Properties of the Orthogonal Projection: (Firm) Nonexpansivness
Let $C$ be a nonempty closed convex set Then:
1. For any $v,w\in\mathbb{R}^n$:
$\begin{aligned} (P_C(v)-P_C(w))^T(v-w)\geq||P_C(v)-P_C(w)||^2 \end{aligned}$
2.(Non-expansiveness) for any $v,w\in\mathbb{R}^n$:
$\begin{aligned} ||P_C(v)-P_C(w)||\leq||v-w|| \end{aligned}$
I've learnt this theorem in convex analysis course and I would like to get some geometric intuition for both of them
Orthogonal projection in ${\mathbb R}^n$ can be thought of as shining a light onto an object and looking at the shadow it casts on the far wall. That oversimplifies a little too much, because the "far wall" is a plane so we're projecting down into at most two dimensions, but it's not a bad image to start with.
When we project we are reducing the number of dimensions we are considering (this is sometimes referred to as 'setting co-ordinates to zero') in order to get a feel for the macroscopic properties of the object. Your second point expresses this: when we reduce the number of dimensions that we consider an object in, we are losing part of the object and so the end-result cannot grow, it can stay the same or shrink. (Mathematically: if $v = (v_1, v_2, \ldots ,v_n)\in {\mathbb R}^n$ is our vector and $Pv = (w_1, w_2, \ldots ,w_n)$ is an orthogonal projection then we have $|w_n| \leq |v_n|$ for each $n \in {\mathbb N}$ so $v$ is "smaller" in some sense than $Pv$).
Your first point expresses the concept that the orthogonal projection preserves orientation: whatever direction the vector $v-w$ points (i.e. what sign it has), the projection $Pv-Pw$ points in the same direction (has the same sign).
