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I find myself unable to start the following problem in Differential Geometry of Curves and Surfaces by Do Carmo, Section 3.3 Problem 24.a

Edit: (Definition) (Local Convexity and Curvature). A surface $S \subset R^3$ is locally convex at a point p ∈ S if there exists a neighborhood V ⊂ S of p such that V is contained in one of the closed half-spaces determined by Tp(S) in R3. If, in addition, V has only one common point with Tp(S), then S is called strictly locally convex at p.

"Prove that S is strictly locally convex at $p$ if the principal curvatures of $S$ at $p$ are nonzero with the same sign (that is, the Gaussian curvature $K(p)$ satisfies $K(p) > 0$)."

I fail to see why this is strictly locally convex, in fact, I fail to see why this must be locally convex.

What if we had a surface that was generated by revolving about the y axis a curve with infinitely many bumps as x approaches to 0 (with decreasing amplitude to bound its derivative)?, but also somehow makesure that this surface is elliptic at (x,y) = (0,0)?

My guts tell me that this would not be a regular surface, but I am unable to prove it.

I would appreciate any hints!

HK Lee
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Ecotistician
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  • What is a "strictly locally convex surface"? – Robert Lewis Apr 26 '18 at 04:00
  • Sorry, I should have given more definitions. added in edit. – Ecotistician Apr 26 '18 at 04:11
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    Well, it would help understand your question and increase the probability of an answer if you added the definition. Cheers! – Robert Lewis Apr 26 '18 at 04:13
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    A small enough neighbourhood $V$ can be written as a graph of some function $f$ over $T_p S,$ after which the curvatures are proportional to the second partial derivatives of $f$ and lying in one of the half-spaces is equivalent to $f$ having constant sign. From there it's just calculus. – Anthony Carapetis Apr 26 '18 at 05:50
  • I guess I'm unstuck at a more elementary level then, do we have proof from calculus that if you have continuous partial derivatives at a point that with second partial derivatives the same sign in every direction, then the surface must be contained in one side of the tangent plane? What's there to prevent something like the example I listed from happening? – Ecotistician Apr 26 '18 at 06:07
  • I understand the part about every curve passing through the point p must have a normal curvature with the same sign => hence curvature of the same sign. But I'm still trying to convince myself that, there exists a neigborhood V of p that prevents these curves from going over to the other side of Tp(S), (namely there are only finitely many local extrema in every neighborhood of V of p) – Ecotistician Apr 26 '18 at 06:12
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    I don't really follow your proposed example. The fact to be used is the multivariable generalization of the following: if $f \in C^2(\mathbb R)$ satisfies $f(0)=0, f'(0)=0, f''(0)>0,$ then $f(x)>0$ for all $x\ne 0$ in some neighbourhood of $0$. – Anthony Carapetis Apr 26 '18 at 06:52
  • My confusion is resolved now, my "example" does't have a positive second derivative at the point x. also I see how you can use taylor's theorem (or a proof by integration on the interval for which second derivative in a neighborhood of x is positive, which exists by continuity) to prove the theorem above. Sorry I was having my chicken brain moment. – Ecotistician Apr 29 '18 at 11:53

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