101 lines
5.2 KiB
TeX
101 lines
5.2 KiB
TeX
\section{Spaces of Linear Maps}
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\label{section:space-linear-map-new}
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\begin{definition}[Space of Bounded Linear Maps]
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\label{definition:bounded-linear-map-space}
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Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in \mathfrak{B}(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
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Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$.
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\end{definition}
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\begin{proposition}[{{\cite[III.3.3]{SchaeferWolff}}}]
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\label{proposition:bounded-linear-map-space-bounded}
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Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $A \subset B_\sigma(E; F)$, then the following are equivalent:
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\begin{enumerate}
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\item $A \subset B_\sigma(E; F)$ is bounded with respect to the $\sigma$-uniform topology.
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\item For each $V \in \cn_F(0)$, $\bigcap_{T \in A}T^{-1}(V)$ absorbs every $S \in \sigma$.
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\item For every $S \in \sigma$, $\bigcup_{T \in A}T(A)$ is bounded in $F$.
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\end{enumerate}
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\end{proposition}
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% Proof omitted because it is obvious.
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\begin{definition}[Strong Operator Topology]
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\label{definition:strong-operator-topology}
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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 $F^E$ is the \textbf{strong operator topology}.
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The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
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\end{definition}
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\begin{proposition}
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\label{proposition:strong-operator-dense}
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Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If
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\begin{enumerate}
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\item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$.
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\item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous.
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\end{enumerate}
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then $T_\alpha \to T$ in $L_s(E; F)$.
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\end{proposition}
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\begin{proof}
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Let $x \in E$, $U \in \cn_F(Tx)$, and $V \in \cn_F(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_E(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_\alpha(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_0 \in A$ such that for all $\alpha \ge \alpha_0$, $T_\alpha y - Ty \in V$. In which case, for any $\alpha \ge \alpha_0$,
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\[
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T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U
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\]
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\end{proof}
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\begin{definition}[Weak Operator Topology]
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\label{definition:weak-operator-topology}
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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 $F_w^E$ is the \textbf{weak operator topology}.
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The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
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\end{definition}
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\begin{definition}[Bounded Convergence Topology]
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\label{definition:bounded-convergence-topology}
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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}.
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The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
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\end{definition}
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\begin{definition}[Topology of Precompact Convergence]
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\label{definition:compact-operator-topology}
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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}.
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The space $L_c(E; F)$ denotes $L(E; F)$ equipped with the topology of precompact convergence.
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\end{definition}
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\begin{proposition}
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\label{proposition:operator-space-completeness}
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
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\begin{enumerate}
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\item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
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\item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): For each $x, y \in E$ and $\lambda \in K$, the mappings
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\[
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\phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
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\]
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and
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\[
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\psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - \lambda Tx
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\]
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are continuous with respect to the product topology. Since
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\[
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\hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
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\]
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and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$.
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(2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
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\end{proof} |