Maximization of average pairwise distance

Question

May be this is some already some standard problem but i do not know.Consider $n$ variable points points $ P_1,P_2,...P_n$ in the closed disc $x^2+y^2 \leq 1.$ Now define the average pairwise distance between them to be: $$ d_{av}(n)= \sum_{\substack{1 \leq i <j \leq n}}\dfrac{2}{n(n-1)} \operatorname{dist}(P_i,P_j) .$$ Is it true that $ d_{av}(n)$ is maximised when all the points lie on the boundary? We can start with the observation if all the points lie inside the circle then we can increase the distance by moving one point $P$ on the boundary.However,when we move some next $Q$ point that is inside to the boundary it might decrease the distance between $P$ and $Q.$Any help/hints/links will be greatly appreciated

Answer 1

It seems to be true, average pairwise distance $d_{av}(n)$ is maximised when all the points lie on the boundary.

Assume that point $P_1$ lies strictly inside the circle. Then we can choose $2$ points $P'_1$ and $P''_1$, such that

• $P_1$ is the midpoint of segment $[P'_1, P''_1]$
• One of the points $P_2, \dots, P_n$ doesn't lie on the same line as $P'_1$ and $P''_1$.

Then, by triangle inequality, $$dist(P_i, P'_1) + dist(P_i, P''_1) \geq 2 dist(P_i, P_1) \ \ \forall 2 \leq i \leq n$$ And fore some $i$, the inequality is strict.

Therefore, if we replace $P_1$ either with $P'_1$, or with $P''_1$, average pairwise distance $d_{av}(n)$ will increase.

Answer 2

Yes it is attained on the boundary, reason is the average distance is convex, because the norm is, the disc is compact and convex, thus the maximum is attained on the extremal points , which is the boundary in the case of a disc, for a simplex for example you need to check the vertices.

Added note as it may not be clear from the above $d_{av}$ is a convex function defined on $\mathbb{D}^{1} \times \cdots \times \mathbb{D} \subseteq \mathbb{R}^{2n}$, which is convex as the cartesian product of convex sets is convex (Here $\mathbb{D}$ is the closed disc).