Bernstein's problem

In differential geometry, Bernstein's problem is as follows: if the graph of a function on Rn&minus;1 is a minimal surface in Rn, does this imply that the function is linear? This is true for n at most 8, but false for n at least 9. The problem is named for Sergei Natanovich Bernstein who solved the case n = 3 in 1914.

Statement
Suppose that f is a function of n &minus; 1 real variables. The graph of f is a surface in Rn, and the condition that this is a minimal surface is that f satisfies the minimal surface equation


 * $$\sum_{i=1}^{n-1} \frac{\partial}{\partial x_i}\frac{\frac{\partial f}{\partial x_i}}{\sqrt{1+\sum_{j=1}^{n-1}\left(\frac{\partial f}{\partial x_j}\right)^2}} = 0$$

Bernstein's problem asks whether an entire function (a function defined throughout Rn&minus;1 ) that solves this equation is necessarily a degree-1 polynomial.

History
proved Bernstein's theorem that a graph of a real function on R2 that is also a minimal surface in R3 must be a plane.

gave a new proof of Bernstein's theorem by deducing it from the fact that there is no non-planar area-minimizing cone in R3.

showed that if there is no non-planar area-minimizing cone in Rn&minus;1 then the analogue of Bernstein's theorem is true for graphs in Rn, which in particular implies that it is true in R4.

showed there are no non-planar minimizing cones in R4, thus extending Bernstein's theorem to R5.

showed there are no non-planar minimizing cones in R7, thus extending Bernstein's theorem to R8. He also showed that the surface defined by


 * $$\{ x \in \mathbb{R}^8 : x_1^2+x_2^2+x_3^2+x_4^2=x_5^2+x_6^2+x_7^2+x_8^2 \}$$

is a locally stable cone in R8, and asked if it is globally area-minimizing.

showed that Simons' cone is indeed globally minimizing, and that in Rn for n≥9 there are graphs that are minimal, but not hyperplanes. Combined with the result of Simons, this shows that the analogue of Bernstein's theorem is true in Rn for n≤8, and false in higher dimensions.