5.21 Continuous Functions Vanishing at Infinity

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

Thus $\bracs{f \not\in U}$ is a closed subset of $\bracs{g \not\in V}$, which is compact by Proposition 5.16.3.

If $E$ is complete, then $BC(X; E)$ is complete by Definition 12.12.3. Since $C_{0}(X; E)$ is a closed subspace, it is complete by Proposition 6.7.3.

(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

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

Proof. Let $\phi \in C_{0}(X; E)$. Using Proposition 5.21.2, assume without loss of generality that $\phi \in C_{c}(X; E)$.

Since $\supp{\phi}$ is compact, so is $\phi(X)$ by Proposition 5.16.3. 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 relatively compact open sets. By Proposition 5.20.8, 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 $x \in E$,

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

Proof. To see that the canonical map is continuous, let $\rho: E \to [0, \infty)$ be a continuous seminorm on $E$. By the Hahn-Banach Theorem, there exists an equicontinuous family $T \subset E^{*}$ such that for each $y \in E$, $\rho(y) = \sup_{\phi \in T}|\dpn{y, \phi}{E}|$.

Let $\lambda = \sum_{j = 1}^{n} f_{j} \otimes y_{j} \in C_{0}(X; K) \otimes_{\eps} E$, then

Since the evaluation maps $\bracsn{\pi_x: C_0(X; K) \to K|x \in X}$ are equicontinuous, the uniform seminorm on $C_{0}(X; E)$ with respect to $\rho$ is bounded above by a cross seminorm of the injective tensor product. Thus the inclusion is continuous.

On the other hand, let $T \subset E^{*}$ be equicontinuous, then there exists a continuous seminorm $\rho: E \to [0, \infty)$ such that $|\phi| \le \rho$ for all $\phi \in T$. In which case, for any $\lambda = \sum_{j = 1}^{n} f_{j} \otimes y_{j} \in C_{0}(X; K) \otimes_{\eps} E$, $I \in C_{0}(X; K)^{*}$, and $\phi \in T$,

Hence the cross seminorm corresponding to $B_{C_0(X; K)^*}(0, 1)$ and $T$ is bounded above by the uniform seminorm on $C_{0}(X; E)$ with respect to $\rho$, so the inclusion is an embedding.

Finally, by Proposition 5.21.2, $C_{0}(X; K)$ is complete. By Proposition 5.21.3, $C_{0}(X; K) \otimes_{\eps} E$ is dense in $C_{0}(X; E)$. Therefore the canonical map extends to an isomorphism through the Linear Extension Theorem.$\square$

Proof. By Theorem 5.21.4.$\square$