\input{"preamble.tex"}

\addbibresource{FunctionalAnalysis.bib}

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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Functional Analysis
  }
    \\ {\normalsize Lectures by Weiwei Hu. University of Georgia, Spring
2022} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd
from a graduate course in Functional Analysis taught by Weiwei Hu at the
University of Georgia in Spring 2022. As such, any errors or
inaccuracies are almost certainly my own. } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2022-05-29 }
\end{flushleft}


\newpage

% Note: addsec only in KomaScript
\addsec{Table of Contents}
\tableofcontents
\newpage

\hypertarget{tuesday-january-11}{%
\section{Tuesday, January 11}\label{tuesday-january-11}}

\begin{remark}

This course: solving \(Lf=g\) for \(L\) a linear operator, in analogy to
solving \(Ax=b\) in matrices. References:

\begin{itemize}
\tightlist
\item
  Hutson-Pym-Cloud, \emph{Applications of Functional Analysis and
  Operator Theory}
\item
  Reed-Simon, \emph{Methods of Modern Mathematical Physics}
\item
  Brezis, \emph{Functional Analysis, Sobolev Spaces, and PDEs}
\end{itemize}

\end{remark}

\begin{remark}

The issue when passing to infinite-dimensional vector spaces: the
topology matters. E.g. the closure of the unit ball is closed and
bounded and thus compact in finite dimensions, but this may no longer be
true in \({\mathbb{R}}^\infty\) or \({\mathbb{C}}^\infty\). Recall that
a Banach space is a complete normed space, and is further a Hilbert
space if the norm is induced by an inner product. See the textbook for a
review of vector spaces, metric spaces, norms, and inner products.

\end{remark}

\begin{example}[?]

Our first example of infinite dimensional vector spaces: sequence spaces
\(\ell\) with elements \(f \coloneqq(f_1, f_2, \cdots )\) with each
\(f_i\in {\mathbb{R}}\).

\end{example}

\begin{remark}

Linear subspaces are subspaces that contain zero, as opposed to affine
subspaces. An example is
\(C_0([0, L]; {\mathbb{R}}) \leq C([0, L]; {\mathbb{R}})\), the subspace
of bounded continuous functionals on \([0, L]\) which vanish at the
endpoints. For any subset \(S \subseteq V\), write \([S]\) or
\(\mathop{\mathrm{span}}S\) for the linear span of \(S\): all finite
linear combinations of elements in \(S\).

\end{remark}

\begin{example}[?]

Let \(V = C([-1, 1])\) and \(x_1\neq x_2\in [-1, 1]\), and set
\(M_i \coloneqq\left\{{f\in V {~\mathrel{\Big\vert}~}f(x_i) = 0}\right\}\).
Then \(M_i \leq V\) is a linear subspace, and in fact \(V = M_1 + M_2\)
but \(V\neq M_1 \oplus M_2\) since the zero function is in both
subspaces.

\end{example}

\begin{remark}

Limits of finite operators are compact. The classical example: set
\((A_N)_{i, i} = {1\over i}\), so
\(A_N = \operatorname{diag}\qty{{1}, {1\over 2}, {1\over 3}, \cdots, {1\over N}}\).
Then \(\operatorname{Spec}A_N = \left\{{1\over n}\right\}_{n\leq N}\),
but \(A\coloneqq\lim_N A_N\) is an operator with
\(0\in \mkern 1.5mu\overline{\mkern-1.5mu\operatorname{Spec}(A)\mkern-1.5mu}\mkern 1.5mu\)
as an accumulation point. Exercise: what is \(\ker A\)? Is it
nontrivial?

\end{remark}

\begin{definition}[Convexity]

A subset \(S \subseteq V\) is \textbf{convex} iff
\begin{align*}
tf + (1-t)g \in S \qquad \forall f, g\in S,\quad \forall t\in (0, 1]
.\end{align*}
Equivalently,
\begin{align*}
{af+bg\over a+b}\in S \qquad \forall f, g\in S,\quad \forall a,b\geq 0
\end{align*}
where not both of \(a\) and \(b\) are zero. The \textbf{convex hull} of
\(S\) is the smallest convex set containing \(S\).

\end{definition}

\begin{remark}

Recall Holder's inequality:
\begin{align*}
{\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_p \cdot {\left\lVert {g} \right\rVert}_q
,\end{align*}
Schwarz's inequality
\begin{align*}
{\left\lvert {{\left\langle {f},~{g} \right\rangle}} \right\rvert} \leq {\left\lVert {f} \right\rVert} {\left\lVert {g} \right\rVert} \qquad {\left\lVert {f} \right\rVert} \coloneqq\sqrt{{\left\langle {f},~{f} \right\rangle}}
,\end{align*}
and Minkowski's inequality
\begin{align*}
{\left\lVert {f+g} \right\rVert}_p \leq {\left\lVert {f} \right\rVert}_p + {\left\lVert {g} \right\rVert}_p
.\end{align*}

A nice proof of Cauchy-Schwarz:

\includegraphics{figures/2022-01-11_15-43-17.png}

\includegraphics{figures/2022-01-11_15-43-24.png}

\end{remark}

\hypertarget{thursday-january-13}{%
\section{Thursday, January 13}\label{thursday-january-13}}

\begin{remark}

My notes:

\begin{itemize}
\tightlist
\item
  \(K \subseteq {\mathcal{H}}\) is \textbf{complete} iff
  \(K^\perp = 0\).
\item
  Bessel: for \(f\in {\mathcal{H}}\) write
  \(f_n \coloneqq{\left\langle {f},~{e_n} \right\rangle} e_n\), then
  \({\left\lVert {(f_n)} \right\rVert}_{\ell^2({\mathbb{C}})} \leq {\left\lVert {f} \right\rVert}_{{\mathcal{H}}}^2\).
\item
  Best estimate: for any other sequence
  \((c_n) \in \ell^2({\mathbb{C}}), {\left\lVert {f - \sum c_n e_n} \right\rVert} \geq {\left\lVert {f - \sum f_n e_n} \right\rVert}\).
\item
  For \(\left\{{e_n}\right\}\) orthonormal,
  \((c_n) \in \ell^2({\mathbb{C}}) \iff \sum c_n e_n\) converges. If the
  series converges, it can be rearranged.
\item
  Differentiating through an integral:
\end{itemize}

\includegraphics{figures/2022-01-13_14-58-19.png}

\begin{itemize}
\tightlist
\item
  Parseval, Plancherel, and Fourier inversion:
\end{itemize}

\includegraphics{figures/2022-01-13_15-02-34.png}

\end{remark}

\begin{remark}

Last time: any norm yields a metric:
\(d(f, g) \coloneqq{\left\lVert {f-g} \right\rVert}\).

\begin{itemize}
\tightlist
\item
  Open/closed balls:
  \(B_r(f) \coloneqq\left\{{x{~\mathrel{\Big\vert}~}{\left\lVert {f-x} \right\rVert} < r}\right\}\).
\item
  Bounded subsets: contained in some ball of finite radius.
\item
  \({\operatorname{diam}}S = \inf_{r, f} {\operatorname{diam}}B_r(f)\)
  is the diameter of the smallest ball containing \(S\).
\item
  \(d(f, S) \coloneqq\inf_{x\in S} {\left\lVert {f-x} \right\rVert}\).
\item
  \(V = {\mathbb{R}}^n\) with
  \({\left\lVert {f} \right\rVert}_2^2 \coloneqq\sum_{k\leq n} f_i^2\),
  \(\mkern 1.5mu\overline{\mkern-1.5muB\mkern-1.5mu}\mkern 1.5mu_0(1)\)
  is a metric space but not a vector space.
\item
  For \(L^2\), there are unique least squares projections, but
  uniqueness may fail for \(L^1\).

  \begin{itemize}
  \tightlist
  \item
    Counterexample: take a line
    \(M = \left\{{{\left[ {\alpha, \alpha} \right]}}\right\}\) in
    \({\mathbb{R}}^2\) of angle \(\pi/4\) with respect to the
    \(x{\hbox{-}}\)axis and consider
    \(f\coloneqq{\left[ {[} \right]}0, 1]\). Then for
    \(g\coloneqq(\alpha, \alpha)\),
    \({\left\lVert {f-g} \right\rVert} = {\left\lvert {1-\alpha} \right\rvert} + {\left\lvert {\alpha} \right\rvert} \geq {\left\lvert {1-\alpha + \alpha} \right\rvert} = 1\),
    and the minimizer occurs for \emph{any} \(\alpha \in [0, 1]\).
  \item
    Similar issues may happen for \(L^\infty\) -- but \(L^1, L^\infty\)
    have sharper tails than \(L^2\), so this can be useful e.g.~in image
    problems.
  \end{itemize}
\item
  If limits of sequences \((f_n)\) exist,
  i.e.~\({\left\lVert {f_n - f_m} \right\rVert}\to 0\), then the
  limiting function \(f_n\to f\) is unique by the triangle inequality.
\item
  Example from last time:
  \(\operatorname{diag}\qty{ 1, {1\over 2}, \cdots, {1\over n}} \to A\)
  a compact self-adjoint operator with
  \(\operatorname{Spec}A = \left\{{1\over n}\right\}_{n\geq 0}\).

  \begin{itemize}
  \tightlist
  \item
    What is \(\ker A\)? Note that \(0\in \sigma(A)\), where
    \(\sigma(A)\) is the set where \((A-I\lambda)^{-1}\) is not defined.
    It turns out \(\ker A = \left\{{0}\right\}\).
  \end{itemize}
\item
  Defining closures of subsets: for \(S \subseteq V\), say
  \(f\in \mkern 1.5mu\overline{\mkern-1.5muS\mkern-1.5mu}\mkern 1.5mu\)
  iff there exists a sequence of not necessarily distinct points
  \(f_n \in S\) with \(f_n\to f\).

  \begin{itemize}
  \tightlist
  \item
    Say \(S_1 \subseteq S_2 \subseteq V\) is closed in \(S_2\) iff
    \(S_1 = C \cap S_2\) for some \(C\) closed in \(V\). The closure of
    \(S_1\) in \(S_2\) is \({ \operatorname{cl}} _V(S_1) \cap S_2\).
  \end{itemize}
\item
  A set that is neither open nor closed:
  \(X \coloneqq[a, b] \cap{\mathbb{Q}}\), and
  \({{\partial}}X = [a,b] \supseteq X\) is actually larger.
\item
  Recall the little \(\ell^p\) norms:
  \({\left\lVert {(f_n)} \right\rVert}_{p} \coloneqq\qty{ \sum {\left\lvert {f_n} \right\rvert}^p }^{1\over p}\)
  and
  \({\left\lVert {(f_n)} \right\rVert}_{\infty} \coloneqq\sup_n {\left\lvert {f_n} \right\rvert}\).
\end{itemize}

\end{remark}

\begin{exercise}[?]

\begin{itemize}
\tightlist
\item
  Prove Jensen's inequality for concave functions.
\item
  Prove Young's inequality.
\item
  Prove Holder's inequality.

  \begin{itemize}
  \tightlist
  \item
    Idea: consider
    \(a = \widehat{f_n} \coloneqq{\left\lvert {f_n} \right\rvert}/{\left\lVert {f_n} \right\rVert}^p, b = \widehat{g_n} \coloneqq{\left\lvert {g_n} \right\rvert}/{\left\lVert {g_n} \right\rVert}_q\)
    and apply Young's after summing over \(n\).
  \end{itemize}
\item
  Prove Minkowski's inequality.

  \begin{itemize}
  \tightlist
  \item
    Idea: use that \((p-1)q=p\) and apply the triangle inequality and
    then Holder to \(\sum {\left\lvert {f_n + g_n} \right\rvert}^p\).
    Also use that \(q^{-1}= 1-p^{-1}\), and divide through this
    inequality at the end. Be sure to check the cases
    \({\left\lVert {f+g} \right\rVert}_p = 0, \infty\).
  \end{itemize}
\end{itemize}

\end{exercise}

\hypertarget{tuesday-january-18}{%
\section{Tuesday, January 18}\label{tuesday-january-18}}

\begin{remark}

Last time:

\begin{itemize}
\tightlist
\item
  \({\left\lVert {f} \right\rVert}_{\ell^p} = \qty{\sum {\left\lvert {f_n} \right\rvert} ^p}^{1\over p}\)
\item
  \({\left\lVert {f} \right\rVert}_{\ell^ \infty} = \sup_{n} {\left\lvert {f_n} \right\rvert}\).
\end{itemize}

Today:

\begin{itemize}
\item
  \(\ell_p = \left\{{f \coloneqq(f_n) {~\mathrel{\Big\vert}~}{\left\lVert {f} \right\rVert}_{\ell^p} < \infty }\right\}\).
\item
  Example: set \(f^k \coloneqq(0, 0, \cdots, 1, 1, \cdots)\) with zeros
  for the first \(k-1\) entires and ones for all remaining entries. Then
  \(f^k_i \overset{k\to\infty}\longrightarrow 0\) for each fixed
  component at index \(i\). So \(f^k \to (0)\) component-wise, but
  \({\left\lVert {f_k} \right\rVert}_{\ell^\infty} = 1\) for every
  \(k\), so this doesn't converge in \(\ell_\infty\).
\item
  Recall the \({\varepsilon}{\hbox{-}}\delta\) and limit definitions of
  continuity.
\item
  Recall the definition of uniform continuity.
\item
  For \(\Omega \subseteq {\mathbb{R}}^n\), write \(C(\Omega)\) for the
  \({\mathbb{R}}{\hbox{-}}\)vector space of continuous bounded
  functionals \(f: \Omega\to {\mathbb{C}}\) with the norm
  \({\left\lVert {f} \right\rVert}_{L^\infty} = \sup_{x\in \Omega} {\left\lvert {f(x)} \right\rvert}\).
\item
  Define \(C^k(\Omega, {\mathbb{C}}^m)\) to be functions with \(k\)
  continuous partial derivatives which are bounded, and set
  \(C^\infty(\Omega, {\mathbb{C}}^m) = \cap_{k\geq 0} C^k(\Omega, {\mathbb{C}}^m)\).
  Define a norm
  \({\left\lVert {f} \right\rVert}_{C^k} = \sum_{j\leq k} \sup_{x\in \Omega} {\left\lvert {f^{(j)} (x)} \right\rvert}\).
\item
  Take \(g(x) = 2-x^2\) and consider \({\mathbb{B}}_{1\over 2}(g)\) in
  \(C[0, 1]\) with \({\left\lVert {{-}} \right\rVert}_{L^\infty}\).
\item
  Show that convergent implies Cauchy-convergent using the triangle
  inequality.
\item
  Lipschitz with \({\left\lvert {c} \right\rvert} < 1\) implies Cauchy.
\item
  Lemma 1.4.2:
  \({\left\lVert {f_{n+k} - f_n} \right\rVert} \leq q^n (1-q)^{-1} {\left\lVert {f_1 - f_0} \right\rVert}\)
  for all \(k\geq 0\). Use
  \begin{align*}
  {\left\lVert {f_{n+k} - f_n} \right\rVert} = {\left\lVert {\sum_{j=1}^k (f_{n+j} - f_{n+j-1} ) } \right\rVert} \leq {\left\lVert {f_1 - f_0} \right\rVert} \sum_{j=1}^k q^{n+j-1} \overset{n\to\infty}\longrightarrow 0
  .\end{align*}
\item
  Counterexample, not all normed spaces are complete: take
  \(V = C[-1, 1]\) with
  \({\left\lVert {f} \right\rVert}_{L^1} \coloneqq\int_{-1}^1 {\left\lvert {f(x) } \right\rvert}\,dx\).
  Define a sequence of functions \((f_n)\):

  \includegraphics{figures/2022-01-18_15-09-38.png}

  Check that
  \({\left\lVert {f_{n+1} - f_n} \right\rVert}_{L^1} \leq q {\left\lVert {f_n - f_{n-1}} \right\rVert}_{L^1}\),
  and pointwise \(f_n \to \chi_{[-1, 0]}\) which is discontinuous and
  not in \(C[-1, 1]\).
\item
  Banach spaces: complete normed vector spaces.
\item
  Series:

  \begin{itemize}
  \tightlist
  \item
    Convergent: \(f \coloneqq\lim_N \sum_{n\leq N} f_n \in V\).
  \item
    Absolute convergence:
    \(\sum {\left\lVert {f_n} \right\rVert} < \infty\).
  \item
    In a Banach space, absolutely convergent series can be rearranged.
  \end{itemize}
\item
  Theorem: A normed space is complete iff absolute convergence
  \(\implies\) convergence. Proof:

  \begin{itemize}
  \tightlist
  \item
    Step 1: show that every Cauchy sequence has a convergent
    subsequence.
  \item
    Set
    \(a_n \coloneqq\sup_{m > n} {\left\lVert {f_n - f_m} \right\rVert}\),
    then Cauchy implies \(a_n\to 0\) in \({\mathbb{R}}\).
  \item
    Get a convergent subsequence \(a_{n_j} \leq j^{-2}\).
  \item
    Set \(g_j \coloneqq f_{n_j} - f_{n_{j+1}}\), then
    \(g\coloneqq\sum g_j\) absolutely converges, say to \(g\).
  \item
    Note \(f_{n_i} - f_{n_{i+1}} = \sum_{j=1}^i g_j\), so the
    subsequence \((f_{n_j})\) is convergent.
  \item
    Step 2: use this to show that the original sequence \((f_n)\)
    converges.
  \item
    Set \(f = \lim f_n\), then
    \({\left\lVert {f_n - f} \right\rVert} \leq {\left\lVert {f_n - f_{n_i}} \right\rVert} + {\left\lVert {f_{n_i} - f} \right\rVert}\),
    using Cauchy for the first \({\varepsilon}\) and the convergent
    subsequence for the second.
  \end{itemize}
\end{itemize}

\end{remark}

\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}

\begin{remark}[Some random notes]

Some theorems that hold in Hilbert spaces but not necessarily Banach
spaces:

\includegraphics{figures/2022-01-18_14-48-04.png}

\includegraphics{figures/2022-01-18_14-48-13.png}

\includegraphics{figures/2022-01-18_14-54-38.png}

Absolute continuity:

\includegraphics{figures/2022-01-18_14-55-21.png}

\includegraphics{figures/2022-01-18_14-56-04.png}

\(L_p^{\mathsf{loc}}\):

\includegraphics{figures/2022-01-18_14-57-54.png}

\(\ker L = 0\) may not be sufficient to guarantee bijectivity in
infinite dimensions:

\includegraphics{figures/2022-01-18_15-27-21.png}

Boundedness:

\includegraphics{figures/2022-01-18_15-28-55.png}

Bounded iff continuous:

\includegraphics{figures/2022-01-18_15-29-25.png}

\end{remark}

\hypertarget{more-banach-spaces-thursday-january-20}{%
\section{More Banach Spaces (Thursday, January
20)}\label{more-banach-spaces-thursday-january-20}}

\begin{remark}

Last time: complete iff absolutely convergent implies convergent. Today:
wrapping up some results on Banach and Hilbert spaces, skipping a review
of \(L^p\) spaces, and starting on operators next week.

\end{remark}

\begin{remark}

Note that \(S \coloneqq(0, 1]\) is open and not complete, but
\({ \operatorname{cl}} _{\mathbb{R}}(S) = [0, 1]\) is both closed and
complete. Generalizing:

\end{remark}

\begin{lemma}[?]

A subset \(S \subseteq B\) of a Banach space is complete iff \(S\) is
closed in \(B\).

\end{lemma}

\begin{proof}[?]

\(\impliedby\): If \(S\) is closed, a Cauchy sequence \((f_n)\) in \(S\)
converges to some \(f\in B\). Since \(S\) is closed in \(B\), in fact
\(f\in S\).

\(\implies\): Suppose that for \(f\in { \operatorname{cl}} _B(S)\),
there is a Cauchy sequence \(f_n\to f\) with \(f_n \in S\) and
\(f\in B\). Since \(S\) is complete, \(f\in S\), so
\({ \operatorname{cl}} _B(S) \subseteq S\) making \(S\) closed.

\end{proof}

\begin{theorem}[?]

For \(\Omega \subseteq {\mathbb{R}}^n\), the space
\(X = (C(\Omega; {\mathbb{C}}), {\left\lVert {{-}} \right\rVert}_\infty)\)
is a Banach space.

\end{theorem}

\begin{warnings}

This is \emph{not} complete with respect to any other \(L^p\) norm with
\(p<\infty\)!

\end{warnings}

\begin{proof}[of theorem]

Use the lemma -- write \(B\) for the space of all bounded (not
necessarily continuous) functions on \(\Omega\), which is clearly a
normed vector space, so it suffice to show

\begin{itemize}
\tightlist
\item
  \(X \subseteq B\) is closed,
\item
  \(B\) is a Banach space (i.e.~complete).
\end{itemize}

\textbf{Step 1}: we'll show \({ \operatorname{cl}} _B(X) \subseteq X\).
Take \(f\) to be a limit point of \(X\), then for every
\({\varepsilon}>0\) there is a \(g\in X\) with
\({\left\lVert {f-g} \right\rVert} < {\varepsilon}\). Apply the triangle
inequality:
\begin{align*}
{\left\lVert {f} \right\rVert} \leq {\left\lVert {f-g+g} \right\rVert}\leq {\left\lVert {f-g} \right\rVert} + {\left\lVert {g} \right\rVert} = {\varepsilon}+ C < \infty
,\end{align*}
so \(f\in { \operatorname{cl}} _B(X) \subseteq B\) since it is bounded.
It remains to show \(f\) is continuous. Use that
\({\left\lVert {f-g} \right\rVert}\infty <{\varepsilon}\) and continuity
of \(g\) to get
\({\left\lvert {g(x) - g(x_0)} \right\rvert} < {\varepsilon}\) for
\({\left\lvert {x-x_0} \right\rvert}<{\varepsilon}\). Now
\begin{align*}
{\left\lvert {f(x) - f(x_0)} \right\rvert} 
&= {\left\lvert {f(x) - g(x) + g(x) -g(x_0) + g(x_0) - f(x_0)} \right\rvert} \\
&\leq {\left\lvert {f(x) - g(x)} \right\rvert} + {\left\lvert {g(x) - g(x_0)} \right\rvert} + {\left\lvert {f(x_0) - g(x_0)} \right\rvert} \\
&\leq 3{\varepsilon}
.\end{align*}
So \(X\) is closed in \(B\).

\textbf{Step 2}: Let \(f_n\) be Cauchy in \(B\), and note that we have a
pointwise bound
\({\left\lvert {f_n(x) - f_n(x_0)} \right\rvert} \leq {\left\lVert {f_n - f_m} \right\rVert} \to 0\).
So pointwise, \(f_n(x)\) is a Cauchy sequence in \({\mathbb{C}}\) which
is complete, so \(f_n(x) \to f(x)\) for some
\(f: \Omega\to {\mathbb{C}}\). We now want to show \(f_n\to f\) in
\(X\). Using that \(f_n\) is Cauchy in \(X\), produce an \(N_0\) such
that
\(n, m\geq N_0 \implies {\left\lVert {f_n - f_m} \right\rVert}< {\varepsilon}\).
Now
\begin{align*}
{\left\lVert {f - f_n} \right\rVert} 
&= \sup_{x\in \Omega} {\left\lvert {f(x) - f_n(x)} \right\rvert} \\
&\leq \sup_{x\in \Omega} \sup_{m\geq N_0} {\left\lvert {f_m(x) - f_n(x)} \right\rvert} \\
&= \sup_{m\geq N_0} \sup_{x\in \Omega} {\left\lvert {f_m(x) - f_n(x)} \right\rvert} \\
&= \sup_{m\geq N_0} {\left\lVert {f_m - f_n} \right\rVert} \\
&\leq {\varepsilon}
.\end{align*}
Now use the reverse triangle inequality to show \(f_n\) is bounded
\begin{align*}
{\left\lVert {f} \right\rVert} - {\left\lVert {f_n} \right\rVert} \leq {\left\lVert {f-f_n} \right\rVert} < {\varepsilon}\implies {\left\lVert {f} \right\rVert} < \infty
.\end{align*}

Now by problem 1.13, every Cauchy sequence is bounded, so
\(f_n \to f\in B\).

\end{proof}

\begin{remark}

Extending to vector-valued functions: for
\(\Omega \subseteq {\mathbb{R}}^n\), take
\(\mathbf{x} = {\left[ {x_1, \cdots, x_n} \right]}\) and
\(F = {\left[ {f_1, \cdots, f_m} \right]}: {\mathbb{C}}^n\to {\mathbb{C}}^m\).
Then there is a Banach space
\begin{align*}
X = C^(\Omega, {\mathbb{C}}^m), \qquad {\left\lVert {f} \right\rVert}_{C_1(\Omega)} \coloneqq\sum_{i\leq m} \sup_{x\in \Omega} {\left\lvert {f(x)} \right\rvert} + \sum_{i\leq m, j\leq n} \sup_{x\in \Omega} {\left\lvert {{\frac{\partial f_i}{\partial x_j}\,}(x) } \right\rvert}
.\end{align*}
Similarly define \(L^p(\Omega)\), noting that
\({\left\lVert {f} \right\rVert}_{L^\infty(\Omega)}\) is the
\emph{essential supremum}.

\end{remark}

\begin{theorem}[?]

For \(p\in (1, \infty)\), the sequence space
\(X = (\ell^p, {\left\lVert {{-}} \right\rVert}_{\ell^p})\) is a Banach
space.

\end{theorem}

\begin{definition}[Closed subspaces]

A \textbf{closed subspace} of a Banach space is a linear subspace
\(M \leq B\) which is norm-closed in \(B\).

\end{definition}

\begin{example}[?]

\begin{align*}
M \coloneqq\ker { \nabla\cdot }= \left\{{f\in C(\Omega) {~\mathrel{\Big\vert}~}{ \nabla\cdot }f = 0}\right\} \leq C(\Omega)
\end{align*}
is closed, where \({ \nabla\cdot }f\) is the divergence of a function
\(f\).

For any \(S \subseteq B\), one can also take the corresponding closed
subspace
\(\mkern 1.5mu\overline{\mkern-1.5mu[S]\mkern-1.5mu}\mkern 1.5mu \coloneqq{ \operatorname{cl}} _B \mathop{\mathrm{span}}_{\mathbb{C}}\left\{{s\in S}\right\}\),
i.e.~all linear combinations of elements in \(S\) and their limits. This
is called the \textbf{closed linear span} of \(S\).

\end{example}

\begin{exercise}[?]

Let \(B = ( C[a, b], {\left\lVert {{-}} \right\rVert}_{L^\infty}\) and
for \(x_0 \in [a, b]\) define
\begin{align*}
M \coloneqq\left\{{f\in C[a, b] {~\mathrel{\Big\vert}~}f(x_0) = 0}\right\}, \qquad
N \coloneqq\left\{{f\in C[a, b] {~\mathrel{\Big\vert}~}f(x_0) \leq c}\right\}
.\end{align*}
Show that these are closed subspaces with no nontrivial open subsets of
\(B\), since any \(f\in M\) can be perturbed to be nonzero at \(x_0\)
with an arbitrarily small norm difference.

\end{exercise}

\begin{remark}

Recall that for \(S_1 \subseteq S_2 \subseteq B\), \(S_1\) is
\textbf{dense} in \(S_2\) iff
\({ \operatorname{cl}} _{S_2}(S_1) = S_2\). Recall Weierstrass' theorem:
for \(\Omega \subseteq {\mathbb{R}}^n\) is closed and bounded and write
\({\mathcal{O}}\coloneqq{\mathbb{R}}[x_1, \cdots, x_n]\) for the
polynomials in \(n\) variables. Then
\({\mathcal{O}}\subseteq C(\Omega)\), and
\({ \operatorname{cl}} _{C(\Omega)} {\mathcal{O}}= C(\Omega)\),
i.e.~\({\mathcal{O}}\) is a dense subspace in the \(L^\infty\) norms. In
fact, piecewise linear functions are dense.

\end{remark}

\begin{remark}

Norms are equivalent iff
\(c_1 {\left\lVert {f} \right\rVert}_a \leq {\left\lVert {f} \right\rVert}_b\leq c_2 {\left\lVert {f} \right\rVert}_a\)
for some constants \(c_i\). All norms on \({\mathbb{R}}^n\) (resp.
\({\mathbb{C}}^n\)) are equivalent.

\end{remark}

\begin{example}[?]

For \(a>0, f\in C[0, 1]\), define
\begin{align*}
{\left\lVert {f} \right\rVert} \coloneqq\sup_{x\in I} {\left\lvert {e^{-ax} f(x)} \right\rvert}
,\end{align*}
which can be thought of as a weighting on the uniform norm which
de-emphasizes the tails of functions near the endpoints. This is
equivalent to \({\left\lVert {{-}} \right\rVert}_\infty\), since
\begin{align*}
e^a \sup {\left\lvert {f} \right\rvert} \leq \sup {\left\lvert {e^{-ax} f} \right\rvert} \leq 1\cdot \sup {\left\lvert {f} \right\rvert}
.\end{align*}

\end{example}

\begin{remark}

Note that a basis for a norm can be used as a basis with respect to an
equivalent norm in finite dimensions. In infinite dimensions this may
not hold -- e.g.~for Fourier series,
\(\left\{{e_k(x)}\right\}_{k\in {\mathbb{Z}}}\) is not a basis for
\(C[0, 2\pi]\) with the sup norm.

\end{remark}

\begin{definition}[Separable Banach spaces]

An \(B\in \mathcal{B}\) is \textbf{separable} iff \(X\) contains a
countable dense subset \(S = \left\{{f_k}\right\}_{k\geq 0}\) such that
for each \(f\in B\) and \({\varepsilon}>0\), there is an \(f_k\in S\)
with \({\left\lVert {f-f_k} \right\rVert} < {\varepsilon}\).

\end{definition}

\begin{example}[?]

Show that

\begin{itemize}
\tightlist
\item
  \(\Omega \subseteq {\mathbb{R}}^n\) a bounded subset,
  \(C(\Omega), {\left\lVert {{-}} \right\rVert}_{\sup}\) is separable.
\item
  \(\ell^p\) is separable for \(p\in (1,\infty)\).
\item
  \(\ell^\infty\) is \emph{not} separable.
\end{itemize}

\end{example}

\hypertarget{random-notes}{%
\subsection{Random Notes}\label{random-notes}}

\begin{remark}

\includegraphics{figures/2022-01-20_14-48-53.png}

\includegraphics{figures/2022-01-20_14-49-39.png}

\end{remark}

\hypertarget{tuesday-january-25}{%
\section{Tuesday, January 25}\label{tuesday-january-25}}

\begin{remark}

\begin{itemize}
\tightlist
\item
  Inner products:

  \begin{itemize}
  \tightlist
  \item
    \({\left\langle {f},~{f} \right\rangle} \geq 0\) and
    \({\left\langle {f},~{f} \right\rangle} = 0 \iff f=0\)
  \item
    \({\left\langle {f},~{g+h} \right\rangle} = {\left\langle {f},~{g} \right\rangle} + {\left\langle {f},~{h} \right\rangle}\)
  \item
    \({\left\langle {f},~{g} \right\rangle} = \mkern 1.5mu\overline{\mkern-1.5mu{\left\langle {g},~{f} \right\rangle}\mkern-1.5mu}\mkern 1.5mu\)
  \item
    \({\left\langle {\alpha f},~{g} \right\rangle} = \alpha {\left\langle {f},~{g} \right\rangle}\)
  \end{itemize}
\item
  A pre-Hilbert space is an inner-product space.
\item
  Example:
  \({\left\langle {f},~{g} \right\rangle} \coloneqq\int_\Omega w(x) f(x) \mkern 1.5mu\overline{\mkern-1.5mug(x)\mkern-1.5mu}\mkern 1.5mu\,dx\)
  for \(f,g\in C(\Omega)\), where \(w\) is any weighting function.
\item
  Example:
  \({\left\langle {f},~{g} \right\rangle} = \sum f_i \mkern 1.5mu\overline{\mkern-1.5mug_i\mkern-1.5mu}\mkern 1.5mu\)
  for \(f,g\in {\mathbb{C}}^n\).
\item
  Inner products induce norms:
  \({\left\lVert {f} \right\rVert} \coloneqq\sqrt{{\left\langle {f},~{f} \right\rangle}}\)

  \begin{itemize}
  \tightlist
  \item
    Orthogonality: write \(f\perp g\) iff
    \({\left\langle {f},~{g} \right\rangle} = 0\) and
    \(S^\perp = \left\{{f\in {\mathcal{H}}{~\mathrel{\Big\vert}~}f\perp s \, \forall s\in S}\right\}\).
  \end{itemize}
\item
  Definition of a Hilbert space: a pre-Hilbert space complete with
  respect to the norm induced by its inner product.
\item
  Recall \(C(\Omega)\) with
  \({\left\langle {f},~{g} \right\rangle} \coloneqq\int_\Omega f\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu\)
  is not complete, thus not a Hilbert space.
\item
  Example: a common optimization problem,
  \(\mathrm{argmin} {\left\lVert {x} \right\rVert}\) such that \(Ax=0\).
\end{itemize}

\end{remark}

\begin{theorem}[Cauchy-Schwarz]

\begin{align*}
{\left\lvert {{\left\langle {f},~{g} \right\rangle}} \right\rvert}\leq {\left\lVert {f} \right\rVert} {\left\lVert {g} \right\rVert}
.\end{align*}

\end{theorem}

\begin{proof}[?]

Use
\({\left\langle {f},~{g} \right\rangle} = {\left\lvert {f} \right\rvert} {\left\lvert {g} \right\rvert} \cos \theta_{fg}\)
where \({\left\lvert { \cos \theta } \right\rvert} \leq 1\). - Assume
\(g\neq 0\), then STS
\({\left\langle {f},~{g\over {\left\lVert {g} \right\rVert}} \right\rangle} \leq {\left\lVert {f} \right\rVert}\).
- Use
\begin{align*}
0 
&\leq {\left\lVert {f - {\left\langle {f},~{g} \right\rangle} g} \right\rVert}^2 \\
&= {\left\langle {f - {\left\langle {f},~{g} \right\rangle} g},~{f - {\left\langle {f},~{g} \right\rangle} g} \right\rangle} \\
&= {\left\langle {f},~{f} \right\rangle} - {\left\langle {f},~{g} \right\rangle} {\left\langle {g},~{f} \right\rangle} - \mkern 1.5mu\overline{\mkern-1.5mu{\left\langle {f},~{g} \right\rangle}\mkern-1.5mu}\mkern 1.5mu {\left\langle {f},~{g} \right\rangle} + {\left\langle {f},~{g} \right\rangle} \mkern 1.5mu\overline{\mkern-1.5mu{\left\langle {f},~{g} \right\rangle} \mkern-1.5mu}\mkern 1.5mu \\
&= {\left\lVert {f} \right\rVert}^2 - {\left\lvert { {\left\langle {f},~{g} \right\rangle}} \right\rvert}^2
.\end{align*}

\end{proof}

\begin{theorem}[Closest approximations]

Let \(f\in M\leq {\mathcal{H}}\) be a closed (and thus complete)
subspace of a Hilbert space \({\mathcal{H}}\). Then there is a unique
element \(g\) in \({\mathcal{H}}\) closest to \(M\) in the norm.

\end{theorem}

\begin{proof}[?]

Let \(d\coloneqq\operatorname{dist}(f, M)\), choose a sequence
\(g_n \in M\) such that \({\left\lVert {f-g_n} \right\rVert}\to d\),
which is possible since
\(d = \inf_{g\in M}{\left\lVert {f-g} \right\rVert}\). Apply the
parallelogram law to write
\begin{align*}
{\left\lVert {g_n - g_m} \right\rVert}^2 
&= {\left\lVert {(g_n - f) - (g_m - f) } \right\rVert}^2 \\
&= 2{\left\lVert {g_n - f} \right\rVert}^2 + 2{\left\lVert {g_m - f} \right\rVert}^2 - 4{\left\lVert { {1\over 2} (g_n + g_m) - f} \right\rVert}^2 \\
&\leq 2{\left\lVert {g_n - f} \right\rVert}^2 + 2{\left\lVert {g_m - f} \right\rVert}^2 - 4d^2 \\
&\leq 2d^2 + 2d^2 - 4d^2 \\
&= 0
,\end{align*}
so the \(g_n\) are Cauchy. Here we've used that
\({1\over 2}(g_n + g_m) = m \in M\) since \(M\) is a subspace, and
\({\left\lVert {m-f} \right\rVert} \geq d\). Since \(M\) is complete,
\(g_n\to g\in M\) and moreover \({\left\lVert {f-g} \right\rVert} = d\).
For uniqueness, if \({\left\lVert {f-g'} \right\rVert} = d\) then
\begin{align*}
{\left\lVert { f - {1\over 2} (g+g')} \right\rVert}^2 = d^2 - {\left\lVert {g-g'} \right\rVert}^2 < d^2 \qquad \contradiction
.\end{align*}

\end{proof}

\begin{theorem}[Projection theorem]

Let \(M\leq {\mathcal{H}}\) be a closed subspace of a Hilbert space.
Then \(M^\perp \leq {\mathcal{H}}\) is closed, and
\({\mathcal{H}}= M \oplus M^\perp\). In the decomposition \(f= g+h\), in
fact \(g\in M\) is the closest approximation to \(f\) in \(M\), making
this decomposition unique.

\end{theorem}

\begin{proof}[?]

STS \({\mathcal{H}}= M + M^\perp\) by the exercises. If \(f\in M\), take
\(f=g+h\) where \(g=f\) and \(h=0\), so suppose \(f\not\in M\). Let
\(g = \mathrm{argmin} \operatorname{dist}(f, M) \in M\), and we claim
\(f-g\in M^\perp\), so \({\left\langle {f-g},~{h} \right\rangle} = 0\)
for any \(h\in M\). For all \(h\in M\) and \(\alpha > 0\), we have
\(g + \alpha h\in M\), so
\begin{align*}
{\left\lVert {f - g} \right\rVert}^2 
&\leq {\left\lVert {f - (g+\alpha h)} \right\rVert}^2 \\
&= {\left\lVert {f-g} \right\rVert}^2 -2\Re\alpha {\left\langle {h},~{f-g} \right\rangle} + \alpha^2 {\left\lVert {h} \right\rVert}^2
,\end{align*}
so
\begin{align*}
2\Re \alpha {\left\langle {h},~{f-g} \right\rangle} \leq \alpha^2 {\left\lVert {h} \right\rVert}^2 \implies 2\Re {\left\langle {h},~{f-g} \right\rangle} \leq \alpha {\left\lVert {h} \right\rVert}^2 \overset{\alpha\to 0}\longrightarrow 0
,\end{align*}
so \(\Re{\left\langle {h},~{f-g} \right\rangle} = 0\). Similarly
\(\Im{\left\langle {h},~{f-g} \right\rangle} = 0\).

\end{proof}

\begin{exercise}[?]

Show \(S^\perp\) is closed for any \(S\in {\mathcal{H}}\), and in fact
\(S^\perp = \mathop{\mathrm{span}}_{\mathbb{C}}({ \operatorname{cl}} _{\mathcal{H}}(S))^\perp\),
and if \(f\in S \cap S^\perp\) then \(f=0\).

\end{exercise}

\begin{exercise}[?]

Prove the parallelogram law
\begin{align*}
{\left\lVert {f-g} \right\rVert}^2 + {\left\lVert {f+g} \right\rVert}^2 = 2{\left\lVert {f} \right\rVert}^2 + 2{\left\lVert {g} \right\rVert}^2
.\end{align*}

\end{exercise}

\hypertarget{thursday-january-27}{%
\section{Thursday, January 27}\label{thursday-january-27}}

\begin{remark}

Notes:

\begin{itemize}
\tightlist
\item
  Bessel:
  \begin{align*}
  \sum_n {\left\lvert { {\left\langle {f},~{\phi_n} \right\rangle} } \right\rvert} \leq {\left\lVert {f} \right\rVert}^2,\qquad {\left\lVert {f} \right\rVert} \coloneqq\sqrt{{\left\langle {f},~{f} \right\rangle}}
  .\end{align*}

  \begin{itemize}
  \tightlist
  \item
    Prove using the fact that
    \begin{align*}
    0\leq {\left\lVert {f - \sum_{n\leq m} {\left\langle {f},~{\phi_n} \right\rangle} \phi_n } \right\rVert}^2 = {\left\lVert {f} \right\rVert}^2 - \sum_{n\leq m} {\left\lvert {{\left\langle {f},~{\phi_n} \right\rangle}} \right\rvert}^2
    .\end{align*}
  \end{itemize}
\item
  Best fit:
  \begin{align*}
  {\left\lVert {f - \sum_{n\leq m} c_n \phi_n } \right\rVert} \geq {\left\lVert {f - \sum_{n\leq m} {\left\langle {f},~{\phi_n} \right\rangle} \phi_n } \right\rVert}
  ,\end{align*}
  so define projections
  \(P_M(f) \coloneqq\sum {\left\langle {f},~{\phi_n} \right\rangle} \phi_n\)
  for \(\mathop{\mathrm{span}}\left\{{\phi_n}\right\} = M\).

  \begin{itemize}
  \tightlist
  \item
    Prove using
    \begin{align*}
    {\left\lVert {f - \sum_{n\leq m} c_n \phi_n } \right\rVert} 
    &= {\left\lVert {f} \right\rVert}^2 - \sum_{n\leq m} {\left\lvert {{\left\langle {f},~{\phi_n} \right\rangle}} \right\rvert}^2 + \sum_{n\leq m} {\left\lvert {{\left\langle {f},~{\phi_n} \right\rangle} - c_n } \right\rvert}^2 \\
    &\geq {\left\lVert {f} \right\rVert}^2 - \sum_{n\leq m} {\left\lvert {{\left\langle {f},~{\phi_n} \right\rangle}} \right\rvert}^2
    .\end{align*}
  \end{itemize}
\item
  Hilbert spaces are separable: have countable dense subsets.

  \begin{itemize}
  \tightlist
  \item
    \(\ell^\infty({\mathbb{Z}})\) is not separable.
  \end{itemize}
\item
  For \(\left\{{\phi_n}\right\}\) orthonormal and
  \(\left\{{c_n}\right\}\) scalars, \(\sum c_n \phi_n\) is convergent
  iff \(\left\{{a_n}\right\}\in \ell^2({\mathbb{Z}})\), so
  \(\sum{\left\lvert {c_n} \right\rvert}^2<\infty\). If it converges, it
  can be rearranged, and
  \begin{align*}
  {\left\lVert {\sum c_n \phi_n} \right\rVert}^2 = {\left\lVert {\left\{{c_n}\right\}} \right\rVert}_{\ell^2({\mathbb{Z}})}^2 = \sum {\left\lvert {c_n} \right\rvert}^2
  .\end{align*}

  \begin{itemize}
  \tightlist
  \item
    To prove, use that
    \({\left\lVert { \sum_{i\leq n\leq j}c_n \phi_n} \right\rVert}^2 = \sum_{i\leq n\leq j}{\left\lvert {c_n} \right\rvert}^2\),
    so the sequence \(\left\{{S_m}\right\}_{m\geq 0}\) where
    \(S_m \coloneqq\sum_{n\leq m} c_n\phi_n\) is Cauchy since
    \(\sum{\left\lvert {c_n} \right\rvert}^2\) converges.
  \item
    If \(g = \sum c_n \phi_n\) and \(f = \sum c_{m_n} \phi_{m_n}\) is a
    rearrangement, if \(\left\{{c_n}\right\}\in \ell^2\) then
    \({\left\lVert {f} \right\rVert}^2 = {\left\lVert {g} \right\rVert}^2 = \sum {\left\lvert {c_n} \right\rvert}^2\).
    Then
    \({\left\lVert {f-g} \right\rVert}^2 = {\left\lVert {f} \right\rVert}^2 + {\left\lVert {g} \right\rVert}^2 - 2\Re{\left\langle {f},~{g} \right\rangle}\),
    but
    \({\left\langle {f},~{g} \right\rangle} = \sum {\left\lvert {c_n} \right\rvert}^2\),
    so \({\left\lVert {f-g} \right\rVert} = 0\).
  \end{itemize}
\item
  If \(K = \left\{{\phi_n}\right\} \subseteq {\mathcal{H}}\) is a proper
  subset, so
  \(\mathop{\mathrm{span}}\left\{{\phi_n}\right\}\neq {\mathcal{H}}\),
  write
  \(P_K(f) = \sum {\left\langle {f},~{\phi_n} \right\rangle}\phi_n\).
  Then \(P_K(f) = 0\) if \(f\in { \operatorname{cl}} (K)^\perp\),
  \(P_K(f) = f\) if \(f\in { \operatorname{cl}} (K)\), and if
  \(f\not \in K\), since \({ \operatorname{cl}} (K)\leq {\mathcal{H}}\)
  is closed then there exists a \(g\in { \operatorname{cl}} (K)\) where
  \({\left\lVert {f-g} \right\rVert} = \operatorname{dist}(f, K)\).
  Write
  \(f = h + (f-g) \in { \operatorname{cl}} (K) \oplus { \operatorname{cl}} (K)^\perp\),
  so \(f = P_K f + (I-P_K)f\).
\item
  Recall complete subspaces of Banach spaces are closed.
\item
  Next theorem: every separable Hilbert space admits an orthonormal
  basis.
\end{itemize}

\end{remark}

\begin{theorem}[?]

If \({\mathcal{H}}\) is a separable Hilbert spaces and
\(K = \left\{{\phi_n}\right\}\) is an orthonormal set, then TFAE

\begin{itemize}
\tightlist
\item
  \(K\) is complete, i.e.~\(K^\perp = 0\) (taking the closure is not
  needed),
\item
  \({ \operatorname{cl}} {\mathop{\mathrm{span}}K} = {\mathcal{H}}\),
\item
  \(K\) is an orthonormal \emph{basis},
\item
  For all \(f\in {\mathcal{H}}\),
  \({\left\lVert {f} \right\rVert} = \sum_{n\geq 0} {\left\lvert {{\left\langle {f},~{\phi_n} \right\rangle}} \right\rvert}^2\)
  (Parseval).
\end{itemize}

\end{theorem}

\begin{remark}

Note \({\left\langle {f},~{\phi_n} \right\rangle}\) is the \(n\)th
Fourier coefficient
\(\widehat{f}(\xi) = \sum {\left\langle {f},~{\phi_n} \right\rangle}\phi_n(\xi)\),
and Parseval says
\({\left\lVert {f} \right\rVert}^2 = {\left\lVert {\widehat{f}} \right\rVert}^2\).

\end{remark}

\begin{proof}[?]

\(1\implies 2\): Let
\(f\in {\mathcal{H}}\setminus{ \operatorname{cl}} \mathop{\mathrm{span}}(K)\)
and project, so \(f= g+h\) with \(g,h\neq 0\). But
\(g\in { \operatorname{cl}} \mathop{\mathrm{span}}(K)\) and
\(h\in { \operatorname{cl}} \mathop{\mathrm{span}}(K)^\perp = 0\),
forcing \(K^\perp = { \operatorname{cl}} (K)^\perp \neq 0\).
\(\contradiction\)

\(2\implies 3\): Follows directly from previous lemma that
\(f = P_Kf + (I-P_K)f\).

\(3\implies 4\): Write \(f\in {\mathcal{H}}\) as
\(f = \sum {\left\langle {f},~{\phi_n} \right\rangle} \phi_n\) by
sending \(m\to \infty\) in the previous lemma.

\(4\implies 1\): Toward a contradiction, suppose \(f\neq 0\in K^\perp\).
Then \({\left\lVert {f} \right\rVert} \neq 0\) but
\({\left\langle {f},~{\phi_n} \right\rangle}=0\) for all \(n\),
contradicting Parseval. \(\contradiction\)

\end{proof}

\hypertarget{tuesday-february-01}{%
\section{Tuesday, February 01}\label{tuesday-february-01}}

\begin{remark}

Notes:

\begin{itemize}
\tightlist
\item
  Assume \({\mathcal{H}}\) is a \textbf{separable} Hilbert space: there
  exists a countable set of vectors
  \(\left\{{v_i}\right\}_{i\in {\mathbb{Z}}}\) which span a subspace
  that is dense in \({\mathcal{H}}\), so
  \({ \operatorname{cl}} (\mathop{\mathrm{span}}\left\{{v_i}\right\}) = {\mathcal{H}}\).
\item
  Complete subspaces: \(M\leq {\mathcal{H}}\) with \(M^\perp = 0\).
\item
  Show that for any \(S \subseteq {\mathcal{H}}\), \(S^\perp\) is closed
  in \({\mathcal{H}}\),
  \(S^\perp = \qty{{ \operatorname{cl}} \mathop{\mathrm{span}}S}^\perp\),
  and \(f\in S \cap S^\perp \implies f=0\).
\item
  If \(K = \left\{{\phi_k}\right\}_{k\in {\mathbb{Z}}}\) is an
  orthonormal set in \({\mathcal{H}}\), then TFAE

  \begin{itemize}
  \tightlist
  \item
    \(K \leq {\mathcal{H}}\) is a complete subspace, so \(K^\perp = 0\),
    i.e.~\({\left\langle {f},~{\phi_k} \right\rangle} = 0\) for all
    \(k\) implies \(f=0\).
  \item
    \({ \operatorname{cl}} \mathop{\mathrm{span}}K = {\mathcal{H}}\), so
    every \(f\in {\mathcal{H}}\) is the limit of a sequence of vectors
    from \(\mathop{\mathrm{span}}K\).
  \item
    \(K\) is an orthonormal \emph{basis}
  \item
    Parseval: equality in Bessel,
    i.e.~\({\left\lVert {f} \right\rVert} = \sum {\left\lvert { {\left\langle {f},~{\phi_k} \right\rangle} } \right\rvert}^2\)
  \end{itemize}
\item
  Lemma: if \(M, N\leq {\mathcal{H}}\) with \(\dim M < \dim N\), then
  \(M^\perp \cap N \neq 0\).

  \begin{itemize}
  \tightlist
  \item
    Without loss of generality assume \(\dim N = n + 1\) where
    \(n\coloneqq\dim M\), take a basis
    \(\left\{{\psi_k}\right\}_{k\leq n+1}\) for \(N\).
  \item
    Try to find \(f\in N\) with \(f\perp M\), i.e.~coefficients
    \(\left\{{b_i}\right\}_{i\leq n}\) with
    \(\sum b_i \psi_i \perp \phi_k\) for every \(\phi_k\) basis elements
    of \(M\).
  \item
    This is a linear system of \(n\) equations in \(n+1\) unknowns, so
    it has a nontrivial solution.
  \end{itemize}
\item
  Theorem, orthonormal bases are stable: if \(\left\{{\phi_k}\right\}\)
  is an orthonormal basis and \(\left\{{\psi_k}\right\}\) is an
  orthonormal \emph{system}, if
  \(\sum {\left\lVert {\phi_k - \psi_k} \right\rVert}^2 < \infty\) then
  \(\left\{{\psi_k}\right\}\) is a basis.

  \begin{itemize}
  \tightlist
  \item
    Assume note, then find a \(\psi_0 \in {\mathcal{H}}\) with
    \({\left\lVert {\psi_0} \right\rVert} = 1\) and
    \({\left\langle {\psi_0},~{\psi_j} \right\rangle} = 0\) for all
    \(j\).
  \item
    Choose \(N\gg 1\) so that
    \(\sum_{k\geq N} {\left\lVert {\psi_k - \phi_k} \right\rVert} < 1\).
  \item
    Use previous lemma to produce
    \(w\in \mathop{\mathrm{span}}\left\{{\psi_0, \psi_1,\cdots, \psi_N}\right\}\)
    with \(w\neq 0\) and \(w\perp \phi_j\) for all \(j\leq N\).
  \item
    Note
    \(w\perp \mathop{\mathrm{span}}\left\{{\psi_n}\right\}_{n > N}\).
  \item
    Apply Parseval:
    \begin{align*}
    0 
    &< {\left\lVert {w} \right\rVert}^2 \\
    &= \sum_{n\geq 1} {\left\lvert {{\left\langle {w},~{\phi_n} \right\rangle}} \right\rvert}^2 \\
    &= \sum_{n\geq N+1}{\left\lvert {{\left\langle {w},~{\phi_n} \right\rangle}} \right\rvert}^2 \\
    &= \sum_{n\geq N+1} {\left\lvert {{\left\langle {w},~{\phi_n - \psi_n} \right\rangle}} \right\rvert}^2 \\
    &\leq {\left\lVert {w} \right\rVert}^2 \sum_{n\geq N+1}{\left\lVert {\phi_n - \psi_n} \right\rVert}^2 \\
    &< {\left\lVert {w} \right\rVert}^2 \cdot 1
    ,\end{align*}
    where we've used that
    \({\left\langle {w},~{\psi_n} \right\rangle} = 0\) for \(n\geq N\).
    \(\contradiction\)
  \end{itemize}
\item
  \({\mathcal{H}}\) admits a countable orthonormal basis iff
  \({\mathcal{H}}\) is separable.

  \begin{itemize}
  \tightlist
  \item
    \(\implies\): clear, since the basis is countable, and every element
    is a limit of partial sums against the basis.
  \item
    \(\impliedby\): Gram-Schmidt.

    \begin{itemize}
    \tightlist
    \item
      \(h_1 = \psi_1\) and
      \(\phi_1 = h_1/{\left\lVert {h_1} \right\rVert}\)
    \item
      \(h_n = \psi_n - \sum_{1\leq k\leq n-1} {\left\langle {\psi_k},~{\phi_k} \right\rangle} \phi_k\)
      and normalize.
    \end{itemize}
  \end{itemize}
\item
  Exercise: a closed subspace of a separable Hilbert space is separable.
\item
  Linearly isometric inner product spaces: \(E\sim F\) iff there is a
  map \(A: E\twoheadrightarrow F\) with

  \begin{itemize}
  \tightlist
  \item
    \(A(au + bv) = aAu + bAv\)
  \item
    \({\left\lVert {Au} \right\rVert}_F = {\left\lVert {u} \right\rVert}_E\)
  \end{itemize}
\item
  Theorem: if \({\mathcal{H}}_1, {\mathcal{H}}_2\) are infinite
  dimensional separable Hilbert spaces, then
  \({\mathcal{H}}_1 \sim {\mathcal{H}}_2\). Thus for any Hilbert space
  \({\mathcal{H}}\) over \({\mathbb{C}}\),
  \({\mathcal{H}}\sim \ell^2({\mathbb{C}})\).

  \begin{itemize}
  \tightlist
  \item
    Pick orthonormal bases
    \(\left\{{\phi_k}\right\} \subseteq {\mathcal{H}}_1, \left\{{\psi_k}\right\} \subseteq {\mathcal{H}}_2\).
  \item
    For \(u\in {\mathcal{H}}_1\), define
    \(Au \coloneqq\sum {\left\langle {u},~{\phi_k} \right\rangle}\psi_k\),
    which converges -- this will be the linear isometry, and satisfies
    condition (i).
  \item
    Check
    \({\left\lVert {Au} \right\rVert}_{{\mathcal{H}}_2}^2 = \sum_{k\geq 1}{\left\lvert {{\left\langle {u},~{\phi_k} \right\rangle}} \right\rvert}^2 = {\left\lVert {u} \right\rVert}_{{\mathcal{H}}_2}\),
    which is condition (ii).
  \item
    Check \(A\) is surjective: for \(y\in {\mathcal{H}}_2\), write
    \(y = \sum_{k} {\left\langle {y},~{\psi_k} \right\rangle} \psi_k = Av\)
    for
    \(v\coloneqq\sum_k {\left\langle {y},~{\psi_k} \right\rangle}\phi_k \in {\mathcal{H}}_1\).
  \end{itemize}
\item
  Non-separable spaces: look at \emph{almost-periodic functions}.

  \begin{itemize}
  \tightlist
  \item
    E.g. \(\sum_{k\leq n} c_k \exp(i\lambda_k t)\) for
    \(\lambda_k \in {\mathbb{R}}\).
  \end{itemize}
\end{itemize}

\end{remark}

\hypertarget{tuesday-february-08}{%
\section{Tuesday, February 08}\label{tuesday-february-08}}

\begin{remark}

Motivating question: when is an operator equation solvable? Today:
relation between boundedness and continuity for linear operators.
Nonlinear operators next week.

\begin{itemize}
\tightlist
\item
  A map of vector spaces \(V\to W\) is a linear map defined on some
  domain \(D(A) \subseteq V\), where \(D(A)\) need not equal \(V\).

  \begin{itemize}
  \tightlist
  \item
    Notation: \(A(f) = Af = g\).
  \item
    \(Af \subseteq W\) is the image of \(f\), and
    \(R(A) \coloneqq\left\{{Af{~\mathrel{\Big\vert}~}f\in D(A)}\right\} \subseteq W\)
    is the range. Preimages of \(S \subseteq W\):
    \(A^{-1}(S) = \left\{{f{~\mathrel{\Big\vert}~}f\in D(A) \text{ and } Af\in S}\right\}\).
  \end{itemize}
\item
  We distinguish operators with different domains,
  e.g.~\(Af \coloneqq f'\) can be the formula for distinct operators
  \(A_1, A_2\) where \(D(A_1) = C^\infty[0, 1] \subseteq C^0[0, 1]\) or
  \(D(A_2) = C^1[0, 1] \subseteq C^0[0, 1]\), so \(A_1\neq A_2\).

  \begin{itemize}
  \tightlist
  \item
    I.e. \(A_1 = A_2 \iff D(A_1) = D(A_2)\) and \(A_1 f = A_2 f\) for
    all \(f\in D(A_1)=D(A_2)\).
  \item
    If \(A_1 f = A_2 f\) with \(D(A_1) \subseteq D(A_2)\), say \(A_2\)
    is an extension of \(A_1\). The extension is proper iff
    \(D(A_1)\neq D(A_2)\).
  \end{itemize}
\item
  Example operator: the Laplace equation \(\Delta f= g\) where
  \(\Delta= {\partial}_{xx} + {\partial}_{yy}\). We can take domains
  \(g\in C[0, 1], L^2[0, 1], H^2[0, 1] = \left\{{f\in L^2(0, 1) {~\mathrel{\Big\vert}~}{\partial}_{xx} f, {\partial}_{yy} f\in L^2(0, 1) }\right\}\).

  \begin{itemize}
  \tightlist
  \item
    Why domains matter: boundary conditions affect what eigenfunctions
    you get. Examples where \(A_1\neq A_2\):
  \item
    Dirichlet boundary conditions:
    \(\Delta f = g, { \left.{{f}} \right|_{{{{\partial}}\Omega}} } = 0\).
    The relevant solution spaces is
    \(D(A_1) = \left\{{\phi\in C^2[0,1]^2 {~\mathrel{\Big\vert}~}{ \left.{{\phi }} \right|_{{{{\partial}}\Omega}} } = 0}\right\}\)
    for \(A_1 \phi \coloneqq\Delta\phi\).
  \item
    Neumann boundary conditions:
    \(\Delta f = g, { \left.{{ {\frac{\partial f}{\partial \mathbf{n}}\,}}} \right|_{{{{\partial}}\Omega}} } = 0\),
    i.e.~there is no flux across the boundary. The relevant solution
    space is
    \(D(A_2) = \left\{{\psi \in C^2[0,1]^2{~\mathrel{\Big\vert}~}{ \left.{{ {\frac{\partial \psi}{\partial \mathbf{n}}\,} }} \right|_{{{{\partial}}\Omega}} } = 0 }\right\}\)
    for \(A_2 \psi \coloneqq\Delta\psi\).
  \end{itemize}
\item
  Injectivity: for \(A: V\to W\), for every \(g\in R(A)\) there is
  exactly one \(f\in D(A)\) with \(Af=g\). Equivalently for linear
  operators, \(Af = 0 \implies f=0\).
\item
  Surjectivity: \(R(A) = W\).
\item
  Example: \(A\coloneqq x\mapsto \sin(x)\) regarded as a function
  \(A:{\mathbb{R}}\to {\mathbb{R}}\) is neither surjective nor
  injective: \(R(A) = [-1, 1] \subsetneq {\mathbb{R}}\), and
  \(\sin(\pi {\mathbb{Z}}) = 0\).
\item
  Linearity: for \(Lf = g\), \(L\) is linear if
  \(L(af + bg) = aLf + bLg\).
\end{itemize}

\end{remark}

\begin{exercise}[?]

Show that the following are equivalent conditions for continuity of
\(A: V\to W\) at \(f_0\in D(A)\):

\begin{itemize}
\tightlist
\item
  \({\left\lVert {Af - Af_0} \right\rVert}_W < {\varepsilon}\) for all
  \(f\in D(A)\) with \({\left\lVert {f-f_0} \right\rVert}_V < \delta\)
\item
  For every sequence \(\left\{{f_k}\right\} \to f_0\),
  \(Af_k \to Af_0\).
\item
  Preimages of open sets in \(W\) are again open in \(V\).
\end{itemize}

\end{exercise}

\hypertarget{tuesday-february-15}{%
\section{Tuesday, February 15}\label{tuesday-february-15}}

\begin{remark}

Last time:

\begin{itemize}
\tightlist
\item
  Continuous operators are bounded:

  \begin{itemize}
  \tightlist
  \item
    If \({\left\lVert {Lf_n} \right\rVert} = 1\) and
    \({\left\lVert {f_n} \right\rVert} \to 0\), check
    \(\lim (Lf_n) = L(\lim f_n) = L0 = 0\).
  \item
    Take norms to contract \({\left\lVert {Lf_n} \right\rVert} = 1\).
  \end{itemize}
\end{itemize}

\end{remark}

\begin{theorem}[3.4.4]

If \(L: B\to C\) with dense image (so
\({ \operatorname{cl}} _B(D(L)) = B\)), if \(L\) is continuous on
\(D(L)\) then it has a unique extension \(\tilde L\) to all of \(B\), so
\(D(\tilde L) = B\), with
\({\left\lVert {L} \right\rVert} = {\left\lVert {\tilde L} \right\rVert}\).

\end{theorem}

\begin{proof}[of theorem]

In steps:

\begin{itemize}
\tightlist
\item
  Defining the extension:

  \begin{itemize}
  \tightlist
  \item
    For \(f\in B\), pick \(f_n \to f\) with \(f_n \in D(L)\) using
    density.
  \item
    Convergent implies Cauchy, so estimate:
    \begin{align*}
    {\left\lVert {Lf_n - Lf_m} \right\rVert} = {\left\lVert {L(f_n - f_m)} \right\rVert} \leq {\left\lVert {L} \right\rVert} {\left\lVert {f_n - f_m} \right\rVert} \to 0
    .\end{align*}
  \item
    Thus \(Lf_n\) is Cauchy, by completeness \(Lf_n \to g\) for some
    \(g\).
  \end{itemize}
\item
  Preservation of norm:

  \begin{itemize}
  \tightlist
  \item
    Define the extension as \(\tilde L f \coloneqq g\), by continuity it
    is independent of the sequence \(\left\{{f_k}\right\}\to f\).
  \item
    Check that \(\tilde L\) is a bounded operator:
    \begin{align*}
    {\left\lVert {\tilde L f} \right\rVert} \coloneqq{\left\lVert {g} \right\rVert} = {\left\lVert {\lim L f_n} \right\rVert} = \lim {\left\lVert {Lf_n} \right\rVert} \leq \lim {\left\lVert {L} \right\rVert} {\left\lVert {f_n} \right\rVert} = {\left\lVert {L} \right\rVert} {\left\lVert {f} \right\rVert} \\
    \implies {\left\lVert {\tilde L} \right\rVert} \leq {\left\lVert {L} \right\rVert}
    .\end{align*}
  \item
    Since \({\left\lVert {A} \right\rVert}_{^{\operatorname{op}}}\) is
    defined in terms of sups over test functions in \(D(A)\) for any
    operator \(A\) and here \(D(\tilde L) \supseteq D(L)\) is a larger
    set, we have
    \({\left\lVert {\tilde L} \right\rVert} \geq {\left\lVert {L} \right\rVert}\)
    by definition, yielding
    \({\left\lVert {\tilde L} \right\rVert} = {\left\lVert {L} \right\rVert}\).
  \end{itemize}
\item
  Uniqueness of the extension:

  \begin{itemize}
  \tightlist
  \item
    Take \(\tilde L_1, \tilde L_2\) extending \(L\), then
    \begin{align*}
    \tilde L_1 f = \lim \tilde L_1 f_n = \lim L f_n = \lim \tilde L_2 f_n = \tilde L_2 f
    .\end{align*}
  \item
    Use linearity:
    \begin{align*}
    \tilde L_1 f - \tilde L_2 f = (\tilde L_1 - \tilde L_2)f = 0 \implies \tilde L_1 - \tilde L_2 = 0
    .\end{align*}
  \end{itemize}
\end{itemize}

\end{proof}

\begin{example}[?]

Let \({\mathcal{L}}\in L({\mathbb{C}}^n, {\mathbb{C}}^n)\) be defined in
coordinates by
\(({\mathcal{L}}f)_i \coloneqq\sum_{1\leq j\leq n} \alpha_{ij} f_j\) for
\(1\leq i\leq n\). Take
\({\left\lVert {{-}} \right\rVert}_{\ell^\infty}\) and check
\begin{align*}
{\left\lVert {Lf} \right\rVert}_\infty 
&\coloneqq\sup_i {\left\lvert {\sum \alpha_{ij} f_j} \right\rvert} \\
&\leq \qty{ \sup_i \sum {\left\lvert {\alpha_{ij} } \right\rvert} } \sup_j {\left\lvert {f_j} \right\rvert} \\
&\coloneqq m {\left\lVert {f} \right\rVert}_{\ell^\infty}
.\end{align*}
So \({\left\lVert {L} \right\rVert} \leq m\), where \(m\) is the largest
row sum. Is there an \(f\) for which equality holds? In this case, we'd
need
\begin{align*}
{\left\lVert {Lf} \right\rVert}_{\ell^\infty} \geq m {\left\lVert {f} \right\rVert}_{\ell^\infty}
.\end{align*}
Identify the row \(k\) so that
\(m = \sum_{1\leq j\leq n} {\left\lvert {\alpha_{kj}} \right\rvert}\).
Set \(f\) to be a unit vector with coefficients
\((f)_j = \mkern 1.5mu\overline{\mkern-1.5mu\alpha_{kj}\mkern-1.5mu}\mkern 1.5mu / {\left\lvert {\alpha_{kj}} \right\rvert}\).
Then
\begin{align*}
{\left\lVert {Lf} \right\rVert}_\infty
&= \sup_i {\left\lvert { \sum_j \alpha_{ij} f_j } \right\rvert} \\
&\geq {\left\lvert {\sum_j \alpha_{kj} f_j } \right\rvert} \\
&= {\left\lvert {\sum_j \alpha_{kj} {\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu_{kj} / {\left\lvert {\alpha_{k j}} \right\rvert} } } \right\rvert} \\
&= \sum_j {\left\lvert {\alpha_{kj}} \right\rvert} \\
&= m {\left\lVert {f} \right\rVert}_{\ell^\infty}
.\end{align*}

So the answer is yes in this case. Does this also work for
\({\left\lVert {{-}} \right\rVert}_{\ell^p}\) with \(p\in (1, \infty)\)?
Recall Holder's inequality:
\begin{align*}
{\left\lvert {\sum \alpha_{ij} f_j } \right\rvert} 
&\leq \qty{ \sum {\left\lvert {\alpha_{ij}} \right\rvert}^q }^{1\over q} \qty{\sum {\left\lvert {f_j} \right\rvert}^p }^{1\over p} \\
&= \qty{ \sum {\left\lvert {\alpha_{ij}} \right\rvert}^q }^{1\over q} {\left\lVert {f} \right\rVert}_{\ell^p}
.\end{align*}
Check that
\begin{align*}
{\left\lVert {Lf} \right\rVert}_{\ell^p}^p
&= \sum_i {\left\lvert {(Lf)_i } \right\rvert}^p \\
&= \sum_i {\left\lvert {\sum_j \alpha_{ij} f_j } \right\rvert}^p \\
&\leq \sum_i \qty{ \sum_j {\left\lvert {\alpha_{ij} } \right\rvert} }^{p\over q} {\left\lVert {f} \right\rVert}_{\ell^p}^p
,\end{align*}
where we've applied Holder in the last line. Thus
\begin{align*}
{\left\lVert {L} \right\rVert} \leq \qty{ \sum_i \qty{ \sum_j {\left\lvert { \alpha_{ij}} \right\rvert} }^{p\over q} }^{1\over p}
.\end{align*}

\begin{exercise}[?]

Is there an \(f\) that attains this bound in the \(\ell^p\) case?

\end{exercise}

\end{example}

\begin{remark}

For \({\mathcal{L}}\in L({\mathbb{C}}^\infty, {\mathbb{C}}^\infty)\)
defined by \((Lf)_i = \sum_{j\geq 11} \alpha_{ij} f_j\) for \(j\geq 1\),
one needs a notion of convergence of the coordinates \(\alpha_{ij}\) in
order for \({\mathcal{L}}\) to be bounded. A sufficient condition is
\(m\coloneqq\sup_i \sum_{j\geq 1} {\left\lvert {\alpha_{ij}} \right\rvert} < \infty\).

\end{remark}

\begin{definition}[?]

Some notation:

\begin{align*}
{\left\lVert \left\lVert {\alpha} \right\rVert\right\rVert}_1 &\coloneqq\sup_j \sum_i {\left\lvert { \alpha_{ij}} \right\rvert} \\
{\left\lVert \left\lVert {\alpha} \right\rVert\right\rVert}_p &\coloneqq\qty{ \sum_i \qty{ \sum_j {\left\lvert {\alpha_{ij} } \right\rvert}^{q} }^{p\over q} }^{1\over p} \\
{\left\lVert \left\lVert {\alpha} \right\rVert\right\rVert}_\infty &\coloneqq\sup_i \sum_j {\left\lvert { \alpha_{ij}} \right\rvert}
.\end{align*}

\end{definition}

\begin{remark}

Note that if \({\mathcal{L}}: \ell^p\to \ell^p\), then
\({\left\lVert {L} \right\rVert} \leq {\left\lVert \left\lVert {\alpha} \right\rVert\right\rVert}_p\)
for \(p\in [1, \infty)\) and for \(p=\infty\) this is an equality.

\end{remark}

\begin{example}[Kernels]

Consider \(C[a, b]\) with \({\left\lVert {{-}} \right\rVert}_\infty\)
and
\(k\in C^0( [a,b]{ {}^{ \scriptscriptstyle\times^{2} } }, {\mathbb{C}})\).
Define
\begin{align*}
K: C[a, b] &\to C[a, b] \\
f &\mapsto \int_a^b k(x, y) f(y) \,dy
.\end{align*}
What is \({\left\lVert {K} \right\rVert}\)? Estimate
\begin{align*}
{\left\lVert {Kf} \right\rVert} 
&\leq \sup_{y\in [a, b]}{\left\lvert {f(y)} \right\rvert}  \sup_{x\in [a, b]} \int_a^b {\left\lvert {k(x, y)} \right\rvert} \,dy\\
&\leq {\left\lVert {f} \right\rVert}_\infty {\left\lVert {k} \right\rVert}_\infty
,\end{align*}
so
\({\left\lVert {K} \right\rVert} \leq {\left\lVert {k} \right\rVert}_\infty\).

Define
\begin{align*}
{\left\lVert \left\lVert {k} \right\rVert\right\rVert}_1 &\coloneqq\sup_y \int {\left\lvert {k} \right\rvert} \,dx\\
{\left\lVert \left\lVert {k} \right\rVert\right\rVert}_p &\coloneqq\qty{\int \qty{ \int {\left\lvert {k} \right\rvert}^q \,dy}^{p\over q} \,dx}^{1\over p} \\
{\left\lVert \left\lVert {k} \right\rVert\right\rVert}_\infty &\coloneqq\sup_x \int {\left\lvert {k} \right\rvert} \,dy
.\end{align*}

\end{example}

\addsec{ToDos}
\listoftodos[List of Todos]
\cleardoublepage

% Hook into amsthm environments to list them.
\addsec{Definitions}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={definition}, numwidth=3.5em]
\cleardoublepage

\addsec{Theorems}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em]
\cleardoublepage

\addsec{Exercises}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={exercise}, numwidth=3.5em]
\cleardoublepage

\addsec{Figures}
\listoffigures
\cleardoublepage



\newpage
\printbibliography[title=Bibliography]


\end{document}
