# On the difference of two Lipschitz functions defined on a Carnot group

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!) $G$ and two functions $f,g: U \subset G \rightarrow G$, where $U$ is an open set in $G$. (We may consider instead two Carnot groups $G$ and $H$ (both non commutative) and two functions $f,g: U \subset G \rightarrow H$.)

Denote by $h$ the difference of these functions: for any $x \in U$ $h(x) = f(x) (g(x))^{-1}$  (here the group operations  and inverses are denoted multiplicatively, thus if $G = \mathbb{R}^{n}$ then $h(x) = f(x) - g(x)$; but I shall suppose further that we work only in groups which are NOT commutative).

1.  Suppose $f$ and $g$ 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 $h$ Lipschitz?

NO! Indeed, consider the Lipschitz functions $f(x) = x$, the identity function,  and $g(x) = u$ a constant function, with $u$ not in the center of $G$. Then $h$ is a right translation, notoriously NOT Lipschitz with respect to a CC left invariant distance.

2. Suppose instead that $f$ and $g$ are everywhere Pansu differentiable and let us compute the Pansu “finite difference”:

$(D_{\varepsilon} h )(x,u) = \delta_{\varepsilon^{-1}} ( h(x)^{-1} h(x \delta_{\varepsilon} u) )$

We get that $(D_{\varepsilon} h )(x,u)$ is the product w.r.t. the group operation of two terms: the first is the conjugation of the finite difference $(D_{\varepsilon} f )(x,u)$  by $\delta_{\varepsilon^{-1}} ( g(x) )$ and the second term is the finite difference   $(D_{\varepsilon} g^{-1} )(x,u)$.  (Here  $Inn(u)(v) = u v u^{-1}$ is the conjugation of $v$ by $u$ in the group $G$.)

Due to the non commutativity of the group operation, there should be some miracle in order for the finite difference of $h$ to converge, as $\varepsilon$ 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.