# 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.

• What do you mean with smooth in this case? You begin with a closed subset $M$ and then you move to "the sphere". Why? Which is the sphere you are talking about? What is its relation to $M$? Jun 11, 2013 at 7:50
• @Avitus: user39490 is asking about an infinite-dimensional analog of the Gauß map sending a point of a smooth oriented hypersurface (the boundary of $M$) to the unit normal vector of its tangent plane (an element of the unit sphere). Jun 11, 2013 at 8:50
• The problem with infinite dimensional spaces is its weird behavior, for example the identity in the sphere is contractible in $H-\{0\}$. I think I can prove the hypothesis if $M$ is convex, maybe star-convex, but I want a stronger result. I thought I could prove some stability of the property by homotopies, but once again I run into dificulties because of the infinite dimension. Jun 11, 2013 at 21:36
• If you are saying smooth boundary I would assume the logical choice of definition would be that $\partial M \subset H$ is a smooth Hilbert manifold with model space $H$ (Manifolds, Tensors and Applications by Marsden, Ratiu and Abraham gives a full definition of this). Apr 8, 2018 at 22:31
• The first point of contact needs not exist for closed sets that are not weakly closed. For example, if $H=L^2(\mathbb R)$ and $M$ is the closed set $\{ f(\cdot + h)\ :\ h\in\mathbb R\}$, where $f>0$ pointwise and $\|f\|_2=1$, then the map $$h\mapsto \langle -f| f(\cdot +h)\rangle = \int_{-\infty}^\infty (-f(x)) f(x+h)\, dx$$ has sup $0$ and it has no maximum (because the maximum is at $h\to \infty$). So the first point of contact of $M$ with the plane having normal $-f$ is the origin, which is not an element of $M$. This example is not 100% satisfactory because the interior of $M$ is empty. Aug 9, 2018 at 12:08

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.