When is $x\mapsto |x|^{s-1}x$ a diffeomorphism?

Question

Consider the function $f:B^n\rightarrow B^n$ from the disk to itself $$f(x)=\vert x\vert^{s-1}x$$ where $s>0$ and we are considering the euclidean norm (we define the function to be $0$ in the origin if $s<1$). This function defines a homemorphism from the disk to itself. My question is why if $f$ is a diffeomorphism, then $s=1$? This is part of an exercise in Lee's Introduction to Smooth Manifolds chapter 1.

Answer 1

Here's a quantitative way. Every differentiable function on $B^n$ is Lipschitz continuous: that is, there exists $L$ such that $$|f(a)-f(b)|\le L|a-b|,\quad a,b\in B^n \tag1$$ This follows from the Mean Value Theorem.

If $f:B^n\to B^n$ is a diffeomorphism, then its inverse $f^{-1}$ is also Lipschitz, for the same reason. Hence, such $f$ is bi-Lipschitz, meaning there is $L$ such that $$L^{-1}|a-b|\le |f(a)-f(b)|\le L|a-b|,\quad a,b\in B^n \tag1$$

It remains to observe that $|f(x)-f(0)|/|x| = |x|^{s-1}$.