Completeness of the space of bounded bilinear maps 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?
 A: It is the completeness of $Z$ that matters.
Since $B_n$ is Cauchy, you have $\|B_n\| \leq M$, for some $M < \infty$.
If $B_n$ is Cauchy, then this allows you to define a $B$ pointwise (by completeness of $Z$), since for any $x,y$, $\|B_n(x,y)-B_m(x,y)\| \leq \|B_n-B_m\| \|x\| \|y\| $, and so $B_n(x,y)$ is Cauchy too, and converges to some $B(x,y) \in Z$.
Since each $B_n$ is multilinear, it follows that $B$ is also. To illustrate, since $B_n(x_1+x_2,y) = B_n(x_1,y)+B_n(x_2,y)$, it follows by taking limits that $B(x_1+x_2,y) = B(x_1,y)+B(x_2,y)$, and similarly for homogeneity and the other variable. 
To show that $B$ is bounded, we have
$\|B(x,y)\| \leq \|B_n(x,y)\| + \|B(x,y)-B_n(x,y)\| \leq M \|x\| \|y\| + \|B(x,y)-B_n(x,y)\|$. By definition of $B$, for any $\epsilon>0$, there exists an $n$ such that $\|B(x,y)-B_n(x,y)\| < \epsilon$, hence $\|B(x,y)\| \leq M \|x\| \|y\| + \epsilon$. It follows that $\|B(x,y)\| \leq M \|x\| \|y\|$, and hence that $\|B\| \leq M$.
To show that $B_n \to B$, choose $x,y$ in the unit ball and consider the bound $$\|B(x,y)-B_n(x,y)\| \leq \|B_n(x,y)-B_m(x,y)\| + \|B(x,y)-B_m(x,y)\|$$
$$ \leq \|B_n-B_m\| \|x\| \|y\|+ \|B(x,y)-B_m(x,y)\| \leq \|B_n-B_m\| + \|B(x,y)-B_m(x,y)\|.$$
Choose $N$ such that if $n,m>N$, then $\|B_n-B_m\| < \frac{\epsilon}{2}$. Choose $m>N$ such that $\|B(x,y)-B_m(x,y)\| < \frac{\epsilon}{2}$. Then we have $\|B(x,y)-B_n(x,y)\| < \epsilon$ for all $x,y$ in their respective unit balls. Thus, for $n>N$, $\|B-B_n\| \leq \epsilon$.
The same proof, mutatis mutandis, applies to multilinear operators.
