Hello, I have a question which is related to a partial order in a set of self-adjoint operators.

Let $\mathcal{M}$ be a semifinite von Neumann algebra with a faithful semi-finite normal trace $\tau$. Let $T$ and $S$ be two self-adjoint operators (possibly unbounded) $\tau$-measurable (here probably the assumption that they are affiliated with $\mathcal{M}$ is enough) such that $0 \leq T \leq S$ i.e. $S-T$ is positive. How to get that $$E_{(s, \infty)}(|T|) \preceq E_{(s, \infty)}(|S|), \ \ s \geq 0,$$ where $E_I(|T|)$ (resp. $E_I(|S|)$) stands for a spectral projection of $T$ (resp. $S$) corresponding to the interval $I$ and $\preceq$ means sub-equivalence relation in Murray-von Neumann sense.

I am looking also for some good references which describe the relation between $U|T|$ the elements of the polar decomposition of closed densely defined (possibly unbounded) operator $T$ affiliated with some von Neumann algebra $\mathcal{M}$. I mean that $U$ and each spectral projection of $|T|$ are in this von Neumann algebra. Probably, I can find this in Takesaki vol 2 or vol 3.

I will be really grateful for any help.

Thank you, VdM

  • $\begingroup$ I don't quite understand the bit about "describe the relation between ..." as it's not clear what is between what, as it were. $\endgroup$ – Matthew Daws Jun 14 '11 at 18:42
  • $\begingroup$ Sorry, my mistake I mean the relation between $U|T|$ and von Neumann algebra $\mathcal{M}$ i.e. that $U$ and each spectral projection of $|T|$ are in this von Neumann algebra. I know it suffices to show that $U$ and $\textbf{1}(|T|)$ are in $\mathcal{M}$, because by virtue of Double Commutant Theorem the spectral projection of $f(|T|)$ are there since $\textbf{1}(|T|)$ is. I am looking for some good references for the theory of the operators affiliated with some von Neumann algebra. $\endgroup$ – Romanov Jun 14 '11 at 18:52
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    $\begingroup$ In Takesaki II Problem IX.7 there is a outline of a proof to the statment that $\tau(f(S))\leq \tau(f(T))$ for any $f\geq 0$ continuous with $f(0)=0.$ My guess is that one can work with this a little bit to show that the same inequality holds for $f=\chi_{(s,\infty)}$ which at least handles the factor case. The general case you may be able to do by working with the extended center-valued trace but I don't really know. $\endgroup$ – Benjamin Hayes Jun 15 '11 at 3:20
  • $\begingroup$ The first part of my question is in particular a part of this problem in Takesaki. Because $\mu_t(T) \leq \mu_s(T)$ iff $\lambda_s(T)= \tau(E_{(s,\infty)}(|T|)) \leq \tau(E_{(s,\infty)}(|S|))=\lambda_s(S)$ iff $E_{(s,\infty)}(|T|) \preceq E_{(s,\infty)}(|S|)$. $\endgroup$ – Romanov Jun 16 '11 at 15:03
  • $\begingroup$ Another property of $s$-numbers is that $\mu_t(f(T))= f(\mu_t(T))$ for increasing continuous $f$ on $[0,\infty) with f(0) \geq 0$ $\tau(T) = \int_{0}^{\infty}\mu_t(T) dt$ for positive $\tau$ measurable $T$ we have $$\tau(f(S)) = \int_{0}^{\infty} f(\mu_t(S)) dt \leq \int_{0}^{\infty} f(\mu_t(T)) dt = \int_{0}^{\infty} \mu_t(f(T))= \tau(f(T)).$$ So this is not a good point. $\endgroup$ – Romanov Jun 16 '11 at 15:03

I assume you are following the proof in Fack-Kosaki (if you are not, we are talking here about Proposition 2.2 and 2.5 there).

Note that there is no need for absolute value bars since both $T,S$ are positive.

The key fact is that $E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)=0$ (to be proven afterwards). Using this, we have (using Kaplansky's formula) \begin{align} E_{(s,\infty)}(T)&=E_{(s,\infty)}(T)-E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)\sim E_{(s,\infty)}(T)\vee E_{[0,s]}(S)-E_{[0,s]}(S)\\ &\leq I-E_{[0,s]}(S)=E_{(s,\infty)}(S) \end{align}

So we only need to prove that $E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)=0$. Now, if $\xi\in E_{(s,\infty)}(T)H \cap E_{[0,s]}(S)H$ with $\|\xi\|=1$, the following happens: \begin{align} \langle T\xi,\xi\rangle&=\langle TE_{(s,\infty)}(T)\xi,E_{(s,\infty)}(T)\xi\rangle =\|T^{1/2}E_{(s,\infty)}(T)\xi\|^2>s, \end{align} \begin{align} \langle T\xi,\xi\rangle&=\langle E_{[0,s]}(S)TE_{[0,s]}(S)\xi,\xi\rangle \leq\langle E_{[0,s]}(S)SE_{[0,s]}(S)\xi,\xi\rangle=\|S^{1/2}E_{[0,s]}(S)\xi\|^2\leq s \end{align} The contradiction implies that $\xi$ cannot exist.

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  • $\begingroup$ I definitely agree. It was the key point! Thank you very much! $\endgroup$ – Romanov Jun 16 '11 at 17:00
  • $\begingroup$ Look at Marianne Tarp's lecture notes on noncommutative $L^p$ spaces. Here is a link fuw.edu.pl/~kostecki/scans/terp1981.pdf. $\endgroup$ – A beginner mathmatician Aug 13 '19 at 5:36
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    $\begingroup$ @Abeginnermathmatician Terp's work is mostly needed/relevant dealing with the non-semifinite case; noncommutative L^p-spaces associated to semifinite von Neumann algebras go back much further and are explained well in older sources. Moreover, you do not indicated why Terp's notes should be helpful for the question asked by the OP $\endgroup$ – Yemon Choi Aug 14 '19 at 0:43

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