# If $\mathcal{B}(A,E)$ is banach then $E$ is banach ($A$ is a set)

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.

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

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

• thanks, I'll try to elaborate – mate89 Jul 14 at 6:46
• If $x_n$ is Cauchy in $E$. Define $f_n: A \to E: f_n(a) = x_n$... – Henno Brandsma Jul 14 at 6:51
• 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 $x\in A$ ?? – mate89 Jul 14 at 6:59
• @mate89 Yes, call that constant map $f: A \to E$, $F(e) \in \mathcal{B}(A,E)$ and note that $\|F(e) - F(e')\|_\infty = \|e-e'\|$. – Henno Brandsma Jul 14 at 7:01
• 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 $f(x) = e$ – mate89 Jul 14 at 7:27