Let $f: \mathbb{R}^{n}\to \mathbb{R}^{n}$ defined by $\Vert x \Vert^{2} x$.
(a) Show that $f$ is of class $C^{\infty}$.
(b) Show that $g=f\mid_{B_{1}(0)}$ is injective and determines the range of $g$.
(c) Prove that $g^{-1}$ (defined on $g(B_{1}(0))$) is not differentiable in $0$.
My attempt.
(a) Write $f = \phi\cdot \psi$ where $\phi(x) = \Vert x \Vert^2$ and $\psi = x$. But $D_{x}\phi(h) = 2\langle x : h \rangle$, that is,$D_{x}\phi(h) \in C^{\infty}$ and obviously, $\psi(x) \in C^{\infty}$. Therefore, $f \in C^{\infty}$
(b) Take $x, y \in B_{1}(0)\setminus\{0\}$ such that $g(x) = g(y)$, that is, $\Vert x \Vert^{2}x = \Vert y \Vert^{2}y$. But $\Vert x^{2} \Vert = k_{1} \leq 1$ and $\Vert y \Vert^{2} = k_{2} \leq 1$, then $\displaystyle x_{i} = \frac{k_{2}}{k_{1}}y_{i}$. Is there a way to show that $\displaystyle \frac{k_{2}}{k_{1}} = 1$ or is there a better way to show injectivity?
Im not sure about the range of $g$. We have $\Vert x \Vert \leq 1$ so, $\Vert \Vert x \Vert^{2} x \Vert = \Vert x \Vert^{3} \leq 1$, then $g(B_{1}(0)) \subset B_{1}(0)$?
I have not tried it since Im not sure about the range of $g$. But I think I should try to show that some of the partial derivatives are not continuous at $0$.
The range is all of $\mathbb R^{n}$. Clearly, $0$ is in the range. If $y \neq 0$ let $x=\|y\|^{-2/3} y$. Then $\|x||^{2}x=y$. Also, if $y \in B_1(0) \setminus \{0\}$ then $x \in B_1(0)$ (and $g(x)=y$) so $g$ maps $B_1(0)$ onto itself. Injectivity of $f$ and $g$: let $\|x\|^{2}x=\|y\|^{2}y$. Take norm on both sides. We get $\|x\|^{3}=\|y\|^{3}$ which implies $\|x\|=\|y\|$. If $x=0$ then so is $y$. Otherise, we can divide by $\|x\|^{3}$ to get $x=y$.