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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

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I don't quite understand the bit about "describe the relation between ..." as it's not clear what is between what, as it were. –  Matthew Daws Jun 14 '11 at 18:42
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. –  Romanov Jun 14 '11 at 18:52
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. –  Benjamin Hayes Jun 15 '11 at 3:20
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 $E_{(s,\infty)}(|T|) \preceq E_{(s,\infty)}(|S|)$. –  Romanov Jun 16 '11 at 15:03
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. –  Romanov Jun 16 '11 at 15:03
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1 Answer

up vote 3 down vote accepted

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) \[ E_{(s,\infty)}(T)=E_{(s,\infty)}(T)-E_{(s,\infty)}(T)\wedge E_{[0,s]}(S)\sim E_{(s,\infty)}(T)\vee E_{[0,s]}(S)-E_{[0,s]}(S)\leq I-E_{[0,s]}(S)=E_{(s,\infty)}(S) \]

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: \[ \langle T\xi,\xi\rangle=\langle TE_{(s,\infty)}(T)\xi,E_{(s,\infty)}(T)\xi\rangle =\|T^{1/2}E_{(s,\infty)}(T)\xi\|^2>s, \] \[ \langle T\xi,\xi\rangle=\langle E_{[0,s]}(S)TE_{[0,s]}(S)\xi,\xi\rangle \leq\langle E_{[0,s]}(S)SE_{[0,s]}(S)\xi,\xi\rangle=\|S^{1/2}E_{[0,s]}(S)\xi\|^2\leq s \] The contradiction implies that $\xi$ cannot exist.

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I definitely agree. It was the key point! Thank you very much! –  Romanov Jun 16 '11 at 17:00
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