Let $\mathbf R^n$ be endowed with the standard Euclidean structure and let $\mathcal X$ be the collection of (nonempty) compact subspaces of $\mathbf R^n$. Let $\mathcal B\subset \mathcal X$ be the sub-collection of all closed ball in $\mathbf R^n$. Then we can endow $\mathcal X$ (and hence $\mathcal B$) with the Hausdorff metric defined by
$$d_H(C,D)=\max\left\{\max_{x\in C}d(x,D),\max_{x\in D}d(x,C)\right\}.$$
Given a compact subspace $K\subset \mathbf R^n$, there is a unique closed ball $B_K$ of least radius containing $K$. This yields a map
$$F:\mathcal X\longrightarrow\mathcal B.$$
I would like to understand the regularity of $F$. More precisely :
Is $F$ Lipschitz or locally Lipschitz ?
I was able to show that $F$ is continuous. The key thing to note is that the center $x_K$ of $B_K$ is characterized by the fact that it minimizes the distance to $K$ :
$$d_H(\{x_K\},K)< d_H(\{x\},K) ~~\forall x\in\mathbf{R}^n,~x\neq x_K.$$
A compactness argument shows that $K\mapsto x_K$ is continuous.
Also if $r_K$ denotes the radius of $B_K$, then for $K,L\in\mathcal X$ we have $$K\subset L +d_H(K,L)\subset \bar{B}(x_L,r_L)+d_H(K,L)=\bar{B}(x_L,r_L+d_H(K,L)),$$ so $r_K\leq r_L+d_H(K,L)$ and therefore $K\mapsto r_K$ is 1-Lipschitz.
Finally if we endow $\mathbf{R}^n\times\mathbf{R}_+$ with the product distance, then the natural map $(x,r)\mapsto \bar{B}(x,r)$ is an isometry of $\mathbf{R}^n\times\mathbf{R}_+$ onto $\mathcal B$, so $F$ is continuous by composition. In fact we just have to show that $K\mapsto x_K$ is Lipschitz in order to prove that $F$ is.
I tried proving that $K\mapsto x_K$ is locally Lipschitz by geometric arguments without success. Any help will be greatly appreciated.
Proof that $K\mapsto x_K$ is continuous :
Let $(K_n)_n$ be a sequence of compact subspaces converging to $K$. Write $x_n = x_{K_n}$. We want to show that $(x_n)_n$ converges to $x_K$. For $n\geq 0$ we have
$$d_H(\{x_n\},K)\leq d_H(\{x_n\},K_n)+d_H(K_n,K)\leq \underbrace{d_H(\{0\},K_n)+d_H(K_n,K)}_{\text{bounded since }(K_n)_n\text{ converges}},$$
so the sequence $(x_n)_n$ lives in a compact set. Hence we just need to show that $x_K$ is the only possible limit for a sub-sequence of $(x_n)_n$. Assume that
$$x_{\varphi(n)}\underset{n\to +\infty}{\longrightarrow} y$$
for some $y\in \mathbf R^n$. For every $x\in \mathbf R^n$ and every $n\geq 0$ we have $$d_H(\{x_{\varphi(n)}\},K_{\varphi(n)})\leq d_H(\{x\},K_{\varphi(n)}).$$
By passing to the limit we get $$d_H(\{y\},K)\leq d_H(\{x\},K)$$ which is true for any $x$, so $y=x_K$.
Best Answer
I think I found a counterexample for local Lipschitz-continuity of $F$ and $K\mapsto x_K$, where $n=2$.
In this example, we still get Hölder estimates with exponent $1/2$, and I would conjecture that $F$ is Hölder continuous with exponent $1/2$.
construction of counterexample:
Let $n=2$. We define $$ K= B_1((0,0))\cap \{(x,y) : x\geq 0\}, K_\varepsilon := B_1((2\varepsilon,0)) \cap K, $$ where $\varepsilon>0$ is a small number.
Then one can calculate that $$ x_K=(0,0) \qquad x_{K_\varepsilon}=(\varepsilon,0) $$ and $$ d_H(K,K_\varepsilon) = \sqrt{1+4\varepsilon^2}-1 = O(\varepsilon^2) $$ Also, one can calculate that $r_K=1$ and $r_\varepsilon := r_{K_\varepsilon}=\sqrt{1-\varepsilon^2}$ holds.
However, $F$ cannot be Lipschitz continuous, because of $$ d_H(F(K),F(K_{\varepsilon})) = d_H(B_1((0,0)),B_{r_\varepsilon}((\varepsilon,0))) = \varepsilon + (1- \sqrt{1-\varepsilon^2}) \geq \varepsilon. $$
(sorry that I did not include detailed calculations for all the Hausdorff metrics. One can draw the sets on a piece of paper and estimate that the calculations are approximately right.)