I've tried to find counterexamples or results in this direction, but I haven't found what I'm after (except for the $\mathbb{R}^2$ case).


Allard's regularity theorem guarantees that $(\Lambda,r\_0)$-perimeter minimisers are $C^{1,\frac{1}{2}-\epsilon}$ for all $0<\epsilon<\frac{1}{2}$. But as I understand it, Allard's regularity theorem can never prove a set is $C^{1,1}$. The sets I'm interested in are calibrable, but they do not minimise any obvious variational equations (e.g. the Cheeger set for a square is a square with rounded edges, but I'm not sure which variational equation that minimises).


I want to apply Ketterer's [Heintze-Karcher inequality for metric measure spaces](https://doi.org/10.1017/CBO9781139108133.001) to a $(\Lambda,r\_0)$-perimeter minimising set $E$, but this requires a certain outer curvature to be finite everywhere, corresponding to an interior ball condition. The interior and exterior ball conditions together are satisfied if and only if $E$ has $C^{1,1}$ boundary. Some way of proving an interior ball condition for calibrable sets would also be greatly appreciated.



Are $(\Lambda,r_0)$-perimeter minimising sets $C^{1,1}$?
No.
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Let $f(x,y):=(x^2-y^2)\log(x^2+y^2)$. Then $f$ has bounded mean curvature on bounded sets, and $f\in C^{1,1-\epsilon}$ for all $\epsilon>0$, but $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are not Lipschitz.


