Prove that the space of continuous linear functionals B(X,Y) is complete iff Y is complete

Question

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?

Answer 1

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

Answer 2

It's a little surprising to find this question remain unanswered, so although the original poster must have lost use of it, for the sake of completeness here's a partial one following the book Functional Analysis by B.V. Limaye (Third edition 2014, New Age International Publishers).

Fix a non-zero element $ f \in X^* $ and an ${\mathbf x}_0 \in \mathbf X $ where $f$ does not vanish. Given a Cauchy sequence $({\mathbf y}_n )$ in $\mathbf Y$, define $ T_n: \mathbf X \to \mathbf Y $ by $ T_n (\mathbf x) = f(\mathbf x) \cdot \mathbf y_n $.

Now argue that $(T_n)$ is Cauchy in $ \mathbf B (\mathbf X, \mathbf Y) $, call the limit $T$ and conclude that $ \mathbf y_n = T_n (\mathbf x_0) / f(\mathbf x_0) \to T (\mathbf x_0) / f(\mathbf x_0) $ in $\mathbf Y $ as $ n \to \infty $.