Order of Product of Entire Function with Polynomial

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

Let $P: \C \to \C$ be a nonzero polynomial.

Then $f \cdot P$ has order $\omega$.

Proof
If $f=0$, then $f \cdot P = 0$.

Thus the claim is trivial.

Therefore suppose that $f$ is not identically zero.

By Order of Product of Entire Functions and Polynomial has Order Zero, $f\cdot P$ has order at most $\omega$.

By Limit at Infinity of Polynomial, there exist $r, \delta > 0$ such that $\size {\map P z} \ge \delta$ for $\size z \ge r$.

$\ds \map \log {\max_{\size z \mathop \le R} \size {\map f z \map P z} } = \map \OO {R^\beta}$ for some $\beta < \omega$.

By the Maximum Modulus Principle:
 * $\ds \max_{\size z \mathop \le R} \size {\map f z} \le \dfrac 1 \delta \max_{\size z \mathop \le R} \size {\map f z \map P z}$

for $R \ge r$.

Thus:
 * $\ds \map \log {\max_{\size z \mathop \le R} \size {\map f z} } = \map \OO {R^\beta}$

By, this means:
 * $\omega \le \beta$

This is a contradiction.

Thus $f \cdot P$ has order $\omega$.

Also see

 * Order of Product of Entire Functions
 * Polynomial has Order Zero