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If $\Pi_C(\cdot):\mathbb{R}^n \to C$ is a projection operator for a set $C\subseteq \mathbb{R}^n$ under some norm $||\cdot||_p$ ie. $\Pi_C(x) = \underset{y \in C}{\mathrm{argmin}}||y-x||_p$. If $\Pi_C$ is a contraction map (under the same norm) is it necessary that $C$ is convex?

My intuition here is that if its not convex, then $\exists x,y \in C$ such that two points $x',y' \notin C$ can be chosen on the line segment $l(x,y)$ joining $x,y$ such that $||x'-y'||_p \leq ||\Pi_C(x')-\Pi_C(y')||_p$ leading to a contradiction. Although the non-convexity can be very pathological like $C = C'\setminus l(x,y)$ where $C'$ is a convex set.

Also what if $\Pi_C$ is non-expansion instead?

I_am_AM
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    If the set $C$ is not convex, the projection is not well defined for all $x \in \mathbb R^n$, see, e.g. https://math.stackexchange.com/questions/1130393/chebyshev-sets-in-finite-dimension-are-closed-and-convex. – gerw Jan 10 '20 at 07:04

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Thanks to @gerw for the reference about ill defined projection for non-convex $C$.

I think then the counter example for the case of contraction map $\Pi_C$ for a non-convex set $C$ would simply be one such point where the projection $\Pi_C(x)$ is ill defined ie. $\underset{y \in C}{\mathrm{argmin}}||y-x||_p = S$ is a set, if we try to arbitrary break the tie by choosing a point $x'\in S$ and define $\Pi_C(x) =x'$ then $\exists y' \in S, y'\neq x'$. Let the angle $\angle x'xy' =\theta >0$ and $\min_{y\in C}||y-x||_p =d$. Picking any $y''\notin C$ on the line segment $xy'$ with $||x-y''||_p = d\sin(\frac{\theta}{2})$, we have $\Pi_C(y'') = y'$ by the triangle inequality and $||x-y''||_p< ||\Pi_C(x)-\Pi_C(y'')||_p = 2d\sin(\frac{\theta}{2})$.

This counterexample also seems to work for the case where $\Pi_C$ is non-expansion.

I_am_AM
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