Let $f$ be an entire function. If there exist a positive number $\rho$ and constants $A,B>0$ such that
$$
|f(z)|\leq Ae^{B|z|^{\rho}}\ \text{for all}\ z\in{\Bbb C}
$$
then we say that $f$ has an order of growth $\leq\rho$. We define the order of growth of $f$ as
$$
\rho_f=\inf\rho
$$
where the infimum is over all $\rho>0$ such that $f$ has an order of growth $\leq\rho$.
Here is my question:
Using the definition above, how can I find the order of $f(z)=e^z-1$?
It's not hard to show that $$
|e^z-1|\leq e^{|z|}
$$
and thus $f$ has an order of growth $\leq 1$. I guess the order should be $1$. Then for any $\varepsilon>0$, and $A,B>0$, I need a $z\in{\Bbb C}$ such that
$$
|e^z-1|> Ae^{B|z|^{1-\varepsilon}}.
$$
How can I go on?
Best Answer
As you said it has order of growth $\leq 1$.
Now if it had order of growth $\leq \rho<1$, we would have $$ |e^z-1|\leq Ae^{B|z|^\rho} \quad\forall z\in\mathbb{C}\qquad\Rightarrow\qquad |e^x-1|\leq Ae^{Bx^\rho} \quad\forall x>0. $$ Hence $$ e^x\leq |e^x-1|+1\leq(A+1)e^{Bx^\rho}\quad\Rightarrow\quad e^{x-Bx^\rho}\leq A+1$$ $$ \Rightarrow\quad x\left(1 -Bx^{\rho-1}\right)= x-Bx^\rho\leq\log(A+1)\quad\forall x>0. $$ Now the lhs clearly tends to $+\infty$ as $x$ tends to $+\infty$. Contradiction.