Equivalence of Definitions of Order of Entire Function

Theorem

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

Let $\ln$ denote the natural logarithm.

The following definitions of the concept of Order of Entire Function are equivalent:

Definition 1

The order $\alpha \in \closedint 0 {+\infty}$ of $f$ is the infimum of the $\beta \ge 0$ for which:

$\map f z = \map \OO {\map \exp {\size z^\beta} }$

or $\infty$ if no such $\beta$ exists, where $\OO$ denotes big-$\OO$ notation.

Definition 2

Let $f$ be not identically zero.

The order $\alpha \in \closedint 0 {+\infty}$ of $f$ is the infimum of the $\beta \ge 0$ for which:

$\ds \map \ln {\max_{\size z \mathop \le R} \size {\map f z} } = \map \OO {R^\beta}$

or $\infty$ if no such $\beta$ exists, where $\OO$ denotes big-$\OO$ notation

The order of $0$ is $0$.

Definition 3

Let $f$ be non-constant.

The order $\alpha \in \closedint 0 {+\infty}$ of $f$ is the limit superior:

$\ds \limsup_{R \mathop \to \infty} \frac {\ds \ln \ln \max_{\cmod z \mathop \le R} \cmod f} {\ln R}$

The order of a constant function is $0$.

Proof

Let:

$\alpha_1 = \ds \limsup_{R \mathop \to \infty} \frac {\ds \ln \ln \max_{\cmod z \mathop \le R} \cmod f} {\ln R}$

$\alpha_2 = \inf \set {\beta \ge 0: \ds \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^\beta} }$

$\alpha_3 = \inf \set {\beta \ge 0: \map f z = \map \OO {\map \exp {\cmod z^\beta} } }$

$3$ less than $2$

Let $\beta > \alpha_2$.

Then there exists $K > 0$ such that:

$\ds \ln \max_{\cmod z \mathop \le R} \cmod f \le K \cdot R^\beta$

for $R$ sufficiently large.

That is:

$\dfrac {\ds \ln \ln \max_{\cmod z \le R} \cmod f} {\ln R} \le \dfrac K {\ln R} + \beta$

for $R$ sufficiently large.

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

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

$\Box$

$2$ less than $3$

Let $\beta > \alpha_1$.

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

Thus $\beta \ge \alpha_2$.

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

$\Box$

$1$ less than $2$

Let $\beta > \alpha_2$.

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

Then:

$\ds \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } = \map \OO {R^{\beta - \epsilon} }$

Then there exists $R_0 > 0$ such that:

$\ds \map \ln {\max_{\cmod z \mathop \le R} \cmod {\map f z} } \le R^\beta$

for $R \ge R_0$.

By Exponential is Strictly Increasing, this implies:

$\ds \max_{\cmod z \mathop \le R} \cmod {\map f z} \le \map \exp {R^\beta}$

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

Then:

$\ds \cmod {\map f z} \le \max_{\cmod w \mathop \le \cmod z} \cmod {\map f w} \le \map \exp {\cmod z^\beta}$

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:

$\alpha_2 \ge \alpha_3$

$\Box$

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