Question

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.

Answer 1

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 $x\ge 0$.

Answer 2

This is pretty similar to what fedja is doing... Let $\sum_n a_n x^{np}$ be the (possibly nonconvergent) Taylor series of $f(x)$ at $x = 0$. By Borel's theorem, let $g(x)$ be a smooth function on a neighborhood of $x = 0$ whose Taylor expansion is $\sum_n a_n x^n$. Then since $f(x^{1 \over p}) = f(x^{1 \over p}) - g(x) + g(x)$, it suffices to prove that $f(x^{1 \over p}) - g(x)$ is smooth. Writing $h(x) = f(x) - g(x^p)$, we want to show $h(x^{1 \over p})$ is smooth. $h(x)$ has the advantage that its derivatives are all zero at $x = 0$.

By the chain rule, as $x$ goes to zero, any derivative of $h(x^{1 \over p})$ goes to zero faster than any power of $x$ since the same is true for $h(x)$. Thus successively taking difference quotients at $x = 0$, one gets that the $k$th derivative of $h(x^{1 \over p})$ at $x = 0$ is zero for all $k$. Hence $h(x^{1 \over p})$ is smooth.