\documentclass{exam}

%% IMPORTANT:

%% This example dates from the old course scheme and therefore has the
%% command \mmathfalse. If you're using the headers from this file as a
%% template, you probably want to delete this line.

%\usepackage{times,mathptm}

\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\K}{\mathbb{K}}
\renewcommand{\L}{\mathcal{L}}
\newcommand{\ip}[1]{\left<#1\right>}
\newcommand{\bbar}[1]{\overline{#1}}
\newcommand{\inv}{^{-1}}
\newcommand{\eps}{\varepsilon}
\newcommand{\hsnorm}[1]{\left\|#1\right\|_{\operatorname{H-S}}}
%\newcommand{}{}

\DeclareMathOperator{\Span}{span}
\DeclareMathOperator{\spanbar}{\overline{span}}
\DeclareMathOperator{\lin}{lin}
\DeclareMathOperator{\clin}{clin}
\DeclareMathOperator{\im}{im}
%\DeclareMathOperator{}{}

\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}

\mmathfalse

\begin{document}

\codenumber{12345}
\year{1996}
\title{Functional Analysis Sample Paper}
\duration{$1\half$ hours}

\begin{rubric}
  Answer not more than \underline{3} questions. \\
  Calculators may be used but should be unnecessary.
\end{rubric}

\maketitle

\romantrue

\begin{question}
  Let $H$ be a complex vector space. What is meant by an \emph{inner
    product} on $H$? Define the norm $\|x\|$ of a vector in $H$. What does 
  it mean to say that $H$ is a \emph{Hilbert space}?

  Consider the vector space $\ell^2(\C)$ of square-summable complex
  sequences, with
  $$\ip{x,y}=\sum_{j=1}^\infty x_j\bbar{y_j}$$
  Show that this sum converges absolutely for all $x,y\in\ell^2(\C)$ and
  defines an inner product which makes $\ell^2(\C)$ into a Hilbert
  space. (You may assume the finite-dimensional Cauchy-Schwarz inequality.)
\end{question}

\begin{question}
  Let $U$ be a closed subspace of a Hilbert space $H$. Define the
  \emph{orthogonal complement} $U^\perp$ of $U$ and show that $U^\perp$
  is also a closed subspace of $H$, with $U\cap U^\perp=\{0\}$.
  
  Given that for any $x\in H$ there exist vectors $u\in U$ and $v\in U^\perp$ 
  with $x=u+v$, show that these are the unique vectors in $U$ and
  $U^\perp$ whose sum is $x$, and show that $(U^\perp)^\perp=U$.
  
  Define the \emph{orthogonal projections} $P$ and $Q$ of $H$ onto $U$
  and $U^\perp$. Show that $P+Q=I$, that $P$ and $Q$ are linear, that 
  $P^2=P$, $Q^2=Q$ and $PQ=QP=0$.
\end{question}

\begin{question}
  Let $H$ be a Hilbert space and let $(e_j)$ be a sequence in $H$. What
  does it mean to say that $(e_j)$ is a \emph{complete orthonormal sequence}?
  State the Riesz-Fischer Theorem.
  
  In the Hilbert space $L^2([-\pi,\pi],\C)$, what is the
  \emph{exponential basis}? Show that it is orthonormal.
  
  Compute the Fourier coefficients of the function $f$ given by $f(t)=t$
  with respect to the exponential basis and use the Riesz-Fischer
  Theorem to show that
  $$\sum_{j=1}^\infty \frac{1}{j^2}=\frac{\pi^2}{6}$$
\end{question}

\begin{question}
  Let $T$ be a linear operator on a separable Hilbert space $H$.
  What does it mean to say that $T$ is a \emph{Hilbert-Schmidt operator}?
  Define the \emph{Hilbert-Schmidt norm} $\hsnorm{T}$ of $T$.

  Let $M$ be the multiplication operator on $\ell^2(\C)$ defined by
  $$(Mx)_j=m_jx_j$$
  where $(m_j)$ is a bounded sequence of complex numbers. Show that $M$ is
  a Hilbert-Schmidt operator if and only if $m\in\ell^2(\C)$, in which case
  $$\hsnorm{M}^2=\|m\|_2^2=\sum_{j=1}^\infty|m_j|^2$$
  
  Now suppose $T$ is a compact self-adjoint operator on $H$.
  Define the terms \emph{eigenvalue} and \emph{eigenvector} of $T$ and
  state without proof the Spectral Theorem for $T$. Use the Spectral
  Theorem to show that if $(\lambda_j)$ is the sequence of eigenvalues 
  of $T$, repeated according to multiplicity, then $T$ is a
  Hilbert-Schmidt operator if and only if $\lambda\in\ell^2(\R)$, in
  which case
  $$\hsnorm{T}^2=\sum_{j=1}^\infty\lambda_j^2=\|\lambda\|_2^2$$
\end{question}

\begin{solution}
  An inner product on $H$ is a map $\ip{\cdot,\cdot}:H\times H\to\C$ 
  satisfying
  \begin{enumerate}
  \item For $x,y\in H$, $\ip{x,y}=\bbar{\ip{y,x}}$;
  \item for $x,y,z\in H$ and $\lambda,\mu\in\C$,
    $$\ip{\lambda x+\mu y,z}=\lambda\ip{x,z}+\mu\ip{y,z}$$
  \item for $x\in H$, $\ip{x,x}\geq 0$ and $\ip{x,x}=0$ if and only if
    $x=0$
  \end{enumerate}
  The norm is defined by
  $$\|x\|=\sqrt{\ip{x,x}}$$
  taking the non-negative square root. $H$ is a Hilbert space if it is 
  complete in this norm.

  Write $\|x\|^2=\sum_{j=1}^\infty|x_j|^2$, and similarly $\|y\|^2$. For any 
  $N\in\N$, the finite-dimensional Cauchy-Schwarz inequality implies that
  $$\sum_{j=1}^N|x_j\bbar{y_j}| 
    \leq \left(\sum_{j=1}^N|x_j|^2\right)^{1/2} 
         \left(\sum_{j=1}^N|y_j|^2\right)^{1/2}
    \leq \|x\|\|y\|$$
  so the partial sums of $\sum|x_j\bbar{y_j}|$ are bounded above. Since
  it is a sum of non-negative terms, it is convergent and hence the
  original sum is absolutely convergent for any $x,y\in\ell^2(\C)$.
  
  To show it is an inner product,
  \begin{enumerate}
  \item $\ip{x,y}=\sum_{j=1}^\infty x_j\bbar{y_j}=
    \sum_{j=1}^\infty\bbar{y_j\bbar{x_j}}=
    \bbar{\sum_{j=1}^\infty y_j\bbar{x_j}}=\bbar{\ip{y,x}}$
    since conjugation is continuous.
  \item $\ip{\lambda x+\mu y,z}=\sum_{j=1}^\infty(\lambda x_j+\mu y_j)\bbar{z_j}
    =\lambda\sum_{j=1}^\infty x_j\bbar{z_j}+\mu\sum_{j=1}^\infty y_j\bbar{z_j}
    =\lambda\ip{x,z}+\mu\ip{y,z}$
  \item $\ip{x,x}=\sum_{j=1}^\infty|x_j|^2$; this is a sum of
    non-negative terms, so $\ip{x,x}\geq 0$. If $\ip{x,x}=0$ then each
    term must be zero, so $x_j=0$ for all $j$ showing that $x=0$.
  \end{enumerate}
  
  Completeness: let $(x^{(n)})$ be a Cauchy sequence in $\ell^2(\C)$,
  so given $\eps>0$ there exists $N\in\N$ such that if $n,m>N$ then
    \begin{equation}
      \sum_{j=1}^\infty|x^{(n)}_j-x^{(m)}_j|^2<\eps^2 \tag{*}
    \end{equation}
    so in particular for each $j$,
    $$|x^{(n)}_j-x^{(m)}_j|^2<\eps^2$$
    This shows that if $j$ is fixed, then $(x^{(n)}_j)$ is a Cauchy
    sequence in $\C$, hence convergent to $x_j$ say. From (*), for 
    any $M\in\N$,
    $$\sum_{j=1}^M|x^{(n)}_j-x^{(m)}_j|^2<\eps^2$$
    Since this is a finite sum, we can let $m\to\infty$, to deduce that
    $$\sum_{j=1}^M|x^{(n)}_j-x_j|^2\leq\eps^2$$
    and now it follows that
    \begin{equation}
      \sum_{j=1}^\infty|x^{(n)}_j-x_j|^2\leq\eps^2 \tag{**}
    \end{equation}
    so for $n>N$, $x^{(n)}-x\in\ell^2(\K)$ and $\|x^{(n)}-x\|\leq\eps$.
    Thus, $x^{(n)}\to x$ in the $\ell^2(\C)$ metric.
    
    Finally, $x\in\ell^2(\C)$ since $x=(x-x^{(n)})+x^{(n)}$,
    $x^{(n)}\in\ell^2(\C)$ by choice, $x-x^{(n)}\in\ell^2(\C)$ by (**)
    and $\ell^2(\C)$ is a vector space, hence closed under addition.
\end{solution}

\begin{solution}
  The orthogonal complement of $U$ is defined by
  $$U^\perp=\{x\in U:\ip{x,u}=0 \text{ for all } u\in U\}$$
  Clearly, $0\in U^\perp$ so $U^\perp\neq\emptyset$. 
  Suppose $x,y\in U^\perp$ and $\lambda,\mu$ are scalars. Then for $u\in U$,
  $$\ip{\lambda x+\mu y,u}=\lambda\ip{x,u}+\mu\ip{y,u}=0$$
  so $U^\perp$ is a subspace. 
  
  Now let $(x_n)$ be a sequence in $U^\perp$ converging to $x\in H$. 
  For any $u\in U$, $\ip{x_n,u}=0$. But $x_n\to x$ and for fixed $u$ the 
  function $\phi$ defined by $\phi(v)=\ip{v,u}$ is continuous (by the
  Cauchy-Schwarz inequality), so $\ip{x_n,u}\to\ip{x,u}$, showing that 
  $\ip{x,u}=0$. Since $u$ was an arbitrary element of $U$, we have 
  $x\in U^\perp$ showing that $U^\perp$ is closed.

  Clearly $0\in U\cap U^\perp$. Suppose $x\in U\cap U^\perp$. Since
  $x\in U^\perp$, it is orthogonal to every element of $U$. But $x\in U$, 
  so $\ip{x,x}=0$ and hence $x=0$, showing that $U\cap U^\perp=\{0\}$.

  Suppose $x=u+v$ and $x=u'+v'$ where $u,u'\in U$ and $v,v'\in U^\perp$.
  Since $u+v=u'+v'$, we have $u-u'=v'-v=y$ say. But $U$ and $U^\perp$ are both
  subspaces, so $u-u'\in U$ and $v'-v\in U^\perp$. Thus $y\in U\cap U^\perp$
  so $y=0$ by the previous part. Thus, $u-u'=v'-v=0$ so $u'=u$ and
  $v'=v$ showing that $u$ and $v$ are the only elements of $U$ and
  $U^\perp$ which sum to $x$.
  
  Suppose $u\in U$. Then for all $x\in U^\perp$, $\ip{x,u}=0$ so 
  $x\in(U^\perp)^\perp$ and hence $U\subseteq(U^\perp)^\perp$.

  Now suppose $x\in(U^\perp)^\perp$. Since $x\in H$, we have $x=u+v$ where 
  $u\in U$ and $v\in U^\perp$. Since $x\in(U^\perp)^\perp$, we must have 
  $\ip{u+v,v}=0$, so $\ip{u,v}+\|v\|^2=0$. But $\ip{u,v}=0$, so $v=0$
  and $x=u$. This shows that $x\in U$, so $(U^\perp)^\perp\subseteq U$.

  The orthogonal projection of $H$ onto $U$ is defined by $Px=u$ where
  $u$ is the unique element of $U$ for which $x=u+v$ for some $v\in U^\perp$.
  Similarly, the orthogonal projection onto $U^\perp$ is defined by $Qx=w$
  where $w$ is the unique element of $U^\perp$ for which $x=w+z$ for some
  $z\in(U^\perp)^\perp$.
  
  If $x\in H$ then $x=u+v$ where $u\in U$ and $v\in U^\perp$. By
  definition, $Px=u$. But since $(U^\perp)^\perp=U$, we also have
  $x=v+u$ with $v\in U^\perp$ and $u\in(U^\perp)^\perp=U$. Again by
  definition, $Qx=v$. Thus $Px+Qx=u+v=x$, or equivalently $P+Q=I$.
  
  If $x,y\in H$ and $\lambda,\mu$ are scalars, then $x=Px+Qx$ and
  $y=Py+Qy$, so 
  $$\lambda x+\mu y=\lambda(Px+Qx)+\mu(Py+Qy)=(\lambda Px+\mu Py)+
    (\lambda Qx+\mu Qy)$$
  Since $U$ and $U^\perp$ are both subspaces, $\lambda Px+\mu Py\in U$
  and $\lambda Qx+\mu Qy\in U^\perp$. But the only representation of
  $\lambda x+\mu y$ as a sum of elements from $U$ and $U^\perp$ is
  $$\lambda x+\mu y=P(\lambda x+\mu y)+Q(\lambda x+\mu y)$$
  so we must have
  $$P(\lambda x+\mu y)=\lambda Px+\mu Py; \qquad
    Q(\lambda x+\mu y)=\lambda Qx+\mu Qy$$
  showing that $P$ and $Q$ are linear.
  
  For any $x$, $Px=Px+0$. Since $Px\in U$ and $0\in U^\perp$, the 
  uniqueness of the decomposition implies that $P(P(x))=P(x)$ and $Q(P(x))=0$ 
  so $P^2=P$ and $QP=0$. Similarly, $Qx=0+Qx$ so $Q(Q(x))=Q(x)$ and $P(Q(x))=0$.
\end{solution}

\begin{solution}
  $(e_j)$ is an orthonormal sequence in $U$ if $\ip{e_j,e_k}=0$
  if $j\neq k$ and $\ip{e_j,e_j}=1$ for all $j$. It is complete if in
  addition the only vector $x\in U$ for which $\ip{x,e_j}=0$ for all $j$
  is $x=0$.
  
  The Riesz-Fischer Theorem states that if $(e_j)$ is an orthonormal
  sequence then the following are equivalent:
  \begin{enumerate}
  \item $(e_j)$ is complete;
  \item for all $x\in H$, $x=\sum_{j=1}^\infty\ip{x,e_j}e_j$;
  \item for all $x\in H$, $\|x\|^2=\sum_{j=1}^\infty|\ip{x,e_j}|^2$;
  \item the linear span of $\{e_j:j\in\N\}$ is dense in $H$.
  \end{enumerate}

  The exponential basis in $L^2([-\pi,\pi],\C)$ is given by the functions
  $$\frac{1}{\sqrt{2\pi}}e^{int}$$
  for $n\in\Z$.
  
  For any $n,m\in\Z$,
  $$\ip{e_n,e_m}=\frac{1}{2\pi}\int_{-\pi}^\pi e^{int}\bbar{e^{imt}}dt
    =\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(n-m)t}dt$$
  If $n\neq m$ this gives
  $$\ip{e_n,e_m}=\frac{1}{2\pi i(n-m)}\left[(-1)^{n-m}-(-1)^{m-n}\right]=0$$
  and if $n=m$ it gives
  $$\ip{e_n,e_n}=\frac{1}{2\pi}\int_{-\pi}^\pi dt=1$$
  so $(e_n)$ is orthonormal.
  
  The $n$th Fourier coefficient $c_n$ of $f$ with respect to this basis is
  \begin{align*}
    \sqrt{2\pi}c_n &= \int_{-\pi}^\pi te^{-int}dt \\
                   &= \left[\frac{te^{-int}}{-in}\right]_{-\pi}^\pi
                      -\int_{-\pi}^\pi\frac{e^{-int}}{-in}dt \\
                   &= \frac{2\pi(-1)^n}{-in}
                      -\left[\frac{e^{-int}}{n^2}\right]_{-\pi}^\pi \\
                   &= \frac{2\pi i(-1)^n}{n}
  \end{align*}
  provided $n\neq 0$; $c_0$ satisfies
  $$\sqrt{2\pi}c_0=\int_{-\pi}^\pi tdt=0$$
  so $c_0=0$. By the Riesz-Fischer Theorem,
  $$\int_{-\pi}^\pi|f(t)|^2dt=\sum_{n=-\infty}^\infty|c_n|^2$$
  But $c_{-n}=c_n$ and $c_0=0$, so
  $$\int_{-\pi}^\pi t^2dt=2\sum_{n=1}^\infty \frac{2\pi}{n^2}$$
  $$\frac{2\pi^3}{3}=4\pi\sum_{n=1}^\infty\frac{1}{n^2}$$
  $$\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}$$
\end{solution}

\begin{solution}
  $T$ is a Hilbert-Schmidt operator if there exists a complete
  orthonormal sequence $(e_j)$ such that
  $$\sum_{j=1}^\infty\|Te_j\|^2<\infty$$
  The Hilbert-Schmidt norm is defined to be
  $$\hsnorm{T}^2=\sum_{j=1}^\infty\|Te_j\|^2$$
  which can be shown to be independent of the choice of $(e_j)$.
  
  Let $e^{(n)}$ be the $n$th standard basis sequence in $\ell^2(\C)$,
  i.e.\ $e^{(n)}$ is $1$ in the $n$th term and zero elsewhere. This is a
  complete orthonormal sequence in $\ell^2(\C)$, so
  $$\hsnorm{M}^2=\sum_{n=1}^\infty\|Me^{(n)}\|^2$$
  but $\|Me^{(n)}\|=\|m_ne^{(n)}\|=|m_n|$, so
  $$\hsnorm{M}^2=\sum_{n=1}^\infty|m_n|^2=\|m\|_2^2$$

  $\lambda\in\C$ is an eigenvalue of $T$ if $Tu=\lambda u$ for some non-zero 
  $u\in H$; such a vector $u$ is an eigenvector of $T$.

  The Spectral Theorem for compact self-adjoint operators states that there is
  an complete orthonormal sequence $(e_j)$ in $H$ consisting of eigenvectors 
  of $T$ and a sequence $(\lambda_j)$ of corresponding eigenvalues such that 
  for all $x$,
  $$Tx=\sum_{j=1}^\infty\lambda_j\ip{x,e_j}e_j.$$
  Each eigenvalue of $T$ occurs exactly $m$ times, where $m$ is its
  multiplicity.

  Using $(e_j)$ as the complete orthonormal sequence in the definition
  of the Hilbert-Schmidt norm,
  $$\hsnorm{T}^2=\sum_{j=1}^\infty\|Te_j\|^2=\sum_{j=1}^\infty\|\lambda_je_j\|^2
                =\sum_{j=1}^\infty|\lambda_j|^2=\sum_{j=1}^\infty\lambda_j^2
                =\|\lambda\|_2^2$$
  using the fact that $\lambda_j$ is real, so $T$ is Hilbert-Schmidt if
  and only if $\lambda\in\ell^2(\R)$ in which case $\hsnorm{T}=\|\lambda\|_2$.
\end{solution}

\end{document}
