# Absolute value of $\sigma$-weakly convergent sequence of self-adjoint operators

Let $H$ be a Hilbert space and denote by $L(H)$ the space of all bounded linear operators of $H$. If $T_n\to 0$ in the $\sigma$-weak topology, it is not necessarily true that $\lvert T_n\rvert\to 0$ $\sigma$-weakly. The standard example is the following: Let $S$ be the shift on $\ell^2(\mathbb{N})$ and $T_n=S^n$. Then $T_n\to 0$ $\sigma$-weakly, but $\lvert T_n\rvert=((S^\ast)^nS^n)^{1/2}=1$.

My question is the following: If $(T_n)$ is a sequence of self-adjoint operators such that $T_n\to 0$ $\sigma$-weakly, is it true that $\lvert T_n\rvert\to 0$ $\sigma$-weakly?

Remark: A sequence $(T_n)$ converges $\sigma$-weakly to $0$ if $\mathrm{tr}(S T_n)\to 0$ for all trace class operator $S$.

Let $\{E_{kj}\}$ be the canonical matrix units. Define $T_n=E_{1n}+E_{n1}$. We have $$\langle E_{1n}x,y\rangle=x_n\overline{y_1}\xrightarrow[n\to\infty]{}0.$$ So $E_{1n}\to0$ in the weak operator topology, and so does $E_{n1}$, and thus so does $T_n$. But the weak operator topology agrees with the $\sigma$-weak topology on bounded sets, so $T_n\to0$ in the $\sigma$-weak.
But $|T_n|^2=(E_{1n}+E_{n1})^2=E_{11}+E_{nn}$, so $|T_n|=E_{11}+E_{nn}$. Thus $T_n$ is selfadjoint for all $n$, $$T_n\to0,\ \ \ \ |T_n|\to E_{11}.$$
Besides the concrete example, here is a more general fact: $$\overline{\{T\in L(H):\ \|T\|\leq1,\ T^*=T\}}^{\sigma-\text{weak}} =\{T\in L(H):\ \|T\|\leq1\}.$$ Actually, one can take even less than all the selfadjoints: the set of orthogonal projections is already $sigma$-weak dense in the unit ball.