Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

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

1 Answer 1

up vote 4 down vote accepted

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.

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.

share|improve this answer
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
...because you can separate a point from a convex open set. Right. –  Nik Weaver Feb 13 '13 at 3:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.