Existence of a Diffeomorphism with $f(\phi(x)) = x_1$

Question

Let $f \in C^1(\mathbb{R}^n)$ a function with $f(0) = 0$ and $\partial_1f(0) \neq 0$

Show that neighbourhoods $U, V$ of $x = 0$ as well as a diffeomorphism $\phi: U \to V$ exist so that $f(\phi(x)) = x_1$ for all $ x = (x_1,...,x_n) \in U$.

The existence can be proven be the inverse functon theorem, but how do I show that $f(\phi(x)) = x_1$ holds?

Has Anyone an idea?

Kind Regards

Answer 1

Consider the function $g : \mathbb R^n \to \mathbb R^n$ defined by $$ g(x_1, \dots, x_n) = \begin{bmatrix} f(x_1, \dots, x_n) \\ x_2 \\ x_3 \\ \vdots \\ x_n\end{bmatrix}.$$ I suggest you compute the Jacobian matrix for $g$, and verify that its determinant is non-vanishing at $(x_1, \dots, x_n) = (0,\dots, 0)$.

Next, I suggest you apply the inverse function theorem to the function $g$ to show that there exists a diffeomorphism $\phi : U \to \phi(U) \subset \mathbb R^n$, where $U$ is a sufficiently small open neighbourhood of $(0, \dots, 0)$, such that $$ g (\phi(y_1, \dots, y_n)) = (y_1, \dots, y_n).$$

You can verify that $\phi$ obeys all the required properties.