---
tag: NP948D6FR0
kind: problem
name: Bernstein's problem
msc: [53A10, 35J93, 49Q05]
uses: [3GV2E2YTZX, 41X0YW2R9N]
about: [3GV2E2YTZX]
sources:
  - url: https://en.wikipedia.org/wiki/Bernstein%27s_problem
same_as:
  wikidata: Q4894564
  wikipedia: Bernstein's problem
history:
  posed_by: "Sergei Bernstein"
  first_stated: "1915"
---

## Problem

Let $m \ge 1$ and let $u \colon \R^m \to \R$ be a twice continuously differentiable function that solves the [[minimal surface equation|3GV2E2YTZX]]
$$\operatorname{div} \left( \frac{\nabla u}{\sqrt{1 + \abs{\nabla u}^2}} \right) = 0$$
on all of $\R^m$.
Must $u$ be an affine function, so that its graph is a hyperplane in $\R^{m+1}$?

## Remarks

The answer is yes for $m \le 7$ and no for $m \ge 8$.
Sergei Bernstein proved the case $m = 2$: an entire [[minimal surface|41X0YW2R9N]] graph over the plane is a plane.
Wendell Fleming gave a new proof for $m = 2$ based on area-minimizing cones.
This approach led to the cases $m = 3$ by Ennio De Giorgi, $m = 4$ by Frederick Almgren and $m \le 7$ by James Simons.
Enrico Bombieri, Ennio De Giorgi and Enrico Giusti found a non-affine entire solution for $m = 8$, and so for every $m \ge 8$.
The dividing line comes from Simons' cone $\set{x \in \R^8 : x_1^2 + x_2^2 + x_3^2 + x_4^2 = x_5^2 + x_6^2 + x_7^2 + x_8^2}$, which is area-minimizing.
For $m = 1$ the statement is elementary: the equation says that $u'/\sqrt{1 + u'^2}$ is constant, so $u'$ is constant.

## Explanations

### L1 human

A soap film that spreads over the whole plane as the graph of a function must be flat.
Bernstein proved this, and the question was whether the same holds in higher dimensions.
It does up to dimension seven, but in dimension eight and beyond there are curved minimal graphs over the whole space.
The change at dimension eight is one of the surprising facts of geometric analysis.