Problem. Let $X$ be a Hilbert space and $\emptyset \neq K \subseteq X$ be closed and convex. Then, $$ \|P_Kx - P_Ky \| \leq \|x-y \|$$ for all $x,y \in X$. Here, $P_K$ is the projection from $X$ onto $K$; that is the unique nearest element in $K$.
It doesn't look too difficult but I can't really see how to prove it. Geometrically (in low-dimensional cases at least) it makes sense but I'm not sure how to connect these objects. How should I approach this?
On
Let the ground field be $\mathbb{R}$. By the variational inequality $$ \langle P_K(x)-x, P_K(y)-P_K(x) \rangle \geq 0 \quad \text{and} \quad \langle P_K(y) - y, P_K(x) - P_K(y) \rangle \geq 0. $$ Summing them up gives $$ \langle x-y, P_K(x)-P_K(y) \rangle \geq \|P_K(x)-P_K(y) \|^2 $$ Cauchy-Schwarz finishes. $$ \|x-y \| \cdot \|P_K(x)-P_K(y) \| \geq \langle x-y, P_K(x)-P_K(y) \rangle \geq \|P_K(x)-P_K(y) \|^2. $$
Remark. The proof is identical on $\mathbb{C}$. Only the variational inequality is slightly different.
Hint: The projection is characterized by the variational inequality (VI) $$ ( P_K(x) - x, z - P_K(x) ) \ge 0 \qquad\forall z \in K.$$ (If you do not have seen this formula, you should try to prove it.) Now, you have to use this VI for $P_K(x)$ and for $P_K(y)$ and you have to use clever choices for $z$.