Let $d\in\mathbb N$, $u\in C^{0,\:1}(\mathbb R^d,\mathbb R^d)$ and $c:=|u|_{C^{0,\:1}(\mathbb R^d,\:\mathbb R^d)}$ (the semi-norm given by the Lipschitz constant).
I would like to show that there is a $\tau>0$ such that $$T_t(x):=x+tu(x)\;\;\;\text{for }x\in\mathbb R^d$$
satisfies
- $T_t$ is a bijection for all $t\in[0,\tau]$;
- $T,T^{-1}\in C^{0,\:1}(\mathbb R^d,C^0([0,\tau],\mathbb R^d))$.
It's easy to see that $$\left\|T(x)-T(y)\right\|_{C^0([0,\:\tau],\:\mathbb R^d)}\le(1+\tau c)\left\|x-y\right\|\;\;\;\text{for all }x,y\in\mathbb R^d.$$ Assuming $\tau\le(2c)^{-1}$ (I've read at other places that we need to assume $\tau\le\min(1,(2c)^{-1})$ but I don't understand why), we can show that $T_t$ is bijective for all $t\in[0,\tau]$:
In fact, let $t\in[0,\tau]$, $y\in\mathbb R^d$ and $$f(x):=y-(T_t(x)-x)\;\;\;\text{for }x\in\mathbb R^d.$$ Then \begin{equation}\begin{split}\left\|f(x_1)-f(x_2)\right\|&=\left\|(x_1-x_2)-(T_t(x_1)-T_t(x_2))\right\|\\&\le tc\left\|x_1-x_2\right\|\end{split}\tag2\end{equation} for all $x_1,x_2\in\mathbb R^d.$ If $\tau\le(2c)^{-1}$, then $tc\le2^{-1}<1$ and hence $f$ is a strict contraction so that, by the Banach fixed-point theorem, there is a unique $x\in\mathbb R^d$ with $f(x)=x$, which is equivalent to $y=T_t(x)$.
From the reverse triangle inequality and $(2)$ we see that $$\left\|x_1-x_2\right\|\le2\left\|T_t(x_1)-T_t(x_2)\right\|\tag3$$ for all $t\in[0,\tau]$ and $x_1,x_2\in\mathbb R^d$ and hence $$\left\|T_t^{-1}(x)-T_t^{-1}(y)\right\|\le2\left\|T_t(T_t^{-1}(x))-T_t(T_t^{-1}(y))\right\|=2\left\|x-y\right\|\tag4$$ for all $t\in[0,\tau]$ and $x,y\in\mathbb R^d$.
How can we show the remaining claims? And, if $u\in C^1(\mathbb R^d,\mathbb R^d)$, under which condition can we show that ${\rm D}T_t=\operatorname{id}_{\mathbb R^d}+t{\rm D}u$ has a nonnegative determinant?