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:
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.