I am stuck at one point in my research, where I need to prove something which appears trivial to me, but cannot find a rigorous proof. I describe it below. Whenever I will say projection, I will mean the $L^2$ projection in Euclidean spaces.

We all know that the projection of a point on a closed, convex subset of the Euclidean space is unique. Now, if a set $A \subseteq \mathbb{R}^n$ is closed, we can easily show that the projection of a point $p \in \mathbb{R}^n$ on $A$ exists, but can find counterexamples, such that it is not unique. So, here is my first question (assume henceforth, that $A$ is a closed set):

  1. Is it true that the set of all points $p \in \mathbb{R}^n$ that have more than one projection on the set $A$ (let us call that set $\mathcal{P}(A)$), has Lebesgue measure $0$?

It seems too much for question 1 to have an affirmitive answer for any arbitrary closed set $A$ (I do not even know whether the set $\mathcal{P}(A)$ is measurable, but if not, I can work with outer measures). So, here comes a simpler question.

  1. If $A$ is the union of two convex sets, is it true that $\mathcal{P}(A)$ has Lebesgue measure $0$?

Even if this seems too much, I would really be happy to have an affirmitive answer (with a proof) to the following even simpler question:

  1. If $A$ is the union of two polyhedra (a polyhedron is a finite intersection of half-spaces, and hence, is convex), does $\mathcal{P}(A)$ have Lebesgue measure $0$?

I intuitively feel that 3 must be correct, the reason being as follows. If a point $p$ has two distinct projections in $A$, then these projections must lie on two different faces of the union of the polyhedra, must be projections of $p$ on the respective faces too, and must be equidistant from $p$. The set of such point seems to be a lower dimensional hyperplane, which has Lebesgue measure $0$. But I cannot make this further rigorous. The problem is, I cannot seek help from linear algebra, as these faces are flats, and not even subspaces.

Any help (at least with answering question 3) will be greatly appreciated!


That is true. The set of points with non-unique projection has measure zero. The proof is a beautiful application of the Rademacher theorem. You can find comments and the link to a proof here:

Set of points with a unique closest point in a compact set.

See also: https://mathoverflow.net/a/324877/121665.

  • 2
    $\begingroup$ Thanks a lot, Piotr! I will add your post as a reference in my research paper :) $\endgroup$ – abcd Oct 7 at 1:58
  • $\begingroup$ wait, why did you start off with "actually"? OP said they thought that was true $\endgroup$ – mathworker21 Oct 10 at 2:13
  • 1
    $\begingroup$ @mathworker21 Good point. No clue why I did say "actually". I am changing my answer now. $\endgroup$ – Piotr Hajlasz Oct 10 at 4:25

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