diff --git a/src/fa/norm/ap.tex b/src/fa/norm/ap.tex new file mode 100644 index 0000000..e98471c --- /dev/null +++ b/src/fa/norm/ap.tex @@ -0,0 +1,45 @@ +\section{The Approximation Property} +\label{section:approximation-property} + +\begin{definition}[Approximation Property] +\label{definition:approximation-property} + Let $E$ be a separated locally convex space over $K \in \RC$, then the following are equivalent: + \begin{enumerate} + \item The closure of $E^* \otimes E$ in $L_c(E; E)$ contains the identity map. + \item $E^* \otimes E$ is dense in $L_c(E; E)$. + \item For each locally convex space $F$ over $K$, $E^* \otimes F$ is dense in $L_c(E; F)$. + \item For each locally convex space $F$ over $K$, $F^* \otimes E$ is dense in $L_c(F; E)$. + \end{enumerate} + + If the above holds, then $E$ has the \textbf{approximation property}. +\end{definition} +\begin{proof} + (1) $\Rightarrow$ (2): Let $T \in L_c(E; E)$ and $A \subset E$ be precompact, then $T(A)$ is also precompact by \autoref{proposition:totally-bounded-image}. Let $U \in \cn_E(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in U$ for all $x \in T(A)$. In which case, $STx - Tx \in U$ for all $x \in A$. + + (1) $\Rightarrow$ (3): Let $T \in L_c(E; F)$ and $A \subset E$ be precompact, and $U \in \cn_F(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in T^{-1}(U)$ for all $x \in A$. In which case, $TS \in E^* \otimes F$ and $TSx - Tx \in U$ for all $x \in A$. + + (1) $\Rightarrow$ (4): Let $T \in L_c(F; E)$ and $A \subset F$ be precompact, then $T(A)$ is also precompact. Let $U \in \cn_E(0)$, then there exists $S \in E^* \otimes E$ such that $Sx - x \in U$ for all $x \in T(A)$. Thus $STx - Tx \in U$ for all $x \in A$. +\end{proof} + +\begin{proposition} +\label{proposition:approximation-property-associated} + Let $E$ be a locally convex space over $K \in \RC$. If there exists a fundamental system of convex and circled neighbourhoods $\fB \subset \cn_E(0)$ such that for each $V \in \fB$, $\wh E_V$ has the approximation property, then $E$ has the approximation property. +\end{proposition} +\begin{proof} + Let $V \in \fB$, $\pi_V: E \to \wh E_V$ be the canonical projection, and $A \subset E$ be precompact, then $\pi_V(A)$ is precompact as well. Since $\wh E_V$ has the approximation property, there exists $T \in E_V^* \otimes \wh E_V$ such that $Tx - x \in \pi_V(V)$ for all $x \in \pi_V(A)$. As $E_V$ is dense in $\wh E_V$, there exists $S \in E_V^* \otimes E_V$ such that $Sx - Tx \in \pi_V(V)$ for all $x \in \pi_V(A)$. In which case, $Sx - x \in 2\pi_V(V)$ for all $x \in \pi_V(A)$, and $S \circ \pi_V(x) - \pi_V(x) \in 2\pi_V(V)$. + + Write $S = \sum_{j = 1}^n \phi_j \otimes y_j$. For each $1 \le j \le n$, choose any representative $x_j \in \pi_V^{-1}(y_j)$, then for any $x \in A$, + \[ + \pi_V \braks{x - \sum_{j = 1}^n x_j\dpn{x, \phi_j \circ \pi_V}{E}} = S \circ \pi_V(x) - \pi_V(x) \in -2\pi_V(V) = 2\pi_V(V) + \] + + Finally, since $\ker(\pi_V) = \bigcap_{\lambda > 0}\lambda V \subset V$, $x - \sum_{j = 1}^n x_j\dpn{x, \phi_j \circ \pi_V}{E} \in -3V = 3V$. Therefore if $R = \sum_{j = 1}^n (\phi_j \circ \pi_V) \otimes x_j \in E^* \otimes E$, then $Rx - x \in 3V$. +\end{proof} + +\begin{corollary} +\label{corollary:approximation-property-hilbert} + Every subspace of a product of Hilbert spaces has the approximation property. Every subspace of a projective limit of Hilbert spaces has the approximation property. +\end{corollary} + + + diff --git a/src/fa/norm/compact.tex b/src/fa/norm/compact.tex new file mode 100644 index 0000000..d2f7905 --- /dev/null +++ b/src/fa/norm/compact.tex @@ -0,0 +1,28 @@ +\section{Compact Operators} +\label{section:compact-operator} + +\begin{definition}[Compact Operator] +\label{definition:compact-operator} + Let $E, F$ be locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then $T$ is \textbf{compact} if there exists $U \in \cn_E(0)$ such that $T(U)$ is relatively compact in $F$. + + The set $\mathcal{K}(E; F)$ is the \textbf{space of compact operators} from $E$ to $F$. +\end{definition} + +\begin{proposition} +\label{proposition:compact-normed-complete} + Let $E$ be a normed space over $K \in \RC$ and $F$ be a complete Hausdorff topological vector space over $K$, then $\mathcal{K}(E; F)$ is a closed subspace of $L_b(E; F)$. +\end{proposition} +\begin{proof} + Let $T \in \ol{\mathcal{K}(E; F)}$ and $U \in \cn_F(0)$ be circled, then there exists $S \in \mathcal{K}(E; F)$ such that $Sx - Tx \in U$ for all $x \in B_E(0, 1)$. Since $S$ is compact, there exists $Y \subset F$ finite with $S(B_E(0, 1)) \subset Y + U$. In which case, $T(B_E(0, 1)) \subset Y + U + U = Y + 2U$. Therefore $T(B_E(0, 1))$ is totally bounded, and as $F$ is complete, relatively compact in $F$. +\end{proof} + +\begin{theorem}[Schauder] +\label{theorem:compact-adjoint} + Let $E, F$ be normed vector spaces over $K \in \RC$ with $F$ being complete and $T \in L(E; F)$, then $T$ is compact if and only if $T^* \in L(F^*; E^*)$ is compact. +\end{theorem} +\begin{proof} + ($\Rightarrow$): Let $\cf \subset F^*$ be bounded, then $\cf$ is equicontinuous and hence relatively compact in the weak* topology by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Since $F$ is complete, $\ol{T(B_E(0, 1))}$ is compact. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\cf$ is relatively compact with respect to the topology of uniform convergence on $\ol{T(B_E(0, 1))}$. Thus $T^*(\cf) = \bracsn{\phi \circ T|\phi \in \cf}$ is relatively compact with respect to the topology of uniform convergence on $B_E(0, 1)$. In other words, $T^*(\cf)$ is relatively compact in $E^*$, and $T^*$ is a compact operator. + + ($\Leftarrow$): By \autoref{proposition:operator-space-completeness}, $E^*$ is complete. The preceding case then implies that $T^{**} \in L(E^{**}; F^{**})$ is compact. As such, its restriction to $E$, being identified with $T$, is also compact. +\end{proof} + diff --git a/src/fa/norm/index.tex b/src/fa/norm/index.tex index 503fcba..a7a7b70 100644 --- a/src/fa/norm/index.tex +++ b/src/fa/norm/index.tex @@ -7,3 +7,5 @@ \input{./separable.tex} \input{./multilinear.tex} \input{./hilbert.tex} +\input{./compact.tex} +\input{./ap.tex} diff --git a/src/fa/norm/separable.tex b/src/fa/norm/separable.tex index 1e8c57c..979cc64 100644 --- a/src/fa/norm/separable.tex +++ b/src/fa/norm/separable.tex @@ -56,4 +56,31 @@ then $\phi_x$ is Borel measurable with respect to the weak topology, so $B(x, r) = \bracs{\phi_x < r}$ is a Borel set with respect to the weak topology. \end{proof} +\begin{lemma} +\label{lemma:compact-embed} + Let $E$ be a normed vector space over $K \in \RC$ and $A \subset [0, 1]$ be closed, then $C(A; E)$ embeds isometrically into $C([0, 1]; E)$. +\end{lemma} +\begin{proof} + First note that if $0 \not\in A$ or $1 \not\in A$, $C(A; E)$ embeds isometrically into $C(A \cup \bracs{0, 1}; E)$ through extension by $0$. Thus assume without loss of generality that $A$ contains the endpoints $0$ and $1$. + + Let $U = [0, 1] \setminus A$, then there exists $\seq{(a_n, b_n)} \subset [0, 1]^2$ such that $U = \bigsqcup_{n \in \natp}(a_n, b_n)$. For each $f \in C(A; E)$, let + \[ + Tf: [0, 1] \to E \quad x \mapsto \begin{cases} + f(x) &x \in A \\ + \frac{b_n - x}{b_n - a_n}f(a_n) + \frac{x - a_n}{b_n - a_n}f(b_n) &x \in (a_n, b_n) \subset [0, 1] + \end{cases} + \] + + then the mapping $f \mapsto Tf$ is an isometric embedding into $E^{[0, 1]}$ with respect to the uniform norm. + + Since $U$ is open and $Tf$ is affine on each component of $U$, $Tf$ is continuous on $U$. It remains to show that $Tf$ is continuous on $A$. Let $x \in A$ and $\eps > 0$, then there exists $\delta > 0$ such that $\norm{f(y) - f(x)}_E < \eps$ for all $y \in (x -\delta, x + \delta) \cap A$. Now, a case analysis: + \begin{enumerate} + \item If there exists $y \in (x - \delta, x) \cap A$, then for any $z \in U \cap (y, x)$, there exists $n \in \natp$ such that $(a_n, b_n) \subset (y, x)$ and $z \in (a_n, b_n)$. In which case, since $\norm{f(a_n) - f(x)}_E < \eps$ and $\norm{f(b_n) - f(x)}_E < \eps$, $\norm{Tf(z) - Tf(x)}_E < \eps$. Thus $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x)$. + \item Otherwise, $x = 0$ or $Tf|_{(x - \delta, x)}$ is an affine function. Either way, there exists $y \in (x - \delta, x)$ such that $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x) \cap [0, 1]$. + \end{enumerate} + + Thus there exists $y \in (x - \delta, x)$ with $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (y, x) \cap [0, 1]$. Similarly, there exists $y' \in (x, x + \delta)$ with $\norm{Tf(z) - Tf(x)}_E < \eps$ for all $z \in (x, y') \cap [0, 1]$. Therefore $Tf$ is continuous at $x$. Since this holds for all $x \in U$ and $x \in A$, $Tf \in C([0, 1]; E)$. +\end{proof} + + diff --git a/src/fa/notation.tex b/src/fa/notation.tex index 0fd76d9..0184d46 100644 --- a/src/fa/notation.tex +++ b/src/fa/notation.tex @@ -17,6 +17,8 @@ $L_s(E; F)$ & $L(E; F)$ with strong operator topology. & \autoref{definition:strong-operator-topology} \\ $L_w(E; F)$ & $L(E; F)$ with weak operator topology. & \autoref{definition:weak-operator-topology} \\ $L_b(E; F)$ & $L(E; F)$ with topology of bounded convergence. & \autoref{definition:bounded-convergence-topology} \\ + $L_c(E; F)$ & $L(E; F)$ with topology of precompact convergence. & \autoref{definition:compact-operator-topology} \\ + $\mathcal{K}(E; F)$ & Space of compact operators from $E$ to $F$. & \autoref{definition:compact-operator} \\ $\widehat{E}$ & Hausdorff completion of TVS $E$. & \autoref{definition:tvs-completion} \\ % ---- Locally Convex ---- $\mathrm{Conv}(A)$ & Convex hull of $A$. & \autoref{definition:convex-hull} \\ diff --git a/src/fa/tvs/space-of-linear.tex b/src/fa/tvs/space-of-linear.tex index 38e4d47..dc56d1f 100644 --- a/src/fa/tvs/space-of-linear.tex +++ b/src/fa/tvs/space-of-linear.tex @@ -120,11 +120,19 @@ \begin{definition}[Bounded Convergence Topology] \label{definition:bounded-convergence-topology} - Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}. + Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}, or the \textbf{uniform topology}. The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence. \end{definition} +\begin{definition}[Topology of Precompact Convergence] +\label{definition:compact-operator-topology} + Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{K} \subset 2^E$ be the collection of precompact subsets of $E$, then the $\mathfrak{K}$-uniform topology on $L(E; F)$ is the \textbf{topology of precompact convergence}. + + The space $L_c(E; F)$ denotes $L(E; F)$ equipped with the topology of precompact convergence. +\end{definition} + + \begin{proposition} \label{proposition:operator-space-completeness} Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then: diff --git a/src/topology/dst/index.tex b/src/topology/dst/index.tex new file mode 100644 index 0000000..2a11c8d --- /dev/null +++ b/src/topology/dst/index.tex @@ -0,0 +1,4 @@ +\chapter{Polish Spaces and Analytic Sets} +\label{chap:polish-spaces} + +\input{./polish.tex} diff --git a/src/topology/dst/polish.tex b/src/topology/dst/polish.tex new file mode 100644 index 0000000..2aa147c --- /dev/null +++ b/src/topology/dst/polish.tex @@ -0,0 +1,65 @@ +\section{Polish Spaces} +\label{section:polish} + + + + +\begin{definition}[Polish Space] +\label{definition:polish-space} + Let $X$ be a topological space, then $X$ is \textbf{Polish} if it is completely metrisable and second countable. +\end{definition} + +\begin{proposition} +\label{proposition:polish-space-extension} + The following spaces are Polish: + \begin{enumerate} + \item Closed subspace of a Polish space. + \item Open subspace of a Polish space. + \item Countable products of Polish spaces. + \item Countable disjoint union of Polish spaces. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): Let $X$ be a Polish space with complete metric $d$ and $A \subset X$ be a closed subset, then $A$ is second countable. By \autoref{proposition:complete-closed}, $A$ is complete with respect to $d$. + + (2): Let $X$ be a Polish space with complete metric $d$ and $U \subset X$ be open. Assume without loss of generality that $U \subsetneq X$ and $d(X \times X) \subset [0, 1]$. Define + \[ + d_U: U \times U \to [0, \infty] \quad (x, y) \mapsto d(x, y) + \abs{\frac{1}{d(x, U^c)} - \frac{1}{d(y, U^c)}} + \] + + then $d_U$ is a metric on $U$. Since $d \le d_U$, the topology induced by $d_U$ is finer than the topology induced by $d$. On the other hand, the mapping $x \mapsto d(x, U^c)$ is continuous, so the topology induced by $d_U$ is coarser than the topology induced by $d$. Therefore $d_U$ induces the subspace topology of $U$. + + Now, let $\seq{x_n} \subset U$ be a Cauchy sequence with respect to $d_U$, then there exists $N \in \natp$ such that $d_U(x_m, x_n) \le 1$ for all $m, n \ge N$. Thus + \[ + \frac{1}{d(x_n, U^c)} \le \frac{1}{d(x_N, U^c)} + d(x_n, x_N) + \abs{\frac{1}{d(x_n, U^c)} - \frac{1}{d(x_N, U^c)}} \le \frac{1}{d(x_N, U^c)} + 1 + \] + + and $\delta = \inf_{n \in \natp}d(x_n, U^c) > 0$. Since $d \le d_U$, $\seq{x_n}$ is Cauchy with respect to $d$ as well. Thus as $\bracsn{x \in X|d(x, U^c) \ge \delta} \subset U$ is a closed subset of $X$, there exists $x \in U$ such that $x_n \to x$ as $n \to \infty$. Therefore $U$ is complete with respect to $d_U$. + + (3): By \autoref{proposition:separable-product}, \autoref{proposition:product-complete}, and \autoref{theorem:uniform-metrisable}. +\end{proof} + +\begin{proposition} +\label{proposition:polish-subspace} + Let $X$ be a Polish space and $Y \subset X$, then $Y$ is Polish if and only if it is $G_\delta$ in $X$. +\end{proposition} +\begin{proof}[Proof, {{\cite[Proposition 8.1.5]{CohnMeasure}}}. ] + ($\Rightarrow$): Suppose that $Y$ is Polish. Let $d_X: X^2 \to [0, 1]$ and $d_Y: Y^2 \to [0, 1]$ be complete metrics on $X$ and $Y$, respectively. For each $n \in \natp$, let $\mathcal{U}_n \subset 2^X$ be the collection of subsets of $X$ such that for each $U \in \mathcal{U}_n$, + \begin{enumerate}[label=(\roman*)] + \item $U$ is a non-empty open subset of $X$. + \item $\sup_{x, y \in U}d_X(x, y) \le 1/n$. + \item $\sup_{x, y \in U \cap Y}d_Y(x, y) \le 1/n$. + \end{enumerate} + + Let $U_n = \bigcup_{U \in \mathcal{U}_n}U$, then $Y \subset \ol{Y} \cap \bigcap_{n \in \natp}U_n$. On the other hand, let $x \in \ol{Y} \cap \bigcap_{n \in \natp}U_n$. Let $V_1 \in \mathcal{U}_1 \cap \cn_X(x)$ and $x_1 \in V_1$. For each $n \in \natp$ with $n \ge 2$, let $V_n \in \mathcal{U}_n \cap \cn_X(x)$ with $V_n \subset V_{n-1}$ and $x_n \in V_n \cap Y$, then by (ii) and (iii), $\seq{x_n}$ is Cauchy with respect to $d_X$ and $d_Y$. In particular, there exists $y \in Y$ such that $x_n \to y$ with respect to $d_Y$ as $n \to \infty$. Since $d_Y$ induces the subspace topology on $Y$, $x_n \to y$ with respect to $d_X$ as $n \to \infty$ as well. Therefore $x = y$, and $Y \supset \ol{Y} \cap \bigcap_{n \in \natp}U_n$. + + As every closed subset of $X$ is $G_\delta$, $Y = \ol{Y} \cap \bigcap_{n \in \natp}U_n$ is also $G_\delta$. + + ($\Leftarrow$): Suppose that $Y$ is $G_\delta$ in $X$. Let $\seq{U_n} \subset 2^X$ be open sets such that $Y = \bigcap_{n \in \natp}U_n$. In which case, $Y$ is homeomorphic to the diagonal + \[ + \Delta = \bracs{x \in \prod_{n \in \natp}U_n \bigg | x_m = x_n \forall m, n \in \natp} + \] + + For each $n \in \natp$, $U_n$ is Polish by (2) of \autoref{proposition:polish-space-extension}. As a closed subspace of a product of Polish spaces, $\Delta$ is Polish by (1) and (3) of \autoref{proposition:polish-space-extension}. Therefore $Y$ is also Polish. +\end{proof} + diff --git a/src/topology/metric/metric.tex b/src/topology/metric/metric.tex index 03dd888..4457af8 100644 --- a/src/topology/metric/metric.tex +++ b/src/topology/metric/metric.tex @@ -68,11 +68,6 @@ (2) $\Rightarrow$ (3): Since $\fF$ is Cauchy, there exists $\seq{E_n} \subset \fF$ such that for each $n \in \natp$, $E_n \supset E_{n+1}$ and $\sup_{y, z \in E_n}d(y, z) \le 1/n$. For each $n \in \natp$, let $x_n \in E_n$, then there exists a subsequence $\seq{n_k}$ and $x \in X$ such that $x = \limv{n}x_n$. In which case, $x \in \bigcap_{n \in \natp}\overline{E_n}$. For each $n \in \natp$, $\sup_{y, z\in E_n}d(y, z) \le 1/n$, so $B_X(x, 2/n) \supset E_n$. Therefore $\fF \to x$. \end{proof} -\begin{definition}[Polish Space] -\label{definition:polish-space} - Let $X$ be a topological space, then $X$ is \textbf{Polish} if it is completely metrisable and second countable. -\end{definition} - \begin{theorem}[Banach's Fixed Point Theorem]