# Increasing Function bounded below by a polynomious

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.

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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

## 2 Answers

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.

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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.

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Sorry, I forgot one hypothesis, I edited it. –  Tomás Mar 7 '13 at 13:41