Let $p,q$ be orthogonal projection operators in a Hilbert space.
Does the following equation hold?
$$ \langle p(1-q)x,p(1-q)x\rangle = \langle p(1-q)x,x\rangle$$
It's clear to me that $ \langle p(1-q)x,p(1-q)x\rangle = \langle p(1-q)x,(1-q)x\rangle$ but it's not clear if I can go from $\langle (1-q)^\ast p(1-q)x,x\rangle$ to $\langle p(1-q)x,x\rangle$.
Edit
Is the image of $1-q$ invariant with respect to $p$? That is, is it true that
$$ p(\mathrm{im}(1-q))\subseteq \mathrm{im}(1-q)$$
I think the answer is yes because I can see that it's true in finite dimension. It's not 100% clear to me how to prove it in infinite dimension.
If it was indeed true and $\mathrm{im}(1-q)$ was an invariant space then $p(1-q)p=(1-q)p$ would hold.
Edit 2
What I wrote in the previous edit appears to be wrong: applying an orthogonal projection $p$ to the image of an orthogonal projection $q$ is not necessarily again in the image of $q$.
Edit 3
If neither of this works: is there another way to derive
$$ \|p(1-q)(x)\|\le \|q(1-q)(x)\|$$
given $\forall x \in H: \|p(x)\|\le \|q(x)\|$?
Here's a counter-example to the claim: let $H = \mathbb{R}^2$, let $p$ and $r$ be the orthogonal projections on to $y = x$ and the $x$-axis respectively. Take $q = 1 - r$. For $x = (1,-1)$, we then have $p(1-q)x = prx = p(1,0) = (\frac{1}{2},\frac{1}{2})$ so that $$\langle p(1-q)x,p(1-q)x\rangle = \frac{1}{2}$$ but $\langle p(1-q)x,x\rangle = 0$.