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Let $M$ be a von Neumann algebra and $\varphi$ be a normal weight on it, and let $A$ be its definition subalgebra. We still denote $\varphi$ the extension to $A$ as a linear positive functional. It is known that $\varphi$ is lower-ultraweakly-semicontinuous on $M^+$ (the positive elements of $M$).

Questions:

  • Is $\varphi$ ultraweakly continuous on $A$ (as a linear positive functional), where $A$ has the induced ultraweak topology of $M$ ? (Clearly, if $A=M$ this assertion is classical, right ?)

  • If we fix a Hilbert space representation of $M$ (so that $M\subseteq \mathcal{B}(H)$). Do we have that $\varphi$ is strongly or weakly continuous on A ?

(If it helps, one can suppose $\varphi$ to be a trace, semifinite and faithful).

  • A third (somewhat related) question: Suppose that $B$ is a subalgebra of a von Neumann algebra, and that $f:B\to M$ is a positive linear map, such that $f$ is normal in the following sense : for any increasing net in $B^+$ with supremum in $B^+$, the image (by $f$) of this supremum is the supremum of the image of the net (the usual notion of normality, but with the hypothesis that the supremum lies in $B^+$). Do we have that $f$ is continuous (for the ultraweak topologies)?

These questions seem natural to me, but I haven't been able to locate any reference about them.

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1) if $A=M$ the assertion is indeed classical: normal states are exactly those ultraweakly continuous. But consider the case where $M=B(H)$ and $\varphi$ is the trace. Then the definition subalgebra $A$ is exactly the trace-class operators. Let $\{p_k\}\subset A$ be a maximal net of pairwise orthogonal projections of trace 1 (i.e. $\{e_{kk}\}$ for any choice of matrix units). Then $p_k\to0$ ultraweakly, but $\mbox{Tr}(p_k)=1$ for all $k$. So the trace is not ultraweakly continuous on the definition algebra, only ultraweakly lower-semicontinuous.

2) The example on 1) is already explicitly represented, so no.

3) Still thinking about it.

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    $\begingroup$ For (3), doesn't the same example work, considered as the map $T(H)\rightarrow B(H), x\mapsto \tr(x) 1$? I think this follows, as the question insists that the supremum of our increasing net in $B^+$ exists in $B^+$. $\endgroup$
    – Matthew Daws
    Commented Jan 29, 2012 at 9:38
  • $\begingroup$ I first answered that same example to 3, but then I thought that maybe the question required $B$ to be a von Neumann subalgebra and the map to be defined everywhere. So I stopped to consider cases like when $B$ is a monotone complete C$^*$-subalgebra of $B(H)$. Maybe Oliver can clarify what he expects in 3). $\endgroup$
    – Martin Argerami
    Commented Jan 29, 2012 at 13:28
  • $\begingroup$ I have just added a possible answer to 3). $\endgroup$
    – Martin Argerami
    Commented Jan 29, 2012 at 14:02
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    $\begingroup$ Never mind. Let's wait for Oliver to clarify question 3. $\endgroup$
    – Martin Argerami
    Commented Jan 29, 2012 at 14:46
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    $\begingroup$ Indeed, this answers also my question 3. Thanks! But I am actually also curious about the case you mentioned where $f$ is everywhere defined and $B$ is a von Neumann subalgebra. $\endgroup$
    – Oliver
    Commented Jan 29, 2012 at 20:42

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