LEC ANA2 19

Proposition 359.1(Proposition).

Let $K\subseteq L^2[-\pi ,\pi ]$ be compact. Then

$$\underset{|n|\to \infty }{\lim }\underset{f\in K}{\sup }|{\hat{f}}_n|=0.$$

Proof.

We will denote ${\hat{f}}_n$ by $\hat{f}(n)$ for national convenience. Let $ϵ>0$. Use totally boundedness of $K$ to write $K\subseteq {\bigcup }_{i=1}^{n}B(f_i,ϵ)$. There exists $N_i$ such that $|{\hat{f}}_i(n)|<ϵ$ for $n⩾N_i$; let $N=\max N_i$. For $f\in K$, and $n⩾N$,

$$\begin{array}{rl}|\hat{f}(n)|& ⩽|\hat{f}(n)-{\hat{f}}_i(n)|+|{\hat{f}}_i(n)|\\ & =|(\widehat{f-f_i})(n)|+|{\hat{f}}_i(n)|\\ & =\Vert \widehat{f-f_i}{\Vert }_2+|{\hat{f}}_i(n)|\\ & =\Vert f-f_i{\Vert }_2+|{\hat{f}}_i(n)|\\ & ⩽2ϵ,\end{array}$$

where the fourth equality follows from Prp 352.1.□

5d576d

! Not really sure what I’ve done here.

Exercise 359.2(Exercise).

Let $f\in C^0[-\pi ,\pi ]$.

$${\Delta }_f(t):=\underset{s,r\in [-\pi ,\pi ],|s-r|<t}{\sup }|f(s)-f(r)|.$$

Assume

$${\int }_{-\pi }^{\pi }\frac{{\Delta }_f(t)}{|t|}dt<\infty .$$

Prove that $S_{N}f\to f$ uniformly.

Proof.

We will mimic the proof of Prp 355.4.

$$\begin{array}{rl}|(S_{N}f)(t_0)-f(t_0)|& =\frac{1}{2\pi }\left|{\int }_{-\pi }^{\pi }{D}_N(t-t_0)f(t)dt-{\int }_{-\pi }^{\pi }{D}_N(t)f(t_0)dt\right|\\ & =\frac{1}{2\pi }\left|{\int }_{-\pi }^{\pi }{D}_N(t)f(t+t_0)dt-{\int }_{-\pi }^{\pi }{D}_N(t)f(t_0)dt\right|\\ & =\frac{1}{2\pi }\left|{\int }_{-\pi }^{\pi }\left(\frac{\sin (N+1/2)t}{\sin t/2}\right)(f(t+t_0)-f(t_0))dt\right|\\ & ⩽\frac{1}{2\pi }\left[\underset{=:A^{\delta }(t_0)}{\underbrace{{\int }_{-\delta }^{\delta }\frac{|f(t+t_0)-f(t_0)|}{|\sin t/2|}dt}}+\underset{=:B_N^{\delta }(t_0)}{\underbrace{\left|{\int }_{(-\delta ,\delta {)}^c}\left(\frac{\sin (N+1/2)t}{\sin t/2}\right)(f(t+t_0)-f(t_0))dt\right|}}\right]\end{array}$$

Define $M=\sup \{(t/2)/(\sin (t/2)):t\in [-\pi ,\pi ]\}$. Then, $|\sin (t/2)|⩾|t|/2M$. For all $t_0\in [-\pi ,\pi ]$,

$$\begin{array}{r}{A}^{\delta }(t_0)⩽2M{\int }_{-\delta }^{\delta }\frac{|f(t+t_0)-f(t_0)|}{|t|}dt⩽2M{\int }_{-\delta }^{\delta }\frac{{\Delta }_f(t)}{|t|}dt\to 0\text{as}\delta \to 0.\end{array}$$

Now, consider the family of functions $\mathcal{F}=\{f(t+t_0)-f(t_0):t_0\in [-\pi ,\pi ]\}$. Denote $f(t+t_0)-f(t_0)$ by $f_{t_0}(t)$.

1. $\mathcal{F}$ is clearly bounded by $2\Vert f{\Vert }_{\infty }$.
2. For $ϵ>0$, and $\delta >0$ be chosen by the uniform continuity of $f$. Then, if $|t-t^{\prime }|<\delta$, $|f(t+t_0)-f(t^{\prime }+t_0)|<ϵ$ for all $t_0$. Thus, $\mathcal{F}$ is equicontinuous.
3. If $\{f_{t_k}{\}}_{k=1}^{\infty }\subseteq \mathcal{F}$ is a Cauchy sequence, let $t_0$ be a limit point of $\{t_k{\}}_{k=1}^{\infty }\subseteq S_1$. For $ϵ>0$, choose $N$ such that $|t_0-t_k|<\delta$ for all $k⩾N$, where $\delta$ is chosen by the uniform continuity of $f$. Then, for all $k⩾N$, $|(f(t+t_0)-f(t+t_k))+(f(t_k)-f(t_0))|⩽2ϵ$ for all $t\in [-\pi ,\pi ]$. Thus, $\{f_{t_k}{\}}_{k=1}^{\infty }\rightrightarrows f_{t_0}$.

By ^d4059b, $\mathcal{F}$ is compact in $(C^0[-\pi ,\pi ],\Vert \cdot {\Vert }_{\infty })$.

Fix $\delta >0$ and set $K:=(-\delta ,\delta {)}^c$. Define two families of continuous functions on $K$ indexed by $t_0\in [-\pi ,\pi ]$:

$$\begin{array}{rl}{\mathcal{G}}_1& :=\left\{g_{t_0}(t):=f_{t_0}(t)\left(\frac{\cos t/2}{\sin t/2}\right)\right\},\\ {\mathcal{G}}_2& :=\{h_{t_0}(t):=f_{t_0}(t)\}.\end{array}$$

${\mathcal{G}}_2$, being the restriction of $\mathcal{F}$ to $K$, is compact in $C(K)$. Because $\sin t/2$ is bounded away from $0$ on $K$, the map

$$u(t)\mapsto u(t)\left(\frac{\cos t/2}{\sin t/2}\right)$$

is a continuous linear operator $C(K)\to C(K)$. It follows that ${\mathcal{G}}_2$, being the image of ${\mathcal{G}}_1$ under this map, is compact.

Since the identity map $I:(C(K),\Vert \cdot {\Vert }_{\infty })\to (C(K),\Vert \cdot {\Vert }_2)$ is continuous (see Exm 163.7), ${\mathcal{G}}_1$ and ${\mathcal{G}}_2$ are compact in $L^2(K)$. As before, write:

$$\begin{array}{rl}{B}_N^{\delta }(t_0)& =\left|{\int }_K{g}_{t_0}(t)\sin Ntdt+{\int }_K{h}_{t_0}(t)\cos Ntdt\right|.\end{array}$$

From Prp 1, we have

$$\underset{|n|\to \infty }{\lim }\underset{g_{t_0}\in {\mathcal{G}}_1}{\sup }|{\hat{g}}_{t_0}(n)|=0,\underset{|n|\to \infty }{\lim }\underset{h_{t_0}\in {\mathcal{G}}_2}{\sup }|{\hat{h}}_{t_0}(n)|=0.$$

It follows that

$$\underset{|N|\to \infty }{\lim }\underset{t_0\in [-\pi ,\pi ]}{\sup }{B}_N^{\delta }=0.$$

Finally, given $ϵ>0$, choose $\delta$ such that $A^{\delta }(t_0)<ϵ/2$ for all $t_0\in [-\pi ,\pi ]$, and choose $M$ such that ${\sup }_{t_0\in [-\pi ,\pi ]}{B}_N^{\delta }<ϵ/2$ for all $N⩾M$. It follows that $|(S_{N}f)(t_0)-f(t_0)|<ϵ$ for all $t_0$ for all $N⩾M$, i.e, $\Vert S_{N}f-f{\Vert }_{\infty }\to 0$ as $N\to \infty$.□

---

Convolution

For $f,g\in C^0[0,2\pi ]$, define

$$(f\star g)(t):=\frac{1}{2\pi }{\int }_0^{2\pi }f(t-s)g(s)ds=\frac{1}{2\pi }{\int }_0^{2\pi }f(s)g(t-s)ds.$$

Thm 238.3 can be used to show that $f\star g$ is continuous. Recall that $S_{N}f=D_N\star f$. Define

$$({\sigma }_{N}f)(t):=\frac{1}{N}\sum _{n=0}^{N-1}(S_{N}f)(t).$$

Proposition 359.3(Proposition).

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

$$(\widehat{f\star g})(n)=\hat{f}(n)\cdot \hat{g}(n).$$

Proof.

$$\begin{array}{r}(\widehat{f\star g})(n)=\frac{1}{4{\pi }^2}{\int }_0^{2\pi }\left({\int }_0^{2\pi }f(t-s)g(s)ds\right)e^{-int}dt\end{array}$$

To use Fubini’s theorem to swap the integrals, we need to show that the absolute value of the integrand is integrable:

$$\begin{array}{r}{\int }_0^{2\pi }{\int }_0^{2\pi }|f(t-s)||g(s)|dsdt\end{array}$$

$|f(t-s)||g(s)|$ is non-negative and measurable, so Tonelli lets you swap the order of integration:

$$\begin{array}{rl}& {\int }_0^{2\pi }\left({\int }_0^{2\pi }|f(t-s)|dt\right)|g(s)|ds\\ & ={\int }_0^{2\pi }\left({\int }_0^{2\pi }|f(u)|du\right)|g(s)|ds\\ & =\left({\int }_0^{2\pi }|f(u)|du\right)\left({\int }_0^{2\pi }|g(s)|ds\right)\\ & =\Vert f{\Vert }_1\Vert g{\Vert }_1.\end{array}$$

Thus, by Fubini,

$$\begin{array}{rl}(\widehat{f\star g})(n)& =\frac{1}{4{\pi }^2}{\int }_0^{2\pi }\left({\int }_0^{2\pi }f(t-s)e^{-in(t-s)}dt\right)g(s)e^{-ins}ds\\ & =\hat{f}(n)\cdot \hat{g}(n).\end{array}$$ □

Definition 359.4(Approximate identity).

$\{g_N{\}}_{N=1}^{\infty }$ is said to be an approximate identity if

1. $g_N⩾0$ for all $N$
2. $\frac{1}{2\pi }{\int }_0^{2\pi }{g}_N(s)ds=1$ for all $N$
3. ${\lim }_{N\to \infty }{\int }_{ϵ}^{2\pi -ϵ}{g}_N(s)ds=0$ for all $ϵ>0$.

Proposition 359.5(Proposition).

Let $\{g_N{\}}_{N=1}^{\infty }$ be an approximate identity. Then $g_N\star f\to f$ uniformly for all $f\in C^0[0,2\pi ]$.

Proof.

$$\begin{array}{rl}|(g_N\star f-f)(t)|& =\frac{1}{2\pi }\left|{\int }_0^{2\pi }{g}_N(s)f(t-s)ds-{\int }_0^{2\pi }{g}_N(s)f(t)ds\right|\\ & ⩽\frac{1}{2\pi }\left|{\int }_0^{2\pi }|g_N(s)||f(t-s)-f(t)|ds\right|.\end{array}$$

Evaluating the integral over $[\delta ,2\pi -\delta ]$, we get

$${\int }_{\delta }^{2\pi -\delta }|g_N(s)||f(t-s)-f(t)|ds⩽2\Vert f{\Vert }_{\infty }{\int }_{\delta }^{2\pi -\delta }|g_N(s)|ds\to 0\text{as}N\to \infty .$$

Evaluating the integral over $[\delta ,2\pi -\delta {]}^c$, we get

$$\begin{array}{r}{\int }_{[\delta ,2\pi -\delta {]}^c}|g_N(s)||f(t-s)-f(t)|ds⩽\underset{s\in [\delta ,2\pi -\delta {]}^c}{\sup }|f(t-s)-f(t)|\end{array}$$

Given $ϵ>0$, choose $\delta >0$ such that $|f(s)-f(r)|<ϵ$ for all $|s-r|<2\delta$.□

Proposition 359.6(Proposition).

$({\sigma }_{N}f)(t)=(F_N\star f)(t)$, where

$$F_N(t)=\frac{1}{N}\frac{{\sin }^2(Nt/2)}{{\sin }^2(t/2)}.$$

Proof.

$$\begin{array}{rl}({\sigma }_{N}f)(t)& =\frac{1}{N}\sum _{n=0}^{N-1}(S_{N}f)(t)\\ & =\frac{1}{N}\sum _{n=0}^{N-1}(D_N\star f)(t)\end{array}$$

It follows that

$$\begin{array}{rl}{F}_N& =\frac{1}{N}\sum _{n=0}^{N-1}{D}_n\\ & =\frac{1}{N}\sum _{n=0}^{N-1}\frac{\sin (N+1/2)t}{\sin t/2}\\ & =\frac{1}{N\sin t/2}\sum _{n=0}^{N-1}\mathrm{Im}{e}^{i(N+1/2)t}\\ & =\frac{1}{N\sin t/2}\mathrm{Im}\left(\frac{e^{iNt}-1}{e^{it/2}-e^{-it/2}}\right)\\ & =\frac{-1}{2N{\sin }^{2}t/2}\mathrm{Re}(e^{iNt}-1)\\ & =\frac{{\sin }^2(Nt/2)}{N{\sin }^{2}t/2}.\end{array}$$ □

Theorem 359.7(Fejér).

Let $f\in C^0[0,2\pi ]$. Then, ${\sigma }_{N}f$ converges to $f$ uniformly.

Proof.

It suffices to verify

$${\left\{g_N(t)=\frac{1}{N}\frac{{\sin }^2(Nt/2)}{{\sin }^{2}t/2}\right\}}_{N=0}^{\infty }$$

is an approximate identity.□

Lemma 359.8(Lemma).

Let $\{x_n{\}}_{n=1}^{\infty }\subseteq \mathbb{C}$. Let ${\overline{x}}_n:=(x_1+x_2+\cdots +x_n)/n$. If $x_n\to x$, then ${\overline{x}}_n\to x$.

Proof.

There exists $N_{ϵ}$ such that $|x_n-x|<ϵ$ for all $n⩾N_{ϵ}$. Then,

$$\begin{array}{rl}|{\overline{x}}_n-x|& ⩽\sum _{k=1}^n\frac{|x_k-x|}{n}\\ & =\sum _{k=1}^{N_{ϵ}}\frac{|x_k-x|}{n}+\underset{⩽ϵ}{\underbrace{\sum _{k=N_{ϵ}+1}^n\frac{|x_k-x|}{n}}}.\end{array}$$

Choose $n$ such that $n⩾{\sum }_{k=1}^{N_{ϵ}}|x_k-x|/ϵ$ and $n⩾N_{ϵ}$.□

Corollary 359.9(Corollary).

If $S_{N}f$ converges, then it must converge to $f$.