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Hello,

I have a question about trace measurable operators and I think it's not a hard one. However, I'm quite confused because I cannot prove it.

Let $\mathcal{M}$ be a semi-finite von Neumann algebra with a faithful normal semi-finite trace $\tau$. Let $T$ be a $\tau$- measurable operator (densely defined closed (possibly unbounded) operator affiliated with $\mathcal{M}$ such that $$ \forall_{\varepsilon >0} \ \exists_{E - \text{a projection in} \text{M}} \ \mbox{Range}(E) \subset D(T) \ \& \ \tau(1-E) \leq \varepsilon.)$$

Let $E_{(s,\infty)}(|T|)$ be a spectral projection of $|T|$ corresponding to the interval $(s, \infty)$, $s \geq 0$.

How do we know that $\| |T|E_{(s,\infty)}(|T|) \| > s$ or $\| |T|E_{[0,s]}(|T|) \| \leq s$.

Thank you in advance for any help.

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  • $\begingroup$ I don't think this has anything to do with "measurable operators"-- it just follows from the definition of the spectral projections... $\endgroup$ – Matthew Daws Jun 16 '11 at 9:10
  • $\begingroup$ Ok, I didn't know that because this is a part of the problem IX.2.7 in Takesaki vol. 2 and this problem concerns "measurable operators". I can't see this from the definition of the spectral projection. $\endgroup$ – Romanov Jun 16 '11 at 14:23
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Like Matthew said, this doesn't look like it has to do with measurability at all. Those two inequalities follow from the operator inequalities \[ |T|\,E_{(s,\infty)}(|T|) \geq s\,E_{(s,\infty)}(|T|), \ \ \ |T|\,E_{[0,s]}(|T|) \leq s\,E_{[0,s]}(|T|). \] In turn, these inequalities follow from the corresponding inequalities for real functions, \[ t\,1_{(s,\infty)}(t) \geq s\,1_{(s,\infty)}(t), \ \ \ t\,1_{[0,s]}(t)\leq s\,1_{[0,s]}(t), \] and the fact that functional calculus is positive.

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  • $\begingroup$ It seems that $f$ is missing on the right hand side, isnt it? $\endgroup$ – Tomek Kania Jun 16 '11 at 16:45
  • $\begingroup$ That's for the observation, Tomasz. Actually, there was no need for $f$, so I've removed it. $\endgroup$ – Martin Argerami Jun 16 '11 at 16:54
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Maybe I should give an answer, not just a comment. The following maybe isn't the "correct" way to think about spectral projections, but I find that it helps my intuition. Basically, if $T$ is a normal, possibly closed unbounded, operator on a Hilbert space $H$, then $H$ is unitarily equivalent to a measure space $L^2(\mu)$ and under this equivalence, $T$ is just a multiplication operator by a complex function $f$ (which is bounded if and only if $T$ is bounded).

If now $T$ is positive, then so is $f$. The spectral projection $E_{(s,\infty)}(T)$ then corresponds to the projection given by the indicator function of the set $\{ x : f(x)>s \}$. It's then clear that $T E_{(s,\infty)}(T)$ is multiplication by the function $g$ defined by $g(x)=f(x)$ if $f(x)>s$, and $g(x)=0$ otherwise. If $T$, and hence $f$, is unbounded, then so will be $g$.

Anyway, I find this mental picture of "it's just multiplication operators" to be handy when dealing with spectral stuff. Martin's arguments are probably "cleaner" if you want a polished version...

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  • $\begingroup$ Thank you for that, this is a really nice application of the spectral theorem. $\endgroup$ – Romanov Jun 16 '11 at 21:33

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