Conclusion about Banach space 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!
 A: 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).
A: 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. 
