Timeline for Measurable functions and unbounded operators in von Neumann algebras
Current License: CC BY-SA 2.5
5 events
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| Nov 16, 2009 at 4:35 | comment | added | Semyon Dyatlov | I have read my own question and realized that it did not ask what I had in mind... sorry about that. Question updated. | |
| Nov 14, 2009 at 22:14 | comment | added | Dave Penneys | To show $M_f$ has the properties stated, see Section VIII.3 of Reed and Simon's Functional Analysis. For the second part, GNS with respect to what state? Are you talking about the semi-cyclic representation from the trace? In that case, yes, you get $B(H)$ acting by left multiplication on the Hilbert-Schmidts. The set of unbounded operators on $H$ is just the set of pairs $(D(T),T)$ where $D(T)$ is a subspace and $T\colon D(T)\to H$ is a linear transformation. Unbounded really means not necessarily bounded. | |
| Nov 14, 2009 at 4:52 | comment | added | Semyon Dyatlov | Thanks! I still don't see, though, why any measurable function $f(x)$ will give a measurable operator. It is required that the restriction of $f$ to the set $\\{f>N\\}$ is integrable for some $N$, which is wrong, for example, for $f(x)=1/x$ with the Lebesgue measure. Also, for the algebra of bounded operators on a Hilbert space $H$, doesn't GNS construction + affiliated operators give unbounded operators on the space of all Hilbert-Smidt operators on $H$? What I am looking for is how to get all unbounded operators on $H$, or an explanation why this question does not make sense. | |
| Nov 14, 2009 at 2:52 | history | edited | Dave Penneys | CC BY-SA 2.5 |
deleted 35 characters in body
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| Nov 13, 2009 at 18:25 | history | answered | Dave Penneys | CC BY-SA 2.5 |