Center of convex figure Let us denote by $|F-G|_H$ the Hausdorff distance between compact sets $F$ and $G$ in the plane.

Is it possible to choose a point $p_F\in F$ in any non-empty compact convex figure $F\subset\mathbb{R}^2$ such that
$$|p_F-p_G|\leqslant |F-G|_H?$$

Comments

*

*The barycentre does not work.


*The center of minimal ball containing the figure does not work. Look at the two triangles shown below (this example was suggested by Saúl RM).



*Now we see that there is no such choice (thanks to Saúl RM and Fedja). Let me mention that it also implies that there is no short retraction $\mathrm{Haus}\,\mathbb{R^2}\to \mathbb{R^2}$. Here $\mathrm{Haus}\,\mathbb{R^2}$ denotes the set of all nonempty compact sets in the plane equipped with Hausdorff metric and $\mathbb{R^2}$ is considered as a subspace of $\mathrm{Haus}\,\mathbb{R^2}$ via the distance-preserving embedding that sens a point $x$ to the one-point set $\{x\}$.


*It was proved that the best choice is the so-called Steiner center $s_F$ --- the  center of mass of curvature of $\partial F$. Namely, we have
$$|s_F-s_G|\leqslant \tfrac4\pi\cdot|F-G|_H.$$
This statement (plus the higher-dimensional version) was proved by E. D. Positzelskii in his "Lipschitz mappings..." (1971); it was rediscovered by Krzysztof Przesławski and David Yost in their "Continuity properties of selectors..." (1989) and it is a simple corollary of the results of Denis Rutovitz (1965) and Igor Daugavet (1968). It also appear as Proposition 2.21 in "Geometric nonlinear functional analysis. Vol. 1" by Yoav Benyamini and Joram Lindenstrauss.
 A: There does not exist any function $p:F\mapsto p_F$ as in the question, to prove it by contradiction suppose such a function $p$ exists. In the following let $k+A=\{k+a;a\in A\}$ for any $k\in\mathbb{R}^2$, $A\subseteq\mathbb{R}^2$
To see why, consider the triangle $T_0$ with vertices $(-1,0),(1,0)$ and $(0,1)$. Let $r:(x,y)\mapsto(x,-y)$ be the reflection respect to the $x$ axis.
Now consider for each natural $n$ the triangle $T_n=r^n(T_0+10(n,0))$, and let $p_{T_n}=(x_n,y_n)$. Note that $d_H(T_n,T_{n+1})=10$, so $\sqrt{(x_{n+1}-x_n)^2+(y_{n+1}-y_n)^2}=d(p_{T_n},p_{T_{n+1}})\leq 10$. This, of course, implies that $x_{n+1}-x_n\leq10$. Moreover, $(y_{n+1}-y_n)^2=(|y_{n+1}|+|y_n|)^2$, since $y_n$ and $y_{n+1}$ have opposite signs.
This will allow us to deduce by contradiction that $y_n\to 0$: if not, we would have some $\varepsilon>0$ and infinitely many $n$ such that $|y_n|>\varepsilon$, so $x_{n+1}-x_n<\sqrt{100-\varepsilon^2}$. This implies that for big enough $n$, we will have $x_n-x_0<10n-2$, which is impossible since $10n-2=d(T_n,T_0)$.
Similarly, let $S_0$ be the triangle with vertices $(-1,1),(1,1)$ and $(0,0)$, let $s$ be the reflection around the line $y=1$ and let $S_n=s^n(S_0+10(n,0))$. Then letting $p_{S_n}=(z_n,w_n)$, we can prove that $w_n\to 1$ as before. So for big enough even $n$ we have $d_H(T_n,S_n)=\frac{\sqrt{2}}{2}$ and $d(p_n,q_n)\geq0.9$, the contradiction we were looking for.
A: OK, here is why $Lip1$ is impossible.
Suppose we have such choice.
Then consider all $K$ whose minimal containing box is a unit square.
Let $[x(K),x(K)+1]\times[y(K),y(K)+1]$ be the minimal containing box of $K$. Let $e=(1,0)$. Note that if $L$ is another such domain, then for fixed $t$ and $T\to+\infty$, we have $d_H(te+K,Te+L)=(T-t)+x(L)-x(K)+o(1)$ (the fixed vertical shift becomes of no importance).
Thus, we conclude that if $p(te+K).x-x(te+K)=w$, then $p(Te+L).x-x(Te+L)\le w+o(1)$ as $T\to\infty$, whence the difference $p(Te+K).x-x(Te+K)$ has a (usual) limit $w_x$ independent of $K$ as $T\to+\infty$.
Now if we define
$$
p_T(K)=p(Te+K)-Te
$$
and put $p_1(K)=\lim_T p_T(K)$ where the limit is taken with respect to some ultrafilter containing the usual filter of rays going to $+\infty$, we shall get a choice whose $x$-coordinate is $x(K)+w_x$ for every $K$ in our class and that choice still is $Lip1$. Now we can run the same construction in the $y$-direction and conclude that we can make a $Lip1$ choice $p_2$ such that $p(K).y=y(K)+w_y$ as well. But then for this class of domains, choosing the box center works as well. However if we take $K=conv((-1,0),(0,-1),(0,0))$ and $L=-K$ and observe that $d_H(K,L)=1$ but the box shift is of length $\sqrt2$.
Remarks:

*

*The ultrafilter is not really needed but it makes the proof very transparent, so I decided to use it. Those who dislike AC are welcome to recast it as an argument with classical limits only (or, better, without limits at all using the pigeonhole principle instead) as an exercise.


*While this argument shows that the minimal possible Lipschitz constant is strictly greater than $1$ (if we could get it arbitrarily close to $1$, we could get $1$ as well by taking the limit along, erm, some ultrafilter on positive integers containing the standard filter for the usual limit), it does not provide any explicit lower bound. So finding a reasonable lower bound $1+\delta$ for the Lipschits constant remains an open problem (some quantification is actually possible even using just finitely many shifts of the last two triangles, but I would rather not write down the ridiculously small $\delta$ it yields).


*I hope that is all correct, but check everything carefully because it is quite late here now and I'm not at my best today :-)
A: Almost complete proof (the final argument still needs to be formalized).
We work in any Euclidean space $E$.
Informally, I take $p_F$ equal to the center of the closed ball with minimum radius containing $F$ (existence and uniqueness are shown below).
For $\delta \ge 0$, call $B_\delta$ the closed ball with radius $\delta$ centered at $0$.
By definition, given two non-empty compact sets $F$ and $G$, the Hausdorff distance  $|F-G|_H$ is the least $\delta$ such that $F \subset G + B_\delta$ and $G \subset F + B_\delta$.
For every non-empty compact convex set $F$, and every $x \in E$, set $f_F(x) = \sup\{d(x,y) : y \in F\}$. Observe that for each $r \ge 0$, $f_F(x) \le r$ if and only if $F \subset \overline{B}(x,r)$.
The function $f_F$ is $1$-Lipschitz hence continuous. Its infimum on $F$ is achieved (by compactness) at only one point (by convexity). Indeed, if $F$ is contained in two closed ball with same radius with centers $x_1 \ne x_2$, it is contained in some ball centered at $(x_1+x_2)/2$ with smaller radius. Hence $$r_F := \min_{x \in F} f_F(x) \text{ and } p_F := \arg\min_{x \in F} f_F(x)$$
are well-defined.
Now consider another non-empty compact convex set $G$, and set $\delta = |F-G|_H$. Then
$$G \subset F+B_\delta \subset \overline{B}(p_F,r_F) + B_\delta = \overline{B}(p_F,r_F+\delta),$$
so $$r_G \le f_G(p_F) \le r_F+\delta.$$
Of course, we can reverse the roles of $F$ and $G$. By symmetry, we may and we do assume that $r_G \ge r_F$. One has
$$G \subset \overline{B}(p_G,r_G) \cap \overline{B}(p_F,r_F+\delta).$$
(To be formalized, looks clear on a picture) I claim that $\overline{B}(p_G,r_G) \subset \overline{B}(p_F,r_F+\delta)$, so $|p_F-p_G| \le r_F+\delta-r_G \le \delta$. Otherwise, the intersection of these two balls would be contained in some ball with center in $G$ (starting from $p_G$, move a bit the center in the direction of $p_F$) and radius $<r_G$, and this would contradict the definition of $r_G$.
