SOT limit of Self-adjoint operators is self-adjoint?

Question

Say $T_n$ is a sequence of self-adjoint operators on a Hilbert space and converges in the strong operator topology to $\mathcal{T}$, must $\mathcal{T}$ be self-adjoint? Since $T_nx$ converges to $\mathcal{T}x$ in norm, it converges weakly, and so I figured that

$\langle\mathcal{T}x,x\rangle-\langle x,\mathcal{T}x\rangle=\lim_{n\to \infty}\langle T_nx, x\rangle-\langle x, T_nx\rangle=0,$

which does the job, but there's a chance that my limit operators aren't justified.

Thanks.

You've got no reason to worry; everything's correct in your solution. — Mateusz Wasilewski, Mar 14 '13 at 19:45
Does this statement remain true when SOT is replaced by the norm topology? — eatfood, Apr 25 '20 at 10:59
@eatfood Yes, since convergence in norm topology implies convergence in SOT, simply because $|T_nx - Tx|\leq|T_n-T| \ |x|$. — Rodrigo Dias, Jan 12 '21 at 19:34

score 4 · Answer 1 · answered Mar 14 '13 at 19:48

That's correct. Your assumptions are $(T_nx,y)\longrightarrow (Tx,y)$ for all $x,y$ (sot hence wot limit). And $(T_nx,y)=(x,T_ny)$ for all $x,y$ and all $n$ (self-adjoint). So $$(Tx,y)=\lim (T_nx,y)=\lim (x,T_ny)= (x,Ty)$$ for all $x,y$. That is: $T$ is self-adjoint. The first and the third equalities are from the wot condition. The middle equality is by self-adjointness of each $T_n$.

SOT limit of Self-adjoint operators is self-adjoint?

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