Let $H$ be a Hilbert space and $B(H)$ be the space of all bounded operators on $H$. Let $p$ and $q$ are two projections on $B(H)$. Now define an operator $u$ on $B(H)$ by: $$u:=q-\inf\{q,1-p\}.$$ I want to prove that $u$ is a projection on $B(H)$ and $\overline{\text{Ran}(qp)}=\overline{\text{Ran}(u)},$ where ${\text{Ran}(qp)}$ and $\text{Ran}(u)$ are ranges of $qp$ and $u$ respectively.
In order to show that $u$ is a projection, I need to show that $u^*=u$ and $u^2=u$. Please help me to solve this and also $\overline{\text{Ran}(qp)}=\overline{\text{Ran}(u)}$. Thank you for your help.
Def: For any projections $p,q \in B(H)$, the operator $\inf\{p,q\}$ is the projection from $H$ to $\left(pH\cap qH\right)$.
It is well-known that $$ \inf\{q,1-p\}=q\wedge (1-p)=\lim_{\rm sot} (q(1-p)q)^n. $$ Thus $u$ (which is a projection by definition), satisfies $$\tag1 u=\lim_{\rm sot}q-(q(1-p)q)^n. $$ This shows that $qu=u$; in particular, taking adjoints, $uq=u$. Also, from $(1)$, $$ u=\lim_{\rm sot}q-(q(1-p)q)^n =\lim_{\rm sot}q(1-((1-p)q(1-p))^n). $$ Thus $up=qp$. This shows that $\operatorname{ran}qp\subset\operatorname{ran}u$.
Now suppose that $\xi\in\operatorname{ran} u\ominus \overline{\operatorname{span}}qp$. This means that $$\tag2 \xi=q\xi-(q\wedge(1-p))\xi $$ and $$\tag3 \langle \xi,qp\eta\rangle=0,\ \eta\in H. $$ We can rewrite $(3)$ as $\langle pq\xi,\eta\rangle=0$ for all $\eta$, so $$\tag5 pq\xi=0. $$ Then $$ p\xi=-p(q\wedge(1-p))\xi=0 $$ and $$ q\xi=pq\xi+(1-p)q\xi=(1-p)q\xi. $$ Iterating this equality, $$ (q(1-p)q)^n\xi=q\xi, $$ and taking the limit we get $$ (q\wedge(1-p))\xi=q\xi. $$ Thus, by $(2)$, $$ \xi=u\xi=q\xi-(q\wedge(1-p))\xi=q\xi-q\xi=0, $$ and hence $\overline{\operatorname{ran}}qp=\operatorname{ran}u$.