What does convergence in the $L^2$ sense to a constant mean curvature surface imply? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:25:18Z http://mathoverflow.net/feeds/question/89980 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89980/what-does-convergence-in-the-l2-sense-to-a-constant-mean-curvature-surface-imp What does convergence in the $L^2$ sense to a constant mean curvature surface imply? Dorian 2012-03-01T18:09:10Z 2012-04-04T11:33:12Z <p>I have been thinking about the following question and have been unable to find any literature on the subject. </p> <p><em>Question:</em> Assume I have a sequence of <strong>smooth, simply connected, compact</strong> domains $\Omega_s \subset \mathbb{R}^d$ such that $|\Omega_s|=1$ and</p> <p>$\int_{\partial \Omega_s} (\kappa - \bar \kappa)^2 dS(y) \to 0$ as $s \to +\infty$,</p> <p>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 <strong>two</strong> cases:</p> <ol> <li>All of the sets $\Omega_s$ are convex. or</li> <li>I assume the uniform bound $\limsup_{s \to +\infty} |\partial \Omega_s| + \int_{\partial \Omega_s} \kappa^2 dS &lt; +\infty$. </li> </ol> <p>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.</p>