Let $x$ and $Y$ be normed vector spaces and assume that $X\ne\{0\}$. Prove that $B(X,Y)$ – space of continuous linear functionals $A:X\rightarrow Y$ – is complete with respect to the norm $\|A\|:=\sup_{\|x\|\le1}{\|Ax\|}$ iff $Y$ is complete.
My attempt:
"$\implies$"
Let $(A_n)$ be a Cauchy sequence of functionals from $B(X,Y)$. We have that
$$\forall_{\epsilon>0}\exists_{N>0}\forall_{n,m>N} \sup_{\|x\|\le1}{\|(A_n-A_m)x\|} <\epsilon$$
and
$$\exists A\in B(X,Y) : \sup_{\|x\|\le1}{\|(A_n-A)x\|} \rightarrow 0$$
I notice that $\|(A_n-A_m)x\|\le\sup_{\|x\|\le1}{\|(A_n-A_m)x\|}$ when $\|x\|\le1$ and therefore $(A_nx)$ is a Cauchy sequence in $Y$ when $\|x\|\le1$. I don't know how can I move from this point. Any suggestions?
Best Answer
See this answer for the proof $Y$ is complete$\implies$$\mathcal{B}(X,Y)$ is complete
See this answer for the proof $Y$ is not complete$\implies$ $\mathcal{B}(X,Y)$ is not complete