Consider $\mathcal{B}(A,E)$ the space of bounded functions from $A$ to $E$ where $A$ is a set and $E$ a linear normed vector space. I need to proof that if $\mathcal{B}(A,E)$ is Banach then $E$ is Banach.

I found proofs of this in the case that $A$ is also a linear normed vector space, and those proofs involved linear functionals. But given a Cauchy sequence in $E$ how can I construct a function from the set $A$? Thanks in advance.

  • $\begingroup$ Try realizing $E$ as a (closed) subspace of $\mathcal{B}(A, E)$. $\endgroup$ – user125932 Jul 14 at 6:18
  • $\begingroup$ @user125932 thanks! I am going to give it a try $\endgroup$ – mate89 Jul 14 at 6:19

The constant functions form a closed subspace of $\mathcal{B}(A,E)$ isometric to $E$.

  • $\begingroup$ thanks, I'll try to elaborate $\endgroup$ – mate89 Jul 14 at 6:46
  • $\begingroup$ If $x_n$ is Cauchy in $E$. Define $f_n: A \to E: f_n(a) = x_n$... $\endgroup$ – Henno Brandsma Jul 14 at 6:51
  • $\begingroup$ thanks. For proving the isometric part, is that as trivial as saying for a given $e\in E$ take $f:A\to E$ such that $f(x)=e$ for all $x\in A$ ?? $\endgroup$ – mate89 Jul 14 at 6:59
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    $\begingroup$ @mate89 Yes, call that constant map $f: A \to E$, $F(e) \in \mathcal{B}(A,E)$ and note that $\|F(e) - F(e')\|_\infty = \|e-e'\|$. $\endgroup$ – Henno Brandsma Jul 14 at 7:01
  • $\begingroup$ Thanks a lot sir. Finally, for proving this set is closed, can you confirm this idea of proof is correct?: given a convergent sequence of functions $f_n$ in that space of constant functions, we have for a given $\varepsilon > 0$ $|f_n(x)-f(x)| < ||f_n-f||_\infty < \varepsilon$ for all $x\in E$, but this is the same as a sequence $(e_n)$ in E that should converge to a point $e$ and by uniqueness of limit $f(x) = e$ $\endgroup$ – mate89 Jul 14 at 7:27

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