LEC ANA2 18

Recall that $f\in L^2[0,2\pi ]$ can be expressed as

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

It follows from Prp 344.11 that ${\sum }_{n\in \mathbb{Z}}|{\hat{f}}_n{|}^2<\infty$, and hence ${\hat{f}}_n\to 0$ as $|n|\to \infty$. This immediately yields

Proposition 355.1(Riemann-Lebesgue).

Let $f\in L^2[0,2\pi ]$. Then,

$$\begin{array}{rl}& \int f(t)\cos (nt)dt\to 0\text{ and }\int f(t)\sin (nt)dt\to 0\end{array}$$

as $n\to \infty$.

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Convergence of Fourier series

We have seen that

• $S_{N}f$ converges to $f$ in $L^2[0,2\pi ]$, and
• this convergence is not pointwise in general.

We will now explore hypotheses under which $S_{N}f$ converges uniformly.

Suppose $f\in C^0[0,2\pi ]$ such that

$${\int }_0^{2\pi }f(t)dt=0.$$

Define

$$F(t):={\int }_0^{t}f(s)ds.$$

It follows from the fundamental theorem of calculus that $F\in C^0[0,2\pi ]$ ¹. We see that

$$\begin{array}{rl}{\hat{F}}_n& =\frac{1}{2\pi }{\int }_0^{2\pi }\left({\int }_0^{t}f(s)ds\right)e^{-int}dt\\ & =\frac{1}{2\pi }\left({\left[\left({\int }_0^{t}f(s)ds\right)\frac{e^{-int}}{-in}\right]}_0^{2\pi }-\frac{1}{-in}{\int }_0^{2\pi }f(t)e^{-int}dt\right)\\ & =\frac{{\hat{f}}_n}{-in}.\end{array}$$

It follows that if $f\in C^0[0,2\pi ]$ is continuously differentiable, then

$$\begin{array}{r}{\hat{f}}_n=\frac{\widehat{f_n^{\prime }}}{-in}.\end{array}$$

If $f^{\prime }$ is also continuously differentiable (i.e, $f$ is continuously twice differentiable), then we have

$$\begin{array}{r}{\hat{f}}_n=-\frac{\widehat{f_n^{\prime \prime }}}{n^2}.\end{array}$$

This leads to the following

Proposition 355.2(Proposition).

If $f\in C^0[0,2\pi ]$ is twice continuously differentiable, then $S_{N}f$ converges uniformly.

Proof.

Note that for general $h\in L^2[0,2\pi ]$,

$$\begin{array}{r}|{\hat{h}}_n|⩽\frac{1}{2\pi }{\int }_0^{2\pi }|h(s)|ds=\Vert h{\Vert }_1.\end{array}$$

In particular,

$$|{\hat{f}}_n|=\frac{1}{n^2}|\widehat{f_n^{\prime \prime }}|⩽\frac{1}{n^2}\Vert f^{\prime \prime }{\Vert }_1.$$

Thus, ${\sum }_{n\in \mathbb{Z}}|{\hat{f}}_n|$ converges. Now consider the sequence $\{S_{N}f{\}}_{N=1}^{\infty }$. For $M⩾N$, we have

$$\begin{array}{rl}\Vert S_{M}f-S_{N}f{\Vert }_{\infty }& ={\Vert \sum _{n=-M}^{-N-1}{\hat{f}}_n{e}^{int}+\sum _{n=N+1}^M{\hat{f}}_n{e}^{int}\Vert }_{\infty }\\ & ⩽\sum _{n=-M}^{-N-1}|{\hat{f}}_n|+\sum _{n=N+1}^M|{\hat{f}}_n|\to 0\text{ as }N,M\to \infty .\end{array}$$

Thus, $\{S_{N}f{\}}_{N=1}^{\infty }$ converges uniformly.□

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Of course, since $C^0[0,2\pi ]$ is complete ($S_1$ is compact; see Rmk 163.6), it follows that $S_{N}f$ converges uniformly to some $\tilde{f}\in C^0[0,2\pi ]$. We will show later that $\tilde{f}$ is in fact equal to $f$.

The hypotheses of Prp 2 can be relaxed:

Proposition 355.3(Proposition).

If $f\in C^0[0,2\pi ]$ is continuously differentiable, then $S_{N}f$ converges uniformly.

Proof.

As in the proof of Prp 355.2, It suffices to show that ${\sum }_{n\in \mathbb{Z}}|{\hat{f}}_n|$ converges.

$$\begin{array}{rl}\sum _{n\in \mathbb{Z}}|{\hat{f}}_n|& =\sum _{n\in \mathbb{Z}}\frac{1}{|n|}|{\hat{f}}_n^{\prime }|\\ & ⩽\sqrt{\sum _{n\in \mathbb{Z}}\frac{1}{n^2}}\sqrt{\sum _{n\in \mathbb{Z}}|{\hat{f}}_n^{\prime }{|}^2}\\ & ⩽\infty ,\end{array}$$

where the first inequality follows from Prp 344.12.5, Prp 161.3, and Thm 87.8.□

Proposition 355.4(Proposition).

Let $f\in C^0[-\pi ,\pi ]$. If $t_0\in [-\pi ,\pi ]$ is such that

$${\int }_{-\pi }^{\pi }\frac{|f(t)-f(t_0)|}{|t-t_0|}dt<\infty ,$$

then $(S_{N}f)(t_0)\to f(t_0)$ as $N\to \infty$.

Proof.

First, note that since $e^{in(t+\pi )}=e^{int}(-1{)}^n$, $\{e^{int}:n\in \mathbb{Z}\}$ remains an orthonormal basis for $L^2[-\pi ,\pi ]$. WLOG, we can assume $t_0=0$. Using Lem 352.4,

$$\begin{array}{rl}|(S_{N}f)(0)-f(0)|& =\left|\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{D}_N(t)f(t)dt-f(0)\right|.\end{array}$$

Observe that ${\int }_{-\pi }^{\pi }{D}_N(t)dt={\int }_{-\pi }^{\pi }{\sum }_{-N}^N{e}^{int}dt=2\pi$. Thus, we can write

$$\begin{array}{rl}|(S_{N}f)(0)-f(0)|& =\frac{1}{2\pi }\left|{\int }_{-\pi }^{\pi }{D}_N(t)f(t)dt-{\int }_{-\pi }^{\pi }{D}_N(t)f(0)dt\right|\\ & =\frac{1}{2\pi }\left|{\int }_{-\pi }^{\pi }\left(\frac{\sin (N+1/2)t}{\sin t/2}\right)(f(t)-f(0))dt\right|\\ & ⩽\frac{1}{2\pi }\left[\underset{=:A^{\delta }}{\underbrace{{\int }_{-\delta }^{\delta }\frac{|f(t)-f(0)|}{|\sin t/2|}dt}}+\underset{=:B_N^{\delta }}{\underbrace{\left|{\int }_{[-\delta ,\delta {]}^c}\left(\frac{\sin (N+1/2)t}{\sin t/2}\right)(f(t)-f(0))dt\right|}}\right]\end{array}$$

We will individually bound $A^{\delta }$ and $B_N^{\delta }$.

$$\begin{array}{rl}{B}_N^{\delta }& =\left|{\int }_{[-\delta ,\delta {]}^c}\underset{=:g_1(t)}{\underbrace{\left(\frac{(f(t)-f(0))\cos t/2}{\sin t/2}\right)}}\sin Ntdt+{\int }_{[-\delta ,\delta {]}^c}\underset{=:g_2(t)}{\underbrace{(f(t)-f(0))}}\cos Ntdt\right|.\end{array}$$

Define both $g_1$ and $g_2$ to be zero on $[-\delta ,\delta ]$. Note that $g_1$ and $g_2$ are bounded and Riemann integrable on $[-\pi ,\pi ]$, and hence in $L^2[-\pi ,\pi ]$ by Rmk 352.3. It follows from Prp 1 that

$$\begin{array}{r}{\int }_{[-\delta ,\delta {]}^c}{g}_1(t)\sin Ntdt={\int }_{-\pi }^{\pi }{g}_1(t)\sin Ntdt\to 0\text{as}N\to \infty .\end{array}$$

Ditto for $g_2$. Thus, $B_N^{\delta }\to 0$ as $N\to \infty$.

Now, $A^{\delta }$‘s turn. Define $M=\sup \{(t/2)/(\sin (t/2)):t\in [-\pi ,\pi ]\}$. Then, $|\sin (t/2)|⩾|t|/2M$.

$$\begin{array}{r}{A}^{\delta }⩽2M{\int }_{-\delta }^{\delta }\frac{|f(t)-f(0)|}{|t|}dt.\end{array}$$

By hypothesis, ${\int }_{-\pi }^{\pi }|f(t)-f(0)|/|t|dt$ exists. It follows that ${\int }_{-\delta }^{\delta }|f(t)-f(0)|/|t|dt\to 0$ as $\delta \to 0$.

Now, let $ϵ>0$ be given. Choose $\delta$ such that $A^{\delta }<ϵ/2$, and $N$ such that $B_N^{\delta }<ϵ$.□

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Example 355.5(Example).

If $f$ and $t_0$ satisfy

$$|f(t)-f(t_0)|⩽k|t-t_0{|}^{\alpha }$$

for all $t\in [-\pi ,\pi ]$ for some $\alpha >0$, then $(S_{N}f)(t_0)\to f(t_0)$.

Footnotes

1. In fact, $F$ is continuously differentiable - I just can’t think of a way to extend the godawful notation of $C^0[0,2\pi ]$ to convey that. ↩