Motivation for this post: the paper “Lipschitz and biLipschitz Maps on Carnot Groups” by William Meyerson. I don’t get it, even after several readings of the paper.
The proof of Fact 2.10 (page 10) starts by the statement that the difference of two Lipschitz functions is Lipschitz and the difference of two Pansu differentiable functions is differentiable.
Let us see: we have a Carnot group (which I shall assume is not commutative!) and two functions , where is an open set in . (We may consider instead two Carnot groups and (both non commutative) and two functions .)
Denote by the difference of these functions: for any (here the group operations and inverses are denoted multiplicatively, thus if then ; but I shall suppose further that we work only in groups which are NOT commutative).
1. Suppose and are Lipschitz with respect to the respective CC left invariant distances (constructed from a choice of euclidean norms on their respective left invariant distributions). Is the function Lipschitz?
NO! Indeed, consider the Lipschitz functions , the identity function, and a constant function, with not in the center of . Then is a right translation, notoriously NOT Lipschitz with respect to a CC left invariant distance.
2. Suppose instead that and are everywhere Pansu differentiable and let us compute the Pansu “finite difference”:
We get that is the product w.r.t. the group operation of two terms: the first is the conjugation of the finite difference by and the second term is the finite difference . (Here is the conjugation of by $u$ in the group .)
Due to the non commutativity of the group operation, there should be some miracle in order for the finite difference of to converge, as goes to zero.
We may take instead the sum of two differentiable functions, is it differentiable (in the sense of Pansu?). No, except in very particular situations, because we can’t get rid of the conjugation, because the conjugation is not a Pansu differentiable function.