Are all points x of the boundary of a convex set C of a Hilbert space H projections onto C of a point different than x? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T22:39:51Z http://mathoverflow.net/feeds/question/121526 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121526/are-all-points-x-of-the-boundary-of-a-convex-set-c-of-a-hilbert-space-h-projectio Are all points x of the boundary of a convex set C of a Hilbert space H projections onto C of a point different than x? Henri Th 2013-02-11T20:40:25Z 2013-02-12T23:02:18Z <p>It appears that this is not true if H is of infinite dimension. My question is therefore the following: does anyone have a counter-example? Is there a caracterisation for the points of the boundary that are (non trivial) projections? Thanks in advance.</p> http://mathoverflow.net/questions/121526/are-all-points-x-of-the-boundary-of-a-convex-set-c-of-a-hilbert-space-h-projectio/121532#121532 Answer by Nik Weaver for Are all points x of the boundary of a convex set C of a Hilbert space H projections onto C of a point different than x? Nik Weaver 2013-02-11T21:32:32Z 2013-02-12T23:02:18Z <p>I'm a little concerned that my original incorrect answer was accepted and the only feedback on the edit was a comment that it didn't make sense ... here's a slightly simpler counterexample.</p> <p>Let $H = l^2$ and let $C$ be the set of sequences $(a_n)$ satisfying $|a_n| \leq \frac{1}{n}$ for all $n$. $C$ is clearly closed and convex.</p> <p>$0$ is a boundary point of $C$ (in fact $C$ has no interior), but it is not the nearest element of $C$ to any point outside of $C$. If $(a_n)$ is any nonzero sequence in $l^2$ then some entry $a_{n_0}$ must be nonzero, and then we can find a point in $C$ that is closer to $(a_n)$ than $0$ is. For sufficiently small $\epsilon$, the point $\epsilon a_{n_0}e_{n_0}$ (where $e_n$ is the standard basis) works. Another way to say this is that $0$ has no support hyperplane.</p>