TITLE: Completeness of Operator space.
QUESTION [6 upvotes]: Assume $(X,\|\cdot\|_X),(Y,\|\cdot\|_Y)$ are normed spaces and $\dim X\geq 1$. The following holds:
$Y$ complete $\iff$ $\mathscr L(X,Y)$ complete. 
The latter denotes the space of bounded operators between $X,Y$. The right implication is not hard to prove, but I was unable to find a proof of the left implication. Assistance would be greatly appreciated.

REPLY [3 votes]: Let $\{y_n\}\subset Y$ be a Cauchy sequence which is not convergent. Fix $f\colon X\to \Bbb R$ a non-zero continuous linear functional and define $T_n(x):=f(x)y_n$. Then $T_n$ is a sequence of linear bounded operators and $\lVert T_n-T_m\rVert=\lVert f\rVert_{X'}\lVert y_n-y_m\rVert$, so the sequence $\{T_n\}$ is Cauchy for the operator norm. If it converged to some $T$, we would have a contradiction taking $x$ such that $f(x)\neq 0$.
We need to ensure the existence of $f$. It's possible to do that thanks to Hahn-Banach theorem.