I originally asked this on MSE, but did not get an answer there.


Let $M$ be a von Neumann algebra. Let $\varphi: M\_+ \to [0, \infty]$ be a weight on $M$. Consider
\begin{align\*}&\mathfrak{p}\_\varphi:= \{x\in M\_+: \varphi(x) < \infty\}\\
&\mathfrak{n}\_\varphi:= \{x \in M: \varphi(x^\*x) < \infty\}\\
& \mathfrak{m}\_\varphi:=\mathfrak{n}\_\varphi^\* \mathfrak{n}\_\varphi=
\left\{\sum\_{j=1}^ny\_j^\*x\_j: x\_j ,y\_j \in \mathfrak{n}\_\varphi\right\}\end{align\*}


In Takesaki's second volume, chapter VII, (the proof of) lemma 1.9, the following is claimed:


If $x =x^\* \in \mathfrak{m}\_\varphi$ and $x= u|x|$ is its polar decomposition, then $|x| \in \mathfrak{m}\_\varphi$.


**Question**: Why is this the case?


**Attempt**: We have $|x|= x^+ + x^-$, so I tried to show that $x^+, x^- \in \mathfrak{m}\_\varphi$. Now, we can write $x=p-q$ where $p,q \in \mathfrak{p}\_\varphi= \mathfrak{m}\_\varphi^+$ so if we would have $p \ge x^+$ and $q \ge x^-$, we would be done by the hereditary property. I think these inequalities are true when $p$ and $q$ commute, but I don't know if we can choose this decomposition with $pq = qp$.


Thanks in advance for any help or suggestions!



Decomposition of an element as a difference of positive elements in the definition subalgebra of a weight (Takesaki)
I don't know what Takesaki had in mind for the proof, but what you're asking is incorrect. Here is a counterexample where $\phi$ in the counterexample is a normal faithful semifinite (n.f.s.) weight.


Let $H$ be a separable infinite dimensional Hilbert space, and $M = B(H\oplus H)$. Fix a faithful normal state $\psi$ on $B(H)$ and let $\mathrm{Tr}$ be the standard trace on $B(H)$. Define $\phi ((x\_{ij})\_{i,j=1,2}) = \mathrm{Tr}(x\_{11}) + \psi(x\_{22})$. This is clearly a n.f.s. weight.


Fix a positive contraction $a\in B(H)$ which is trace-class but so that $a^{1/2}$ is not trace-class. Let $p = \left( \begin{array}{cc} a & (a-a^2)^{1/2} \\ (a-a^2)^{1/2} & 1-a \end{array} \right)$ and $q= \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)$ which are projections that are both in $\mathfrak m\_{\phi}$. Hence $x=p-q \in \mathfrak m\_\phi$ is self-adjoint. A straightforward computation gives $x^2 = \left( \begin{array}{cc} a & 0 \\ 0 & a \end{array} \right)$ and thus $|x| = \left( \begin{array}{cc} a^{1/2} & 0 \\ 0 & a^{1/2} \end{array} \right)$. As $a^{1/2}$ is not trace-class we have $\phi(|x|) = \infty$ so $|x| \notin \mathfrak m\_\phi$.


