Comparison of surfaces by curvatures (I)

We address the following problem: suppose two surfaces tangent at a point $p$ and a surface lies above the other one around $p$. Could we say something about the curvatures of the surfaces at $p$?

Since the sign of the principal curvatures depend on the orientation, suppose that $S_1$ and $S_2$ are 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 $S_i$. After a rigid motion, we suppose that $p=(0,0,0)$,  $T_pS_i$ is the horizontal plane of equation $z=0$ and $N_i(p)=(0,0,1)$. By this condition, $S_i$ is locally the graph of a function on the tangent plane $T_pS_i$ (exercise!). Let $S_1$ and $S_2$ be the graph of $u$ and $v$ respectively.

Because the tangent plane is $z=0$, then with the usual parametrization $X^i(x,y)$ of the surface, 

$X^1_x(0,0)=(1,0,u_x(0,0))$, $X^1_y(0,0)=(0,1,u_y(0,0))$. Since they belong to the $xy$-plane then $u_x(0,0)=u_y(0,0)=0$. Similarly, $v_x(0,0)=v_y(0,0)=0$. By the expression of the mean curvature $H$, we obtain

$$H_1(p)=\frac12(u_{xx}+u_{yy})(0,0),\ H_2(p)=\frac12(v_{xx}(0,0)+v_{yy}(0,0)\ \ (*).$$

Suppose that $S_1$ lies above $S_2$ around $p$ and with respect to the direction that indicates $N_i(p)$. This is equivalent to $u\geq v$ around $q=(0,0)$. Thus the function $u-v$ attains a minimum at $q$. In particular, the Hessian of $(u-v)$ at $q$ is positive semidefinite. But the Hessian is

$$\left(\begin{array}{cc} (u-v)_{xx}&(u-v)_{xy}\\ (u-v)_{xy} &(u-v)_{yy}\end{array}\right)(q).$$

Thus $$(u-v)_{xx}(q)\geq 0,\ \ (u-v)_{yy}(q)\geq 0$$

and (*) implies $H_1(p)\geq H_2(p)$. Summarizing, the result is:

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 $H_1(p)\geq H_2(p)$.