Interior $C^{1,\alpha}$ regularity for infinity-harmonic functions

The explanation

Start with Aronsson's function u(x) = |x₁|^{4/3} − |x₂|^{4/3}. Its slope in the x₁ direction is (4/3)·|x₁|^{1/3}, with the sign of x₁. At the origin the slope is 0. A step h away along x₁, the slope is (4/3)·h^{1/3}. For h = 0.001 that is 0.1333. For h = 0.000001 it is 0.01333. So the slope is continuous, but not smooth. Divide the change by h^{1/3} and you always get 4/3. Divide by h^{1/2} instead and you get 4.22, then 13.33, without bound.

This u is infinity-harmonic. That means it solves Δ∞u = 0, where Δ∞u is the second derivative of u along its own slope direction, times the slope squared. Only the second derivative in that one direction enters the equationSource 1. For Aronsson's u, the two terms are 64/81 and −64/81, so they cancel. These functions solve a problem of filling in values between fixed boundary data with the smallest possible steepest slope. Aronsson found the equation for smooth solutionsSource 2. Jensen gave the right weak meaning, called viscosity solutions, and proved uniquenessSource 3.

The hard question is how smooth these solutions are. A Hölder estimate with exponent α says the slope changes by at most C·h^α over a distance h. Aronsson's example shows α can never exceed 1/3Source 4. Savin proved the slope is continuous in the planeSource 5. Evans and Savin then gave a Hölder estimate in the planeSource 6. In higher dimensions, Evans and Smart proved each solution has a slope at every pointSource 7. Whether that slope changes continuously was a further questionSource 8.

The paper claims an interior Hölder estimate for the slope in every dimension d ≥ 3Source 9. On the half ball, the slope and its Hölder size are at most a constant times the spread of u on the full ball. So these functions have continuous slopes inside any open setSource 10. The exponent is positive but not given as a numberSource 11. It comes from an argument by contradiction with limits, called compactnessSource 12. If the estimate failed, rescaled solutions would converge to a limit on all of space. A Liouville theorem, a result that all-space solutions must be simple, shows that limit is affine, a flat planeSource 13. That contradicts the failure.

How sure is this? There is no Lean check. A computer proof checker has not looked at any part of it. The proof uses delicate limit and scaling arguments. No specialist review of the paper has been published.

What it doesn't show. This does not give the exponent as a number, and it does not show the exponent reaches the best possible value 1/3. It says nothing about regularity at the boundary.