Measure Theory – Proving the Sphere Minimizes Surface Area to Volume Ratio

differential-geometrymeasure-theory

It is well known that a sphere minimizes the surface area to volume ratio since it reaches equality in the Isoperimetric Inequality. I'm trying to prove that no other closed surface has this property.

Since the inequality says that $36\pi V^2 \leq A^3$ we can state the problem as:

Let $\mathcal{S}$ be a simple closed surface in $\mathbb{R}^3$ with surface area $4\pi$. Prove that if its volume is $\dfrac{4}{3}\pi$ then $\mathcal{S} = \mathbf{S}^2$.

My hypothesis:

If there is a surface $\mathcal{S}\neq \mathbf{S}^2$ but with the same surface area to volume ratio then there is a function $F(t,x):I\times \mathbf{S}^2\to \mathbb{R}^3$ so that:

$$F(t,x_0) \text{ is continuous }\forall x_0\in\mathbf{S}^2$$
$$F(t_0,\mathbf{S}^2) \text{ is a simple closed surface }\forall t_0\in I$$
$$F(0,\mathbf{S}^2) = \mathbf{S}^2\qquad F(1,\mathbf{S}^2) = \mathcal{S}$$
$$F(0,\mathbf{S}^2) \cong F(t,\mathbf{S}^2)\qquad \forall t\in I$$
$$V(F(t,\mathbf{S}^2)) = \frac{4}{3}\pi\qquad A(F(t,\mathbf{S}^2)) = 4\pi\qquad \forall t\in I$$

Where $\cong$ refers to equivalence via a bijective function, and $V(X)$ and $A(X)$ are, respectively, the volume and surface area functions.

In other words, there is some kind of continuous homeomorphism between both surfaces that conserves both volume and surface area.

Then by using variation calculus we can prove that if a surface is "almost" a sphere, it has a higher $S/V$, thus there cannot be such function $F$ and therefore there is no such surface $\mathcal{S}$.

Perhaps my hypothesis is false. Feel free to prove the statement without using it.

Best Answer

If I am not mistaking, the statement

Sphere is the only closed surface in $\Bbb R^3$ that minimizes the surface area to volume ratio.

is equivalent to either of the following statements:

  1. Sphere is the only closed surface in $\Bbb R^3$ that minimizes surface area, enclosed volume fixed.
  2. Sphere is the only closed surface in $\Bbb R^3$ that maximizes enclosed volume, surface area fixed.

I am going to try to sketch proof of the first statement.

Statement:

Assume that $\mathcal S$ is 2D closed (Riemannian) manifold embedded in $\Bbb R^3$ with enclosed volume $V$. Then $\mathcal S$ has minimal possible surface area $\implies\mathcal S = S^2$.

Proof (sketch):

  1. Take an arbitrary compact Riemannian 2-manifold $\mathcal{M}$ of enclosed volume $V$ embedded in $\Bbb R^3$.
  2. Recall that a surface is called minimal if it minimizes area.
    • Minimal surfaces are the minimizers of mean curvature flow, i.e. they have constant mean curvature [1]. (This relation is supported by the First variation of area formula [2], which describes the relation between total surface area and its mean curvature).
  3. Under the volume-presuming mean curvature flow any compact surface will evolve into a compact surface with constant mean curvature.
    • The mean curvature of compact surface in $\Bbb R^3$ cannot be zero, because the maximum principle for harmonic functions implies that there are no compact complete surfaces of zero mean curvature in $\Bbb R^3$.
  4. The only surface in $\Bbb R^3$ with constant nonzero curvature is sphere $S^2$ [2].
    • Closed surfaces in $\Bbb R^3$ cannot have zero mean curvature, therefore the only surface minimizing its area for a fixed volume is $S^2$.

In fact, these conclusions are supported by the formal derivations in [1, sec. 9-10], where it is proved that under volume-presuming mean curvature flow any compact surface will evolve into sphere.

In relation to your proof, you can argue that if the surface $\mathcal S$ has the same volume and surface area as $S^2$, then it must also have the same (constant nonzero) mean curvature. Otherwise it would not be a minimizer of the volume-preserving mean curvature flow, i.e. it could have been deformed into something with the same volume but smaller surface area. But if it has the same constant nonzero mean curvature as $S^2$, then it must coincide with $S^2$ by Alexandrov's theorem them. Contradiction.

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