Let $p,q:H\to H$ bounded, linear operator on a Hilbert space $H$, such that $p^*=p=p^2$ and $q^*=q=q^2$ ($p^*$ is the adjoint of $p$, for q the same. YOu call $p$ and $q$ a projection). Let $q(H)\subseteq p(H)$. Why is $\langle (p-q)(h),h\rangle \ge 0$ for all $h\in H$?
Here (bounded linear) orthogonal projections on Hilbert spaces is a discussion that $p$ is the identity on $im(p)$, zero on $im(p)^\perp$ and $q$ is the identity on $im(q)$, zero on $im(q)^\perp$. If $im(q)\subseteq im(p)$ and if you take $h\in H$ such that $h=h_1+h_2$ with $h_1\in im(p)$ and $h_2\in im(p)^\perp$, I have to do distinctions of cases now, right?
I think the proof shouldn't be difficult, but I'm stuck here. Can you explain me how to continue the proof?