# Weak limit of continuous operator

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...

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.