Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that $f,g:\mathbb{R}^{2}\rightarrow \mathbb{R}$ are smooth functions, with $g(u,v)\geq f(u,v)\geq 0$ for all $u, v \in\mathbb{R}$ and with f(0,0)=g(0,0)=0. Let $κ_{f},κ_{g}$ be respectively the Gauss curvatures of the graphs of f and g at the origin. Show that $κ_{g}≥κ_{f}≥0$.

share|cite|improve this question
up vote 0 down vote accepted

Since $g$ and $f$ agree up to first-order derivatives, the inequality $g\ge f$ implies the corresponding inequality for Hessians: $D^2g \ge D^2f$ in the sense of positive semidefinite order. The inequality then propagates to eigenvalues (ordered, of course): $\lambda_j(D^2g)\ge \lambda_j(D^2f)$ for $j=1,\dots,n$. The proof for $n=2$ is not hard: see the comment by Jonas Meyer here.

So, not only the Gaussian curvatures are related by such an inequality, but the principal and mean curvatures as well.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.