Inequality in proof concerning trace class norm

Question

Consider the following fragment from Murphy's '$C^*$- algebras and operator theory':

Why is the marked inequality true? I.e. why is $\Vert w'' \Vert \leq \Vert v \Vert?$

Answer 1

$$\|w''\|=\|w^{\prime*}vw\|\le\|w^{\prime*}\|\|v\|\|w\|=\|v\|$$

P.S. $\|w\|=1$ for partial isometries since $\|wx\|\le\|x\|$ for all $x$, with equality on some non-trivial subspace. A partial isometry on a Hilbert space is defined by $(w-1)Y=0$ on some closed subspace $Y$ and $wY^\perp=0$.

Even if one works purely in $C^*$-algebra terms, so a partial isometry is defined by $(w^*w)^2=w^*w$, then $\|w\|^2=\|w^*w\|=\|(w^*w)^2\|=\|w\|^4$.