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I don't believe this to be true but i am having trouble in finding a counterexample of this: let $T_n$, $n \in \mathbb{N}$ be a sequence of self adjoint operators in a Hilbert space $H$ that converges in the WOT topology to $T$. Show that $T_n$ converges in the SOT topology.

I'll show all the reasonings that i did and how i am trying to find a counterxample.

Clearly i cannot use the classical counterexample of the shift operator since it is not self adjoint. What i know is that the WOT limit of self adjoint operators is self adjoint, hence $T$ is self adjoint.
Given this, in case i want to prove that $T_n$ converges SOT it's enough for me to prove that, given an Hilbert Basis $\{e_i\}_{i \in \mathbb{N}}$ (writing $T(Te_i)$ as $T^2e_i$)

$(T_n^2 e_i \cdot e_i) \rightarrow (T^2e_i \cdot e_i)$

If this helds true then i can easily deduce the SOT convergence by definition. If there were some hypotesis of monotonicity of the operators (or maybe some kind of non-negativity) i could also prove this limit.

So in searching for a counterxample i am looking for some sequence of self adjoint operators such that $(T_n e_i \cdot e_i) \rightarrow (T e_i \cdot e_i)$ but $(T_n^2 e_i \cdot e_i) \nrightarrow (T^2 e_i \cdot e_i)$

My main problem is that most of the self adjoint operators that i am trying to use all end up to converge SOT too. Maybe this is true and i am missing some crucial step in the proof? I am pretty sure that the core element is proving $(T_n^2 e_i \cdot e_i) \rightarrow (T^2e_i \cdot e_i)$.

PS: I've found some results that show that the closure in the SOT topology of self adjoint operators is the same as the WOT topology but in the proof it's clear that the successions in the closure can be different.

d.polv
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  • If $S$ is the shift operator, can you just take its real part? Try $T = (S + S^*)/2$. I haven’t checked the details, but that would be my first guess if you wanted to fix up the shift example. – Josh Keneda Dec 31 '23 at 18:43
  • Is your sequence norm-bounded? Otherwise, it is not sufficient to check strong convergence on an orthonormal basis. – J. De Ro Dec 31 '23 at 23:40
  • If you are able to find a counterexample $T_n$ not necessarily selfadjoint, then $T_n=R_n+iS_n$ where $R_n$ and $S_n$ are selfadjoint. Then either $R_n$ or $S_n$ will provide a selfadjoint counterexample. – Ryszard Szwarc Jan 01 '24 at 01:22
  • @J.DeRo yes the sequence is norm bounded by uniform boundedness principle. All the $T_n$ are bounded and weakly converging. – d.polv Jan 01 '24 at 11:06
  • @JoshKeneda I'm checking the real part suggestion on the sequence of the shift. The main problem is that even though it's self-adjoint when you iterate the real part it's really hard to check the convergence. If i write R right shift and S the left shift (which are adjointed) then $(S + R)^n = T_n$ should be the sequence but since S and R do not commute it gets really hard to check the convergence of this sequence – d.polv Jan 01 '24 at 11:06
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    Consider the sequence $S^n+(S^*)^n.$ This sequence is convergent to $0,$ weakly, but doesn't tend to $0$ strongly. By the way the sequence the sequence $(S+R)^n$ is unbounded hence it cannot be weakly convergent. However $T_n={1\over 2^n}(S+R)^n$ tends to $0$ strongly and $|T_n|=1.$ – Ryszard Szwarc Jan 01 '24 at 14:17
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    @d.polv Ryszard’s suggestion to consider $S^n + (S^*)^n$ eliminates the need to consider all of the complicated terms. That seems like the cleanest approach to me. – Josh Keneda Jan 01 '24 at 19:43
  • I.e. it’s better to consider the real parts of the powers of $S$, not the (more complicated) powers of the real part of $S$. – Josh Keneda Jan 01 '24 at 20:00

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