How to show that $f : \mathbb{R}^n → \mathbb{R}^n$, $f(x) = \frac{h(\Vert x \Vert)}{\Vert x \Vert} x$, is a diffeomorphism onto the open unit ball?

Question

Could anyone help me with the following problem?

The problem

Fix $\varepsilon \in (0, 1)$ and choose a smooth function $h$ on $[0,\infty)$ such that $h'(t) > 0$ for all $t ≥ 0$, $h(t) = t$ for $t \in [0, \varepsilon]$, $h(t) = 1 − \frac{1}{\ln t}$ for all $t$ large enough.

(You don’t have to explain why such a function exists.)

Consider the map $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$, $f(a) = \frac{h(\lVert a \rVert)}{\lVert a \rVert} a$, where $\lVert a \rVert = \sqrt{ (a^1)^2 + \cdots + (a^n)^2}$ is the usual Euclidean norm.

Show that $f$ is a diffeomorphism of $\mathbb{R}^n$ onto the open unit ball $B \subset \mathbb{R}^n$.

What I have so far (see edit below)

In order to show that $f$ is a diffeomorphism it is enough to show that

1. $f \in C^{\infty}$,
2. the Jacobian is nowhere zero,
3. $f$ is a bijection.

I've shown that $f$ is a bijection onto $B$.

I've also shown that $f = id$, the identity function, on $\overline{B_{\varepsilon}(0)}$ (i.e. the closed ball with radius $\varepsilon$ centered at the origin).

But what about outside of $\overline{B_{\varepsilon}(0)}$? There we don't know that much about $h$, and so finding the Jacobian is not so easy.

Should I look at the following partial derivatives in order to figure out the Jacobian? $$\frac{\partial f^i}{\partial x^j} = \frac{\partial \frac{h(\lVert x \rVert)}{\lVert x \rVert} x^i}{\partial x^j}$$

Any help is greatly appreciated!

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Edit:

We have that $h^{-1}$ is differentiable (even smooth since $h$ is smooth) and its derivative by the Inverse Function Theorem. We also have that $\Vert f(x) \Vert = h(\Vert x \Vert )$, and $\frac{f(x)}{\Vert f(x)\Vert } = \frac{x}{\Vert x\Vert }$. So $f$ is a bijection, with $f^{-1}(x) = h^{-1}(\Vert x\Vert )\frac{x}{\Vert x\Vert }$.

Now, I was just wondering, since we have that

1. $f$ is a bijection,
2. $f \in C^{\infty}$ (since it is a composition of $C^{\infty}$-functions), and
3. $f^{-1} \in C^{\infty}$ (since it also is a composition of $C^{\infty}$-functions),

does it not follow that $f$ is a diffeomorphism?

Answer 1

That $f$ is $\mathcal C^\infty$ you've already done: on a ball around the origin, it is (=identity), off that ball $\|x\|$ is smooth never vanishing and then $f$ is the conposite of smooth functions. To check $f$ is local diffeo, that is, its derivative at every point $a$ is a linear iso, I suggest the following strategy. Fix a point $a\in\mathbb R^n$, which we can suppose $a\ne0$ (because around the origin $f$ is the identity, hence smooth there). Now think of $\mathbb R^n$ as the sum of the line $L$ through $a$ and the origin and the hyperplane $H$ orthogonal to $a$. Note that $L$ is the tangent space to that line at $a$ and $H$ is the tangent space to the sphere $S:\|x\|=\|a\|$ at $a$. Hence you can describe the derivative $d_af$ of $f$ at $a$ using $d_a(f|S)$ and $d_a(f|L)$. Thus:

(i) For $x\in S$ we have $f(x)=\frac{h(\|a\|)}{\|a\|}x$ which a homothety, hence its own derivative on $H$,

(ii) For $x=ta\in L$, $t>0$, we have $g(t)=f(x)=\frac{h(t\|a\|)}{\|a\|}t\|a\|=th(\|a\|t)$. This one variable function can be easily derived to get $g'(t)=h(\|a\|t)+\|a\|h'(\|a\|t)>0$ by the hypothesis on $f$.

Thus $d_af$ is a homothety on $H$ and another on $L$, all in all a linear iso on $\mathbb R^n$.