Trace class operator is an ideal

Question

To show that trace class operators space is an ideal, we need to show that $\|uv\|_1\leq \|u\|\|v\|_1$ where $u,v \in B(H)$ and $\|u\|_1 = tr(|u|)$.

Murphy in his book (C*-algebras and operator theory) prove it as below:

Let $u=w|u|$ and $vu = w'|vu|$ be the polar decomposition of $u$ and $vu$ respectively and $w''=w'vw$ . Then $|vu| = w'^*vu=w''|u|$. Hence, $|vu|^2=|u| w''^*w''|u| \leq |u|^2\|w''\|^2 \leq |u|^2\|v\|^2$, so $|vu|\leq |u|\|v\|$.

and he continues his proof to get the desired inequality. But I can not understand how he concludes $|u| w''^*w''|u| \leq |u|^2\|w''\|^2$. Please help me. Thanks.

Answer 1

Note first that $w''^*w''$ is a positive semidefinite operator such that $w''^*w''\leq \|w''^*w''\|$ (on the right is a positive scalar times the identity operator). Since in $C^*$-algebra $\|w''^*w''\|=\|w''\|^2$ we have $w''^*w''\leq \|w''\|^2$. Multiply this inequality from the left and the right side by a positive semidefinite $|u|$ and you have the desired inequality.