02 Dec 2021


## Introduction

This post is a follow up to this previous post wherein we established that Laplace transforms of complex measures converge absolutely in half-planes. More generally, in fact, this result holds for local complex measures $\mu$ whose magnitude grows at most exponentially— that is to say, the measure $\vert\mu\vert$ has a cumulative distribution function $F(t)=\vert\mu\vert([0,t])$ which is of exponential order, viz. $F(t)=O(\exp(At))$ for some $A>0$ as $t\to\infty$.

As a corollary, we obtained that a generalized Dirichlet series (realized as a Laplace transform of a discrete measure) converges absolutely in regions shaped like half-planes.

Here, we shall show that a generalized Dirichlet series can converge absolutely in any half-plane $$\HH_x=\{ s\in\CC : \Re(s) > x\}$$, $x\in\mathbb{R}$. We will prove more in fact: we shall demonstrate a examples of a generalized Dirichlet series whose abscissa of convergence $\sc$ and whose abscissa of absolute convergence $\sac$ can be any two real numbers, provided that $\sc\leq \sac$.

We will not make comment about the convergence behavior at points in the boundaries of such half-planes, only on the open interior of such half-planes.

## Dirichlet Series with Prescribed Half-Planes of Convergence

#### The Starting Example

We shall start by considering the following (classical) Dirichlet series:

$f(s)= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$

This is sometimes known as the alternating zeta function, or the Dirichlet eta function.

By elementary techniques, we are able to deduce the following:

#### Theorem

The Dirichlet series $f$ converges in the half plane $\HH_0$ and converges absolutely in the half-plane $\HH_1$. In other words, $\sc=0$ and $\sac=1$.

#### Proof

We have that the absolute version of the series takes the form:

$\sum_{n=1}^\infty \left\vert\frac{(-1)^{n+1}}{n^s}\right\vert = \sum_{n=1}^\infty \frac{1}{n^\sigma}$

The left hand side is a $p$-series with parameter $p=\sigma$, so the series $f$ converges absolutely when $\sigma>1$.

Meanwhile, when $\sigma>0$ we have that the sequence $a_n=n^{-\sigma}=\vert n^{-s}\vert$ is monotonically decreasing to zero as $n\to\infty$. Since the terms are alternating in sign, Leibniz’s alternating series test ensures the the series converges.

$\blacksquare$

#### Modification of the Example

We now write the series above as a generalized Dirichlet series:

$f(s) = \sum_{n=1}^\infty (-1)^{n+1}e^{-s\log n}$

Apply the transformation linear transformation:

$s\mapsto \frac{s-x_0}{x_1-x_0}$

to obtain the new series, $g(s) := f\left(\frac{s-x_0}{x_1-x_0}\right)$.

#### Theorem (Arbitrary Abscissa of Convergence, $\sac>\sc$)

Let $x_1,x_0$ be any two real numbers with $x_1>x_0$. The series $g$ as defined is a generalized Dirichlet series of the form:

$g(s) = \sum_{n=1}^\infty b_n e^{-\lambda_n s}$

Further, we have that abscissa of convergence $\sc=x_0$ and abscissa of absolute convergence $\sc=x_1$.

#### Proof

We first compute that $g$ remains a generalized Dirichlet series:

\begin{align*} g(s) &= \sum_{n=1}^\infty (-1)^{n+1}e^{-\left(\frac{s-x_0}{x_1-x_0}\right)\log n} \\ &= \sum_{n=1}^\infty (-1)^{n+1}n^{\frac{x_0}{x_1-x_0}}e^{-s\frac{\log n}{x_1-x_0}} \\ &= \sum_{n=1}^\infty b_n e^{-\lambda_n s} \end{align*}

where we make the assignments $b_n=(-1)^{n+1}n^{\frac{x_0}{x_1-x_0}}$ and $\lambda_n=\frac{\log n}{x_1-x_0}$.

We remark that the condition $x_1> x_0$ ensures that each $\lambda_n>0$ (so that they forms an increasing sequence). When $x_1-x_0=1$, the series above reduces to a classical Dirichlet series with coefficients $a_n=(-1)^{n+1} n^{x_0}$.

Since $f(t) = g((x_1-x_0)t+x_0))$ absolutely converges when $\Re(t)>1$ and converges when $\Re(t)>0$, $g$ converges under these same conditions. In the variable $s=(x_1-x_0)t+x_0$, it follows that $g(s)$ converges absolutely when $\Re(s)>x_1$ and converges when $\Re(s)>x_0$.

$\blacksquare$

Thus, we have shown that the convergence of a generalized Dirichlet series may have arbitrary abscissa of convergence $\sc=x_0$ and abscissa of absolute convergence $\sac=x_1> x_0$. Note that this construction involves dividing by $x_1-x_0$, which fails when they are equal.

#### Equal Abscissae Case

We next proceed similarly but with a different example to prove the theorem when the abscissa of convergence coincide.

In the spirit of the previous proof, we shall also take a known example and modify it with a linear transformation. In this case, the ordinary Riemann zeta function will do:

$\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$

This series is absolutely convergent when $\Re(s)>1$, and it diverges when $\Re(s)<1$. These can be seen, for instance, by using the integral test and comparison with the harmonic series, respectively. (Any other technique to prove this result for such a $p$-series would do.) Thus, the abscissae of convergence coincide and equal to one.

Note that the absolute convergence for real valued $s$ is automatic from convergence since the coefficients are positive, and absolute convergence of the series with complex $s=\sigma+i\tau$ follows from comparison to the series with real parameter $\sigma$.

#### Theorem (Arbitrary Abscissa of Convergence, $\sac=\sc$)

For any real number $x_0$, there is a classical Dirichlet series $g$ whose abscissa of convergence and of absolute convergence coincide and equal this number: $\sac=\sc=x_0$.

#### Proof

Define the new Dirichlet series $g(s) = \zeta(s-(x_0-1))$ for $x_0\in\mathbb{R}$, which converges when $\Re(s)-x_0+1>1$, or $\Re(s)>x_0$. Thus $\sc=x_0$ for $g$. The abcissa of absolute convergence is the same for the $\zeta$ function, and hence for $g$ as well.

Since this transformation is only a shift by $x_0-1$, the series $g$ remains a classical Dirichlet series:

\begin{align*} g(s) &= \sum_{n=1}^\infty \frac{1}{n^{s-x_0+1}} = \sum_{n=1}^\infty \frac{n^{x_0-1}}{n^{s}} \end{align*}

Since $x_0$ is a arbitrary real number, we are done.

$\blacksquare$

In summary, we have shown that for any two real numbers $x_1,x_0\in\mathbb{R}$ with $x_1\geq x_0$, there is a generalized Dirichlet series $g$ whose abcissae satisfy $\sc=x_0$ and $\sac=x_1$.