The definition, used in the book I study, of the order of an entire function is as follows :
An entire function $f$ is of finite order if there is a positive constant $a$ and an $r_{0}$ such that
$$\vert f(z)\vert <\exp \left(\vert z\vert ^{a}\right)$$
for $\vert z\vert >r_{0}$. If $f$ is of finite order then the number
$$\lambda =\inf \ \lbrace a :\vert f(z)\vert <\exp \left( \vert z\vert ^{a}\right) \ \text{for} \ \vert z\vert \ \text{sufficiently large} \rbrace $$ is called the order of $f$.
I’m trying to show that this definition is equivalent to the definition found in another book which is:
An entire function $f$ is of order finite $\leq a$ if there is positive constants $A,B,\alpha$ such that there is a $r_{0}$ with
$$\vert f(z)\vert \leq \exp(A\vert z\vert ^{a} +B)$$
for$\vert z\vert >r_{0}$.
I would like to show that the second definition is equivalent to the first and for that I have tried to develop the exponential of the second definition but I cannot come back directly to my definition. Is it necessary to use a specific theorem or lemma to show this result?
Best Answer
Assume first that $f$ is of finite order $\le a$ according to the second definition, that is, $$\lvert f(z)\rvert\le\exp(A\lvert z\rvert^\alpha+B)$$ for all $z$. Let $\varepsilon>0$. Then for $\lvert z\rvert\ge r_0$, there holds $$\lvert z\rvert^{a+\varepsilon}\ge r^\varepsilon\lvert z\rvert^a\ge\frac{r_0^{\varepsilon}}2\lvert z\rvert^a+\frac{r_0^{\varepsilon}}2r_0^a\text.$$ So if $r_0>0$ is sufficiently large (depending on $\varepsilon$), there holds $$\lvert z\rvert^{a+\varepsilon}\ge A\lvert z\rvert^a+B\text,$$ and so it follows for this $r_0$ and $\lvert z\rvert\ge r_0$ that $$\lvert f(z)\rvert\le\exp(\lvert z\rvert^{a+\varepsilon})\text.$$ Thus, $f$ is of finite order at most $a+\varepsilon>0$ according to the first definition; since $\varepsilon>0$ was arbitrary, $f$ is of finite order at most $a$ according to the first definition.
Conversely, if $f$ is of finite order according to the first definition with finite order $a$ and $\varepsilon>0$, then note that $f$ is bounded on the compact set $\{z:\lvert z\rvert\le r_0\}$, and so for sufficiently large $B$ there holds $$\lvert f(z)\rvert\le\max\{\exp(\lvert z\rvert^{a+\varepsilon}),\exp B\}\le\exp(\lvert z\rvert^{a+\varepsilon}+B)$$ Thus $f$ is of finite order according to the second definition with order $\le a+\varepsilon$. Since this holds for every $\varepsilon>0$, $f$ must be of finite order according to the second definition with order $\le a$.