Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Suppose $g:[0,\infty)\rightarrow [0,\infty)$ is a cotinuous increasing function with $g(0)=0$. Is it possible to find two constants $C>0$ and $q>0$ such that $$g(s)\geq Cs^q,\ \forall\ s\in [0,\delta)$$

where $\delta>0$ is an small number.

Thank you.

Edit: Sorry, I forgot one hypothesis.

share|cite|improve this question
On which ultra common functions did you test your conjecture? – Did Mar 7 '13 at 13:55
@Did, what do you have in mind? – Tomás Mar 7 '13 at 14:03
Nothing to add to my first comment: surely you tested your conjecture on some functions, which ones? – Did Mar 7 '13 at 15:15
It was clear for me that no polynomious could be a counter example, and that the function must go to zero more fast than any polynomious. THen I got stuck – Tomás Mar 7 '13 at 15:19
up vote 1 down vote accepted

This is still not true. Recall the classical example of a smooth non-zero function with all Taylor coefficients vanishing at 0 (you can construct it with $e^{-1/x}$ for $x>0$ and 0 otherwise).

The Taylor Young theorem says that you won't be able to find such constants for that function.

share|cite|improve this answer
addendum : note that your claim stand if you assume the function to be analytical near 0 – Glougloubarbaki Mar 7 '13 at 13:57
Thank you. That's the example I was looking for. – Tomás Mar 7 '13 at 14:05

No,. this need not be true. One simple way it could fail is if $g$ is bounded (since $Cs^q$ is unbounded). So for example $g(s)=\arctan(s)$ is a counterexample.

share|cite|improve this answer
Sorry, I forgot one hypothesis, I edited it. – Tomás Mar 7 '13 at 13:41

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.