Does $g$ behave like $t^k$ near the origin? I asked a similar question early ago, however, my function satisfies more hypothesis, so I am gonna ask it again, please be patient with me. Suppose $G:[0,\infty]\to [0,\infty]$ defined by $G(t)=\int_0^t g(s)ds$ is a (Young or N) function with $g:[0,\infty]\to [0,\infty]$ and $g\in C^1(0,\infty)$ (for na definition of Young function see here page 13). Assume that there exist constants $b_1,b_2>1$, $b_1<b_2$ such that (condition $\Delta_2$) $$b_1\leq \frac{tg(t)}{G(t)}\leq b_2$$ 
Then, we can conclude that 
$$G(1) t^{b_2}\leq G(t)\leq G(1) t^{b_1},\ \forall\ t\in [0,1]$$
Is it possible to find constants $c_1,c_2,k>0$ such that $$c_1 t^k\leq G(t)\leq c_2 t^k$$
for small $t$?
Update 1: I added some hypothesis that fits better in what I want. Also I would like to point out, that this problem come from the paper of Montenegro:  Uniqueness of nonnegative solutions of quasilinear elliptic equations. Dynam. Systems Appl. 6 (1997), no. 1, 125–137. The statement is in inequality $(22)$ of the paper.
Update 2: I have asked this question. I think that in our case here, $k=k_1=k_2$.
Thanks
 A: I assume you are asking "whether $G(t)$ always behaves like $t^k$" (as in your question) instead of "whether $g(t)$ always behaves like $t^k$" (as in the title) for small $t$.
The answer is No.
The question is equivalent to given a monotonic increasing $C^2$ function $G : [0,\infty] \to [0,\infty]$ satisfying:
$$G(0) = 0 \quad\quad\text{and}\quad\quad 1 < b_1 \le \frac{d \log G(t)}{d \log t} \le b_2 < \infty\text{ for }t \in (0,\infty)$$
Is it always true that there exists some $k_G > 0$ such that:
$$k_G = \lim_{t\to 0+} \frac{\log G(t)}{\log t} \quad\text{ and }\quad \limsup_{t\to 0+} | \log G(t) - k_G \log t| < \infty$$
Define $G(t)$ by:
$$G(t) = \begin{cases}t^3 e^{f(\log t)}, & t > 0\\0, & t = 0\end{cases}$$
where $f(s) = \sqrt[4]{s^2+1}$. It is not hard show for any $s \in (-\infty,\infty)$,
$$|f'(s)| = |\frac{s}{2(s^2+1)^{\frac34}}| \le \frac{1}{\sqrt[4]{108}} \sim 0.3102 $$
This implies for $t \in (0,\infty)$,
$$\frac{d \log G(t)}{d \log t} = 3 + f'(\log t) \implies 2.6 < \frac{d \log G(t)}{d \log t} < 3.4 $$
Furthermore, the $t^3$ factor in $G(t)$ make it falls to zero fast enough as $t$ approaches $0$ and turn $G(t)$ into a $C^2$ function over $[0,\infty)$. More precisely, this mean:
$$\begin{align}
\lim_{t\to0+} G'(t)  &= 0 = G'(0)  \stackrel{def}{=} \lim_{h\to 0+}\frac{G(h)}{h}\\
\lim_{t\to0+} G''(t) &= 0 = G''(0) \stackrel{def}{=} \lim_{h\to 0+}\frac{G'(h)}{h}
\end{align}$$ 
Notice 
$$\lim_{t\to0+}\frac{\log G(t)}{\log t} = \lim_{s\to-\infty}\frac{3s+f(s)}{s} = 3$$
$k_G$ for this particular $G$ is $3$. However,
$$\limsup_{t\to0+}|\log G(t) - k_G \log t| = \limsup_{s\to-\infty}|f(s)| \ge \limsup_{s\to-\infty}\sqrt{|s|} = \infty$$ 
This means in general, $G(t)$ need not behave like any $t^k$ for small $t$.
A: You can think of your statement as find the constants $c_{1}, c{2}, k>0$ such that $c_{1}t^k \leq a_{1}t^{b_{2}} \leq a_{2}t^{b_{1}} \leq c_{2}t^k$ $\forall t\in[0,1]$
$c_{1} \leq a_{1}t^{b_{2}-k} \leq a_{2}t^{b_{1}-k} \leq c_{2}$  $\forall t\in[0,1]$
Now if $k < b_{2} \Rightarrow \lim_{t\to0}a_{1}t^{b_{2}-k}=0 \Rightarrow c_{1}=0$ so impossible.
Then $k \geq b_{2}$.
Now if $k > b_{1} \Rightarrow \lim_{t\to0}a_{2}t^{b_{1}-k}=\infty \Rightarrow c_{2}=\infty$ so again impossible.
Then $k \leq b_{1}$.
Since $b_{2} > b_{1}$ no such $k$ exists.
And to give you a hint how to imagine a counterexample, think if $g(t)$ oscilates between the 2 boundry functions without braking its convexity.
