10.12 Vector-Valued Function Spaces

Proof. (1): Let $U \subset F \times F$ be an entourage and $S \in \sigma$. Using Proposition 10.1.6, 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.

(2): Let $f, g, f', g' \in \cf$, $S \in \sigma$, 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 $(f + g, f' + g') \in E(S, U) \cap \cf$.

(T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-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.

(T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,

Let $U_{0} \in \cn_{F}(0)$. By (TVS1) and Proposition 10.1.11, 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

for all $x \in S$.$\square$

Proof. (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 Proposition 10.4.2, so $B(T; E)$ is closed under addition and scalar multiplication.

Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by Proposition 10.12.1.

(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,

so $f \in B(T; E)$.

If $E$ is complete, then $E^{T}$ with the uniform topology is complete by Proposition 7.2.5. Thus $B(T; E)$ is also complete by Proposition 6.7.3.$\square$

Proof. (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 Definition 10.12.2.

(2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^{X}$ by Definition 10.12.2, $BC(X; E)$ is a closed subspace.

If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by Definition 10.12.2. Therefore $BC(X; E)$ is also complete.$\square$