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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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...

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.