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.

  • $\begingroup$ 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 $M$? $\endgroup$
    – Avitus
    Jun 11, 2013 at 7:50
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    $\begingroup$ @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). $\endgroup$
    – Martin
    Jun 11, 2013 at 8:50
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    $\begingroup$ 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. $\endgroup$
    – user90189
    Jun 11, 2013 at 21:36
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    $\begingroup$ 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). $\endgroup$
    – S. Dewar
    Apr 8, 2018 at 22:31
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    $\begingroup$ 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. $\endgroup$ Aug 9, 2018 at 12:08

1 Answer 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.

  • $\begingroup$ Thank you @Giuseppe for your interest in this question. Curiously, I was recently engaged in a question you posed in MO about maximizers and weakly compactness. I got nothing. Coming back to this post, I suspect that the set behaves much like a convex set near to the contact point. In finite dimension, these points are locally the graph of a non-negative quadratic form. $\endgroup$
    – user90189
    Aug 11, 2018 at 16:13
  • $\begingroup$ @user90189: Thank you for your interest in that question on MO! (I also didn't get much, but I collected what I got in an answer, in case you are interested). Concerning the problem of this question, I conjecture that the Gauss map can fail to be surjective if the set is not weakly closed. The difficult part in building counterexamples is the construction of smooth open sets in Hilbert space whose closure is not weakly closed. Any convex set won't do, as you know. $\endgroup$ Aug 12, 2018 at 15:39

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