Is every weakly closed bounded convex set of $B(H)$ is weakly compact? I saw that in many papers, and I don't know why they concluded the weak compactness. e.g. Page 382 .
Is every weakly closed bounded convex set of $B(H)$ is weakly compact?
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functional-analysis
operator-theory
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0What is $H$? To what weak topology on $B(H)$ do you refer? With respect to the weak topology of $B(H)$ as a Banach space, or to the weak operator topology? – 2017-01-09
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0If you refer to a Hilbert space $H$ and the weak operator topology, see http://math.stackexchange.com/questions/1092119/closed-unit-ball-of-bh-with-wot-topology-is-compact – 2017-01-09
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0@martini not WOT, I meant Weak Banach topology. I know the unit ball of B(H) is WOT compact, however, in these papers, they must use Weak topology (not necessarily Banach, could be the weak topology of semi-norms). I am totally confused. – 2017-01-09
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0As far as I see it, the author refers to the WOT by "weak topology". First, otherwise $B_{B(H)}$ is not weakly compact, and second: In the introduction he writes "ultraweak, weak, strong and ultrastrong", this for me only makes sense if he is refering to the WOT with "weak topology". – 2017-01-09
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0http://www.sciencedirect.com/science/article/pii/0022123678900290?np=y However, in this paper, theorem 5.1. The author also refers to the reference I gave above. But in this paper, the author used Ryll-Nardzewski theorem, which requires Weak topology rather than WOT. @martini – 2017-01-09
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0Also, in a paper published in DUKE MATH J, I also saw that the authors used Ryll-Nardzewski theorem in a WOT compact set. @martini http://projecteuclid.org/euclid.dmj/1077307046 section 7.4 – 2017-01-09