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.


1 Answer 1


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.

In particular, for Hilbert space application of polynomials is continuous in the weak operator topology if and only if the Hilbert space is finite-dimensional. An interesting question would be if the same is true for Banach spaces, i.e., if this continuity fails for all infinite-dimensional Banach 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 $T_n^2=I$.

  • $\begingroup$ Thanks for the answer! I'll accept this tomorrow in case anyone has anything to say about the Banach space case (but this does answer my question, so I will accept this). $\endgroup$
    – csch2
    Jul 20, 2021 at 17:48

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