Difference of two functions with constant mean curvature

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.)