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I'm trying to prove that, on a Hilbert space $H$, convergence of a sequence of bounded operators $a_n$ to $a$ w.r.t. the weak operator topology (i.e. $\langle a_n x ,y \rangle \to \langle ax,y \rangle$ for all $x,y \in H$) implies that $a_n$ is norm-bounded. I have managed to prove the analogous statement for the strong operator topology using the Uniform Boundedness Principle, but I can't quite get it to work in the weak case. Could someone provide a hint?

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Fix $x$. Consider the linear functionals $f_x:y\longmapsto \overline{\langle a_nx,y\rangle}$. Because the sequence of numbers $\{\langle a_nx,y\rangle\}$ is convergent, it is bounded. That is, $$\tag1 \sup_n|f_n(y)|=\sup_n|\langle a_nx,y\rangle|<\infty. $$ By the Uniform Boundedness Principle, we get that $$\tag2 \sup_n\|a_nx\|=\sup_n\sup_{\|y\|=1}|\langle a_nx,y\rangle|<\infty. $$ As $(2)$ holds for all $x$, applying the UBP again we get $$ \sup_n\|a_n\|<\infty $$

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  • $\begingroup$ Thanks for the detailed answer. Though I guess you need to put a conjugate in the first line, to make the functional really linear, instead of conjugate-linear. $\endgroup$ – Wizact Apr 17 at 6:28
  • $\begingroup$ Indeed. Or I could say they are conjugate-linear ;) $\endgroup$ – Martin Argerami Apr 17 at 9:01

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