Suppose that $V$ is a locally convex topological vector space and $f:V^2 \to V$ is a bilinear map. Suppose that $C \subseteq V$ is compact and convex, $f$ maps $C^2$ into $C$ and $f \restriction C^2$ is separately continuous. Must $f \restriction C^2$ be jointly continuous?

In the particular application I have in mind, $V = \ell_\infty^*$ with the weak* topology. Moreover the function $f$ is injective. I suspect even in this setting that this is false.

I am also interested in a good reference for the optimal results of concerning separate and joint continuity of bilinear maps. Ideally this would be written for someone who is not a functional analyst.


A personal obsession is (weakly) almost period functions. Let $G$ be a discrete group (you can work more generally) and for $f\in \ell^\infty(G)$ let $O(f)$ be the set of translate of $f$ by the group action. Set $$ AP(G) = \{ f\in \ell^\infty(G) : O(f)\text{ is relatively compact}\} \\$$ $$WAP(G) = \{ f\in \ell^\infty(G) : O(f)\text{ is weakly relatively compact}\}$$

Then these are unital sub-$C^*$-algebras of $\ell^\infty(G)$ and so have character spaces $G^{AP}$ and $G^{WAP}$. You can extend the product from $G$ to these: then the product on $G^{AP}$ is jointly continuous, but that on $G^{WAP}$ only separately jointly continuous. Translated back to algebra, this means that the measure spaces $M(G^{AP})$ and $M(G^{WAP})$ become Banach algebras for the convolution product. The continuity translates to say that the product on $M(G^{WAP})$ is separately weak$^*$-continuous, and that on $M(G^{AP})$ is even jointly weak$^*$-continuous on bounded sets.

So that gives an example: if $C$ is the closed unit ball of $M(G^{WAP})$ and $f$ is the product map, then $f$ satisfies your requirements, but $f$ is not jointly continuous unless $WAP(G) = AP(G)$ (I think-- this last claim needs a little chasing of definitions). If $G$ is abelian then $G^{AP}$ is the classical Bohr compacitification, but $G^{WAP}$ is much larger. There are books by Berglund (and coauthors) on this topic; they do, IMHO, need a functional analysis background.

Surely there are easier examples for what you want though. (And $f$ is not injective in my example).

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  • $\begingroup$ Thanks. This is in the spirit of what I was lookng for. $\endgroup$ – Justin Moore Apr 7 '11 at 16:37

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