Completeness of Operator space. 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.
 A: Let $x_0\in X$ be a fixed unit vector, and choose a direct complement of it:
$$X=\langle x_0\rangle\oplus X_1$$
using e.g. Hahn-Banach theorem.
Then, let $(y_n)$ be a Cauchy sequence in $Y$, and consider $\phi_n:X\to Y$ mapping $x_0\mapsto y_n$ and $\phi_n|_{X_1}=0$. Then $(\phi_n)$ is also Cauchy, so $\phi_n\to\phi$ in operator norm for some bounded $\phi:X\to Y$. Pointwise convergence follows, so we have $\phi|_{X_1}=0$, and let $y:=\phi(x_0)$, this is going to be the limit of $(y_n)$.
A: 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. 
