# Function as sum of distances over a connected, compact metric space

If $$X$$ is a connected, compact metric space with distance function $$d : X^2 \rightarrow \mathbb{R}^+$$, is it true that there exists a positive real number $$a$$, dependent on $$X$$ and $$d$$, such that for any $$n$$ and for any $$x_1, x_2, \cdots, x_n \in X$$, there exists $$y$$ such that $$\frac{1}{n}\left(d(x_1,y) + d(x_2,y) + \cdots + d(x_n,y)\right) = a?$$

Motivation: trivial when $$X$$ is the boundary of a circle, a tricky contest problem when $$X$$ is the boundary of a square (these examples all use Euclidean distance)

• the "boundary of a circle"???
– YCor
Nov 2 '18 at 22:55

It's a classic theorem of O. Gross from 1964. The number $$a$$ is also unique for a given space $$(X,d)$$.

There is an exposition at https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Cleary-Morris-Yost260-275.pdf

The original paper is: O. Gross, "The rendezvous value of a metric space", Advances in Game Theory, Ann. of Math. Studies no. 52, Princeton, 1964, 49-53

Arguing by contradiction assume there is no such number. By 1-dimesional Helly's theorem, there is a pair of point-arrays $$\{x_1,\dots,x_n\}$$ and $$\{y_1,\dots,y_m\}$$ such that for their average distance functions $$f(z)=\tfrac1n\cdot\sum_i|x_i-z|$$ and $$h(z)=\tfrac1m\cdot\sum_j|y_j-z|$$, we have $$f(p)>h(q)$$ for any two points $$p,q$$.
Note that $$\tfrac1m\cdot\sum_if(y_i)=\tfrac1{m\cdot n}\cdot\sum_{i,j}|x_i-y_j|=\tfrac1n\cdot\sum_ih(x_i),$$ a contradiction.
• Why is it enough to consider only two sets $X$ and $Y$? By compactness it is possible to choose a finite number, but how do you reduce it to two?