Convex sets and projections

2010-10-09T15:57:31Z

Hello!

I recently started (it's purely self-education) reading a "Mathematical programming and optimizations" book, did a vast part of the exercises related to the theoretical part and at one moment I got the following question about convex sets:

I'm almost sure this statement is correct, but unfortunately, couldn't find something similiar on the internet and I tried to prove it, but I couldn't.

Assume we have some set $S \subset \mathbb{R_n}$ and for this set: $S = \overline{S}$ (set closure equals the set itself).

Now, there exists only one projection of arbitrary point $y$ which doesn't belong to the set $S :$

$\forall y \in \mathbb{R_n}, \space y \notin S: \space \exists ! \space p = \pi_S(y) $

This should mean that $S$ is a convex set.

Could someone please point me if I'm wrong (or right, but with some limitations for this statement) and help me proving it if I'm right.

I also understand that this question be a too "basic" to post here, but I've just started educating myself in this sphere and hope that sometimes I'll get smart enough to ask really bright questions :)

Thank you.

Answer by Robin Chapman for Convex sets and projections

2010-10-09T16:13:34Z

I presume what you want to prove is the following. Let $S$ be a nonempty closed subset of $\mathbb{R}^n$. Then if there is a point $y\in\mathbb{R}^n$ and there are at least two points $p$ and $q$ in $S$ with Euclidean distance $d$ from $y$ (where $d$ is the distance of $y$ from $S$), then $S$ is not convex. To see this, note that the midpoint $r$ of the line segment $pq$ is closer to $y$ than $p$ of $q$ is, and so cannot lie in $S$. Hence $S$ isn't convex.

Convex sets and projections - MathOverflow

Convex sets and projections

Answer by Robin Chapman for Convex sets and projections