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Let $\mathcal{A}$ be the von Neumann algebra of bounded operators on a Hilbert Space and $\mathcal{A}_{*}$ its predual. Further consider a weakly convergent sequence of continuous and bounded operators $V^{m}:[0,T]\rightarrow\mathcal{A}_{*}.$ I know that the limit $V$ is bounded (with that I mean that every element in $V([0,T])$ is bounded) by the Uniform boundedness principle but is the limit continuous as well? This might be an easy consequence or just plain wrong. I feel like I'm missing something...

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You're going to need some notion of uniform convergence to ensure that the limit $V$ is continuous. For example, fix $a\in\mathcal A_*$ non-zero, and define $V^m:[0,1]\to \mathcal A_*$ by $V^m(t)=t^ma$. Then the sequence $(V^m)$ converges pointwise to $V:[0,1]\to\mathcal A_*$ defined by \begin{align*} V(t)=\left\{\begin{array}{lcl}0&:&t\in[0,1),\\ a&:&t=1, \end{array}\right. \end{align*} which is certainly not continuous.

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  • $\begingroup$ I see. Thank you :) $\endgroup$
    – Alvo
    Jul 21 '19 at 20:04
  • $\begingroup$ You're welcome, glad to help! $\endgroup$
    – Aweygan
    Jul 21 '19 at 20:12

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