Let $f : \mathbb{R}^n\to \mathbb{R}^n$ be given by the equation $f(x) = ||x||^2·x$. Show that $f$ is of class $C^{\infty}$ and that $f$

Question

Let $f : \mathbb{R}^n\to \mathbb{R}^n$ be given by the equation $f(x) = ||x||^2·x$. Show that $f$ is of class $C^{\infty}$ and that $f$ carries the unit ball $B(0, 1)$ onto itself in a one-to-one fashion. Show, however, that the inverse function is not differentiable at $0$.

I know there are answers to many of the questions of the exercise here:Function $f(\textbf{x})=\|\textbf{x}\|^2\cdot \textbf{x}$ is of class $C^\infty$ , How to differentiate the function $f(\mathbf x) = \|\mathbf x\|^2 \mathbf x$? , To show, that the inverse function is not differentiable at $\mathbf 0$.

But I have doubts in showing that $f$ carries the unit ball $B(0, 1)$ onto itself in a one-to-one fashion, how could I do this? I know I do $||x||<1$ then $||f(x)||<||x||^3<1$, but I do not know how to prove that $f$ is one-to-one in $B(0,1)$. Could someone help me please? Thank you very much.

Answer 1

• $f$ maps $B(0,1)$ into $B(0,1)$: $\|f(x)\| = \| \|x\|^2 x \| ≤ 1$.

• $f$ is onto: If $y \in B(0,1)$, then note $$f( c y) = c^3 \|y\|^2 y$$ so if $c^3 = \frac1{\|y\|^2}$, $\|cy\| = \|y\|^{-2/3} \|y\| = \|y\|^{1/3} ≤ 1$ so $cy\in B(0,1)$, and $f(cy) = y$.

• $f$ is one-to-one: suppose $\|x\|^2x = \|y\|^2y$. Then $x,y$ are collinear; if $x=0$ there is nothing to show; if $x≠0$, then $y≠0$, and we can write $y=\lambda x$, giving $$ x = \lambda^3 x $$ which implies $\lambda =1$ and hence $x=y$.

Answer 2

Hints:

1. Observe that if $x$ and $y$ are not multiples of each other, then $f(x)\not=f(y)$ since $f(x)$ is a scaled copy of $x$ and $f(y)$ is a scaled copy of $y$.

2. Consider $f(tx)=t^3\|x\|^2x$ for $t>0$. Since the cube function is injective, so is $f$. Moreover, since the cube function is surjective, every possible positive scaled copy of $x$ occurs as such an image.

3. Observe that $\|f(x)\|=\|x\|^3$. Therefore, $\|f(x)\|\leq 1$ iff $\|x\|\leq 1$.