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(This question was previously posted on MSE and I decided to post it here too.)

Let $f:\mathbb{R}\times \mathbb{R}^n \to \mathbb{R}$ be a smooth function, such that $$f(0,0)=0,\ \frac{\partial f}{\partial t} (0,0) = 0,\ldots, \frac{\partial^{k-1} f}{\partial t^{k-1}} (0,0) = 0,\ \frac{\partial^{k} f}{\partial t^{k}} (0,0) \neq 0,$$ then the Malgrange preparation theorem, states that there exists smooth functions $c,a_0,...,a_{k-1}:\mathbb{R^n}\to\mathbb{R}$, such that near the origin $${\displaystyle f(t,x)=c(t,x)\left(t^{k}+a_{k-1}(x)t^{k-1}+\cdots +a_{0}(x)\right)},$$ and $c(0,0)\neq 0$.

I'm reading the paper "S. M Vishik - Vector Fields Near the Boundary of a Manifold", and on page $17$, the author says that the following theorem is a corollary of the Magrange preparation theorem.

Corollary: Let $\varphi:\mathbb{R}\times \mathbb{R}^n\to \mathbb{R}$ a smooth function such that $$\varphi(0,0)=0,\ \frac{\partial \varphi}{\partial t} (0,0) = 0,\ldots, \frac{\partial^{k-1} \varphi}{\partial t^{k-1}} (0,0) = 0,\ \frac{\partial^{k} \varphi}{\partial t^{k}} (0,0) \neq 0,$$ then there exists smooth functions $b,a_1,...,a_k:\mathbb{R}^n\times\mathbb{R} \to \mathbb{R}$ such that $$t^{k+1}+\sum_{i=1}^{k} a_i(x,\varphi(t,x))\cdot t^i = b(x,\varphi(t,x)), $$ in a neighborhood of the origin.

Does anyone know this corollary and could you give me information on how to demonstrate it or let me know where I can find its demonstration?


Just some commentaries

This is the part that the author claims such corollary, $M$ is a manifold, $Q$ is a codimension 1 submanifold of $M$ and "shoots" is a fancy name for "germs". enter image description here

And the numbered reference:

enter image description here however, I was not able to find this paper.

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Part I, II, III, and IV of the paper of Malgrange (in French) can be found at numdam.

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  • $\begingroup$ Thank you for your references, I'm reading the papers right now. I don't speak French, but as Doctor Strange says "I'm fluent in google translator". If I find a proof of my problem I will accept your answer. $\endgroup$ – Matheus Manzatto Feb 23 at 16:53

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