1
$\begingroup$

Let $\Omega$ be a bounded smooth convex domain in $\mathbb{R}^3$, then consider the following minimization problem:

$$\inf_{c \in \overline{\Omega}} f(c), \quad f(c) := \int_{\partial \Omega} |x-c| d \sigma_x,$$ where $\sigma$ is the surface measure.

If $\Omega$ is a ball, then it is easy to prove that the infimum is attained only at the center of the ball. I guess for a generic fixed convex domain $\Omega$, the infimum should be attained at the "center" of $\Omega$, but I've no idea how to prove or disprove it, and also I don't know if the "center" is the standard meaning, that is, $c=\frac{\int_{\Omega} x dx}{|\Omega|} $.

Any comment or hint would be really appreciated.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ It is highly unlikely that it will be in the center. Hint: differentiate and write that all partial derivatives at c are equal to 0. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 3, 2016 at 23:43
  • $\begingroup$ @Alexandre Eremenko, we can always try to find critical point. Here using $\nabla |x-c|=\frac{c-x}{|x-c|}$, I get $\int_{\partial \Omega}\frac{c-x}{|x-c|} d\sigma_x=0$ , but I cannot solve for the $c$. I agree that the critical number may not be at the center, but I believe it should be a point related to the geometry of $\Omega$. Like when $\Omega$ is a ball, one can really prove $c$ is the center of the ball. $\endgroup$
    – student
    Commented Nov 4, 2016 at 0:00
  • $\begingroup$ Of course this is difficult to solve for $c$, but this equation indicates that this is not the center as you conjectured. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 4, 2016 at 0:08
  • $\begingroup$ @Alexandre Eremenko, you are right. Thank you for your comment and time! $\endgroup$
    – student
    Commented Nov 4, 2016 at 0:09
  • $\begingroup$ It is not even clear that the critical point (or absolute minimum) is unique, the thing one would like to know before solving for $c$. $\endgroup$
    – Alexandre Eremenko
    Commented Nov 4, 2016 at 0:09

0

Reset to default

You must log in to answer this question.