From a704806321232e1ae041f8c9c1b4e0e8c487abb8 Mon Sep 17 00:00:00 2001 From: Bokuan Li <47512608+Jerry-licious@users.noreply.github.com> Date: Tue, 6 Jan 2026 17:25:01 -0500 Subject: [PATCH] Finished basic topologies on function spaces. --- src/fa/tvs/bounded.tex | 3 -- src/fa/tvs/continuous.tex | 38 --------------------- src/fa/tvs/definition.tex | 4 +-- src/fa/tvs/index.tex | 1 + src/fa/tvs/spaces-of-linear.tex | 60 +++++++++++++++++++++++++++++++++ 5 files changed, 63 insertions(+), 43 deletions(-) create mode 100644 src/fa/tvs/spaces-of-linear.tex diff --git a/src/fa/tvs/bounded.tex b/src/fa/tvs/bounded.tex index 5cf8b35..13f619d 100644 --- a/src/fa/tvs/bounded.tex +++ b/src/fa/tvs/bounded.tex @@ -1,9 +1,6 @@ \section{Bounded Sets} \label{section:bounded} -\subsection{Bounded Sets} -\label{subsection:tvs-bounded} - \begin{definition}[Bounded] \label{definition:bounded} Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$. diff --git a/src/fa/tvs/continuous.tex b/src/fa/tvs/continuous.tex index a1b2996..60c223a 100644 --- a/src/fa/tvs/continuous.tex +++ b/src/fa/tvs/continuous.tex @@ -84,41 +84,3 @@ \end{enumerate} The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$. \end{definition} - -\begin{definition}[Uniform/Bounded Operator Topology] -\label{definition:uniform-op-topo} - Let $E, F$ be TVSs over $K \in \RC$. For each bounded set $B \subset E$ and entourage $V \subset F \times F$, let - \[ - U(B, V) = \bracs{(S, T) \in L(E; F) \times L(E; F)|(Sx, Tx) \in V \forall x \in B} - \] - then - \[ - \fB = \bracs{U(B, V)| B \subset E \text{ bounded}, U \subset F \times F \text{ entourage}} - \] - is a fundamental system of entourages for a vector space uniformity on $L(E; F)$, which induces the \textbf{uniform/bounded operator topology} on $L(E; F)$. -\end{definition} -\begin{proof} - Firstly, - \begin{enumerate} - \item[(FB1)] Let $U(B, V), U(B', V') \in \fB$, then $U(B, V) \cap U(B', V') \supset U(B \cup B', V \cap V') \in \fB$. - \item[(UB1)] For any $U(B, V) \in \fB$, the diagonal is contained in $V$, so the diagonal is also contained in $U(B, V)$. - \item[(UB2)] For any $U(B, V) \in \fB$, there exists an entourage $W \subset F \times F$ such that $W \circ W \subset V$. So $U(B, W) \in \fB$ with - \[ - U(B, W) \circ U(B, W) \subset U(B, W \circ W) \subset U(B, V) - \] - \end{enumerate} - By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a uniformity on $L(E; F)$. - - TODO SCHAEFER WOLFF PAGE 79 -\end{proof} - - -\begin{definition}[Strong Operator Topology] -\label{definition:strong-op-topo} - Let $E, F$ be TVSs over $K \in \RC$, then the \textbf{strong operator topology} on $L(E; F)$ is the initial topology generated by the maps $\bracs{T \mapsto Tx| x \in E}$, where $F$ is equipped with the strong topology. -\end{definition} - -\begin{definition}[Weak Operator Topology] -\label{definition:weak-op-topo} - Let $E, F$ be TVSs over $K \in \RC$, then the \textbf{weak operator topology} on $L(E; F)$ is the initial topology generated by the maps $\bracs{T \mapsto Tx| x \in E}$, where $F$ is equipped with the weak topology. -\end{definition} diff --git a/src/fa/tvs/definition.tex b/src/fa/tvs/definition.tex index 3a9e676..52a38b0 100644 --- a/src/fa/tvs/definition.tex +++ b/src/fa/tvs/definition.tex @@ -130,12 +130,12 @@ \label{proposition:tvs-good-neighbourhood-base} Let $E$ be a topological vector space over $K \in \RC$, then \begin{enumerate} - \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of balanced and absorbing sets. + \item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of circled and absorbing sets. \item The fundamental system of neighbourhoods in $(1)$ can be taken to be open or closed. \end{enumerate} \end{proposition} \begin{proof} - Firstly, (TVS2) implies that every neighbourhood of $0$ is balanced. + Firstly, (TVS2) implies that every neighbourhood of $0$ is circled. By \ref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets. diff --git a/src/fa/tvs/index.tex b/src/fa/tvs/index.tex index 9e5b021..42827b5 100644 --- a/src/fa/tvs/index.tex +++ b/src/fa/tvs/index.tex @@ -6,3 +6,4 @@ \input{./src/fa/tvs/dual.tex} \input{./src/fa/tvs/continuous.tex} \input{./src/fa/tvs/completion.tex} +\input{./src/fa/tvs/spaces-of-linear.tex} diff --git a/src/fa/tvs/spaces-of-linear.tex b/src/fa/tvs/spaces-of-linear.tex new file mode 100644 index 0000000..b193de2 --- /dev/null +++ b/src/fa/tvs/spaces-of-linear.tex @@ -0,0 +1,60 @@ +\section{Spaces of Vector-Valued Maps} +\label{section:spaces-linear-map} + +\begin{proposition}[{{\cite[3.3.1]{SchaeferWolff}}}] +\label{proposition:tvs-set-uniformity} + Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be an upward-directed system of sets, $F$ be a TVS over $K \in \RC$, then + \begin{enumerate} + \item The $\mathfrak{S}$-uniformity on $F^T$ (\ref{definition:set-uniform}) is translation invariant. + \item The composition defined by + \[ + T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x) + \] + is continuous. + \end{enumerate} + For any vector subspace $\cf \subset F^T$, the following are equivalent: + \begin{enumerate} + \item[(3)] The $\mathfrak{S}$-uniform topology on $\cf \subset F^E$ is a vector space topology. + \item[(4)] For each $S \in \mathfrak{S}$ and $f \in \cf$, $f(S) \subset E$ is bounded. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \ref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant. + + (2): Let $f, g, f', g' \in \cf$, $S \in \mathfrak{S}$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$. + + (3) $\Rightarrow$ (4): Let $f \in \cf$, $S \in \mathfrak{S}$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\mathfrak{S}$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded. + + (4) $\Rightarrow$ (3): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \mathfrak{S}$, then for any $x \in X$, + \begin{align*} + \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\ + &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x)) + \end{align*} + Let $U_0 \in \cn_F(0)$. By (TVS1) and \ref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then + \[ + \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0 + \] + for all $x \in S$. +\end{proof} + + +\begin{definition}[Strong Operator Topology] +\label{definition:strong-operator-topology} + Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F)$ is the \textbf{strong operator topology}. + + The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology. +\end{definition} + +\begin{definition}[Weak Operator Topology] +\label{definition:weak-operator-topology} + Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F_w)$ is the \textbf{weak operator topology}. + + The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology. +\end{definition} + +\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}. + + The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence. +\end{definition}