Define the set $\Omega := (-\epsilon,\epsilon) \times (-1,1)^{n-1}$, and define $\Gamma := \{-\epsilon,\epsilon\} \times (-1,1)^{n-1} \subset \partial \Omega$. Suppose I have two functions $u,v \in C^\infty(\Omega) \cap C^1(\overline{\Omega})$, with the following properties:

- $u = v$ and $\nabla u=\nabla v$ on $\Gamma$
- $u > v$ on $\Omega$
- $u$ has constant positive mean curvature $K$
- $v$ has constant negative mean curvature $-K$.

That is, $u$ and $v$ curve 'away from' each other, but they still agree along with their first derivatives on $\Gamma$. (I would be happy if I could prove no such pair of functions exists, but I suspect they do exist, possibly resembling the surfaces in Figure 4 of Triply periodic constant mean curvature surfaces.)

My question is, can I find a lower bound on $\sup_\Omega |u-v|$, in terms of $n$, $\epsilon$ and $K$, which doesn't converge to 0 as $\epsilon \to 0$? My intuition is that the mean curvature difference implies $u-v$ has at least some positive second derivatives. When those are integrated over the intervals $(-1,1)$, there should be some lower bound on the maximum distance between them.

More generally, does there exist a similar lower bound if we allow $\Omega$ to range over all subsets of $(-1,1)^n$ such that $\Omega \cap (-1/2,1/2)^n \not=\emptyset$, but we restrict the $n$-dimensional volume of $\Omega$ to be less than $\epsilon$? (If it helps, this paper's Notation and Methods section gives a formula for the mean curvature of the graph of a function.)