If $\varphi$ is analytic and $t^k + \sum_{i=1}^{k-1}a_i(\varphi(t,x),x)t^i = b(\varphi(t,x),x)$ , then $a_i$ and $b$ are analytic functions.

Question

Let $\varphi: \mathbb{R}\times \mathbb{R}^n \to \mathbb{R}$ be a real analytic function, such that

$$\varphi(0,0)=0,\ \frac{\partial \varphi}{\partial t}(0,0)=0,\ \ldots\ ,\frac{\partial^{k-1} \varphi}{\partial t^{k-1}}(0,0)=0\ \text{and}\ \frac{\partial^k \varphi}{\partial t^k}(0,0) \neq 0. $$

Now, suppose that there exist smooth functions $a_1,...,a_k,b: \mathbb{R}\times U\subset \mathbb{R}\times \mathbb{R}^n\to \mathbb{R}$ such that $$t^k + \sum_{i=1}^{k-1} t^i \cdot a_i(\varphi(t,x),x) = b(\varphi(t,x),x),\ \forall\ t\in \mathbb{R} \ \text{and}\ x\in U, $$ where $U$ is an open neighboorhood of $0$ in $\mathbb{R}^n$.

Question: Is it possible to guarantee that the functions $a_1,...,a_k,b$ are real analytic functions?

Answer 1

It is enough to assume that $a_j = 0$ for $j >1$ to give a counterexample. Then choosing $a_1$ to be non-analytic, it is "unlikely" that $b$ be analytic.

Take for example $\varphi(t,x) = t^k$, and $$ a_1(t,x) = \begin{cases} e^{-\frac{1}{t^2}}, & t > 0\\ 0, & t \leq 0. \end{cases} $$

Defining $b(t,x) = t-t^\frac{1}{k}a_1(t,x)$, one sees that $a_1$ and $b$ are smooth, but non-analytic.