Comparison of curves by curvatures (III)

By the theorem of the previous entry, we prove:
Theorem. Any compact surface has points with positive Gauss curvature.

Proof. Take $p_0\in S$ the fairest point of $S$ from the origin of ${\mathbb R}^3$: this point exists because $S$ is compact and the distance function to a fixed point is a continuous function. We do the next steps.

Take ${\mathbb S}^2(r)$ the sphere centered at the origin and radius $r=|p_0|$: this number is positive because on the contrary is only one point. 

The surfaces $S$ and ${\mathbb S}^2(r)$ are tangent at $p_0$. For ${\mathbb S}^2(r)$ we know that the tangent plane is orthogonal to the position vector $p_0$. For $S$, consider the function $f(p)=|p|^2$. Because $p_0$ is a maximum, it is a critical point, so $df_{p_0}=0$. But it is is immediate that $df_{p_0}(v)=2\langle p_0,v\rangle$ for any $v\in T_{p_0}S$. Thus $T_{p_0}S$ is orthogonal to $p_0$. 

We orient ${\mathbb S}^2(r)$ according the orientation pointing inside, so the normal curvature for any tangent vector is $1/r$. Consider the orientation on $S$ so $N(p_0)=-p_0/|p_0|$, that is, the same than ${\mathbb S}^2(r)$. Moreover, $S$ lies above ${\mathbb S}^2(r)$ around $p_0$.

By the Theorem in the previous day, $\kappa_n(v)\geq 1/r$, in particular, in along the principal directions, $\kappa_i(p_0)\geq 1/r$, so $K(p_0)\geq 1/r^2$.

In particular, we have an estimate of the Gauss curvature at the fairest point  $p_0$ from the origin: $$K(p_0)\geq\frac{1}{|p_0|^2}.$$

As a consequence of the inequality $H^2\geq K$, we have:

Corollary. There are no compact minimal surfaces.

Comparison of curves by curvatures (II)

The definition of the mean curvature is the average of the principal curvatures. But it is also the average of the normal curvatures in two orthogonal directions, that is, 

$$H(p)=\frac12\left(\kappa_n(\theta)+\kappa_n(\theta+\frac{\pi}{2})\right),\ \ (*)$$ where $\kappa_n(\theta)$ is the normal curvature in the direction that makes an angle $\theta$ with a fix direction of the tangent plane.

Suppose again  $S_1$ and $S_2$  two surfaces with $p\in S_1\cap S_2$, $T_pS_1=T_pS_2$ and $N_1(p)=N_2(p)$, where $N_i$ are the orientation in each surface $S_i$. Take all normal sections through $p$, that is, the intersection of the planes formed by $N_i(p)$ and the tangent vectors. The set of these planes in both surfaces is the same because $N_1(p)=N_2(p)$. Fix $v\in T_p S_i$ and $\Pi_v$ the corresponding normal section. This plane meets $S_i$ in two curves $\alpha_i^v$. We are computing the normal curvature with respect to the normal (of the curve) coincides with $N_i(p)$ at $p$ and this normal curvature is the curvature of the curve

Suppose that $S_1$ lies above $S_2$ around $p$. Then for each $v\in T_pS_i$, the curve $\alpha_1^v$ lies above $\alpha_2^v$ at $p$. But we know by the theory of curves that the curvature of $\alpha_1^v$ is greater or equal to the one of $\alpha_2^v$ at $p$. Thus, all normal curvatures of $S_1$ at $p$ are greater or equal to the ones of $S_2$ at $p$. We state this remarkable result:

Theorem. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then for avery $v\in T_pS_i$, we have the inequality
$$\kappa_n^1(v)\geq\kappa_n^2(v).$$

As a consequence,  it is immediate from (*) 

Corollary. Suppose $S_1$ and $S_2$ two surfaces tangent at a point $p$ and that the orientations of the surfaces coincide at $p$, that is, $N_1(p)=N_2(p)$. If $S_1$ lies above $S_2$ around $p$, then $H_1(p)\geq H_2(p)$.