Let $M$ be a von Neumann algebra acting on the Hilbert space $\mathcal H$. Let $0 \ne p \in M$ be a projection. Now consider a projection $q \in M$ defined by $$q:=\sup \{upu^*: u \text{ is an unitary element of } M\}.$$ I want to show that $q$ is in the center of $M$. Note that $q$ is a projection from $\mathcal H$ onto $\overline{\cup_{u\in \mathcal U(M)} upu^*(\mathcal H)}$.
Basically, in order to show that $q$ is in the center of $M$, it suffices to show $uqu^*=q$ for all $u \in \mathcal U(M),$ where $\mathcal U(M)$ denotes the set of all unitary elements of $M$. Now $q \ne 0$ as $p \ne 0$ and also we have $q\mathcal H= \overline{\cup_{u\in \mathcal U(M)} upu^*(\mathcal H)}.$ So, $q \ge upu^*$ for all $u \in \mathcal U(M)$. Let $u$ be an arbitrary element of $\mathcal U(M)$, then $q=u^*uqu^*u$.
But I am unable to show that $q=uqu^*$. Please help me to solve this. Thank you for your time and help.
Note that if $u$ is unitary, then $u^\ast(\mathcal H)=\mathcal H$. Thus, if $v\in \mathcal U(M)$, then $$ vqv^\ast(\mathcal H)=vq(\mathcal H)=v\overline{\bigcup_{u\in\mathcal U(M)}up(\mathcal H)}=\overline{\bigcup_{u\in\mathcal U(M)}vup(\mathcal H)}=q(\mathcal H). $$ The last step follows since $v\mathcal U(M)=\mathcal U(M)$.