Let $H$ be a Hilbert space and assume that $(\gamma_n)_n\subset B(H)^*$, $\gamma_n\overset{*}{\rightharpoonup}\gamma$, and $(A_n)_n\subset B(H)$, $\|A_nx-Ax\|\to 0$ for all $x\in H$. As it is easily seen that $\gamma_n(A_n)\to\gamma(A)$ if $A_n\to A$ in norm, I was wondering whether this holds also true for the strong operator convergence.


No, it's not true. For an easy example fix an ortonormal basis $\{e_n\}$ and let $E_{1n}$ be the rank-one operator that sends $e_n$ to $e_1$. For any $x\in H$, $$ \|E_{1n}x\|^2=\langle E_{1n}x,E_{1n}x\rangle=|x_n|^2\to0. $$ So $E_{1n}\to0 $ sot. Let $$\gamma_n(T)=\langle Te_n,e_1\rangle.$$ Then $$ \gamma_n(T)=\langle e_n,T^*e_1\rangle\to0, $$ since $\|T^*e_1\|^2=\sum_n|\langle T^*e_1,e_n\rangle|^2=\sum_n|\langle e_n,T^*e_1\rangle|^2$.

So $\gamma_n\to0$ in the weak$^*$-topology, $E_{1n}\to0$ sot, and $$\gamma_n(E_{1n})=\langle E_{1n}e_n,e_1\rangle=\langle e_1,e_1\rangle=1$$ for all $n$.

  • $\begingroup$ Nice counterexample. Thanks. $\endgroup$ – julian May 25 '18 at 4:14
  • $\begingroup$ Glad I could help. $\endgroup$ – Martin Argerami May 25 '18 at 4:46

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