Finite rank operators spanned by rank-one projections proof.

Question

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

Could someone explain the equality $pu = u^{1/2} p u^{1/2}$. I can see that this holds when $u$ is invertible. Thanks in advance.

Answer 1

You have $u=pu$. Since $u$ is selfadjoint, taking adjoints you get $u=up$. So $pu=up$. Then $pu^n=u^np$ for all $n\in\mathbb N$ and thus $pf(u)=f(u)p$ for all polynomials $f$. Since $u^{1/2}$ is a limit of polynomials on $u$ (it's actually a polynomial and not limit because $u$ is finite-rank, but we don't need that here), $pu^{1/2}=u^{1/2}p$.

Answer 2

Note that $u^{1/2}(H)=u(H)$ so $p$ is identity on the image of $u^{1/2}$ and $u^{1/2}pu^{1/2}=u^{1/2}u^{1/2}=u$.