Used "separated" instead of Hausdorff in the context of topological vector spaces.
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@@ -47,7 +47,7 @@
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\begin{proposition}
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\label{proposition:bornologic-continuous-complete}
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Let $E$ be a bornologic space and $F$ be a complete Hausdorff locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
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Let $E$ be a bornologic space and $F$ be a complete separated locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
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\end{proposition}
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\begin{proof}
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By \autoref{proposition:bornologic-bounded}, $L_b(E; F) = B(E; F)$. By \autoref{proposition:operator-space-completeness}, $B(E; F)$ is complete, so $L_b(E; F)$ is complete as well.
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@@ -136,7 +136,7 @@
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\item $\dpb{x, \phi}{E} = \inf_{y \in M}\rho(x + y)$.
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\end{enumerate}
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\item For any $x \in E$ and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ with $|\phi| \le \rho$ and $\dpb{x, \phi}{E} = \rho(x)$.
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\item If $E$ is Hausdorff, then for any $x, y \in E$, there exists $\phi \in E^*$ with $\dpb{x, \phi}{E} \ne \dpb{y, \phi}{E}$.
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\item If $E$ is separated, then for any $x, y \in E$, there exists $\phi \in E^*$ with $\dpb{x, \phi}{E} \ne \dpb{y, \phi}{E}$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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@@ -25,7 +25,7 @@
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\begin{proposition}[{{\cite[II.5.4]{SchaeferWolff}}}]
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\label{proposition:complete-lc-projective-limit}
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Let $E$ be a Hausdorff complete locally convex space over $K \in \RC$, $\mathcal{B} \subset \cn_E(0)$ be a fundamental system of neighbourhoods consisting of convex, circled, and radial sets, directed under inclusion.
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Let $E$ be a separated complete locally convex space over $K \in \RC$, $\mathcal{B} \subset \cn_E(0)$ be a fundamental system of neighbourhoods consisting of convex, circled, and radial sets, directed under inclusion.
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For each $U \in \mathcal{B}$, let $[\cdot]_U$ be its gauge, $M_U = \bracs{x \in E|[\cdot]_U = 0}$, $E_U = E/M_U$, and $\norm{\cdot}_U: E_U \to [0, \infty)$ be the quotient of $[\cdot]_U$ by $M_U$, then
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\begin{enumerate}
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@@ -51,7 +51,7 @@
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(3): Let $\lim E_U$ be the projective limit. For each $U \in \mathcal{B}$, let $p_U: \lim E_U \to E_U$ be the canonical map.
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Let $x \in E$. Since $E$ is Hausdorff and $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$, there exists $U \in \mathcal{B}$ such that $\pi_U(x) \ne 0$. In which case, $p_U \circ \pi(x) = \pi_U(x) \ne 0$, so $\pi$ is injective.
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Let $x \in E$. Since $E$ is separated and $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$, there exists $U \in \mathcal{B}$ such that $\pi_U(x) \ne 0$. In which case, $p_U \circ \pi(x) = \pi_U(x) \ne 0$, so $\pi$ is injective.
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Let $x \in \lim E_U$. For each $U \in \mathcal{B}$, let $x_U \in E$ such that $\pi_U(x_U) = p_U(x)$. For any $V \in \cn_E(0)$, there exists $W \in \mathcal{B}$ with $W \subset V$. In which case, for any $U \in \mathcal{B}$ with $U \subset W$,
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\[
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@@ -5,7 +5,7 @@
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\label{definition:tvs-completion}
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Let $E$ be a TVS over $K \in \RC$, then there exists $(\wh E, \iota)$ such that:
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\begin{enumerate}
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\item $\wh E$ is a complete Hausdorff TVS.
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\item $\wh E$ is a complete separated TVS.
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\item $\iota \in L(E; \wh E)$.
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\item[(U)] For any $(F, T)$ satisfying (1) and (2), there exists a unique $\ol{T} \in L(\wh E; F)$ such that the following diagram commutes:
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\end{enumerate}
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@@ -54,7 +54,7 @@
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\begin{theorem}[Linear Extension Theorem (TVS)]
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\label{theorem:linear-extension-theorem-tvs}
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Let $E$ be a TVS over $K \in \RC$, $F$ be a complete Hausdorff TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
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Let $E$ be a TVS over $K \in \RC$, $F$ be a complete separated TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
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\begin{enumerate}
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\item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$.
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\item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
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@@ -40,7 +40,7 @@
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\begin{proposition}
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\label{proposition:tvs-quotient-hausdorff}
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Let $E$ be a TVS over $K \in \RC$, $M \subset E$ be a subspace, then $E/M$ is Hausdorff if and only if $M$ is closed.
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Let $E$ be a TVS over $K \in \RC$, $M \subset E$ be a subspace, then $E/M$ is separated if and only if $M$ is closed.
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\end{proposition}
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\begin{proof}
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The space $M$ is closed if and only if
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@@ -48,5 +48,5 @@
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M = \bigcap_{V \in \cn(0)}M + V
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\]
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which is equivalent to $E/M$ being Hausdorff.
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which is equivalent to $E/M$ being separated.
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\end{proof}
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