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



A minimal surface with a local extremum in normal direction is a plane
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.


