diff --git a/functional_analysis.pdf b/functional_analysis.pdf new file mode 100644 index 0000000..f6adde4 Binary files /dev/null and b/functional_analysis.pdf differ diff --git a/functional_analysis.tex b/functional_analysis.tex new file mode 100644 index 0000000..d52fbed --- /dev/null +++ b/functional_analysis.tex @@ -0,0 +1,294 @@ +\documentclass{article} + +% preamble +\def\npart{III} +\def\nyear{2023} +\def\nterm{Michaelmas} +\def\nlecturer{Dr Andr\'as Zs\'ak} +\def\ncourse{Functional Analysis} +\def\draft{Incomplete} + +\usepackage{mathrsfs} +\usepackage{imakeidx} +\usepackage{marginnote} +\usepackage{mathdots} +\usepackage{tabularx} +\usepackage{ifthen} + +\input{header} +\swapnumbers +\reversemarginpar + +\usetikzlibrary{positioning, decorations.pathmorphing, decorations.text, calc, backgrounds, fadings} +\tikzset{node/.style = {circle,draw,inner sep=0.8mm}} + +\makeindex[intoc] + +\setcounter{section}{-1} + +% and here we go! +\begin{document} +\maketitle + +\tableofcontents + +\clearpage + +\section{Introduction} + +\subsection*{Prerequisites} + +\begin{itemize} + \item some basic functional analysis + \item a bit of measure theory + \item a bit of complex analysis +\end{itemize} + +\subsection*{Books} + +Books relevant to the course are: +\begin{itemize} + \item Bollob\'as, \textit{Linear Analysis} + \item Murphy, \textit{$C^*$-algebras} + \item Rudin + \item Graham-Allan +\end{itemize} + +\subsection*{Notation} + +We will use $\K$ to mean "either $\R$ or $\C$". + +For $X$ a normed space, we define +$$B_X = \{x \in X | \norm x \le 1\}$$ +$$S_X = \{x \in X | \norm x = 1\}$$ + +For $X, Y$ normed spaces, we write $X \sim Y$ if $X, Y$ are isomorphic, ie there +exists a linear bijection $T : X \mor Y$ such that $T$ and $T^{-1}$ are continuous. We write $X \cong Y$ if $X, Y$ are isometrically isomorphic, ie there exists a surjective linear map $T : X \mor Y$ such that $\norm{T x} = \norm x$ for all $x$. + +\clearpage + +\section{Hahn-Banach extension theorems} + +\newlec + +Let $X$ be a normed space. The {\bf dual space} of $X$ is the space $X^*$ of +bounded linear functionals on $X$. +$X^*$ is always a Banach space in the operator norm: for $f \in X^*$, +$$\norm f = \sup_{x \in B_X} |f(x)|$$ + +\begin{eg} + For $1 < p, q < \infty$, $p^{-1} + q^{-1} = 1$, $\ell_p^* \cong \ell_q$. \\ + We also have $\ell_1^* \cong \ell_\infty$, $c_0^* \cong \ell_1$. \\ + If $H$ is a Hilbert space, then $H^* \cong H$ (the isomorphism is conjugate-linear in the complex case). +\end{eg} + +For $x \in X, f \in X^*$, we write $\langle x, f\rangle = f(x)$. Note that +$$\langle x, f\rangle = |f(x)| \le \norm f\norm x$$ + +\begin{defi} + Let $X$ be a {\it real} vector space. A functional $p : X \mor \R$ is + \begin{itemize} + \item {\bf positive homogeneous} if $p(tx) = tp(x)$ for all $x \in X$, $t \ge 0$ + \item {\bf subadditive} if $p(x + y) \le p(x) + p(y)$ for all $x, y \in X$ + \end{itemize} +\end{defi} + +\begin{defi} + Let $P$ be a preorder, $A \subseteq P, x \in P$. We say + \begin{itemize} + \item $x$ is an {\bf upper bound} for $A$ if $\for a \in A, a \le x$. + \item $A$ is a {\bf chain} if $\for a, b \in A, a \le b \lor b \le a$. + \item $x$ is a {\bf maximal element} if $\for y \in P, x \not < y$ + \end{itemize} +\end{defi} + +\begin{fact}[Zorn's lemma] + A nonempty preorder in which all nonempty chains have an upper bound has a maximal element. +\end{fact} + +\begin{thm}[Hahn-Banach, positive homogeneous version]\label{thm:hb-positive} + Let $X$ be a real vector space and $p : X \mor \R$ be positive homogeneous and subadditive. Let $Y$ be a subspace of $X$ and $g : Y \mor \R$ be linear such that $\for y \in Y, g(y) \le p(y)$. Then there exists $f : X \mor \R$ linear such that $f\restriction_Y = g$ and $\for x \in X, f(x) \le p(x)$. +\end{thm} +\begin{proof} + Let $P$ be the set of pairs $(Z, h)$ where $Z$ is a subspace of $X$ with $Y \subseteq Z$ and $h : Z \mor \R$ linear, $h\restriction_Y = g$ and $\for z \in Z, h(z) \le p(z)$. $P$ is nonempty since $(Y, g) \in P$, and is partially ordered by + $$(Z_1, h_1) \le (Z_2, h_2) \iff Z_1 \subseteq Z_2 \land h_2\restriction_{Z_1} = h_1$$ + If $\{(Z_i, h_i) | i \in I\}$ is a chain with $I$ nonempty, then we can define + $$Z := \bigcup_{i \in I} Z_i, h\restriction_{Z_i} = h_i$$ + The definition of $h$ makes sense thanks to the chain assumption. $(Z, h) \in P$ is therefore an upper bound for the chain. \\ + Hence find by Zorn a maximal element $(Z, h)$ of $P$. If $Z = X$, we won. So assume there is some $x \in X \ Z$. Let $W = \Span(Z \cup \{x\})$ and define $f : W \mor \R$ by + $$f(z + \lambda x) = h(z) + \lambda\alpha$$ + for some $\alpha \in \R$. Then $f$ is linear and $f\restriction_Z = h$. We now look for $\alpha$ such that $\for w \in W, f(w) \le p(w)$. We would then have $(W, f) \in P$ and $(Z, h) < (W, f)$, contradicting maximality of $(Z, h)$. \\ + We need + $$h(z) + \lambda\alpha \le p(z + \lambda x) \for z \in Z, \lambda \in \R$$ + Since $p$ is positive homogeneous, this becomes + \begin{align} + h(z) + \alpha \le p(z + x) + h(z) - \alpha \le p(z - x) + \end{align} + ie + $$h(z) - p(z - x) \le \alpha \le p(z + x) - h(z) \for z \in Z$$ + The existence of $\alpha$ now amounts to + $$h(z_1) - p(z_1 - x) \le \alpha \le p(z_2 + x) - h(z_2) \for z_1, z_2 \in Z$$ + But indeed + $$h(z_1) + h(z_2) = h(z_1 + z_2) \le p(z_1 + z_2) \le p(z_1 - x) + p(z_2 + x)$$ +\end{proof} + +\begin{defi} + Let $X$ be a $\K$-vector space. A {\bf seminorm} on $X$ is a functional $p : X \mor \R$ such that + \begin{itemize} + \item $\for x \in X, p(x) \ge 0$ + \item $\for x \in X, \lambda \in \K, p(\lambda x) = |\lambda| p(x)$ + \item $\for x, y \in X, p(x + y) \le p(x) + p(y)$ + \end{itemize} +\end{defi} + +\begin{rmk} + $$\text{norm} \implies \text{seminorm} \implies \text{positive homogeneous}$$ +\end{rmk} + +\newlec + +\begin{thm}[Hahn-Banach, absolute homogeneous version]\label{thm:hb-absolute} + Let $X$ be a real of complex vector space and $p$ a seminorm on $X$. Let $Y$ be a subspace of $X$, $g$ a linear functional on $Y$ such that $\for y \in Y, |g(y)| \le p(y)$. Then there exists a linear functional $f$ on $X$ such that $f\restriction_Y = g$ and $\for x \in X, |f(x)| \le p(x)$. +\end{thm} +\begin{proof}~\\ + {\bf Real case} + $$\for y \in Y, g(y) \le |g(y)| \le p(y)$$ + By Theorem \ref{thm:hb-positive}, there exists $f : X \mor \R$ such that $f\restriction_Y = g$ and $\for x \in X, f(x) \le p(x)$. We also have + $$\for x \in X, -f(x) = f(-x) \le p(-x) = p(x)$$ + Hence $|f(x)| \le p(x)$ \\ + {\bf Complex case} \\ + $\Re g : Y \mor \R$ is real-linear. + $$\for y \in Y, |\Re g(y)| \le |g(y)| \le p(y)$$ + By the real case, find $h : X \mor \R$ real-linear such that $h\restriction_Y = \Re g$ + \begin{claim} + There exists a unique complex-linear $f : X \mor \C$ such that $h = \Re f$. + \end{claim} + \begin{proof}~\\ + {\bf Uniqueness} \\ + If we have such $f$, then + \begin{eqnarray*} + f(x) + & = & \Re f(x) + i\Im f(x) \\ + & = & \Re f(x) - i\Re f(ix) \\ + & = & h(x) - ih(ix) + \end{eqnarray*} + {\bf Existence} \\ + Define $f(x) = h(x) - ih(ix)$. Then $f$ is real-linear and $f(ix) = if(x)$, so $f$ is complex-linear with $\Re f = h$. + \end{proof} + We now have $f : X \mor \C$ such that $\Re f = h$. + $$\Re f\restriction_Y = h\restriction_Y = \Re g$$ + So, by uniqueness, $f\restriction_Y = g$. \\ + Given $x \in X$, find $\lambda$ with $|\lambda| = 1$ such that + \begin{eqnarray*} + |f(x)| + & = & \lambda f(x) \\ + & = & f(\lambda x) \\ + & = & \Re f(\lambda x) \\ + & = & h(\lambda x) \\ + & \le & p(\lambda x) \\ + & = & p(x) + \end{eqnarray*} +\end{proof} + +\begin{rmk} + For a complex vector space $X$, if we write $X_\R$ for $X$ considered as a real vector space, the above proof shows that + $$\Re : (X^*)_\R \mor X_\R^*$$ + is an isometric isomorphism. +\end{rmk} + +\begin{cor}\label{cor:hb-point} + Let $X$ be a $\K$-vector space, $p$ a seminorm on $X$, $x_0 \in X$. Then there exists a linear functional $f$ on $X$ such that $f(x_0) = p(x_0)$ and $\for x \in X, |f(x)| \le p(x)$. +\end{cor} +\begin{proof} + Let $Y = \Span(x_0)$, + \begin{eqnarray*} + g : Y & \mor & \K \\ + \lambda x_0 & \mapsto & \lambda p(x_0) + \end{eqnarray*} + We see that $\for y \in Y, g(y) \le p(y)$. Hence find by Theorem \ref{thm:hb-absolute} a linear functional $f$ on $X$ such that $f\restriction_Y = g$ and $\for x \in X, |f(x)| \le p(x)$. We check that $f(x_0) = g(x_0) = p(x_0)$. +\end{proof} + +\begin{thm}[Hahn-Banach, existence of support functionals] + Let $X$ be a real or complex normed space. Then + \begin{enumerate} + \item If $Y$ is a subspace of $X$ and $g \in Y^*$, then there exists $f \in X^*$ such that $f\restriction_Y = g$ and $\norm f = \norm g$. + \item Given $x_0 \ne 0$, there exists $f \in S_{X^*}$ such that $f(x_0) = \norm{x_0}$. + \end{enumerate} +\end{thm} +\begin{proof}~ + \begin{enumerate} + \item Let $p(x) = \norm g \norm x$. Then $p$ is a seminorm on $X$ and + $$\for y \in Y, |g(y)| \le \norm g \norm y = p(y)$$ + Find by Theorem \ref{thm:hb-positive} a linear functional $f$ on $X$ such that $f\restriction_Y = g$ and $\for x \in X, |f(x)| \le p(x) = \norm g \norm x$. So $\norm f \le \norm g$. Since $f\restriction_Y = g$, we also have $\norm g \le \norm f$. Hence $\norm f = \norm g$. + \item Apply Corollary \ref{cor:hb-point} with $p(x) = \norm x$ to get $f \in X^*$ such that + $$\for x \in X, |f(x)| \le \norm x \text{ and } f(x_0) = \norm{x_0}$$ + It follows that $\norm f = 1$. + \end{enumerate} +\end{proof} + +\begin{rmks}~ + \begin{itemize} + \item Part 1 is a sort of linear version of Tietze's extension theorem: Given $K$ compact Hausdorff, $L \subseteq K$ closed, $g : L \mor \K$ continuous, there exists $f : K \mor \K$ such that $f\restriction_L = g$ and $\norm f_\infty = \norm g_\infty$. + \item Part 2 shows that for all $x \ne y$ in $X$ there exists $f \in X^*$ such that $f(x) \ne f(y)$, namely $X^*$ {\bf separates points} of $X$. This is a sort of linear version of Urysohn: $C(K)$ separates points of $K$. + \item The $f$ in part 2 is called a {\bf norming functional}, aka {\bf support functional}, for $x_0$. The existence of support functionals shows that + $$x_0 = \max_{g \in B_{X^*}} \langle x_0, g\rangle$$ + Assuming $X$ is a real normed space and $\norm{x_0} = 1$, we have $B_X \subseteq \{x \in X| f(x) \le 1\}$. Visually, TODO: insert tangency diagram + \end{itemize} +\end{rmks} + +\subsection{Bidual} + +Let $X$ be a normed space. Then $X^{**}$ is called the {\bf bidual} or {\bf second dual} of $X$. + +For $x \in X$, define $\hat x : X^* \mor \K$, the {\bf evaluation at $x$}, by $\hat x(f) = f(x)$. $\hat x$ is linear and $|\hat x(f)| = |f(x)| \le \norm f \norm x$, so $\hat x \in X^{**}$ and $\norm{\hat x} \le \norm x$. + +The map $x \mapsto \hat x : X \mor X^{**}$ is called the {\bf canonical embedding} of $X$ into $X^{**}$. + +\begin{thm}\label{thm:can-emb} + The canonical embedding is an isometric embedding. +\end{thm} +\begin{proof}~\\ + {\bf Linearity} + \begin{eqnarray*} + \widehat{x + y}(f) & = & f(x + y) = f(x) + f(y) = \hat x(f) + \hat y(f) \\ + \widehat{\lambda x}(f) & = & f(\lambda x) = \lambda f(x) = \lambda \hat x(f) + \end{eqnarray*} + {\bf Isometry} \\ + If $x \ne 0$, there exists a support functional $f$ for $x$. Then + $$ \norm{\hat x} \ge |\hat x(f)| = |f(x)| = \norm x$$ +\end{proof} + +\begin{rmks}~ + \begin{itemize} + \item In bracket notation, $\langle f, \hat x\rangle = \langle x, f\rangle$ + \item Let $\hat X$ be the image of $X$ in $X^{**}$. Theorem \ref{thm:can-emb} says + $$X \cong \hat X \subseteq X^{**}$$ + We often identify $\hat X$ with $X$ and think of $X$ as living isometrically inside $X^{**}$. Note that + $$X \text{ complete } \iff \hat X \text{ closed in } X^{**}$$ + \item More generally, $\bar{\hat X}$ is a Banach space containing an isometric copy of $X$ as a dense subspace. We proved that normed spaces have completions! + \end{itemize} +\end{rmks} + +\begin{defi} + A normed space $X$ is {\bf reflexive} if the canonical embedding $X \mor X^{**}$ is surjective. +\end{defi} + +\begin{egs}~ + \begin{itemize} + \item Some reflexive spaces are Hilbert spaces, finite-dimensional spaces, $\ell_p$ and $L_p(\mu)$ for $1 < p < \infty$. + \item Some non-reflexive spaces are $c_0, \ell_1, \ell_\infty, L_1[0, 1]$. + \end{itemize} +\end{egs} + +\begin{rmks}~ + \begin{itemize} + \item If $X$ is reflexive, then $X \cong X^{**}$, so $X$ is complete. + \item There are Banach spaces $X$ such that $X \cong X^{**}$ but $X$ is not reflexive, eg {\bf James' space}. Any isomorphism to the bidual is then necessarily not the canonical embedding. + \end{itemize} +\end{rmks} + +\newlec + +\printindex +\end{document} \ No newline at end of file