Uniform Lipschitz constants of family of bilipschitz homeomorphisms

Question

Let's asume that we are given a family of linear maps $\lbrace{A_i,B_i \rbrace}_{i \in I, A_i,B_i \in Gl_n(\mathbb{Z})}$ and a bilipschitz homeomorphism $F: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$such that for each $i$ there exists a positive constant $C_i$ and a $(K,\lambda)$ quasi isometry ($K,\lambda$ are fixed) $G_i$ such that $sup_{x \in \mathbb{R}^{n}}( ||A_i \circ F \circ B_i(x) - G_i(x)|| < C_i$ (which means that for all $(x,y) \in \mathbb{R}^2 $ we have that $||A_i \circ F \circ B_i(x) - A_i \circ F \circ B_i(y)|| \le C_i +K ||x-y|| $). Does it mean that family of bilipschitz homeomorphisms $\lbrace{ A_i \circ F \circ B_i \rbrace}_{i \in I}$ has uniform lipschitz constant - lispchitz constants in the sense of real analysis ? It's certainly true for $F$ affine or if either $\lbrace{A_i \rbrace}_{i \in I}$ or $\lbrace{B_i \rbrace}_{i \in I}$ have uniform quasi-isometry constants. The weaker question is - can we choose two constants $(K',C')$ in such a way that all functions $\lbrace{ A_i \circ F \circ B_i \rbrace}_{i \in I}$ are $(K',C')$ coarse (large scale) Lipschitz?

Answer 1

Let $f: R^2\to R^2$ be a diffeomorphism which equals the $90$ degree rotation on the unit disk $B(0, 1)$ and is the identity map outside of the radius 2 disk $B(0, 2)$. Let $A\in SL(2,Z)$ be a symmetric matrix with distinct positive eigenvalues $$ \lambda> \lambda^{-1}>0, $$ e.g. $$ A=\left[\begin{array}{cc} 2&1\\ 1&1\\ \end{array}\right]. $$ Let $E_\lambda, E_{\lambda^{-1}}$ be the eigenspaces of $A$. I will regard $A$ as a linear map $R^2\to R^2$. Consider the sequence $f_n=A^n \circ f \circ A^{-n}$. Then each $f_n$ is the identity outside of a sufficiently large disk in $R^2$ (namely, the disk $B(0, 2\lambda^n)$). Thus, for each $n$ there exists a constant $C_n<\infty$ such that $$ || f_n -id||\le C_n. $$

We also have $$ ||Df_n(0)|| = \lambda^{2n}, $$ which diverges to infinity as $n\to\infty$. Hence, the sequence $(f_n)$ is not uniformly Lipschitz.
Furthermore, for each $n$ there exist $x_n\in B(0,1)\cap E_{\lambda^{-1}}$ such that $||A^{-n}x_n||=1$. (Namely, take a unit vector $u\in E_{\lambda^{-1}}$ and set $x_n:= \lambda^{-n}u$.) Then $$ ||f_n(x_n)||=\lambda^n, f_n(0)=0. $$ Hence, the sequence $(f_n)$ is not uniformly coarse Lipschitz, since $$ \lim_{n\to\infty} ||f_n(x_n) - f_n(0)||=\infty, $$ while $||x_n - 0||\le 1$.