Given a nonempty closed convex set $A\subset\mathbb R^n$, we know that for each $x\in\mathbb R^n$ there is a unique $p_A(x)\in A$ such that $$\|x-p_A(x)\|\le \|x-y\| \quad\forall y\in A.$$ The map $p_A:\mathbb R^n\to A$ is called the metric projection of $A$, and sends each $x$ to its unique nearest point in $A$.
One notable property of metric projections is that they are contracting, meaning that $$\|p_A(x)-p_A(y)\| \le \|x-y\|,\quad\forall x,y\in\mathbb R^n.$$ Geometrically, this is pretty clear from figures such as this:
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One way to prove this, following Schneider's book (Theorem 1.2.1), is the following:
- Define $v\equiv p_A(y)-p_A(x)$, and assume $v\neq \mathbf 0$.
- Define $f(t)\equiv\|x-(p_A(x)+tv)\|^2$ for $t\ge0$.
- Observe that $f$ has a minimum at $t=0$ (clear from the definition of $p_A$) and thus $f'(0)=2\langle p_A(x)-x,v\rangle\ge0$.
- Doing the same replacing $x\to y$ we show that $\langle p_A(y)-y,v\rangle\le0$.
- Conclude that the segment $[x,y]$ crosses the two hyperplanes orthogonal to $v$ and passing through $p_A(x)$ and $p_A(y)$. The distance between these two hyperplanes is $\|p_A(x)-p_A(y)\|$, so this implies that $\|x-y\|$ must be larger than this, QED.
While this proof is fine, I was wondering if there is a "better" way to prove it that only relies on geometrical arguments. In particular, a proof that doesn't require to introduce a function such as $f$ and reason on its first derivative.
All the proofs that I have seen rely on similar arguments. However, there is one geometric element that can be better highlighted.
It is the fact that for $x \in V$ and $z \in A$ you have: $$\langle z - x, z - y \rangle \le 0$$ for any $y \in A$ if and only if $z$ is the orthogonal projection of $x$ onto $A$. Meaning that all the points in $A$ are on one side of the hyperplane passing through $p_A(x)$ and orthogonal to $p_A(x) -x$.
It is a pity that this is hidden in the proof provided.
From this result, you can get that the projection is contracting in the following way. You have
$$\begin{aligned} \Vert x - y \Vert^2 &= \Vert p_A(x) - p_A(y) + r \Vert^2 \text{ where } r= (x-y) + (p_A(x) - p_A(y)\\ &= \Vert p_A(x) - p_A(y) \Vert^2 + \Vert r \Vert^2 + 2 \langle r, p_A(x)-p_A(y) \rangle\\ &= \Vert p_A(x) - p_A(y) \Vert^2 + \Vert r \Vert^2\\ &+ \langle x-p_A(x), p_A(x)-p_A(y) \rangle + \langle y-p_A(y), p_A(y)-p_A(x) \rangle \end{aligned}$$
And you can conclude as $$\Vert r \Vert^2\\ + \langle x-p_A(x), p_A(x)-p_A(y) \rangle + \langle y-p_A(y), p_A(y)-p_A(x) \rangle \ge 0.$$
An image (courtesy of glS) describing what happens: