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:= \{y^*x: x ,y \in \mathfrak{n}_\varphi\}\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!

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!