if $A_n$ weakly converges to $A$, does $|A_n| \rightarrow _{wo} |A|$?

Question

Suppose that $A_n,A$ are self-adjoint operators in $B(H)$. If $A_n$ weakly converges to $A$, does $|A_n| \rightarrow _{wo} |A|$?

From Proposition. 2.3.2 of Pederson'book, I know the result holds in the strong case.

Answer 1

When $H$ is infinite-dimensional, the set of unitaries is weakly dense in the unit ball. So there exists a net of unitaries such that $u_n\to \frac12\,I$. As $|u_n|=I$ for all $n$, the net $|u_n|$ does not converge to $|\frac12\,I|=\frac12\,I$.

For a proof of the density of the unitaries, let $x\in B(H)$ be a contraction. Given $\varepsilon>0$ and $\xi_1,\ldots,\xi_n\in H$, we need to find a unitary $u$ such that $|\langle (u-x)\xi_j,\xi_j\rangle|<\varepsilon$ for all $j$. Basically one considers the range projection $p$ onto the span of $\xi_1,\ldots,\xi_n$, a projection $q$ equivalent to $p$ and orthogonal to it; then we can consider the operators in $B(H)$ as $3\times 3$ matrices in terms of $p,q,1-p-q$. We can now define $u$ to be $$ u=\begin{bmatrix}x&(1-xx^*)^{1/2}&0\\ (1-x^*x)^{1/2}&-x^*&0\\0&0&1 \end{bmatrix}. $$ In that case, $$\langle u\xi,\xi\rangle=\langle up\xi,p\xi\rangle=\langle pup\xi,\xi\rangle=\langle x\xi,\xi\rangle.$$