with $\abs{f} +
\abs{\hat f} \leq C(1 + \abs\cdot)^{-2}$ suffices). Then

$$
\sum_{n\in\Z} f(n) \;=\; \sum_{k\in\Z}\hat f(2\pi k) .
$$

**Proof.** Let $F(x) = \sum_{n\in\Z}f(x + n)$: the series converges normally on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) (decay of $f$), so $F$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), and it is $1$-periodic. Its Fourier coefficients (period $1$: $c_k(F) = \int_0^1F(t)\eu^{-2\iu\pi kt}\dd t$):

$$
c_k(F) = \sum_n\int_0^1 f(t + n)\,\eu^{-2\iu\pi kt}\dd t
= \int_\R f(t)\,\eu^{-2\iu\pi kt}\dd t = \hat f(2\pi k)
$$

(normal convergence justifies the interchange; the phase is $1$-periodic). The series $\sum_k\abs{c_k(F)}$ converges (decay of $\hat f$), so the Fourier series of $F$ converges normally; its sum is a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) function with the same Fourier coefficients as $F$, hence equals $F$ (injectivity on the circle: the difference has zero coefficients, and [Theorem 13.9](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-fourier) gives zero in $L^2$, hence everywhere by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)). Evaluate at $x = 0$. ∎

**Example 14.10 (The theta identity).**

Applying Poisson to $f(x) = \eu^{-\pi tx^2}$ ($t > 0$), whose transform is $\hat f(\xi) = t^{-1/2}\eu^{-\xi^2/4\pi t}$ ([Example 14.2](#ex-b3-fouriertransform-gaussian) with $a = \pi t$):

$$
\sum_{n\in\Z}\eu^{-\pi n^2t}
= \frac1{\sqrt t}\sum_{k\in\Z}\eu^{-\pi k^2/t} :
$$

the functional equation of Jacobi’s theta function, key to the functional equation of Riemann’s $\zeta$ — and a spectacular numerical accelerator: for $t$ small, the left side converges slowly, the right side blazingly fast.

**Method 14.11.**

Working ranges: $L^1$ — transform defined pointwise, inversion needs $\hat f \in L^1$; $\mathcal S$ — everything is legal, prove here first; $L^2$ — transform defined by [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) (not by the integral!), [perfect](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#prop-b3-galois-perfect) symmetry, [Parseval](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-parseval) bookkeeping. To compute a transform: reduce to the table (indicator, exponential, Gaussian) by the rules of [Proposition 14.1](#prop-b3-fouriertransform-basic); to prove an identity: establish it on $\mathcal S$ (or $\mathcal C_c^\infty$) and extend by [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) and [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ([Method 12.13](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#met-b3-lp-toolkit)); to solve a linear PDE or ODE with constant coefficients: transform, divide, invert.

![The heat kernel g_t(x) = 12√π t\, -x2/4t at three times: total mass 1 forever, height t-1/2, width √ t. Convolving initial data with this spreading Gaussian is the entire content of the weekend problem; in frequency, the same picture reads g_t( ) = -t 2 — high frequencies die first, and that asymmetry is the arrow of time.](https://one-course.com/images/onecourse/chapters/math-5/b3-fouriertransform/fig-22725c4b9b19.svg)

*The [heat kernel](#pb-b3-fouriertransform-1) $g_t(x) =
\frac1{2\sqrt{\pi t}}\,\eu^{-x^2/4t}$ at three times: total mass $1$ forever, height $\sim t^{-1/2}$, width $\sim
\sqrt t$. Convolving initial data with this spreading Gaussian is the entire content of the weekend problem; in frequency, the same picture reads $\hat g_t(\xi) = \eu^{-t\xi^2}$ — high frequencies die first, and that asymmetry is the arrow of time.*

## 14.5 Exercises

**Exercise 14.1 ★.**

Compute the Fourier transforms of: $\mathbf 1_{\intcc{-a}a}$; $\eu^{-a\abs x}$ ($a > 0$); the tent function $\max(0, 1 -
\abs x)$; $\frac1{x^2 + a^2}$ *(use inversion on the second)*. Record the emerging table.

**Solution of Exercise 14.1.**

$\widehat{\mathbf 1_{\intcc{-a}a}}(\xi) =
\int_{-a}^a\eu^{-\iu\xi x}\dd x = \frac{2\sin(a\xi)}{\xi}$ (value $2a$ at $0$). $\widehat{\eu^{-a\abs x}}(\xi) = \int_0^\infty\eu^{-(a +
\iu\xi)x} + \eu^{-(a - \iu\xi)x}\,\dd x = \frac1{a + \iu\xi} +
\frac1{a - \iu\xi} = \frac{2a}{a^2 + \xi^2}$. Tent: $\max(0, 1 - \abs x) = \mathbf 1_{\intcc{-1/2}{1/2}} *
\mathbf 1_{\intcc{-1/2}{1/2}}$, so its transform is $\bigl(\frac{2\sin(\xi/2)}\xi\bigr)^2 =
\bigl(\frac{\sin(\xi/2)}{\xi/2}\bigr)^2$. Last: $\frac{2a}{a^2+\xi^2} \in L^1$, so inversion ([Theorem 14.5](#thm-b3-fouriertransform-inversion)(2)) applied to $\eu^{-a\abs x}$ gives, after renaming variables,

$$
\widehat{\Bigl(\frac1{x^2 + a^2}\Bigr)}(\xi) =
\frac{\pi}{a}\,\eu^{-a\abs\xi} .
$$

**Exercise 14.2 ★.**

Let $f \in L^1$. Express in terms of $\hat f$ the transforms of: $f(x - a)$, $f(x)\cos(bx)$, $f(ax + b)$, $\overline{f(-x)}$, $(f * f)(x)$. Verify each rule on the Gaussian.

**Solution of Exercise 14.2.**

From [Proposition 14.1](#prop-b3-fouriertransform-basic): $\widehat{f(\cdot - a)} = \eu^{-\iu a\xi}\hat f(\xi)$; $\widehat{f\cos(b\cdot)} = \frac12\bigl(\hat f(\xi - b) + \hat
f(\xi + b)\bigr)$; $\widehat{f(a\cdot + b)}(\xi) = \frac1a\,\eu^{\iu
b\xi/a}\,\hat f(\xi/a)$ ($a > 0$); $\widehat{\overline{f(-\cdot)}} = \overline{\hat f}$; $\widehat{f * f} = \hat f^2$. On the Gaussian ($\widehat{\eu^{-x^2}} = \sqrt\pi\eu^{-\xi^2/4}$) each rule is a one-line check — e.g. $\eu^{-(x-a)^2}$ has transform $\sqrt\pi\,\eu^{-\iu a\xi}\eu^{-\xi^2/4}$, which the direct computation ([complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) the square) confirms.

**Exercise 14.3 ★★.**

(a) Show that $\mathbf 1_{\intcc{-1}1} * \mathbf
1_{\intcc{-1}1}$ has transform $\bigl(\frac{2\sin\xi}\xi
\bigr)^2$, and deduce $\int_\R\bigl(\frac{\sin\xi}\xi\bigr)^2\dd\xi = \pi$ by Plancherel — or by inversion at $0$. Compare [Problem 10.1](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#pb-b3-lebesgue-1). (b) Compute $\int_\R\frac{\dd x}{(x^2+1)^2}$ via Plancherel applied to $\eu^{-\abs x}$.

**Solution of Exercise 14.3.**

(a) $h = \mathbf 1_{\intcc{-1}1}*\mathbf 1_{\intcc{-1}1}$ has $\hat h = \bigl(\frac{2\sin\xi}\xi\bigr)^2 \in L^1$; inversion at $x = 0$, where $h(0) = \lambda(\intcc{-1}1\cap\intcc{-1}1)
= 2$:

$$
2 = \frac1{2\pi}\int_\R\Bigl(\frac{2\sin\xi}\xi\Bigr)^2
\dd\xi
\ \Longrightarrow\
\int_\R\Bigl(\frac{\sin\xi}\xi\Bigr)^2\dd\xi = \pi ,
$$

consistent with $\int_0^\infty\frac{\sin^2}{\xi^2} =
\frac\pi2$ ([Problem 10.1](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#pb-b3-lebesgue-1)).

(b) Plancherel for $f = \eu^{-\abs x}$: $\int\abs{\hat f}^2 =
2\pi\int\abs f^2$ reads $\int\frac{4\,\dd\xi}{(1 + \xi^2)^2} =
2\pi\int\eu^{-2\abs x}\dd x = 2\pi$: $\int_\R\frac{\dd\xi}{(1+\xi^2)^2} = \frac\pi2$.

**Exercise 14.4 ★★.**

([Heat kernel](#pb-b3-fouriertransform-1) algebra) With $g_t(x) =
\frac1{2\sqrt{\pi t}}\eu^{-x^2/4t}$: (a) verify $\hat g_t(\xi) = \eu^{-t\xi^2}$; (b) deduce the semigroup law $g_t * g_s = g_{t+s}$ without any integral computation; (c) show $\norm{g_t}_1 = 1$ and $\norm{g_t}_2^2 =
(8\pi t)^{-1/2}$.

**Solution of Exercise 14.4.**

(a) $g_t(x) = \frac1{2\sqrt{\pi t}}\eu^{-x^2/4t}$: by [Example 14.2](#ex-b3-fouriertransform-gaussian) with $a = \frac1{4t}$, $\hat g_t(\xi) = \frac1{2\sqrt{\pi t}}\sqrt{4\pi
t}\,\eu^{-t\xi^2} = \eu^{-t\xi^2}$. (b) $\widehat{g_t * g_s} = \eu^{-t\xi^2}\eu^{-s\xi^2} =
\widehat{g_{t+s}}$, and the transform is injective on $L^1$ ([Theorem 14.5](#thm-b3-fouriertransform-inversion)(3)): $g_t * g_s =
g_{t+s}$. (c) $\norm{g_t}_1 = 1$ (Gaussian integral); $\norm{g_t}_2^2 = \frac1{4\pi t}\int\eu^{-x^2/2t}\dd x =
\frac{\sqrt{2\pi t}}{4\pi t} = \frac1{\sqrt{8\pi t}}$.

**Exercise 14.5 ★★.**

Show that if $f \in L^1$ is even and real, $\hat f$ is even and real; if $f$ is odd and real, $\hat f$ is odd and purely imaginary. What does $\hat f(0)$ compute? Deduce that $f \geq
0$ forces $\norm{\hat f}_\infty = \hat f(0) = \int f$, and interpret for probability densities ([Chapter 23](https://one-course.com/books/math/5/en/chapter/23-characteristic-functions-and-the-central-limit-theorem#ch-b3-clt): a characteristic function has modulus $\leq 1$, attained at $0$).

**Solution of Exercise 14.5.**

For real even $f$: $\hat f(\xi) = \int f\cos(\xi x)\dd x$ (the sine part cancels): real and even. Odd: $\hat f(\xi) =
-\iu\int f\sin(\xi x)$: odd, purely imaginary. $\hat f(0) =
\int f$: the total mass. If $f \geq 0$: $\abs{\hat f(\xi)}
\leq \int\abs f = \int f = \hat f(0)$, so the sup is attained at $0$. For a probability [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma), $\hat f(-\xi)$ is the characteristic function of [Chapter 23](https://one-course.com/books/math/5/en/chapter/23-characteristic-functions-and-the-central-limit-theorem#ch-b3-clt): modulus $\leq 1$ everywhere, $= 1$ at the origin.

**Exercise 14.6 ★★★.**

(Non-surjectivity) Show that $\widehat{\phantom f}\colon L^1
\to \mathcal C_0$ is injective and [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), but *not* surjective, in three steps. (i) Injectivity ([Theorem 14.5](#thm-b3-fouriertransform-inversion)) and [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ($\norm{\hat f}_\infty \leq \norm f_1$), and $\mathcal C_0$ is a Banach space (closed in $\norm\cdot_\infty$). (ii) If the map were surjective, it would be bijective, and the open mapping theorem ([Theorem 8.12](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#thm-b3-banach-openmapping)) would give a constant $C$ with $\norm f_1 \leq C\norm{\hat
f}_\infty$ for all $f \in L^1$. (iii) Contradict this with $f_n(x) = \frac{\sin
x}{x}\cdot\frac{\sin(x/n)}{x/n}$: its transform is (up to constants) the convolution $\mathbf 1_{\intcc{-1}1} *
\mathbf 1_{\intcc{-1/n}{1/n}}$-type trapezoid — show $\norm{\hat f_n}_\infty \leq \pi$ uniformly, while $\norm{f_n}_1 \geq c\ln n$ by counting the arches of $\frac{\abs{\sin x}}x$ on $[1, n]$ (where the second factor is bounded below), as in [Theorem 8.11](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#thm-b3-banach-fourierdiverge).

**Solution of Exercise 14.6.**

(i) Injectivity is [Theorem 14.5](#thm-b3-fouriertransform-inversion)(3); [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) is $\norm{\hat f}_\infty \leq \norm f_1$ (with values in $\mathcal C_0$ by Riemann–Lebesgue); $\mathcal C_0$ is closed in the sup norm (uniform limits of vanishing-at-infinity functions vanish at infinity): Banach.

(ii) A [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) bijection between Banach spaces has [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) inverse ([Theorem 8.12](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#thm-b3-banach-openmapping)): there would be $C$ with $\norm f_1 \leq C\norm{\hat f}_\infty$.

(iii) Let $f_n(x) = \frac{\sin x}x\cdot\frac{\sin(x/n)}{x/n}$: a product of two $L^2$ functions, and $O(x^{-2})$ at infinity, so $f_n \in L^1\cap L^2$. Since $\bigl(\frac{\sin(ax)}{ax}\bigr)$ has $L^2$-transform $\frac\pi a\mathbf 1_{\intcc{-a}a}$, the product formula $\widehat{gh} = \frac1{2\pi}\hat g * \hat h$ (valid for $g, h
\in L^2$ with $gh \in L^1$; check it on Schwartz functions by Fubini and extend by $L^2$-continuity of both sides via Plancherel) gives

$$
\hat f_n = \frac1{2\pi}\,\bigl(\pi\mathbf
1_{\intcc{-1}1}\bigr) * \bigl(\pi n\,\mathbf
1_{\intcc{-1/n}{1/n}}\bigr):
$$

a trapezoid of height $\frac{\pi n}2\cdot\frac2n = \pi$: $\norm{\hat f_n}_\infty = \pi$ for every $n$. But on $[1,
n]$, $\frac{\sin(x/n)}{x/n} \geq \sin 1 > 0$, so

$$
\norm{f_n}_1 \geq \sin 1\int_1^n\frac{\abs{\sin x}}x\dd x
\geq c\ln n
$$

(arch-counting, as in [Theorem 8.11](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#thm-b3-banach-fourierdiverge)). The bound $\norm{f_n}_1 \leq C\pi$ fails for large $n$: not surjective. (The image is a dense — by Stone–Weierstrass-type arguments — but proper subspace of $\mathcal C_0$.)

**Exercise 14.7 ★★.**

(Smoothness $\leftrightarrow$ decay dictionary) Prove: $f \in
L^1$ with $\hat f(\xi) = O(\abs\xi^{-k-1-\delta})$ for some $\delta > 0$ implies $f$ has a $\mathcal C^k$ representative. Conversely $f \in \mathcal C^k_c$ implies $\hat f(\xi) =
O(\abs\xi^{-k})$. Illustrate both directions on the tent function.

**Solution of Exercise 14.7.**

If $\hat f(\xi) = O(\abs\xi^{-k-1-\delta})$: then $\xi^j\hat f
\in L^1$ for $0 \leq j \leq k$ ([integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) at infinity by the decay, locally by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $\hat f$). Inversion ([Theorem 14.5](#thm-b3-fouriertransform-inversion)(2)) represents $f$ a.e. by $x \mapsto \frac1{2\pi}\int\hat f(\xi)\eu^{\iu
x\xi}\dd\xi$, and differentiation under the integral (dominators $\abs{\xi^j\hat f}$) makes this representative $\mathcal C^k$. Conversely for $f \in \mathcal C_c^k$: iterating [Proposition 14.1](#prop-b3-fouriertransform-basic)(4), $(\iu\xi)^k\hat f = \widehat{f^{(k)}}$, so $\abs{\hat f} \leq
\norm{f^{(k)}}_1\abs\xi^{-k}$. Tent function: [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support ($k = 0$: transform bounded), and its transform $\sim \xi^{-2} = O(\abs\xi^{-0-1-1})$ returns, by the first direction, a $\mathcal C^0$ representative — both sharp: the tent is not $\mathcal C^1$, and its transform decays no faster than $\xi^{-2}$.

**Exercise 14.8 ★★★.**

(Heisenberg’s inequality) For $f \in \mathcal S(\R)$ real with $\norm f_2 = 1$, prove

$$
\Bigl(\int x^2f(x)^2\dd x\Bigr)\cdot
\Bigl(\frac1{2\pi}\int \xi^2\abs{\hat f(\xi)}^2\dd\xi\Bigr)
\;\geq\; \frac14 ,
$$

with equality for Gaussians. *(Write $1 = \int f^2 =
-\int x\,(f^2)'$ by parts, bound by Cauchy–Schwarz, and convert $\norm{f'}_2$ by Plancherel.)* Interpretation: a signal and its spectrum cannot both be concentrated.

**Solution of Exercise 14.8.**

Integration by parts ($f \in \mathcal S$; boundary terms vanish):

$$
1 = \int f^2 = \bigl[xf^2\bigr]_{-\infty}^{\infty} - \int
x\,(f^2)' = -2\int xff' \leq 2\,\norm{xf}_2\,\norm{f'}_2 .
$$

Plancherel and $\widehat{f'} = \iu\xi\hat f$: $\norm{f'}_2^2 = \frac1{2\pi}\int\xi^2\abs{\hat f}^2$. Squaring the display:

$$
\frac14 \leq \norm{xf}_2^2\cdot\frac1{2\pi}
\int\xi^2\abs{\hat f}^2\dd\xi .
$$

Equality requires equality in Cauchy–Schwarz: $f' =
\lambda xf$ with $\lambda < 0$ ([integrability](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1)), i.e. $f(x) =
c\,\eu^{\lambda x^2/2}$: Gaussians. A signal concentrated in $x$ (small $\norm{xf}_2$) must have spread-out spectrum, and conversely: the uncertainty principle.

**Exercise 14.9 ★★.**

Justify [Example 14.10](#ex-b3-fouriertransform-theta) in detail (hypotheses of Poisson for the Gaussian), and use the identity to evaluate $\sum_{n\in\Z}\eu^{-\pi n^2}$ to six decimals with three terms. How many terms of the *defining* series would the same accuracy require at $t = 10^{-2}$, versus the transformed series?

**Solution of Exercise 14.9.**

The Gaussian $f(x) = \eu^{-\pi tx^2}$ is Schwartz, so [Theorem 14.9](#thm-b3-fouriertransform-poisson) applies, and $\hat f(\xi) = t^{-1/2}\eu^{-\xi^2/4\pi t}$; at $\xi = 2\pi
k$ the right side becomes $t^{-1/2}\eu^{-\pi k^2/t}$: the theta identity. At $t = 1$:

$$
\sum_{n\in\Z}\eu^{-\pi n^2} = 1 + 2\eu^{-\pi} + 2\eu^{-4\pi} +
\cdots \approx 1 + 0.0864278 + 0.0000070 = 1.0864348,
$$

accurate to $6$ decimals with three terms ($\eu^{-9\pi}
\approx 5\cdot10^{-13}$). At $t = 10^{-2}$: the defining series needs $\eu^{-\pi n^2/100} < 10^{-7}$, i.e. $n \gtrsim
23$ — about $47$ terms — while the transformed series is $10\sum_k\eu^{-100\pi k^2}$, where already the $k = 1$ term is $\sim 10^{-136}$: one term suffices.

**Exercise 14.10 ★★.**

(Band-limited functions) Let $f \in L^2(\R)$ with $\mathcal Ff$ supported in $\intcc{-\pi}\pi$. Show that $f$ has a representative extending each of whose values is recoverable from samples: prove the *Shannon interpolation* at the integers,

$$
f(x) = \sum_{n\in\Z} f(n)\,
\frac{\sin\bigl(\pi(x - n)\bigr)}{\pi(x - n)}
\quad\text{in } L^2,
$$

by expanding $\mathcal Ff$ in the Fourier basis of $L^2(\intcc{-\pi}\pi)$ ([Theorem 13.9](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-fourier)) and transforming back term by term.

**Solution of Exercise 14.10.**

$\mathcal Ff \in L^2(\intcc{-\pi}\pi) \subseteq
L^1(\intcc{-\pi}\pi)$ (finite [measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-measure)), so inversion gives the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) representative $f(x) =
\frac1{2\pi}\int_{-\pi}^\pi\mathcal Ff(\xi)\eu^{\iu
x\xi}\dd\xi$, with

$$
f(n) = \frac1{2\pi}\int_{-\pi}^{\pi}\mathcal
Ff(\xi)\,\eu^{\iu n\xi}\dd\xi = \langle e_{-n}, \mathcal
Ff\rangle
$$

in the notation of [Theorem 13.9](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-fourier). Expanding in that [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb): $\mathcal Ff = \sum_nf(n)\,\eu^{-\iu
n\xi}$ in $L^2(\intcc{-\pi}\pi)$. Apply the $L^2$-continuous $\mathcal F^{-1}$ term by term:

$$
\mathcal F^{-1}\bigl(\mathbf
1_{\intcc{-\pi}\pi}\eu^{-\iu n\xi}\bigr)(x)
= \frac1{2\pi}\int_{-\pi}^{\pi}\eu^{\iu\xi(x - n)}\dd\xi
= \frac{\sin\bigl(\pi(x-n)\bigr)}{\pi(x - n)} ,
$$

giving $f = \sum_nf(n)\operatorname{sinc}(\cdot - n)$ in $L^2$: a band-limited signal is determined by its integer samples — Shannon’s sampling theorem.

**Exercise 14.11 ★★.**

(The transform as an operator of order four) On $\mathcal
S(\R)$, let $\mathcal F f = \hat f$. (a) Using the inversion formula, show $(\mathcal F^2f)(x) =
2\pi\,f(-x)$, and deduce $\mathcal F^4 = (2\pi)^2\,
\mathrm{id}$. (b) Deduce that every eigenvalue of $\mathcal F$ on $\mathcal S$ belongs to $\{\pm\sqrt{2\pi},
\pm\iu\sqrt{2\pi}\}$, and exhibit an eigenfunction for $+\sqrt{2\pi}$ *(which function of this chapter is proportional to its own transform?)*. (c) Show that even functions satisfy $\mathcal F^2f = 2\pi
f$ and odd ones $\mathcal F^2f = -2\pi f$; produce an eigenfunction for the eigenvalue $-\iu\sqrt{2\pi}$ from $x\eu^{-x^2/2}$ by computing its transform (differentiate the Gaussian’s transform).

**Solution of Exercise 14.11.**

(a) Inversion on $\mathcal S$: $f(x) =
\frac1{2\pi}\int\hat f(\xi)\eu^{\iu x\xi}\dd\xi =
\frac1{2\pi}(\mathcal F\hat f)(-x)$, i.e. $(\mathcal
F^2f)(x) = 2\pi f(-x)$. Applying twice: $\mathcal F^4f =
2\pi\,\mathcal F^2f(-\cdot) = (2\pi)^2f$.

(b) If $\mathcal Ff = \lambda f$ with $f \neq 0$: $(2\pi)^2f = \mathcal F^4f = \lambda^4f$, so $\lambda^4 =
(2\pi)^2$: $\lambda \in \{\pm\sqrt{2\pi},
\pm\iu\sqrt{2\pi}\}$. The Gaussian $g(x) = \eu^{-x^2/2}$ has $\hat g = \sqrt{2\pi}\,g$ ([Example 14.2](#ex-b3-fouriertransform-gaussian) at $a = \frac12$): eigenfunction for $+\sqrt{2\pi}$.

(c) $\mathcal F^2f = 2\pi f(-\cdot)$ equals $\pm2\pi f$ according to parity. For $h(x) = x\eu^{-x^2/2}$: differentiating $\hat g(\xi) = \sqrt{2\pi}\eu^{-\xi^2/2}$ with the rule $\widehat{xf} = \iu\frac{\dd}{\dd\xi}\hat f$:

$$
\hat h(\xi) = \iu\,\frac{\dd}{\dd\xi}\bigl(\sqrt{2\pi}
\eu^{-\xi^2/2}\bigr) = -\iu\sqrt{2\pi}\,\xi\eu^{-\xi^2/2}
= -\iu\sqrt{2\pi}\,h(\xi) :
$$

an eigenfunction for $-\iu\sqrt{2\pi}$. (The Hermite functions continue the pattern, cycling through the four eigenvalues — the discrete Fourier clock.)

**Exercise 14.12 ★★.**

(Autocorrelation and Wiener’s lemma) For $f \in L^2(\R)$ define $\tilde f(x) = \overline{f(-x)}$ and the *autocorrelation* $A_f = f * \tilde f$. (a) Show that $A_f$ is a bounded [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) function with $A_f(0) = \norm f_2^2 \geq \abs{A_f(x)}$ for all $x$ ([Exercise 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#exo-b3-lp-6) and Cauchy–Schwarz). (b) Show, first for $f \in L^1\cap L^2$, that $\widehat{A_f} = \abs{\hat f\,}^2 \geq 0$: the autocorrelation has nonnegative transform — spectra of autocorrelations are power spectra. (c) Deduce the identity $\int_\R\abs{\hat
f(\xi)}^2\eu^{\iu x\xi}\,\dd\xi = 2\pi A_f(x)$ (inversion; justify its applicability when $\hat f \in L^2$ has $\abs{\hat f}^2 \in L^1$), and evaluate it for $f = \mathbf
1_{\intcc{-1/2}{1/2}}$ at $x = 0$: recover $\int_\R\bigl(\frac{\sin u}u\bigr)^2\dd u = \pi$.

**Solution of Exercise 14.12.**

(a) $\tilde f \in L^2$ with $\norm{\tilde f}_2 = \norm f_2$; [Exercise 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#exo-b3-lp-6) (conjugate exponents $p = q = 2$) makes $A_f = f * \tilde f$ bounded and uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), with

$$
A_f(x) = \int f(y)\,\overline{f(y - x)}\,\dd y,
\qquad A_f(0) = \norm f_2^2,
\qquad \abs{A_f(x)} \leq \norm f_2\,\norm{f(\cdot -
x)}_2 = A_f(0)
$$

by Cauchy–Schwarz.

(b) For $f \in L^1\cap L^2$: $\tilde f \in L^1$ too, and the convolution theorem gives $\widehat{A_f} = \hat f\,
\widehat{\tilde f}$; computing, $\widehat{\tilde f}(\xi) =
\int\overline{f(-x)}\eu^{-\iu\xi x}\dd x =
\overline{\int f(u)\eu^{-\iu\xi u}\dd u} =
\overline{\hat f(\xi)}$: $\widehat{A_f} = \abs{\hat f}^2
\geq 0$.

(c) When $\abs{\hat f}^2 \in L^1$, inversion applies to the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $A_f$ (its transform is [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1); [Theorem 14.5](#thm-b3-fouriertransform-inversion)):

$$
A_f(x) = \frac1{2\pi}\int\abs{\hat f(\xi)}^2
\eu^{\iu x\xi}\,\dd\xi .
$$

For $f = \mathbf 1_{\intcc{-1/2}{1/2}}$: $\hat f(\xi) =
\frac{2\sin(\xi/2)}\xi = \frac{\sin(\xi/2)}{\xi/2}$, and at $x = 0$:

$$
1 = \norm f_2^2 = \frac1{2\pi}\int_\R
\Bigl(\frac{\sin(\xi/2)}{\xi/2}\Bigr)^2\dd\xi
= \frac1{2\pi}\cdot2\int_\R\Bigl(\frac{\sin u}u\Bigr)^2\dd u
$$

($\xi = 2u$), i.e. $\int_\R\bigl(\frac{\sin u}u\bigr)^2\dd
u = \pi$ — Plancherel’s favorite integral, recovered by autocorrelation.

## 14.6 Problem: the heat equation on the line

**Problem 14.1.**

Weekend problem — $\partial_tu =
\partial^2_{xx}u$, solved end to end

Heat spreads; the equation $\partial_tu = \partial_{xx}^2u$ says its [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) diffuses at a rate given by the local curvature of the temperature profile. We solve the Cauchy problem on $\R$ — given $f$, find $u(t, x)$ for $t > 0$ with $u(0, \cdot) = f$ — prove the solution’s remarkable properties, and see why time cannot be reversed. Throughout, $g_t(x) = \frac1{2\sqrt{\pi t}}\eu^{-x^2/4t}$ is the *heat kernel* and $u(t, \cdot) = g_t *
f$.

**Part I — Deriving the kernel.** Work formally first: suppose $u(t, \cdot) \in \mathcal S$ solves the equation, and let $\hat u(t, \xi)$ be the transform in $x$.

1. Show (formally) $\partial_t\hat u = -\xi^2\hat u$ , hence $\hat u(t,\xi) = \eu^{-t\xi^2}\hat f(\xi)$ , and recognize $u(t) = g_t * f$ ( [Exercise 14.4](#exo-b3-fouriertransform-4) ). This motivates the *definition* of $u$ ; everything is now proved directly, for $f \in \mathcal C_b(\R)$ (bounded [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ) or $f \in L^p$ .

**Part II — Verification.**

2. Show that for $t > 0$ , $u(t, x) = \int  g_t(x-y)f(y)\dd y$ is well defined for $f \in \mathcal  C_b$ , and that $u$ is $\mathcal C^\infty$ in $(t, x)$ on $\intoo0\infty\times\R$ *(differentiate under the integral; dominate Gaussian derivatives locally uniformly in $(t,x)$)* .
3. Verify $\partial_tg_t = \partial^2_{xx}g_t$ by direct computation, and deduce $\partial_tu =  \partial^2_{xx}u$ for $t > 0$ .
4. (Initial condition) Show that for $f \in \mathcal  C_b$ , $u(t, x) \to f(x)$ as $t \to 0^+$ , uniformly on [compacts](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) *(approximate identity: split $\abs y  \leq \delta$, $\abs y > \delta$)* ; for $f \in L^p$ ( $p < \infty$ ), show $\norm{u(t) - f}_p \to 0$ .
5. (Instant smoothing) Conclude: even for merely bounded [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$ , the solution is $\mathcal C^\infty$ for every $t > 0$ — heat instantly erases roughness. Compute $u(t, \cdot)$ explicitly for $f =  \mathbf 1_{\intoo0\infty}$ (an error function) and sketch its profile for three values of $t$ .

**Part III — Structural properties.**

6. (Positivity and comparison) If $f \geq 0$ then $u > 0$ for all $t > 0$ (strictly, unless $f = 0$ a.e.); if $f_1 \leq f_2$ then $u_1 \leq u_2$ . A cold spot warms instantly: comment.
7. (Conservation) For $f \in L^1$ : $\int u(t, x)\dd x =  \int f$ for all $t$ *(Tonelli)* — total heat is conserved.
8. (Dissipation) For $f \in L^1\cap L^2$ , show via Plancherel that $t \mapsto \norm{u(t)}_2$ is nonincreasing, strictly unless $f = 0$ , and compute its limit as $t \to \infty$ . Show moreover $\norm{u(t)}_\infty \leq \frac{\norm  f_1}{2\sqrt{\pi t}} \to 0$ : heat spreads and dies.
9. (Uniqueness, $L^2$ class) Let $u$ be a solution with $u(t) \in L^2$ for all $t$ , $u \in \mathcal  C^1(\intoo0\infty, L^2)$ in the natural sense and $u(t) \to f$ in $L^2$ as $t\to0$ ; admitting that the transform converts it to $\partial_t\hat u =  -\xi^2\hat u$ pointwise a.e. in $\xi$ for a.e. $t$ *(justified by testing against $\mathcal  C_c^\infty$ in $\xi$ — outline this)* , show $\hat u(t,\xi) = \eu^{-t\xi^2}\hat f(\xi)$ , hence uniqueness in this class.

**Part IV — The arrow of time.**

10. Show that the *backward* problem is ill-posed: for the solution to exist at time $-s$ ( $s > 0$ ) with data $f$ at time $0$ — i.e. for $f = g_s * h$ to have a solution $h \in L^2$ — it is necessary that $\eu^{s\xi^2}\hat f(\xi) \in L^2$ : an extreme decay condition on $\hat f$ . Exhibit an explicit smooth $f  \in L^2$ for which no backward solution exists on any time interval: take the function with $\hat f(\xi) =  \eu^{-\abs\xi}$ — identify $f$ ( [Exercise 14.1](#exo-b3-fouriertransform-1) ) and show $\eu^{s\xi^2}\eu^{-\abs\xi} \notin L^2$ for every $s >  0$ .
11. (Smoothing vs. information) Explain in a short paragraph, using questions 5, 9 and 10, why the heat semigroup $(f \mapsto g_t * f)_{t\geq0}$ is injective but not surjective on $L^2$ , and why this expresses the irreversibility of diffusion.

**Part V — Shannon’s sampling theorem.** A function $f \in L^2(\R)$ is *band-limited* to $\Omega$ if $\hat f = 0$ a.e. outside $\intcc{-\Omega}\Omega$; write $PW_\Omega$ (Paley–Wiener space) for these.

12. Show that every $f \in PW_\Omega$ agrees a.e. with the $\mathcal C^\infty$ function $\frac1{2\pi}\int_{-\Omega}^{\Omega}\hat  f(\xi)\eu^{\iu x\xi}\,\dd\xi$ (justify smoothness and the identification), with all derivatives bounded: band limitation is an extreme form of regularity. From now on $f$ denotes this representative.
13. Expand $\hat f \in L^2(\intcc{-\Omega}\Omega)$ in the Fourier basis of that interval and identify the coefficients as *samples* of $f$: $$\hat f(\xi) = \frac\pi\Omega\sum_{n\in\Z}  f\Bigl(\frac{n\pi}\Omega\Bigr)\,  \eu^{-\iu n\pi\xi/\Omega}  \quad\text{in } L^2(\intcc{-\Omega}\Omega) .$$
14. Deduce the *sampling theorem*: for $f \in  PW_\Omega$, $$f(x) = \sum_{n\in\Z}f\Bigl(\frac{n\pi}  \Omega\Bigr)\,\operatorname{sinc}(\Omega x - n\pi),  \qquad \operatorname{sinc}t = \frac{\sin t}t,$$ with convergence in $L^2(\R)$ and uniformly on $\R$ *(inject the series of question 13 into the inversion formula and compute the elementary integral)*: a band-limited signal is entirely determined by its values on a grid of step $\pi/\Omega$ — the Nyquist rate.
15. Show that the functions $x \mapsto  \operatorname{sinc}(\Omega x - n\pi)$ , $n \in \Z$ , form an orthogonal family in $L^2(\R)$ with constant norm $\sqrt{\pi/\Omega}$ , and deduce the energy identity $\norm f_2^2 =  \frac\pi\Omega\sum_n\abs{f(n\pi/\Omega)}^2$ .
16. (Aliasing) Exhibit a nonzero $g \in PW_{2\Omega}$ vanishing at every sample point $\frac{n\pi}\Omega$ *(consider $g(x) =  \sin(\Omega x)\operatorname{sinc}(\Omega x)$ and check its band)* : sampling below the Nyquist rate loses information — two different signals can share all samples: the stroboscopic wagon-wheel effect, mathematized.
17. (Degrees of freedom) Using questions 14–15, justify the engineering rule: a signal band-limited to $\Omega$ whose energy is essentially carried by a time window of length $T$ is described by approximately $\frac{\Omega T}\pi$ real samples — make “essentially” precise through the energy identity and the tail $\sum_{\abs{n\pi/\Omega} >  T/2}$ .
18. (Consistency checks) Verify the sampling theorem by hand on two members of $PW_\Omega$ : (a) $f =  \operatorname{sinc}(\Omega\,\cdot)$ , whose samples are $\delta_{n0}$ ; (b) $f(x) = \cos(\omega x)  \operatorname{sinc}(\varepsilon x)$ -type narrowband signals — more precisely, show that for $f \in  PW_{\Omega'}$ with $\Omega' < \Omega$ , the $\Omega$ -rate series also reconstructs $f$ (oversampling is harmless), by embedding $PW_  {\Omega'} \subseteq PW_\Omega$ .

**Part VI — Uncertainty, twice more.** Heisenberg’s inequality ([Exercise 14.8](#exo-b3-fouriertransform-8)) bounds *how* concentrated $f$ and $\hat f$ can jointly be; here are its all-or-nothing sibling and its exact saturation.

19. Let $f \in L^1$ with $\operatorname{supp}f \subseteq  \intcc{-A}A$. Show that $\hat f$ is the sum of an everywhere-convergent power series: $$\hat f(\xi) = \sum_{k\geq0}\frac{(-\iu\xi)^k}{k!}  \,m_k, \qquad m_k = \int_{-A}^{A}x^kf(x)\,\dd x,  \quad \abs{m_k} \leq A^k\norm f_1$$ *(expand $\eu^{-\iu\xi x}$ and justify the interchange by normal convergence)*: the transform is *real-analytic*, with infinite radius of convergence at every point.
20. Deduce the support dichotomy: a real-analytic function vanishing on a nonempty open interval vanishes identically *(the set where all derivatives vanish is open and closed — work out the Taylor argument)* ; conclude that no nonzero $f$ has both $f$ and $\hat f$ compactly supported, and that $PW_\Omega$ contains no nonzero compactly supported function — band-limited signals last forever, and time-limited signals leak into all frequencies.
21. (Saturation of Heisenberg) On the Gaussian family $f  = \eu^{-ax^2}$ , compute both concentration factors and verify that the normalized product $\bigl(\int x^2\abs f^2\bigr)\bigl(\frac1{2\pi}\int  \xi^2\abs{\hat f}^2\bigr)\big/\norm f_2^4$ equals $\frac14$ for *every* $a$ — the equality family of [Exercise 14.8](#exo-b3-fouriertransform-8) in the flesh; explain by a scaling argument why the product must be constant along the family.
22. Explain the physical reading (position/momentum densities of a quantum state; $\hbar$ in the normalization gives $\sigma_x\sigma_p \geq  \frac\hbar2$ ), and connect across the chapter: instant smoothing (Part II), irreversibility (Part IV), sampling (Part V), Heisenberg and the support dichotomy are five expressions of one law — the behavior of $\hat f$ at infinity legislates what $f$ may do anywhere.

**Part VII — The kernel’s algebra, and one [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) example.**

23. (Semigroup) Prove the *Chapman–Kolmogorov identity* $g_t * g_s = g_{t+s}$ for $t, s > 0$ (via the convolution theorem and injectivity of the transform on $L^1$), and deduce $u(t + s) = g_s *  u(t)$: evolving for time $t + s$ is evolving for $t$, then for $s$. Sharpen the dissipation of question 8: writing $\norm{u(t)}_2^2 = \frac1{2\pi}\int  \eu^{-2t\xi^2}\abs{\hat f(\xi)}^2\dd\xi$, show by Cauchy–Schwarz that $$t \longmapsto \ln\,\norm{u(t)}_2  \quad\text{is convex on } \intoo0{+\infty} :$$ the $L^2$ energy does not merely decrease, it decreases in a log-convex way.
24. (Where the heat goes) Let $f \geq 0$, $f \in L^1$, with $\int x^2f(x)\dd x < \infty$. Show that for all $t > 0$ $$\int_\R x\,u(t, x)\,\dd x = \int_\R x f(x)\,\dd x,  \qquad  \int_\R x^2u(t, x)\,\dd x  = \int_\R x^2f(x)\,\dd x + 2t\int_\R f :$$ the [center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) of heat never moves, and the variance grows *linearly* in time — the diffusive scaling $x \sim \sqrt{2t}$, to be reread when Brownian motion appears in [Chapter 22](https://one-course.com/books/math/5/en/chapter/22-probability-foundations-and-the-law-of-large-numbers#ch-b3-probability). *(Compute the first two moments of $g_t$ and use Tonelli on the convolution.)*
25. (The Gaussian, solved end to end) For $f(x) =  \eu^{-x^2}$, establish the closed form $$u(t, x) = \frac1{\sqrt{1 + 4t}}\,  \exp\Bigl(-\frac{x^2}{1 + 4t}\Bigr),$$ and verify on it, by hand: the equation $\partial_tu  = \partial^2_{xx}u$; conservation $\int u(t) =  \sqrt\pi$; the dissipation law $\norm{u(t)}_2 =  (\pi/2)^{1/4}(1 + 4t)^{-1/4}$ (compare its $t^{-1/4}$ decay with the $t^{-1/2}$ sup-norm decay of question 8); and the exact variance growth of question 24. At $t = 6$: the peak has dropped to $\frac15$ of its initial height while the profile is five times wider — same heat, spread out.

**Solution of Problem 14.1.**

**1.** Transforming the equation in $x$ (formally): $\partial_t\hat u(t,\xi) = \widehat{\partial^2_{xx}u} =
(\iu\xi)^2\hat u = -\xi^2\hat u$, an ODE in $t$ for each frequency: $\hat u(t,\xi) = \eu^{-t\xi^2}\hat f(\xi)$. Since $\eu^{-t\xi^2} = \hat g_t$ ([Exercise 14.4](#exo-b3-fouriertransform-4)), the product is the transform of $g_t * f$.

**2.** $\abs{u(t,x)} \leq \norm f_\infty\int g_t = \norm
f_\infty$: well defined. On $[t_0, T]\times[-A, A]$: each mixed derivative $\partial^m_t\partial^n_xg_t(x - y)$ is a polynomial in $(x - y)$ and $t^{-1}$ times $\eu^{-(x-y)^2/4t}$, bounded for $\abs y \geq 2A$ by $C\,(1 + y^2)^N\eu^{-(\abs y - A)^2/4T}$, an [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) dominator independent of $(t, x)$ in the window (and bounded for $\abs y \leq 2A$): repeated differentiation under the integral ([Theorem 10.15](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#thm-b3-lebesgue-paramdiff)) applies: $u \in
\mathcal C^\infty(\intoo0\infty\times\R)$.

**3.** With $g_t(x) = \frac1{2\sqrt{\pi
t}}\eu^{-x^2/4t}$:

$$
\partial_tg_t = g_t\Bigl(\frac{x^2}{4t^2} - \frac1{2t}\Bigr)
= \partial^2_{xx}g_t
$$

(differentiate twice in $x$: $\partial_xg_t = -\frac
x{2t}g_t$, $\partial^2_{xx}g_t = \bigl(\frac{x^2}{4t^2} -
\frac1{2t}\bigr)g_t$). By question 2 the derivatives pass under the integral: $\partial_tu = \partial^2_{xx}u$.

**4.** $u(t,x) - f(x) = \int g_t(y)\bigl(f(x - y) -
f(x)\bigr)\dd y$. Given a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K$ and $\varepsilon$: uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $f$ on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $K$ gives $\delta$ with $\abs{f(x-y) - f(x)} < \varepsilon$ for $x \in
K$, $\abs y \leq \delta$; the tail contributes $\leq 2\norm
f_\infty\int_{\abs y > \delta}g_t(y)\dd y = 2\norm
f_\infty\,\P$-mass beyond $\delta$, which is $\frac2{\sqrt\pi}\int_{\delta/2\sqrt t}^\infty\eu^{-z^2}\dd z
\to 0$ as $t \to 0$. For $f \in L^p$: $\norm{u(t) - f}_p \leq
\int g_t(y)\norm{\tau_yf - f}_p\dd y$ (Minkowski/Jensen as in [Theorem 12.9](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-regularization)), split the same way using [Theorem 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-density)(3).

**5.** Instant smoothing is question 2 ($u(t)$ is $\mathcal C^\infty$ for $t > 0$ with no smoothness of $f$ used). For $f = \mathbf 1_{\intoo0\infty}$:

$$
u(t, x) = \int_0^\infty g_t(x - y)\dd y
= \frac1{\sqrt\pi}\int_{-x/2\sqrt t}^{\infty}\eu^{-z^2}\dd z
= \frac12\Bigl(1 +
\operatorname{erf}\Bigl(\frac{x}{2\sqrt t}\Bigr)\Bigr),
\qquad \operatorname{erf}(s) =
\frac2{\sqrt\pi}\int_0^s\eu^{-z^2}\dd z :
$$

a smoothed step whose transition zone widens like $\sqrt t$ (profiles at $t_1 < t_2 < t_3$: ever flatter ramps through $(0, \frac12)$).

**6.** The integrand $g_t(x-y)f(y)$ is $\geq 0$ and the kernel is strictly positive: $u(t,x) = 0$ would force $f = 0$ a.e. Monotonicity in $f$ is monotonicity of the integral. A spot where $f = 0$ on an interval still has $u(t, \cdot) > 0$ there for every $t > 0$: heat propagates at infinite speed (any positivity anywhere is felt everywhere instantly).

**7.** Tonelli ($g_t(x-y)\abs{f(y)}$ is [integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) on $\R^2$): $\int u(t,x)\dd x = \int f(y)\bigl(\int g_t(x -
y)\dd x\bigr)\dd y = \int f$.

**8.** Plancherel: $2\pi\norm{u(t)}_2^2 =
\int\eu^{-2t\xi^2}\abs{\hat f(\xi)}^2\dd\xi$, nonincreasing in $t$ (pointwise), strictly unless $\hat f = 0$ a.e. ($= f =
0$), with limit $0$ as $t\to\infty$ by DCT. And $\norm{u(t)}_\infty \leq \norm{g_t}_\infty\norm f_1 =
\frac{\norm f_1}{2\sqrt{\pi t}} \to 0$.

**9.** For $\varphi \in \mathcal C_c^\infty$, $t \mapsto
\langle\varphi, \hat u(t)\rangle$ is $\mathcal C^1$ with derivative $\langle\varphi, \partial_t\hat u\rangle =
\langle\varphi, \widehat{\partial_{xx}u}\rangle =
\langle\xi^2\varphi\dots\rangle$ — precisely, $\widehat{\partial^2_{xx}u} = -\xi^2\hat u$ transfers the equation. Then for a.e. $\xi$, the absolutely [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) function $t \mapsto \eu^{t\xi^2}\hat u(t,\xi)$ has derivative $\eu^{t\xi^2}(\xi^2\hat u + \partial_t\hat u) = 0$ in the integrated sense: it is constant, and letting $t \to 0$ ($\hat u(t) \to \hat f$ in $L^2$, a.e. along a subsequence): $\hat u(t, \xi) = \eu^{-t\xi^2}\hat f(\xi)$ a.e. Two solutions in the class have the same transform: they are equal.

**10.** $f = g_s * h$ with $h \in L^2$ forces $\hat f =
\eu^{-s\xi^2}\hat h$, i.e. $\hat h = \eu^{s\xi^2}\hat f \in
L^2$. Take $\hat f(\xi) = \eu^{-\abs\xi}$: then $f(x) =
\frac1\pi\cdot\frac1{1 + x^2}$ ([Exercise 14.1](#exo-b3-fouriertransform-1), inversion), a perfectly smooth $L^2$ function; but $\eu^{2s\xi^2 - 2\abs\xi} \to
\infty$: $\eu^{s\xi^2}\hat f \notin L^2$ for every $s > 0$. The Cauchy profile is *never* the result of prior diffusion.

**11.** The heat semigroup multiplies transforms by $\eu^{-t\xi^2}$, which vanishes nowhere: injective — formally, no information is destroyed. But its range consists of functions whose transforms decay like $\eu^{-t\xi^2}$: a tiny, dense-but-proper subspace of $L^2$ (question 10 shows even excellent functions lie outside). Inverting would amplify the frequency $\xi$ by $\eu^{t\xi^2}$: unbounded, hence unstable against any perturbation. Diffusion is irreversible not because the map forgets, but because its inverse cannot be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) — an arrow of time made of functional analysis.

**12.** $\hat f \in L^2(\intcc{-\Omega}\Omega)
\subseteq L^1$ (Cauchy–Schwarz on a bounded interval), so $F(x) = \frac1{2\pi}\int_{-\Omega}^\Omega\hat
f(\xi)\eu^{\iu x\xi}\dd\xi$ is defined everywhere, and differentiation under the integral (dominated by $\Omega^k\abs{\hat f} \in L^1$ on the band) makes it $\mathcal C^\infty$ with $\abs{F^{(k)}} \leq
\frac{\Omega^k}{2\pi}\norm{\hat f}_{L^1}$ everywhere. And $F
= f$ a.e.: both sides have the same transform, and the transform is injective on $L^2$ ([Theorem 14.8](#thm-b3-fouriertransform-plancherel) and its $L^2$ extension).

**13.** The exponentials $\xi \mapsto \eu^{-\iu
n\pi\xi/\Omega}$, $n \in \Z$, form a [Hilbert basis](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#def-b3-hilbert-onb) of $L^2(\intcc{-\Omega}\Omega)$ ([Theorem 13.9](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#thm-b3-hilbert-fourier), rescaled). The coefficient of $\hat f$ along the $n$-th one is

$$
\frac1{2\Omega}\int_{-\Omega}^\Omega\hat f(\xi)\,
\eu^{\iu n\pi\xi/\Omega}\dd\xi
= \frac{2\pi}{2\Omega}\cdot
\frac1{2\pi}\int_{-\Omega}^{\Omega}\hat f(\xi)\,
\eu^{\iu(n\pi/\Omega)\xi}\dd\xi
= \frac\pi\Omega\,f\Bigl(\frac{n\pi}\Omega\Bigr),
$$

by question 12’s formula at $x = \frac{n\pi}\Omega$: the stated expansion holds in $L^2$ of the band.

**14.** Insert the expansion into the inversion formula of question 12; the exchange of sum and integral is the [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of the $L^2$ pairing against $\frac1{2\pi}
\eu^{\iu x\xi}\mathbf 1_{\abs\xi\leq\Omega}$ (of $L^2$ norm $\frac{\sqrt{2\Omega}}{2\pi}$, independent of $x$ — whence uniformity):

$$
f(x) = \sum_nf\Bigl(\frac{n\pi}\Omega\Bigr)\cdot
\frac1{2\Omega}\int_{-\Omega}^\Omega
\eu^{\iu(x - n\pi/\Omega)\xi}\dd\xi
= \sum_nf\Bigl(\frac{n\pi}\Omega\Bigr)
\operatorname{sinc}(\Omega x - n\pi),
$$

since $\frac1{2\Omega}\int_{-\Omega}^\Omega\eu^{\iu
u\xi}\dd\xi = \frac{\sin(\Omega u)}{\Omega u}$.

**15.** Reading question 14’s computation backwards, the transform of $s_n = \operatorname{sinc}(\Omega\cdot -
n\pi)$ is $\hat s_n = \frac\pi\Omega\,\eu^{-\iu
n\pi\xi/\Omega}\,\mathbf 1_{\intcc{-\Omega}\Omega}$. Plancherel:

$$
\langle s_n, s_m\rangle = \frac1{2\pi}
\Bigl(\frac\pi\Omega\Bigr)^2\int_{-\Omega}^\Omega
\eu^{\iu(n-m)\pi\xi/\Omega}\dd\xi =
\frac\pi\Omega\,\delta_{nm} :
$$

an orthogonal family of constant norm $\sqrt{\pi/\Omega}$. Taking norms in question 14’s expansion: $\norm f_2^2 =
\frac\pi\Omega\sum_n\abs{f(n\pi/\Omega)}^2$.

**16.** $g(x) = \sin(\Omega x)\operatorname{sinc}
(\Omega x) = \frac{\sin^2(\Omega x)}{\Omega x}$ vanishes at every grid point $\frac{n\pi}\Omega$ (including $0$, by the limit) and is not identically zero. Its band: write $g =
\frac1{2\iu}\bigl(\eu^{\iu\Omega x} - \eu^{-\iu\Omega
x}\bigr)\operatorname{sinc}(\Omega x)$; modulation by $\eu^{\pm\iu\Omega x}$ shifts the transform by $\mp\Omega$, so $\hat g$ is supported in $\intcc{-2\Omega}{2\Omega}$ (indeed in the union of two shifted bands): $g \in PW_{2\Omega}$, invisible to $\Omega$-rate sampling — aliasing incarnate.

**17.** By question 15 the samples carry the energy democratically: $\norm f^2 = \frac\pi\Omega\sum
\abs{f(n\pi/\Omega)}^2$. If the signal’s energy outside the time window $\intcc{-T/2}{T/2}$ is $\leq \varepsilon^2\norm
f^2$, the samples outside the window satisfy (up to boundary terms controlled by the uniform bound of question 12) $\frac\pi\Omega\sum_{\abs{n\pi/\Omega} > T/2}
\abs{f(n\pi/\Omega)}^2 \approx \norm{f\,\mathbf 1_{\abs x >
T/2}}^2 \leq \varepsilon^2\norm f^2$: truncating the sampling series to the $\approx \frac{\Omega T}\pi$ in-window indices reconstructs $f$ up to relative error $\approx\varepsilon$. Hence the time–bandwidth product $\frac{\Omega T}{\pi}$ counts the effective real degrees of freedom of the signal — the rule behind every audio format.

**18.** (a) $\operatorname{sinc}(\Omega x)$ has samples $f(n\pi/\Omega) = \operatorname{sinc}(n\pi) =
\delta_{n0}$: the series reduces to its $n = 0$ term, $\operatorname{sinc}(\Omega x)$ — the theorem reproduces its own kernel. (b) If $\hat f$ is supported in $\intcc{-\Omega'}{\Omega'} \subseteq
\intcc{-\Omega}\Omega$, every step of questions 13–14 runs verbatim with the larger band $\Omega$ (the expansion of $\hat f$ on the bigger interval is still legitimate): sampling faster than one’s own Nyquist rate changes nothing in the reconstruction — oversampling is harmless, and in practice beneficial (faster-decaying reconstruction kernels can then be used).

**19.** Expand $\eu^{-\iu\xi x} =
\sum_k\frac{(-\iu\xi x)^k}{k!}$ inside the integral; on $\intcc{-A}A$ the series converges normally ($\sum_k\frac{\abs{\xi}^kA^k}{k!}\abs f \in L^1$), so integration term by term is legitimate:

$$
\hat f(\xi) = \sum_{k\geq0}\frac{(-\iu\xi)^k}{k!}m_k,
\qquad \abs{m_k} \leq A^k\norm f_1 .
$$

The bound makes the series converge for every complex $\xi$; around any point $\xi_0$, regrouping (absolute convergence) gives a power series in $\xi - \xi_0$: $\hat f$ is real-analytic with infinite radius everywhere.

**20.** Let $g$ be real-analytic on $\R$ (Taylor series converging to $g$ near each point) and $Z = \{\xi :
g^{(k)}(\xi) = 0\ \forall k\}$. $Z$ is closed (intersection of closed sets); it is open, since at $\xi_0 \in Z$ the local Taylor expansion of $g$ is the zero series, so $g$ vanishes identically near $\xi_0$, together with all derivatives. If $g$ vanishes on an interval, $Z \neq \varnothing$; by [connectedness](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) of $\R$, $Z = \R$: $g \equiv 0$. Now if $f
\neq 0$ had [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) support together with $\hat f$: question 19 makes $\hat f$ real-analytic, vanishing outside a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact), hence on intervals: $\hat f \equiv 0$, so $f = 0$ a.e. by injectivity — contradiction. Likewise a nonzero $f \in
PW_\Omega$ cannot be compactly supported (swap the roles of $f$ and $\hat f$ via inversion): band-limited signals never die, time-limited signals occupy unbounded spectrum.

**21.** For $f = \eu^{-ax^2}$: $\norm f_2^2 =
\sqrt{\frac\pi{2a}}$ and $\int x^2\abs f^2 =
\frac1{4a}\sqrt{\frac{\pi}{2a}}$ (Gaussian second moment); $\hat f = \sqrt{\frac\pi a}\,\eu^{-\xi^2/4a}$ ([Example 14.2](#ex-b3-fouriertransform-gaussian)) and

$$
\frac1{2\pi}\int\xi^2\abs{\hat f}^2\dd\xi =
\frac1{2\pi}\cdot\frac\pi a\int\xi^2
\eu^{-\xi^2/2a}\dd\xi = \frac1{2a}\cdot a\sqrt{2\pi a}
= \frac{\sqrt{2\pi a}}2 .
$$

Normalized product: $\frac1{4a}\sqrt{\frac\pi{2a}}\cdot
\frac{\sqrt{2\pi a}}2\big/\frac{\pi}{2a} = \frac14$, independent of $a$. Scaling explains the constancy: replacing $f$ by $f(\lambda\cdot)$ multiplies $\int x^2\abs f^2/\norm
f^2$ by $\lambda^{-2}$ and $\frac1{2\pi}\int\xi^2\abs{\hat
f}^2/\norm f^2$ by $\lambda^{2}$: the product is a dilation invariant, and the Gaussians form one dilation [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action).

**22.** With $\abs f^2$ the position [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) and $\frac1{2\pi}\abs{\hat f}^2$ the momentum [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) of a quantum state (physical units insert $\hbar$), [Exercise 14.8](#exo-b3-fouriertransform-8) reads $\sigma_x\sigma_p
\geq \frac\hbar2$: no state is sharp in both observables. Across the chapter, one law wears five suits: heat instantly smooths because $\eu^{-t\xi^2}$ annihilates high frequencies (Part II); the flow cannot run backward because restoring them is unbounded (Part IV); a band-limited signal is rigid enough to live on a countable grid (Part V); no function beats Heisenberg’s floor; and no function is compactly supported on both sides of the transform (questions 19–20). What $\hat f$ does at infinity governs what $f$ may do anywhere.

**23.** Both $g_t$ and $g_s$ are in $L^1$ with $\widehat{g_t}(\xi) = \eu^{-t\xi^2}$ (question 1’s computation), so the convolution theorem gives $\widehat{g_t * g_s} = \eu^{-t\xi^2}\eu^{-s\xi^2} =
\eu^{-(t+s)\xi^2} = \widehat{g_{t+s}}$; two $L^1$ functions with the same transform agree a.e. (injectivity, via the inversion theorem — here both sides are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), so they agree everywhere): $g_t * g_s = g_{t+s}$. Consequently $u(t + s) = g_{t+s} * f = g_s * (g_t * f) = g_s * u(t)$ (associativity of convolution, Tonelli). Log-convexity: let $N(t) = \norm{u(t)}_2^2 = \frac1{2\pi}\int
\eu^{-2t\xi^2}\abs{\hat f}^2\dd\xi$ (Plancherel, question 8). For $t = \frac{t_1 + t_2}2$, write

$$
\eu^{-2t\xi^2}\abs{\hat f}^2
= \Bigl(\eu^{-2t_1\xi^2}\abs{\hat f}^2\Bigr)^{1/2}
\Bigl(\eu^{-2t_2\xi^2}\abs{\hat f}^2\Bigr)^{1/2},
$$

and Cauchy–Schwarz gives $N\bigl(\frac{t_1+t_2}2\bigr) \leq
\sqrt{N(t_1)\,N(t_2)}$: $\ln N$ is midpoint-convex, and being [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (dominated convergence in $t$), convex; so is $\ln\norm{u(t)}_2 = \frac12\ln N(t)$. Decay with a convex logarithm: the heat flow cannot lose energy in a burst and then stall.

**24.** The kernel’s moments: $\int g_t = 1$ (question 7 with $f = g_s$, or directly the Gaussian integral), $\int
x\,g_t(x)\dd x = 0$ (odd integrand), and, substituting $x =
2\sqrt t\,v$,

$$
\int_\R x^2g_t(x)\,\dd x
= \frac{4t}{\sqrt\pi}\int_\R v^2\eu^{-v^2}\dd v = 2t .
$$

Substituting $x = z + y$ in the convolution and noting $\iint(\abs z + \abs y)^2g_t(z)f(y)\,\dd z\,\dd y < \infty$ (each of $\int\abs z^kg_t$, $\int\abs y^kf$ is finite for $k
\leq 2$, using $\abs y \leq \frac{1 + y^2}2$), Fubini and Tonelli apply to the moment integrals below:

$$
\int x\,u(t,x)\dd x = \iint (z + y)\,g_t(z)f(y)\,\dd z\,\dd
y = 0\cdot\!\int\! f + 1\cdot\!\int\! yf(y)\dd y,
$$

which is the first claim; and

$$
\iint (z+y)^2g_t(z)f(y)\,\dd z\,\dd y
= 2t\int f + 2\cdot0\cdot\!\int\! yf + \int y^2f(y)\dd y ,
$$

the second. Means add, variances add, and the kernel contributes mean $0$ and variance $2t$: after time $t$ the heat has spread over a width of order $\sqrt{2t}$ — distance grows like the square root of time, the signature of diffusion (and of the Brownian [paths](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-pathconnected) of [Chapter 22](https://one-course.com/books/math/5/en/chapter/22-probability-foundations-and-the-law-of-large-numbers#ch-b3-probability)).

**25.** Transform side: $\hat f(\xi) =
\sqrt\pi\,\eu^{-\xi^2/4}$, so $\hat u(t,\xi) =
\sqrt\pi\,\eu^{-(t + \frac14)\xi^2}$, which is the transform of $(1 + 4t)^{-1/2}\exp\bigl(-x^2/(1+4t)\bigr)$ (the Gaussian dictionary $\eu^{-ax^2} \mapsto
\sqrt{\pi/a}\,\eu^{-\xi^2/4a}$ with $a = \frac1{1+4t}$): the closed form. Direct check, with $\sigma = 1 + 4t$:

$$
\partial_tu = \sigma^{-1/2}\eu^{-x^2/\sigma}
\Bigl(-\frac2\sigma + \frac{4x^2}{\sigma^2}\Bigr)
= \partial^2_{xx}u ,
$$

both sides computed from $\partial_xu =
-\frac{2x}\sigma\,u$. Conservation: $\int u(t) =
\sigma^{-1/2}\sqrt{\pi\sigma} = \sqrt\pi$ for all $t$. Dissipation:

$$
\norm{u(t)}_2^2 = \frac1\sigma\int\eu^{-2x^2/\sigma}\dd x
= \frac1\sigma\sqrt{\frac{\pi\sigma}2}
= \sqrt{\frac\pi2}\,(1+4t)^{-1/2},
$$

so $\norm{u(t)}_2 = (\pi/2)^{1/4}(1+4t)^{-1/4}$, nonincreasing with convex logarithm (question 23); the $L^2$ norm decays like $t^{-1/4}$, exactly half the $t^{-1/2}$ exponent of $\norm{u(t)}_\infty$ — consistent with $\norm u_2^2 \leq \norm u_\infty\norm u_1$ and conservation of $\norm u_1$. Variance: $\int x^2u(t) =
\frac1\sigma\cdot\frac{\sigma^{3/2}\sqrt\pi}2 =
\frac{\sqrt\pi}2\,(1 + 4t) = \int x^2f + 2t\sqrt\pi$, as question 24 predicts ($\int x^2f = \frac{\sqrt\pi}2$, $\int
f = \sqrt\pi$). At $t = 6$: $\sigma = 25$, peak height $u(6, 0) = \frac15$ versus $u(0,0) = 1$, width scale $\sqrt\sigma = 5$ times the initial one, and $\int u =
\sqrt\pi \approx 1.7725$ throughout: the spot is five times lower, five times wider, and not a calorie is missing.