Let $X$, $Y$ and $Z$ be Banach spaces, and consider the space $Bil(X\times Y,Z)$ of all bilinear maps $B:X\times Y\to Z$ such that $x\mapsto B(x,y)$ and $y\mapsto B(x,y)$ are bounded maps from $X$ to $Z$ and $Y$ to $Z$ respectively. Endow this space with the following norm:
$\|B\|=\sup\{\|B(x,y)\|:\|x\|\leq 1,\|y\|\leq 1\}$
(In order to show that this norm is always defined, we require the principle of uniform boundedness.)
Why is this space complete under this norm? Here has been my approach thus far:
If we let $(B_n)$ be a Cauchy sequence, and put $\phi^y_n(x)=B_n(x,y)=\psi^x_n(y)$, then each $(\phi^y_n)$ and $(\psi^x_n)$ is a Cauchy sequence in $\mathcal{B}(X,Z)$ and $\mathcal{B}(Y,Z)$ respectively, so we have bounded linear maps $\phi^y$ and $\psi^x$ such that $\phi^y_n\to\phi^y$ and $\psi^x_n\to\psi^x$. Moreover, it is easy to check that $\phi^y(x)=\psi^x(y)$. Put $B(x,y)=\phi^y(x)=\psi^x(y)$, which we now know is in $Bil(X\times Y,Z)$. I claim $B_n\to B$.
The problem seems to be that I can make $\|B_n(x,y)-B(x,y)\|$ small for fixed $x$, or for fixed $y$, but I don't know how to uniformize this. I imagine it involves another application of the principle of uniform boundedness. Any suggestions?