193 lines
12 KiB
TeX
193 lines
12 KiB
TeX
\section{Vector Space Topologies}
|
|
\label{section:tvs-topology}
|
|
|
|
|
|
\begin{definition}[Topological Vector Space]
|
|
\label{definition:tvs}
|
|
Let $E$ be a vector space over $K \in \bracs{\real, \complex}$ and $\topo \subset 2^E$ be a topology. If
|
|
\begin{enumerate}
|
|
\item[(TVS1)] $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
|
|
\item[(TVS2)] $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
|
|
\end{enumerate}
|
|
then the pair $(E, \topo)$ is a \textbf{topological vector space}.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Translation-Invariant Topology]
|
|
\label{definition:translation-invariant-topology}
|
|
Let $E$ be a vector space and $\topo$ be a topology on $E$, then $\topo$ is \textbf{translation-invariant} if for any $U \in \topo$ and $y \in E$, $U + y \in \topo$.
|
|
\end{definition}
|
|
|
|
\begin{lemma}
|
|
\label{lemma:tvs-translation-invariant}
|
|
Let $E$ be a TVS over $K \in \RC$, then the topology of $E$ is translation-invariant.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Let $U \subset E$ open and $y \in E$, then $U + y$ is the preimage of $U$ by the map $x \mapsto x - y$. By (TVS1), $U + y$ is open.
|
|
\end{proof}
|
|
|
|
\begin{definition}[Translation-Invariant Uniformity]
|
|
\label{definition:translation-invariant-uniformity}
|
|
Let $E$ be a vector space, $\fU$ be a uniformity on $E$, and $U \in \fU$, then $U$ is \textbf{translation-invariant} if for every $z \in E$,
|
|
\[
|
|
U = \bracs{(x + z, y + z)|(x, y) \in U}
|
|
\]
|
|
and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages.
|
|
\end{definition}
|
|
|
|
\begin{lemma}
|
|
\label{lemma:translation-invariant-symmetric}
|
|
Let $E$ be a vector space and $\fU$ be a translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \ref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
|
|
\end{proof}
|
|
|
|
|
|
\begin{proposition}[{{\cite[1.1.4]{SchaeferWolff}}}]
|
|
\label{proposition:tvs-uniform}
|
|
Let $E$ be a TVS over $K \in \bracs{\real, \complex}$, then:
|
|
\begin{enumerate}
|
|
\item There exists a unique translation-invariant uniformity $\fU$ on $E$ that induces the topology on $E$.
|
|
\item For each neighbourhood $V \in \cn(0)$, let $U_V = \bracs{(x, y) \in E^2| x - y \in V}$, then for any fundamental system of neighbourhoods $\fB_0$ at $0$, $\fB = \bracs{U_V| V \in \fB_0}$ is a fundamental system of entourages for $\fU$.
|
|
\end{enumerate}
|
|
The space $E$ will always be assumed to be equipped with its translation-invariant uniformity.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
(2): Firstly, for any $V \in \fB_0$, $U_V$ is translation-invariant.
|
|
\begin{enumerate}
|
|
\item[(FB1)] For any $V, V' \in \fB_0$, there exists $W \in \fB_0$ such that $W \subset V \cap V'$. In which case, $U_V \cap U_{V'} \supset U_W \in \fB$.
|
|
\item[(UB1)] For any $V \in \fB_0$, $0 \in V$, so $\Delta \subset U_V$.
|
|
\item[(UB2)] Let $V \in \fB_0$, then by (TVS1), there exists $W \in \fB_0$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_W$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_W \circ U_W \subset U_V$.
|
|
\end{enumerate}
|
|
|
|
By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
|
|
|
|
(1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.
|
|
|
|
Let $W \in \cn(0)$, then by \ref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
|
|
|
|
Let $V \in \mathfrak{V}$. Using \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
|
|
\end{proof}
|
|
|
|
\begin{proposition}
|
|
\label{proposition:tvs-closure}
|
|
Let $E$ be a TVS over $K \in \RC$, $A \subset E$, and $\fB \subset \cn(0)$ be a fundamental system of neighbourhoods, then
|
|
\[
|
|
\ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB}
|
|
\]
|
|
\end{proposition}
|
|
\begin{proof}
|
|
Let $V \in \cn(0)$ be balanced and $U_V = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_V(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_V$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_V(A) = A + V$.
|
|
|
|
Assume without loss of generality that $\fB$ consists of symmetric entourages. By \ref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \ref{proposition:uniformclosure} implies that
|
|
\[
|
|
\ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A
|
|
\]
|
|
\end{proof}
|
|
|
|
\begin{proposition}[{{\cite[1.1.1]{SchaeferWolff}}}]
|
|
\label{proposition:tvs-set-operations}
|
|
Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$, then:
|
|
\begin{enumerate}
|
|
\item If $A$ is open, then $A + B$ is open.
|
|
\item If $A$ is closed and $B$ is compact, then $A + B$ is closed.
|
|
\end{enumerate}
|
|
\end{proposition}
|
|
\begin{proof}
|
|
$(1)$: For every $x \in B$, $A + x$ is open by \ref{definition:translation-invariant-topology}, so
|
|
\[
|
|
A + B = \bigcup_{x \in B}(A + x)
|
|
\]
|
|
is open.
|
|
|
|
$(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
|
|
\[
|
|
y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U}
|
|
\]
|
|
|
|
By \ref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
|
|
\[
|
|
y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U]
|
|
\]
|
|
Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
|
|
\[
|
|
y \in \bigcap_{U \in \fF}[(-B) + U + U] \subset \bigcap_{V \in \cn(0)}[(x-B) + V] = \overline{x - B} = x - B
|
|
\]
|
|
so $x \in y + B \subset A + B$.
|
|
\end{proof}
|
|
|
|
\begin{definition}[Balanced/Circled]
|
|
\label{definition:balanced}
|
|
Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then $A$ is \textbf{balanced/circled} if $\lambda A \subset A$ for all $\lambda \in K$ with $\abs{\lambda} \le 1$.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Absorbing/Radial]
|
|
\label{definition:absorbing}
|
|
Let $E$ be a vector space over $K \in \RC$ and $A, B \subset E$, then $A$ \textbf{absorbs} $B$ if there exists $\lambda \in K$ such that $\lambda A \supset B$, and $A$ is \textbf{absorbing/radial} if it absorbs every point in $E$.
|
|
\end{definition}
|
|
|
|
\begin{proposition}
|
|
\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 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 circled.
|
|
|
|
By \ref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
|
|
|
|
Let $U \in \cn^o(0)$ be open. By (TVS2), there exists $r > 0$ such that $\lambda U \subset U$ for all $\lambda \in K$ with $\abs{\lambda} \le r$. Define
|
|
\[
|
|
V = \bigcup_{\substack{\lambda \in K \\ \abs{\lambda} \le r}} \lambda U \subset U
|
|
\]
|
|
then for any $x \in V$, there exists $\lambda \in K$ with $\abs{\lambda} \le r$ and $y \in U$ such that $x = \lambda y$. In which case, for any $\mu \in K$ with $\abs{\mu} \le 1$, $\mu(\lambda y) = (\mu \lambda) y \in \mu\lambda U$. Since $\abs{\mu \lambda} \le r$, $\mu \lambda U \subset V$. Thus $V \subset U$ is balanced.
|
|
|
|
Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^o(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda} \le 1$, $\lambda \overline{V} = \overline{\lambda V} \subset \overline{V}$ by (TVS2). Therefore $\overline{V} \subset U$ is balanced as well.
|
|
\end{proof}
|
|
|
|
\begin{proposition}[{{\cite[1.2]{SchaeferWolff}}}]
|
|
\label{proposition:tvs-0-neighbourhood-base}
|
|
Let $E$ be a vector space of $K \in \RC$, and $\topo$ be a vector space topology on $E$, then there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ such that:
|
|
\begin{enumerate}
|
|
\item[(TVB1)] For each $U \in \fB$, there exists $V \in \fB$ such that $V + V \subset U$.
|
|
\item[(TVB2)] For each $U \in \fB$, $U$ is circled and radial.
|
|
\end{enumerate}
|
|
Conversely, if $\fB \subset 2^E$ is a family of sets that contain $0$ and satisfies (TVB1) and (TVB2), then there exists a unique topology $\topo$ on $E$ for which $\fB$ is a fundamental system of neighbourhoods at $0$.
|
|
\end{proposition}
|
|
\begin{proof}
|
|
(1): By \ref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
|
|
|
|
(2): For each $V \in \fB$, let $U_V = \bracs{(x, y) \in E|x - y \in V}$, then $U_V$ is symmetric and translation-invariant by (TVB1). Let
|
|
\[
|
|
\mathfrak{V} = \bracs{U_V|V \in \fB}
|
|
\]
|
|
then
|
|
\begin{enumerate}
|
|
\item[(FB1)] For any $V, V' \in \fB$, there exists $W \in \fB$ with $W \subset V \cap V'$. In which case, $U_{V} \cap U_{V'} \supset U_W \in \mathfrak{V}$.
|
|
\item[(UB1)] For any $x \in E$ and $V \in \fB$, $x - x = 0 \in V$, so $\Delta \subset U_V$.
|
|
\item[(UB2)] For any $V \in \fB$, by (TVB1), there exists $W \in \fB$ such that $W + W \subset V$. In which case, for any $x, y, z \in E$ with $x - y, y - z \in W$, $x - z \in V$. Therefore $U_W \circ U_W \subset U_V$.
|
|
\end{enumerate}
|
|
By \ref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.
|
|
|
|
It remains to verify that $\mathfrak{V}$ induces a vector space topology on $E$.
|
|
\begin{enumerate}
|
|
\item[(TVS1)] Let $V \in \fB$, then there exists $W \in \fB$ such that $W + W \subset V$ by (TVB1). In which case, for any $x, x', y, y'$ with $x - x' \in W$ and $y - y' \in W$, $(x + y) - (x' + y') \in W + W \subset V$.
|
|
\item[(TVS2)] Let $V \in \fB$, $\eps > 0$, $x, x' \in E$ with $x - x' \in V$, and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'} < \eps$, then
|
|
\begin{align*}
|
|
\lambda x - \lambda' x' &= \lambda x - \lambda x' + \lambda x' - \lambda' x' \\
|
|
&= \lambda(x - x') + (\lambda - \lambda')x' \in \lambda V + (\lambda - \lambda')x'
|
|
\end{align*}
|
|
Since $V$ is radial, there exists $\mu \in K$ such that $x \in \mu V$. Given that $V$ is circled, $x' = x + (x - x') \in \mu V + V \subset (\abs{\mu} + 1)V$, and
|
|
\[
|
|
\lambda x - \lambda' x' \in \lambda V + \eps(\abs \mu + 1)V \subset \abs{\lambda} V + \eps(\abs \mu + 1)V
|
|
\]
|
|
|
|
Let $W \in \fB$ and $\eps > 0$ such that $\eps(\abs{\mu}+1) \le 1$, then by repeated application of (TVB1), there exists $V \in \fB$ such that $\abs{\lambda} V + V \subset W$. Therefore scalar multiplication is jointly continuous.
|
|
\end{enumerate}
|
|
|
|
Finally, by definition of the uniform topology, $\fB = \bracs{U_V(0)|V \in \fB}$ is a fundamental system of neighbourhoods at $0$.
|
|
\end{proof}
|