Conclusion about Banach space

Question

Let $X,Y$ be normed spaces and $f \in X^*$ a linear functional with $||f||=1$.

Let $B(X,Y)$ be the set of all bounded linear transformations from $X$ to $Y$.

Define $T:Y \to B(X,Y)$ by $(Ty)(x) = f(x)y$.

I have showed that $T$ is an isometry (That is - linear and preserves norm).

I need to conclude that if $B(X,Y)$ is Banach then so is $Y$ .

Im not sure how - one approach is to say that $T(Y)$ is a closed subset of $B(X,Y)$ but im not sure how to do that(if it's true).

Another approach is to take $\{y_n\} \subset Y$ cauchy sequence, then $T(y_n) $ is also a cuachy sequence since $T$ is isometry, so it converges, but this does not imply that $\{y_n\}$ also converge.

Any ideas?

Thanks for helping!

Answer 1

Here is a proof without using completion. Let $(y_n)$ be a Cauchy sequence in $Y$. Then $T(y_n)$ is a Cauchy sequence in $B(X,Y)$ and because it is a Banach space we know there is a function $\phi\in B(X,Y)$ such that $T(y_n)\to\phi$. Also, there is a vector $x_0\in X$ such that $f(x_0)=1$. And finally, convergence in norm implies strong convergence. Hence:

$\phi(x_0)=\lim_{n\to\infty} T(y_n)(x_0)=\lim_{n\to\infty} f(x_0)y_n=\lim_{n\to\infty} y_n$

So the sequence $y_n$ has a limit.

Answer 2

Let $\widetilde{Y}$ be the completion of $Y$, which is a Banach space. We have

$$B(X,Y) \subset B(X, \widetilde{Y})$$

as a normed subspace. Let $y_n$ be a Cauchy sequence in $Y$, converging to some $y \in \widetilde{Y}$. You want to show that $y$ is actually in $Y$.

The operators $T(y_n)$ converge to $T(y)$ in $B(X,\widetilde{Y})$. But if $B(X,Y)$ is a Banach space, then $T(y)$ must actually lie in $B(X,Y)$. Hence $y$ must be in $Y$.

(I would be interested in seeing a proof where we don't make use of $Y$ being contained in a completion).