0
$\begingroup$

I'm currently struggling with concluding a proof and need a hint. So the first part of the exercise was that given an open subset $\Omega \subset \mathbb{R}^2$ and a harmonic function $f: \Omega \to \mathbb{R}$ with a (local) maximum or minimum in $\Omega$, then $f$ is constant. This part is done.

Now I have to show that when a minimal surface has a (local) minimum or maximum which points in normal direction, then the surface has to be the plane.

It would be awesome if someone could give me a hint. I was first thinking about the Frenet-Serret formulas but this is probably not the correct way, so here I am. Any ideas/hints are welcome!

Cheers, Pinch

EDIT1: typo

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Presumably you're supposed to combine harmonic function result with the fact that coordinate functions restricted to minimal surface are harmonic $\endgroup$
    – Otis Chodosh
    Commented Jan 4, 2022 at 18:00
  • 1
    $\begingroup$ But I guess there's a missing step, either you need to introduce isothermal coordinates or prove the fact about harmonic functions for arbitrary metrics not just flat one $\endgroup$
    – Otis Chodosh
    Commented Jan 4, 2022 at 18:01

1 Answer 1

Reset to default
0
$\begingroup$

I don't completely understand what you're asking (what is point in the normal direction?) but minimal surfaces must have non-positive curvature, because $H = k_1 + k_2 = 0 \implies K = k_1k_2 \leq 0$. Local maximal and minima would necessarily be in the interior of the domain, as the domain is open, and at such a point, the curvature of the graph (which seems to be what you're considering, speaking of maxima of a surface is a priori nonsense) is positive unless the graph of the function in consideration is already constant, in which case, you are graphing a plane.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .