Let $L$ and $M$ be closed subspaces of a Hilbert space $H$ and $P_L,P_M\in \mathcal{L}(H,H)$ are orthoprojections to $L$ and $M$. I want to prove that then $$P_M\geq P_L \Leftrightarrow L\subset M.$$
Partial order $P_M\geq P_L$ is defined as $$P_M\geq P_L \Leftrightarrow ((P_M-P_L)x,x)\geq 0 \ \ \ \forall x\in X.$$
Any ideas?
First note that a projection $P\in\mathcal L(H)$ is self-adjoint iff $P$ is an orthogonal projection. For if $P^2=P$ and $\ker P\perp P(H)$, then for $f,g\in H$ we have \begin{align} \langle Pf, g \rangle &= \langle Pf, Pg + (g - Pg) \rangle\\ &= \langle Pf, Pg \rangle\\ &= \langle Pf + (f-Pf), Pg \rangle\\ &= \langle f, Pg \rangle, \end{align} so that $P=P^*$. (The converse is not necessary for this exercise so I will omit the proof.)
Suppose $P_M-P_L$ is a positive operator and let $f\in L$ be nonzero. Then $P_Lf=f$, so $$0\leqslant \langle (P_m-P_L)f,f\rangle=\langle P_mf-f,f\rangle.$$ It follows that \begin{align} \|P_Mf-f\|^2 &= \langle P_M-f,P_Mf-f\rangle\\ &= \langle P_Mf-f, P_Mf\rangle - \langle P_Mf-f, f\rangle\\ &\leqslant \langle P_Mf-f, P_Mf\rangle\\ &= \langle P_Mf,P_Mf\rangle -\langle f,P_Mf\rangle\\ &= \langle f, P_M^2f\rangle -\langle f,P_Mf\rangle\\ &=0, \end{align} and hence $P_Mf=f$, that is, $f\in M$.