\newpage
\section{Normed Spaces and Linear Operators}

\begin{flashcard}[norm-defn]
\begin{definition*}[Norm]
  \cloze{
  Let $X$ be a real or complex vector space. A
  \emph{norm} on $X$ is a function $\|\bullet\| :
  X \to \RR^+$ such that:
  \begin{enumerate}[(i)]
    \item $\|x\| = 0 \iff x = 0$
    \item $\|\lambda x\| = |\lambda| \|x\|$ for
      all scalars $\lambda$ and for all $x \in X$
    \item $\|x + y\| \le \|x\| + \|y\|$ for all
      $x, y \in X$ (triangle inequality)
  \end{enumerate}
  A \emph{normed space} is a pair $(X,
  \|\bullet\|)$ where $X$ is a vector space and
  $\|\bullet\|$ is a norm on $X$.
}
\end{definition*}
\end{flashcard}

\vspace{-1em}

The norm induces a metric space structure on $X$.

\begin{flashcard}[Banach-defn]
\begin{definition*}[Banach space]
  \cloze{A normed space $X$ is \emph{Banach} if it is
  complete as a metric / topological space.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[unit-ball-defn]
\begin{definition*}[Unit ball]
  \cloze{It often helps to look at the \emph{unit ball}
  of $X$, which is defined by
  \[ B = B(X) = B_X = \{x \in X \st \|x\| < 1\}
  .\]}
\end{definition*}
\end{flashcard}

\begin{remark*}
  The unit ball is always convex.
\end{remark*}

\begin{remark*}
  Any set $B \subseteq \RR^n$ which is closed,
  bounded, convex, symmetric about $0$ and a
  neighbourhood of $0$ defines a norm by:
  \[ \|x\| = \inf\{t > 0 \st x \in tB\} .\]
  $B$ is the unit ball of that norm.
\end{remark*}

\begin{flashcard}[youngs-ineq]
\begin{theorem*}[Young's inequality]
  \cloze{Let $1 < p, q < \infty$ with $\frac{1}{p} +
  \frac{1}{q} = 1$ and let $a, b \ge 0$. Then:
  \[ ab \le \frac{a^p}{p} + \frac{b^q}{q} .\]}
\end{theorem*}
\end{flashcard}

\begin{warning*}
  Note that subspaces of normed vector spaces need
  not be closed topological subspaces!
\end{warning*}

\begin{flashcard}[separable-defn]
\begin{definition*}[Separable topological space]
  \cloze{A topological space is \emph{separable} if it
  has a countable dense subset.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[cts-bound-tfae]
\begin{theorem*}
  Let $X, Y$ be normed spaces and $T : X \to Y$ be
  linear. Then the following are equivalent:
  \begin{enumerate}[(i)]
    \item \cloze{$T$ is continuous}
    \item \cloze{$T$ is continuous at $0$}
    \item \cloze{$\exists k > 0$ such that $\|T x\| \le
      k\|x\|$ for all $x \in X$ (this condition is
      called \emph{bounded}).}
  \end{enumerate}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[BXY-notation]
\begin{notation*}
  We will write:
  \[ \cloze{L}(X, Y) = \cloze{\{T : X \to Y \st
  \text{$T$ continuous and linear}\}} \]
\end{notation*}
\end{flashcard}

\begin{note*}
  The operator norm gives a norm on $L(X, Y)$, so
  $L(X, Y)$ is a normed vector space.
\end{note*}

\begin{flashcard}[Banach-LXY-thm]
\begin{theorem*}
  Let $X, Y$ be normed spaces \cloze{with $Y$
  Banach}.
  Then $L(X, Y)$ is Banach. \cloze{In particular, $X^*$
  is always Banach.}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[lp-dual-thm]
\begin{theorem*}[Dual of $l_p$]
  \cloze{If $1 \le p < \infty$, then the dual of $l_p$ is
  \emph{isometrically isomorphic to} $l_q$, where
  $q$ is the conjugate index.

  We also have $c_0^* \cong l_1$, but $l_\infty^*$
  does not have a particularly nice expression.}
\end{theorem*}
\end{flashcard}

\begin{lemma*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Every finite dimensional normed space is
      Banach.
    \item Every finite-dimensional subspace of a
      normed space is closed (useful to know!).
  \end{enumerate}
\end{lemma*}

\begin{flashcard}[Riesz-lemma]
\begin{proposition*}[Riesz's Lemma]
  \cloze{Let $X$ be a normed space, and $Y$ a closed
  subspace of $X$ with $Y \neq X$. Then:
  \begin{enumerate}[(i)]
    \item $\forall \eps > 0, \exists x \in X$ with
      $\|x\| = 1$ and $d(x, Y) \ge 1 - \eps$
    \item $\dim X < \infty$ gives $\exists x \in
      X$ with $\|x\| = 1$ and $d(x, Y) = 1$
  \end{enumerate}}
\end{proposition*}
\end{flashcard}

\begin{corollary*}
  $X$ an infinite dimensional normed space implies
  that there exists $x_n$ in $X$ with $\|x_n\| =
  1$ and $\|x_n - x_m\| \ge 1$ for all $n \neq m$.
  In particular, $B(X)$ is not compact.
\end{corollary*}

\begin{lemma*}
  This is a fact about metric spaces: for a subset
  $A$ of a metric space $X$,
  \[ \text{$A$ is totally bounded} \iff
  \text{$\ol{A}$ is totally bounded} .\]
  In particular if $X$ is complete, then
  \[ \text{$A$ is totally bounded} \iff
  \text{$\ol{A}$ is compact} .\]
\end{lemma*}

\subsection{Compact operators}

\begin{flashcard}[compact-operator-defn]
\begin{definition*}[Compact operator]
  \cloze{Let $X$ and $Y$ be normed spaces. We say that $T : X \to
  Y$ linear is \emph{compact} if $\ol{T(B_X)}$ is
  compact.}
\end{definition*}
\end{flashcard}

\begin{remark*}
  $T$ compact implies $T(B_X)$ (totally) bounded,
  which implies that $T$ is continuous.
\end{remark*}

\begin{remark*}
  $T : X \to Y$ is compact if and only if every
  bounded sequence $(x_n)$ in $X$ has a
  subsequence $(x_{n_i})$ such that $(Tx_{n_i})$
  is convergent.
\end{remark*}

\begin{remark*}
  If $Y$ is Banach, then $T : X \to Y$ is compact
  if and only if $T(B_X)$ is totally bounded.
\end{remark*}

\begin{flashcard}[compact-ops-are-closed-subspace]
\begin{theorem*}
  Let $X$, $Y$ be normed vector spaces, with $Y$
  Banach. Then the compect operators from $X$ to
  $Y$ form \cloze{a closed subspace of $L(X, Y)$.}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[compact-op-composition-fact]
\begin{proposition*}
  Let $X$, $Y$ and $Z$ be normed vector spaces and
  $T : X \to Y$, $S : Y \to Z$ linear maps.
  \begin{enumerate}[(i)]
    \item \cloze{If $S$ is compact, $T$ continuous, then
      $S \circ T$ is compact.}
    \item \cloze{If $T$ is compact, $S$ is continuous
      then $S \circ T$ is compact.}
  \end{enumerate}
\end{proposition*}
\end{flashcard}

\subsection{Open mapping lemma}

\begin{flashcard}[open-mapping-lemma]
\begin{lemma*}[Open mapping lemma]
  \cloze{Let $X, Y$ be normed, $X$ Banach and let $T \in
  L(X, Y)$. Suppose that $B_Y \subseteq
  \ol{T(B_X)}$. Then:
  \begin{enumerate}[(i)]
    \item $B_Y \subseteq T(2B_X)$ (thus $T$ is
      surjective)
    \item $Y$ is Banach
  \end{enumerate}}
\end{lemma*}
\end{flashcard}

\begin{note*}
  $B_Y \subseteq \ol{T(B_X)}$ says that $T(B_X)$
  is dense in $B_Y$. 
\end{note*}

\newpage
\section{The Baire Category Theorem and Applications}

\begin{flashcard}[PoUB-thm]
\begin{theorem*}[Principle of Uniform Boundedness]
  \cloze{Let $X$ Banach and $Y$ normed and let $\tau
  \subseteq L(X, Y)$. Suppose $\tau$ is pointwise
  bounded (i.e. $\forall x \in X$ there exists $k$
  such that $\|T(x)\| \le k$ for all $T \in
  \tau$). Then $\tau$ is uniformly bounded (i.e.
  there exists $k$ such that $\|T\| \le k$ for all
  $T \in \tau$).}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[Banach-Steinhaus-thm]
\begin{corollary*}[Banach Steinhaus Theorem]
  \cloze{Let $X$ Banach, $Y$ normed. Let $T_1, T_2,
  \ldots \in L(X, Y)$ such that $(T_n x)$ is
  convergent for all $x$: say $T_n x \to T x$.
  Then $T$ is a continuous linear map.}
\end{corollary*}
\end{flashcard}

\begin{flashcard}[open-mapping-thm]
\begin{theorem*}[Open Mapping Theorem]
  \cloze{Let $X, Y$ be Banach, and $T \in L(X, Y)$
  surjective. Then $T$ is an open mapping, i.e.
  there exists $k$ such that $\forall y \in Y$,
  $\exists x \in X$ with $Tx = y$ and $\|x\| \le
  k\|y\|$.}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[Inversion-thm]
\begin{corollary*}[Inversion Theorem]
  \cloze{Let $X, Y$ be Banach, and $T \in L(X, Y)$. Then
  if $T$ is bijective, then $T^{-1}$ is continuous
  (hence $T$ is an isomorphism).}
\end{corollary*}
\end{flashcard}

\begin{remark*}
  If $X, Y$ are Banach, $T \in L(X, Y)$
  surjective, then Inversion Theorem gives that
  $\ol{T} : X / \ker T \to Y$ is an isomorphism.
\end{remark*}

\begin{flashcard}[Comparable-Banach-norms-are-equiv]
\begin{corollary*}[Comparable Banach norms are
  equivalent]
  \cloze{Let $\|\bullet\|_1$ and $\|\bullet\|_2$ be
  complete norms on a vector space $V$. Suppose
  that there exists $c > 0$ such that $\|x\|_2 \le
  c\|x\|_1$ for all $x \in V$. Then
  $\|\bullet\|_1$ and $\|\bullet\|_2$ are
  equivalent. ``Comparable Banach norms are
  equivalent''.}
\end{corollary*}
\end{flashcard}

\subsection{The Closed Graph Theorem}

\begin{flashcard}[closed-graph-thm]
\begin{theorem*}[Closed Graph Theorem]
  \cloze{Let $X, Y$ be Banach and $T : X \to Y$ linear.
  Then $T$ is continuous if and only if $T$ has
  closed graph.}
\end{theorem*}
\end{flashcard}

\newpage
\section{Spaces of Continuous Functions}

\newpage
\section{Hilbert Spaces}

\begin{flashcard}[hilbert-space-defn]
\begin{definition*}[Hilbert space]
  \cloze{A \emph{Hilbert space} is a complete inner
  product space.}
\end{definition*}
\end{flashcard}

\begin{flashcard}[parallelogram-law]
\begin{proposition*}[Parallelogram law]
  \cloze{Let $X$ be an inner product space and $x, y \in
  X$. Then:
  \[ \|x + y\|^2 + \|x - y\|^2 = 2\|x\|^2 +
  2\|y\|^2 .\]}
\end{proposition*}
\end{flashcard}

\begin{flashcard}[perp-perp-coro]
\begin{corollary*}
  Let $H$ be a Hilbert space. Then:
  \begin{enumerate}[(i)]
    \item \cloze{$F$ a closed subspace of $H$
      implies}
      that $(F^\perp)^\perp = F$
    \item \cloze{$S \subseteq H$ $\implies$
      $(S^\perp)^\perp = \ol{\langle S \rangle}$}
    \item \cloze{$S \subseteq H$ has dense linear span
      if and only if $S^\perp = \{0\}$}
  \end{enumerate}
\end{corollary*}
\end{flashcard}

\begin{flashcard}[Riesz-representation-thm]
\begin{theorem*}[Riesz Representation Theorem]
  \cloze{Let $H$ be a Hilbert space, $f \in H^*$. Then
  there exists $y \in H$ such that $f = \theta_y$
  (i.e. $f = \langle \bullet, y\rangle$).}
\end{theorem*}
\end{flashcard}

\begin{flashcard}[Hilbert-self-dual]
\begin{corollary*}[Hilbert are self dual]
  \cloze{Let $H$ be a Hilbert space. Then the map $\theta
  : y \mapsto \theta_y$ is an isometric
  conjugate-linear isomorphism from $H$ to $H^*$.
  ``$H$ is self-dual.''}
\end{corollary*}
\end{flashcard}

\begin{flashcard}[separable-inner-prod-ortho-basis]
\begin{corollary*}
  Let $X$ be a \cloze{separable inner product
  space}. Then
  $X$ has an orthonormal basis.
\end{corollary*}
\end{flashcard}

\textbf{Aim:} We will show that for an orthonormal
basis, we have
\[ x = \sum_n (x, e_n) e_n \]
for all $x \in X$.

\begin{remark*}
  This is false for a general Banach space. For
  example, $1, t, t^2, \ldots$ have dense linear
  space in $C_{\RR}([-1, 1])$ but $|t| \neq \sum_k
  c_k t^k$.
\end{remark*}

\newpage
\section{Spectral Theory}

A useful corollary of the Inversion Theorem for
Banach spaces:

\begin{corollary}
  For a Banach space $X$,
  $T \in L(X)$ invertible $\iff$ $T$ injective and
  $T$ surjective.
\end{corollary}

\begin{theorem*}
  Let $X$ be a Banach space, $T \in L(X)$. Then
  $\|T\| < 1$ implies $I - T$ is invertible with
  \[ (I - T)^{-1} = \sum_{n = 0}^\infty T^n .\]
\end{theorem*}

\begin{notation*}
  For a Banach space $X$, we write
  \[ G(X) = \{T \in L(X) \st \text{$T$ is
  invertible}\} .\]
\end{notation*}

\begin{theorem*}
  Let $X$ be a Banach space. Then:
  \begin{enumerate}[(i)]
    \item $G$ is open in $L(X)$
    \item The function $T \mapsto T^{-1}$ from $G$
      to $G$ is continuous
    \item Let $(T_n)$ in $G$, $T \in L(X)$ with
      $T_n \to T$ but $T \notin G$. Then
      $\|T_n^{-1}\| \to \infty$
  \end{enumerate}
\end{theorem*}

\begin{proposition*}
  Let $X$ be a complex Banach space, and $T \in
  L(X)$. Then $\sigma(T)$ is a closed subset of
  \[ \{z \in \CC \st |z| \le \|T\|\} .\]
  In particular, $\sigma(T)$ is compact.
\end{proposition*}

\begin{proposition*}
  Let $X$ be a complex Banach space, and $T \in
  L(X)$, $\lambda \in \CC$. Then:
  \begin{enumerate}[(i)]
    \item $\lambda$ is an eigenvalue of $T$
      implies $\lambda$ is an approximate
      eigenvalue of $T$
    \item $\lambda$ an approximate eigenvalue of
      $T$ implies $\lambda \in \sigma(T)$
  \end{enumerate}
\end{proposition*}

\begin{theorem*}
  Let $X$ be a complex Banach sace, $T \in L(X)$.
  Then $\partial \sigma(T) \subseteq
  \sigma_{ap}(T)$.
\end{theorem*}

\begin{theorem*}[Spectral Mapping Theorem]
  Let $X$ be a complex Banach space, $T \in L(X)$.
  Let $P$ be a non-constant polynomial. Then
  $\sigma(P(T)) = P(\sigma(T))$.
\end{theorem*}

\begin{warning*}
  We did not use the ``fact'' that $A$ not
  invertible $\implies$ $AB$ not invertible (or
  $BA$ not invertible).

  This is false (for example, left and right
  shift).
\end{warning*}

\begin{definition*}[Spectral value]
  Let $X$ be a complex Banach space, $T \in L(X)$.
  The \emph{spectral value} of $T$ is:
  \[ r(T) = \sup\{|\lambda| \st \lambda \in
  \sigma(T)\} .\]
  Certainly have that $r(T) \le \|T\|$.
\end{definition*}

\begin{corollary*}
  Let $X$ be a complex Banach space and $T \in
  L(X)$. Then $r(T) \le \inf_{n \ge 1}
  \|T^n\|^{1/n}$.
\end{corollary*}

\begin{theorem*}[Non-emptiness of the spectrum]
  Let $H$ be a non-zero Hilbert space, and $T \in
  L(H)$. Then $\sigma(T) \neq \emptyset$.
\end{theorem*}

\subsection{Spectral Theory of Hermitian Operators
(all spaces are complex here)}

\begin{proposition*}
  Let $H$ be a Hilbert space, $T \in L(H)$. Then
  $\sigma(T^*) = \{\ol{\lambda} \st \lambda \in
  \sigma(T)\}$.
\end{proposition*}

\begin{theorem*}
  Let $H$ be a Hilbert space, $T \in L(H)$
  Hermitian. Then $\sigma(T) \subseteq \RR$.
\end{theorem*}

\begin{remark*}
  We know that any operator $T$ on a (non-zero)
  Hilbert space $H$ has an approximate eigenvalue
  ($\sigma(T) \neq \emptyset$ so $\partial
  \sigma(T) \neq \emptyset)$. But if $T$ is
  Hermitian, then in fact every $\lambda \in
  \sigma(T)$ is an approximate eigenvalue
  ($\sigma(T) \subseteq \RR$, so $\sigma(T) =
  \partial \sigma(T)$).
\end{remark*}

\begin{proposition*}
  Let $H$ be a Hilbert space, $T \in L(H)$
  Hermitian. Then $r(T) = \|T\|$.
\end{proposition*}

\begin{proposition*}
  Let $H$ be a Hilbert space, $T \in L(H)$, and
  $Y$ a subspace of $H$. Then if $T$ acts on $Y$,
  then $T^*$ acts on $Y^\perp$. In particular for
  $T$ Hermitian, if $T$ acts on $Y$ then $T$ acts
  on $Y^\perp$.
\end{proposition*}

\subsection{Spectral Theory of Compact Operators}

\begin{proposition*}
  Let $X$ be an infinite dimensional Banach space,
  and $T \in L(X)$ compact. Then $0$ is an
  approximate eigenvalue of $T$.
\end{proposition*}

\begin{note*}
  $0$ need not be an eigenvalue of $T$, for
  example $T : l_2 \to l_2$ defined by
  \[ T \left( \sum_n x_n e_n \right) = \sum_n
  \frac{1}{2^n} x_n e_n .\]
\end{note*}

\begin{proposition*}
  Let $X$ be a Banach space, $T \in L(X)$ compact,
  and $\lambda \in \CC$ with $\lambda \neq 0$.
  Then if $\lambda \in \sigma_{ap}(T)$, then
  $\lambda \in \sigma_{p}(T)$.
\end{proposition*}

\begin{proposition*}
  Let $X$ be a Banach space, $T \in L(X)$ compact.
  Then every eigenspace $E(\lambda)$ for $\lambda
  \neq 0$ is finite dimensional.
\end{proposition*}

\begin{lemma*}
  Let $X$ be a Banach space, $T \in L(X)$ compact.
  Then $\forall \delta > 0$, $T$ has only finitely
  many eigenvalues with $|\lambda| > \delta$.
\end{lemma*}

\begin{theorem*}[Spectral Theorem for Compact
  Operators]
  Let $X$ be a Banach space, $T \in L(X)$
  compact. Then:
  \begin{enumerate}[(i)]
    \item Either $\sigma(T)$ is finite or
      $\sigma(T) = \{0, \lambda_1, \lambda_2,
      \ldots\}$ where $\lambda_n \to 0$
    \item $\lambda \in \sigma(T)$, $\lambda \neq
      0$ $\implies$ $\lambda \in \sigma_{ap}(T)$
    \item $\lambda \in \sigma(T)$, $\lambda \neq
      0$ $\implies$ $\dim E(\lambda) < \infty$
  \end{enumerate}
\end{theorem*}

\begin{theorem*}[Spectral Theorem for Compact
  Hermitian Operators]
  Let $H$ be a separable Hilbert space, $T \in
  L(H)$ compact Hermitian. Then:
  \begin{enumerate}[(i)]
    \item $H$ has an orthonormal basis $(e_n)$
      consisting of eigenvectors of $T$
    \item The corresponding eigenvalues $\lambda_n
      \to 0$ (if $\dim H = \infty$)
  \end{enumerate}
\end{theorem*}

\begin{theorem*}
  Let $H$ be a Hilbert space, $T \in L(H)$ compact
  Hermitian. Then there exists a closed subspace
  $Y$ of $H$ and an orthonormal basis $(e_n)$ for
  $Y$ and $(\lambda_n)$ in $\RR$ such that
  $\forall x \in H$:
  \[ x = \sum_n x_n e_n + z; z \in Y^\perp
  \implies T x = \sum_n \lambda_n x_n e_n .\]
\end{theorem*}
