Let $f_t:U_t\rightarrow V_t$ be an analytic family of homeomorphisms, where $0\leq t\leq 1$, and $U_t, V_t\subset \mathbb{R}^n$ open and bounded domains. Moreover, there exist $c,C>0$ such that for all x,y,t:
$$c\cdot \vert x-y \vert\leq \vert f_t(x)-f_t(y) \vert\leq C\cdot \vert x-y \vert.$$
I want to show that all derivatives $\frac{d^n}{dt^n} f _{\vert t=t_0}:U_{t_0}\rightarrow \mathbb{R}^n$ are Lipschitz continuous for all $n\in\mathbb{N}$.
Unfortunately, I do not know how to prove this claim. Since $f$ is analytic, there is an interval $I_{\epsilon}:=(t_0-\epsilon, t_0+\epsilon)$ such that for all $t\in I_{\epsilon}$:
$$f_t(x)=\sum_{n=0}^{\infty} \frac{d^n}{dt^n} f _{\vert t=t_0}(x)\cdot (t-t_0)^n.$$
I think this might help.But I do not know exactly what to do next. If you know how to prove the result, please help me.
Best regards
Best Answer
This is false in general. Let $U_t=V_t=\mathbb{R}^2$ for all $t$. For $0\le t\le 1$ define $f_t$ by $$ f_t(x_1, x_2) = (x_1, x_2 + \sin (tx_1)) $$ This is an analytic function of $x_1, x_2, t$. It is Lipschitz continuous with respect to $(x_1, x_2)$ with Lipschitz constant at most $3$. The same holds for its inverse, which is $$ f_t^{-1}(x_1, x_2) = (x_1, x_2 - \sin (tx_1)) $$ However, $$ \frac{d}{dt}f_t(x_1, x_2) = (x_1, x_2 + x_1\cos (tx_1)) $$ is not Lipschitz continuous, as its derivative with respect to $x_1$ is unbounded.