Suppose $\Sigma\subset \mathbb{R}^{n+1}$ is a closed embedded hypersurface. We know that when $n=1$

$$ \int_{\Sigma} |H|^2 \geq \frac{4 \pi^2}{|\Sigma|} $$ by Gauss-Bonnet and that this is saturated on the round circle -- here $|\Sigma|$ is the length of $\Sigma$ and $H$ the mean curvature.

Likewise, if $n=2$ we have $$ \int_{\Sigma} |H|^2\geq 16\pi $$ which is also saturated on the round sphere (of course as we now know this can be improved for positive genus surfaces).

I'm wondering to what extent one can get sharp lower bounds in dimensions $n+1>3$. Specifically, something like $$ \int_{\Sigma} |H|^2 \geq C_n |\Sigma|^{(n-2)/n}. $$

Ideally, this bound would be sharp on the round sphere at least amongst convex competitors (I suspect otherwise it wouldn't be true).

Any references would be appreciated.

  • $\begingroup$ Isn't it $\ge 4\pi$ for dimension $n=2$? which is realized by round sphere? $\endgroup$ – J. GE May 15 '13 at 11:27
  • $\begingroup$ I think that is a just using a different convention for the mean curvature (I use the trace of the second fundamental form, not the average). However, I think I miscomputed... $\endgroup$ – Rbega May 15 '13 at 12:31
  • 2
    $\begingroup$ The exponent at $|\Sigma|$ should be $(n-2)/n$ for scale invariance. $\endgroup$ – Sergei Ivanov May 15 '13 at 20:56

I have no idea about the general case but in the convex case the sphere is indeed optimal. Moreover the $L^1$ norm of $H$ attains its minimum at the sphere (among the convex surfaces with the same area). To deduce the result for the $L^2$ norm, just apply Cauchy-Schwarz,

Let $A$ be the convex body bounded by $\Sigma$ and $B$ the unit ball in $\mathbb R^{n+1}$. Then the area $|\Sigma|$ is proportional to $V_1(B,A)$ and the integral of the mean curvature is proportional to $V_2(B,A)$, where $V_k(B,A)$ is the mixed volume of $k$ copies of $B$ and $n+1-k$ copies of $A$.

By the Alexandrov-Fenchel inequality, $\log V_k(A,B)$ is a concave function of $k$ ($k\in\{0,1,\dots,n+1\}$). This fact yields a lower bound for $V_2(B,A)$ in terms of $V_1(B,A)=C(n)|\Sigma|$ and $V_{n+1}(B,A)=V(B)$, a constant. Namely $$ C(n)\int_\Sigma H = V_2(B,A) \ge V(B)^{1/n} V_1(B,A)^{(n-1)/n} = C_1(n)|\Sigma|^{(n-1)/n} $$ If $A$ is a ball, the inequality turns to equality because so does the Alexandrov-Fenchel inequality.

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