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Let $A\subset\mathbb R^n$ be such that:

  1. every non-zero linear functional is maximized by a unique point of $A$

  2. every point of $A$ is a point where some linear functional achieves its maximum over $A$ (i.e., every point of $A$ is an exposed point, though normally exposed points are only defined for convex sets)

  3. the boundary of $A$'s convex hull is contained in the closure of $A$.

Does it follow that $A$ is closed, and hence equal to the boundary of its convex hull?

It doesn't follow in the absence of (3). Let $A = \{ e^{i\theta}: 0\le\theta<\pi \} \cup \{ e^{i\theta} - i : \pi\le\theta <2\pi \} \subseteq \mathbb C = \mathbb R^2$.

Alexander Pruss
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  • Could you alternatively to (3) ask that the point where the maximum is achieved in $A$ vary continuously with the linear functional? – Olivier Bégassat Feb 12 '21 at 21:00
  • Not in dimension 3: Take a smooth almost strictly convex set with just one straight segment on the boundary and leave one endpoint in the segment, removing its other points. – fedja Feb 12 '21 at 22:14
  • @Olivier Bégassat: If we have continuity, then A is the image of a compact set under a continuous function, and we get closedness trivially. Do you think you can show that 1-3 imply continuity? – Alexander Pruss Feb 12 '21 at 22:42
  • @fedja: In your example, if A is the boundary of the set minus all the points on the segment except for one endpoint, then my condition (3) is not satisfied. Indeed, (3) is crafted precisely to rule out such cases.But we also don't have continuity.Take a linear functional that is maximized precisely on the straight segment. If you slightly "rock" the functional back and forth, the maximum point will jump from the included endpoint to points in the neighborhood of the excluded endpoint. Am I missing something? (BTW, on another note, do you know an easy construction of a set like you describe?) – Alexander Pruss Feb 12 '21 at 22:47
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    @AlexanderPruss I believe that (3) is satisfied: you remove a one-dimensional interval from the two-dimensional surface, which hardly affects the closure. And you never required continuity: it doesn't hold in your own example either. Am I missing something? – fedja Feb 12 '21 at 22:52
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    @fedja: Oh, I see now! Very nice. Why don't you write it up as an official answer? – Alexander Pruss Feb 12 '21 at 23:13
  • @fedja: Though I still don't know how to explicitly prove the existence of a set like you describe. Note, too, that you need something a bit stronger than just strict convexity everywhere except on the removed line segment. You need to control the normals in the vicinity of the line segment to ensure that (1) is satisfied. – Alexander Pruss Feb 13 '21 at 18:27
  • @AlexanderPruss I believe that the union of 3D balls centered at $(x,1-x^2,0)$ of radii $1-x^2$ where $x$ runs over $[-0.5,0.5]$ is such a set. Am I missing anything? – fedja Feb 14 '21 at 00:22
  • That looks right. You don't happen to have a reference for such a set? I have a paper where I want to give an example, but I don't want to spend a lot of time proving the example, as it's not super important to the main purposes. – Alexander Pruss Feb 26 '21 at 18:24
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    @OlivierBégassat: If we have (1), (2) and your continuous variation condition, then A is compact, being the continuous image of $S^{n-1}$. Given fedja's construction (assuming it works -- I still don't have a proof written up), (1), (2) and (3) will not imply closedness of A. Hence, (3) does not imply your continuous variation condition even given (1) and (2). – Alexander Pruss Mar 14 '23 at 14:51

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