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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 $M^{**}\times M^*\to M^*$.

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    $\begingroup$ Are you sure that for joint continuity it suffices to show that $a_if_i \to af$? I would have expected that one needs to consider something like the product net $(a_if_j)_{i,j\in (I\times J})$ with a suitable ordering. Could you please clarify whether you need joint continuity, or the seemingly weaker requirement that you have stated $\endgroup$
    – Yemon Choi
    Commented Feb 6, 2016 at 15:54
  • $\begingroup$ A first start would be to see what happens if $M=\ell^\infty$ or $M=L^\infty[0,1]$... $\endgroup$
    – Yemon Choi
    Commented Feb 6, 2016 at 15:54
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    $\begingroup$ @YemonChoi: I think the statement of joint continuity is correct. $\endgroup$
    – Nik Weaver
    Commented Feb 6, 2016 at 16:08
  • $\begingroup$ @NikWeaver oh, because you can enlarge the indexing set or refine to a subnet, or similar? $\endgroup$
    – Yemon Choi
    Commented Feb 6, 2016 at 16:13
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    $\begingroup$ Joint continuity means: continuous from the product topology. So take an arbitrary convergent net in $M\times M^*$. That is a net of the form $(a_i, f_i)$ which converges to some $(a,f)$, i.e., $a_i \to a$ and $f_i \to f$. And you want its image to converge, i.e., you want $a_if_i \to af$. $\endgroup$
    – Nik Weaver
    Commented Feb 6, 2016 at 16:17

1 Answer 1

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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 $l^2$.

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  • $\begingroup$ Sorry for delay. $\endgroup$
    – ABB
    Commented Feb 8, 2016 at 13:19

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