\input{"preamble.tex"} \addbibresource{FunctionalAnalysis.bib} \let\Begin\begin \let\End\end \newcommand\wrapenv[1]{#1} \makeatletter \def\ScaleWidthIfNeeded{% \ifdim\Gin@nat@width>\linewidth \linewidth \else \Gin@nat@width \fi } \def\ScaleHeightIfNeeded{% \ifdim\Gin@nat@height>0.9\textheight 0.9\textheight \else \Gin@nat@width \fi } \makeatother \setkeys{Gin}{width=\ScaleWidthIfNeeded,height=\ScaleHeightIfNeeded,keepaspectratio}% \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}