$\newcommand{\R}{\mathbb{R}}$ I am trying to understand a step in a proof of the following lemma (the proof is here, lemma 2.7, pg 6):
(This lemma provides a way of proving Kirszbraun theorem).
Lemma: Let $\{x_1 , \dots, x_k \}$ be a finite collection of points in $\R^n$, and let $\{y_1 , \dots, y_k \}$ be a collection of points in $\R^m$ such that $$|y_i − y_ j | ≤ |x_ i − x_ j | \, \text{ for all } \, i, j \in \{1, . . . , k\}.$$ If $r_ 1 , \dots , r_ k$ are positive numbers such that $$ \cap_{i=1}^k \bar B(x_ i , r _i ) \neq \emptyset, $$ then $$ \cap_{i=1}^k \bar B(y_ i , r _i ) \neq \emptyset .$$
Here is what happens in the proof:
Define $$G:\R^m \to \R, \,G(y)= \max_{i=1,\dots,k} \frac{|y − y _i |}{r _i}$$
$G: \R^m \to \R$ is a continuous function satisfying $G(y) \to \infty$ as $|y| \to \infty$. Hence, $G$ achieves its minimum at a point $w \in \R^ m$ , and we need to show that $G(w) \le 1$.
Assume by contradiction that $G(w) := \lambda >1$. Let $J$ denote those indices $j \in \{1, . . . , k\}$ for which $|w − y _j | = r _j λ$. Pick a point $$x \in\cap_{j \in J} \bar B(x _j , r _j ) ,$$
and consider the following two sets: $$D=\{ \frac{x _j − x}{|x _j − x|} |\, j \in J\} \subseteq \mathbb{S}^{n−1},$$ $$D'=\{ \frac{y _j − w}{|y _j − w|} |\, j \in J\} \subseteq \mathbb{S}^{m−1}.$$
The author than says it is easy to see that the natural map* $D \to D'$ strictly decreases the Euclidean distances.
Why is it so?
*The natural map means we keep the indices.
Problem : In $V:=\mathbb{R}^m$, $$1\leq i\leq m,\ p^i,\ q^i\in V,\ |p^ip^j|\geq |q^iq^j| $$ Then $$ \bigcap\ B_{R_i}(p^i)\ni w_0 \Rightarrow \bigcap\ B_{R_i}(q^i)\neq \emptyset\ \ast $$
Note : This is a special case of Breham : In the above, there is piecewise distance preserving $f$ s.t. $f(p^i)=q^i$ However, we do not use his proof.
Reference : Alexandrov meets Kirszbraun - S. Alexander, V. Kapovitch, A. Petrunin. It treats the more general case
Proof :
Notation : In $\mathbb{R}^n$, $$ w\preceq z \Leftrightarrow w_i\leq z_i $$
$\mathcal{X}=\mathbb{R}^m$ is a metric space with standard Euclidean metric. Define $$ {\bf f} :=(f^0,\cdots, f^k) : \mathcal{X}\rightarrow \mathbb{R}^{k+1} $$
Let $f^i(w)=\frac{1}{2}|w-p^i|^2$, which is convex
Define $$Q\subset V,\ {\rm SupSet}\ Q : =\bigg\{ z\in V\bigg| \exists w\in Q\ {\rm s.t.}\ z\succeq w \bigg\} $$
(1) ${\rm SupSet}\ {\bf f}\ (\mathcal{X})$ is convex
Proof : Convexity of $f^i$ implies that \begin{align*} {\bf f} (tw+(1-t)z) &\preceq t {\bf f} (w) + (1-t) {\bf f} (z) \end{align*}
(2) If $g^i(w) :=\frac{1}{2} |w-q^i|^2$ we have a claim that $$ {\rm SupSet }\ {\bf g}(\mathcal{X})\ni {\bf f}(w_0) $$ where $w_0$ is the vector in $\ast$
Proof : If not there is a supporting hyperplane $$\sum_i\alpha_ix_i=c$$ to ${\rm SupSet} \ {\bf g}(\mathcal{X})$ separating it from ${\bf f}(w_0)$
Hence $$ \sum_i \alpha_i f^i(w_0) < c={\rm inf}\ \bigg\{ \sum_i\alpha_i g^i(w)|w\in \mathcal{X} \bigg\}\ \ast\ast $$
Hence if $$ F:=\sum_i \alpha_i f^i,\ G:= \sum_i\alpha_i g^i $$ we have a claim that $$ F(w_F)\geq G(w_G)$$ which contradicts to $\ast\ast$, where $w_F,\ w_G$ are minimum points respectively
Proof of Claim : If $${\rm Dir}\ p^i -w_F =\frac{p^i -w_F}{|p^i -w_F |},$$ then \begin{align*} 0 &=d_{w_F} F\ {\rm Dir}\ (p^i -w_F) \\&= -\frac{1}{2|p^i -w_F|} \sum_j \alpha_j \ \{ |p^i-w_F|^2+ |p^j-w_F|^2 - |p^i-p^j |^2 \} \ ({\rm cf.\ Cosine\ Law}) \\&\\ 2 F(w_F)&= \sum_{i,j} \alpha_i\alpha_j |p^i-p^j|^2 \end{align*}