65 lines
3.9 KiB
TeX
65 lines
3.9 KiB
TeX
\section{Continuous Functions Vanishing at Infinity}
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\label{section:vanish-at-infinity}
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\begin{definition}[Vanish at Infinity]
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\label{definition:vanish-at-infinity}
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Let $X$ be a topological space, $E$ be a TVS over $K \in \RC$, and $f \in C(X; E)$, then $f$ \textbf{vanishes at infinity} if for every $U \in \cn_E^o(0)$, $\bracs{f \not\in U}$ is compact.
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The set $C_0(X; E)$ is the space of all functions that vanish at infinity, equipped with the uniform topology.
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\end{definition}
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\begin{proposition}
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\label{proposition:c0-properties}
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Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then:
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\begin{enumerate}
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\item $C_0(X; E) \subset BC(X; E)$.
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\item $C_0(X; E)$ is a closed subspace of $BC(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $C_0(X; E)$.
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\item If $X$ is a LCH space, then $C_c(X; E)$ is a dense subspace of $C_0(X; E)$ with respect to the uniform topology.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $U \in \cn_E^o(0)$ be balanced, then $f(X) = \bracs{f \in U} \sqcup \bracs{f \not\in U}$. Since $f(\bracs{f \not\in U})$ is compact, there exists $\lambda > 0$ such that $f(\bracs{f \not\in U}) \subset \lambda U$. In which case, $f(X) \subset (1 \vee \lambda)(U)$.
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(2): Let $f \in \ol{C_0(X; E)}$ and $U \in \cn_E^o(0)$, then there exists $V \in \cn_E^o(0)$ balanced such that $V + V \subset U$.
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Let $g \in C_0(X; E)$ such that $(f - g)(X) \subset V$. Since $g \in C_0(X; E)$, $\bracs{g \not\in V}$ is compact, so
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\[
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f(\bracs{g \in V}) \subset (f - g)(X) + g(\bracs{g \in V}) \subset V + V = U
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\]
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Thus $\bracs{f \not\in U}$ is a closed subset of $\bracs{g \not\in V}$, which is compact by \autoref{proposition:compact-extensions}.
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If $E$ is complete, then $BC(X; E)$ is complete by \autoref{definition:bounded-continuous-function-space}. Since $C_0(X; E)$ is a closed subspace, it is complete by \autoref{proposition:complete-closed}.
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(3): Let $f \in C_0(X; E)$ and $U \in \cn_E^o(0)$ be balanced. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\phi \in C_c(X; [0, 1])$ such that $\phi|_{\bracs{f \not\in U}} = 1$. In which case, $\phi f \in C_c(X; E)$ with
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\[
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(\phi f - f)(X) = \underbrace{(\phi f - f)(\bracs{f \not\in U})}_{0} + \underbrace{(\phi f - f)(\bracs{f \in U})}_{\in U} \in U
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\]
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so $f \in \ol{C_c(X; E)}$.
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\end{proof}
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\begin{proposition}
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\label{proposition:c0-tensor}
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Let $X$ be a LCH space and $E$ be a locally convex space over $K \in \RC$. Identify $C_0(X; K) \otimes E$ as a subspace of $C_0(X; E)$ under the natural map
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\[
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C_0(X; K) \otimes E \to C_0(X; E) \quad \sum_{j = 1}^n \phi_j \otimes x_j \mapsto \sum_{j = 1}^n x_j \cdot \phi_j
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\]
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then $C_0(X; K) \otimes E$ is a dense subspace of $C_0(X; E)$.
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\end{proposition}
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\begin{proof}
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Let $\phi \in C_0(X; E)$. Using \autoref{proposition:c0-properties}, assume without loss of generality that $\phi \in C_c(X; E)$.
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Since $\supp{\phi}$ is compact, so is $\phi(X)$ by \autoref{proposition:compact-extensions}. Let $U \in \cn_E^o(0)$ be balanced, then there exists $\seqf{y_j} \subset E \setminus \bracs{0}$ such that $\bigcup_{j = 1}^n (y_j + U) \supset \phi(X)$. For each $1 \le j \le n$, let $V_j = \phi^{-1}(y_j + U)$, then $\seqf{V_j}$ is an open cover of $\supp{\phi}$ consisting of precompact open sets. By \autoref{proposition:lch-partition-of-unity}, there exists a partition of unity $\seqf{\phi_j} \subset C_c(X; [0, 1])$ on $\supp{\phi}$ subordinate to $\seqf{V_j}$. For any $x \in E$,
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\begin{align*}
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\phi(x) - \sum_{j = 1}^n y_j \phi_j(x) &= \sum_{j = 1}^n \phi(x) \phi_j(x) - \sum_{j = 1}^n y_j \phi_j(x) \\
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&= \sum_{j = 1}^n \phi_j(x)[\phi(x) - y_j] \in \sum_{j = 1}^n \phi_j(x)U \subset U
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\end{align*}
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Therefore $(\phi - \sum_{j = 1}^n y_j \phi_j)(X) \subset U$.
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\end{proof}
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