Surjectivity of the Gauss map in Hilbert spaces

Suppose that $H$ is a Hilbert space and $M\subset H$ is a closed subset with non-empty interior and smooth boundary, whatever smooth boundary could mean. I wonder if the Gauss map on $\partial M$ is onto on the sphere, I hope yes.

The finite dimensional case, say dimension n, is easier (but not easy for me, I lack of enough training) and the problem can be stated better, using a n-dimensional manifold instead a closed subset, but I can not figure out what happens in $l^2$, is the hypothesis true or false?

The hypothesis holds obviously if $M$ is a closed ball.

We should assume that $M$ is bounded. For the finite dimensional case, take any unit vector $v$ and a plane with $v$ as a normal vector. Sweep the plane through the space until it touches $M$, the Gauss map at the first point of contact with $M$ is $v$. The problem in the infinite dimensional case is to verify the existence of the first point of contact. If $M$ is convex, that point exits.

To define smoothness I propose the following: suppose that $x\in\partial M$ and that there exists a functional $F$ defined in a neighborhood $U$ of $x$, such that $F=0$ on $\partial M\cap U$ and $dF(x)$ is a non-zero bounded linear functional. I don't know if I should define it differently.

Solution 1:

This is only a partial answer for lack of an adequate counterexample, see below.

If $M$ is weakly closed and bounded, then for all $f\in H$ with $\lVert f \rVert=1$ the map $$\tag{1} g\in M\mapsto \langle f|g\rangle $$ attains its maximum on $M$, because it is manifestly weakly continuous. This corresponds to the existence of the first point of contact in the OP's language. If $M$ has nonempty interior, such maximum must be attained on the boundary, because the differential of (1) is clearly nonzero. And so, if the boundary satisfies a smoothness assumption that enables the use of Lagrange multipliers, then there exists a point $g$ on the boundary of $M$ such that the unit normal at $g$ is precisely $f$. In conclusion, the Gauss map is surjective, just like in the finite-dimensional case.

Remark. If $M$ is convex and closed, then it is weakly closed. We have thus re-proved that the Gauss map is surjective for convex sets, as stated in the original question.

It would be interesting to find an example of a closed set that is not weakly closed, that has nonempty interior and is such that the Gauss map is not surjective. The example in the comments to the original question has empty interior and so it does not qualify. However, I am convinced that a suitable modification of that example can work.