Exponent of Convergence is Less Than Order

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

Let $\omega$ be its order.

Let $\tau$ be its exponent of convergence.

Then $\tau \le \omega$.

Proof
We may assume $\map f 0 \ne 0$.

Let $f$ have finitely many zeroes.

Then:
 * $\tau = 0 \le \omega$

Let $f$ have infinitely many zeroes.

Let $\sequence {a_n}$ be the sequence of nonzero zeroes of $f$, repeated according to multiplicity and ordered by increasing modulus.

Let $r_n = \size {a_n}$ and $R_n = 2 \size {a_n}$.

By Jensen's Inequality:
 * $n \le \dfrac {\map \log {\max_{\size z \le R_n} \size f} - \log \size {\map f 0} } {\log 2}$

for all $n \in \N$.

Let $\epsilon > 0$.

Because $f$ has order $\omega$:
 * $n \ll_\epsilon \size {a_n}^{\omega + \epsilon}$

Thus:
 * $\size {a_n}^{-\paren {\omega + \epsilon} } \ll_\epsilon n^{-1}$

By Convergence of P-Series:
 * $\omega + \epsilon \ge \tau$

Because $\epsilon$ is arbitrary:
 * $\omega \ge \tau$

Also see

 * Order is Maximum of Exponent of Convergence and Degree