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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?

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Write the fact that your sequence is Cauchy for the involved norm, and let tend to infinity one of the two integers which appear in it. It gives you exactly the convergence needed. This is the same trick that you cand find in a lot of proofs of completeness. – Ahriman Aug 21 '12 at 17:22
Right, this is exactly what I was forgetting. Thanks! – Iian Smythe Aug 21 '12 at 18:43
up vote 1 down vote accepted

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.

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You didn't prove that $B_n$ converves to $B$ with respect to the norm, which was exactly the point lacking to @ismythe. You just have proven the pointwise convergence to a bounded bilinear application. – Ahriman Aug 21 '12 at 18:08
It's fixed now. Thanks for catching my omission. – copper.hat Aug 21 '12 at 18:46

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