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Let $C$ be a convex compact subset of a finite-dimensional normed vector space and let $f:C \to \mathbb R$ be strictly convex and uniformly continuous function (e.g it is sufficient that $f$ be Lipschitz continuous).

Question. Is it true that $f$ has exactly one minimizer ?

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    $\begingroup$ The set of minimizers of a convex function is convex, because it is a sublevel set. So you may assume wlog that your function is constant... $\endgroup$
    – Pietro Majer
    Commented Nov 2, 2020 at 23:42
  • $\begingroup$ Makes sense. Thanks. $\endgroup$
    – dohmatob
    Commented Nov 3, 2020 at 0:09

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First, since $f$ is continuous and $C$ compact, there is $x \in C$ with $f(x) \leq f(y)$ for all $y \in C$. Assume that $z \in C$ is a second minimizer with $x \not= z$. Then $f(x) = f(z)$ and since $f$ is strictly convex we get $f(u) < \frac{1}{2} f(x) + \frac{1}{2} f(z)$ for $u := \frac{1}{2} x + \frac{1}{2} z \in C$, a contradiction.

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  • $\begingroup$ Thanks (upvoted). Someone has voted the question be closed. Donno if I should be accepting this answer or closing the question. This place is weird at times... $\endgroup$
    – dohmatob
    Commented Nov 3, 2020 at 0:10

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