Let $M$ be a W\*-algebra and consider the following map: 
$$\gamma: M\times M^\*\to M^\*: (a,f)\to af$$
where $af(b)=f(ba)$. Let us consider $M$ under the weak topology $\sigma(M,M^\*)$ and $M^\*$ under the weak-star topology $\sigma(M^\*,M)$. 


Q: Is $\gamma$ weakly-weak-star jointly continuous? I mean, assume that $\{a\_i\}$ is weakly convergent to $a$ in $M$ and $\{f\_i\}$ is weak-star convergent to $f$ in $M^\*$. Then can we conclude that $a\_if\_i$ is weak-star convergent to $af$?


Naturally this question may be also asked for the bilinear mapping $M^{\*\*}\times M^\*\to M^\*$. 



Is the module action $M\times M^*\to M^*$ jointly continuous?
Before I answer, let me comment that duals of von Neumann algebras are highly pathological objects, and when you get to that level I suspect most operator algebraists would see you as doing set theory, not operator algebras. Just FYI. (Would the predual $M\_\*$ suit your purposes? It's a lot nicer.)


The answer is no, by the usual sort of counterexample. Let $M = B(l^2)$ and let $(e\_n)$ be the standard basis of $l^2$. For each $n$ let $a\_n$ be the operator $v \mapsto \langle v, e\_n\rangle e\_1$ and let $f\_n$ be the linear functional $a \mapsto \langle ae\_n, e\_1\rangle$. Then $a\_nf\_n(I) = f\_n(a\_n) = 1$, so $a\_nf\_n \not\to 0$ weak\*. However, $f\_n \to 0$ weak\* and $a\_n \to 0$ weakly. Both of these claims follow from the fact that the "first row" operator space (all $a \in B(l^2)$ whose range is contained in ${\rm span}(e\_1)$) is isometrically isomorphic to $l^2$. Thus $f\_n \to 0$ weak\* because $f\_n(a) $ reads off the $n$th entry of the first row of $a$, which goes to $0$ since the row belongs to $l^2$. And $a\_n \to 0$ weakly because it lies in a subspace of $B(l^2)$ that is isometrically isomorphic to $l^2$, where it corresponds to $e\_n \in l^2$, and the sequence $(e\_n)$ goes to $0$ weakly in $l^2$.


