So I found this inequality in a paper:
$|| P(u)-P(v) ||^2 \leq (u-v)^T(P(u)-P(v))$ with $u,v \in \mathbb{R}^n$, where $P$ is a projection and $||\cdot||$ is the Euclidean norm.
But I have no clue how to get this inequality. Thanks for any information or pointers to resources.
I assume that $P$ is onto a linear subspace $V$ of $H=\mathbb R^n$. By the Kolmogorov criterion,
$$\langle P(x)-x, v-P(x) \rangle \geq 0 \ \ \text{ for every } v \in V $$ where $ \langle \cdot , \cdot \rangle$ is the usual inner product in $H$.
Since $P(u) $ and $P(v) $ belong in $V$, $$ \langle P(u)-u,P(v)-P(u) \big \rangle \geq 0$$ and $$\langle - P(v)+v, -P(u)+ P(v) \rangle \geq 0.$$ Hence, $$ \langle P(u) - P(v), P(v) - P(u) \big \rangle + \langle v-u, P(v) - P_C(u) \big \rangle \geq 0 .$$ This then implies that $$||P(u) -P(v)||^2 \leq \langle v-u,P(v) -P(u) \big \rangle. $$
Edit: The Kolmogorov criterion states that if $P $ is an orthogonal projection onto a closed subspace $V$ of a Hilbert space $H$ and $x \in H$, then $$P(x)=y \iff y \in V \text{ and } \langle y-x, v-y \rangle \geq 0 \ \ \text{ for every } v \in V . $$ For a proof see Theorem 4.3-1 in "Linear and nonlinear Functional Analysis with Applications" by Philippe G. Ciarlet.