\classheader{2012-11-20}

Recall that last time, we defined the Poisson kernel
\[P(\rho,\delta)=\frac{1-\rho^2}{1-2\rho\cos(\delta)+\rho^2}.\]
We know that $P$ is harmonic.
\begin{theorem}
If $u$ is harmonic on $\{|z|<1\}$ and continuous on $\{|z|\leq 1\}$, then
\[u(a)=u(\rho\cos(\nu),\rho\sin(\nu))=\frac{1}{2\pi}\int_0^{2\pi}u\cdot P(\rho,\varphi-\nu)\,d\varphi.\]
\end{theorem}
\begin{theorem}
If $f=f(\cos(\varphi),\sin(\varphi))$ is continuous, then
\[u(a)\overset{\text{\emph{def}}}{=}\frac{1}{2\pi}\int_0^{2\pi}f\cdot P(\rho,\varphi-\nu)\,d\varphi\]
is harmonic on the open disc $\{|z|<1\}$. and
\[u^*(a)=\begin{cases}
u(a) & \text{ if }|a|<1,\\
f(a) & \text{ if }|a|=1
\end{cases}\]
is continuous. Thus, the Poisson kernel preserves continuity.
\end{theorem}
\begin{proof}
Because $P$ is harmonic, and finite sums of harmonic functions are harmonic, then in the limit, the integral defining $u(a)$ is harmonic.

Now, we want to show that $u^*(a)$ is continuous, i.e. that $u(a)$ is close to $f(a')$ if $a$ is close to $a'$:
\begin{center}
\begin{tikzpicture}[scale=0.6]
\draw[thick] (0,0) circle (2);
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle,label={[shift={(-0.35,-0.55)}] $a$}] at (45:1.5) {};
\node[inner sep=0pt,outer sep=0pt,minimum size=4pt,fill,circle,label={[shift={(0.45,-0.1)}] $a'$}] at (45:2) {};
\end{tikzpicture}
\end{center}
This is because the Poisson kernel will weight values close to $a'$ more than further points. Thus, we are taking the weighted average concentrated around $a'$. When $a$ is close to $a'$, we have that $\rho\approx 1$ and $\nu\approx\varphi$, so that the Poisson kernel is approximately
\[\frac{1-\rho^2}{2[1-\cos(\nu-\varphi)]]}\approx\frac{\text{small}}{\text{small}}.\]
\end{proof}
\begin{homework}
Formally prove that $u^*(a)$ is continuous.
\end{homework}

For the following material, you should read the proofs somewhere yourself.

\begin{definition}
Given $f\in L^1([0,2\pi])$, the Fourier coefficients of $f$ are defined to be
\[c_n=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx},\qquad -\infty<n<+\infty.\]
\end{definition}
\begin{theorem}
$\text{}$

\begin{enumerate}
\item $f\in L^1$ implies that $c_n\to 0$ as $n\to \pm\infty$.
\item $f\in L^2$ implies that $\sum |c_n|^2<\infty$. Moreover, $\sum|c_n|^2=\|f\|_2$.
\end{enumerate}
\end{theorem}
\begin{remark}
In the measure space $(\Z,P(\Z),\#)$ (i.e., $\Z$ with the counting measure), then $c(n)\in L^2$ if and only if $\sum|c_n|^2<\infty$. $L^2$ of this measure space is usually referred to as $\ell^2$. The Fourier transform is a bijection between $L^2$ functions on $\S^1$ and $L^2$ functions on $\Z$.
\end{remark}
\begin{remark}
We can define the partial sum functions
\[S_N=\sum_{n=-N}^Nc_ne^{int}.\]
The sequence of $S_N$'s is Cauchy in $L^2$, and therefore it converges in $L^2$.
\end{remark}
Note that
\[\frac{1}{2\pi}\int_0^{2\pi}e^{inx}e^{-imx}\,dx=\begin{cases}
1 & \text{ if }n=m,\\
0 & \text{ if }n\neq m.
\end{cases}\]
In fact, defining an inner product on $L^2$ by $\langle f,g\rangle=\int f\overline{g}$, we have that the functions $\{\frac{e^{inx}}{\sqrt{2\pi}}\}$ form an orthonormal basis of $L^2$ (note that we have normalized them with a factor of $\frac{1}{\sqrt{2\pi}}$). We also see that
\[\langle f,f\rangle=\int |f|^2=\|f\|_2.\]
\subsection*{Convolution}

Integrating against the Poisson kernel is a special case of something called convolution.

Given $f:\Z\to\C$ such that $\sum_{n=-\infty}^\infty |f(n)|<\infty$, then
\[\widehat{f}(x)=\sum_{n=-\infty}^\infty f(n)e^{ixn},\]
($\widehat{\Z}$ is the unit circle, i.e. $\R/[0,2\pi]$).

Now consider
\[\widehat{f}(x)\cdot \widehat{g}(x)=\bigg(\sum_{n=-\infty}^\infty f(n)e^{ixn}\bigg)\bigg(\sum_{m=-\infty}^\infty g(m)e^{imx}\bigg).\]
By Fubini's theorem, this is equal to
\[\sum_{n+m}f(n)g(m)e^{ix(n+m)}=\sum_{k}e^{ikx}\sum_{k=m+n}f(m)g(n)=(\widehat{f\ast g})(x),\]
where $f\ast g$ is the convolution of $f$ and $g$.

\begin{proposition}
If $f,g\in L^1(\Z)$, then $(f\ast g)\in L^1(\Z)$.
\end{proposition}
\begin{proof}
\begin{align*}
\sum_k |(f\ast g)(k)| &=\sum_k\bigg|\sum_{k=m+n}f(n)g(m)\bigg|\\
&\leq \sum_k\sum_{k=m+n}|f(n)||g(m)|\\
&=\bigg(\sum_n |f(n)|\bigg)\bigg(\sum_k |g(k-m)|\bigg)=\|f\|_1\|g\|_1\qedhere
\end{align*}
\end{proof}

Now, instead of considering the functions $n\mapsto e^{inx}$, let's look at $t\mapsto e^{ixt}$, where $x,t\in\R^d$ and $xt$ means the dot product of $x$ and $t$.
\begin{definition}
For $f,g\in L^1(\R^d)$, define their convolution to be
\[(f\ast g)(x)=\int_{\R^d}f(y)g(x-y)\,dy.\]
\end{definition}
\begin{theorem}
For any $f,g\in L^1(\R^d)$,
\begin{enumerate}
\item $f\ast g$ exists at almost every $x$.
\item $f\ast g\in L^1$ and $\|f\ast g\|_1\leq\|f\|_1\|g\|_1$.
\item $\ast$ is commutative and associative.
\end{enumerate}
\end{theorem}
\begin{proof}
Observe that
\begin{align*}
\int_{\R^d}\int_{\R^d}|f(y)||g(x-y)|\,dy\,dx&=\int_{\R^d}|f(y)|\bigg(\int_{\R^d}|g(x-y)|\,dx\bigg)dy\\
&=\int |f(y)|\cdot\|g\|_1\,dy=\|f\|_1\|g\|_1,
\end{align*}
which proves claim 2. This also implies that
\[\int|f(y)||g(x-y)|\,dy<\infty\]
at almost every $x$, so indeed $f\ast g$ exists a.e., which is claim 1.

Claim 3 is clear; for example, swap $f$ and $g$ in the sum for commutativity.
\end{proof}
\begin{homework}
If $f\in L^1$ and $g\in L^p$, does it follow that $f\ast g\in L^p$ and $\|f\ast g\|_p\leq\|f\|_1\|g\|_p$?
\end{homework}
\begin{homework}
If $\frac{1}{p}+\frac{1}{q}=1$ and $f\in L^p$, $g\in L^q$, show that $f\ast g$ is continuous and tends to 0 as $|x|\to \infty$.
\end{homework}
\begin{homework}
Prove that there is no ``unit'' function, i.e. that there is no $f\in L^1$ such that $f\ast g=g$ for every $g\in L^1$.
\end{homework}
\begin{lemma}
Let $f\in L^1$ and $y\in \R^d$. Define $f_y(x)=f(x+y)$. Then
\[\lim_{y\to 0}\|f_y-f\|_1=0.\]
\end{lemma}
\begin{proof}
If $f$ is continuous and compactly supported, then this is trivial; if we integrate a small $\epsilon$ on a bounded set, we'll get something small.

Now, take an arbitrary $f\in L^1$, and choose a $g$ which is continuous and compactly supported such that $\|f-g\|_1<\epsilon$. Then
\[f_y-f=(f_y-g_y)+(g_y-g)+(g-f),\]
Note that $f_y-g_y=(f-g)_y$, so that
\[\|f_y-f\|_1\leq\|f_y-g_y\|_1+\|g_y-g\|_1+\|g-f\|_1<3\epsilon.\qedhere\]
\end{proof}
For $t\in\R$, let $f_t$ (\textbf{not} translation) be a non-negative function such that $\int f_t=1$ and $f_t(x)=0$ for $|x|>t$. These functions are ``approximate units''.
\begin{center}
\begin{tikzpicture}[scale=0.5]
\begin{scope}
\clip (-4,-1) rectangle (4,4);
\begin{scope}
\clip (-5,0) rectangle (5,3.5);
\draw [style={thick,line cap=round}] plot [smooth,samples=20] coordinates {(-4,0.0) (-1,0.25) (0,3) (1,0.25) (4,0.00)};
\end{scope}
\begin{scope}[xscale=0.5,yscale=1.3]
\clip (-5,0) rectangle (5,3.5);
\draw [style={thick,line cap=round}] plot [smooth,samples=20] coordinates {(-4,0.0) (-1,0.25) (0,3) (1,0.25) (4,0.00)};
\end{scope}
\begin{scope}[xscale=2,yscale=0.5]
\clip (-5,0) rectangle (5,3.5);
\draw [style={thick,line cap=round}] plot [smooth,samples=20] coordinates {(-4,0.0) (-1,0.25) (0,3) (1,0.25) (4,0.00)};
\end{scope}
\draw[thick] (-4,0) -- (4,0);
\node[fill,circle,inner sep=0pt,outer sep=0pt,minimum size=0.15cm] at (0,0) {};
\draw[thick,<->] (-0.7,-0.4) -- (0.7,-0.4);
\end{scope}
\node at (0,-0.9) {$t$};
\end{tikzpicture}
\end{center}
We can think of $f_t\ast g$ as a weighted average of $g$ around 0.
\begin{theorem}
For any $g\in L^1$, we have $\|f_t\ast g-g\|_1\to 0$ as $t\to 0$.
\end{theorem}
\begin{proof}
First, note that
\begin{align*}
\|f_t\ast g-g\|_1&=\int\bigg|\int f_t(y)g(x-y)\,dy-g(x)\bigg|\,dx\\
&=\int\bigg|\int f_t(y)(g(x-y)-g(x))\,dy\,\bigg|\,dx\\
&\leq \int_{-t}^t\int_{\R^d}|g(x-y)-g(x)|\,dx\, f_t(y)\,dy
\intertext{and, by the lemma, once $t$ is sufficiently small we have that }
&\leq \epsilon\int_{|y|<t}f_t(y)\,dy=\epsilon.\qedhere
\end{align*}
\end{proof}
\begin{example}
We could take
\[f_t(y)=\frac{\chi_{B(0,t)}(y)}{\lambda(B(0,t))},\]
in which case
\[(f_t\ast g)(x)=\int g(x-y)f_t(y)\,dy\overset{\text{commutativity}}{=}\frac{1}{\lambda(B(0,t))}\int_{\R^d}g(y)\chi_{B(0,t)}(x-y)\,dy=\frac{1}{\lambda(B(0,t))}\int_{B(0,t)}g,\]
which goes to $g(x)$ a.e.
\end{example}
\begin{example}
Let $\frac{1}{p}+\frac{1}{q}=1$. We could take $f\in L^p$, and map $g\in L^q$ to $\int f\overline{g}$. We know that
\[\|f\|_p=\|\Lambda_f\|=\sup_{\substack{g\in L^q\\ \|g\|_q=1}}|\Lambda_f(g)|.\]
\end{example}
\begin{theorem}
Let $\frac{1}{p}+\frac{1}{q}=1$, with $p<\infty$. For any $f\in L^p$ and $g\in L^q$,
\[(f\ast g)(x)=\int f(x-y)g(y)\,dy\]
exists for every $x$, and in fact is continuous. Moreover, $\|f\ast g\|_\infty\leq\|f\|_p\|g\|_q$.
\end{theorem}
\begin{theorem}
Let $f_t$ be approximate units, and let $g\in L^p$ with $1<p<\infty$. Then $\|f_t\ast g-g\|_p\to 0$ as $t\to 0$.
\end{theorem}
\begin{proof}
It will suffice to prove that for any $h\in L^q$,
\[\int_{\R^d}(f_t\ast g-g)h\leq(\text{small constant})\|h\|_q,\]
because $L^p$ is the dual of $L^q$ and this would show that the operator norm of the integrate-against-$(f_t\ast g-g)$ operator on $L^q$ (which is equal to the $L^p$ norm of $f_t\ast g-g$) is small.
We know that
\[g(x)=\int f_t(y)g(x)\,dy,\]
so that
\[\int\int(g(x-y)-g(x))f_t(y)\,h(x)\,dy\,dx\leq\int_{|y|<t}\|g_y-g\|_p\|h\|_q\,dy.\]
Therefore, it will suffice to show that
\[\int_{|y|<t}\|g_y-g\|_p\to 0\]
as $t\to 0$; but this is clear.
\end{proof}





