Why is a graph of a function in $\mathbb{R}^n$ satisfying the minimal surface equation actually area minimizing?

Asked Feb 23 '17 at 10:52

Active Jul 21 '20 at 21:34

Viewed 263 times

I am reading the Colding and Minicozzi book "A course in Minimal Surfaces".
I have a question regarding one of the points mentioned in the book. I want to prove that a minimal hypersurface which is a graph in $\mathbb{R}^n$ is area minimizing.

My idea is - if $\omega$ is the volume form in $\mathbb{R}^n$ and $N$ is the normal vector then ${\iota_N\omega}$ must be a calibration and hence the hypersurface will be a calibrated submanifold and hence area minimizing.

Is it correct ?

edited Jul 21 '20 at 21:34

Arctic Char

16,972

asked Feb 23 '17 at 10:52

inquisitive

Are you trying to prove the stability or that they are actually area minimizing? – Edu Feb 23 '17 at 10:58
1

I need only stability though but I think we can easily see that if we can prove that it is area minimizing. – inquisitive Feb 23 '17 at 12:47

Why is a graph of a function in $\mathbb{R}^n$ satisfying the minimal surface equation actually area minimizing?

0 Answers0