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\begin{document}
\title[18.102 Final Exam, Spring 2015]
{Short solutions for 18.102 Final Exam, Spring 2015}

\maketitle
\vspace{15pt}
\section*{Problem 1} Consider the subspace $H\subset\mathcal{C}[0,2\pi]$ consisting of
those continuous functions on $[0,2\pi]$ which satisfy 
%
\begin{equation}
u(x)=\int_0^x U,\ \forall\ x\in[0,2\pi]
\label{18.102-Prob.13}\end{equation}
%
for some $U\in L^2(0,2\pi)$ (depending on $u$ of course). Show that the
function $U$ is determined by $u$ (given that it exists) and that 
%
\begin{equation}
\|u\|_{H}^2=\int_{(0,2\pi)}|U|^2
\label{18.102-Prob.14}\end{equation}
%
turns $H$ into a Hilbert space.
\vspace{10pt}

Solution: If $U\in L^2([0,2\pi]$ then the integral \eqref{18.102-Prob.13}
defines a continuous function since 
%
\begin{equation*}
|u(x)-u(y)|\le\int_y^x|U|\le |x-y|^{\ha}\|U\|_{L^2},\ \sup|u|\le (2\pi)^{\ha}\|U\|_{L^2}
\end{equation*}
so in fact $I:L^2[0,2\pi]\longrightarrow \cC([0,2\pi])$ is a bounded linear
map. To say that $U,$ if it exists, is determined by $u$ is to say that
this map in injective. The vanishing of $u$ means precisely that $\langle
\chi_{[0,x]},U\rangle_{L^2}=0.$ Taking linear combination, this means that
$U$ is orthogonal to all step functions. However the step functions are
dense in $\cC([0,2\pi])$ in the supremum norm and hence in $L^2[0,2\pi],$
so this imples $U=0$ in $L^2.$ Since $I$ is injective, it is a bijection
onto its range, $H$ and this gives a bijection to $L^2[0,2\pi],$ making $H$
into a Hilbert space.

Other arguments that work include computing the Fourier coefficients of $U$
to shows that they are determined by $u.$ In general a measurable set
(where $U>0$ for instance) does not contain a close measurable set of
positive measure, so that sort of approach is hard.

\vspace{15pt}
\section*{Problem 2} Consider the space of those complex-valued functions
on $[0,1]$ for which there is a constant $C\ge0$
(depending on the function) such that
%
\begin{equation}
|u(x)-u(y)|\le C|x-y|^{\frac12}\ \forall\ x,y\in[0,1].
\label{18.102-Prob.15}\end{equation}
%
Show that this is a Banach space with norm 
%
\begin{equation}
\|u\|_{\frac12}=\sup_{[0,1]}|u(x)|+\inf_{\text{\eqref{18.102-Prob.15} holds}}C.
\label{18.102-Prob.16}\end{equation}
%
\vspace{10pt}

Solution: These are the H\"older-$\ha$ functions, $\cC^{\ha}[0,1].$ If
\eqref{18.102-Prob.15} holds for some constant $C\ge0$ then
$$
\|u\|'=\sup_{x\not=y\in[0,1]}\frac{|u(x)-u(y)}{|x-y|^{\ha}}<\infty
$$
is the smallest such constant and the putative norm is
$$
\|u\|_{\frac12}=\sup_{[0,1]}|u(x)|+\|u\|'.
$$
I expected you to quickly check that this is a norm and that the space of
functions $\cC^{\ha}[0,1]$ is linear. The inequality \eqref{18.102-Prob.15}
implies that the elements of $\cC^{\ha}$ are continuous and if $u_n$ is a
Cauchy sequence it follows that it is Cauchy with respect to the supremum
norm, $\|u\|_{\infty}\le\|u\|_{\ha}$ by definition. Since this space is
complete, $u_n\to u$ uniformly with $u:[0,1]\longrightarrow \bbC$
continuous. A Cauchy sequence is bounded in norm so
$$
|u_n(x)-u_n(y)|\le C|x-y|^{\ha}
$$
with $C$ independent of $n.$ Passing to the limit $n\to\infty$ shows that
$u\in\cC^{\ha}.$  The Cauchy condition itself implies that given $\epsilon
>0$ there exists $N$ such that
$$
|(u_n(x)-u_m(x))-(u_n(y)-u_m(y))|\le C\epsilon |x-y|^{\ha}\ \forall\ n,m>N.
$$
Taking $m\to\infty$ and using the convergence in supremum norm it follows
that $\|u-u_n\|_{\ha}\to0.$ 

Generally well done.
\vspace{15pt}


\section*{Problem 3} Let $A_j\subset\bbR$ be a
sequence of subsets with the property that the characteristic function,
$\chi_j$ of $A_j,$ is integrable for each $j.$ Show that the characteristic
function of $\bbR\setminus A,$ where $A=\bigcup_{j}A_j$ is locally integrable.
\vspace{10pt}
Since for each $j,$ $\chi_j\in L^1(\bbR)$ are real functions it follows
that $\chi_{[k]},$ the characteristic function of $\bigcup_{j\le k}A_j$ is
in $L^1(\bbR)$ as the supremum of a finite number of $L^1$ functions and so
is $\chi_{[-R,R]}\chi_{[k]}$ for each $R>0.$ The $L^1$ integral of this increasing
sequence if bounded by $4R$ so by Monotone Convergence,
$\chi_{[-R,R]}\chi_{A}\in L^1(\bbR)$ where $\chi_{A}$ is the
characteristic function of $A=\bigcup_jA_j.$ The difference
$\chi_{[-R,R]}(1-\chi_{[\infty]})$ is therefore also integrable and this is
$\chi_{[-R,R]}\chi_B$ where $B=\bbR\setminus A,$  so $\chi_B$ is locally integrable.
\vspace{15pt}

\section*{Problem 4} Let $A$ be a Hilbert-Schmidt operator on a separable
Hilbert space $H,$ which means that for some orthonormal basis $\{e_i\}$  
%
\begin{equation}
\sum\limits_{i}\|Ae_i\|^2<\infty.
\label{18.102-Prob.6}\end{equation}
%
Using Bessel's identity to expand $\|Ae_i\|^2$ with respect to another
orthonormal basis $\{f_j\}$ show that 
$\sum\limits_{j}\|A^*f_j\|^2=\sum\limits_{i}\|Ae_i\|^2.$
Conclude that the sum in \eqref{18.102-Prob.6} is independent of the othornormal basis used
to define it and that the Hilbert-Schmidt operators form a Hilbert space.
\vspace{10pt}

\vspace{15pt}
\section*{Problem 5} Let $A$ be a compact self-adjoint operator on a
separable Hilbert space and suppose that for \emph{every} orthonormal basis 
%
\begin{equation}
\sum\limits_{i}|(Ae_i,e_i)|<\infty.
\label{18.102-Prob.11}\end{equation}
%
Show that the eigenvalues of $A,$ if infinite in number, form a sequence in
$l^1.$
\vspace{10pt}
Solution: Every compact self-adjoint operator has an orthonormal basis of
eigenvectors so if the eigenvalues are listed with multiplicity then
$$
\sum\limits_{i}|\lambda _i|=\sum\limits_{i}|\langle Ae_i,e_i\rangle |<\infty
$$
from \eqref{18.102-Prob.11}. If the eigenvalues are listed without
multiplicity, the sum is smaller so still in $l^1.$ [Either interpretation
  is acceptable.]

\vspace{15pt}
\section*{Problem 6} For $u\in L^2(0,1)$ show that
%
\begin{equation*}
Iu(x)=\int_0^xu(t)dt,\ x\in(0,1)
\label{18.102-Prob.21}\end{equation*}
%
is a bounded linear operator on $L^2(0,1).$ If $V\in\cC([0,1]),$ is
real-valued and $V\ge0,$ show that there is a bounded linear operator $B$
on $L^2(0,1)$ such that
%
\begin{equation}
B^2u=u+I^*M_VIu\ \forall\ u\in L^2(0,1)
\label{18.102-Prob.22}\end{equation}
%
where $M_V$ denotes multiplication by $V.$  
\vspace{10pt}

Solution: $Iu$ is continuous if $u\in L^2(0,1)$ (see Problem 1 ...) since
$$
|u(x)-u(y)|\le \int_x^y|u|\le |x-y|^{\ha}\|u\|_{L^2}
$$
by Cauchy-Schwartz. By the linearity of the integral, this is a linear may
from $L^2(0,1)$ to $\cC([0,1])$ and
$$
\|Iu\|_{L^2}\le\sup|u|\le\|u\|_{L^2}
$$
so it is bounded on $L^2.$ The image of the unit ball in $L^2(0,1)$ is a
uniformly bounded and equicontinuous set in $\cC(0,1)$ so has compact
closure by Arscoli-Arzela. The image under the inclusion into $L^2(0,1)$ is
therefore also precompact and hence $I$ is a compact operator.

Multiplication by a continuous function $V\ge0$ gives a bounded and
self-adjoint operator on $L^2(0,1),$
$$
\|M_V\|\le\sup V,\ \langle M_vu,v\rangle =\int Vu\overline{v}=\langle u,M_Vv\rangle 
$$
so $I^*M_VI$ is compact (since the compact operators form a $*$-ideal) and
self-adjoint, since $(ABC)^*=C^*B^*A^*.$ It follows that $L^2(0,1)$ has an
orthornomal basis of eigenfunctions $e_i$ for $I^*M_VI$ with eigenvalues
$$
\lambda _i=\langle I^*M_VIe_i,e_i\rangle =\langle M_VIe_i,Ie_i\rangle \ge0
$$
by the positivity of $V.$ So
$$
Be_i=(1+\lambda_i)^{\ha}e_i
$$
defines, by continuous extension, a bounded operator on $L^2(0,1)$ such that
$$
B^2=\Id+I^*M_VI.
$$

Or, without the compactness of $I$ (which can also be proved by checking tails
in the Fourier basis) one needs to show that the spectrum of $A=\Id+I^*M_VI$
is contained in $[0,\|A\|].$ This is not quite obvious, but follows from
the positivity. Namely the operator $A-\ha\|A\|\Id$ satisfies
$$
\ha\|A\|\|u\|\ge\langle (A-\ha\|A\|\Id)u,u\rangle\ge-\ha\|A\|
$$
so its spectrum is contained in $[-\ha\|A\|,\ha\|A\|].$ Then $B$ is
well-defined by the functional calculus.
\end{document}
