Let $\Omega \subset \mathbb{R}^3$ be a smooth, bounded domain. Let $x_1,\ldots,x_n \in \overline{\Omega}$ be chosen so as to minimize $$ \sum_{1\leq i<j\leq n} \frac{1}{|y_i - y_j|} $$ over all possible choices of $y_1,\ldots,y_n \in \overline\Omega$.

Is it then true that $\min \{|x_i - x_j| : 1\leq i < j \leq n \}$ equals the maximum of $\min\{|y_i - y_j| : 1\leq i < j \leq n\}$ over all possible choices of $y_1,\ldots,y_n \in \overline\Omega$?

I know this is related to Fekete points and the Thomson problem. Maybe it's easy. References are appreciated as well.

  • 3
    $\begingroup$ Have you looked at the case of $4$ or more points on an interval of length $1$ (the limiting case of thin parallelepipeds). There the configuration maximizing the minimal distance between points is unique and it is a simple exercise in calculus to check if it is stationary for the energy functional. $\endgroup$ – fedja May 30 at 2:30
  • 1
    $\begingroup$ I very much doubt. In the limit as $n$ goes to infinity, and after appropriate re-scaling, this should converge to the Newtonian capacity of $\Omega$: the minimal Newtonian energy of a probability measure supported on $\Omega$. The minimizing measure (a.k.a. equilibrium measure) is known to be concentrated on $\partial \Omega$, so I suspect for large $n$ the points will tend to concentrate on the boundary. $\endgroup$ – Mateusz Kwaśnicki May 30 at 8:10
  • 1
    $\begingroup$ The answer is "no", by the example suggested by @fedja. For the segment, the $\max \min |y_i-y_j|$ is achieved for uniform distribution while the extremal configuration for the Newton energy is not uniform. $\endgroup$ – Alexandre Eremenko May 30 at 16:18
  • $\begingroup$ @AlexandreEremenko (and @fedja) Even though the segment is not smooth or open? $\endgroup$ – Ben Ciotti May 30 at 18:19
  • $\begingroup$ I realize the idea would be to represent the segment as a limit of smooth domains, i.e. letting $\epsilon$ go to zero in $\{x^2 + (y/\epsilon)^2 + (z/\epsilon)^2 <1\}$. And it is not obvious to me why the extremal configuration for the Newton energy should not be uniform. $\endgroup$ – Ben Ciotti May 30 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.