If range of a projection is contained in another, then the first one is dominated by the second

Question

I want to prove the converse of my previous question. Namely, if $H$ is a Hilbert space and $P,Q$ are projections in $\mathcal{B}(H)$ satisfying $\text{range }P\subseteq \text{range }Q$, then $P\leq Q$, i.e., $\langle x,Px\rangle \leq \langle x,Qx\rangle$ for all $x\in H$.

My attempt: Using the hypothesis, I'm able to show that $QP=P$ and so $PPP=PQP$. Therefore $P(Q-P)P=0$. But how to show $P\leq Q$ from this?

Answer 1

Note that if we prove that $Q-P$ is a projection, then $\langle x,Qx\rangle-\langle x,Px\rangle=\langle x,(Q-P)x\rangle\geq 0$.

Now, $Q-P$ is obviously self-adjoint so all we have to do is check that $Q-P$ is idempotent:

$$ (Q-P)^2=Q^2+P^2-QP-PQ=Q-PQ, $$ using what you've already proven. Hence, we're done if we can prove that $PQ=P$. However, $(PQ)^*=QP=P$, so $PQ=P^*=P.$

Answer 2

$x=Qy+z$ for some $y$ and some $ z\perp range (Q)$. Note that $z \perp range (P)$ also. Now $Px=PQy+Pz$ and $Pz=0$ because $z \perp range (P)$. Hence, $Px=PQy$. Also, $Qx=Q(Qy+z)=Q^2y=Qy$ so we have $Px=PQy=PQx$. Since $\|Q\| \leq 1$ we get $\|Px\|^{2} \leq \|Qx\|^{2}$ which gives $ \langle Px, x \rangle \leq \langle Qx, x \rangle$.