6
$\begingroup$

Given a CAT(0) space $X$ and a compact, convex subset $A$ of $X$. One can define its midpoint $m(A)$ as the point, at which the following function attains its minimum.

$f:A\rightarrow \mathbb{R}\qquad x\mapsto \sup\{d(x,y)|y\in A\}$

One can show, that there is a unique such point. So my question is:

Given two compact,convex subset $A_1,A_2\subset X$. Is $d(m(A_1),m(A_2))$ less or equal to the Hausdorff distance between $A_1$ and $A_2$?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
9
$\begingroup$

No, even if $X=\mathbb R^2$.

Let $A_1$ be (the convex hull of) 4 points with coordinates $(\pm 1,\pm 1)$. Then $m(A_1)=(0,0)$, as the 4 points are on the circle $S_1$ of radius $\sqrt 2$ centered at $(0,0)$. Shift $S_1$ a small distance $\varepsilon$ in the horizontal direction, denote the resulting circle by $S_2$. For each vertex of $A_1$, mark its nearest point on $S_2$. The marked points are vertices of a convex quadrangle $A_2$ inscribed in $S_2$ and containing its center $(\varepsilon,0)$. Hence $m(A_2)=(\varepsilon,0)$ but the Hausdorff distance between $A_1$ and $A_2$ is $\approx\varepsilon/\sqrt 2$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ In my figure the marked points are the vertices of a trapezoid instead of a rectangle. But still its midpoint is the midpoint of $S_2$, I think. $\endgroup$
    – HenrikRüping
    Commented Aug 30, 2010 at 9:56
  • $\begingroup$ If one wants to get rid off the $\approx$ sign, one could also just take $A_1$ as above and $A_2:=cH((0,0),(2,-2),(2,2))$ $\endgroup$
    – HenrikRüping
    Commented Aug 30, 2010 at 11:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .