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/Part 3: Functional Analysis/Chapter 8: Topological Vector Spaces/Section 8.11: Vector-Valued Function Spaces

Proposition 8.11.10. Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:

  1. $\hom(E; F)$ is a closed subspace of $F^{E}$ with respect to the product topology.

  2. $B(E; F)$ is a closed subspace of $F^{E}$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.

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

\[\phi_{x, y}: F^{E} \to F \quad T \mapsto Tx + Ty - T(x + y)\]

and

\[\psi_{x, \lambda}: F^{E} \to F \quad T \mapsto T(\lambda x) - \lambda Tx\]

are continuous with respect to the product topology. Since

\[\hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0}\cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}\]

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

(2): By Definition 8.11.2 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.$\square$

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