Let $X$ denote a Banach space, $A$ a bounded linear map: $X \to X$. Let $\{M_{v}\}$ be a collection of bounded maps from $X$ to itself such that, $\forall x\in X,$ $AM_{v}x=M_{v}Ax$ (That is, the map $A$ conmutes with every $M_{v}$). Show that A conmutes with every bounded linear map belonging to the closed linear span of the set $\{M_{v}\}$ in the weak topology.
Earlier in the book he defines:
The weak topology in $L(X,U)$ is the topology generated by the functions of the form $M \to Mx$, $\forall x\in X$ (i.e, the weakest topology in wich all such linear maps are continuous).
I am able to prove that the result is true in the linear span (Just use linearity). But how can I extend to the weak closure? I don't even know if this topology is first countable, or sequential (If so, I can do it, too). And I wasn't able to prove it or give a counterexample.
Let $\phi \in X^*$, then $\phi_1,\phi_2$ defined by $\phi_1(M) = \phi(AM), \phi_2(M) = \phi(MA)$ are also in $X^*$.
We see that $(\phi_1-\phi_2)(M) = 0$ for all $M \in \operatorname{sp} \{ M_v\}$.
It follows that $(\phi_1-\phi_2)(M) = 0$ for all $M \in \overline{\operatorname{sp}^w} \{ M_v\}$ (weak closure).
It follows from this that $AM-MA = 0$.
Note: There is a simpler proof based on the fact that a convex set is strongly closed iff it is weakly closed.
Addendum: If $M \in \overline{\operatorname{sp}^w} \{ M_v\}$, then if $C$ is any weakly closed set containing $\operatorname{sp} \{ M_v\}$, we must have $M \in C$.
The set $\ker (\phi_1-\phi_2)$ is a weakly closed set that contains $\operatorname{sp} \{ M_v\}$, hence $\overline{\operatorname{sp}^w} \{ M_v\} \subset \ker (\phi_1-\phi_2)$.
More stuff: Since the weak topology is defined slightly differently we adapt the approach above:
Choose $x$ and define $\phi_{x}(M) = (AM-MA)x$. Then $\ker \phi_x$ is weakly closed for all $x$ and so $C=\cap_x \ker \phi_x$ is weakly closed. We see that $M \in C$ iff $AM=MA$.
We see that $M_v \in C$ for all $v$, hence $\overline{\operatorname{sp}^w} \{ M_v\} \subset C$.