Let $\mathcal{B}_0(X, Y)$ is a Banach space. Does this imply that $Y$ is a Banach space?

Question

Consider $$\mathcal{B}_0(X, Y) =\\{T\in\mathcal{B}(X,Y): \overline{T(B_X[0,1])}\subset Y \text{compact}\\}$$

where $B_X[0, 1]=\{x\in X: \|x\|\le 1\}$

Claim: $\mathcal{B}_0(X, Y)$ is a Banach space ( or closed subspace of $(\mathcal{B}(X, Y), \|•\|_{op}) $ iff $Y$ is Banach.

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If $Y$ is a Banach space then it can be proved easily that the operator limit of a sequence of compact operator is compact.

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I guess the other implication is not true!

Let $\mathcal{B}_0(X, Y)$ is a Banach space. Does this imply that $Y$ is a Banach space?

Can we characterize all normed linear spaces $X$ such that $\mathcal{B}_0(X)=\mathcal{B}_0(X,X) $ is closed?

Answer 1

Suppose that $B_0(X,Y)$ is complete. Let $\{y_n\}$ be a Cauchy sequence in $Y$. Fix a nonzero $g\in X^*$ with $\|g\|=1$ and define operators $$ T_nx=g(x)y_n. $$ These are rank-one operators, so they are bounded and compact. We also have $$ \|(T_n-T_m)x\|=|g(x)|\,\|y_n-y_m\|\leq\|x\|\,\|y_n-y_m\|. $$ This shows that $$ \|T_n-T_n\|\leq\|y_n-y_m\| $$ and so the sequence $\{T_n\}$ is Cauchy. By the completeness assumption, there exists $T=\lim T_n$, with $T\in B_0(X,Y)$.

Because $g$ is a linear functional, there exists $x_1$ such that $g(x_1)=1$ and $X=\mathbb C x_1\oplus\ker g$. Let $y=Tx_1$. Now $$ \|y-y_n\|=\|g(x_1)y-g(x_1)y_n\|=\|(T-T_n)x_1\|\leq\|T-T_n\|\,\|x_1\|. $$ Then $y=\lim_ny_n$ and $Y$ is complete.