## Area of convex surface

This is the sequel of the post about Bean’s theorem on convex curve.

I will state the smooth version of Bean’s theorem in the general dimension.

Theorem 1 (Bean’s theorem, 2011, Smooth version) Suppose ${\Sigma}$ is a ((n-1)-dimensional) smooth convex hypersurface contained in the interior of another smooth compact hypersurface $S$ in ${\mathbb{R}^n}$, then the area of ${\Sigma}$ is not greater than the area of $S$.

First of all let us make some remarks. By a smooth convex hypersurface we mean ${\Sigma}$ is compact, connected and such that for all ${p}$, we can find a (outward) normal vector ${\nu(p)}$ such that all ${q\in \Sigma}$, ${\langle q-p, \nu(p)\rangle\leq0}$. i.e. ${\Sigma }$ lies on the non-positive side of ${\nu}$.

Since ${\Sigma}$ (and also $S$) is a compact hypersurface in ${\mathbb{R}^n}$, it is orientable (see also here). Thus we can find a global outward normal field ${\nu}$ of ${\Sigma}$, and thus we can globally define its second fundamental form ${h(X, Y):=\langle \nabla^{\mathbb{R}^n}_X \nu, Y \rangle}$ without any ambiguity, once ${\nu}$ is fixed (here $\nabla^{\mathbb{R}^n}_X \nu$ is only the ordinary directional derivative of the vector field $\nu$ in $\mathbb{R}^n$ in direction $X$). In our convention, the second fundamental form of the unit sphere with respect to its outward normal is ${h(X, X)=|X|^2}$.

We claim that

Lemma 2 If ${\Sigma}$ is convex, then its second fundamental form with respect to the outward normal ${\nu}$ is always non-negative definite.

Proof: Since ${\Sigma}$ is convex, it is locally convex in the sense that at any point ${p}$, ${\Sigma}$ can be locally represented by a graph of function ${f:T_p\Sigma\cong \mathbb{R}^{n-1} \rightarrow \mathbb{R}}$ with ${f(0)=\nabla f(0)=0}$ and ${f\geq 0}$. (What does that mean precisely, and how is ${f}$ related to ${\nu}$? )

Since ${f\geq 0}$, by Taylor’s theorem, at ${0}$ (corresponding to the point ${p}$), for any tangent vector ${X\in T_p\Sigma\cong\mathbb{R}^{n-1}}$

$\displaystyle 0\leq \text{Hess} \, f(X, X)$

But it is easy to see that the hessian of $f$ is equal to the second fundamental form:

$\displaystyle 0\leq \text{Hess} \, f(X, X)=h(X, X).$

$\Box$

Remark: Under some additional assumptions, the converse is also true (but much harder) by a result of Sacksteder.

Lemma 3 For any hypersurface ${S}$ in ${\mathbb{R}^n}$, suppose ${V}$ is a unit vector field defined on a neighborhood of ${S}$ such that ${V|_S}$ is a normal vector field to ${S}$. Then at each point ${p}$ of ${S}$, we have

$\displaystyle div(V)=H(p)$

where ${H}$ is the mean curvature of ${S\subset \mathbb{R}^n}$ w.r.t. ${V|_S}$ (i.e. sum of eigenvalues of the second fundamental form w.r.t. ${V|_S}$, which is symmetric). In our notation, the mean curvature of the unit sphere in ${\mathbb{R}^3}$ is 2.

Proof: Since ${\langle V, V\rangle\equiv1}$, differentiating it gives ${\nabla ^{\mathbb{R}^n}_V V\perp V}$ and thus at ${p}$

$\displaystyle H(p)=\sum_{i=1}^{n-1}\langle \nabla ^{\mathbb{R}^n}_{e_i}V, e_i\rangle=\sum_{i=1}^{n-1}\langle \nabla ^{\mathbb{R}^n}_{e_i}V, e_i\rangle+\langle \nabla ^{\mathbb{R}^n}_V V, V\rangle=div(V)$

where ${e_i}$ forms an orthonormal basis of $T_p\Sigma$. $\Box$

Now for convex ${\Sigma}$ and ${\rho \geq 0}$, we define the surface ${\Sigma ^\rho}$ by

$\displaystyle \Sigma ^\rho:=\{x+\rho \nu(x): x\in \Sigma\}.$

By our previously result (Lemma 2 in here), ${\Sigma^\rho}$ is a regular surface (i.e. without any self-intersection). We claim

Lemma 4 The mean curvature of ${\Sigma^\rho}$ is non-negative if ${\Sigma}$ is convex.

Proof: Let ${X=X(u^1, \cdots, u^{n-1})}$ be a local parametrization of ${\Sigma }$ around ${p=X(0)}$, let us for simplicity denote ${\nu(X(u^1, \cdots, u^{n-1}))}$ simply by ${\nu(u^1, \cdots, u^{n-1})}$. Then for fixed ${\rho}$

$\displaystyle \tilde X(u^1, \cdots, u^{n-1}):=X(u^1, \cdots, u^{n-1})+\rho\vec\nu(u^1, \cdots, u^{n-1})$

gives a natural parametrization for ${\Sigma ^\rho}$. It is easy to see that ${\nu(x)}$ is also the normal vector of ${\Sigma^\rho}$ at ${\tilde X(x)}$ and that ${\frac{\partial \nu}{\partial u^i}\in T_p\Sigma}$. Let us compute the second fundamental form ${\tilde h}$ of ${\Sigma^\rho}$ (denote ${\frac{\partial X}{\partial u^i}}$ by ${X_i}$ etc) at ${\tilde X(0)}$:

$\displaystyle \tilde h(\tilde X_i, \tilde X_i)=\langle \nabla^{\mathbb{R}^n} _{\tilde X_i}\nu, \tilde X_i\rangle=\langle \nabla^{\mathbb{R}^n} _{X_i+\rho \nu_i}\nu, X_i+\rho\nu_i\rangle=h(X_i+\rho \nu_i, X_i+\rho \nu_i)\geq0.$

Thus the second fundamental form of ${\Sigma^\rho}$ is also non-negative and so its mean curvature is non-negative. $\Box$

Proof:(Proof of Bean’s theorem)
Extend the outward normal ${\nu}$ of ${\Sigma}$ to the region ${\Omega}$ bounded between ${\Sigma}$ and ${S}$ by parallel translating ${\nu}$ along its outward direction. Then by our previous result (Lemma 2 in here), this defines a vector field ${X}$ on ${\Omega}$, extending ${\nu}$.

It is easy to see that at each point ${x\in \Omega}$, ${X(x)}$ is the normal to the surface ${\Sigma^{\rho}}$ containing ${x}$ where ${\rho=d(x, \Sigma)}$. Thus by Lemma 4 and Lemma 3,

$\displaystyle div X\geq 0.$

Now, we apply divergence theorem to ${X}$ in ${\Omega}$,

$\displaystyle \int_{S}\langle X, \nu_S\rangle dS- \int_{\Sigma}\langle X, \nu\rangle dS=\int_\Omega div X dx\geq 0$

where ${\nu_S}$ is the unit outward normal of ${S}$. But clearly ${\langle X, \nu_S\rangle \leq 1}$, and ${\langle X, \nu \rangle=1}$, so

$\displaystyle Area(S)- Area (\Sigma)\geq 0.$

$\Box$