# Hausdorff-Lipschitz continuity of cone correspondence

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 $$$$f: \mathcal{X} \rightarrow \mathbb{R}, \quad g: \mathcal{X} \rightarrow \mathbb{R}^{1\times m},\quad h: \mathcal{X} \rightarrow \mathbb{R}_+, \quad S: \mathcal{X} \rightarrow \mathbb{R}^{m\times m}$$$$ be continuously differentiable functions and assume $$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 $$$$\Pi(x) = \left\{p \in \mathcal{P} \ \Big\vert \ f(x) + g(x)p \geq \sqrt{h(x) + p^\top S(x) p \ \ } \ \right\},$$$$ and assume that $$\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 $$$$d_H( \Pi(x), \Pi(x')) \leq L\Vert x-x'\Vert_2$$$$ hold for every $$x, x'\in \mathcal{X}$$, where $$d_H$$ denotes the Hausdorff distance?

$$\newcommand\P{\mathcal P}\newcommand\X{\mathcal X}$$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 $$0$$.
• @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 $f(x)=c+x^2$. Apr 15 at 19:36