MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
Did you see ? – Davide Giraudo Dec 17 '11 at 19:26
Thanks, but Iwould like to have at least a references. – arc Dec 18 '11 at 8:11
up vote 2 down vote accepted

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 $x\ge 0$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.