# Show that the perturbation of identity satisfies certain continuity and Lipschitz properties

Let $$dinmathbb N$$, $$uin C^{0,:1}(mathbb R^d,mathbb R^d)$$ 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 $$T_t(x):=x+tu(x);;;text{for }xinmathbb R^d$$

satisfies

1. $$T_t$$ is a bijection for all $$tin[0,tau]$$;
2. $$T,T^{-1}in C^{0,:1}(mathbb R^d,C^0([0,tau],mathbb R^d))$$.

It’s easy to see that $$left|T(x)-T(y)right|_{C^0([0,:tau],:mathbb R^d)}le(1+tau c)left|x-yright|;;;text{for all }x,yinmathbb R^d.$$ Assuming $$taule(2c)^{-1}$$ (I’ve read at other places that we need to assume $$taulemin(1,(2c)^{-1})$$ but I don’t understand why), we can show that $$T_t$$ is bijective for all $$tin[0,tau]$$:

In fact, let $$tin[0,tau]$$, $$yinmathbb R^d$$ and $$f(x):=y-(T_t(x)-x);;;text{for }xinmathbb R^d.$$ Then $$begin{equation}begin{split}left|f(x_1)-f(x_2)right|&=left|(x_1-x_2)-(T_t(x_1)-T_t(x_2))right|\&le tcleft|x_1-x_2right|end{split}tag2end{equation}$$ for all $$x_1,x_2inmathbb R^d.$$ If $$taule(2c)^{-1}$$, then $$tcle2^{-1}<1$$ and hence $$f$$ is a strict contraction so that, by the Banach fixed-point theorem, there is a unique $$xinmathbb R^d$$ with $$f(x)=x$$, which is equivalent to $$y=T_t(x)$$.

From the reverse triangle inequality and $$(2)$$ we see that $$left|x_1-x_2right|le2left|T_t(x_1)-T_t(x_2)right|tag3$$ for all $$tin[0,tau]$$ and $$x_1,x_2inmathbb R^d$$ and hence $$left|T_t^{-1}(x)-T_t^{-1}(y)right|le2left|T_t(T_t^{-1}(x))-T_t(T_t^{-1}(y))right|=2left|x-yright|tag4$$ for all $$tin[0,tau]$$ and $$x,yinmathbb R^d$$.

How can we show the remaining claims? And, if $$uin C^1(mathbb R^d,mathbb R^d)$$, under which condition can we show that $${rm D}T_t=operatorname{id}_{mathbb R^d}+t{rm D}u$$ has a nonnegative determinant?

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