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Let $\mathbb{R}_+$ denote the strictly positive real numbers, let $\mathcal{X} \subset \mathbb{R}^n$ and $\mathcal{P} \subset \mathbb{R}^m$ be compact and convex subsets, let \begin{equation} f: \mathcal{X} \rightarrow \mathbb{R}, \quad g: \mathcal{X} \rightarrow \mathbb{R}^{1\times m},\quad h: \mathcal{X} \rightarrow \mathbb{R}_+, \quad S: \mathcal{X} \rightarrow \mathbb{R}^{m\times m} \end{equation} be continuously differentiable functions and assume $S(x)$ corresponds to a strictly positive definite matrix for every $x\in\mathcal{X}$.

Consider the correspondence $\Pi :\mathcal{X}\rightarrow 2^{\mathcal{P}}$, given by \begin{equation} \Pi(x) = \left\{p \in \mathcal{P} \ \Big\vert \ f(x) + g(x)p \geq \sqrt{h(x) + p^\top S(x) p \ \ } \ \right\}, \end{equation} and assume that $\Pi(x)$ is non-empty for every $x\in\mathcal{X}$.

$\textbf{Question: Is the correspondence }$ $\Pi$ $\textbf{Hausdorff-Lipschitz in}$ $\mathcal{X}$, i.e., does there exist a positive scalar $L$, such that \begin{equation} d_H( \Pi(x), \Pi(x')) \leq L\Vert x-x'\Vert_2 \end{equation} hold for every $x, x'\in \mathcal{X}$, where $d_H$ denotes the Hausdorff distance?

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$\newcommand\P{\mathcal P}\newcommand\X{\mathcal X}$A counterexample is as follows: $n=m=1$, $\X=[-1,1]$, $\P=[0,2]$, $f(x)=\sqrt2-1/\sqrt2+|x|^{3/2}$, $g(x)=1/\sqrt2$, and $h(x)=1=S(x)$ (for all $x\in\X$).

Indeed, then $$\Pi(x)=\{p\in[0,2]\colon d(p)\le|x|^{3/2}\},$$ where $$d(p):=\sqrt{1+p^2}-(\sqrt2-1/\sqrt2+p/\sqrt2),$$ so that $d$ is a convex function with $d(1)=0=d'(1)$ and $d''(1)=2^{-3/2}$. So, $d(p)\sim2^{-5/2}(p-1)^2$ for $p\to1$ and hence $$\Pi(x)=[p_-(x),p_+(x)]$$ for some functions $p_\pm$ such that $p_\pm(x)-1\sim2^{-5/4}|x|^{3/4}$ as $x\to0$. Since $3/4<1$, the function $\Pi$ is not Lipschitz in any neighborhood of $0$.

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  • $\begingroup$ Thanks. Does it help if all functions are smooth? $\endgroup$
    – Heinrich A
    Apr 15 at 19:33
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    $\begingroup$ @HeinrichA : I don't think so. I think you can modify this example to have $(p-1)^4$ in place of $(p-1)^2$ and then use something like $f(x)=c+x^2$. $\endgroup$ Apr 15 at 19:36

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