Show $f(x)=||x||^2\cdot x$ carries unit ball injectively onto its image

Question

For the problem below, I want to prove that the unit ball is carried onto itself in a $1$-$1$ fashion. I gather from the unit ball condtiion that each $x_i$ in the domain is restricted by $x_i<1$, $i=1,...,n$. Now I need to prove that the map $||x^2||\cdot x$, on the restricted domain, is injective. I've seen similar problems, such as $f(x,y)=(x^2-y^2,2xy)$ where I could write $(x,y)$ as a complex number $z$ and get that the image was $z^2$, but finding this representation of $f$ took some work.

If the trick to proving that this function is injective is by writing the expresion in complex form, then could you assist me in writing this expression in such form? If that is not the direction to go in to solve this problem then I'd appreciate insight into the right direction.

Thanks!

Answer 1

It is clear that $f(x)=0$ if and only if $x=0$. Now, suppos that $x,y\neq0$ and that $f(x)=f(y)$. Then $\|x\|^2x=\|y\|^2y$. But\begin{align}\|x\|^2x=\|y\|^2y\implies&\bigl\|\|x\|^2x\bigr\|=\bigl\|\|y\|^2y\bigr\|\\\iff&\|x\|^3=\|y\|^3\\\iff&\|x\|=\|y\|\end{align}and therefore, since $\|x\|=\|y\|\neq0$, it follows from the equality $\|x\|^2x=\|y\|^2y$ that $x=y$.

Besides, $f'(0)$ is the null function. Therefore, by the chain rule, $f^{-1}$ cannot be differentiable at $0(=f(0))$.

Answer 2

Hint If $f(x)=f(y)$ then $$\| f(x) \| = \|f(y)\|$$

What is the norm of $f(x)$?

Answer 3

Look for an inverse function of the form $$f^{-1}(x)=||x||^kx$$ finding $k\in \mathbb R$ such that it is indeed the inverse, that is, such that $$f\big(f^{-1}(x)\big)=f^{-1}\big(f(x)\big)=x.$$

Do not forget to complete the definition to the whole $B(0,1)$ if $k$ turns out to be negative.