Ingham's Theorem on Convergent Dirichlet Series

Theorem

Suppose $|a_n| \le 1$, and form the series $\displaystyle \sum_{n=1}^\infty a_n n^{-z}$ which converges to an analytic function $F(z)$ for $\Re \left({z}\right) > 1$, where $\Re \left({z}\right)$ is the real part of $z$.

If $F(z)$ in analytic throughout $\Re \left({z}\right) \ge 1$, then $\displaystyle \sum_{n=1}^\infty a_n n^{-z}$ converges throughout $\Re \left({z}\right) \ge 1$.

Proof

Fix a $w$ in $\Re \left({w}\right) \ge 1$. Then $F(z+w)$ is analytic in $\Re \left({z}\right) \ge 0$.

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

Choose some $R \ge 1$. Then, since $F(z+w)$ is analytic on such an open set, we can determine $\delta = \delta (R) > 0, \delta \le \dfrac 1 2$ such that $F(z+w)$ is analytic in $\Re(z) \ge - \delta, | \Im (z) | \le R$.

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

Now form the counterclockwise contour $\Gamma$ as the arc $|z|=R, \Re \left({z}\right) > - \delta$ and the segment $\Re \left({z}\right) = -\delta, |z| \le R$. We denote by $A, B$ respectively, the parts of $\Gamma$ in the right and left half-planes.

By the Residue Theorem:

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

Since $F(z+w)$ converges to its series on $A$, we may split it into the partial sum and remainder after $N$ terms, $s_N(z+w), r_N(z+w)$ respectively, and find that, again, by the residue theorem:

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

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

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

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

Combining these results gives:

$\displaystyle 2 \pi i \left({F(w) - s_N(w) }\right) = \int_\Gamma F(z+w)N^z \left({\frac 1 z + \frac z {R^2} }\right) dz - \int_A s_N(z+w) N^z \left({ \frac 1 z + \frac z {R^2} }\right) dz - \int_A s_N(w-z) N^{-z} \left({ \frac 1 z + \frac z {R^2} }\right) dz$

$\displaystyle = \int_A F(z+w) N^z \left({\frac 1 z + \frac z {R^2} }\right) dz + \int_B F(z+w)N^z \left({\frac 1 z + \frac z {R^2} }\right) dz - \int_A s_N(z+w) N^z \left({ \frac 1 z + \frac z {R^2} }\right) dz - \int_A s_N(w-z) N^{-z} \left({\frac 1 z + \frac z {R^2} }\right) dz$

$\displaystyle = \int_A \left({ r_N(z+w)N^z - s_N(w-z)N^{-z} }\right) \left({ \frac 1 z + \frac z {R^2} }\right) dz + \int_B F(z+w) N^z \left({\frac 1 z + \frac z {R^2} }\right) dz$

For what follows, allow $z = x + i y$ and observe that on $A, |z|=R$, so:

$\displaystyle \frac 1 z + \frac z {R^2} = \frac{\overline z}{|z|^2} + \frac z {R^2} = \frac{x-iy}{R^2} + \frac{x+iy}{R^2} = \frac{2x}{R^2}$

and on $B$, we have:

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

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

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