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\begin{document}

\centerline {\bf Math 523b - Final Exam Solutions- Spring 2013}

\bigskip

\noindent 
{\bf Do four of the following five problems.} 

\medskip

\noindent We use $dx$ to denote Lebesgue measure.

\bigskip

\begin{enumerate}

\item Let $X$ and $Y$ be locally compact topological spaces. Prove
that $X \times Y$ with the product topology is locally compact.

\noindent{\bf Solution:} 
Let $(x,y) \in X \times Y$. Since 
$X$ and $Y$ are locally compact there is a compact neighborhood
$U$ of $x$ in $X$ and a compact neighborhood $V$ of $y$ in $Y$. 
So $x \in int(U)$, $y \in int(V)$. 
By the definition of product toplogy, $int(U) \times int(V)$ is 
open in $X \times Y$. So $U \times V$ is a neighborhood of $(x,y)$ 
in $X \times Y$. 

Since $U$ is compact in $X$, $U$ with the relative topology is a compact
topological space. Likewise for $V$. So by Tychynoff's theorem, 
$U \times V$ with its product topology is compact. Some definition chasing 
shows that this product topology on $U \times V$ is the same as
the relative topology it gets from the product topology on $X \times Y$. 
So $U \times V$ is  compact in $X \times Y$ and so is a compact neighborhood
of $(x,y)$. 

\item Let $1 < p < \infty$. Suppose $f_n \in L^p([0,1],dx)$ with 
$||f_n||_p \le 1$ for all $n$. Suppose also that for all $\alpha \ge 0$ 
\beann
\lim_{n \ra \infty} \int_0^1 f_n(x) e^{-\alpha x} \, dx =0
\eeann
Prove that $f_n$ converges to $0$ weakly in $L^p$. 
(You may assume all functions are real valued.)

\noindent{\bf Solution:} 
Let $q$ be the conjugate exponent to $p$. 
To show $f_n$ converges weakly to $0$, we must show
$\int_0^1 f_n g dx$ converges to $0$ for all $g \in L^q$.
We are given that this is true for $g$ of the form $e^{-\alpha x}$.
Let $A$ be the set of functions of the form $\sum_{j=1}^n c_j \exp(-\alpha_j x)$
where $n$ is any positive integer, $c_j \in \reals$ and $\alpha_j \ge 0$. 
By linearity we have the desired convergence for $g \in A$. 

Since $\exp(-\alpha_1 x) \exp(-\alpha_2 x) =\exp(-(\alpha_1+\alpha_2) x)$,
$A$ is an algebra. It contains the constant functions (take $\alpha=0$). 
It contains $e^{-x}$ which separates all point in $[0,1]$. 
So by the Stone-Weierstrass theorem, $A$ is dense in $C[0,1]$ with 
the sup norm. This implies it is dense in $C[0,1]$ with 
respect to the $L^q$ norm since the measure is finite.
And since $C[0,1]$ is dense in $L^q$, $A$ is dense in $L^q$. 

Now let $h \in L^q$ and let $\epsilon > 0$. Pick $g \in A$ such that 
$||h-g||_q < \epsilon$. Then by Holder's inequality, 
\beann
|\int_0^1 f_n g dx - \int_0^1 f_n h dx| 
\le ||f_n||_p ||g-h||_q \le ||g-h||_q < \epsilon 
\eeann
So
\beann
\limsup_{n \ra \infty} |\int_0^1 f_n(x) h(x) \, dx| \le \epsilon
\eeann
This holds for all $\epsilon >0$, so the limsup is $0$ and 
so 
\beann
\lim_{n \ra \infty} \int_0^1 f_n(x) h(x) \, dx=0
\eeann

\item Let $M$ be a closed subspace of a Hilbert space $H$.
Let $F$ be a bounded linear functional on $M$. The Hahn Banach
theorem says that there is an extension of $F$ to a linear 
functional on all of $H$ which has the same norm. Prove
that such an extension is unique.

\noindent{\bf Solution:} 
We recall for $y \in H$, $x \ra <x,y>$ defines a bounded linear
functional on $H$ whose norm is $||y||$. This gives an isometry 
between the dual of a Hilbert space and the Hilbert space itself. 
The closed subspace $M$ is itself a Hilbert space. So we can restrict
$F$ to $M$ and get a bounded linear functional on $M$. 
Then there is a unique $y \in M$ such that $F(x) = <x,y>$ for $x \in M$. 

Any extension of $\bar{F}$ of $F$ to $H$ must be of the form
$\bar{F}(x) = <x,z>$ for some $z$ in $H$. 
By the projection theorem $z$ has a unique decomposition 
$z=z_1+z_2$ where $z_1 \in M$ and $z_2 \in M^\perp$.
Since it is an extension we must have $<x,y>=<x,z_1+z_2>=<x,z_1>$ for 
all $x \in M$. So $<x,y-z_1>=0$ for all $x \in M$. But $y-z_1 \in M$, 
so $y-z_1$ must be $0$, i.e.,  $z_1=y$. 
Finally we note that the norm of the original $F$ is $||y||$
while the norm of the extension is 
$||z_1+z_2||=||y+z_2||=\sqrt{||y||^2+||z_2||^2}$
So $||z_2||=0$, i.e., $z_2=0$. Thus the only possible 
extension is given by $z=y$. 

\item
Let $C_0=C_0(\reals)$ be the real-valued 
continuous functions on the real line that 
go to zero at $\pm \infty$. We use the usual sup norm.
If $g \in L^1(\reals,dx)$ is a real-valued Lebesgue integrable 
function, then we can define a bounded linear function $\phi_g$
on $C_0$ by 
\beann 
\phi_g(f)= \int f(x) g(x) \, dx
\eeann
Let $F$ be the set of all such linear functionals:
\beann
F= \{\phi_g: g \in L^1(\reals,dx)\}
\eeann
Prove that $F$ is a closed subset of $C_0^*$.

\noindent{\bf Solution One:} The dual $C_0^*$ is identified with the 
space of finite Borel measures on $\reals$ in the usual way. 
For the linear functionals $\phi_g$, the measure that represent 
the functional is $g(x) dx$. The norm of $\phi_g$ as a linear
functional is the norm of the measure which is the $L^1$ norm 
of $g$. So if $\phi_{g_n}$ converges to some bounded linear 
functional $\phi$ in $C_0^*$, then $g_n$ is a Cauchy sequence in $L^1$ 
and so by the completeness of $L^1$ converges to some $g \in L^1$. 
This means that $\phi_{g_n}$ converges to $\phi_g$ in $C_0$, and 
so $F$ is closed.  

\smallskip
\noindent{\bf Solution Two:} Again, we identify $C_0^*$ with the 
space of finite Borel measures. By the Randon-Nikodym theorem, 
$F$ corresponds with the Borel measures that are absolutely 
continuous with respect to Lebesgue measure. The norm on $C_0$
corresponds to the total variation norm of the measures. 
So we need to show that if $||\mu_n - \mu|| \ra 0$ for 
Borel measures $\mu_n$ and $\mu$ and $\mu_n << m$ for all $n$
where $m$ is Lebesgue measure, then $\mu << m$. 
Suppose $E$ is a Borel set with $m(E) =0$. Then $\mu_n(E)=0$
for all $n$. Since 
\beann 
|\mu_n(E)-\mu(E)|=|(\mu_n-\mu)(E)| 
\le ||\mu_n-\mu||
\eeann 
we conclude that $\mu(E)=0$. So $\mu << m$ and thus $F$ is closed.


\item Let $X_n$ be an independent, identically distributed 
sequence of random variables with $X_1 \ge 0$ a.e. and 
$E[X_1]=\infty$. This implies that for all $c>0$,
\beann
\sum_{n=1}^\infty P(X_1 \ge c n) = \infty
\eeann
You may assume this fact without proving it. 

(a) Prove that $P(X_n \ge c \, n \quad infinitely \quad often)=1$ 
for all $c>0$. 

(b) Prove that for all $c>0$
\beann 
P(\limsup_{n \ra \infty} \frac{1}{n} \sum_{j=1}^n X_j \ge c) =1
\eeann 

(c) Prove that 
\beann 
P(\limsup_{n \ra \infty} \frac{1}{n} \sum_{j=1}^n X_j =\infty) =1
\eeann 

\noindent{\bf Solution:} 

(a) Since they are identically distributed, 
$P(X_1 \ge cn)=P(X_n \ge cn)$. So
\beann
\sum_{n=1}^\infty P(X_n \ge c n) = \infty
\eeann
By the Borel Cantelli Lemma the statement in (a) holds. 

(b) If $\omega$ is in the event 
$\{X_n \ge c \, n \quad infinitely \quad often\}$
then there is a subsequence $n_k$ (which depends on $\omega$)
such that $X_{n_k} \ge c n_k$. So we have for all $k$
\beann
\frac{1}{n_n} \sum_{j=1}^{n_k} X_j \ge c
\eeann
Hence the statement in (b) holds. 

(c) 
\beann 
\{\limsup_{n \ra \infty} \frac{1}{n} \sum_{j=1}^n X_j =\infty\} 
= \bigcap_{N=1}^\infty 
\{\limsup_{n \ra \infty} \frac{1}{n} \sum_{j=1}^n X_j \ge N \} 
\eeann 
Each event in this countable intersection has probability one, so 
the intersection has probability one. (Countable union of 
measure zero sets is measure zero.)

\end{enumerate}



\end{document}
\end


