If $u \in B(H)$ is a bounded linear operator on a Hilbert space and $u=w|u|$ is it's polar decomposition: I.e. w is a partial isometry ($w=ww^*w$) and also $|u|=w^*u.$
I'm trying to prove that then $|u*|=w|u|w*$ By arguing that $uu^*=w|u|^2 w^*$, and thus $|u^*|^2=uu^*=w|u|^2 w^*$ What I am wondering is why does $w|u|^2 w^*=(w|u|w^*)^2$? This seems to use $w^2=id$ which I believe doesn't hold in general.
Note that it's not $w^2=1$ (which can fail badly, you can even have $w^2=0$) that you need but rather $w^*w=1$, and not even that; what you need is that the projection $w^*w$ leaves the range of $|u|$ untouched.
If you are using "the" polar decomposition (that is, the unique version) then it is a given that $w^*w$ is the range projection of $u^*$ and $|u|$ (that is the projection onto the closure of the range of these operators). Then $$ (w|u|w^*)^2=w|u|w^*w|u|w=w|u|\,|u|w^*=w|u|^2w^*. $$ And even if you don't have the condition as a given, from $u=w|u|$ you get $$ |u|^2=u^*u=|u|w^*w|u|. $$ That is, $$|u|(1-w^*w)|u|=0.$$ This gives you, as long as you know that $w$ is a contraction, $$(1-w^*w)^{1/2}|u|=0.$$ Now multiply by $(1-w^*w)^{1/2}$ to get $w^*w|u|=|u|$.