Is projection onto a convex set strictly contractive when one point is in the set?

Question

Given a nonempty closed convex set $S\in\mathbb{R}^n$, its projection operator, defined as $$ p(x)= \textrm{argmin}_{z \in S} \| x - z\|_2 $$ for $x \in \mathbb{R}^n$, is unique and sends each $x$ to its unique nearest point in $S$. We know that projection onto such $S$ is a nonexpansive map, i.e., $$ \|p(x) - p(x)\|_2 \leq \|x-y\|_2, $$ for all $x,y \in \mathbb{R}^n$. However, if one of the point $y$ is already in $S$ (i.e., $p(y)=y$), and the other point $x$ is not in $S$, then is $p(\cdot)$ (strictly) contractive? In other words, do we have $$ \|p(x)-y\|_2 < \|x-y\|_2 $$ when $y\in S$ and $x \notin S$?

Answer 1

True. Projections onto closed convex sets in Hilbert spaces are firmly non-expansive, i.e., \begin{equation} \| p(x) - p(y) \|^2+\| (I-p) (x) - (I-p) (y)\|^2 \leq \| x-y \|^2 \end{equation} So that if $ y \in S $, we have \begin{equation} \| p(x) - y \|^2 \leq \| x-y \|^2-\| x-p(x) \|^2 \end{equation} which means \begin{equation} \| p(x) - y \| < \| x-y \| \end{equation} whenever $x \not\in S$