10.11 Vector-Valued Function Spaces

Proof. (1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. 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 $f, g, f', g' \in \cf$, $S \in \mathfrak{S}$, 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 $(f + g, f' + g') \in E(S, U) \cap \cf$.

(3) $\Rightarrow$ (4): Let $f \in \cf$, $S \in \mathfrak{S}$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\mathfrak{S}$-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.

(4) $\Rightarrow$ (3): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \mathfrak{S}$, 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 10.3.2, 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 10.11.1.

(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 7.1.5. Thus $B(T; E)$ is also complete by Proposition 6.5.3.$\square$

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 10.11.2.

(2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^{X}$ by Definition 10.11.2, $BC(X; E)$ 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 10.11.2. Therefore $BC(X; E)$ is also complete.$\square$

Proof. (1): To see that $I$ is surjective, let $T \in B_{\mathfrak{S}}^{k+1}(E; F)$ and

Let $(x_{1}, \cdots, x_{k}) \in E^{k}$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_{1}^{k} \subset S$. In which case,

by assumption. Thus $I^{-1}T(x_{1}, \cdots, x_{k}) \in B_{\mathfrak{S}}(E; F)$.

In addition, for any $S_{1} \in \mathfrak{S}$ and entourage $E(S_{2}, U)$ of $B_{\mathfrak{S}}(E; F)$ where $S_{2} \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ with $S \supset S_{1} \cup S_{2}$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^{k}) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^{k}) \in B(B_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in B^{k}_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F))$.

It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \mathfrak{S}$ and $U$ be an entourage of $F$, then for any $T \in E(S^{k}, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.

On the other hand, let $S_{1} \in \mathfrak{S}$ and $E(S_{2}, U)$ be an entourage of $B_{\mathfrak{S}}(E; F)$ where $S_{2} \in \mathfrak{S}$ and $U$ is an entourage of $F$. Let $S \in \mathfrak{S}$ with $S \supset S_{1} \cup S_{2}$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.

(2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then

Thus (2) holds for all $k \in \natp$.$\square$

Proof. Let $x \in E$, $U \in \cn_{F}(Tx)$, and $V \in \cn_{F}(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_{E}(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_{\alpha}(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_{0} \in A$ such that for all $\alpha \ge \alpha_{0}$, $T_{\alpha} y - Ty \in V$. In which case, for any $\alpha \ge \alpha_{0}$,

$\square$

Proof. (1): For each $x, y \in E$ and $\lambda \in K$, the mappings

and

are continuous with respect to the product topology. Since

and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^{E}$.

(2): By Definition 10.11.2 and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^{E}$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.$\square$