Let $A_n$ be a sequence of positive operators on the Hilbert space $\mathcal{H},$ weakly convergent to an operator $A$ (necessarily positive). Does it imply the strong convergence ?
By the weak convergence we mean $$\lim_n\langle A_nx,y\rangle =\langle Ax,y\rangle,\qquad x,y\in \mathcal{H}$$ while the strong convergence
(or pointwise convergence) means
$$\lim_n\|A_nx-Ax\|=0,\qquad x\in \mathcal{H}$$
If $A_n\to 0$ weakly, then $A_n\to 0$ strongly
as $$\|A_n^{1/2}x\|^2=\langle A_nx,x\rangle \to 0$$ Hence $A_n^{1/2}\to 0$ strongly, and consequently $A_n=A_n^{1/2}A_n^{1/2}\to 0$ strongly (as the norms $\|A_n^{1/2}\|$ are uniformly bounded).
According to post, the condition $\lim\|A_nx\|=\|Ax\|$ and weak convergence imply the strong convergence for general class of operators.
Besides, if the sequence $A_n$ is monotonic, increasing $(A_n\le A_{n+1})$ or decreasing $(A_n\ge A_{n+1})$ then $A_n$ is convergent to $A$ strongly.
I was unable to come up with any example such that $A_n\to A$ weakly but not strongly and $A_n\ge 0$. Perhaps the weak convergence implies the strong convergence for positive operators ?
Best Answer
No, this does not hold. Let $S$ be the unilateral shift on $\ell^2(\mathbb N)$ and $A_n=S^n+(S^\ast)^n+2I$. Since $S$ is a contraction, $A_n\geq (-\lVert S\rVert^n-\lVert S^\ast\rVert^n+2)I\geq 0$. Clearly, $A_n\to 2I$ weakly, but $(A_n)$ does not converge to $2I$ strongly. In fact, if $n$ sufficiently large, then $A_n\delta_k=\delta_{k+n}+2\delta_k$, which has norm $\sqrt 5$.