Let $A$ be a finite von Neumann algebra acting on $B(H)$ and $p \in A$ be a non-zero projection such that $p \notin A'$. Let $u =E(p)\in A \cap A'$ such that $u$ is not a projection, where $A'$ is the commutant of $A$ and $E$ is the center-valued trace. Since $u \in A\cap A'$, then the spectral projection $z=\mathbf 1_{\left(0,\frac{1}2\right]}(u) \in A \cap A'$ $($using bounded Borel functional calculus$).$ Also assuming that $z \ne 0$ and let $q=E(pz)$, where $E$ is the center-valued trace. Then prove that $pz$ is a projection with $q \le \frac{1}2$.
Clearly $pz$ is a projection, since both $p,z$ are projections and $z$ commutes with $p$. Now in order to show that $q \le \frac{1}2$, it enough to show $\left\langle \left( 1-2q\right)\xi,\xi\right\rangle \ge 0,$ for all $\xi \in H$. Only thing I know about $(1-2q)$ is that it i self adjoint and unable to prove that $q \le \frac{1}2$. Please help me to solve this. Thanking you.
Note that, $\mathbf 1_{\left(0,\frac{1}2\right]}$ is the characteristic function of $\left(0,\frac{1}2\right]$.
Since $z$ is in the centre, $$ E(pz)=E(p)z=u\,1_{(0,\frac12]}(u). $$ Since $t\,1{(0,\frac12]}(t)\leq\frac12$ for all nonnegative $t$, functional calculus gives you $$ E(pz)\leq\frac12. $$