$u$ is a projection on $B(H)$ and $\overline{\text{Ran}(qp)}=\overline{\text{Ran}(u)}$ 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 $\left(pH\cap qH\right)$.
 A: 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 $\overline{\operatorname{ran}}qp=\operatorname{ran}u$.
