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I am reading Lecture Notes on Minimal Surfaces, which can be accessed here and I am trying to find a proof of the following:

Mean curvature of a surface $S$ is zero everywhere if and only if surface $S$ has locally minimal area.

In the section Why a Minimal Surface is Minimal (or Critical) of the mentioned notes, it is shown, that if $H = 0$ everywhere, then area function has a critical point at a chosen point. I wonder what this means; do there exist surfaces, which have mean curvature $H = 0$, but do not have locally minimal area? What is an example of such function? If not, how do we show that this extrema is in fact a local minimum?

It is also shown that if $S$ minimizes area, then its mean curvature vanishes everywhere. According to these two results, the Local least area definition and Mean curvature definition on the wikipedia page are not equivalent; mean curvature implies local least area, but not the other way around. What is happening? What am I understanding wrong?

How can I complete the proof to give me the result I desire (the equivalence of two definitions), does it even hold? Perhaps, you can refer me to a completely unrelated proof of this equivalence, I am having trouble finding it proven anywhere.

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    $\begingroup$ Suggestion: first consider and find answers to the analogous questions for geodesics. Answers and proofs are essentially the same. $\endgroup$
    – Deane
    Mar 17, 2023 at 16:56
  • $\begingroup$ The lecture notes you mentioned have shown that the minimization ensures that mean curvature vanishes everywhere, but actually, they seem to haven't illustrated the converse case. I'm also considering this problem... $\endgroup$ Nov 15, 2023 at 16:38
  • $\begingroup$ @一団和気 I wrote an answer now, see if it addresses your concerns. $\endgroup$
    – Jesus
    Nov 15, 2023 at 17:24

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I am writing an answer to my question, as I was reminded of it by a new comment and it has no answer so far. Please correct me if there are any mistakes.

In fact, after reviewing the Wikipedia, there are four (probably even more) equivalent definitions of minimal surfaces. Let me state them as they are stated on Wikipedia.

Local least area definition: A surface $M \subset \mathbb{R}^3$ is minimal if and only if every point $p \in M$ has a neighbourhood, bounded by a simple closed curve, which has the least area among all surfaces having the same boundary.

Variational definition: A surface $M \subset \mathbb{R}^3$ is minimal if and only if it is a critical point of the area functional for all compactly supported variations.

Mean curvature definition: A surface $M \subset \mathbb{R}^3$ is minimal if and only if its mean curvature is equal to zero at all points.

Differential equation definition: A surface $M \subset \mathbb{R}^3$ is minimal if and only if it can be expressed as the graph of a solution of $$ \left( 1 + u_x^2 \right) u_{yy} - 2u_x u_y u_{xy} + \left( 1 + u_y^2 \right) u_{xx} = 0. $$

What the notes in the question actually prove is that the variational definition is equivalent to the mean curvature definition, but the question is looking for the proof that this is also equivalent to the local least area definition.

This is instead shown in the beautiful paper Minimal Surfaces for Undergraduates by Franc Forstnerič. In Theorem 4.1 he shows that the mean curvature definition is equivalent to the differential equation definition.

Now it remains to consider why the local least area definitions is equivalent to the other three. This is also evident from Forstnerič's article, from the second chapter on graphs with minimal areas, where he derives the differential equation as the equation of minimal graphs. Combined with Theorem 4.1, we can conclude that a surface $M \subset \mathbb{R}^3$ has vanishing mean curvature if and only if small pieces of the surface are minimal graphs over affine planes. This is exactly what the question was asking for.

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