Is it true that the differentiable function $f:\mathbb R\to\mathbb R$ such that $f(x)=o(x^n)$ for every positive $n$ is infinitely differentiable at $0$?

Given the differentiability constraint, it seems to rule out all the pathological examples I'm aware of, yet, I don't know how to prove it.

  • $\begingroup$ Little o notation only means that $f$ is eventually dominated by $x^n$. We can still have that $f$ is erratic at $0$ as long as $f$ eventually becomes bounded. $\endgroup$ – Robert Wolfe Mar 14 '14 at 2:16
  • $\begingroup$ @Bryan I understood that $o(x^n)$ refers to $x\to 0$; but the OP should have been more clear at this point. $\endgroup$ – user127096 Mar 14 '14 at 2:17
  • $\begingroup$ Ok thanks. That would be a more interesting problem. $\endgroup$ – Robert Wolfe Mar 14 '14 at 2:18

No. A counterexample: $$f(x)=e^{-1/x^2} \sin( e^{1/x^2})$$ extended by $f(0)=0$. It is once differentiable everywhere, including $0$. It has the required decay at $0$. But
$$f'(x) = \frac{2}{x^3}e^{-1/x^2}\sin(e^{1/x^2}) - \frac{2}{x^3}\cos (e^{1/x^2})$$ is not differentiable, or even continuous, at $0$.

Perhaps related: Does $f(x)=o(x^n) \forall n$ imply $f^{(n)}(0)=0$?


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