1
$\begingroup$

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 $S$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.