Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have been thinking about the following question and have been unable to find any literature on the subject.

Question: Assume I have a sequence of smooth, simply connected, compact domains $\Omega_s \subset \mathbb{R}^d$ such that $|\Omega_s|=1$ and

$\int_{\partial \Omega_s} (\kappa - \bar \kappa)^2 dS(y) \to 0$ as $s \to +\infty$,

where here $\kappa$ is the mean curvature of the surface $\partial \Omega_s$ and $\bar \kappa$ denotes the average mean curvature over $\partial \Omega_s$. I can prove that the limit is in fact a ball in the following two cases:

  1. All of the sets $\Omega_s$ are convex. or
  2. I assume the uniform bound $\limsup_{s \to +\infty} |\partial \Omega_s| + \int_{\partial \Omega_s} \kappa^2 dS < +\infty$.

I would however like to remove these restrictions since they seem quite artificial. I have been able to rule out the standard "pinching" counter examples of a long rod with capped ends, but am not sure if there could exist other pathologies. Any direction to results in this direction would be appreciated.

share|improve this question
    
Sorry I deleted the question before by accident. –  Dorian Mar 1 '12 at 18:09
    
When you say "two cases", do you mean you can prove the statement using each assumption alone or do you need both assumptions at the same time? –  Deane Yang Mar 1 '12 at 18:22
    
Each assumption alone, without one another. –  Dorian Mar 1 '12 at 18:22
    
Is the second assumption so unreasonable? You need to make sure the domain doesn't expand or shrink indefinitely, and these seem to be reasonable assumptions for making sure that doesn't happen. What would be a weaker assumption that you need? –  Deane Yang Mar 1 '12 at 18:36
    
I think finite perimiter and an $L^2$ curvature bound should follow from the convergence. –  Dorian Mar 1 '12 at 21:04

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.