Fourier Analysis

InteractiveEtingof et al. §3 (the abelian case); Grafakos, Classical Fourier Analysis, §3.1

Abelian groups and the Fourier transform

For an abelian group every irreducible representation is 1-dimensional, and breaking the space of functions on the group into those pieces is exactly the Fourier transform. The discrete Fourier transform, Fourier series and the Fourier integral are the same theorem for three different groups.

The idea

Characters of an abelian group. A character of a finite abelian group

G

is a homomorphism

\chi:G\to\mathbb C^\times

; by Exercise 2.2 of the representation theory page these are all its irreducible representations. They form a group

\hat G

under pointwise multiplication, the dual group, and

\hat G\cong G

(Exercise 4.4). For

G=\mathbb Z/n

,

\chi_k(x)=e^{2\pi ikx/n}

,

k=0,\dots,n-1

.

Fourier transform and inversion. For

f:G\to\mathbb C

put

\hat f(\chi)=\frac1{|G|}\sum_xf(x)\overline{\chi(x)}

. Since the characters are an orthonormal basis of functions on

G

for

\langle f,g\rangle=\frac1{|G|}\sum f\bar g

(Exercise 4.1),

f=\sum_{\chi\in\hat G}\hat f(\chi)\,\chi,\qquad \frac1{|G|}\sum_x|f(x)|^2=\sum_\chi|\hat f(\chi)|^2\quad\text{(Plancherel)}.

In the language of representations: functions on

G

are the regular representation, and

\hat f(\chi)\chi

is the component of

f

in the copy of

\chi

.

Convolution becomes multiplication. With

(f*g)(x)=\sum_yf(y)g(x-y)

, one has

\widehat{f*g}=|G|\,\hat f\,\hat g

(Exercise 4.2). Translation-invariant linear operators (convolutions, circulant matrices) are intertwiners of the regular representation, so by Schur's lemma each acts on each character by a scalar: the Fourier transform diagonalizes all of them at once.

Three groups, one theorem. For the circle

\mathbb T=\mathbb R/2\pi\mathbb Z

the continuous characters are

e^{ik\theta}

,

k\in\mathbb Z

, and the transform is the sequence of Fourier coefficients (Grafakos §3.1). For

\mathbb R

they are

e^{2\pi ix\xi}

,

\xi\in\mathbb R

, and the transform is the Fourier integral. Sums become integrals against Haar measure; the algebra is the same. For non-abelian compact groups the irreducibles are no longer 1-dimensional, and the Peter–Weyl theorem replaces characters by matrix entries of irreducible representations.

Explore: a signal and its Fourier transform on $\mathbb Z/n$

Signal f

n Then convolve with a box of width

f

(drag the bars) , and

f*b

for a box

b

of the chosen width

|\hat f(\chi_k)|

and

|\widehat{f*b}(\chi_k)|=n|\hat f||\hat b|

, for

k=0,\dots,n-1

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Dual group: The collection of all "pure frequencies" of an abelian group (homomorphisms to the unit circle), which is itself a group under pointwise multiplication.
Fourier transform: Writing a function on a group as a combination of pure frequencies, and recording how much of each frequency it contains.
Plancherel identity: The total energy of a signal equals the total energy of its frequency components (with the right normalisation): nothing is lost or gained by changing to frequencies.
Convolution: Sliding one function along another and adding up the products: a weighted moving average. In frequencies it becomes plain multiplication.
Circulant matrix: A square matrix in which each row is the previous row shifted one place to the right, wrapping around. All of them share the same eigenvectors, the pure frequencies.
Peter–Weyl theorem: The version of Fourier analysis for compact groups that need not be abelian: functions are expanded using the matrix entries of irreducible representations instead of single frequencies.

Examples

Example 4.1 (

\mathbb Z/2

).

The characters are $\chi_0=(1,1)$ and $\chi_1=(1,-1)$ , so $\hat f(\chi_0)=\frac{f(0)+f(1)}2$ and $\hat f(\chi_1)=\frac{f(0)-f(1)}2$ : the average and half the difference: the splitting of a representation of $\mathbb Z/2$ into the parts where the generator acts by $+1$ and by $-1$ .

Example 4.2 (the DFT matrix).

For $\mathbb Z/n$ the matrix $F_{kx}=\frac1n\omega^{-kx}$ , $\omega=e^{2\pi i/n}$ , sends $f$ to $\hat f$ . Orthogonality of characters says $\sqrt nF$ is unitary. The fast Fourier transform computes it in $O(n\log n)$ steps instead of $O(n^2)$ , by splitting $\mathbb Z/2m$ into even and odd residues.

Example 4.3 (a spike and a constant).

The spike $\delta_0$ has $\hat\delta_0(\chi)=\frac1{|G|}$ for every $\chi$ : it contains every frequency equally. The constant 1 has $\hat1=\delta_{\chi_0}$ . Concentrated on one side means spread out on the other (Exercise 4.7). Try "spike" in the figure.

Example 4.4 (moving averages).

Convolving with a box $b=\frac1w\mathbf 1_{\{0,\dots,w-1\}}$ replaces $f$ by its average over a window. Its transform $n\hat b(\chi_k)=\frac1w\sum_{j<w}\omega^{-jk}$ is close to 1 for small $k$ and small for most large $k$ , and vanishes exactly at the $k\ne0$ with $n\mid kw$ : averaging keeps low frequencies and damps high ones.

Example 4.5 (Boolean functions).

For $G=(\mathbb Z/2)^n$ , written multiplicatively as $\{\pm1\}^n$ , the characters are $\chi_S(x)=\prod_{i\in S}x_i$ for $S\subseteq\{1,\dots,n\}$ . The transform is the Walsh–Hadamard transform, the basic tool of the analysis of Boolean functions. For example $\mathrm{Maj}(x_1,x_2,x_3)=\frac12(x_1+x_2+x_3)-\frac12x_1x_2x_3$ .

Example 4.6 (Fourier series).

On the circle, $\hat f(k)=\frac1{2\pi}\int_0^{2\pi}f(\theta)e^{-ik\theta}\,d\theta$ , and Plancherel reads $\frac1{2\pi}\int|f|^2=\sum_k|\hat f(k)|^2$ . The finite theory has no convergence questions; on the circle, whether and in what sense $\sum\hat f(k)e^{ik\theta}$ converges back to $f$ is the subject of Grafakos Chapters 3 and 4.

Check your understanding

Why does the discrete Fourier transform diagonalize every circulant matrix?

Circulant matrices commute with translation, i.e. they intertwine the regular representation of $\mathbb Z/n$ , and each character spans its own 1-dimensional piece, so each is an eigenvector.
Because circulant matrices are symmetric.Most are not, e.g. the shift itself.
Because the DFT matrix is unitary.Unitarity alone doesn't make it diagonalize anything in particular.

The pieces of the regular representation of an abelian group are distinct 1-dimensional representations, so an intertwiner must map each into itself, acting by a scalar (Exercise 4.5).

How many characters does $\mathbb Z/2\times\mathbb Z/4$ have?

6That is $2+4$ ; characters of a product are products of characters.
8
4An abelian group has as many characters as elements.

Every element is its own conjugacy class, so there are

|G|=8

irreducibles, all 1-dimensional:

\chi(a,b)=(-1)^{ja}i^{kb}

.

Exercises

Exercise 4.1Characters are an orthonormal basis

For a finite abelian group, show directly that distinct characters are orthonormal for $\langle f,g\rangle=\frac1{|G|}\sum f\bar g$ , that there are $|G|$ of them, and hence that Fourier inversion holds.

Hint

If $\chi\ne\psi$ , then $\chi\bar\psi$ is a nontrivial character $\eta$ ; pick $a$ with $\eta(a)\ne1$ and compare $\sum_x\eta(x)$ with $\sum_x\eta(ax)$ . For the count, use $\sum d_i^2=|G|$ with all $d_i=1$ .

Solution

$\langle\chi,\chi\rangle=1$ since $|\chi|=1$ . If $\eta=\chi\bar\psi$ is nontrivial, $S=\sum_x\eta(x)$ satisfies $S=\sum_x\eta(ax)=\eta(a)S$ with $\eta(a)\ne1$ , so $S=0$ . By Schur's lemma (Exercise 2.2 on the representation theory page) all irreducibles are characters, and $\sum d_i^2=|G|$ gives $|G|$ of them. $|G|$ orthonormal functions in a $|G|$ -dimensional space are a basis, and the coefficient of $\chi$ in $f$ is $\langle f,\chi\rangle=\hat f(\chi)$ .

Exercise 4.2The convolution theorem

With the conventions above, prove $\widehat{f*g}=|G|\,\hat f\,\hat g$ .

Hint

Substitute $x=y+z$ and use $\chi(y+z)=\chi(y)\chi(z)$ .

Solution

$\widehat{f*g}(\chi)=\frac1{|G|}\sum_{x,y}f(y)g(x-y)\overline{\chi(x)}=\frac1{|G|}\sum_yf(y)\overline{\chi(y)}\sum_zg(z)\overline{\chi(z)}=\frac1{|G|}\cdot|G|\hat f(\chi)\cdot|G|\hat g(\chi)$ .

Exercise 4.3Plancherel

Prove $\sum_x|f(x)|^2=|G|\sum_\chi|\hat f(\chi)|^2$ , and the polarized form for $\sum_xf(x)\overline{g(x)}$ .

Hint

Expand $f=\sum\hat f(\chi)\chi$ and use orthonormality.

Solution

$\sum_x|f(x)|^2=\sum_{\chi,\psi}\hat f(\chi)\overline{\hat f(\psi)}\sum_x\chi(x)\overline{\psi(x)}=|G|\sum_\chi|\hat f(\chi)|^2$ , since the inner sum is $|G|\delta_{\chi\psi}$ . The same computation with $g$ in the second slot gives $\sum f\bar g=|G|\sum\hat f\,\overline{\hat g}$ .

Exercise 4.4The dual group

Show that $\hat G$ is a group under pointwise multiplication, that the dual group of $\mathbb Z/n$ is isomorphic to $\mathbb Z/n$ , and that $\widehat{G\times H}\cong\hat G\times\hat H$ . Deduce $\hat G\cong G$ for every finite abelian group (using the structure theorem). Note that the isomorphism is not canonical, while $G\cong\hat{\hat G}$ is.

Hint

A character of $\mathbb Z/n$ is determined by $\chi(1)$ , an $n$ th root of unity.

Exercise 4.5Eigenvalues of circulant matrices

Let $C_{xy}=c(x-y)$ on $\mathbb Z/n$ . Show $C\chi_k=n\hat c(\chi_k)\,\chi_k$ .

Hint

$(C\chi)(x)=\sum_yc(x-y)\chi(y)$ ; substitute $z=x-y$ .

Solution

$(C\chi)(x)=\sum_zc(z)\chi(x-z)=\chi(x)\sum_zc(z)\overline{\chi(z)}=n\hat c(\chi)\,\chi(x)$ , using $\chi(-z)=\overline{\chi(z)}$ . So $C$ is $c*\cdot$ , and Exercise 4.2 is the same statement.

Exercise 4.6Majority on three bits

Verify the expansion of $\mathrm{Maj}$ in Example 4.5 by computing $\hat f(S)=\frac18\sum_xf(x)\chi_S(x)$ for every $S\subseteq\{1,2,3\}$ .

Hint

By symmetry the coefficients depend only on $|S|$ ; even-size sets give 0 because $\mathrm{Maj}(-x)=-\mathrm{Maj}(x)$ .

Exercise 4.7An uncertainty principle

Show that for $f\ne0$ on a finite abelian group, $|\mathrm{supp}\,f|\cdot|\mathrm{supp}\,\hat f|\ge|G|$ . When is it an equality?

Hint

$\lVert f\rVert_\infty\le\sum_\chi|\hat f(\chi)|\le|\mathrm{supp}\,\hat f|^{1/2}\big(\sum|\hat f|^2\big)^{1/2}$ , then Plancherel and $\sum|f|^2\le|\mathrm{supp}\,f|\,\lVert f\rVert_\infty^2$ . Equality holds for indicator functions of subgroups (and their translates and modulations).

Exercise 4.8Characters of the circle

Show that the continuous homomorphisms $\mathbb R/2\pi\mathbb Z\to\mathbb C^\times$ are $\theta\mapsto e^{ik\theta}$ , $k\in\mathbb Z$ .

Hint

A continuous homomorphism $\phi:\mathbb R\to\mathbb C^\times$ is $t\mapsto e^{ct}$ (smooth it by integrating: $\int_0^\varepsilon\phi(t+s)ds=\phi(t)\int_0^\varepsilon\phi(s)ds$ ). Being $2\pi$ -periodic forces $c=ik$ .

InteractiveRudin, Ch. 7: Sequences and Series of Functions

Uniform convergence

Pointwise limits of continuous functions can be discontinuous. Uniform convergence is the fix, and it is just convergence in the metric $d(f,g)=\sup|f-g|$ : the story of continuity, applied to a space whose points are functions.

The idea

Pointwise vs uniform (Rudin 7.1, 7.7).

f_n\to f

pointwise if

f_n(x)\to f(x)

for each

x

(the

N

may depend on

x

). Uniformly if for every

\varepsilon

one

N

works for all

x

: equivalently

\sup_x|f_n(x)-f(x)|\to0

(Rudin 7.9).

Uniform limits keep continuity (Rudin 7.12). If continuous

f_n\to f

uniformly,

f

is continuous. Uniform limits also commute with integrals (Rudin 7.16), but not with derivatives without more hypotheses (Rudin 7.17).

In the language of the story: $C([a,b])$ with the sup metric is a complete metric space (Rudin 7.15), and the Weierstrass theorem (Rudin 7.26) says polynomials are dense in it. Pointwise convergence, by contrast, is convergence in a topology that comes from no metric.

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Pointwise convergence: At each single input the values settle down eventually, but some inputs may take far longer than others.
Uniform convergence: The whole graph settles down together: from some step on, every input is within $\varepsilon$ of the limit at once.
Sup norm: The largest gap between two functions anywhere on the domain. Convergence in it is exactly the "whole graph settles down together" kind.
Uniform limit theorem: If continuous functions converge with the whole graph settling down together, the limit is continuous too. Limits taken one point at a time can break continuity, as $x^n$ on $[0,1]$ shows.
Weierstrass approximation theorem: Any continuous function on a closed interval can be matched as closely as you like, everywhere at once, by a polynomial.
Complete metric space: A space with no holes: every sequence whose terms bunch up together has a limit inside the space.

Explore

Sequence f_n

n ε

The limit f in ink, with an ε-tube around it. f_n is in blue where it stays in the tube and red where it leaves.

Worked example

Prove Rudin 7.12: a uniform limit of continuous functions is continuous.

Fix $p$ and $\varepsilon$ . Which $n$ do you use?

One with $\sup|f_n-f|<\varepsilon/3$ . This is where uniformity matters: the same $n$ is good at $p$ and at every nearby $x$ .
Now use continuity of that $f_n$ at $p$ .

Pick $\delta$ with $|f_n(x)-f_n(p)|<\varepsilon/3$ when $|x-p|<\delta$ . Then $|f(x)-f(p)|\le|f(x)-f_n(x)|+|f_n(x)-f_n(p)|+|f_n(p)-f(p)|<\varepsilon$ . This " $\varepsilon/3$ argument" is the template for many proofs in analysis.

Examples

Example 7.1 (a discontinuous limit).

$x^n$ on $[0,1]$ converges pointwise to the function that is 0 on $[0,1)$ and 1 at 1. The limit is discontinuous, so the convergence cannot be uniform: $\sup_{x<1}x^n=1$ for every $n$ .

Example 7.2 (uniform on bounded sets only).

$x/n\to0$ uniformly on every bounded interval but not on $\mathbb R$ , where $\sup|x/n|=\infty$ . Many convergence statements are uniform only on compact sets.

Example 7.3 (limits and integrals disagree).

$f_n(x)=2n^2xe^{-n^2x^2}$ tends to 0 at every $x\in[0,1]$ , but $\int_0^1f_n=1-e^{-n^2}\to1$ . The bump escapes towards 0, which uniform convergence would forbid.

Example 7.4 (the M-test).

$\sum_n2^{-n}\cos(3^nx)$ converges uniformly by comparison with $\sum2^{-n}$ , so it is continuous; by Hardy's theorem it is differentiable nowhere. Uniform limits keep continuity but not differentiability.

Check your understanding

$f_n(x)=x^n$ on $[0,1]$ converges pointwise to $f$ with $f(1)=1$ and $f=0$ elsewhere. Is the convergence uniform?

Yes, since each $f_n$ is continuous.Continuity of the $f_n$ doesn't force uniform convergence.
No: a uniform limit of continuous functions would be continuous.

Rudin 7.12 in reverse:

f

is discontinuous at 1, so the convergence can't be uniform. Directly,

\sup_{[0,1)}x^n=1

for every

n

. On

[0,0.9]

it is uniform, with

\sup=0.9^n

.

$f_n(x)=nxe^{-nx}$ on $[0,\infty)$ tends to 0 pointwise. Then $\sup_x f_n(x)$

tends to 0.Look at $x=1/n$ .
equals $1/e$ for every $n$ .

The maximum is at

x=1/n

with value

e^{-1}

. The bump slides toward 0 and never shrinks: pointwise but not uniform convergence, even though the limit is continuous.

Practice

Exercise 7.1The Weierstrass M-test (Rudin 7.10)

Show that $\sum_{n\ge1}\dfrac{\sin(nx)}{n^2}$ converges uniformly on $\R$ to a continuous function.

Solution

$|\sin(nx)/n^2|\le1/n^2$ and $\sum1/n^2<\infty$ . The partial sums are uniformly Cauchy: $\sup_x|\sum_{k=m}^n\dots|\le\sum_{k=m}^n1/k^2\to0$ . So they converge uniformly, and the limit is continuous by Rudin 7.12.

Exercise 7.2Limits and integrals: when can you swap them?

For $f_n(x)=n^2xe^{-nx}$ on $[0,1]$ , compute $\lim\int_0^1f_n$ and $\int_0^1\lim f_n$ .

Solution

$f_n\to0$ pointwise, so the second is 0. But $\int_0^1n^2xe^{-nx}dx=\int_0^nue^{-u}du\to1$ . The convergence is not uniform ( $\sup f_n=n/e$ ), so Rudin 7.16 doesn't apply, and the swap fails.

From Rudin, Principles of Mathematical Analysis (3rd ed., McGraw–Hill, 1976)

Numbers match the book. Each question is restated in our own words; the hints and solutions are written for this page.

Rudin 7.1Uniform limits of bounded functions

Prove that every uniformly convergent sequence of bounded functions is uniformly bounded.

Hint

Use one $N$ from the uniform Cauchy condition, then handle the finitely many earlier functions separately.

Solution

Let $f_n\to f$ uniformly on $E$ . Choose $N$ with $|f_n(x)-f_N(x)|<1$ for all $n\ge N$ and all $x$ (the uniform Cauchy condition, Rudin 7.8). $f_N$ is bounded by some $B$ , so $|f_n|<B+1$ for $n\ge N$ . The finitely many $f_1,\dots,f_{N-1}$ are bounded by $B_1,\dots,B_{N-1}$ . Then $\max(B+1,B_1,\dots,B_{N-1})$ bounds all $f_n$ .

Rudin 7.2Sums and products

If $f_n\to f$ and $g_n\to g$ uniformly on $E$ , prove $f_n+g_n\to f+g$ uniformly. If in addition the $f_n$ and $g_n$ are bounded, prove $f_ng_n\to fg$ uniformly.

Hint

For the product, add and subtract $f_ng$ and use Rudin 7.1.

Solution

Sums: $\sup|(f_n+g_n)-(f+g)|\le\sup|f_n-f|+\sup|g_n-g|\to0$ .

Products: by Rudin 7.1 there is $B$ with $|f_n|,|g_n|\le B$ , and then $|f|,|g|\le B$ too. Then $|f_ng_n-fg|\le|f_n||g_n-g|+|g||f_n-f|\le B\sup|g_n-g|+B\sup|f_n-f|\to0$ .

Without boundedness it fails: on $\R$ , $f_n=g_n=x+1/n$ converge uniformly to $x$ , but $f_ng_n-x^2=2x/n+1/n^2$ is unbounded for each $n$ .

Rudin 7.9Uniform convergence and moving points

Let $f_n$ be continuous and $f_n\to f$ uniformly on $E$ . Prove $f_n(x_n)\to f(x)$ whenever $x_n\to x$ in $E$ . Is the converse true?

Hint

Split $|f_n(x_n)-f(x)|$ through $f(x_n)$ , using that $f$ is continuous (Rudin 7.12).

Solution

$|f_n(x_n)-f(x)|\le|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|\le\sup|f_n-f|+|f(x_n)-f(x)|$ . The first term tends to 0 by uniformity, the second by continuity of $f$ .

The converse fails. Take $f_n(x)=x/n$ on $E=\R$ . If $x_n\to x$ , then $f_n(x_n)=x_n/n\to0=f(x)$ , since $x_n$ is bounded. But $\sup_{\R}|x/n|=\infty$ , so the convergence is not uniform. (On a compact $E$ the converse is true: this condition is called continuous convergence.)

InteractiveGrafakos, Classical, §§1.2.2–1.2.4

Convolution and approximate identities

Convolving with a narrow bump of total mass 1 replaces a function by local averages. As the bump narrows, the averages return the function: uniformly if it is uniformly continuous, in $L^p$ if it is in $L^p$ . Almost every approximation theorem in Fourier analysis, Fejér's included, is a case of this one.

The idea

Convolution (Grafakos §1.2.2).

(f*g)(x)=\int f(y)g(x-y)\,dy

, on

\mathbb R^n

or on the torus

\mathbb T^n=[0,1]^n

with periodic functions. It is commutative and associative, commutes with translations, and

\partial(f*g)=f*\partial g

when

g

is smooth: convolution makes things as smooth as the smoother factor.

Basic inequalities (Grafakos §1.2.3). Minkowski's integral inequality gives

\lVert f*g\rVert_p\le\lVert f\rVert_1\lVert g\rVert_p

for

1\le p\le\infty

; Hölder gives

\lVert f*g\rVert_\infty\le\lVert f\rVert_p\lVert g\rVert_{p'}

. Young's inequality interpolates:

\lVert f*g\rVert_r\le\lVert f\rVert_p\lVert g\rVert_q

when

1+\frac1r=\frac1p+\frac1q

.

Approximate identities (Grafakos §1.2.4). A family

K_\varepsilon

with

\int K_\varepsilon=1

,

\sup_\varepsilon\lVert K_\varepsilon\rVert_1<\infty

, and

\int_{|x|\ge\delta}|K_\varepsilon|\to0

as

\varepsilon\to0

for each

\delta>0

. Then

K_\varepsilon*f\to f

uniformly if

f

is bounded and uniformly continuous, in

L^p

if

f\in L^p

with

1\le p<\infty

, and at each point of continuity of a bounded

f

. The proof is the

\varepsilon/3

argument from the uniform convergence module: split the integral into

|y|<\delta

and

|y|\ge\delta

.

The standard kernels. The box

\frac1{2\varepsilon}\mathbf 1_{[-\varepsilon,\varepsilon]}

; the Gaussian (heat) kernel

\varepsilon^{-n}e^{-\pi|x/\varepsilon|^2}

; the Poisson kernel

P_\varepsilon(x)=\frac1\pi\frac{\varepsilon}{x^2+\varepsilon^2}

on

\mathbb R

; and on the torus, the Fejér kernel of module 4. The Dirichlet kernel is not on the list, and that is the whole difficulty of Fourier series.

Explore: smoothing a function

Function f

Kernel K

Width ε

f

and

K_\varepsilon*f

The kernel

K_\varepsilon

, total area 1

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Convolution: Replacing the value at each point by a weighted average of nearby values, with the weights given by a second function slid along the first.
Approximate identity: A family of averaging weights, each of total mass one, that concentrate more and more tightly around the origin, so that averaging with them gets closer and closer to doing nothing.
Young's inequality: The rule that tells you how integrable a smoothed function is, given how integrable the function and the smoothing weight are.
Mollifier: A smooth bump with compact support and total mass one, used to turn rough functions into infinitely differentiable ones by averaging.
Poisson kernel: The averaging weight that produces the harmonic function in the upper half-plane with given boundary values; its width is the height above the boundary.

Examples

Example 3.1 (moving averages).

With the box kernel, $(K_\varepsilon*f)(x)=\frac1{2\varepsilon}\int_{x-\varepsilon}^{x+\varepsilon}f$ . By the fundamental theorem of calculus it tends to $f(x)$ at every point of continuity, and it is continuous even when $f$ is not.

Example 3.2 (the heat equation).

$u(x,t)=(G_t*f)(x)$ with $G_t(x)=(4\pi t)^{-1/2}e^{-x^2/4t}$ solves $u_t=u_{xx}$ with $u(\cdot,0)=f$ , the initial value attained in the sense of approximate identities as $t\to0$ .

Example 3.3 (the Poisson kernel).

$\frac1\pi\frac y{x^2+y^2}=\frac1\pi\,\mathrm{Im}\frac{-1}{x+iy}$ is harmonic in $(x,y)$ , has integral 1 in $x$ , and is an approximate identity as $y\to0$ . So $P_y*f$ is the harmonic extension of $f$ to the upper half-plane. Its slow decay, like $1/x^2$ , shows in the figure as long tails.

Example 3.4 (smoothing).

If $\phi\in C_c^\infty$ with $\int\phi=1$ and $\phi_\varepsilon(x)=\varepsilon^{-n}\phi(x/\varepsilon)$ , then $\phi_\varepsilon*f\in C^\infty$ for every $f\in L^1_{\mathrm{loc}}$ , and $\phi_\varepsilon*f\to f$ in $L^p$ . Hence $C_c^\infty$ is dense in $L^p$ for $p<\infty$ .

Example 3.5 (Weierstrass by convolution).

The Landau kernels $L_n(x)=c_n(1-x^2)^n$ on $[-1,1]$ , normalized to integral 1, form an approximate identity. For continuous $f$ on $[0,1]$ vanishing at the ends, $L_n*f$ is a polynomial on $[0,1]$ and converges uniformly to $f$ : the proof of the Weierstrass theorem behind the uniform convergence module.

Example 3.6 (jumps).

A continuous function cannot be uniformly close to a step function across its jump. So $K_\varepsilon*\mathbf 1_{[0,\infty)}\to\mathbf 1_{[0,\infty)}$ uniformly away from 0 but not on any neighbourhood of 0. At 0 itself a symmetric kernel gives exactly $\frac12$ , the average of the two sides (Exercise 3.5).

Check your understanding

For $f\in L^1(\mathbb R)$ and $g\in L^2(\mathbb R)$ , which bound holds?

$\lVert f*g\rVert_2\le\lVert f\rVert_1\lVert g\rVert_2$
$\lVert f*g\rVert_1\le\lVert f\rVert_1\lVert g\rVert_2$ $f*g$ need not be integrable.
$\lVert f*g\rVert_\infty\le\lVert f\rVert_1\lVert g\rVert_2$ That needs $g\in L^\infty$ , or $f\in L^2$ .

Minkowski's integral inequality:

f*g

is an average of translates of

g

with weights

|f|

, and each translate has the same

L^2

norm.

Is there an $f\in L^1(\mathbb R)$ with $f*g=g$ for every $g\in L^1$ ?

Yes, the delta function. $\delta$ is not a function (module 7).
No: taking Fourier transforms, $\hat f=1$ , contradicting the Riemann–Lebesgue lemma.

That is why approximate identities are families: the identity itself is not in

L^1

(Exercise 3.8).

Exercises

Exercise 3.1Convolving with an integrable function

Prove $\lVert f*g\rVert_p\le\lVert f\rVert_1\lVert g\rVert_p$ for $1\le p\le\infty$ .

Hint

$f*g=\int f(y)\,\tau_yg\,dy$ with $\tau_yg(x)=g(x-y)$ , an $L^p$ -valued integral; $\lVert\tau_yg\rVert_p=\lVert g\rVert_p$ .

Solution

By Minkowski's integral inequality, $\big\lVert\int f(y)g(\cdot-y)\,dy\big\rVert_p\le\int|f(y)|\,\lVert g(\cdot-y)\rVert_p\,dy=\lVert f\rVert_1\lVert g\rVert_p$ . For $p=\infty$ it is direct: $|f*g(x)|\le\int|f(y)|\,\lVert g\rVert_\infty dy$ .

Exercise 3.2Uniform convergence

Let $K_\varepsilon$ be an approximate identity and $f$ bounded and uniformly continuous. Show $K_\varepsilon*f\to f$ uniformly.

Hint

$K_\varepsilon*f(x)-f(x)=\int K_\varepsilon(y)(f(x-y)-f(x))\,dy$ . Split at $|y|=\delta$ .

Solution

Let $A=\sup\lVert K_\varepsilon\rVert_1$ . Given $\eta>0$ , choose $\delta$ with $|f(x-y)-f(x)|<\eta$ for $|y|<\delta$ and all $x$ . Then $|K_\varepsilon*f(x)-f(x)|\le A\eta+2\lVert f\rVert_\infty\int_{|y|\ge\delta}|K_\varepsilon|$ , uniformly in $x$ , and the last integral tends to 0.

Exercise 3.3

L^p

convergence

Show $K_\varepsilon*f\to f$ in $L^p$ for $f\in L^p$ , $1\le p<\infty$ .

Hint

For $f\in C_c$ , use Exercise 3.2 plus a uniform bound on the tails. For general $f$ , approximate by $g\in C_c$ and use Exercise 3.1: $\lVert K_\varepsilon*(f-g)\rVert_p\le A\lVert f-g\rVert_p$ .

Exercise 3.4The Poisson kernel

Show $\int P_\varepsilon=1$ and that $P_\varepsilon$ is an approximate identity.

Hint

$\int\frac{\varepsilon\,dx}{x^2+\varepsilon^2}=\arctan(x/\varepsilon)\big|_{-\infty}^\infty=\pi$ . Tails: $\int_{|x|\ge\delta}P_\varepsilon=1-\frac2\pi\arctan(\delta/\varepsilon)\to0$ .

Exercise 3.5Symmetric kernels at a jump

If $K_\varepsilon$ is an even approximate identity and $f$ has one-sided limits $f(0^\pm)$ at 0, show $K_\varepsilon*f(0)\to\frac12(f(0^+)+f(0^-))$ .

Hint

Evenness gives $\int_0^\infty K_\varepsilon=\int_{-\infty}^0K_\varepsilon=\frac12$ . Treat the two half-lines separately, as in Exercise 3.2.

Exercise 3.6Convolution of

L^p

and

L^{p'}

is continuous

Show that for $f\in L^p$ , $g\in L^{p'}$ , $1\le p<\infty$ , $f*g$ is uniformly continuous.

Hint

$|f*g(x+h)-f*g(x)|\le\lVert\tau_{-h}f-f\rVert_p\lVert g\rVert_{p'}$ , and translation is continuous in $L^p$ for $p<\infty$ (approximate by $C_c$ ).

Exercise 3.7Weierstrass via Landau kernels

Carry out Example 3.5: show $L_n$ is an approximate identity on $[-1,1]$ , and that $L_n*f$ restricted to $[0,1]$ is a polynomial when $f$ is continuous on $[0,1]$ , zero at the ends, and extended by 0.

Hint

$c_n\le\sqrt n$ (bound $(1-x^2)^n\ge1-nx^2$ on $|x|\le1/\sqrt n$ ). For $\delta\le|x|\le1$ , $L_n\le\sqrt n(1-\delta^2)^n\to0$ . For the polynomial: $\int_0^1f(y)(1-(x-y)^2)^ndy$ is a polynomial in $x$ . This is Rudin's proof of Theorem 7.26.

Solution

Since $(1-x^2)^n\ge1-nx^2$ , $\int_{-1}^1(1-x^2)^n\ge2\int_0^{1/\sqrt n}(1-nx^2)=\frac4{3\sqrt n}>\frac1{\sqrt n}$ , so $c_n<\sqrt n$ . The kernels are positive with integral 1, and $\sup_{\delta\le|x|\le1}L_n\le\sqrt n(1-\delta^2)^n\to0$ , so the tails vanish. For $x\in[0,1]$ and $y\in[0,1]$ , $x-y\in[-1,1]$ , so $L_n*f(x)=c_n\int_0^1f(y)(1-(x-y)^2)^ndy$ , and expanding $(1-(x-y)^2)^n$ shows it is a polynomial in $x$ . Exercise 3.2 gives uniform convergence (the extension of $f$ by 0 is uniformly continuous because $f(0)=f(1)=0$ ).

Exercise 3.8No identity element

Show there is no $f\in L^1(\mathbb R)$ with $f*g=g$ for all $g\in L^1$ .

Hint

Take $g$ a Gaussian and Fourier transform: $\hat f\hat g=\hat g$ forces $\hat f\equiv1$ , but $\hat f(\xi)\to0$ (Exercise 6.8).

InteractiveGrafakos, Classical, §§3.1.3, 3.2.1, 3.4, 3.5.4

Fourier series: Dirichlet, Fejér and Gibbs

The partial sums of a Fourier series are convolution with the Dirichlet kernel, which is not an approximate identity: its $L^1$ norm grows like $\log N$ . Averaging the partial sums gives the Fejér kernel, which is one. That single difference explains Fejér's theorem, the Gibbs overshoot, and why pointwise convergence of Fourier series is hard.

The idea

Coefficients and partial sums (Grafakos §§3.1.2, 3.2.1). For

f\in L^1(\mathbb T)

,

\mathbb T=[0,1]

:

\hat f(m)=\int_0^1f(x)e^{-2\pi imx}dx

, and

S_Nf(x)=\sum_{|m|\le N}\hat f(m)e^{2\pi imx}=(D_N*f)(x)

with the Dirichlet kernel

D_N(x)=\sum_{|m|\le N}e^{2\pi imx}=\frac{\sin((2N+1)\pi x)}{\sin(\pi x)}.

(Grafakos uses

[0,1]

and

e^{2\pi imx}

; the first module used

e^{ik\theta}

on

[0,2\pi]

. Only the bookkeeping differs.)

Fejér's theorem (Grafakos §3.1.3). The Cesàro means

\sigma_Nf=\frac1{N+1}\sum_{j=0}^NS_jf=F_N*f

use the Fejér kernel

F_N(x)=\frac1{N+1}\Big(\frac{\sin((N+1)\pi x)}{\sin(\pi x)}\Big)^2\ge0,

an approximate identity. So

\sigma_Nf\to f

uniformly for continuous

f

, and in

L^p

for

f\in L^p

,

p<\infty

. Consequences: trigonometric polynomials are dense, and a continuous function is determined by its Fourier coefficients.

Why partial sums are harder (Grafakos §§3.4.3–3.4.4).

\lVert D_N\rVert_{L^1}\sim\frac4{\pi^2}\log N

(the Lebesgue constants), so the uniform boundedness principle produces continuous functions whose Fourier series diverge at a point. Positive results need more regularity: if

f

is Hölder continuous at

x_0

, or merely satisfies Dini's condition there, then

S_Nf(x_0)\to f(x_0)

. Carleson's theorem (for

L^2

; Hunt for

L^p

,

p>1

) gives almost-everywhere convergence.

The Gibbs phenomenon (Grafakos §3.5.4). At a jump of size

J

,

S_Nf

overshoots by about

0.0895J

on each side, however large

N

is:

\lim_N\max S_N(\text{square wave})=\frac2\pi\int_0^\pi\frac{\sin t}t\,dt\approx1.17898

. The overshoot moves closer to the jump but does not shrink. Fejér means never overshoot, because

F_N\ge0

.

Explore: partial sums against Fejér means

Function f on [0, 1]

N

Show kernel

f

, partial sum

S_Nf

, Fejér mean

\sigma_Nf

The chosen kernel on

[-\frac12,\frac12]

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Dirichlet kernel: The sum of all pure frequencies up to a cutoff: averaging against it gives the truncated Fourier series. It oscillates and takes negative values, and its total absolute area grows without bound.
Fejér kernel: The average of the first few truncation kernels: a non-negative bump of area one that narrows as more terms are used.
Cesàro mean: The average of the first $N$ partial sums of a series, which can converge even when the partial sums themselves don't.
Lebesgue constants: The total absolute areas of the truncation kernels, which grow like the logarithm of the number of terms; their growth is why some continuous functions have divergent Fourier series.
Gibbs phenomenon: The overshoot of about nine percent of the jump that truncated Fourier series make next to a jump discontinuity, which narrows but never shrinks.
Dini's test: A mild condition on how a function behaves near one point, which is enough for its Fourier series to converge to the right value at that point.

Examples

Example 4.1 (the square wave).

$f=1$ on $(0,\frac12)$ , $-1$ on $(\frac12,1)$ has $\hat f(m)=\frac{2}{\pi im}$ for odd $m$ and 0 for even $m\ne0$ , so $f(x)\sim\frac4\pi\sum_{k\ge0}\frac{\sin(2\pi(2k+1)x)}{2k+1}$ . At $x=\frac14$ this is Leibniz's series: $1=\frac4\pi(1-\frac13+\frac15-\cdots)$ .

Example 4.2 (the Dirichlet kernel changes sign).

$D_N(0)=2N+1$ , $\int_{\mathbb T}D_N=1$ , but $D_N$ oscillates with zeros at $x=\frac k{2N+1}$ . Its positive and negative lobes both have total area of order $\log N$ , cancelling except for the net 1.

Example 4.3 (the Fejér kernel).

$F_N\ge0$ , $\int F_N=1$ , and $F_N(x)\le\frac1{(N+1)\sin^2(\pi\delta)}$ for $\delta\le|x|\le\frac12$ : the three conditions of an approximate identity, visible in the right panel.

Example 4.4 (Gibbs, numerically).

For the square wave, $\max S_9f\approx1.1823$ and $\max S_{59}f\approx1.1791$ , against the limit $1.17898$ : about $9\%$ of the jump of 2 on each side, for every large $N$ . The sawtooth shows the same $9\%$ at its single jump; the continuous triangle wave shows none.

Example 4.5 (Weierstrass on the circle).

Fejér's theorem says every continuous periodic function is a uniform limit of trigonometric polynomials, the periodic version of the Weierstrass theorem in the uniform convergence module.

Example 4.6 (convergence at a smooth point).

The square wave is smooth on $(0,\frac12)$ , so Dini's test gives $S_Nf(x)\to1$ there, and at the jump $S_Nf(0)=0$ for all $N$ , the average of the two sides. Convergence holds everywhere, but not uniformly.

Check your understanding

Why is the Fejér kernel an approximate identity while the Dirichlet kernel is not?

$F_N\ge0$ has $\lVert F_N\rVert_1=1$ , while $\lVert D_N\rVert_1\to\infty$ .
$\int D_N\ne1$ .Both integrate to 1.
$D_N$ is not even.Both are even.

An approximate identity needs

\sup\lVert K\rVert_1<\infty

; the Lebesgue constants

\lVert D_N\rVert_1

grow like

\log N

.

As $N\to\infty$ , the Gibbs overshoot of $S_Nf$ at a jump:

tends to 0.Increase N in the figure.
tends to about $9\%$ of the jump, while moving closer to it.
grows like $\log N$ .That's $\lVert D_N\rVert_1$ , not the overshoot.

The overshoot converges to

\big(\frac1\pi\int_0^\pi\frac{\sin t}t\,dt-\frac12\big)J\approx0.0895J

.

Exercises

Exercise 4.1The Dirichlet kernel

Prove the closed form of $D_N$ .

Hint

Geometric series: $\sum_{m=-N}^Nz^m=\frac{z^{N+1}-z^{-N}}{z-1}$ with $z=e^{2\pi ix}$ ; multiply top and bottom by $z^{-1/2}$ .

Solution

With $z=e^{2\pi ix}$ , $D_N=\frac{z^{N+1}-z^{-N}}{z-1}=\frac{z^{N+1/2}-z^{-N-1/2}}{z^{1/2}-z^{-1/2}}=\frac{2i\sin((2N+1)\pi x)}{2i\sin(\pi x)}$ .

Exercise 4.2The Fejér kernel

Prove the closed form of $F_N$ , and that $F_N$ is an approximate identity on $\mathbb T$ .

Hint

$\sin(\pi x)\,D_j(x)=\sin((2j+1)\pi x)$ , and $2\sin(\pi x)\sin((2j+1)\pi x)=\cos(2j\pi x)-\cos(2(j+1)\pi x)$ telescopes.

Solution

$2\sin^2(\pi x)\sum_{j=0}^ND_j=\sum_j\big(\cos(2j\pi x)-\cos(2(j+1)\pi x)\big)=1-\cos(2(N+1)\pi x)=2\sin^2((N+1)\pi x)$ . Divide by $2(N+1)\sin^2(\pi x)$ . Then $F_N\ge0$ , $\int F_N=\frac1{N+1}\sum\int D_j=1$ , and on $\delta\le|x|\le\frac12$ , $F_N\le\frac1{(N+1)\sin^2(\pi\delta)}\to0$ .

Exercise 4.3Fejér's theorem and uniqueness

Deduce that $\sigma_Nf\to f$ uniformly for continuous $f$ on $\mathbb T$ , and that if all $\hat f(m)=0$ for a continuous $f$ , then $f=0$ .

Hint

Exercises 4.2 and 3.2 (on the torus, continuous means uniformly continuous). If all coefficients vanish, every $\sigma_Nf$ is 0.

Exercise 4.4Lebesgue constants

Show $\lVert D_N\rVert_{L^1(\mathbb T)}\ge c\log N$ for some $c>0$ .

Hint

Use $|\sin\pi x|\le\pi|x|$ and integrate $|\sin((2N+1)\pi x)|/(\pi|x|)$ over the intervals $\big[\frac k{2N+1},\frac{k+1}{2N+1}\big]$ , $k=0,\dots,N-1$ . Each contributes about $\frac c{k+1}$ .

Solution

$\lVert D_N\rVert_1\ge2\int_0^{1/2}\frac{|\sin((2N+1)\pi x)|}{\pi x}dx$ . On $I_k=\big[\frac k{2N+1},\frac{k+1}{2N+1}\big]$ , $x\le\frac{k+1}{2N+1}$ , and $\int_{I_k}|\sin((2N+1)\pi x)|dx=\frac2{\pi(2N+1)}$ . So the integral over $I_k$ is at least $\frac{2}{\pi^2(k+1)}$ , and summing over $k<N$ gives $\frac4{\pi^2}\sum_{k\le N}\frac1k\ge\frac4{\pi^2}\log N$ .

Exercise 4.5Coefficients of the square wave

Verify the coefficients in Example 4.1, and derive Leibniz's series.

Hint

$\hat f(m)=\int_0^{1/2}e^{-2\pi imx}dx-\int_{1/2}^1e^{-2\pi imx}dx=\frac{1-(-1)^m}{\pi im}$ . Use Dini's test at $x=\frac14$ .

Exercise 4.6The Gibbs constant

Show that the first maximum of $S_{2K-1}f$ for the square wave is at $x=\frac1{4K}$ , with value tending to $\frac2\pi\int_0^\pi\frac{\sin t}t\,dt$ .

Hint

$\frac d{dx}S_{2K-1}f=8\sum_{k<K}\cos(2\pi(2k+1)x)=\frac{4\sin(4\pi Kx)}{\sin(2\pi x)}$ , first zero at $\frac1{4K}$ . The value there is a Riemann sum for $\frac2\pi\int_0^\pi\frac{\sin t}t\,dt$ .

Exercise 4.7Dini's test

Show that if $\int_{-1/2}^{1/2}\frac{|f(x_0-t)-f(x_0)|}{|t|}dt<\infty$ , then $S_Nf(x_0)\to f(x_0)$ .

Hint

$S_Nf(x_0)-f(x_0)=\int g(t)\sin((2N+1)\pi t)\,dt$ with $g(t)=\frac{f(x_0-t)-f(x_0)}{\sin\pi t}\in L^1$ . Apply the Riemann–Lebesgue lemma (module 5).

InteractiveGrafakos, Classical, §§3.2.2, 3.3

Decay, smoothness and L²

How fast Fourier coefficients decay measures how smooth the function is: a jump gives $1/m$ , a corner $1/m^2$ , and an analytic function decays exponentially. In $L^2$ there is no loss at all: the coefficients are the coordinates of $f$ in an orthonormal basis, exactly as for the finite groups of the first module.

The idea

Riemann–Lebesgue lemma (Grafakos §3.3.1). For

f\in L^1(\mathbb T)

,

\hat f(m)\to0

as

|m|\to\infty

. Nothing better holds in general: for any sequence

a_m\to0

there is an

f\in L^1

whose coefficients do not decay faster than

a_m

.

Smoothness gives decay (Grafakos §3.3.2). Integrating by parts,

\widehat{f'}(m)=2\pi im\hat f(m)

, so if

f\in C^k(\mathbb T)

then

|\hat f(m)|\le\frac{\lVert f^{(k)}\rVert_1}{(2\pi|m|)^k}

, indeed

o(|m|^{-k})

. Conversely, if

\sum|m|^k|\hat f(m)|<\infty

, the series can be differentiated

k

times term by term and

f\in C^k

. A function of bounded variation has

\hat f(m)=O(1/|m|)

.

Plancherel on the circle (Grafakos §3.2.2).

\{e^{2\pi imx}\}_{m\in\mathbb Z}

is an orthonormal basis of

L^2(\mathbb T)

: orthonormality is a direct computation, completeness follows from Fejér's theorem. Hence

\lVert f\rVert_2^2=\sum_m|\hat f(m)|^2

,

S_Nf\to f

in

L^2

, and

f\mapsto\hat f

is a unitary isomorphism

L^2(\mathbb T)\to\ell^2(\mathbb Z)

. This is the infinite version of the finite Plancherel identity.

Explore: decay rates on log–log axes

|\hat f(m)|

for

m=1,\dots,256

on log–log axes , with reference lines

m^{-1}

to

m^{-4}

. A straight line of slope

-k

means decay like

m^{-k}

.

Function

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Riemann–Lebesgue lemma: The statement that high-frequency content of any integrable function fades out, though possibly very slowly.
Bounded variation: Describes a function whose graph has finite total up-and-down movement, so its frequency content decays at least like one over the frequency.
Orthonormal basis: A list of mutually perpendicular unit vectors such that every vector is the (possibly infinite) sum of its components along them.
Parseval's identity: The energy of a periodic function equals the sum of the squared sizes of its frequency coefficients.
Absolutely convergent Fourier series: A series whose coefficient sizes have a finite sum, which then converges uniformly to a continuous function.

Examples

Example 5.1 (a jump).

The square wave has $|\hat f(m)|=\frac2{\pi|m|}$ for odd $m$ : slope $-1$ , the rate for functions of bounded variation with a jump.

Example 5.2 (a corner).

The triangle wave (the integral of a square wave) has $|\hat f(m)|=\frac{c}{m^2}$ for odd $m$ : one integration, one more power, by $\widehat{f'}(m)=2\pi im\hat f(m)$ .

Example 5.3 (analytic).

$e^{\cos2\pi x}$ has $\hat f(m)=I_m(1)$ , a modified Bessel function, which decays like $\frac1{2^m m!}$ : faster than any power. Analytic periodic functions have exponentially decaying coefficients, which is the trapezoid-rule phenomenon of the UTokyo analysis module.

Example 5.4 (Basel by Parseval).

The sawtooth $f(x)=x-\frac12$ on $[0,1)$ has $\hat f(m)=\frac{i}{2\pi m}$ for $m\ne0$ . Parseval: $\int_0^1(x-\frac12)^2dx=\frac1{12}=\sum_{m\ne0}\frac1{4\pi^2m^2}$ , so $\sum_{m\ge1}\frac1{m^2}=\frac{\pi^2}6$ .

Example 5.5 (a cusp).

$|x-\frac12|^{1/2}$ is continuous and monotone on each half, with a cusp at $\frac12$ where the derivative blows up. Its coefficients decay like $|m|^{-3/2}$ , between the jump and the corner: fractional smoothness gives fractional decay.

Example 5.6 (from decay back to smoothness).

If $|\hat f(m)|\le C|m|^{-4}$ , then $\sum|m|^2|\hat f(m)|<\infty$ , so $f\in C^2$ . The figure's $x^2(1-x)^2$ has exactly this decay: it is $C^2$ , and the jump in its third derivative costs the fourth power.

Check your understanding

A periodic function has $|\hat f(m)|\approx m^{-2}$ . What can you say?

The series converges absolutely and uniformly, so $f$ is continuous; a corner (a jump in $f'$ ) is typical of this rate.
$f$ must have a jump.Jumps give $1/m$ .
$f$ is $C^1$ . $\sum|m|\,m^{-2}$ diverges; the triangle wave is not $C^1$ .

\sum m^{-2}<\infty

gives uniform convergence (the Weierstrass M-test from the uniform convergence module).

For $f\in L^2(\mathbb T)$ , $S_Nf\to f$ :

in $L^2$ , always.
uniformly.The square wave is in $L^2$ .
only if $f$ is continuous.Orthonormal expansions converge in norm for every vector.

\lVert f-S_Nf\rVert_2^2=\sum_{|m|>N}|\hat f(m)|^2\to0

. Almost-everywhere convergence is also true but is Carleson's deep theorem.

Exercises

Exercise 5.1The Riemann–Lebesgue lemma

Prove $\hat f(m)\to0$ for $f\in L^1(\mathbb T)$ .

Hint

It holds for trigonometric polynomials (eventually 0). Approximate $f$ in $L^1$ by one, using Fejér, and use $|\hat f(m)-\hat g(m)|\le\lVert f-g\rVert_1$ .

Solution

Given $\varepsilon$ , Fejér's theorem in $L^1$ gives a trigonometric polynomial $g$ of degree $M$ with $\lVert f-g\rVert_1<\varepsilon$ . For $|m|>M$ , $\hat g(m)=0$ , so $|\hat f(m)|=|\widehat{f-g}(m)|\le\lVert f-g\rVert_1<\varepsilon$ .

Exercise 5.2Differentiation multiplies by

2\pi im

Show $\widehat{f'}(m)=2\pi im\hat f(m)$ for $f\in C^1(\mathbb T)$ , and deduce $\hat f(m)=o(|m|^{-k})$ for $f\in C^k$ .

Hint

Integrate by parts; periodicity kills the boundary term. Then apply Riemann–Lebesgue to $f^{(k)}$ .

Solution

$\widehat{f'}(m)=\big[f(x)e^{-2\pi imx}\big]_0^1+2\pi im\int_0^1f(x)e^{-2\pi imx}dx=2\pi im\hat f(m)$ . Iterating, $\hat f(m)=\frac{\widehat{f^{(k)}}(m)}{(2\pi im)^k}$ for $m\ne0$ , and $\widehat{f^{(k)}}(m)\to0$ .

Exercise 5.3Completeness of the exponentials

Show that $\{e^{2\pi imx}\}$ is an orthonormal basis of $L^2(\mathbb T)$ , and deduce Parseval's identity.

Hint

Orthonormal: integrate $e^{2\pi i(m-n)x}$ . Complete: continuous functions are dense in $L^2$ , and Fejér approximates each one uniformly (hence in $L^2$ ) by trigonometric polynomials.

Exercise 5.4Absolutely convergent series

Show that if $\sum|\hat f(m)|<\infty$ for $f\in L^1$ , then $S_Nf$ converges uniformly to a continuous function equal to $f$ almost everywhere.

Hint

The M-test gives a continuous limit $g$ with $\hat g=\hat f$ . Uniqueness of coefficients in $L^1$ (via Fejér in $L^1$ ) gives $f=g$ a.e.

Exercise 5.5

C^1

functions have absolutely convergent series

Show $\sum_m|\hat f(m)|<\infty$ for $f\in C^1(\mathbb T)$ . (So Fourier series of $C^1$ functions converge uniformly.)

Hint

$\sum_{m\ne0}|\hat f(m)|=\sum\frac{|\widehat{f'}(m)|}{2\pi|m|}\le\big(\sum\frac1{4\pi^2m^2}\big)^{1/2}\big(\sum|\widehat{f'}(m)|^2\big)^{1/2}$ , and Parseval for $f'$ .

Exercise 5.6Bounded variation

Show that if $f$ has bounded variation $V$ on $[0,1]$ , then $|\hat f(m)|\le\frac{V}{2\pi|m|}$ .

Hint

Integrate by parts in the Riemann–Stieltjes sense (the Riemann–Stieltjes module on the Real Analysis page): $\hat f(m)=\frac1{2\pi im}\int_0^1e^{-2\pi imx}\,df(x)$ , plus a boundary term that vanishes for periodic $f$ .

Exercise 5.7Basel and beyond

Verify Example 5.4, then apply Parseval to the triangle wave to compute $\sum_{k\ge0}\frac1{(2k+1)^4}=\frac{\pi^4}{96}$ .

Hint

For the sawtooth, integrate by parts. For the triangle wave $t(x)=|x-\frac12|$ on $[0,1]$ you get $|\hat t(m)|=\frac1{\pi^2m^2}$ for odd $m$ ; compare $\int_0^1(t-\frac14)^2=\frac1{48}$ with $2\sum_{\text{odd }m>0}\frac1{\pi^4m^4}$ .

InteractiveGrafakos, Classical, §§2.2, 3.2.3

The Fourier transform on ℝⁿ

On $\mathbb R^n$ the frequencies form a continuum, and sums become integrals. The Schwartz functions, smooth and rapidly decaying, are the setting in which every formula holds without fuss: the transform swaps differentiation with multiplication, translation with modulation, and the Gaussian is its own transform.

The idea

Schwartz functions and the transform (Grafakos §§2.2.1–2.2.2).

\mathcal S(\mathbb R^n)

is the space of smooth

f

with

x^\alpha\partial^\beta f

bounded for all multi-indices. For

f\in L^1

,

\hat f(\xi)=\int f(x)e^{-2\pi ix\cdot\xi}dx

, also written

\mathcal F(f)

. The transform maps

\mathcal S

to itself, with

\widehat{\partial_jf}=2\pi i\xi_j\hat f

,

\widehat{x_jf}=-\frac1{2\pi i}\partial_j\hat f

,

\widehat{f(\cdot-a)}=e^{-2\pi ia\cdot\xi}\hat f

, and

\widehat{f(\lambda\cdot)}=\lambda^{-n}\hat f(\cdot/\lambda)

.

Inversion and Plancherel (Grafakos §§2.2.3–2.2.4).

e^{-\pi|x|^2}

is its own transform. Using Gaussians as an approximate identity one proves

f(x)=\int\hat f(\xi)e^{2\pi ix\cdot\xi}d\xi

for

f\in\mathcal S

, and

\lVert\hat f\rVert_2=\lVert f\rVert_2

. The transform extends to a unitary map of

L^2(\mathbb R^n)

, and

\widehat{f*g}=\hat f\hat g

.

Poisson summation (Grafakos §3.2.3). For nice

f

on

\mathbb R^n

,

\sum_{k\in\mathbb Z^n}f(x+k)=\sum_{m\in\mathbb Z^n}\hat f(m)e^{2\pi im\cdot x}

: periodizing

f

and sampling

\hat f

are the same operation seen from the two sides. At

x=0

,

\sum f(k)=\sum\hat f(m)

. This links the transform on

\mathbb R

, the series on

\mathbb T

, and theta functions.

Uncertainty. For

f\in\mathcal S(\mathbb R)

,

\lVert xf\rVert_2\lVert\xi\hat f\rVert_2\ge\frac1{4\pi}\lVert f\rVert_2^2

, with equality exactly for Gaussians (Exercise 6.6). A function and its transform cannot both be concentrated.

Explore: Gaussians, widths and Poisson summation

Width σ of f(x) = e^{−πx²/σ²} Modulate by e^{2πiax}, a =

\mathrm{Re}\,f

and

|f|

; ticks at the integers, where

\sum f(k)

samples it

\hat f(\xi)=\sigma e^{-\pi\sigma^2(\xi-a)^2}

; ticks at the integers, where

\sum\hat f(m)

samples it

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Schwartz function: A smooth function that, together with all its derivatives, decays faster than any power of $1/|x|$ : the class where every manipulation of the Fourier transform is safe.
Modulation: Multiplying a function by a pure oscillation $e^{2\pi iax}$ , which shifts its frequency content by $a$ .
Fourier inversion: Recovering a function by superposing its frequency components with the conjugate exponentials.
Poisson summation formula: The sum of a function's values at the integers equals the sum of its transform's values at the integers.
Uncertainty principle: A function and its frequency content can't both be sharply concentrated: the product of their spreads has a positive lower bound.

Examples

Example 6.1 (the Gaussian).

$\widehat{e^{-\pi x^2}}=e^{-\pi\xi^2}$ , and by dilation $\widehat{e^{-\pi x^2/\sigma^2}}=\sigma e^{-\pi\sigma^2\xi^2}$ : width $\sigma$ on one side, $1/\sigma$ on the other.

Example 6.2 (box and sinc).

$\mathcal F\big(\mathbf 1_{[-1/2,1/2]}\big)(\xi)=\frac{\sin\pi\xi}{\pi\xi}$ . It is in $L^2$ but not $L^1$ , so its inverse transform exists only in the $L^2$ sense, and the box's jumps reappear as the slow $1/\xi$ decay.

Example 6.3 (the two-sided exponential).

$\widehat{e^{-2\pi|x|}}(\xi)=\frac1\pi\frac1{1+\xi^2}$ : a corner at 0 gives $\xi^{-2}$ decay, and the transform of the Poisson kernel is $e^{-2\pi|\xi|}$ by inversion.

Example 6.4 (the heat equation).

Transforming $u_t=\Delta u$ in $x$ gives $\hat u_t=-4\pi^2|\xi|^2\hat u$ , so $\hat u(\xi,t)=e^{-4\pi^2t|\xi|^2}\hat f(\xi)$ : high frequencies die fastest, and $u=G_t*f$ with the Gaussian of Example 3.2.

Example 6.5 (theta functions).

Poisson summation for $e^{-\pi tx^2}$ gives $\sum_ke^{-\pi tk^2}=t^{-1/2}\sum_me^{-\pi m^2/t}$ , i.e. $\theta(1/t)=\sqrt t\,\theta(t)$ . This modular transformation is the root of the functional equation of the zeta function, and of the Modular Forms page.

Example 6.6 (sampling).

If $\hat f$ vanishes outside $[-\frac12,\frac12]$ , Poisson summation and inversion give $f(x)=\sum_kf(k)\frac{\sin\pi(x-k)}{\pi(x-k)}$ (Shannon): a band-limited function is determined by its integer samples.

Check your understanding

$f$ is a Gaussian of width $\sigma$ . Halving $\sigma$ makes $\hat f$ :

twice as wide and half as tall.
half as wide.Width $\sigma$ on one side means width $1/\sigma$ on the other.
unchanged.Try it in the figure.

\widehat{f(\lambda\cdot)}=\lambda^{-1}\hat f(\cdot/\lambda)

with

\lambda=2

.

What does multiplying $f$ by $e^{2\pi iax}$ do to $\hat f$ ?

Shifts it by $a$ .
Multiplies it by $e^{2\pi ia\xi}$ .That's what translating $f$ does (with a minus sign).

\int f(x)e^{2\pi iax}e^{-2\pi ix\xi}dx=\hat f(\xi-a)

. Translation and modulation are swapped by the transform.

Exercises

Exercise 6.1The Gaussian is its own transform

Prove $\widehat{e^{-\pi x^2}}=e^{-\pi\xi^2}$ .

Hint

Both sides solve $y'=-2\pi\xi y$ with $y(0)=1$ : differentiate under the integral and integrate by parts. Or complete the square and shift the contour.

Solution

Let $g(\xi)=\int e^{-\pi x^2}e^{-2\pi ix\xi}dx$ . Then $g'(\xi)=\int(-2\pi ix)e^{-\pi x^2}e^{-2\pi ix\xi}dx=i\int\big(\tfrac d{dx}e^{-\pi x^2}\big)e^{-2\pi ix\xi}dx=i\cdot(2\pi i\xi)g(\xi)=-2\pi\xi g(\xi)$ , integrating by parts. With $g(0)=\int e^{-\pi x^2}=1$ , $g(\xi)=e^{-\pi\xi^2}$ .

Exercise 6.2The symmetry rules

Prove the rules for translation, modulation, dilation and differentiation stated above.

Hint

Change variables for the first three; integrate by parts for the last.

Exercise 6.3Fourier inversion

Prove inversion for $f\in\mathcal S(\mathbb R)$ .

Hint

Insert $e^{-\pi\varepsilon^2\xi^2}$ : $\int\hat f(\xi)e^{-\pi\varepsilon^2\xi^2}e^{2\pi ix\xi}d\xi=(f*G_\varepsilon)(x)$ with $G_\varepsilon$ the Gaussian approximate identity, by Fubini and Exercise 6.1. Let $\varepsilon\to0$ on both sides.

Solution

By Fubini, $\int\hat f(\xi)e^{-\pi\varepsilon^2\xi^2}e^{2\pi ix\xi}d\xi=\int f(y)\Big(\int e^{-\pi\varepsilon^2\xi^2}e^{2\pi i(x-y)\xi}d\xi\Big)dy$ , and by Exercise 6.1 with a dilation the inner integral is $\varepsilon^{-1}e^{-\pi(x-y)^2/\varepsilon^2}$ , a Gaussian approximate identity. So the right side tends to $f(x)$ (Exercise 3.2), and the left side tends to $\int\hat fe^{2\pi ix\xi}$ by dominated convergence, since $\hat f\in\mathcal S\subseteq L^1$ .

Exercise 6.4Plancherel

Show $\int f\bar g=\int\hat f\bar{\hat g}$ for $f,g\in\mathcal S$ , and explain how this extends $\mathcal F$ to a unitary map of $L^2$ .

Hint

$\int\hat f\,h=\int f\,\hat h$ (Fubini), with $h=\bar{\hat g}$ , whose transform is $\bar g$ by inversion. Then extend by density of $\mathcal S$ in $L^2$ .

Exercise 6.5Poisson summation

Prove Poisson summation for $f\in\mathcal S(\mathbb R)$ .

Hint

$F(x)=\sum_kf(x+k)$ is smooth and 1-periodic. Compute its Fourier coefficients: $\hat F(m)=\int_0^1\sum_kf(x+k)e^{-2\pi imx}dx=\hat f(m)$ . Its Fourier series converges to it by Exercise 5.5.

Solution

The series for $F$ and its derivatives converge uniformly because $f\in\mathcal S$ , so $F\in C^\infty(\mathbb T)$ . Unfolding, $\hat F(m)=\sum_k\int_k^{k+1}f(y)e^{-2\pi imy}dy=\hat f(m)$ . Since $F\in C^1$ , its Fourier series converges to it everywhere (Exercise 5.5): $F(x)=\sum_m\hat f(m)e^{2\pi imx}$ .

Exercise 6.6Heisenberg's inequality

Prove $\lVert f\rVert_2^2\le4\pi\lVert xf\rVert_2\lVert\xi\hat f\rVert_2$ for $f\in\mathcal S(\mathbb R)$ .

Hint

$\lVert f\rVert_2^2=\int|f|^2=-\int x\frac d{dx}|f|^2=-2\mathrm{Re}\int x\bar ff'$ . Use Cauchy–Schwarz, then Plancherel: $\lVert f'\rVert_2=2\pi\lVert\xi\hat f\rVert_2$ .

Exercise 6.7The two-sided exponential

Verify Example 6.3, and deduce $\widehat{P_y}(\xi)=e^{-2\pi y|\xi|}$ for the Poisson kernel.

Hint

$2\int_0^\infty e^{-2\pi x}\cos(2\pi x\xi)dx=\frac{2\cdot2\pi}{(2\pi)^2+(2\pi\xi)^2}$ . Then invert and dilate.

Exercise 6.8Riemann–Lebesgue on ℝ

Show that for $f\in L^1(\mathbb R)$ , $\hat f$ is uniformly continuous, bounded by $\lVert f\rVert_1$ , and tends to 0 at infinity.

Hint

Continuity: dominated convergence. Decay: true for Gaussians or for $\mathcal S$ , and $\mathcal S$ is dense in $L^1$ .

ReadingGrafakos, Classical, §§2.3–2.4

Tempered distributions

The constant function 1 has no Fourier transform as an integral, and $\delta$ is not a function, yet each should be the transform of the other. Distributions make this precise by defining everything through pairing with test functions: a distribution is whatever you can integrate a Schwartz function against.

The idea

Tempered distributions (Grafakos §2.3.3).

\mathcal S'(\mathbb R^n)

is the space of continuous linear functionals on

\mathcal S

:

u

with

|\langle u,\phi\rangle|\le C\sum_{|\alpha|,|\beta|\le N}\sup|x^\alpha\partial^\beta\phi|

. Every

f

with

\int|f|(1+|x|)^{-N}<\infty

defines one by

\langle f,\phi\rangle=\int f\phi

. So do

\delta

(

\langle\delta,\phi\rangle=\phi(0)

) and its derivatives.

Operations by duality (Grafakos §2.3). Every operation is defined so that it agrees with the usual one on functions:

\langle\partial u,\phi\rangle=-\langle u,\partial\phi\rangle

(integration by parts), and

\langle\hat u,\phi\rangle=\langle u,\hat\phi\rangle

(from

\int\hat f\phi=\int f\hat\phi

). So every tempered distribution has derivatives of all orders and a Fourier transform, and all the rules of module 6 continue to hold.

The basic transforms.

\hat\delta=1

,

\hat1=\delta

,

\widehat{e^{2\pi iax}}=\delta_a

,

\widehat{\partial^\alpha\delta}=(2\pi i\xi)^\alpha

. In one dimension, the principal value

\langle\mathrm{p.v.}\tfrac1x,\phi\rangle=\lim_{\varepsilon\to0}\int_{|x|>\varepsilon}\frac{\phi(x)}x\,dx

is a tempered distribution with

\mathcal F\big(\mathrm{p.v.}\tfrac1x\big)=-\pi i\,\mathrm{sgn}\,\xi

(Exercise 7.4). That is the multiplier of the Hilbert transform in module 9.

Support and homogeneity (Grafakos §2.4). A distribution supported at one point is a finite combination of derivatives of

\delta

there.

|x|^{-a}

on

\mathbb R^n

is locally integrable for

a<n

, and its transform is

c_{n,a}|\xi|^{a-n}

: homogeneity of degree

-a

becomes degree

a-n

. With

a=n-2

this gives the fundamental solution of the Laplacian,

-\Delta\frac{1}{(n-2)\omega_{n-1}|x|^{n-2}}=\delta

.

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Test function: A very well-behaved function, smooth and rapidly decaying, used as a probe: a generalized function is known by the numbers it produces when integrated against these.
Tempered distribution: A rule assigning a number to each well-behaved probe function, linearly and continuously, with at most polynomial growth allowed at infinity.
Dirac delta: The generalized function that returns the value of the probe at the origin: a unit mass concentrated at a single point.
Distributional derivative: The derivative defined by moving it onto the probe function with a minus sign, as integration by parts would; it exists for every generalized function.
Principal value: A way of integrating $1/x$ across 0 by cutting out a symmetric gap and letting the gap shrink, so the infinite parts on the two sides cancel.
Fundamental solution: A generalized function whose image under a differential operator is the Dirac delta; convolving with it solves the equation for any right-hand side.

Examples

Example 7.1 (derivatives of kinks and jumps).

$\frac d{dx}|x|=\mathrm{sgn}\,x$ and $\frac d{dx}\mathrm{sgn}\,x=2\delta$ as distributions (Exercise 7.2). A jump of size $J$ differentiates to $J\delta$ , which is why jumps cost $1/m$ decay in module 5.

Example 7.2 (1 and

\delta

).

$\langle\hat1,\phi\rangle=\int\hat\phi=\phi(0)$ by inversion, so $\hat1=\delta$ . Read backwards: the Dirichlet kernels $D_N$ are partial sums of $\sum_me^{2\pi imx}$ , which "equals" the comb $\sum_k\delta_k$ (Example 7.6).

Example 7.3 (principal value).

$1/x$ is not locally integrable, but $\langle\mathrm{p.v.}\tfrac1x,\phi\rangle=\int_{|x|<1}\frac{\phi(x)-\phi(0)}x\,dx+\int_{|x|\ge1}\frac{\phi(x)}x\,dx$ : the first integrand is bounded by $\sup|\phi'|$ , the second by $\sup|x\phi|/x^2$ .

Example 7.4 (the sign function).

$\mathcal F\big(\mathrm{sgn}\big)=\frac1{\pi i}\,\mathrm{p.v.}\frac1\xi$ , dual to the transform of $\mathrm{p.v.}\frac1x$ . Check with Example 7.1: $\mathcal F\big(\mathrm{sgn}'\big)=2\pi i\xi\cdot\frac1{\pi i\xi}=2=\widehat{2\delta}$ .

Example 7.5 (the Laplacian in

\mathbb R^3

).

$-\Delta\frac1{4\pi|x|}=\delta$ . So $u=\frac1{4\pi|x|}*f$ solves $-\Delta u=f$ : the Newtonian potential. On the Fourier side, $-\Delta$ is multiplication by $4\pi^2|\xi|^2$ , and $\widehat{1/(4\pi|x|)}=\frac1{4\pi^2|\xi|^2}$ .

Example 7.6 (the Dirac comb).

Poisson summation (Exercise 6.5) says exactly that the Dirac comb $\text{Ш}=\sum_{k\in\mathbb Z}\delta_k$ satisfies $\hat{\text{Ш}}=\text{Ш}$ .

Check your understanding

The distributional derivative of the indicator $\mathbf 1_{[0,1]}$ is:

$\delta_0-\delta_1$
0, since it is constant almost everywhere.That is the pointwise derivative; the distributional one sees the jumps.
$\delta_0+\delta_1$ The jump at 1 is downwards.

-\int_0^1\phi'=\phi(0)-\phi(1)

.

Is $e^{x}$ a tempered distribution on $\mathbb R$ ?

Yes, every locally integrable function is.Tempered means at most polynomial growth, roughly.
No: $\int e^x\phi$ diverges for some Schwartz $\phi$ , e.g. $\phi=e^{-\sqrt{1+x^2}}$ .

It is a distribution in the larger space

\mathcal D'

, paired with compactly supported test functions, but it has no Fourier transform in

\mathcal S'

.

Exercises

Exercise 7.1The principal value is tempered

Show that the limit defining $\mathrm{p.v.}\frac1x$ exists for $\phi\in\mathcal S$ , equals the formula in Example 7.3, and satisfies $|\langle\mathrm{p.v.}\frac1x,\phi\rangle|\le2\sup|\phi'|+2\sup|x\phi|$ .

Hint

$\int_{\varepsilon<|x|<1}\frac{\phi(0)}x\,dx=0$ by oddness, so you may subtract $\phi(0)$ inside the inner integral.

Solution

$\int_{|x|>\varepsilon}\frac\phi x=\int_{\varepsilon<|x|<1}\frac{\phi(x)-\phi(0)}x+\int_{|x|\ge1}\frac\phi x$ , using oddness. By the mean value theorem the first integrand is bounded by $\sup|\phi'|$ , so it converges as $\varepsilon\to0$ to an integral over $|x|<1$ of size at most $2\sup|\phi'|$ . The second is at most $\int_{|x|\ge1}\frac{\sup|x\phi|}{x^2}=2\sup|x\phi|$ .

Exercise 7.2Derivatives of

|x|

and

\mathrm{sgn}

Prove the two identities of Example 7.1.

Hint

$-\int|x|\phi'=-\int_0^\infty x\phi'+\int_{-\infty}^0x\phi'$ ; integrate by parts on each half-line.

Solution

$-\int_0^\infty x\phi'=\int_0^\infty\phi$ and $\int_{-\infty}^0x\phi'=-\int_{-\infty}^0\phi$ , so $\langle|x|',\phi\rangle=\int\mathrm{sgn}\,\phi$ . Next, $-\int\mathrm{sgn}\,\phi'=-\int_0^\infty\phi'+\int_{-\infty}^0\phi'=\phi(0)+\phi(0)$ .

Exercise 7.3

\hat\delta=1

and

\hat1=\delta

Prove both from the definitions.

Hint

$\langle\hat\delta,\phi\rangle=\hat\phi(0)=\int\phi$ ; $\langle\hat1,\phi\rangle=\int\hat\phi=\phi(0)$ by inversion at 0.

Exercise 7.4The transform of

\mathrm{p.v.}\frac1x

Show $\mathcal F\big(\mathrm{p.v.}\frac1x\big)=-\pi i\,\mathrm{sgn}\,\xi$ .

Hint

$\mathrm{p.v.}\frac1x$ is the limit in $\mathcal S'$ of $\frac{x}{x^2+\varepsilon^2}$ , whose transform is $-\pi i\,\mathrm{sgn}(\xi)e^{-2\pi\varepsilon|\xi|}$ (use Exercise 6.7 and $\frac x{x^2+\varepsilon^2}=\mathrm{Re}\frac1{x-i\varepsilon}$ ). Or check that $x\cdot\mathrm{p.v.}\frac1x=1$ , transform, and fix the constant by oddness.

Exercise 7.5The Dirac comb is its own transform

Show that $\text{Ш}=\sum_k\delta_k$ is tempered and $\hat{\text{Ш}}=\text{Ш}$ .

Hint

$|\sum_k\phi(k)|\le C\sup(1+x^2)|\phi|$ . Then $\langle\hat\Sha,\phi\rangle=\sum_k\hat\phi(k)=\sum_k\phi(k)$ is Poisson summation.

Exercise 7.6

\delta

is not a function

Show there is no $g\in L^1_{\mathrm{loc}}$ with $\int g\phi=\phi(0)$ for all $\phi\in C_c^\infty$ .

Hint

Take $\phi_\varepsilon(x)=\phi(x/\varepsilon)$ with $\phi(0)=1$ , $0\le\phi\le1$ . Then $1=\int g\phi_\varepsilon\le\int_{|x|<\varepsilon}|g|\to0$ .

Exercise 7.7The Newtonian potential

Prove $-\Delta\frac1{4\pi|x|}=\delta$ in $\mathcal S'(\mathbb R^3)$ .

Hint

$\frac1{|x|}$ is harmonic away from 0. Apply Green's identity to $\int_{|x|>\varepsilon}\frac1{|x|}\Delta\phi$ ; the boundary terms on the sphere of radius $\varepsilon$ tend to $-4\pi\phi(0)$ and 0.

Solution

For $r=|x|>0$ , $\Delta\frac1r=0$ . Green's second identity on $\{\varepsilon<r<R\}$ , with $R$ beyond the support of $\phi$ , gives $\int_{r>\varepsilon}\frac1r\Delta\phi=\int_{r=\varepsilon}\big(\frac1r\partial_\nu\phi-\phi\,\partial_\nu\frac1r\big)dS$ , where $\nu$ points towards the origin. The first boundary term is $O(\varepsilon^{-1}\cdot\varepsilon^2)\to0$ . The second is $-\int_{r=\varepsilon}\phi\cdot\frac1{\varepsilon^2}dS\to-4\pi\phi(0)$ . So $\langle\Delta\frac1{4\pi r},\phi\rangle=-\phi(0)$ .

InteractiveGrafakos, Classical, §2.1

The maximal function and Lebesgue differentiation

The Hardy–Littlewood maximal function records, at each point, the largest average of $|f|$ over balls centred there. It is not integrable, but it is "almost" integrable (weak type (1,1)), and that is exactly enough to prove that averages of an integrable function converge back to it almost everywhere.

The idea

The maximal function (Grafakos §2.1.1).

Mf(x)=\sup_{r>0}\frac1{|B(x,r)|}\int_{B(x,r)}|f(y)|\,dy

. It is sublinear,

Mf\ge|f|

almost everywhere, and

\lVert Mf\rVert_\infty\le\lVert f\rVert_\infty

.

The maximal theorem (Grafakos Thm. 2.1.6). Weak type (1,1):

|\{Mf>\lambda\}|\le\frac{3^n}\lambda\lVert f\rVert_1

. Combined with the trivial

L^\infty

bound by Marcinkiewicz interpolation (Grafakos §1.3.1),

\lVert Mf\rVert_p\le C_{n,p}\lVert f\rVert_p

for

1<p\le\infty

. The proof of the weak bound is a covering argument: Vitali's lemma picks disjoint balls covering a third of the measure (Exercise 8.1).

Lebesgue differentiation (Grafakos Cor. 2.1.16). For

f\in L^1_{\mathrm{loc}}

,

\frac1{|B(x,r)|}\int_{B(x,r)}|f-f(x)|\to0

for almost every

x

. The pattern of proof, "true for a dense class, controlled by a maximal function, hence true a.e.", recurs throughout harmonic analysis. In one dimension it says an indefinite integral

F(x)=\int_a^xf

has

F'=f

almost everywhere: the Lebesgue version of the fundamental theorem of calculus.

Controlling approximate identities. If

|K_\varepsilon|

is bounded by a radially decreasing integrable function

\Phi_\varepsilon

, then

\sup_\varepsilon|K_\varepsilon*f|\le\lVert\Phi\rVert_1\,Mf

(Grafakos Thm. 2.1.10). So Gaussian and Poisson means of

L^1

functions converge almost everywhere, not just in norm.

Explore: Mf and its level sets

Function f

Level λ

|f|

,

Mf

, the level

\lambda

, and the set

\{Mf>\lambda\}

shaded on the axis.

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Hardy–Littlewood maximal function: At each point, the largest average of the function's size over all balls centred at that point.
Weak type (1,1): The set where an operator's output exceeds a level has measure at most a constant times the input's total integral divided by the level: integrable "up to a logarithm".
Vitali covering lemma: From any finite collection of balls you can choose some that don't overlap, so that tripling their radii covers all the original balls.
Lebesgue point: A point where the averages of the function over shrinking balls converge to its value there; almost every point is one.
Marcinkiewicz interpolation: A way to get a genuine bound on the spaces in between from two weaker bounds at the ends, such as one in the integrable space and one in the bounded space.

Examples

Example 8.1 (one box).

For $f=\mathbf 1_{[-1,1]}$ , $Mf=1$ on $[-1,1]$ and $Mf(x)=\frac1{|x|+1}$ for $|x|>1$ (the best ball just reaches the far end of the box). So $Mf\notin L^1$ , although $|\{Mf>\lambda\}|=\max\big(2,\frac2\lambda-2\big)\le\frac2\lambda$ for $0<\lambda<1$ , well within the bound $3\lVert f\rVert_1/\lambda=6/\lambda$ .

Example 8.2 (never integrable).

For any $f\ne0$ in $L^1(\mathbb R^n)$ , $Mf(x)\ge\frac c{|x|^n}$ for large $|x|$ : a ball centred at $x$ of radius $2|x|$ captures a fixed part of the mass of $f$ . So the $L^1$ bound genuinely fails, and weak type is the best possible.

Example 8.3 (a tall spike).

$f=N\mathbf 1_{[0,1/N]}$ has $\lVert f\rVert_1=1$ for all $N$ , while $Mf(x)\approx\frac1{2|x|}$ away from the spike. The level sets do not depend on $N$ : the weak bound sees only $\lVert f\rVert_1$ .

Example 8.4 (a singularity).

$f(x)=|x|^{-1/2}\mathbf 1_{|x|<1}$ is in $L^p$ for $p<2$ , and $Mf$ has the same $|x|^{-1/2}$ singularity at 0, so it is in exactly the same $L^p$ spaces near 0, as the $L^p$ bound predicts.

Example 8.5 (the fundamental theorem of calculus).

For $f\in L^1([a,b])$ , $\frac{F(x+h)-F(x)}h=\frac1h\int_x^{x+h}f\to f(x)$ almost everywhere, by Lebesgue differentiation with one-sided intervals. Continuity of $f$ , as in the Riemann version, is not needed.

Example 8.6 (heat and Poisson means).

The Gaussian and Poisson kernels are radially decreasing with integral 1, so $\sup_t|G_t*f|\le Mf$ and $\sup_y|P_y*f|\le Mf$ . Hence $P_y*f\to f$ almost everywhere for $f\in L^1$ : Fatou's theorem on boundary values of harmonic functions.

Check your understanding

For nonzero $f\in L^1(\mathbb R)$ , is $Mf\in L^1(\mathbb R)$ ?

Yes, by the maximal theorem.The theorem gives weak type (1,1), not an $L^1$ bound.
Never: $Mf\ge c/|x|$ for large $|x|$ .

Example 8.2. The failure is at infinity; near any point

Mf

can be integrable or not depending on

f

(Example 8.4).

In the proof of Lebesgue differentiation, what is the maximal function used for?

To show the set where averages misbehave is small, after approximating $f$ by a continuous function, for which the result is clear.
To compute the limit of the averages.It bounds them; the limit comes from the continuous approximation.

Exercise 8.3: "dense class plus maximal inequality" is the template.

Exercises

Exercise 8.1Vitali's covering lemma

Show that from finitely many balls $B_1,\dots,B_k$ in $\mathbb R^n$ one can choose pairwise disjoint ones $B_{i_1},\dots,B_{i_l}$ with $\bigcup_jB_j\subseteq\bigcup_m3B_{i_m}$ , so $|\bigcup B_j|\le3^n\sum_m|B_{i_m}|$ .

Hint

Greedy: take the largest ball, discard every ball meeting it, repeat. A discarded ball meets a chosen one at least as large, so it lies in its triple.

Solution

Order the balls by decreasing radius. Choose $B_{i_1}$ as the largest; remove all balls meeting it; choose the largest remaining; and so on. The chosen balls are disjoint. If $B_j$ was removed because it meets a chosen $B$ with radius $r\ge\mathrm{rad}(B_j)$ , then every point of $B_j$ is within $r+2\,\mathrm{rad}(B_j)\le3r$ of the centre of $B$ , so $B_j\subseteq3B$ . Finally $|3B|=3^n|B|$ .

Exercise 8.2Weak type (1,1)

Prove $|\{Mf>\lambda\}|\le\frac{3^n}\lambda\lVert f\rVert_1$ .

Hint

Each $x$ in $E=\{Mf>\lambda\}$ is the centre of a ball $B_x$ with $\int_{B_x}|f|>\lambda|B_x|$ . Cover a compact $K\subseteq E$ by finitely many, apply Vitali, and use inner regularity.

Solution

For compact $K\subseteq E$ , finitely many $B_x$ cover $K$ . Vitali gives disjoint ones $B_m$ with $|K|\le3^n\sum|B_m|<\frac{3^n}\lambda\sum\int_{B_m}|f|\le\frac{3^n}\lambda\lVert f\rVert_1$ . Take the supremum over $K$ (Lebesgue measure is inner regular, and $E$ is open since $Mf$ is lower semicontinuous).

Exercise 8.3Lebesgue differentiation

Deduce the differentiation theorem from Exercise 8.2.

Hint

Let $\Omega f(x)=\limsup_{r\to0}\frac1{|B|}\int_B|f-f(x)|$ . For continuous $g$ , $\Omega g=0$ . Show $\Omega f\le M(f-g)+|f-g|$ , so $|\{\Omega f>2\lambda\}|\le\frac{3^n+1}\lambda\lVert f-g\rVert_1$ , which is as small as you like.

Exercise 8.4Compute Example 8.1

Verify the formula for $M\mathbf 1_{[-1,1]}$ and the measure of its level sets.

Hint

For $x>1$ the average over $[x-r,x+r]$ is $\frac{r-x+1}{2r}$ for $x-1\le r\le x+1$ , which increases in $r$ , then $\frac1r$ for larger $r$ . Maximum at $r=x+1$ .

Exercise 8.5Not integrable at infinity

Prove Example 8.2.

Hint

Pick $R$ with $\int_{|y|<R}|f|=a>0$ . For $|x|>R$ , the ball $B(x,2|x|)$ contains $B(0,R)$ , so $Mf(x)\ge\frac a{|B(0,1)|(2|x|)^n}$ .

Exercise 8.6From weak (1,1) to

L^p

Prove $\lVert Mf\rVert_p\le C_p\lVert f\rVert_p$ for $1<p<\infty$ directly, without quoting Marcinkiewicz.

Hint

Split $f=f\mathbf 1_{|f|>\lambda/2}+f\mathbf 1_{|f|\le\lambda/2}$ . The second part has $M\le\lambda/2$ , so $\{Mf>\lambda\}\subseteq\{M(f\mathbf 1_{|f|>\lambda/2})>\lambda/2\}$ . Apply the weak bound and $\lVert Mf\rVert_p^p=p\int_0^\infty\lambda^{p-1}|\{Mf>\lambda\}|d\lambda$ , then Fubini.

Exercise 8.7Radial kernels are controlled by

M

Show that if $\Phi\ge0$ is radial, decreasing and integrable on $\mathbb R^n$ , then $|\Phi*f|\le\lVert\Phi\rVert_1Mf$ .

Hint

Write $\Phi$ as a superposition of normalized indicators of balls (the layer-cake formula). Each indicator convolved with $|f|$ is at most $Mf$ times its integral.

InteractiveGrafakos, Classical, §§4.1, 5.1, 5.3

The Hilbert transform and singular integrals

Convolution with $\frac1{\pi x}$ is the simplest operator that is not an average: its kernel is not integrable, and it is defined only as a principal value. It is the multiplier $-i\,\mathrm{sgn}\,\xi$ , a 90° phase shift of every frequency. Whether it is bounded on $L^p$ is equivalent to whether Fourier series converge in $L^p$ .

The idea

Definition (Grafakos §5.1).

Hf(x)=\frac1\pi\,\mathrm{p.v.}\int\frac{f(y)}{x-y}dy

, that is,

Hf=\frac1\pi\mathrm{p.v.}\frac1x*f

. By module 7,

\widehat{Hf}(\xi)=-i\,\mathrm{sgn}(\xi)\hat f(\xi)

. So

H

is an isometry of

L^2

with

H^2=-I

, and

H\cos=\sin

,

H\sin=-\cos

.

Riesz's theorem (Grafakos §§4.1.2, 5.1).

H

is bounded on

L^p

for

1<p<\infty

, but not on

L^1

or

L^\infty

:

H\mathbf 1_{[a,b]}=\frac1\pi\log\big|\frac{x-a}{x-b}\big|

is unbounded and decays only like

1/x

. At

p=1

it is weak type (1,1), proved by the Calderón–Zygmund decomposition.

Complex analysis.

f+iHf

is the boundary value of the holomorphic function

F=(P_y+iQ_y)*f

on the upper half-plane, where

Q_y(x)=\frac1\pi\frac x{x^2+y^2}

is the conjugate Poisson kernel. On the circle the same operator is the conjugate function

\tilde f=\sum_m-i\,\mathrm{sgn}(m)\hat f(m)e^{2\pi imx}

, and the partial sums

S_N

can be written using it. So

S_Nf\to f

in

L^p

for all

f\in L^p

exactly when the conjugate function is bounded on

L^p

: true for

1<p<\infty

, false for

p=1,\infty

(Grafakos §4.1.1).

Calderón–Zygmund decomposition (Grafakos §5.3). Given

f\in L^1

and

\alpha>0

, write

f=g+b

with

|g|\le2^n\alpha

and

b=\sum b_j

supported on disjoint dyadic cubes

Q_j

, each with

\int b_j=0

and

\sum|Q_j|\le\frac1\alpha\lVert f\rVert_1

. Bounded on

g

by

L^2

, and on each

b_j

because of the cancellation, an operator like

H

becomes weak type (1,1). The Riesz transforms

R_j

, with multipliers

-i\xi_j/|\xi|

, are the

n

-dimensional versions.

Explore: the Hilbert transform of a few functions

Function f

a b

f

and

Hf

, computed by a principal-value quadrature (and compared with the exact formula where there is one).

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Hilbert transform: The operator that shifts the phase of every positive frequency by a quarter turn one way and every negative frequency a quarter turn the other way, turning cosines into sines.
Singular integral: An operator given by integrating against a kernel that blows up like one over the distance to the diagonal, too singular to be an average, but tamed by cancellation.
Conjugate function: On the circle, the function whose sum with $i$ times the original is the boundary value of an analytic function; it is the periodic Hilbert transform.
Calderón–Zygmund decomposition: Splitting a function at a height into a bounded part and a bad part made of pieces with zero average, living on disjoint cubes of small total size.
Fourier multiplier: An operator that multiplies each frequency component by its own number; the list of those numbers, as a function of frequency, is the symbol.
Riesz transforms: The higher-dimensional cousins of the phase-shift operator, one per coordinate, with symbols $-i\xi_j/|\xi|$ .

Examples

Example 9.1 (a phase shift).

$\cos(2\pi x)=\frac12(e^{2\pi ix}+e^{-2\pi ix})$ , and multiplying the two frequencies by $-i$ and $+i$ gives $\sin(2\pi x)$ . (These are tempered distributions, not $L^p$ functions, but the multiplier formula applies.)

Example 9.2 (a box).

$H\mathbf 1_{[a,b]}(x)=\frac1\pi\log\big|\frac{x-a}{x-b}\big|$ : logarithmic spikes of opposite signs at the two ends, and a $\frac{b-a}{\pi x}$ tail. So $H$ does not map $L^\infty$ to $L^\infty$ or $L^1$ to $L^1$ , although $H\mathbf 1_{[a,b]}\in L^p$ for every $1<p<\infty$ .

Example 9.3 (from complex analysis).

$F(z)=\frac i{z+i}$ is holomorphic in the upper half-plane, and on the real line $F(x)=\frac1{1+x^2}+i\frac x{1+x^2}$ . So $H\frac1{1+x^2}=\frac x{1+x^2}$ : the Poisson kernel's conjugate is the conjugate Poisson kernel (at height 1).

Example 9.4 (

L^2

norms).

For the box of length $\ell$ , $\lVert\mathbf 1\rVert_2^2=\ell$ , and the figure's numerical $\lVert Hf\rVert_2^2$ agrees, as the multiplier $|-i\,\mathrm{sgn}\,\xi|=1$ predicts.

Example 9.5 (Riesz transforms).

$\sum_jR_j^2=-I$ , since $\sum_j(-i\xi_j/|\xi|)^2=-1$ . And $\partial_j\partial_ku=-R_jR_k\Delta u$ , so $L^p$ bounds for Riesz transforms give $L^p$ bounds for all second derivatives from the Laplacian alone, the starting point of elliptic regularity.

Example 9.6 (Fourier series in

L^p

).

On the circle let $P$ keep the frequencies $m\ge0$ ; then $P=\frac12(I+i\tilde H)+\frac12E_0$ , where $\tilde H$ is the conjugate function and $E_0f=\hat f(0)$ . With $M_af=e^{2\pi iax}f$ , which shifts frequencies by $a$ and preserves $L^p$ norms, $S_N=M_{-N}PM_N-M_{N+1}PM_{-N-1}$ . So a bounded conjugate function gives $\sup_N\lVert S_N\rVert_{L^p\to L^p}<\infty$ , which is $L^p$ convergence of every Fourier series.

Check your understanding

On which spaces is the Hilbert transform bounded?

$L^p$ for $1<p<\infty$ .
$L^p$ for $1\le p\le\infty$ .Example 9.2 rules out $p=1$ and $p=\infty$ .
Only $L^2$ . $L^2$ is the easy case; Riesz proved the rest.

At the endpoints it is weak type (1,1), and it maps

L^\infty

into BMO (module 10).

What is $H(Hf)$ ?

$-f$
$f$ $(-i\,\mathrm{sgn}\,\xi)^2=-1$ .
0 $H$ is an isometry of $L^2$ .

Two quarter-turn phase shifts make a half turn.

Exercises

Exercise 9.1The Hilbert transform of a box

Compute $H\mathbf 1_{[a,b]}$ and show it is not bounded and not integrable.

Hint

For $x\notin[a,b]$ the integral is ordinary: $\frac1\pi\int_a^b\frac{dy}{x-y}$ . For $x\in(a,b)$ , cut out $(x-\varepsilon,x+\varepsilon)$ and note the two logarithms of $\varepsilon$ cancel.

Solution

$\frac1\pi\int_a^b\frac{dy}{x-y}=\frac1\pi\big[-\log|x-y|\big]_a^b=\frac1\pi\log\frac{|x-a|}{|x-b|}$ , and inside $(a,b)$ the principal value gives the same formula. It tends to $\pm\infty$ at $a$ and $b$ , and for large $|x|$ it equals $\frac1\pi\log\big(1+\frac{b-a}{x-b}\big)\approx\frac{b-a}{\pi x}$ , which is not integrable.

Exercise 9.2The multiplier

Using Exercise 7.4, show $\widehat{Hf}=-i\,\mathrm{sgn}(\xi)\hat f$ , and deduce that $H$ is an isometry of $L^2$ with $H^2=-I$ .

Hint

$\widehat{u*f}=\hat u\hat f$ for $u\in\mathcal S'$ and $f\in\mathcal S$ ; $\frac1\pi\cdot(-\pi i\,\mathrm{sgn}\,\xi)$ . Then Plancherel.

Exercise 9.3Phase shifts

Show $H\cos(2\pi kx)=\sin(2\pi kx)$ for $k>0$ , and check Example 9.3 by evaluating the principal value integral with residues.

Hint

For Example 9.3, close the contour in the lower or upper half-plane and add half the residue at the real pole (the principal value contributes $\pi i$ times the residue there).

Exercise 9.4Symmetries of

H

Show $H$ commutes with translations and positive dilations and anticommutes with the reflection $f(x)\mapsto f(-x)$ . Conversely, show that a bounded operator on $L^2$ with these three properties is $cH$ for a constant $c$ .

Hint

Commuting with translations makes it a multiplier $m(\xi)$ (the Fourier transform diagonalizes translations, exactly as in the finite case). Dilations force $m$ to be constant on $(0,\infty)$ and on $(-\infty,0)$ ; reflection forces $m(-\xi)=-m(\xi)$ .

Exercise 9.5Riesz's theorem for

p=4

Prove $\lVert Hf\rVert_4\le C\lVert f\rVert_4$ for real $f\in\mathcal S$ , using the identity $(Hf)^2=f^2+2H(fHf)$ .

Hint

The identity holds because $(f+iHf)^2$ is again a boundary value of a holomorphic function. Then $\lVert Hf\rVert_4^2=\lVert(Hf)^2\rVert_2\le\lVert f\rVert_4^2+2\lVert fHf\rVert_2\le\lVert f\rVert_4^2+2\lVert f\rVert_4\lVert Hf\rVert_4$ , a quadratic inequality in $\lVert Hf\rVert_4/\lVert f\rVert_4$ . Iterating gives all $p=2^k$ , and interpolation and duality give the rest.

Exercise 9.6The Calderón–Zygmund decomposition

Construct the decomposition in one dimension: given $f\in L^1(\mathbb R)$ and $\alpha>0$ , select the maximal dyadic intervals $Q$ with $\frac1{|Q|}\int_Q|f|>\alpha$ , and show $\alpha<\frac1{|Q|}\int_Q|f|\le2\alpha$ , $\sum|Q|\le\frac1\alpha\lVert f\rVert_1$ , and $|f|\le\alpha$ a.e. off $\bigcup Q$ .

Hint

Maximality: the dyadic parent $\tilde Q$ of a chosen $Q$ has average $\le\alpha$ , and $|\tilde Q|=2|Q|$ . Off the union, every dyadic interval containing $x$ has average $\le\alpha$ ; use Lebesgue differentiation with dyadic intervals. Set $g=f$ off $\bigcup Q$ and $g=$ the average of $f$ on each $Q$ .

Exercise 9.7Riesz transforms and second derivatives

Verify $\sum_jR_j^2=-I$ and $\partial_j\partial_ku=-R_jR_k\Delta u$ for $u\in\mathcal S(\mathbb R^n)$ .

Hint

On the Fourier side: $\partial_j\partial_k$ is $-4\pi^2\xi_j\xi_k$ , $\Delta$ is $-4\pi^2|\xi|^2$ , and $R_jR_k$ is $-\xi_j\xi_k/|\xi|^2$ .

InteractiveGrafakos, Classical, §6.1; Grafakos, Modern, §§1.3, 3.1–3.2

Littlewood–Paley, Sobolev spaces and BMO

Split a function into frequency bands an octave wide. In $L^2$ the bands are orthogonal; the Littlewood–Paley theorem says that, in a square-function sense, they remain almost orthogonal in every $L^p$ . Measuring the bands with weights $2^{js}$ gives Sobolev smoothness, and the right endpoint space at $p=\infty$ is BMO, the functions of bounded mean oscillation. This is where Modern Fourier Analysis begins.

The idea

Littlewood–Paley theorem (Grafakos, Classical §6.1). Choose

\psi

with

\hat\psi

supported in

\frac12\le|\xi|\le2

and

\sum_{j\in\mathbb Z}\hat\psi(2^{-j}\xi)=1

for

\xi\ne0

, and put

\widehat{\Delta_jf}=\hat\psi(2^{-j}\xi)\hat f

. Then for

1<p<\infty

,

\lVert f\rVert_p\approx\Big\lVert\Big(\sum_j|\Delta_jf|^2\Big)^{1/2}\Big\rVert_p.

At

p=2

it is Plancherel plus the finite overlap of the bands; for other

p

it is a vector-valued singular integral estimate, the machinery of module 9.

Sobolev spaces (Grafakos, Modern §1.3).

H^s(\mathbb R^n)=\{f:(1+|\xi|^2)^{s/2}\hat f\in L^2\}

: "

s

derivatives in

L^2

", for any real

s

. Equivalently

\sum_j2^{2js}\lVert\Delta_jf\rVert_2^2<\infty

(plus a low-frequency part). The

L^p

versions use the square function. Sobolev embedding:

H^s\subseteq C_0

when

s>n/2

(Exercise 10.1), and it fails at

s=n/2

.

BMO (Grafakos, Modern §§3.1–3.2).

\lVert f\rVert_{BMO}=\sup_Q\frac1{|Q|}\int_Q|f-f_Q|

, with

f_Q

the average over the cube

Q

.

L^\infty\subsetneq BMO

:

\log|x|

is in BMO. Singular integrals such as

H

map

L^\infty

to BMO, BMO is the dual of the Hardy space

H^1

(Fefferman), and the John–Nirenberg inequality says BMO functions are exponentially integrable on each cube.

Explore: dyadic frequency bands and Sobolev norms

Function on the circle

Sobolev exponent s

Energy

\sum_{2^{j-1}\le|m|<2^j}|\hat f(m)|^2

of each band

j

, and the same weighted by

2^{2js}

Partial sums

\sum_{j\le J}2^{2js}(\text{band energy})

against

J

: bounded exactly when

f\in H^s

Key terms in plain words

The ideas above without the symbols. Switch to Quiz me to test yourself: you see an explanation and pick the term it describes.

Littlewood–Paley decomposition: Splitting a function into pieces whose frequencies lie in bands an octave wide, so that each piece oscillates at roughly one scale.
Square function: At each point, the square root of the sum of the squared sizes of all the frequency-band pieces there; its size in any integrable power measures the size of the function.
Sobolev space: The functions with a given number of derivatives, possibly fractional, in a given integrable power; measured by weighting each frequency by a power of its size.
Sobolev embedding: Enough derivatives in an average sense, compared with the dimension, force a function to be continuous and bounded.
Bounded mean oscillation: Describes a function whose average distance from its own mean, over every cube, is bounded, even if the function itself is not, like the logarithm.
John–Nirenberg inequality: The statement that functions of bounded average oscillation deviate from their means on large sets only with exponentially small probability.

Examples

Example 10.1 (a pure frequency).

$e^{2\pi imx}$ lives in the single band $j$ with $2^{j-1}\le|m|<2^j$ , so $\Delta_j$ of it is itself or 0, and the square function is just $|f|$ .

Example 10.2 (a jump costs half a derivative).

The square wave has $|\hat f(m)|\approx\frac1m$ , so band $j$ has energy about $2^{-j}$ , and $\sum_j2^{2js}2^{-j}<\infty$ iff $s<\frac12$ . Functions with jumps are in $H^s$ exactly for $s<\frac12$ . The triangle wave reaches $s<\frac32$ .

Example 10.3 (Sobolev embedding in one dimension).

For $f\in H^1(\mathbb R)$ , $|f(x)-f(y)|=\big|\int_y^xf'\big|\le|x-y|^{1/2}\lVert f'\rVert_2$ by Cauchy–Schwarz: $H^1$ functions are Hölder continuous of order $\frac12$ .

Example 10.4 (

\log|x|

is in BMO).

Its oscillation over $[a,b]$ is unchanged by rescaling $x\mapsto\lambda x$ (which adds a constant) so it suffices to check intervals of length 1. Those near 0 have bounded oscillation because $\log|x|$ is integrable; those far away because it is nearly constant there. On the circle, $\log|\sin\pi x|$ in the figure has the same singularity, $|\hat f(m)|\approx\frac1{2|m|}$ , and is in $H^s$ for $s<\frac12$ but not bounded.

Example 10.5 (

H

maps

L^\infty

into BMO).

$H\mathbf 1_{[a,b]}=\frac1\pi\log\big|\frac{x-a}{x-b}\big|$ from Example 9.2 is a difference of two translated copies of $\frac1\pi\log|x|$ : unbounded, but in BMO.

Example 10.6 (embedding fails at the endpoint).

$\log\log\frac e{|x|}$ , cut off near $|x|=1$ , is in $H^1(\mathbb R^2)$ but unbounded at 0. With $n=2$ and $s=1=\frac n2$ , $H^s\not\subseteq L^\infty$ ; such functions do lie in BMO.

Check your understanding

For which $s$ is the square wave in $H^s(\mathbb T)$ ?

$s<\frac12$
$s\le1$ A jump can't be in $H^1$ : $H^1$ functions are continuous in one dimension.
$s=0$ onlyMove the slider: the partial sums stay bounded for small positive $s$ .

\sum_m m^{2s}m^{-2}

converges iff

2s-2<-1

.

Which of these is in BMO but not in $L^\infty$ ?

$\log|x|$
$|x|^{-1/2}$ Its oscillation over $[0,r]$ grows like $r^{-1/2}$ (Exercise 10.6).
$\mathrm{sgn}\,x$ That is bounded.

Power singularities are too strong for BMO; logarithmic ones are just right, by John–Nirenberg.

Exercises

Exercise 10.1Sobolev embedding

Show $H^s(\mathbb R^n)\subseteq C_0(\mathbb R^n)$ for $s>n/2$ , with $\lVert f\rVert_\infty\le C_{n,s}\lVert f\rVert_{H^s}$ .

Hint

$\lVert f\rVert_\infty\le\lVert\hat f\rVert_1=\int(1+|\xi|^2)^{-s/2}\cdot(1+|\xi|^2)^{s/2}|\hat f|$ , then Cauchy–Schwarz.

Solution

By Cauchy–Schwarz, $\lVert\hat f\rVert_1\le\big(\int(1+|\xi|^2)^{-s}d\xi\big)^{1/2}\lVert f\rVert_{H^s}$ , and the first integral is finite iff $2s>n$ (polar coordinates). So $\hat f\in L^1$ , and $f$ is the inverse transform of an integrable function: continuous and vanishing at infinity by Riemann–Lebesgue (Exercise 6.8).

Exercise 10.2Littlewood–Paley in

L^2

Show $\frac12\lVert f\rVert_2^2\le\sum_j\lVert\Delta_jf\rVert_2^2\le\lVert f\rVert_2^2$ when $0\le\hat\psi\le1$ .

Hint

By Plancherel, $\sum_j\lVert\Delta_jf\rVert_2^2=\int\sum_j\hat\psi(2^{-j}\xi)^2|\hat f|^2$ . Each $\xi\ne0$ lies in at most two bands, and $\sum\hat\psi=1$ with $0\le\hat\psi\le1$ gives $\frac12\le\sum\hat\psi^2\le1$ .

Exercise 10.3Jumps and corners

Verify Example 10.2, and show that the triangle wave is in $H^s(\mathbb T)$ exactly for $s<\frac32$ . Check both with the figure.

Hint

$\sum_m m^{2s-4}$ converges iff $2s-4<-1$ .

Exercise 10.4

\log|x|\in BMO

Prove Example 10.4 on $\mathbb R$ .

Hint

By scaling invariance, take $|I|=1$ . If $I\subseteq[-2,2]$ , compare with the constant 0: $\int_I|\log|x||\le\int_{-2}^2|\log|x||<\infty$ . If $I$ is far from 0, $\log|x|$ varies by at most $\log\frac{|c|+1}{|c|-1}$ on it, where $c$ is the centre. Use $\frac1{|I|}\int_I|f-f_I|\le\frac2{|I|}\int_I|f-a|$ for any constant $a$ .

Solution

First, for any constant $a$ , $\frac1{|I|}\int_I|f-f_I|\le\frac2{|I|}\int_I|f-a|$ (since $|f_I-a|\le\frac1{|I|}\int|f-a|$ ). Replacing $I$ by $\lambda I$ changes $\log|x|$ by the constant $\log\lambda$ , which doesn't change the oscillation, so assume $|I|=1$ . If $I\subseteq[-2,2]$ , take $a=0$ : the oscillation is at most $2\int_{-2}^2|\log|x||\,dx$ . Otherwise $|x|\ge1$ on $I$ with $|x|$ between $|c|-\frac12$ and $|c|+\frac12$ ; take $a=\log|c|$ , and $|\log|x|-\log|c||\le\log 3$ . Both bounds are independent of $I$ .

Exercise 10.5BMO is a seminorm space

Show $\lVert\cdot\rVert_{BMO}$ is a seminorm vanishing exactly on constants, and $\lVert f\rVert_{BMO}\le2\lVert f\rVert_\infty$ .

Hint

If every oscillation is 0, $f$ equals its average on every cube, hence on nested cubes, hence is constant a.e. For the bound, use the inequality from Exercise 10.4 with $a=0$ .

Exercise 10.6Power singularities are not BMO

Show $|x|^{-1/2}\mathbf 1_{|x|<1}\notin BMO(\mathbb R)$ .

Hint

On $I=[0,r]$ , $f_I=2r^{-1/2}$ , and $f>2f_I$ on $[0,r/16]$ , so the oscillation is at least $\frac1r\int_0^{r/16}(f-f_I)\gtrsim r^{-1/2}\to\infty$ .

Exercise 10.7Consequences of John–Nirenberg

John–Nirenberg says $|\{x\in Q:|f-f_Q|>\lambda\}|\le C|Q|e^{-c\lambda/\lVert f\rVert_{BMO}}$ . Deduce that BMO functions are in $L^p_{\mathrm{loc}}$ for all $p<\infty$ , with $\big(\frac1{|Q|}\int_Q|f-f_Q|^p\big)^{1/p}\le C_p\lVert f\rVert_{BMO}$ .

Hint

Layer cake: $\int_Q|f-f_Q|^p=p\int_0^\infty\lambda^{p-1}|\{|f-f_Q|>\lambda\}|d\lambda$ , and the exponential tail makes it finite.

The idea

Explore: a signal and its Fourier transform on ℤ/n\mathbb Z/n

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Key terms in plain words

Explore

Worked example

Examples

Check your understanding

Practice

From Rudin, Principles of Mathematical Analysis (3rd ed., McGraw–Hill, 1976)

The idea

Explore: smoothing a function

Key terms in plain words

Examples

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Exercises

The idea

Explore: partial sums against Fejér means

Key terms in plain words

Examples

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Exercises

The idea

Explore: decay rates on log–log axes

Key terms in plain words

Examples

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Exercises

The idea

Explore: Gaussians, widths and Poisson summation

Key terms in plain words

Examples

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Exercises

The idea

Key terms in plain words

Examples

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Exercises

The idea

Explore: Mf and its level sets

Key terms in plain words

Examples

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Exercises

The idea

Explore: the Hilbert transform of a few functions

Key terms in plain words

Examples

Check your understanding

Exercises

The idea

Explore: dyadic frequency bands and Sobolev norms

Key terms in plain words

Examples

Check your understanding

Exercises

Explore: a signal and its Fourier transform on $\mathbb Z/n$