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A key lemma to Kirszbraun's theorem for $\mathbb{R}^2$ states the following:

Given any two finite collections of points $x_1,\dots,x_n$ and $x_1',\dots,x_n'$ in $\mathbb{R}^2$ such that $|x_i'x_j'|\le |x_ix_j|$ for all $i,j=1,\dots,n$ and any $x\in \mathbb{R}^2$, it's always possible to find $x'$ in the convex hull of $x_1',\dots,x_n'$ such that $|x'x_k'|\le |xx_k|$ for all $k=1,\dots,n$.

Intuitively, it seems to me that this result could be generalized in the following way:

Given any two finite collections $x_1,\dots,x_n$ and $x_1',\dots,x_n'$ with $1<l<m<n$ such that

(1) $|x_i'x_j'|\le |x_ix_j|$ for all $i,j=1,\dots,m$ and

(2) $|x_k'x_h'|\le |x_kx_h|$ for all $k,\ h=1,\dots,l,\ m+1,\dots,n$,

and any point $x\in Conv(x_1,\dots,x_l)\setminus \{x_1,\dots,x_l\}$, it is always possible to find $x'$ in the convex hull of $x_1',\dots,x_l'$ such that $|x'x_t'|\le|xx_t|$ for all $t=1,\dots,n$.

As it's clear, my idea is to take more than one collection of points which satisfy the hypothesis of Kirszbarun's lemma and which have non empty intersection.

I'm not able to prove it, but it seems right to me, since I can not find any counterexample. So I'd like to ask you if what I'm saying could have sense and, if you think so, if you have an idea of a possible strategy for the proof.

Thank you.

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Finally I understood your question; here is a counterexample:

Consider a rhombus $x_1x_3x_2x_4$ and let $x$ be the midpoint of the diagonal. Let $x_1'x_3'x_2'x_4'$ be a rhombus with the same side lengths such that $|x_1'-x_2'|<|x_1-x_2|$.

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    $\begingroup$ I guess you mean $|a-b|<|a'-b'|$, not $|a-b|=|a'-b'|$. Do you think is it still possible to find a counterexample with the additional condition $x\in Conv(x_1,\dots,x_l)\setminus \{x_1,\dots,x_l\}$ (I've edited my question)? $\endgroup$ – user294185 Mar 25 '17 at 19:40
  • $\begingroup$ @user294185 it is corrected now. No, any point $x\in [ac]$ will work as well. $\endgroup$ – Anton Petrunin Mar 25 '17 at 19:43
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    $\begingroup$ I'm sorry but your conterexample doesn't seem correct to me. If I understand correctly in my notation you pose $b=x_l$, $a=x_m$, $c=x_n$ and so the two groups are $\{x_l,x_m\}=\{b,a\}$ and $\{x_l,x_n\}=\{b,c\}$. The inequalities you set are correct, but for $x=x_l=b$ we need to prove the existence of a point $x'$ such that $|x'x_l'|\le |xx_l|=0$, $|x'x_m'|\le |xx_m|=|ab|$ and $|x'x_n'|\le |xx_n|=|bc|$. The point $x'=b'$ satisfies the inequalities. $\endgroup$ – user294185 Mar 26 '17 at 11:21
  • $\begingroup$ @user294185 now it is corrected. $\endgroup$ – Anton Petrunin Mar 26 '17 at 15:41
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    $\begingroup$ @user294185 Your question was formulated correctly, but in a way that hard follow --- if you would divide your set of points clearly in three sets, say $x_i,y_j,z_k$ then there would be easier to understand (altho you will need to write 6 more inequalities) $\endgroup$ – Anton Petrunin Mar 26 '17 at 18:01

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