Compact smooth surfaces in $\mathbb{R}^3$ with constant mean curvature and positive Gaussian curvature is a sphere.

Question

I need to show that a regular ,closed, compact, orientable surface in $\mathbb{R}^3$, with constant mean curvature $H$, and positive Gaussian curvature(not necessarily constant) must be a sphere. The proof given in solution is to show that the two principal curvatures must be equal everywhere on this surface and this surface is umbilical everywhere. I know what's after this, but I don't understant how to show that the surface is umbilical everywhere. How should I show it?

Answer 1

Here's a proof, using the Hilbert's lemma: Let $\lambda_1$ and $\lambda_2$ be principal curvatures such that $\lambda_1\leq\lambda_2$. Since $\lambda_1+\lambda_2=2H$ is constant, if at point $p$ $\lambda_1$ reaches minimum then $\lambda_2$ reaches maximum at $p$. By Hilbert's lemma, if $\lambda_1\neq\lambda_2$ at $p$ then $K(p)<0$, which contradicts with our assumption that $K>0$. Thus $\lambda_1\geq\lambda_2$ at $p$, so $\lambda_1=\lambda_2$ at $p$, it follows that $\lambda_1=\lambda_2$ everywhere on the surface, so $K>0$ is also constant and the surface must be a sphere.