\classheader{2012-11-06}

There will be no class on Thursday.

There were 3 questions on the exam, with a maximum score of 1 each. The grading scale for the midterm is
\begin{center}
\begin{tabular}{c|r}
A & 2\\\hline
B & 1.5\\\hline
C & 1\\\hline
D & 0.5
\end{tabular}
\end{center}

Some people were confused about the definition of a measurable function. Recall that, for measurable spaces $(X,\cal{A})$ and $(Y,\cal{B})$, a function $f:X\to Y$ is $(\cal{A},\cal{B})$-measurable if for all $B\in\cal{B}$, we have $f^{-1}(B)\in\cal{A}$. When $Y$ is a topological space, then we just say $\cal{A}$-measurable, because (unless specified otherwise) $Y$ will be given the Borel $\sigma$-algebra. This is equivalent to requiring that $f^{-1}(G)\in\cal{A}$ for all open $G\subseteq Y$. 

Let's get back to what we talked about last class.

\begin{definition}
Let $B$ be a normed linear space. A linear map $\Lambda:B\to\R$ (or $\Lambda:B\to\C$) is called a linear functional.
\end{definition}

\begin{definition}
Let $B_1,B_2$ be normed linear spaces. A linear map $A:B_1\to B_2$ is called a linear operator. We say that $A$ is bounded if there is some $K$ for which $\|Ax\|\leq K\|x\|$ for all $x\in B_1$. The best possible $K$ is
\[\sup_{\|x\|=1}\|Ax\|.\]
We call this the norm of $A$. It turns out that the bounded linear operators themselves form a normed linear space under this norm; and in particular, we call
\[B^*=\{\text{bounded linear functionals }\Lambda\text{ on }B\}\]
the dual of $B$, which is a normed linear space with $\|\Lambda\|=\sup_{\|x\|=1}|\Lambda(x)|.$
\end{definition}

\begin{example}
Let $B=\R^n$, with norm
\[\|x\|=\|x\|_2=\left(\sum x_j^2\right)^{1/2}.\]
What is $B^*$? Let $e_1,\ldots,e_n$ denote the standard basis for $\R^n$, and given a linear functional $\Lambda$, let $\Lambda(e_i)=c_i$. Then
\[\Lambda(x)=\sum x_jc_j=\langle x,c\rangle\leq\|x\|_2\cdot\|c\|_2,\]
so $\|\Lambda\|\leq\|c\|_2$, and because $|\Lambda(c)|=\|c\|_2^2$ we must have that $\|\Lambda\|=\|c\|_2$. The map identifying $\Lambda$ with $c$ is linear, and it preserves norms. Thus $B^*\cong B$.
\end{example}

\begin{theorem}
For any $1<p,q<\infty$ with $\frac{1}{p}+\frac{1}{q}=1$, or $p=1,q=\infty$, we have $(L^p)^*=L^q$.
\end{theorem}
\begin{remark}
Note that this implies $(L^2)^*=L^2$. Also, $(L^\infty)^*\neq L^1$.
\end{remark}

What do we really mean by this theorem? For any $\Lambda\in(L^p)^*$, there is some $g\in L^q$ such that $\Lambda(f)=\int fg\,d\mu$ for all $f\in L^p$.

\begin{proof}
Suppose that $g\in L^q$. Define $\Lambda(f)=\int fg\,d\mu$. We need to show that this is in fact a bounded linear operator.

It is obvious that $\Lambda$ is linear, and it is bounded by H\"older's inequality:
\[|\Lambda(f)|=\left|\int fg\right|\leq\|f\|_p\|g\|_q.\]
This shows that $\|\Lambda\|\leq\|g\|_q$, and in fact we have equality, because choosing $f=g^{q-1}$, we have $\|f\|_p^p=\|g\|_q^q$, and therefore
\[\left|\int fg\right|=\int|g^q|=\|f\|_p\|g\|_q.\]

Now we want to prove that any element of $(L^p)^*$ can be obtained this way. We will prove it in the case that $\mu$ is finite, i.e. $\mu(X)<\infty$.

\begin{homework}
Prove this claim when $\mu$ is any $\sigma$-finite measure.
\end{homework}

Given a linear functional $\Lambda\in(L^p)^*$, how can we construct the corresponding $g$? For any meausurable set $A$, we have that $\chi_A\in L^\infty\subseteq L^p$ (this inclusion holds because $\mu$ is finite). Now denote $\Lambda(\chi_A)=\varphi(A)$. We claim that $\varphi$ is a measure (note that we don't know $\varphi(A)$ will be positive, so this could be a \textit{signed} measure).

Given pairwise disjoint measurable $A_1,A_2,\ldots$ we have
\[A=\bigcup_{j=1}^\infty A_j=\bigg(\bigcup_{j=1}^nA_j\bigg)\cup\bigg(\underbrace{\bigcup_{j=n+1}^\infty A_j}_{B_n}\bigg).\]
Because $\Lambda$ is linear,
\[\varphi(A)=\sum_{j=1}^n\varphi(A_j)+\varphi(B_n).\]
We need to show that $\varphi(B_n)\to 0$ as $n\to\infty$. Because $\Lambda$ is bounded,
\[|\varphi(B_n)|=|\Lambda(\chi_{B_n})|\leq\|\Lambda\|\cdot \|\chi_{B_n}\|_{L^p}\underset{\substack{\uparrow\\\text{not true}\\\text{for }p=\infty}}{=}\|\Lambda\|\cdot\mu(B_n)^{1/p}.\]
Because
\[\mu(B_n)=\sum_{j=n+1}^\infty\mu(A_j)\to 0\;\;\text{ as }\;\;n\to\infty,\]
we are done. Note that this also demonstrates that $\varphi\ll \mu$. By the Radon-Nykodim theorem, there is some $g\in L^1$ such that 
\[\Lambda(\chi_A)=\varphi(A)=\int g\chi_A\,d\mu.\]
So, we just need to show that $g\in L^q$. In the case that $p=1$, we have
\[\left|\int_Ag\,d\mu\right|=\Lambda(\chi_A)\leq\|\Lambda\|\cdot\mu(A),\]
and therefore
\[\left|\frac{1}{\mu(A)}\int_Ag\,d\mu\right|\leq\|\Lambda\|\]
for any meausurable $A$ with $\mu(A)>0$. This shows that $|g|\leq\|\Lambda\|$ $\mu$-a.e.

For $1<p<\infty$, because
\[\Lambda(h)=\int hg\,d\mu\;\;\text{ for every characteristic function }h,\]
we therefore also have that
\[\Lambda(h)=\int hg\,d\mu\;\;\text{ for every simple function }h\]
because $\Lambda$ and $\int$ are linear, and then
\[\Lambda(h)=\int hg\,d\mu\;\;\text{ for every }L^\infty\text{ function }h\]
because if $h\in L^\infty$ then for any $\epsilon>0$ there is some simple $h_1$ such that $\|h-h_1\|_\infty<\epsilon$ (we've proved this before), and this means that
\[\bigg|\Lambda(h)-\int hg\bigg|=\bigg| \Lambda(h-h_1)-\int(h-h_1)g+\underbrace{\Lambda(h_1)-\int h_1g}_{=\,0}\bigg|\leq \underbrace{|\Lambda(h-h_1)|}_{\substack{\leq\,\|\Lambda\|\cdot\|h-h_1\|_p\\\leq\,\|\Lambda\|\cdot\epsilon\mu(X)^{1/p}}}+\underbrace{\left|\int(h-h_1)g\right|}_{\leq\,\epsilon\cdot\|g\|_1}.\]
Because $g\in L^1$, the sets $A_n=\{x:|g(x)|\leq n\}$ have the property that $\mu(X\setminus A_n)\to 0$ as $n\to\infty$. Let \[f=|g|^{q-1}\cdot\sign(g)\cdot\chi_{A_n}.\]
Because $f\in L^\infty$, we know that $\Lambda(f)=\int fg$. We have
\[\int_{A_n}|g|^q=\Lambda(f)\leq\|\Lambda\|\cdot\left(\int_X|f|^p\right)^{1/p}=\|\Lambda\|\cdot\left(\int_{A_n}|g|^q\right)^{1/p}\]
and therefore
\[\left(\int_{A_n}|g|^q\right)^{1-1/p}=\left(\int_{A_n}|g|^q\right)^{1/q}\leq\|\Lambda\|\]
for all $n$, which implies that $\|g\|_q\leq\|\Lambda\|$. To show that there is actually equality, choose $f$ such that $\|f\|_p^p=\|g\|_q^q$.
\end{proof}

Now we will prove Fubini's theorem. First, we need to discuss what it means to take the product of two measures.

Given two measure spaces $(X,\cal{A},\mu)$ and $(Y,\cal{B},\nu)$, for any $A\in\cal{A}$ and $B\in\cal{B}$, we define $\varphi(A\times B)=\mu(A)\cdot\nu(B)$. We extend $\varphi$ to a measure on the $\sigma$-algebra on $Z=X\times Y$ which is generated by sets of the form $A\times B$ for $A\in\cal{A}$, $B\in\cal{B}$.

\begin{theorem}[Fubini]
If $\varphi$ is $\sigma$-finite and $\int_Zf\,d\varphi$ exists, then the function
\[g(x)\overset{\text{\emph{def}}}{=}\int_Yf(x,y)\,d\nu\]
exists for almost all $x\in X$, and
\[\int_Xg(x)\,d\mu=\int_Zf\,d\varphi.\]
\end{theorem}

\begin{homework}
Let $X=Y=[0,1]$, and define $f:X\times Y\to \R$ to be
\[f(x,y)=\begin{cases}
1 & \text{ if }x= y,\\
0 & \text{ if }x\neq y.
\end{cases}\] Let $\mu$ be Lebesgue measure on $X$, and $\nu$ be counting measure on $Y$. Calculate
\[\int_X\left(\int_Yf(x,y)\,d\nu\right)\,d\mu\]
and
\[\int_Y\left(\int_X f(x,y)\,d\mu\right)\,d\nu,\]
and explain why Fubini fails.
\end{homework}

\begin{homework}
On the midterm, you showed that for measurable $A,B\subseteq[0,1]$, the function
\[f(t)=\lambda(A\cap(B+t))\]
is continuous. Now, find $\int f(t)$.
\end{homework}







