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 $R_n = 2 \size {a_n}$.

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:

$\omega \ge \tau$

$\blacksquare$

Also see

• Order is Maximum of Exponent of Convergence and Degree

Sources

• 1932: A.E. Ingham: The Distribution of Prime Numbers: Chapter III: Further Theory of $\zeta(s)$. Applications: $\S 7$: Integral functions: Theorem $\text F 2$