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.



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?
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.


