Equivalence of Definitions of Order of Entire Function

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

Let $\ln$ denote the natural logarithm.

Proof
Let:
 * $\alpha_1 = \displaystyle \limsup_{R \mathop \to \infty} \frac {\displaystyle \ln \ln \max_{\cmod z \mathop \le R} \cmod f} {\ln R}$
 * $\alpha_2 = \inf \set {\beta \ge 0: \displaystyle \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^\beta} }$
 * $\alpha_3 = \inf \set {\beta \ge 0: \map f z = \map \OO {\map \exp {\cmod z^\beta} } }$

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

Then there exists $K > 0$ such that:
 * $\displaystyle \ln \max_{\cmod z \mathop \le R} \cmod f \le K \cdot R^\beta$

for $R$ sufficiently large.

That is:
 * $\dfrac {\ln \ln \max_{\cmod z \le R} \cmod f} {\ln R} \le \frac K {\ln R} + \beta$

for $R$ sufficiently large.

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

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

2 less than 3
Let $\beta > \alpha_1$.

We have $\displaystyle \ln \max_{\cmod z \mathop \le R} \cmod f \le R^\beta$ for $R$ sufficiently large.

Thus $\beta \ge \alpha_2$.

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

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

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

Then:
 * $\displaystyle \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^{\beta - \epsilon} }$

Then there exists $R_0 > 0$ such that:
 * $\displaystyle \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } \le R^\beta$

for $R \ge R_0$.

By Exponential is Strictly Increasing, this implies:
 * $\displaystyle \max_{\cmod z \mathop \le R} \cmod {\map f z} \le \map \exp {R^\beta}$

Let $z \in \C$ with $\cmod z \ge R_0$.

Then:
 * $\cmod {\map f z} \le \max_{\cmod w \le \cmod z} \cmod {\map f w} \le \map \exp {\cmod z^\beta}$

Thus:
 * $\map f z = \map \OO {\map \exp {\cmod z^\beta} }$

Thus $\beta \ge \alpha_3$.

Taking the limit $\beta \to \alpha_2$, we obtain:
 * $\alpha_2 \ge \alpha_3$