Equivalence of Definitions of Order of Entire Function

Theorem
Let $f: \C \to \C$ be an entire function.

Let $\log$ denote the natural logarithm.

The following definitions of order of entire function are equivalent:

Proof
Let:
 * $\alpha_1 = \displaystyle \limsup_{R \mathop \to \infty} \frac {\log \log \max_{\left\lvert{z}\right\rvert \le R} \left\lvert{f}\right\rvert} {\log R}$
 * $\alpha_2 = \inf \left\{ \beta\geq0 : \displaystyle \log\left(\max_{|z|\leq R}|f(z)|\right) = O(R^\beta) \right\}$
 * $\alpha_3 = \inf \left\{ \beta\geq0 : f \left({z}\right) = \mathcal O \left( \exp \left({\left\lvert{z}\right\rvert^\beta}\right) \right) \right\}$

1 less than 2
Let $\beta > \alpha_2$.

Then there exists $K>0$ such that $\log \max_{\left\lvert{z}\right\rvert \le R} \left\lvert{f}\right\rvert \leq K\cdot R^\beta$ for $R$ sufficiently large.

That is, $\frac{\log \log \max_{\left\lvert{z}\right\rvert \le R} \left\lvert{f}\right\rvert}{\log R} \leq \frac K{\log R} + \beta$ for $R$ sufficiently large.

Taking $\limsup$, we get $\alpha_1 \leq \beta$.

Taking limits, $\alpha_2\geq\alpha_1$.

2 less than 1
Let $\beta>\alpha_1$.

We have $\log \max_{\left\lvert{z}\right\rvert \le R} \left\lvert{f}\right\rvert \leq R^\beta$ for $R$ sufficiently large.

Thus $\beta \geq \alpha_2$.

Taking limits, $\alpha_1\geq \alpha_2$.

3 less than 2
Let $\beta > \alpha_2$.

Let $\epsilon>0$ be such that $\beta-\epsilon>\alpha_1$.

Then $\displaystyle \log\left(\max_{|z|\leq R}|f(z)|\right) = O(R^{\beta-\epsilon})$.

Then there exists $R_0>0$ such that $\log\left(\max_{|z|\leq R}|f(z)|\right) \leq R^\beta$ for $R\geq R_0$.

By Exponential is Increasing, this implies $\max_{|z|\leq R}|f(z)| \leq \exp(R^\beta)$.

Let $z\in\C$ with $|z| \geq R_0$.

Then $|f(z)| \leq \max_{|w|\leq |z|}|f(w)| \leq \exp(|z|^\beta)$.

Thus $f \left({z}\right) = \mathcal O \left( \exp \left({\left\lvert{z}\right\rvert^\beta}\right) \right)$

Thus $\beta\geq\alpha_3$.

Taking the limit $\beta\to\alpha_2$, we obtain, $\alpha_2\geq\alpha_3$.