226 lines
13 KiB
TeX
226 lines
13 KiB
TeX
\section{Vector-Valued Function Spaces}
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\label{section:spaces-linear-map}
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\begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
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\label{proposition:tvs-set-uniformity}
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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
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\begin{enumerate}
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\item The $\mathfrak{S}$-uniformity on $F^T$ (\autoref{definition:set-uniform}) is translation invariant.
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\item The composition defined by
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\[
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T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
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\]
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is continuous.
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\end{enumerate}
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For any vector subspace $\cf \subset F^T$, the following are equivalent:
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\begin{enumerate}
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\item[(3)] The $\mathfrak{S}$-uniform topology on $\cf \subset F^E$ is a vector space topology.
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\item[(4)] For each $S \in \mathfrak{S}$ and $f \in \cf$, $f(S) \subset E$ is bounded.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \autoref{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.
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(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 $(f + g, f' + g') \in E(S, U) \cap \cf$.
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(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.
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(4) $\Rightarrow$ (3): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \mathfrak{S}$, then for any $x \in X$,
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\begin{align*}
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\lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
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&= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
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\end{align*}
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Let $U_0 \in \cn_F(0)$. By (TVS1) and \autoref{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
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\[
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\lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
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\]
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for all $x \in S$.
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\end{proof}
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\begin{definition}[Space of Bounded Functions]
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\label{definition:bounded-function-space}
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Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
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\begin{enumerate}
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\item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$.
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\item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
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\end{enumerate}
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\end{definition}
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\begin{proof}
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(1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication.
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Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
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(2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
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\[
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f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
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\]
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so $f \in B(T; E)$.
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If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
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\end{proof}
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\begin{definition}[Space of Bounded Continuous Functions]
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\label{definition:bounded-continuous-function-space}
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Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
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\begin{enumerate}
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\item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$.
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\item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
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\end{enumerate}
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\end{definition}
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\begin{proof}
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(1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
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(2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace.
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If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete.
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\end{proof}
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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 $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $B_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
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Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\mathfrak{S}}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$.
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\end{definition}
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\begin{proposition}
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\label{proposition:multilinear-identify}
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Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system that contains all singletons, and $k \in \natp$, then
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\begin{enumerate}
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\item The map
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\[
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I: B_{\mathfrak{S}}^k(E; B_{\mathfrak{S}}(E; F)) \to B^{k+1}_{\mathfrak{S}}(E; F)
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\]
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defined by
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\[
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(IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
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\]
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is an isomorphism.
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\item The map
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\[
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I: \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to B^k_{\mathfrak{S}}(E; F)
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\]
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defined by
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\[
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IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
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\]
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is an isomorphism.
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\end{enumerate}
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which allows the identification
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\[
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\underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = B^k_{\mathfrak{S}}(E; F)
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\]
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under the map $I$ in (2).
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\end{proposition}
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\begin{proof}
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(1): To see that $I$ is surjective, let $T \in B_{\mathfrak{S}}^{k+1}(E; F)$ and
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\[
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I^{-1}T: E \to B_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
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\]
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Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
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\[
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T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
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\]
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by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\mathfrak{S}}(E; F)$.
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In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F))$.
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It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \mathfrak{S}$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
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On the other hand, let $S_1 \in \mathfrak{S}$ and $E(S_2, U)$ be an entourage of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
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(2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
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\[
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\underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F)) = B^{k+1}_{\mathfrak{S}}(E; F)
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\]
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Thus (2) holds for all $k \in \natp$.
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\end{proof}
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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}.
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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{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}
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