4.21 Continuous Functions Vanishing at Infinity

Definition 4.21.1 (Vanish at Infinity). Let $X$ be a topological space, $E$ be a TVS over $K \in \RC$, and $f \in C(X; E)$, then $f$ vanishes at infinity if for every $U \in \cn_{E}^{o}(0)$, $\bracs{f \not\in U}$ is compact.

The set $C_{0}(X; E)$ is the space of all functions that vanish at infinity, equipped with the uniform topology.

Proposition 4.21.2. Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then:

  1. $C_{0}(X; E) \subset BC(X; E)$.

  2. $C_{0}(X; E)$ is a closed subspace of $BC(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $C_{0}(X; E)$.

  3. 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.

Proof. (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, $f(X) \subset (1 \vee \lambda)(U)$.

(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 $V + V \subset U$.

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

\[f(\bracs{g \in V}) \subset (f - g)(X) + g(\bracs{g \in V}) \subset V + V = U\]

Thus $\bracs{f \not\in U}$ is a closed subset of $\bracs{g \not\in V}$, which is compact by Proposition 4.16.2.

If $E$ is complete, then $BC(X; E)$ is complete by Definition 8.11.3. Since $C_{0}(X; E)$ is a closed subspace, it is complete by Proposition 5.5.3.

(3): Let $f \in C_{0}(X; E)$ and $U \in \cn_{E}^{o}(0)$ be balanced. By Urysohn’s lemma, 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

\[(\phi f - f)(X) = \underbrace{(\phi f - f)(\bracs{f \not\in U})}_{0}+ \underbrace{(\phi f - f)(\bracs{f \in U})}_{\in U}\in U\]

so $f \in \ol{C_c(X; E)}$.$\square$

Proposition 4.21.3. 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

\[C_{0}(X; K) \otimes E \to C_{0}(X; E) \quad \sum_{j = 1}^{n} \phi_{j} \otimes x_{j} \mapsto \sum_{j = 1}^{n} x_{j} \cdot \phi_{j}\]

then $C_{0}(X; K) \otimes E$ is a dense subspace of $C_{0}(X; E)$.

Proof. Let $\phi \in C_{0}(X; E)$. Using Proposition 4.21.2, assume without loss of generality that $\phi \in C_{c}(X; E)$.

Since $\supp{\phi}$ is compact, so is $\phi(X)$ by Proposition 4.16.2. 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 Proposition 4.20.6, 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 $x \in E$,

\begin{align*}\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) \\&= \sum_{j = 1}^{n} \phi_{j}(x)[\phi(x) - y_{j}] \in \sum_{j = 1}^{n} \phi_{j}(x)U \subset U\end{align*}

Therefore $(\phi - \sum_{j = 1}^{n} y_{j} \phi_{j})(X) \subset U$.$\square$