Progress over the past week.

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Bokuan Li
2026-01-05 20:10:39 -05:00
parent e8d394a2ac
commit abbed78434
33 changed files with 878 additions and 25 deletions

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\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}

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src/fa/tvs/dual.tex Normal file
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\section{The Dual Space}
\label{section:tvs-dual}
\begin{proposition}[Polarisation, {{\cite[Proposition 5.5]{Folland}}}]
\label{proposition:polarisation-linear}
Let $E$ be a vector space over $\complex$, $\phi \in \hom(E; \complex)$, and $u = \re{\phi}$, then
\begin{enumerate}
\item $u \in \hom(E; \real)$ when $E$ is viewed as a vector space over $\real$.
\item For any $x \in E$, $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$.
\end{enumerate}
Conversely, if $u \in \hom(E; \real)$ and $\phi \in \hom(E; \complex)$ is defined by $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$ for all $x \in E$, then $f \in \hom(E; \complex)$.
\end{proposition}
\begin{proof}
(1): Given that $\phi \in \hom(E; \real)$, $u \in \hom(E; \real)$. For any $x \in E$,
\[
\im{\dpb{x, \phi}{E}} = \re{-i \dpb{x, \phi}{E}} = -\dpb{ix, u}{E}
\]
so (2) holds.
(2): Since $u \in \hom(E; \real)$, so is $\phi$. On the other hand, for any $x \in E$,
\[
\dpb{ix, \phi}{E} = \dpb{ix, u}{E} - i\dpb{-x, u}{E} = i(\dpb{x, u}{E} - i \dpb{ix, u}{E}) = i \dpb{x, \phi}{E}
\]
\end{proof}
\begin{definition}[Topological Dual]
\label{definition:topological-dual}
Let $E$ be a TVS over $K \in \RC$, then the \textbf{topological dual} $E^*$ of $E$ is the set of all \textit{continous} linear functionals on $E$.
\end{definition}
\begin{definition}[Weak Topology]
\label{definition:weak-topology}
Let $E$ be a TVS over $K \in \RC$, then the \textbf{weak topology} on $E$ is the initial topology generated by $E^*$. The space $E$ equipped with its weak topology is denoted $E_w$.
\end{definition}

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\input{./src/fa/tvs/definition.tex}
\input{./src/fa/tvs/bounded.tex}
\input{./src/fa/tvs/dual.tex}
\input{./src/fa/tvs/continuous.tex}
\input{./src/fa/tvs/completion.tex}