# If $T_n\to T$ in the weak operator topology, does $p(T_n)\to p(T)$ in the weak operator topology for any polynomial $p$?

Let $$(T_n)_{n=1}^\infty$$ be a sequence of bounded operators mapping a Banach space $$X$$ to itself, and suppose that $$T_n\to T$$ in the weak operator topology; that is, for every $$y^*\in X$$* and any $$x\in X$$, we have $$y^*(T_nx)\to y^*(Tx)$$. Does it follow that for any polynomial $$p:\mathbb{C}\to\mathbb{C}$$, we have $$p(T_n)\to p(T)$$ in the weak operator topology as well?

I'm pretty sure that this is true if we replace weak convergence by strong convergence. In this case we know that the sequence $$(\Vert T_n\Vert)_{n=1}^\infty$$ is bounded by the uniform boundedness principle, and that multiplication is jointly continuous on bounded sets in the strong operator topology, so that $$T_n^k\to T^k$$ for any $$k\in\mathbb{N}$$. However we don't generally have joint continuity in the weak operator topology (as this example on Wikipedia shows).

Can we still show that $$p(T_n)\to p(T)$$ if we only assume weak convergence $$T_n\to T$$? If so, could you please provide a proof or a reference?

Edit: Even if this is false in general, I'd be interested to know if the convergence holds under special conditions - for example, all $$T_n$$ and $$T$$ commuting, the Banach space $$X$$ being reflexive, etc., as these conditions hold in the case I am currently interested in.

Let $$X=\ell_2(\mathbb N)$$, $$S$$ the shift operator that maps $$e_n$$ to $$e_{n+1}$$ and $$T_n=S^n+(S^\ast)^n$$. Then $$T_n\to 0$$ weakly, but $$T_n^2=S^{2n}+(S^\ast)^{2n}+I+P_{n+1}$$ (where $$P_{n+1}$$ is the projection onto the closed linear span of $$\{e_k\mid k\geq n+1\}$$), which does not converge to $$0$$ weakly. This settles the case of (infinite-dimensional) Hilbert spaces.
There are also counterexamples (on Hilbert spaces) where the $$T_n$$ commute. Let $$X=L_2([0,1])$$ and $$T_n$$ the multiplication by $$\operatorname{sgn}\sin nx$$. It is well-known that these functions converge to $$0$$ in the weak$$^\ast$$ topology on $$L^\infty([0,1])$$, which means that $$T_n\to 0$$ weakly. But $$T_n^2=I$$.