Factorisation of smooth function Suppose I have a smooth $C^\infty$ function $f:\Omega \to \mathbb{R}$ on a convex bounded open domain $\Omega \subset \mathbb{R}^n$ with smooth boundary. Suppose $f = 0$ on the boundary. At any point on the boundary, I can use a change of basis to an orthonormal system so that $x_1$ is the variable in the direction of the normal of the boundary.
Is there some way to show that locally near the boundary, I can factor $f$ as
$$f(x) = x_1g(x)$$ for some smooth function $g$?
I can use the fundamental theorem (of line integrals) to get this in the case of the halfspace (not bounded).
Trying to understand a paper for which I need the above factorisation. They use Malgrange's theorem in a way I do not understand, and I would rather avoid such high-powered theory. However, I would also appreciate a method using Malgrange's theorem to explain this.
 A: You can't do this with a linear change of basis. But you can do it with a smooth change of coordinates.
The fact that $\Omega$ has a smooth boundary means that for each $p\in \partial \Omega$, we can choose smooth coordinates $(u_1,\dots,u_n)$ in a neighborhood $U$ of $p$ such that $U\cap \overline\Omega = \{q\in U: u^1(q)\ge 0\}$. In these coordinates, $f$ is a smooth function that vanishes when $u_1=0$, so (one form of) Taylor's theorem shows that there is a smooth function $g\colon U\to \mathbb R$ such that $$f(u) = u_1 g(u).$$
The version of Taylor's theorem we need is really just a dressed-up version of the fundamental theorem of calculus. The fact that $f$ vanishes when $u_1=0$ implies
\begin{align*}
f(u_1,u_2,\dots,u_n) &= \int_0^1\frac{\partial}{\partial t} \big(f(tu_1,u_2,\dots,u_n)\big)\,dt\\
&= u^1\int_0^1\frac{\partial f}{\partial u_1} (tu_1,u_2,\dots,u_n)\,dt.
\end{align*}
We can define $g(u)$ to be the integral expression in the last line; because the integrand is smooth in all variables, differentiation under the integral sign shows that $g$ is a smooth function of $u$.
