LEC ANA2 17

The Fourier Transform

The following algebras can be identified with each other: $C^0([0,2\pi ])$, $C(S_1)$, and $C^0([-\pi ,\pi ])$, with the norm induced by the inner product.

We have shown that $S=\{{\phi }_n:n\in \mathbb{Z}\}\subseteq L^2[0,2\pi ]$ is an orthonormal basis for $L^2[0,2\pi ]$, where ${\phi }_n=e^{int}$. Therefore, by Prp 344.12, $f\in L^2[0,2\pi ]$ can be expressed as

$$f=\sum _{n\in \mathbb{Z}}{\hat{f}}_n{\phi }_n,$$

where ${\hat{f}}_n:=⟨f,{\phi }_n⟩$. Precisely, if we define

$$\begin{array}{rl}{S}_{N}f& :=\sum _{n=-N}^N{\hat{f}}_n{\phi }_n,\end{array}$$

then $S_{N}f$ converges to $f$ in $L^2[0,2\pi ]$:

$$\underset{N\to \infty }{\lim }{\int }_0^{2\pi }|f-S_{N}f{|}^2=0.$$

In a sense, the function $f$ and the sequence $\{{\hat{f}}_n\}$ are duals of each other.

Proposition 352.1(Proposition).

The map $\zeta :L^2[0,2\pi ]\to {\ell }_2(\mathbb{Z})$ defined by $f\mapsto \{{\hat{f}}_n\}$ is an isometrical linear isomorphism.

Proof.

$\zeta$ is injective by Prp 344.12.3 and surjective by Prp 344.11.1&2. It is clear that $\zeta$ is a linear transformation. $\zeta$ preserves the inner product:

$$\begin{array}{r}⟨f,g⟩=\sum _{n\in \mathbb{Z}}⟨f,{\phi }_n⟩⟨{\phi }_n,g⟩=\sum _{n\in \mathbb{Z}}{\hat{f}}_n{\overline{\hat{g}}}_n.\end{array}$$ □

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By Prp 344.12.5, we also have

$$\Vert f{\Vert }_2^2=\sum _{n\in \mathbb{Z}}|{\hat{f}}_n{|}^2.$$E1

Example 352.2(Example).

$$\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}.$$

Proof.

Let $f:[0,2\pi ]\to \mathbb{C}$ be defined by $f(t)=1$ for $t\in [0,\pi )$, and $f(t)=-1$ for $t\in [\pi ,2\pi )$¹. It is clear from Rmk 344.2 that $f\in L^2[0,2\pi ]$. Clearly, $\Vert f{\Vert }_2=1$.

$$\begin{array}{rl}{\hat{f}}_n& =\frac{1}{2\pi }\left({\int }_0^{\pi }{e}^{-int}dt-{\int }_{\pi }^{2\pi }{e}^{-int}dt\right);\\ & \\ \hat{f}(2k)& =0,\\ \hat{f}(2k+1)& =\frac{-i}{\pi }\frac{2}{2k+1}.\end{array}$$

Using (E1),

$$\begin{array}{rl}1=\Vert f{\Vert }_2^2& =\sum _{n\in \mathbb{Z}}|{\hat{f}}_n{|}^2\\ & =\frac{8}{{\pi }^2}\sum _{k=0}^{\infty }\frac{1}{(2k+1{)}^2}\\ & =\frac{6}{{\pi }^2}\sum _{n=1}^{\infty }\frac{1}{n^2}.\end{array}$$ □

Remark 352.3(Remark).

In general, for a bounded Riemann integrable function $f$ on $[0,2\pi ]$, continuous $g$ can be found to satisfy ${\int }_0^{2\pi }|f-g{|}^{2}dx<ϵ$ for every $ϵ>0$, so $f\in L^2[0,2\pi ]$ by Def 344.1. If $f$ is improper Riemann integrable, it may not be in $L^2[0,2\pi ]$ - consider $1/\sqrt{x}$. All square (proper/improper) Riemann integrable functions on $[0,2\pi ]$ are in $L^2[0,2\pi ]$.

5d1611

Failure of pointwise convergence of Fourier series

Since functions that differ on a set of measure zero are identified with each other in $L^2$, convergence in $L^2$ does not imply $S_{N}f(t)\to f(t)$ for all $t\in [0,2\pi ]$. We will now demonstrate the existence of such $f$.

Periodic domains are assumed from here on.

Lemma 352.4(Lemma).

$$\begin{array}{rl}(S_{N}f)(t)& =\frac{1}{2\pi }{\int }_0^{2\pi }{D}_N(t-s)f(s)ds,\end{array}$$

where

$$\begin{array}{rlrl}{D}_N(t)& =\frac{\sin (N+1/2)t}{\sin t/2}& & t\in (0,2\pi )\\ & =2N+1& & t=0,2\pi .\end{array}$$

Proof.

$$\begin{array}{rl}(S_{N}f)(t)& =\sum _{n=-N}^N{\hat{f}}_n{e}^{int}\\ & =\sum _{n=-N}^N\left(\frac{1}{2\pi }{\int }_0^{2\pi }f(s)e^{-ins}ds\right)e^{int}\\ & =\frac{1}{2\pi }{\int }_0^{2\pi }{D}_N(t-s)f(s)ds,\end{array}$$

where

$$\begin{array}{rlr}{D}_N(u)& =\sum _{n=-N}^N{e}^{inu}& \\ & =e^{-iNu}\left(\frac{e^{i(2N+1)u}-1}{e^{iu}-1}\right)& u\ne 0,2\pi \\ & =\frac{e^{i(N+1)u}-e^{-iNu}}{e^{iu}-1}& \\ & =\frac{e^{i(N+1/2)u}-e^{-i(N+1/2)u}}{e^{iu/2}-e^{-iu/2}}& \\ & =\frac{\sin (N+1/2)u}{\sin u/2}.& \end{array}$$ □

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Note that $D_N\in C^0[0,2\pi ]$ for all $N$.

Lemma 352.5(Lemma).

$$\underset{N\to \infty }{\lim }{\int }_0^{2\pi }|D_N(t)|dt=\infty .$$

Proof.

$$\begin{array}{rlr}{\int }_0^{\delta }|D_N(t)|dt& ={\int }_0^{\delta }\left|\frac{\sin \left(\frac{2N+1}{2}\right)t}{\sin t/2}\right|dt& \\ & ⩾2{\int }_0^{\delta }\left|\frac{\sin \left(\frac{2N+1}{2}\right)t}{t}\right|dt& \\ & ={\int }_0^{\delta (2N+1)/2}\left|\frac{\sin s}{s}\right|ds& \\ & ⩾{\int }_0^{n\pi }\left|\frac{\sin s}{s}\right|ds& n=\lfloor \delta (2N+1)/2\pi \rfloor \to \infty \text{ as }N\to \infty \\ & ⩾\sum _{k=1}^n\frac{1}{k}\left(\frac{1}{\pi }{\int }_{(k-1)\pi }^{k\pi }\left|\sin s\right|ds\right)& \\ & =\frac{2}{\pi }\sum _{k=1}^n\frac{1}{k}\to \infty \text{ as }n\to \infty .& \end{array}$$ □

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Define $T_N:(C^0[0,2\pi ],\Vert \cdot {\Vert }_{\infty })\to \mathbb{C}$ by $f\mapsto (S_{N}f)(0)$. $T_N$ is linear.

Lemma 352.6(Lemma).

$$\begin{array}{r}\Vert T_N\Vert =\frac{1}{2\pi }{\int }_0^{2\pi }|D_N(t)|dt.\end{array}$$

Proof.

Denote $1/2\pi {\int }_0^{2\pi }|D_N(s)|ds$ by $M$.

$$\begin{array}{rl}|T_{N}f|& =\left|\frac{1}{2\pi }{\int }_0^{2\pi }{D}_N(-s)f(s)dt\right|\\ & ⩽\frac{1}{2\pi }{\int }_0^{2\pi }\left|D_N(s)f(s)\right|ds\\ & ⩽\frac{\Vert f{\Vert }_{\infty }}{2\pi }{\int }_0^{2\pi }|D_N(s)|ds\end{array}$$ $$⟹\Vert T_N\Vert ⩽M.$$

For $n=1,2,\dots$, define

$$g_n(t):=\frac{D_N(t)}{\sqrt{D_N^2(t)+1/n^2}}.$$

Note that

1. $g_n\in C^0[0,2\pi ]$,
2. $|g_n(t)|<1$ for all $t\in [0,2\pi ]$, and
3. $\{g_n(t){\}}_{n=1}^{\infty }\to D_N(t)/|D_N(t)|$ for all $t\in [0,2\pi ]$.

By Thm 238.3,

$$\begin{array}{r}\underset{n\to \infty }{\lim }{\int }_0^{2\pi }{D}_N(s)g_n(s)ds={\int }_0^{2\pi }|D_N(s)|ds.\end{array}$$

Thus, ${\lim }_{n\to \infty }|T_N{g}_n|=M$.

Now, given any $ϵ>0$, there exists $n$ such that $|T_N{g}_n|>M-ϵ$. Since $\Vert g_n{\Vert }_{\infty }<1$ for all $n$, we have $|T_N{g}_n|/\Vert g_n{\Vert }_{\infty }>M-ϵ$. Therefore, $\Vert T_N\Vert =M$.□

Lem 5 tells us that $\Vert T_N\Vert \to \infty$ as $N\to \infty$. The following proposition follows immediately from Thm 314.3.2.

Proposition 352.7(Proposition).

$$\left\{f\in C^0[0,2\pi ]:\underset{N\in \mathbb{N}}{\sup }|(S_{N}f)(0)|=\infty \right\}$$

is a dense $G_{\delta }$ subset of $C^0[0,2\pi ]$.

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So, there exists a dense subset of $C^0[0,2\pi ]$ for which the Fourier series diverges at $0$! Moreover, we can translate this set to obtain a dense subset for which the Fourier series diverges at any $a\in [0,2\pi ]$: for $f_a(x):=f(x-a)$,

$$\begin{array}{rl}(S_N{f}_a)(t)& =\frac{1}{2\pi }{\int }_0^{2\pi }{D}_N(t-s)f(s-a)ds\\ & =\frac{1}{2\pi }{\int }_a^{2\pi +a}{D}_N((t-a)-u)f(u)du\\ & =\frac{1}{2\pi }{\int }_0^{2\pi }{D}_N((t-a)-u)f(u)du\\ & =(S_{N}f)(t-a),\end{array}$$E2

so if $|(S_{N}f)(0)|$ diverges, $|(S_N{f}_a)(a)|$ will too.

Footnotes

1. Take $f(2\pi )=1$, if you’d like; it doesn’t matter. ↩