Minimal surface terminology

Question

If $f: \Omega \subset \mathbb{R}^2 \rightarrow \mathbb{R}$ is smooth, then its graph $G = \{(x,y, f(x,y) | (x,y) \in \Omega\}$ is a minimiser of area if and only if it satisfies the Euler-Lagrange equation for the area functional. I saw further that if $\Omega$ is convex, then $G$ is absolutely area minimising. What is the definition of absolutely area minimising vs just area minimising? The terminology is used here: https://arxiv.org/abs/math/0511469v1

Answer 1

There are some misunderstanding of the terms. First of all, a graph $G$ which satisfies the Euler Lagrange equation

$$ \text{div} \left( \frac{\nabla f}{\sqrt{1+|\nabla f|^2}}\right) = 0$$

if and only if it is a critical point of the area functional. In this case the graph is a minimal surface. That this is a minimal surface does not mean that it is a local or global minimizer of the area functional.

However, as shown in p.4 in Colding-Minicozzi's note, if the graph $G$ of $f$ is a minimal surface, then it is also a area-minimizer amongst surfaces in the cylinder $\Omega \times \mathbb R$ with the same boundary. That is, if $M$ is a minimal surface in $\Omega \times \mathbb R$ such that $\partial M = \partial G:= \{ (x, y, f(x, y)): (x, y)\in \partial \Omega\}$, then $$\text{Area}(M) \ge \text{Area} (G).$$

In particular, the graph $G$ might not be an absolute area-minimizer in $\mathbb R^3$: there might be some surface $M$ in $\mathbb R^3$ with $\partial M = \partial G$ with strictly smaller area. As shown in the note, $G$ is indeed an absolute area-minimizer when $\Omega$ is convex.