Ingham's Theorem on Convergent Dirichlet Series

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Theorem

Let $\left\vert{a_n}\right\vert \le 1$



For a complex number $z \in \C$, let $\Re \left({z}\right)$ denote the real part of $z$.

Form the series $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z}$ which converges to an analytic function $F \left({z}\right)$ for $\Re \left({z}\right) > 1$.



Let $F \left({z}\right)$ be analytic throughout $\Re \left({z}\right) \ge 1$.

Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z}$ converges throughout $\Re \left({z}\right) \ge 1$.


Proof

Fix a $w$ in $\Re \left({w}\right) \ge 1$.

Then $F \left({z + w}\right)$ is analytic in $\Re \left({z}\right) \ge 0$.



We note that since $F \left({z + w}\right)$ is analytic on $\Re \left({z}\right) = 0$, it must be analytic on an open set containing $\Re \left({z}\right) = 0$.


Choose some $R \ge 1$.



We have that $F \left({z + w}\right)$ is analytic on such an open set.

Thus we can determine $\delta = \delta \left({R}\right) > 0, \delta \le \dfrac 1 2$ such that $F \left({z + w}\right)$ is analytic in $\Re \left({z}\right) \ge -\delta, \left\vert{\Im \left({z}\right)}\right\vert \le R$.



We also choose an $M = M \left({R}\right)$ so that $F \left({z + w}\right)$ is bounded by $M$ in $-\delta \le \Re \left({z}\right), \left\vert{z}\right\vert \le R$.



Now form the counterclockwise contour $\Gamma$ as the arc $\left\vert{z}\right\vert = R, \Re \left({z}\right) > - \delta$ and the segment $\Re \left({z}\right) = -\delta, \left\vert{z}\right\vert \le R$.

We denote by $A, B$ respectively, the parts of $\Gamma$ in the right and left half-planes.


By the Residue Theorem:

$\displaystyle 2 \pi i F \left({w}\right) = \oint_{\Gamma} F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z$


Since $F \left({z + w}\right)$ converges to its series on $A$, we may split it into the partial sum and remainder after $N$ terms:

$s_N \left({z + w}\right), r_N \left({z + w}\right)$

respectively.

Again, by the Residue Theorem:

$\displaystyle \int_A s_N \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z = 2 \pi i s_N \left({w}\right) - \int_{-A} s_N \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z$

where $-A$ is the reflection of $A$ through the origin.


Changing $z \to -z$, we have:

$\displaystyle \int_A s_N \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z = 2 \pi i s_N \left({w}\right) - \int_A s_N \left({w - z}\right) N^{-z} \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z$


Combining these results gives:

\(\displaystyle \) \(\) \(\displaystyle \displaystyle 2 \pi i \left({F \left({w}\right) - s_N \left({w}\right) }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \int_\Gamma F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z - \int_A s_N \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z - \int_A s_N (w-z) N^{-z} \left({ \frac 1 z + \frac z {R^2} }\right) \, \mathrm d z\)
\(\displaystyle \) \(=\) \(\displaystyle \int_A F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z + \int_B F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z\)
\(\displaystyle \) \(\) \(\, \displaystyle - \, \) \(\displaystyle \int_A s_N \left({z + w}\right) N^z \left({ \frac 1 z + \frac z {R^2} }\right) \, \mathrm d z - \int_A s_N(w-z) N^{-z} \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z\)
\(\displaystyle \) \(=\) \(\displaystyle \int_A \left({r_N \left({z + w}\right) N^z - s_N(w-z)N^{-z} }\right) \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z + \int_B F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z\)


For what follows, allow $z = x + i y$ and observe that on $A, \left\vert{z}\right\vert = R$.

So:

\(\displaystyle \frac 1 z + \frac z {R^2}\) \(=\) \(\displaystyle \frac {\overline z} {\left\vert{z}\right\vert^2} + \frac z {R^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {x - i y} {R^2} + \frac {x + i y} {R^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 x} {R^2}\)


and on $B$:

\(\displaystyle \left\vert{\frac 1 z + \frac z {R^2} }\right\vert\) \(=\) \(\displaystyle \left\vert{\frac 1 z \left({1 + \left({\frac z R}\right)^2}\right)}\right\vert\)
\(\displaystyle \) \(\le\) \(\displaystyle \left\vert{\frac 1 \delta \left({1 + 1}\right)}\right\vert\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 2 \delta\)


Already we can place an upper bound on one of these integrals:

$\displaystyle \left\vert{\int_B F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z }\right\vert \le \int_{-R}^R M N^x \frac 2 \delta \, \mathrm d y + 2 M \int_{-\delta}^0 N^x \frac{2x}{R^2} \, \mathrm d x$



Source of Name

This entry was named for Albert Edward Ingham.