# $a_1t^{\alpha_2}\leq g(t)\leq a_2t^{\alpha_1},\ \forall\ t\in [0,1]$ implies $b_1t^k\leq g(t)\leq b_2 t^k$?

Suppose that there exist constants $a_1,a_2>0$ and $\alpha_1,\alpha_2>1$ with $\alpha_1<\alpha_2$ such that $$a_1t^{\alpha_2}\leq g(t)\leq a_2t^{\alpha_1},\ \forall\ t\in [0,1]$$

where $g:[0,1]\to\mathbb{R}$ is a differentiable function. Is it possible to find constants $b_1,b_2,k>0$ such that $$b_1t^k\leq g(t)\leq b_2 t^k$$

for $t$ small?

Thanks

No. Try $g(t)=t^{3+\sin(\log t)}$ for $t\gt0$, $g(0)=0$.
• (+1), nice example. Do you think that there is a counter example, if I ask the additional hypothesis: $g$ is convex and strictly increasing? Thanks you. – Tomás Apr 21 '13 at 19:52