Order of Product of Entire Functions

Theorem
Let $f, g: \C \to \C$ be entire functions of order $\alpha$ and $\beta$.

Then $f g$ has order at most $\map \max {\alpha, \beta}$.

Proof
If $\map \max {\alpha, \beta} = +\infty$, the claim is trivial.

Thus we may assume that $\alpha < +\infty$ and $\beta < +\infty$.

Let $\epsilon > 0$ be arbitrary.

By, we have:
 * $\map f z = \map \OO {\map \exp {\cmod z^{\alpha + \epsilon} } }$

and:
 * $\map g z = \map \OO {\map \exp {\cmod z^{\beta + \epsilon} } }$

By, there are $c_1,c_2,r_1,r_2 \in \R$ such that:
 * $\cmod z \ge r_1 \implies \cmod {\map f z} \le c_1 \map \exp {\cmod z^{\alpha + \epsilon} }$

and:
 * $\cmod z \ge r_2 \implies \cmod {\map g z} \le c_2 \map \exp {\cmod z^{\beta + \epsilon} }$

Let $r_0 := \max \set {r_1, r_2, 2^{1 / \epsilon } }$.

Then, if $\cmod z \ge r_0$, we have:

Therefore the order of $fg$ is:
 * $\le \map \max {\alpha, \beta} + 2 \epsilon$

Letting $\epsilon \to 0$, the claim follows.

Also see

 * Order of Sum of Entire Functions
 * Order of Product of Entire Function with Polynomial