$\varphi_a(x)=a+(1-|a|^2)\frac{x+a}{|x+a|^2}$ is a diffeomorphism from unit ball to unit ball in $\mathbb R^n(|a|>1)$.

Question

I want to prove that $\varphi_a(x)=a+(1-|a|^2)\frac{x+a}{|x+a|^2}$ is a diffeomorphism from unit ball to unit ball in $\mathbb R^2$($a\in \mathbb R^n,|a|>1$, where $||$ is the usual Euclidean norm).

I don't know how to show that $|\varphi_a(x)|<1$ for $|x|<1$ and that $\varphi_a$ is surjective onto the unit ball(is it surjective?). I have tried using a lot of inequalities but they are all fruitless.\

Note that in $\mathbb C$ this is a linear fractional transformation that sends the unit disk to itself.

Separate expression with terms of $x$ and $a$ and take the norm $\phi$. — Nosrati, Sep 27 '17 at 16:22
Do you mean $\varphi_a(x)=a+(1-|a|^2)(\frac{x}{|x+a|^2}+\frac{a}{|x+a|^2})$? I don't see how to continue... — No One, Sep 27 '17 at 16:50
BTW, for the linear fractional transformation case in $\Bbb C$, you need $|a|<1$ and the map is quite different. — Ted Shifrin, Sep 27 '17 at 20:50

score 0 · Answer 1 · answered Sep 27 '17 at 19:36

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HINT: I would recommend that you start by doing this for $n=1$. You should be able to show that $|\phi_a(x)|\le 1$ if and only if $(1-x^2)(a^2-1)\ge 0$. Next do the same argument for $n>1$ by using dot products.

answered Sep 27 '17 at 19:36

Ted Shifrin

115,160

$\varphi_a(x)=a+(1-|a|^2)\frac{x+a}{|x+a|^2}$ is a diffeomorphism from unit ball to unit ball in $\mathbb R^n(|a|>1)$.

1 Answers1