Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a smooth function let $p \in \mathbb{N}$, $p \geq 2$.
Assume that $f^{(k)}(0)=0$ for all $k \notin p \mathbb{N}$. Is it true that then $g(x)=f(\sqrt[p] x)$ for $x\geq 0$ is smooth?


Thanks.



A smoothness of $f(\sqrt[p] x)$
If $P\_{n}$ is the $n$-th order Taylor polynomial of $f$ at $0$, then $g(x)=f(x)-P\_{n}(x)=o(x^{n})$. Since the statement is obvious for polynomials, we are down to the case when $f$ has all derivatives up to some arbitrarily high order vanishing at $0$, which means that all derivatives up to some arbitrarily high order are o-small of some arbitrarily high power of $x$ at $0$. Now just differentiate $f(x^{q})$ and notice that any fixed order derivative is bounded by $x$ to some fixed negative power times the sum of a few derivatives of $f$ at $x^q$, so it tends to $0$ as $x\to 0$. At last, if $g$ is smooth for $x>0$ and the first $k$ derivatives have limits at $0$, then $g\in C^k$ for $x\ge 0$.


