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I have the following conjecture:

Given a closed convex set $S \subseteq \mathbb{R}^n$ and one of its exposed face $F=\{x \in S \mid \pi x = \pi_0\}$, where $\pi x =\pi_0$ is the supporting hyperplane of $S$. For any $\delta>0$, the neighborhood $F_\delta$ of $F$ is defined as: $F_\delta = \{x \in S \mid \exists x' \in F \text{ s.t. } \|x'-x\|_2 < \delta\}$. Then there exists $\epsilon>0$, such that for any $\pi'$ with $\|\pi' - \pi\| < \epsilon, \sup_{x \in S} \langle \pi', x \rangle =\sup_{x \in F_\delta} \langle \pi', x \rangle$.

This conjecture is correct if $S$ is a polyhedron or $F$ is compact. For general (unbounded) set $S$ however, I really have no clue... In fact for $S$ being polyhedron, we have a stronger result: for any vector $\pi'$ close enough to $\pi$, the optimal solution of $\sup_{x \in S} \langle \pi', x \rangle$ with be $\textbf{contained}$ in $F$ (this is also known as "sticky face lemma").

I believe this conjecture is intuitively correct, and I think it should appear somewhere in the literature probably in some other forms. Any guidance or references will be much appreciated!

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  • $\begingroup$ Would you mind explaining what it means for $\pi_0$ to be the supporting hyperplane of a closed convex set - is it just any hyperplane so that the $S$ lies on one side of it? Also, do you orient the planes in a specific way? $\endgroup$
    – Leo Moos
    Commented Apr 3, 2021 at 14:54
  • $\begingroup$ Yes, a supporting hyperplane is simply the one "touches" the boundary of the closed convex set, without passing through it (or like you said, S lies on the one side of it). Here the intersection of this hyperplane with S is denoted as F, which is on the boundary of S. $\endgroup$
    – HAORAN ZHU
    Commented Apr 3, 2021 at 14:57

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