\classheader{2012-11-01}

Today we'll go back over some more basic material that not everyone has seen yet.

Recall the definition of $L^p$ space:
\begin{definition}
Given a measure space $(X,\mu)$, and a function $f$ on $X$ such that $\int |f|^p\,d\mu<\infty$, we say that $f\in L^p(\mu)$. When there exists a $K$ such that $|f(x)|\leq K$ almost everywhere, we say that $f\in L^\infty(\mu)$.
\end{definition}

\begin{definition}
A normed space is a vector space $V$ with a function $\|\cdot\|:V\to\R$ such that
\begin{itemize}
\item $\|f\|\geq 0$ for all $f$, with equality if and only if $f=0$
\item $\|cf\|=|c|\cdot\|f\|$
\item $\|f_1+f_2\|\leq\|f_1\|+\|f_2\|$
\end{itemize}
We say that $V$ is complete if $\rho(f,g)=\|f-g\|$ is a complete metric. A complete normed space is called a Banach space.
\end{definition}

As we will see later, the norms
\[\|f\|_p=\left(\int|f|^p\,d\mu\right)^{1/p},\qquad \|f\|_\infty=\inf\{K:|f(x)|\leq K\text{ a.e.}\}\]
make $L^p$ and $L^\infty$, respectively, into complete normed spaces, but only after we identify functions $f$ and $g$ if $f=g$ a.e. (otherwise we can have $\|f\|=0$ even when $f\neq 0$).

\begin{theorem}[H\"older's Inequality]
For any $f\in L^p$ and $g\in L^q$ where $\frac{1}{p}+\frac{1}{q}=1$, and either $1<p,q<\infty$ or $p=1,q=\infty$, we have $fg\in L^1$ and
\[\|fg\|_1\leq\|f\|_p\|g\|_q.\]
\end{theorem}
\begin{lemma}
For any $a,b\geq 0$ and $0<\lambda<1$, we have $a^\lambda b^{1-\lambda}\leq \lambda a+(1-\lambda)b$.
\end{lemma}
\begin{proof}[Proof of lemma]
Taking logarithms of both sides,
\[\lambda\log(a)+(1-\lambda)\log(b)\leq\log(\lambda a+(1-\lambda)b).\]
But $\log$ is concave,
\begin{center}
\begin{tikzpicture}[scale=1.5,thick]
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\draw[->,name path global=linethree] (-0.2,0) -- (4.2,0) node[right] {$x$};
\draw[->] (0,-1.2) -- (0,1.8) node[above] {$y$};
\draw[name path global=lineone,color=blue,domain=0.3:4] plot function{log(x)} node[right] {$\log(x)$};
\begin{scope}
\clip plot[domain=0.3:4] function{log(x)};
\clip (0,0) rectangle (3,2);
\draw [name path global=linetwo,color=red,domain=0:4] plot function{0.5*x-0.4};
\end{scope}
\begin{scope}
\clip (0,0) rectangle (4,2);
\draw[name path global=linefour,densely dotted,name intersections={of=lineone and linetwo}] (intersection-1) node [fill,circle,minimum size=3pt,inner sep=0pt,outer sep=0pt,label={[label distance=1pt,inner sep=0pt,outer sep=0pt]135:{\small $\log(a)$}}] (fa) {} -- ++(0,-2);
\draw[name path global=linefive,densely dotted,name intersections={of=lineone and linetwo}] (intersection-2) node [fill,circle,minimum size=3pt,inner sep=0pt,outer sep=0pt,label={[label distance=3pt,inner sep=0pt,outer sep=0pt]92:{\small $\log(b)\;$}}] (fb) {} -- ++(0,-3);
\end{scope}'
\begin{scope}[name intersections={of=linethree and linefour}]
\node [fill,circle,minimum size=3pt,inner sep=0pt,outer sep=0pt,label={[label distance=3pt,inner sep=0pt,outer sep=0pt]270:{\small $a$}}] at (intersection-1) {};
\end{scope}
\begin{scope}[name intersections={of=linethree and linefive}]
\node [fill,circle,minimum size=3pt,inner sep=0pt,outer sep=0pt,label={[label distance=3pt,inner sep=0pt,outer sep=0pt]270:{\small $b$}}] at (intersection-1) {};
\end{scope}
%\draw[color=blue,domain=0.3:4] plot function{log(x)} node[right] {$\log$};
\end{tikzpicture}
\end{center}
so this is true.
\end{proof}
\begin{proof}[Proof of H\"older]
Let's do the case when $\|f\|_p=1$ and $\|g\|_q=1$ first. Let $\lambda=\frac{1}{p}$ and $1-\lambda=\frac{1}{q}$, and let $a=|f(x)|^p$ and $b=|g(x)|^q$. The lemma implies that
\[|f(x)|\cdot|g(x)|\leq\frac{1}{p}|f(x)|^p+\frac{1}{q}|g(x)|^q,\]
and therefore
\[\int|f(x)|\cdot|g(x)|\leq\frac{1}{p}\underbrace{\int|f(x)|^p}_{=\,1}+\frac{1}{q}\underbrace{\int|g(x)|^q}_{=\,1}=\frac{1}{p}+\frac{1}{q}=1.\]
In the general case, we can just let $F=\frac{f}{\|f\|_p}$ and $G=\frac{g}{\|g\|_q}$, so that we can apply the special case to see that
\[\int|FG|\leq 1\]
and hence
\[\int\frac{|fg|}{\|f\|_p\|g\|_q}\leq 1\implies \int|fg|\leq\|f\|_p\|g\|_q.\]
Finally, if $p=1$ and $q=\infty$, we have that
\[\int|fg|\leq\int|f|\cdot K=K\int|f|\]
when $|g|\leq K$ almost everywhere.
\end{proof}

\begin{theorem}[Minkowski Inequality]
For any $1\leq p\leq \infty$, we have
\[\|f+g\|_p\leq\|f\|_p+\|g\|_p\]
for any $f,g\in L^p$.
\end{theorem}
\begin{proof}
If $p=1$ or $p=\infty$, this is trivial. Now suppose $1<p<\infty$. First, let's show that $f+g\in L^p$: by the convexity of the logarithm again, we can see that
\[\left|\frac{f+g}{2}\right|^p\leq\frac{|f|^p+|g|^p}{2}\]
and therefore $|f+g|^p\leq 2^{p-1}(|f|^p+|g|^p)$. This shows that $f+g\in L^p$.

Now let $q$ be the solution to $\frac{1}{p}+\frac{1}{q}=1$ and let $F\in L^p$. Then we claim $|F|^{p-1}\in L^q$. This is because $pq-q=p$ implies
\[\left(|F|^{p-1}\right)^q=|F|^p\]
and moreover
\[\||F|^{p-1}\|_q^q=\|F\|_p^p.\]

Now for the final step. Note that
\[\|f+g\|_p^p=\int|f+g|^p=\int \underbrace{|f+g|^{p-1}}_{\in\,L^q}|f+g|\leq\int |f+g|^{p-1}|f|+\int|f+g|^{p-1}|g|\]
\[\overset{\text{H\"older}}{\leq}\left\||f+g|^{p-1}\right\|_q\cdot\|f\|_p+\left\||f+g|^{p-1}\right\|_q\cdot\|g\|_p=\|f+g\|_p^{p/q}\cdot\|f\|_p+\|f+g\|_p^{p/q}\|g\|_p.\]
Therefore
\[\|f+g\|_p^p\leq\|f+g\|_p^{p/q}(\|f\|_p+\|g\|_p),\]
but since $p-p/q=p(1-\frac{1}{q})=1$, this implies
\[\|f+g|_p\leq\|f\|_p+\|g\|_p.\qedhere\]
\end{proof}
Given a normed linear space $B$, and a sequence $b_1,b_2,\ldots\in B$, let $s_n=\sum_{j=1}^nb_j\in B$. Then we say that the series $\sum_{j=1}^\infty b_j$ converges if there is an $s\in B$ such that $\|s_n-s\|\to 0$. We say that it is absolutely convergent if $\sum_{j=1}^\infty\|b_j\|<\infty$.
\begin{theorem}[Riesz-Fisher]
For any $1\leq p\leq \infty$, $L^p$ is complete.
\end{theorem}
\begin{lemma}
For any normed linear space $B$, $B$ is complete if and only if every absolutely convergent sequence converges.
\end{lemma}
\begin{proof}[Proof of lemma]
Suppose $B$ is complete. For any absolutely convergent sequence $b_1,b_2,\ldots,\in B$,
\[\|s_n-s_m\|=\left\|\sum_{j=n+1}^m b_j\right\|\leq\sum_{j=n+1}^m\|b_j\|<\epsilon\]
if $n$ is large enough. Thus the sequence of $s_n$'s is Cauchy, and because $B$ is complete the limit exists.

Now conversely, suppose that $b_1,b_2,\ldots\in B$ is a Cauchy sequence, and assume that every absolutely convergent sequence in $B$ converges. Then for any $k$, there exists an $N_k$ such that for all $n,m\geq N_k$, we have $\|b_m-b_n\|<\frac{1}{2^k}$. Thus,
\[b_{N_1}+(b_{N_2}-b_{N_1})+(b_{N_3}-b_{N_2})+\cdots\]
is absolutely convergent because
\[\|b_{N_1}\|+\|b_{N_2}-b_{N_1}\|+\|b_{N_3}-b_{N_2}\|+\cdots\leq\|b_{N_1}\|+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots<\infty\]
The partial sums of this series are just $b_{N_1},b_{N_2},\ldots$, so that by the assumption that every absolutely convergent series converges, there must be some $b\in B$ such that $b_{N_k}\to b$. But since $b_1,b_2,\ldots$ is a Cauchy sequence, we must also have that $b_n\to b$.
\end{proof}
\begin{homework}
Prove the $p=\infty$ case of the Riesz-Fisher theorem.
\end{homework}
\begin{proof}[Proof of Riesz-Fisher]
Let $1\leq p<\infty$. Then for any sequence $f_1,f_2,\ldots\in L^p$, we want to show
\[\underbrace{\sum\|f_n\|_p}_{\sigma}<\infty\implies \sum f_n\text{ converges in }L^p\text{ to some function }f.\]
Let $g_n=\sum_{j=1}^n|f_j|$, which is in $L^p$. In particular, by Minkowski's inequality,
\[\|g_n\|_p\leq \sum_{j=1}^n\|f_j\|_p\leq \sigma,\]
and therefore $\int |g_n|^p\leq \sigma^p$. The sequence of functions $g_n$ is monotone increasing, so $g_n^p\nearrow g^p$, and therefore
\[\int g^p=\lim\int g_n^p\leq \sigma^p\implies g^p<\infty\text{ a.e. }\implies g<\infty\text{ a.e.}\]
Thus, $\sum |f_n(x)|$ converges for almost all $x$, so
\[f(x)\overset{\text{def}}{=}\sum f_n(x)\]
converges for almost all $x$. But we still need to show that the partial sums $s_n=\sum_{j=1}^n f_j(x)$ satisfy $\|f-s_n\|_p\to 0$. To see this, observe that
\[\int|f-s_n|^p\leq\int \left(|f|+|s_n|\right)^p\leq\int (2g)^p,\]
and because $|f-s_n|^p\in L_1$ and $|f-s_n|^p\to 0$ a.e., we have that $|f-s_n|^p\to 0$ in $L_1$, which is the case if and only if $|f-s_n|\to 0$ in $L^p$.
\end{proof}

\textbf{For the rest of this class, assume we have a finite measure space, so that $\mu(X)<\infty$.}

\begin{proposition}
For any $f\in L^p$ and $\nu<p$, we also have $f\in L^\nu$. More generally,
\[L^\infty\subseteq L^p\subseteq L^\nu\subseteq\cdots\subseteq L^1.\]
\end{proposition}
\begin{proof}
Intuitively, this is because raising to any power higher than 1 is no longer concave, it is convex, making Minkowski's inequality fail.

Letting $X_1=\{x\in X\mid  f(x)\leq 1\}$ and $X_2=\{x\in X\mid f(x)>1\}$, we have that
\[\int\limits_X|f|^p=\underbrace{\int\limits_{X_1}|f|^p}_{\leq \mu(X)<\infty}+\int\limits_{X_2}|f|^p\geq \int\limits_{X_2}|f|^\nu\]
because $|f|^p\geq |f|^\nu$ on $X_2$, so that
\[\int\limits_X|f|^\nu=\underbrace{\int\limits_{X_1}|f|^\nu}_{\leq \mu(X)<\infty}+\int\limits_{X_2}|f|^\nu\]
is finite, and therefore $f\in L^\nu$.
\end{proof}
\begin{proposition}
If $f\in L^\infty$, then $f\in L^p$ for any $p$. Moreover, $\|f\|_p\to\|f\|_\infty$ as $p\to\infty$.
\end{proposition}
\begin{proof}
For any $t<\|f\|_\infty$, we have that $\mu(A)>0$ where $A=\{x\mid f(x)>t\}$. Then
\[\int\limits_X|f|^p\geq\int\limits_A|f|^p\geq\int\limits_A t^p=t^p\mu(A).\]
Thus $\|f\|_p\geq t\mu(A)^{1/p}$. As $p\to\infty$, we have that
\[\liminf_{p\to\infty}\|f\|_p\geq t\cdot 1,\]
and therefore $\liminf\|f\|_p\geq\|f\|_\infty$. In the other direction, $|f|\leq\|f\|_\infty$ a.e., so that
\[\int|f|^p\leq\int\|f\|_\infty^p=\|f\|_\infty^p\cdot\mu(X),\]
and hence $\|f\|_p\leq\|f\|_\infty\cdot\mu(X)^{1/p}$. Therefore, $\limsup_{p\to\infty} \|f\|_p\leq\|f\|_\infty\cdot 1$.
\end{proof}
\begin{homework}
If $\mu(X)<\infty$, is it true that if $f\in L^p$ for all $1<p<\infty$, then $f\in L^\infty$?
\end{homework}
\begin{theorem}
For any $1\leq p\leq \infty$, the simple functions are dense in $L^p$.
\end{theorem}
\begin{proof}
The case of $p=\infty$ is easy - to define a simple function close to $f\in L^\infty$, break up the range of $f$ in steps of size $\epsilon$, and on the set $\{x\mid k\epsilon\leq f(x)<(k+1)\epsilon\}$, set the simple function to be $k\epsilon$. This is a simple function because it will only take on finitely many values (since $\|f\|_\infty$ is finite), and it differs from $f$ by at most $\epsilon$ everywhere.

Now let $1\leq p<\infty$. Note that it is enough to show the claim is true for $f\geq 0$. Fix an $\epsilon>0$, and let $A=\{x\mid f(x)>\delta\}$ where $\delta$ is chosen such that
\[\int_{X/A}|f|^p<\frac{\epsilon}{4}.\]
Choose an $n$ such that $A_n=\{x:f(x)\leq n\}$ satisfies
\[\int_{X\setminus A_n}f^p<\frac{\epsilon}{4}.\]
Lastly, choose $\eta$ such that
\[\frac{\epsilon}{4\mu(A_n)}=\eta^p.\]
Then, again breaking up the range of $f$, we define the set
\[M_\nu=\{x\in A_n\mid (\nu-1)\eta\leq f(x)\leq \nu\eta\},\]
and now we can define the simple function
\[g(x)=\begin{cases}
0 & \text{ if }x\notin A_n,\\
(\nu-1) & \text{ if }x\in M_\nu.
\end{cases}\]
This satisfies
\[\int|f-g|^p=\int_{X\setminus A}|f|^p+\sum_{\nu}\int_{M_\nu}|f-g|^p+\int_{A/A_n}|f|^p<\left(\frac{\epsilon}{4}\right)+\underbrace{\eta^p\cdot\mu(A_n)}_{=\,\epsilon/4}+\left(\frac{\epsilon}{4}\right)<\epsilon.\qedhere\]
\end{proof}

Next time, we'll look more at the special properties of $\frac{1}{p}+\frac{1}{q}=1$. In particular, we'll prove that the dual of $L^p$ is $L^q$ and the dual of $L^1$ is $L^\infty$.