# 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?

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.

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The question is not clear to me at all. What is projected onto what? – Włodzimierz Holsztyński Feb 12 '13 at 4:30

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.
$0$ 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.
Nick: Your new answer seems to make no sense: Your assumption implies that $a=0$. Your original argument (to my best recollection of it) was OK: Take a support hyperplane $H$ to $C$ at the boundary point $x\in C$, let $u$ be the unit normal vector to $H$ pointing away from $C$ and take $y=x+u$. Then the nearest point to $y$ in $C$ is $x$ since $x$ is nearest to $y$ on $H$. – Misha Feb 12 '13 at 3:23
@Misha: apparently you weren't the only one who thought my original argument was okay ... but no, in my example $0$ is a boundary point yet there is no support hyperplane containing it. Even in a Hilbert space! Counterintuitive, isn't it? – Nik Weaver Feb 12 '13 at 5:24
Oh, I see: I was assuming that $C$ has nonempty interior, in which case for every boundary point there exists a support hyperplane. – Misha Feb 13 '13 at 0:44