Let $\varphi$ be a linear functional continuous with respect to strong operator topology of $B(H)$ on the unit ball. Is it imply that $\varphi$ is weak operator continuous on the unit ball of $B(H)$?
2026-03-25 22:41:28.1774478488
Linear functionals continuos on the unit ball with respect to SOT implies it is continuous with respect to WOT on the unit ball
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Yes, this is contained (for states) in Theorem 7.1.12 in volume 2 of Kadison-Ringrose. For arbitrary functionals it can be obtained from states via the polar decomposition for functionals: since $\varphi$ is normal, there exists a partial isometry $v$ and a positive normal $\omega$ such that $\varphi=\omega(v\cdot)$ (see for instance Theorem III.4.2 in Takesaki's Volume I). If $x_j\to0$ wot, then $vx_j\to0$ wot, and $$\varphi(x_j)=\omega(vx_j)\to0$$ by the aforementined result.