I'm looking at the following proof, where $||u||_1 = \sum_{x \in E} \langle |u|(x),x\rangle$ is the trace class norm of an operator and $||u||_2 = \left(\sum_{x \in E} ||u(x)||^2 \right)^{\frac{1}{2}}$, for $E$ an orthonormal basis of $H$.
They use the following inequality while proving (1):
I was under the impression initially what they were doing was $|| |u|^{\frac{1}{2}}w^{*}w''||_2 \leq || |u|^{\frac{1}{2}}||_2 \: ||w^{*}||_2 \: ||w''||_2$ and then using the fact that they were partial isometries, but $w$ being a partial isometry doesn't imply that $||w||_2 \leq 1$ unless $w = 0$ or has rank $1$. What are they doing here?
What's being used here is part (3) of Theorem 2.4.10 in Murphy's book: for operators $u,v$ on $H$, $\|uv\|_2\leq\|u\|_2\|v\|$. Using this, we have $$\||u|^{1/2}w^*w''\|_2\leq\||u|^{1/2}\|_2\|w^*w''\|\leq\|u\|_1^{1/2}\|w^*\|\|w''\|=\|u\|_1^{1/2}.$$ This gives us the first term in the sum, and the inequality for the second term follows the exact same reasoning.