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Let $H$ be a complex separable Hilbert space and $C^1(H)$ the space of trace class compact operators on $H$. My question is:

Is the trace function $\mbox{Tr}:C^1(H)\rightarrow \mathbb{C}$ continuous with respect to the weak topology on $C^1(H)$?. I guess that it is just lower semicontinuous,but I'm not shure.

None
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  • Every norm-continuous linear functional on a normed space is weakly continuous. – Ruy Oct 24 '20 at 22:04
  • While Martin's and Ruy's answer is of course correct, let me add a quick side note: If by "weak topology" you really meant "weak operator topology" then the answer is no. Equipping the trace class with any operator topology weaker than the one induced by the trace norm (in infinite dimensions), unsurprisingly, turns the trace into a discontinuous functional. – Frederik vom Ende Oct 30 '20 at 13:24

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The weak topology is given, by definition, by convergence under the norm-continuous linear functionals. So every norm-continuous linear functional is weakly continuous. By the inequality $$ |\operatorname{Tr}(A)|\leq \operatorname{Tr}(|A|)=\|A\|_1, $$ the trace is continuous, so it is in the dual and it thus weakly continuous.

Martin Argerami
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