Exponent of Convergence is Less Than Order
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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:
- $\tau = 0 \le \omega$
$\Box$
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 $R_n = 2 \size {a_n}$.
- $n \le \dfrac {\map \log {\max_{\size z \le R_n} \size f} - \log \size {\map f 0} } {\log 2}$
for all $n \in \N$.
Let $\epsilon > 0$.
Because $f$ has order $\omega$:
- $n \ll_\epsilon \size {a_n}^{\omega + \epsilon}$
Thus:
- $\size {a_n}^{-\paren {\omega + \epsilon} } \ll_\epsilon n^{-1}$
- $\omega + \epsilon \ge \tau$
Because $\epsilon$ is arbitrary:
- $\omega \ge \tau$
$\blacksquare$
Also see
Sources
- 1932: A.E. Ingham: The Distribution of Prime Numbers: Chapter III: Further Theory of $\zeta(s)$. Applications: $\S 7$: Integral functions: Theorem $\text F 2$