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Let $s,\delta\in\mathbb{R}^{N}$, $S\subseteq\mathbb{R}^{N}$ be a compact convex set, $f:S\rightarrow\mathbb{R}$ be a twice differentiable strictly convex function on $S$ and $$s\left(\delta\right)=\arg\max_{s\in S}s\cdot\delta-f\left(s\right).$$ Notice that this is related to the definition of convex conjugate of $f$: $$f^*\left(\delta\right)\equiv\max_{s\in S}s\cdot\delta-f\left(s\right).$$ Being $f$ strictly convex, $s\left(\cdot\right)$ is single valued, and it is continuous by the maximum theorem. My question: is $s\left(\cdot\right)$ Lipschitz continuous? If not, which additional conditions do I need to get Lipschitz continuity?

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The (global) Lipschitz continuity should be equivalent to $f$ being strongly convex. In order to find a counterexample, you should try some $f$ which is not strongly convex.

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