# Holomorphic functions of finite order generated by Dirichlet series

Andreas Defant  Andreas Defant
Institut für Mathematik,
Carl von Ossietzky Universität,
26111 Oldenburg, Germany<EMAIL_ADDRESS>,  Ingo
Schoolmann  Ingo Schoolmann
Institut für Mathematik,
Carl von Ossietzky Universität,
26111 Oldenburg, Germany<EMAIL_ADDRESS>and  Andreas
Defant and Ingo Schoolmann

###### Abstract.

Given a frequency $\lambda=(\lambda_{n})$ and $\ell\geq 0$, we introduce the
scale of Banach spaces $H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ of
holomorphic functions $f$ on the right half-plane $[\operatorname{Re}>0]$,
which satisfy $(A)$ the growth condition $|f(s)|=O((1+|s|)^{\ell})$, and $(B)$
are such that they on some open subset and for some $m\geq 0$ coincide with
the pointwise limit (as $x\to\infty$) of the so-called $(\lambda,m)$-Riesz
means

$\sum_{\lambda_{n}<x}a_{n}e^{-\lambda_{n}s}\Big{(}1-\frac{\lambda_{n}}{x}\Big{)}^{m},\,x>0$

of some $\lambda$-Dirichlet series $\sum a_{n}e^{-\lambda_{n}s}$. Reformulated
in our terminology, an important result of M. Riesz shows that in this case
the function $f$ for every $k>\ell$ is the pointwise limit of the
$(\lambda,k)$-Riesz means of $D$.

Our main contribution is an extension – showing that ’after translation’ every
bounded set in $H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ is uniformly
approximable by all its $(\lambda,k)$-Riesz means of order $k>\ell$. This
follows from an appropriate maximal theorem, which in fact turns out to be at
the very heart of a seemingly interesting structure theory of the Banach
spaces $H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. One of the many
consequences is that $H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$
basically consists of those holomorphic functions on $[\operatorname{Re}>0]$,
which fulfill the growth condition $|f(s)|=O((1+|s|)^{\ell})$ and are of
finite uniform order on all abscissas $[\operatorname{Re}=\sigma],\,\sigma>0$.

To establish all this and more, we need to reorganize (and to improve) various
aspects and keystones of the classical theory of Riesz summability of general
Dirichlet series as invented by Hardy and M. Riesz.

###### Contents

  1. 1 Introduction
  2. 2 Riesz summation revisted
    1. 2.1 Ordinary summation
    2. 2.2 Riesz means
    3. 2.3 Integral representation
    4. 2.4 Perron formula
    5. 2.5 Bohr-Cahen formula
  3. 3 Summatory functions
    1. 3.1 Integral forms
    2. 3.2 Estimates
    3. 3.3 Two crucial lemmas
    4. 3.4 Proofs of Theorem 2.9 and Theorem 2.16
  4. 4 A new scale of Banach spaces
    1. 4.1 Ordinary and power case
    2. 4.2 Perron formula – a variant
    3. 4.3 A maximal inequality for Riesz means
    4. 4.4 Uniform Riesz approximation
    5. 4.5 Theorem 41 of M. Riesz revisted
    6. 4.6 The bounded case $\ell=0$
    7. 4.7 Boundedness of coefficient functionals
    8. 4.8 A Montel theorem
    9. 4.9 Completeness
    10. 4.10 Behavior far left
    11. 4.11 Behavior far right
    12. 4.12 Finite order
    13. 4.13 Equivalence

††footnotetext: 2010 _Mathematics Subject Classification_ : Primary 43A17,
Secondary 30B50, 43A50††footnotetext: _Key words and phrases_ : general
Dirichlet series, finite order, Riesz summability, almost everywhere
convergence, Hardy spaces. ††footnotetext:

##  1\. Introduction

Recently, the theory of ordinary Dirichlet series $\sum a_{n}n^{-s}=\sum
a_{n}e^{-s\log n}$ and that of general Dirichlet series $\sum
a_{n}e^{-\lambda_{n}s}$, where the frequency $\lambda=(\lambda_{n})$ is a
strictly increasing, non-negative real sequence, saw a remarkable renaissance.
Fixing a frequency $\lambda$, the collection of all such $\lambda$-Dirichlet
series $\sum a_{n}e^{-\lambda_{n}s}$ is denoted by $\mathcal{D}(\lambda)$ .

General Dirichlet series $\sum a_{n}e^{-\lambda_{n}s}$ converge on maximal
half-planes $[\operatorname{Re}>\leavevmode\nobreak\ \sigma]$ in $\mathbb{C}$,
where they define holomorphic functions $f$, and even though in concrete
situations these half-planes might be small, the holomorphic functions $f$
often extend to holomorphic functions on larger half-planes.

The problem of determining whether a holomorphic $f$, defined on some half-
plane, is or is not representable in the form of a $\lambda$-Dirichlet series
$\sum a_{n}e^{-\lambda_{n}s}$, is in general difficult.

An important criterion is that the limit function of a $\lambda$-Dirichlet
series has finite order in each closed half-plane of its domain of
convergence. Moreover, if this limit function can be holomorphically extended
to a larger half-plane, it may happen that its continuation still has finite
order in a larger domain.

Recall that, given a holomorphic function
$f:[\operatorname{Re}>\sigma_{0}]\to\mathbb{C},\,\sigma_{0}\in\mathbb{R}$, its
order $\mu_{f}(\sigma)\in[\sigma_{0},\infty]$ on the line
$[\operatorname{Re}\leavevmode\nobreak\ =\leavevmode\nobreak\
\sigma],\,\sigma>\sigma_{0}$ is the infimum over all $\ell\in[0,\infty]$ such
that

(1) $\exists\leavevmode\nobreak\ C,t_{0}>0\leavevmode\nobreak\
\forall\leavevmode\nobreak\ |t|>t_{0}\colon\leavevmode\nobreak\
|f(\sigma+it)|\leq C|t|^{\ell}\,.$

The function $f$ is said to be of finite order on
$[\operatorname{Re}>\sigma_{0}]$ whenever $\mu_{f}<\infty$ for all
$\sigma>\sigma_{0}$, and in this case
$\mu_{f}:]\sigma_{0},\infty]\to\mathbb{R}_{\geq 0}$ is continuous and convex.

Similarly, the uniform order $\nu_{f}\in[0,\infty]$ of $f$ on the half-plane
$[\operatorname{Re}>\sigma_{0}]$ is given by the infimum over all
$\ell\in[0,\infty]$ such that the above condition holds uniformly, i.e.,

(2) $\exists\leavevmode\nobreak\ C,t_{0}>0\leavevmode\nobreak\
\forall\leavevmode\nobreak\
\sigma>\sigma_{0},|t|>t_{0}\colon\leavevmode\nobreak\ |f(\sigma+it)|\leq
C|t|^{\ell}.$

Moreover, the function $f$ is said to have finite uniform order on
$[\operatorname{Re}>\sigma_{0}]$ whenever $\nu_{f}<\infty$.

To see a famous example, denote by
$\zeta:\mathbb{C}\setminus\\{1\\}\to\mathbb{C}$ the zeta-function, which is
holomorphic with a simple pole in $s=1$, and which on $[\operatorname{Re}>1]$
is the pointwise limit of the zeta-Dirichlet series $\sum n^{-s}$. The famous
Lindelöf conjecture asks whether the order of $\zeta$ on the critical line
$[\operatorname{Re}=\frac{1}{2}]$ equals $0$. Equivalently, consider the
entire function

$\eta:\mathbb{C}\to\mathbb{C}\,,\,\,\,\eta(s)=(1-2^{1-s})\zeta(s)\,,$

which on $[\operatorname{Re}>0]$ is nothing else then the pointwise limit of
the $\eta$-Dirichlet series $\sum(-1)^{n+1}n^{-s}$. It is known that
$\mu_{\eta}(\sigma)=\frac{1}{2}-\sigma$ for $\sigma<0$ and
$\mu_{\eta}(\sigma)=0$ for $\sigma>1$. Hence Lindelöf’s conjecture is
equivalent to the question whether or not we have that

$\mu_{\eta}(\sigma)=\begin{cases}\frac{1}{2}-\sigma&\,\,\,\,0\leq\sigma<\frac{1}{2}\\\
0&\,\,\,\,\frac{1}{2}<\sigma\leq 1.\end{cases}$

The main aim here is to study holomorphic functions of finite uniform order
generated by Dirichlet series from the point of view of functional analysis –
and our main inspiration comes from the classical monograph [13] of Hardy and
Riesz on so-called Riesz summation of general Dirichlet series.

Given a frequency $\lambda$ and a holomorphic function
$f:[\operatorname{Re}>0]\to\mathbb{C}$, we say that the $\lambda$-Dirichlet
series $D=\sum a_{n}(D)e^{-\lambda_{n}s}$ is a $\lambda$-Riesz germ of $f$
whenever $D$ for some $m>0$ and some $\sigma\in\mathbb{R}$ satisfies that

$f(s)=\lim_{x\to\infty}\sum_{x<\lambda_{n}}a_{n}(D)e^{-\lambda_{n}s}\Big{(}1-\frac{\lambda_{n}}{x}\Big{)}^{m}\,\,\,\,\text{for
all $s\in[\operatorname{Re}>\sigma]$}.$

The trigonometric polynomials

$R_{x}^{\lambda,m}(f)(s)=\sum_{\lambda_{n}<x}a_{n}(D)e^{-\lambda_{n}s}\Big{(}1-\frac{\lambda_{n}}{x}\Big{)}^{m}\,,\,\,x>0$

are said to be the $(\lambda,m)$-Riesz means of $f$ of order $m$.

An important fact will be that such $\lambda$-Riesz germs $D$ of $f$, whenever
they exist (!), are unique (Corollary 2.15) – and so in this situation the
coefficients $a_{n}(f):=a_{n}(D)$ uniquely determine the function $f$.

We study a graduated scale of Banach spaces which consist of holomorphic
functions on the right half-plane, which are of finite uniform order on
$[\operatorname{Re}>0]$ and are generated by $\lambda$-Riesz germs. More
precisely, given a frequency $\lambda=(\lambda_{n})$ and $\ell\geq 0$, the
linear space

(3) $H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$

consists of all holomorphic functions $f:\leavevmode\nobreak\
[\operatorname{Re}>0]\to\mathbb{C}$, which are generated by a $\lambda$-Riesz
germ and satisfy the growth condition

(4)
$\|f\|_{\infty,\ell}=\sup_{\operatorname{Re}s>0}\frac{|f(s)|}{(1+|s|)^{\ell}}<\infty\,.$

This defines a scale of Banach spaces (a non-trivial fact from Theorem 4.16),

$\big{(}H_{\infty,\ell}^{\lambda}[\operatorname{Re}\leavevmode\nobreak\
>\leavevmode\nobreak\ 0],\|\boldsymbol{\cdot}\|_{\infty,\ell}\big{)}_{\ell\geq
0}\,,$

which is intimately connected with the notion of finite uniform order (Theorem
4.21):

  * $\bullet$

Each function $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ has finite
uniform order $\nu_{f}\leq\ell$ on $[\operatorname{Re}\leavevmode\nobreak\
>\leavevmode\nobreak\ 0]$, and conversely,

  * $\bullet$

if $f$ is a holomorphic function on $[\operatorname{Re}>0]$, which is
generated by a $\lambda$-Riesz germ and is of finite uniform order on
$[\operatorname{Re}\leavevmode\nobreak\ >\leavevmode\nobreak\ 0]$, then for
every $k>\nu_{f}$ all translations $f_{\sigma}=f(\sigma+\cdot)\in
H_{\infty,k}^{\lambda}[\operatorname{Re}>0]\,,\,\sigma>0$.

More background. Mainly inspired by the classical monograph [13] of Hardy and
M. Riesz from 1915, the recent works [2], [6], [8], [9], [10], [17], and [20]
suggest a modern study of general Dirichlet series $D=\sum
a_{n}e^{-\lambda_{n}s}\,.$ Whereas the dominant tool of the early days of this
theory was complex analysis, the idea now is to implement modern techniques
like functional analysis, Fourier analysis, or abstract harmonic analysis on
compact abelian groups.

An important object of study are Hardy spaces
$\mathcal{H}_{p}(\lambda),\,1\leq p\leq\infty$, which may be defined as
follows: Let $G$ be a compact abelian group, and
$\beta:(\mathbb{R},+)\to\leavevmode\nobreak\ G$ a continuous homomorphism with
dense range. Then $(G,\beta)$ is said to be $\lambda$-Dirichlet group,
provided, for all $n\in\mathbb{N}$, there exists
$h_{\lambda_{n}}\in\widehat{G}$ such that $h_{\lambda_{n}}\leavevmode\nobreak\
\circ\leavevmode\nobreak\ \beta=\leavevmode\nobreak\ e^{-i\lambda_{n}}$. The
Banach space $\mathcal{H}_{p}(\lambda)$ then consists of all Dirichlet series
$D=\sum a_{n}e^{-\lambda_{n}s}$ for which there is $f\in L_{p}(G)$ such that
$\operatorname{supp}{\widehat{f}}\subset\\{h_{\lambda_{n}}\colon
n\in\mathbb{N}\\}$ and $\widehat{f}(h_{\lambda_{n}})=a_{n}$ for all
$n\in\mathbb{N}$, endowed with the norm $\|D\|_{p}=\|f\|_{p}$. It is important
to note that for every frequency $\lambda$ there always exist a
$\lambda$-Dirichlet group, and that $\mathcal{H}_{p}(\lambda)$ does not depend
on the choice of this group. For ordinary Dirichlet series these spaces have
been introduced in [1] and [14], and this has in fact caused a fruitful
renaissance of the analysis of such series. Within general Dirichlet series
various aspects of $\mathcal{H}_{p}(\lambda)$ have been studied in [2], [6],
[8], [9], [10], and [20].

The particular case $p=\infty$ is of special interest, since then
$\mathcal{H}_{\infty}(\lambda)$ may be described in terms of holomorphic
functions on the right half-plane, and in fact we are only interested trying
to develop this case further here.

Recall that $H_{\infty}^{\lambda}[\operatorname{Re}>0]$ (as defined in [9])
denotes the linear space of all holomorphic and bounded functions
$f\colon[\operatorname{Re}>0]\to\mathbb{C}$, which are uniformly almost
periodic function on all vertical lines $[\operatorname{Re}=\sigma]$ (or
equivalently, some line $[\operatorname{Re}=\sigma]$) with Bohr coefficients

$a_{x}(f)=\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^{T}f(\sigma+it)e^{-(\sigma+it)x}dt,\leavevmode\nobreak\
\leavevmode\nobreak\ x>0.$

supported in $\\{\lambda_{n}\mid n\in\mathbb{N}\\}$. Note that here the
integrals are independent of the choice of $\sigma$. Together with the sup
norm on the right half-plane, $H_{\infty}^{\lambda}[\operatorname{Re}>0]$
forms a Banach space, and by [9, Theorem 2.16] there is a coefficient
preserving isometric linear bijection identifying
$\mathcal{H}_{\infty}(\lambda)$ and
$H_{\infty}^{\lambda}[\operatorname{Re}>0]$,

(5)
$\mathcal{H}_{\infty}(\lambda)\,=\,H_{\infty}^{\lambda}[\operatorname{Re}>0]\,.$

The most fundamental theorem of the theory of almost periodic functions on
$\mathbb{R}$ is due to Bohr, and states that a function
$f:\leavevmode\nobreak\ \mathbb{R}\to\mathbb{C}$ is unifomly almost periodic
if and only if it is uniformly approximable by trigonometric polynomials of
the form $p(t)=\sum_{k=1}^{n}c_{k}e^{-i\alpha_{k}t}$, where the
$\alpha_{k}\in\mathbb{R}$ are finitely many frequencies. We refer to [3] for
all needed information on uniformly almost periodic functions.

For bounded, holomorphic functions on half-planes the following result from
[9, Corollary 2.19] (see also [11, Lemma 3.3] for a may be more elementary
proof) should be viewed as an analogue of Bohr’s approximation theorem: If
$f\in H^{\lambda}_{\infty}[\operatorname{Re}>0]$, then for every $k>0$ and
$s\in[\operatorname{Re}>0]$,

(6)
$f(s)=\lim_{x\to\infty}\sum_{\lambda_{n}<x}a_{\lambda_{n}}(f)e^{-\lambda_{n}s}(1-\frac{\lambda_{n}}{x})\,,$

and the convergence is even uniform on every half-plane
$[\operatorname{Re}>\sigma],\,\sigma>0$.

In Corollary 4.11 we characterize $H_{\infty}^{\lambda}[\operatorname{Re}>0]$
in terms of our new setting showing that the canonical embedding

(7)
$H_{\infty}^{\lambda}[\operatorname{Re}>0]=H_{\infty,0}^{\lambda}[\operatorname{Re}>0]\,$

in fact is isometric (obvious), onto and coefficient preserving.

But then in view of Bohr’s approximation theorem for uniformly almost periodic
functions on $\mathbb{R}$, and in view of the approximation theorem from (6)
for almost periodic holomorphic functions on $[\operatorname{Re}>0]$, the
following question arises:

Given $\ell>0$, to which extend is it possible to approximate any function
$f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ by $(\lambda,k)$-Riesz
means

$R_{x}^{\lambda,k}(D)(s)=\sum_{\lambda_{n}<x}a_{n}(f)e^{-\lambda_{n}s}(1-\frac{\lambda_{n}}{x})^{k}\,,\,\,x>0$

pointwise on $[\operatorname{Re}>0]$ or even in the norm of
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$? More precisely, is there
some $k>0$, and if yes for which set of $k$’s, is it possible to approximate
any $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ pointwise on
$[\operatorname{Re}>0]$ or in the norm of
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ by their $(\lambda,k)$-Riesz
means?

A first answer to this question is given by an important theorem of M. Riesz,
first published in [13, Theorem 41], which inspired the definition from (3) as
well as the whole article.

It basically shows that in view of (7) the validity of (6) exceeds the class
of functions in $H_{\infty}^{\lambda}[\operatorname{Re}>0]$ considerably, and
in our terminology it reads as follows:

######  Theorem 1.1.

Let $\lambda$ be any frequency and $\ell\geq 0$. Then every $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ is $(\lambda,k)$-Riesz
summable on $[\operatorname{Re}>0]$ for every $k>\ell$, i.e.

$f(s)=\lim_{x\to\infty}R_{x}^{\lambda,k}(f)(s),\leavevmode\nobreak\
\leavevmode\nobreak\ s\in[\operatorname{Re}>0].$

We reprove this result in Corollary 4.9, where we also carefully compare our
formulation with the original formulation given in [13, Theorem 41].

Main results. Theorem 1.1 is a pointwise approximation theorem of holomorphic
functions on half-planes by Riesz means, and we are going to extend various of
its different aspects – including equivalent reformulations and sufficient
conditions on $\lambda$ for the validity of Theorem 1.1 when $k=\ell>0$
(proved in Theorem 4.25). But in general, it fails for $k=\ell=0$; see the
forthcoming Theorems 4.23 and 4.24.

Since we recover this result of Riesz in terms of the Banach spaces
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$, we in Theorem 4.6 are able
to isolate the corresponding maximal inequality:

Given a frequency $\lambda$ and $k>\ell\geq 0$, there is a constant
$C=C(k,\ell,\lambda)$ such that for all $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$

$\sup_{x>0}\|R_{x}^{\lambda,k}(f)\|_{\infty,\ell}\leq C\|f\|_{\infty,\ell}\,.$

This then in Theorem 4.7 leads to a uniform approximation theorem of bounded
sets in $H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ under Riesz
summation:

For every bounded set $M\subset
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$, $k>\ell$ and
$\varepsilon,u>0$ there exists $x_{0}>0$ such that

$\sup_{x>x_{0}}\sup_{f\in
M}\|f(u+\boldsymbol{\cdot})-R_{x}^{\lambda,k}(f)(u+\boldsymbol{\cdot})\|_{\infty,\ell}\leq\varepsilon.$

Indeed, the preceding two facts form the main body of this article, since they
turn out to be the driving forces for most of the structure theory on the
Banach spaces $H_{\infty,\ell}^{\lambda}[\operatorname{Re}\leavevmode\nobreak\
>\leavevmode\nobreak\ 0]$ we intend to present.

This applies both, to the results on
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ that we have already
mentioned here (like e.g. the facts that these spaces are complete and
intimately linked with the notion of ’finite order’), but also to a few others
to come (like e.g. a Montel theorem for
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ from Theorem 4.15).

Studying problems for general Dirichlet series $\sum a_{n}e^{-\lambda_{n}s}$,
the very first orientation usually comes from

  * •

the frequency $\lambda=(n)$, which after the substitution $e^{-s}=z$ generates
power series $\sum a_{n}z^{n}$ ,

  * •

and the frequency $\lambda=(\log n)$ generating ordinary Dirichlet series
$\sum a_{n}n^{-s}$ .

In Proposition 4.3 we show that the scale
$H_{\infty,\ell}^{(n)}[\operatorname{Re}>0]\,,\,\ell\geq 0$ collapses:

$H_{\infty}^{(n)}[\operatorname{Re}>0]=H_{\infty,\ell}^{(n)}[\operatorname{Re}>0]\,.$

But, as shown in Proposition 4.4 the situation for the frequency
$\lambda=(\log n)$ is very different. For example $\eta\notin
H_{\infty,0}^{(\log n)}[\operatorname{Re}>0]$, but $\eta\notin
H_{\infty,\ell}^{(\log n)}[\operatorname{Re}>0]$ for all
$\ell\geq\frac{1}{2}$. In Proposition 4.4 we show that $f\in
H_{\infty,1}^{(\log n)}[\operatorname{Re}>0]$ if and only if the growth
condition (4) holds, and moreover there exists an ordinary Dirichlet series
which on $[\operatorname{Re}>0]$ is Cesàro summable with limit $f$.

Twofold interest. The proof of our main results and all its consequences for
the structure theory of our new scale of Banach spaces
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ is very much inspired by the
theory of Riesz summation from [13].

Hardy and Riesz write in the preface of their book ’…The task of condensing
any account of so extensive a theory into the compass of one of these tracts
has proved an exceedingly difficult one. Many important theorems are stated
without proof, and many details are left to the reader.’

In fact, at several occasions we even need improvements of the results from
[13], and this has prompted us to devote the first part of our article to a
detailed, differently organized, and in parts more modern presentation of
several key stones of the theory of Hardy and Riesz. We hope that apart from
our new results on Banach spaces of holomorphic functions of finite order
generated by Dirichlet series, this creates an additional benefit of our work.

Outlook. The classical Carleson-Hunt result implies that the Taylor series of
bounded holomorphic functions on the open unit disk converge almost everywhere
on the boundary. In contrast, there exists an ordinary Dirichlet series, which
on the open right half-plane converges to a bounded, holomorphic function –
but diverges at each point of the imaginary line, although its limit function
extends continuously to the closed right half plane. As a continuation of the
present article we in the forthcoming work [12] study Riesz summability of
functions in the spaces
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0],\,\ell\geq 0$, on the
imaginary axis. Again our inspiration comes from an important, surprisingly
deep result of M. Riesz published (without proof) in [13, Theorem 42].

Structure of the article. In the Sections 2 and 3 we develop parts of the
summation theory on Riesz means, which provides us with an important integral
formula of the limit function of $\lambda$-Dirichlet series under Riesz
summation, which then in a second step by ’Laplace inversion’ leads to the so-
called Perron formulas for the summatory function of these series. Then
Section 4 covers the main results of this article including the basic
properties of our new scale of Banach spaces
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0],\ell\geq 0$.

##  2\. Riesz summation revisted

As explained above, the aim of this first section is to give a modern self-
contained presentation of the most crucial ingredients of the theory of Riesz
summation of general Dirichlet series.

###  2.1. Ordinary summation

Every presentation of the classical convergence theory of general Dirichlet
series necessarily starts with the following theorem, and for the sake of
completeness we repeat the standard proof based on Abel smmation (see e.g.
[13, Theorem 1]).

######  Theorem 2.1.

Let $D=\sum a_{n}e^{-\lambda_{n}s}$ be a $\lambda$-Dirichlet series which
converges at $s_{0}\in[\operatorname{Re}\geq 0]$. Then $D$ converges uniformly
on each cone $|\arg(s-s_{0})|\leq\gamma<\frac{\pi}{2}$.

###### Proof.

Assume without loss of generality that $s_{0}=0$, and fix some
$\gamma<\frac{\pi}{2}$ and $0\neq s\in\mathbb{C}$ with $|\arg(s)|\leq\gamma$.
Then by standard Abel summation for all $m<n$

(8)
$\displaystyle\sum_{\nu=m}^{n}a_{n}e^{-\lambda_{n}s}=\sum_{\nu=m}^{n-1}\Big{(}\sum_{\ell=m}^{\nu}a_{\ell}\Big{)}(e^{-\lambda_{\nu}s}-e^{-\lambda_{\nu+1}s})+\Big{(}\sum_{\nu=m}^{n}a_{\nu}\Big{)}e^{-\lambda_{n}s}\,.$

But with $\operatorname{Re}\leavevmode\nobreak\ s=\sigma$

$|e^{-\lambda_{\nu}s}-e^{-\lambda_{\nu+1}s}|\leq\frac{|s|}{\operatorname{Re}s}(e^{-\lambda_{\nu}\sigma}-e^{-\lambda_{\nu+1}\sigma})\leq\sec{\gamma}\,(e^{-\lambda_{\nu}\sigma}-e^{-\lambda_{\nu+1}\sigma})\,,$

and hence, for $m$ and $n$ large, we have that

$\Big{|}\sum_{\nu=n}^{m}a_{n}e^{-\lambda_{n}s}\Big{|}\leq\varepsilon$

uniformly on the cone $|\arg(s)|\leq\gamma$ . ∎

Theorem 2.1 suggest the following basic definition. For each
$\lambda$-Dirichlet series $D=\sum a_{n}e^{-\lambda_{n}s}$ the number

$\sigma^{\lambda}_{c}(D)=\inf\big{\\{}\sigma\in\mathbb{R}\colon
D\,\,\,\text{converges at $\sigma$}\big{\\}}\in\mathbb{R}\cup\\{\pm\infty\\}$

is called abscissa of convergence of $D$. Obviously, $D$ converges on
$[\operatorname{Re}>\sigma^{\lambda}_{c}(D)]$ and diverges on
$[\operatorname{Re}<\sigma^{\lambda}_{c}(D)]$ .

The following integral formula describes the limit function $f$ of a
$\lambda$-Dirichlet series $D=\sum a_{n}e^{-\lambda_{n}s}$ on the half-plane
$[\operatorname{Re}>\sigma^{\lambda}_{c}(D)]$ of convergence in terms of the
Laplace transform of the so-called summatory function given by

$S^{\lambda}_{t}(D)(s)=\sum_{\lambda_{n}<t}a_{n}e^{-\lambda_{n}s}\,,\,\,\,\,t\geq
0\,.$

Our proof is inspired by Helson [17, (2.3)], and serves as a model for the
proof of the much more involved variant Theorem 2.9 on Riesz summation.

######  Theorem 2.2.

Let $D=\sum a_{n}e^{-\lambda_{n}s}$ be convergent at
$s_{0}\in[\operatorname{Re}\geq 0]$. Then $D$ converges on
$[\operatorname{Re}>\operatorname{Re}\leavevmode\nobreak\ s_{0}]$, and the
limit function

(9) $f\colon[\operatorname{Re}>\operatorname{Re}\leavevmode\nobreak\
s_{0}]\to\mathbb{C},\leavevmode\nobreak\ \leavevmode\nobreak\
s\mapsto\lim_{x\to\infty}\sum_{\lambda_{n}<x}a_{n}e^{-\lambda_{n}s}$

is holomorphic and satisfies for all
$s\in[\operatorname{Re}>\operatorname{Re}\leavevmode\nobreak\ s_{0}]$

$\frac{f(s)}{s}=\int_{0}^{\infty}e^{-st}\sum_{\lambda_{n}<t}a_{n}\,\,dt.$

Moreover, the convergence in (9) is uniform on each cone
$|\arg(s-s_{0})|\leq\gamma<\frac{\pi}{2}$.

Note that the preceding theorem shows that the function $s\mapsto f(s)/s$ is
nothing else than the Laplace transform of the summatory function $t\mapsto
S_{t}(D)(0)$.

We prepare the proof with two simple lemmas.

######  Lemma 2.3.

Let $D\in\mathcal{D}(\lambda)$. Then for all $s,w\in\mathbb{C}$ and all $x\geq
0$

$S^{\lambda}_{x}(D)(s+w)=S^{\lambda}_{x}(D)(w)e^{-sx}-\int_{0}^{x}S^{\lambda}_{t}(D)(w)se^{-st}dt.$

###### Proof.

Since $S^{\lambda}_{x}(D)(s+w)=S^{\lambda}_{x}(D_{w})(s)$, we concentrate on
the case $w=0$. Then

$\displaystyle
S^{\lambda}_{x}(D)(s)=\int_{0}^{x}e^{-ts}dS^{\lambda}_{t}(D)(0)=\int_{0}^{x}e^{-ts}\big{(}S^{\lambda}_{\bullet}(D)(0)\big{)}^{\prime}(t)dt\,,$

where the first integral is a Stieltjes integral. Obviously, summatory
functions are of bounded variation, hence by partial integration (see e.g.
Helson [16, Appendix])

$S^{\lambda}_{x}(D)(s)=e^{-xs}S^{\lambda}_{x}(D)(0)-\int_{0}^{x}se^{-ts}S^{\lambda}_{t}(D)(0)dt\,.\qed$

######  Lemma 2.4.

Let $D\in\mathcal{D}(\lambda)$, and $s_{0},s\in\mathbb{C}$. Then for all $x>0$

$|S_{x}^{\lambda}(D)(s)|\leq\frac{|s_{0}|}{\operatorname{Re}s_{0}}e^{x\operatorname{Re}s_{0}}\sup_{y\leq
x}|S_{y}^{\lambda}(s_{0}+s)|\,.$

###### Proof.

Since $S_{x}^{\lambda}(D)(s)=S_{x}^{\lambda}(D_{s})(0)$, we may assume that
$s=0$. By Lemma 2.3

(10)
$S_{x}^{\lambda}(D)(0)=S_{x}^{\lambda}(D)(-s_{0}+s_{0})=e^{s_{0}x}S^{\lambda}_{x}(D)(s_{0})-\int_{0}^{x}S^{\lambda}_{t}(D)(s_{0})(-s_{0})e^{s_{0}t}dt.$

For the first summand we get

$|e^{s_{0}x}S^{\lambda}_{x}(D)(s_{0})|\leq
e^{x\operatorname{Re}s_{0}}\sup_{y\leq
x}|S_{y}^{\lambda}(s_{0})|\leq\frac{|s_{0}|}{\operatorname{Re}s_{0}}e^{x\operatorname{Re}s_{0}}\sup_{y\leq
x}|S_{y}^{\lambda}(s_{0})|\,,$

and for the second

$\Big{|}\int_{0}^{x}S^{\lambda}_{t}(D)(s_{0})s_{0}e^{s_{0}t}dt\Big{|}\leq\frac{|s_{0}|}{\operatorname{Re}s_{0}}e^{x\operatorname{Re}s_{0}}\sup_{y\leq
x}|S_{y}^{\lambda}(s_{0})|\,.\qed$

###### Proof of Theorem 2.2.

Note first that by Lemma 2.4 (with $s=0$) we know that there is a constant
$C=C(s_{0})>0$ such that for all $x>0$

(11) $|S_{x}^{\lambda}(D)(0)|\leq Ce^{x\operatorname{Re}s_{0}}\,.$

Now fix some $s\in[\operatorname{Re}>\operatorname{Re}\leavevmode\nobreak\
s_{0}]$. We start using Lemma 2.3, and see that for all $x>0$

(12)
$S^{\lambda}_{x}(D)(s)=S^{\lambda}_{x}(D)(0)e^{-sx}-\int_{0}^{x}S^{\lambda}_{t}(D)(0)se^{-st}dt.$

By (11) for all $x>0$

$e^{-x\operatorname{Re}s}|S_{x}^{\lambda}(D)(0)|\leq
Ce^{x(\operatorname{Re}s_{0}-\operatorname{Re}s)}\,,$

which converges to $0$ whenever $x\to\infty$. It remains to show that

$\lim_{x\to\infty}\int_{0}^{x}S^{\lambda}_{t}(D)(0)e^{-st}dt=\int_{0}^{\infty}S^{\lambda}_{t}(D)(0)e^{-st}dt\,.$

But this follows from the dominated convergence theorem. Indeed, if we fix
$x>0$, then we conclude from (11) that for all $t>0$

$|S^{\lambda}_{t}(D)(0)e^{-st}\chi_{[0,x]}(t)|\leq
Ce^{t\operatorname{Re}s_{0}}e^{-t\operatorname{Re}s}\chi_{[0,x]}(t)\leq
Ce^{t(\operatorname{Re}s_{0}-\operatorname{Re}s)}\,.$

The last statement of the theorem is then a consequence of Theorem 2.1. ∎

The following formula, which we for historical reasons call Bohr-Cahen
formula, is taken from [13, Theorem 7].

######  Corollary 2.5.

Let $D\in\mathcal{D}(\lambda)$. Then

$\sigma_{c}^{\lambda}(D)\leq\limsup_{x\to\infty}\frac{\log(|S_{x}^{\lambda}(D)(0)|)}{x},$

with equality whenever $\sigma_{c}^{\lambda}(D)$ is non-negative.

###### Proof.

We denote the limes superior by $L$, assume that it is finite, and choose some
$\sigma_{0}>L$. Then there is a constant $C>0$ such that for all $x>0$ we have
that

$|S_{x}^{\lambda}(D)(0)|\leq Ce^{\sigma_{0}x}\,.$

Now, starting as in (12), we show that $D$ converges on
$[\operatorname{Re}>\sigma_{0}]$, so $\sigma^{\lambda}_{c}(D)\leq L$. To
finish, assume that $\sigma^{\lambda}_{c}(D)\geq 0$. Let $\varepsilon>0$, and
define $\sigma_{0}=\sigma^{\lambda}_{c}(D)+\varepsilon$. Then $D$ converges at
$\sigma_{0}$ (Theorem 2.1), and so by Lemma 2.4 for all $x>0$

$|S_{x}^{\lambda}(D)(0)|\leq C(\sigma_{0})e^{\sigma_{0}x}\,,$

implying $L\leq\sigma_{0}=\sigma^{\lambda}_{c}(D)+\varepsilon$ for all
$\varepsilon>0$. ∎

Given a frequency $\lambda$, we also need the abscissa of uniform convergence
of a $\lambda$-Dirichlet series $D=\sum a_{n}e^{-\lambda_{n}s}$ determined by
the number

$\sigma^{\lambda}_{u}(D)=\inf\big{\\{}\sigma\in\mathbb{R}\colon
D\,\,\,\text{converges uniformly on $[\operatorname{Re}>\sigma]$}\big{\\}}\,.$

Clearly, $D$ for each $\varepsilon>0$ converges uniformly on
$[\operatorname{Re}>\sigma^{\lambda}_{u}(D)+\varepsilon]$, whereas it does not
converge uniformly on
$[\operatorname{Re}>\sigma^{\lambda}_{u}(D)-\varepsilon]$ .

We finish with the Bohr-Cahen formula for this abscissa.

######  Corollary 2.6.

Let $D\in\mathcal{D}(\lambda)$. Then

$\sigma_{u}^{\lambda}(D)\leq\limsup_{x\to\infty}\frac{\log\big{(}\sup_{t\in\mathbb{R}}|S_{x}^{\lambda}(D)(it)|\big{)}}{x},$

with equality whenever $\sigma_{u}^{\lambda}(D)$ is non-negative.

In fact this is a relatively simple consequence of Corollary 2.5, once one
realizes that all we proved so far, after proper modifications, also holds for
$\lambda$-Dirichlet series $E=\sum x_{n}e^{-\lambda_{n}s}$ with coefficients
$x_{n}$ in a Banach space $X$.

###### Proof.

Define $X=H_{\infty}[\operatorname{Re}>0]$ (the Banach space of all
holomorphic and bounded functions on the right half-plane) and the $X$-valued
$\lambda$-Dirichlet series

$E=\sum(a_{n}e^{-\lambda_{n}s})e^{-\lambda_{n}w}$

with the coefficients $x_{n}=a_{n}e^{-\lambda_{n}s}\in X$. We easily see that

$\sigma_{u}^{\lambda}(D)=\sigma_{c}^{\lambda}(E)\,,$

and that by the Hahn-Banach theorem and the maximum modulus theorem we, for
all $x>0$, have

$\|S_{x}^{\lambda}(E)(0)\|_{X}=\sup_{\operatorname{Re}s>0}|S_{x}^{\lambda}(D)(s)|=\sup_{t\in\mathbb{R}}|S_{x}^{\lambda}(D)(it)|\,.$

Then by the vector-valued extension of Corollary 2.5 we immediately obtain

$\displaystyle\sigma_{u}^{\lambda}(D)=\sigma_{c}^{\lambda}(E)$
$\displaystyle\leq\limsup_{x\to\infty}\frac{\log(\|S_{x}^{\lambda}(D)(0)\|_{X})}{x}$
$\displaystyle=\limsup_{x\to\infty}\frac{\log\big{(}\sup_{t\in\mathbb{R}}|S_{x}^{\lambda}(D)(it)|\big{)}}{x},$

with equality whenever $\sigma_{u}^{\lambda}(D)=\sigma_{c}^{\lambda}(E)\geq
0$. ∎

###  2.2. Riesz means

The following definitions are again inspired by [13] (see also [9]). Let
$\lambda$ be a frequency, $k\geq 0$, and $\sum c_{n}$ a series in a normed
space $X$. Then we call the series $\sum c_{n}$ $(\lambda,k)$-Riesz summable
whenever the limit

$\lim_{x\to\infty}\sum_{\lambda_{n}<x}\Big{(}1-\frac{\lambda_{n}}{x}\Big{)}^{k}c_{n}$

exists, and the finite sums

$R_{x}^{\lambda,k}\Big{(}\sum
c_{n}\Big{)}:=\sum_{\lambda_{n}<x}\Big{(}1-\frac{\lambda_{n}}{x}\Big{)}^{k}c_{n}\,,\,\,\,x>0$

are said to be the $(\lambda,k)$-Riesz means of $\sum c_{n}$ of first kind.

Hardy and Riesz in [13, Theorem 16, 17, and 21] collected the following basic
properties of Riesz summability (in the order given).

######  Remark 2.7.

Let $\lambda$ be a frequency, $k\geq 0$, and $\sum c_{n}$ a series in a normed
space $X$.

  * (i)

If $\sum c_{n}$ is $(\lambda,k)$-Riesz summable, then it is
$(\lambda,\ell)$-Riesz summable for any $\ell\geq k$, and the associated
limits coincide. In particular, if $\sum c_{n}$ is summable (i.e. the series
converges), then for all $k>0$

$\sum_{n=1}^{\infty}c_{n}=\lim_{x\to\infty}\sum_{\lambda_{n}<x}\Big{(}1-\frac{\lambda_{n}}{x}\Big{)}^{k}c_{n}.$

  * (ii)

If $\sum c_{n}$ is $(e^{\lambda},k)$-Riesz summable, then $\sum c_{n}$ is
$(\lambda,k)$-Riesz summable, and the associated limits coincide.

  * (iii)

If $\sum c_{n}$ is $(\lambda,k)$-Riesz summable summable with limit $C$, then

$\lim_{N\to\infty}\Big{(}\frac{\lambda_{N+1}-\lambda_{N}}{\lambda_{N+1}}\Big{)}^{k}\Big{(}\sum_{n=1}^{N}c_{n}-C\Big{)}=0\,.$

Note that, if $\sum c_{n}$ is $(e^{\lambda},k)$-Riesz summable, then

$\lim_{x\to\infty}R_{x}^{e^{\lambda},k}\big{(}\sum
c_{n}\big{)}=\lim_{x\to\infty}R_{e^{x}}^{e^{\lambda},k}\big{(}\sum
c_{n}\big{)}=\lim_{x\to\infty}T_{x}^{\lambda,k}\big{(}\sum c_{n}\big{)}\,,$

where

(13) $T_{x}^{\lambda,k}\big{(}\sum
c_{n}\big{)}=\sum_{\lambda_{n}<x}\Big{(}1-\frac{e^{\lambda_{n}}}{e^{x}}\Big{)}^{k}c_{n}$

are the $(\lambda,k)$-Riesz means of second kind.

Let us again turn to $\lambda$-Dirichlet series $D=\sum
a_{n}e^{-\lambda_{n}s}$. For such series and $k\geq 0$, the Dirichlet
polynomials

(14)
$R_{x}^{\lambda,k}(D)(s)=\sum_{\lambda_{n}<x}a_{n}e^{-\lambda_{n}s}(1-\frac{\lambda_{n}}{x})^{k},\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ x>0,s\in\mathbb{C}$

are called $(\lambda,k)$-Riesz means of $D$, and $D$ is said to be
$(\lambda,k)$-Riesz summable at $s_{0}\in\mathbb{C}$, whenever the limit

$\lim_{x\to\infty}R_{x}^{\lambda,k}(D)(s_{0})$

exists.

A basic property states that the $(\lambda,k)$-Riesz summability of $D$ at
$s_{0}$ implies the $(\lambda,k+\varepsilon)$-Riesz summability of $D$ at
$s_{0}$ for every $\varepsilon>0$ (see Remark 2.7,(i) and alternatively
Theorem 2.9). As done for $k=0$ above, we define

$\sigma_{c}^{\lambda,k}(D)=\inf\big{\\{}\sigma\in\mathbb{R}\colon
D\,\,\,\text{$(\lambda,k)$-Riesz summable in $\sigma$}\big{\\}}\,,$

which characterizes the largest possible open half-plane on which $D$ is
pointwise $(\lambda,k)$-Riesz summable. Similarly, we put

$\sigma_{u}^{\lambda,k}(D)=\inf\big{\\{}\sigma\in\mathbb{R}\colon
D\,\,\,\text{uniformly $(\lambda,k)$-Riesz summable in $\sigma$}\big{\\}}\,,$

defining the largest possible open half-plane on which $D$ is uniformly
$(\lambda,k)$-Riesz summable.

Finally, we mention without proof the following important result from [13,
Theorem 30] comparing Riesz summability of first and second kind.

######  Theorem 2.8.

For every $\lambda$-Dirichlet series $D$ and every $k\geq 0$

$\sigma_{c}^{\lambda,k}(D)=\sigma_{c}^{e^{\lambda},k}(D)\,.$

.

###  2.3. Integral representation

The following analog of Theorem 2.2 rules the theory of Riesz summation of
general Dirichlet series. The result is taken from [13, Theorem 23]; in fact,
we here need an improvement of its original form which we repeat in Remark
2.11,(i).

Given $k\geq 0$ and a frequency $\lambda$, the theory of $(\lambda,k)$-Riesz
summabilty of $\lambda$-Dirichlet series $\sum a_{n}e^{-\lambda s}$ is ruled
by the so-called summatory functions of order $k$, which for $s\in\mathbb{C}$
and $x>0$ are defined by

(15)
$S_{x}^{\lambda,k}(D)(s)=\sum_{\lambda_{n}<x}a_{n}e^{-\lambda_{n}s}(x-\lambda_{n})^{k}=x^{k}R_{x}^{\lambda,k}(D)(s).$

######  Theorem 2.9.

Let $k\geq 0$, and $D=\sum a_{n}e^{-\lambda_{n}s}\in\mathcal{D}(\lambda)$ be
$(\lambda,k)$-summable at $s_{0}\in[\operatorname{Re}\geq 0]$. Then $D$ is
$(\lambda,k)$-summable on
$[\operatorname{Re}>\operatorname{Re}\leavevmode\nobreak\ s_{0}]$ and the
limit function

(16) $f\colon[\operatorname{Re}>\operatorname{Re}\leavevmode\nobreak\
s_{0}]\to\mathbb{C},\leavevmode\nobreak\ \leavevmode\nobreak\
s\mapsto\lim_{x\to\infty}\sum_{\lambda_{n}<x}a_{n}(1-\frac{\lambda_{n}}{x})^{k}e^{-\lambda_{n}s},$

is holomorphic and satisfies for all
$s\in[\operatorname{Re}>\operatorname{Re}\leavevmode\nobreak\ s_{0}]$

(17)
$\Gamma(1+k)\frac{f(s)}{s^{1+k}}=\int_{0}^{\infty}e^{-st}S^{\lambda,k}_{t}(D)(0)dt,$

where as usual $\Gamma$ denotes the Gamma function. Moreover, the convergence
in (16) is uniform on each cone $|\arg(s-s_{0})|\leq\gamma<\frac{\pi}{2}$.

The proof, although far more complex, is analog to that for the case $k=0$
stated in Theorem 2.2. The proof of the cases $k\in\mathbb{N}_{0}$ and $0<k<1$
follows the lines from [13]. Very much inspired by the ideas given there, the
proof for the general case is given in full detail in Section 3. Finally, in
the rest of this section we collect a few crucial consequences of the
preceding theorem.

######  Remark 2.10.

As above, we like to mention that the integral formula from (17) may be read
as follows: The function $s\mapsto\Gamma(1+k)\frac{f(s)}{s^{1+k}}$ is the
Laplace transform of $t\mapsto S^{\lambda,k}_{t}(D)(0)$, in short

$\mathcal{L}\big{(}t\mapsto
S^{\lambda,k}_{t}(D)(0)\big{)}(s)=\Gamma(1+k)\frac{f(s)}{s^{1+k}}.$

The case $f=1$ of (17) (assuming $\lambda_{1}=0$) reproves that

$\mathcal{L}(t^{k})(s)=\frac{\Gamma(1+k)}{s^{1+k}},\leavevmode\nobreak\
\leavevmode\nobreak\ s\in[\operatorname{Re}>0].$

The following remark collects two useful reformulations of Theorem 2.9 in
terms of the abscissa of $(\lambda,k)$-Riesz summability of $D=\sum
a_{n}e^{-\lambda_{n}s}$.

######  Remark 2.11.

Let $k\geq 0$, and $D=\sum a_{n}e^{-\lambda_{n}s}\in\mathcal{D}(\lambda)$.
Assume that $\sigma_{c}^{\lambda,k}(D)\in\mathbb{R}$, and let
$f:[\operatorname{Re}>\sigma_{c}^{\lambda,k}(D)]\to\mathbb{C}$ be the limit
function of $D$.

  * (i)

For every $s_{0}\in[\operatorname{Re}\geq\sigma_{c}^{\lambda,k}(D)]$ and
$s\in[\operatorname{Re}>0]$

$f(s+s_{0})\frac{\Gamma(1+k)}{s^{1+k}}=\int_{0}^{\infty}e^{-st}S^{\lambda,k}_{t}(D)(s_{0})dt.$

  * (ii)

For every $s_{0}\in[\operatorname{Re}>\sigma_{c}^{\lambda,k}(D)]$ and
$s\in[\operatorname{Re}>0]$

$f(s+s_{0})=f(s_{0})+\frac{s^{1+k}}{\Gamma(1+k)}\int_{0}^{\infty}e^{-st}\big{(}S_{t}^{\lambda,k}(D)(s_{0})-f(s_{0})\big{)}dt.$

###### Proof.

In order to prove $(i)$ assume first that
$\operatorname{Re}s_{0}>\sigma_{c}^{\lambda,k}(D)$, so $D$ is
$(\lambda,k)$-summable in $s_{0}$. Clearly, we obtain $(i)$, if we apply
Theorem 2.9 to the translation $D_{s_{0}}=\sum
a_{n}e^{-\lambda_{n}s_{0}}e^{-\lambda_{n}s}$. If
$\operatorname{Re}s_{0}=\sigma_{c}^{\lambda,k}(D)$, then take $s=\sigma+i\tau$
and $\varepsilon>0$ such that $2\varepsilon<\sigma$. Then we conclude from the
first case that

$f(s+s_{0}+\varepsilon)\frac{\Gamma(1+k)}{s^{1+k}}=\int_{0}^{\infty}e^{-st}S^{\lambda,k}_{t}(D)(s_{0}+\varepsilon)dt,$

and hence by continuity of $f$ the claim follows once we prove that

$\lim_{\varepsilon\to
0}\int_{0}^{\infty}e^{-st}S^{\lambda,k}_{t}(D)(s_{0}+\varepsilon)dt=\int_{0}^{\infty}e^{-st}S^{\lambda,k}_{t}(D)(s_{0})dt.$

Indeed, by (the forthcoming) Lemma 3.6 we have

$|R^{\lambda,k}_{t}(D)(s_{0}+\varepsilon)|\leq
e^{(\sigma-\varepsilon)t}|\sup_{0<y<t}|R^{\lambda,k}_{y}(D)(s_{0}+\varepsilon+(\sigma-\varepsilon))|\leq
C(\sigma)e^{\frac{\sigma}{2}t}\,,$

and hence the dominated convergence theorem does the job. Finally, in order to
see $(ii)$, apply $(i)$ to the function $g(s)=f(s)-f(s_{0})$, assuming without
loss of generality that $\lambda_{1}=0$. ∎

######  Corollary 2.12.

Let $k,D$ and $f$ be as in Theorem 2.9. Then

$\forall\varepsilon,\delta>0\,\,\,\exists
t_{0}>0\,\,\,\forall|t|\geq\tau_{0},\sigma>\delta+\sigma_{c}^{\lambda,k}(D)\colon|f(\sigma+it)|\leq\varepsilon|t|^{1+k}.$

In particular, the function $f$ for every $\delta>0$ has finite uniform order
$\leq k+1$ on $[\operatorname{Re}>\delta+\sigma_{c}^{\lambda,k}(D)]$.

The result is taken from [13, Theorem 38].

###### Proof.

We without loss of generality assume that $D$ is $(\lambda,k)$-Riesz summable
in $s=0$. We choose $\varepsilon,\delta>0$, an arbitrary
$0<\gamma\leq\frac{\pi}{2}$, and put

$M=\sup_{x>0}|R^{\lambda,k}_{x}(D)(0)|<\infty.$

By Theorem 2.9 we know that $f$ is bounded on the cone
$C=[|\arg\bullet|\leq\gamma$]. Hence it suffices to consider $f$ on the
complement of $C$ in $[\operatorname{Re}>0]$ intersected with
$[\operatorname{Re}>\delta]$, i.e. all
$s=\sigma+it\in[\operatorname{Re}>\delta]$ such that
$\frac{|s|}{t}\leq\frac{1}{\sin\alpha}$. Fix such $s$ and choose $x_{0}>0$
such that

$\int_{x_{0}}^{\infty}y^{k}e^{-\delta
y}dy<\frac{\Gamma(1+k)\,\sin\alpha}{M}\varepsilon.$

Applying again by Theorem 2.9, we see that

$\displaystyle f(s)=\frac{s^{k+1}}{\Gamma(1+k)}\bigg{(}\int_{0}^{x_{0}}$
$\displaystyle S_{y}^{\lambda,k}(D)(0)e^{-sy}dy$
$\displaystyle+\int_{x_{0}}^{\infty}S_{y}^{\lambda,k}(D)(0)e^{-sy}dy\bigg{)}=J_{1}(s)+J_{2}(s)\,.$

Then

$J_{2}(s)\leq|s|^{k+1}\frac{M}{\Gamma(1+k)}\int_{x_{0}}^{\infty}y^{k}e^{-\delta
y}dy\leq\varepsilon|t|^{k+1}\,.$

Moreover, by partial integration we have

$J_{1}(s)=\frac{s^{k}}{\Gamma(1+k)}\Big{(}-S_{x_{0}}^{\lambda,k}(D)(0)e^{-sx_{0}}+\int_{0}^{x_{0}}\big{(}S_{\bullet}^{\lambda,k}(D)(0)\big{)}^{\prime}(y)e^{-sy}dy\Big{)}\,,$

and then, since $|e^{-sx_{0}}|\leq 1$, there is $C=C(x_{0})>0$ such

$|J_{1}(s)|\leq|s|^{k}C\leq\frac{C}{\sin\alpha}|t|^{k}\leq\varepsilon|t|^{k+1}\,,$

whenever $\frac{C}{\varepsilon\sin\alpha}\leq|t|$. Combining the fact that $f$
is bounded on $[|\arg\bullet|\leq\gamma]$ with the two estimates for
$J_{1}(s)$ and $J_{2}(s)$, the conclusion follows. ∎

###  2.4. Perron formula

We prove Perron’s formula – an integral formula for the summatory function of
a $\lambda$-Dirichlet series in terms of its limit function. The result was
first presented in [13, Theorem 39]. In fact, it turns out that it follows by
Fourier inversion from the integral formula in Theorem 2.9. The original proof
of Perron’s formula is mainly a consequence of Cauchy’s integral theorem. Our
approach, deeply inspired by ideas used in [15] to cover the case $k=0$,
differs considerable.

######  Theorem 2.13.

For $k\geq 0$ let $D=\sum a_{n}e^{-\lambda_{n}s}$ be somewhere
$(\lambda,k)$-Riesz summable and
$f:[\operatorname{Re}>\sigma_{c}^{\lambda,k}(D)]\to\mathbb{C}$ its limit
function. Then for each $x>0$ and $c>\max\\{\sigma_{c}^{\lambda,k}(D),0\\}$ we
have

$S_{x}^{\lambda,k}(D)(0)=\frac{\Gamma(1+k)}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{f(s)}{s^{1+k}}e^{xs}ds.$

###### Proof.

Let us denote by $\mathcal{F}_{L_{1}(\mathbb{R})}$ the Fourier transform on
$L_{1}(\mathbb{R})$ with the Lebesgue measure as the normalized Haar measure.
Then for all $y\in\mathbb{R}$ by Theorem 2.9

$\displaystyle\frac{f(c+iy)}{(c+iy)^{1+k}}$
$\displaystyle=\frac{1}{\Gamma(1+k)}\int_{0}^{\infty}\big{[}S_{t}^{\lambda,k}(D)(0)e^{-ct}\big{]}e^{-iyt}dt$
$\displaystyle=\frac{1}{\Gamma(1+k)}\mathcal{F}_{L_{1}(\mathbb{R})}\big{(}t\mapsto
S_{t}^{\lambda,k}(D)(0)e^{-ct}\big{)}(y)\,.$

Since the function $g(t)=S_{t}^{\lambda,k}(D)(0)e^{-ct}$ satisfies a Lipschitz
condition at all $t\notin\\{\lambda_{n}\mid n\in\mathbb{N}\\}$, we by [15, p.
8] for all such $t$ obtain

$S_{t}^{\lambda,k}(D)(0)e^{-ct}=\lim_{A\to\infty}\frac{1}{2\pi}\int_{-A}^{A}\Gamma(1+k)\frac{f(c+iy)}{(c+iy)^{1+k}}e^{iyt}dy.$

Hence,

$\displaystyle S_{t}^{\lambda,k}(D)(0)$
$\displaystyle=\frac{\Gamma(1+k)}{2\pi}\lim_{A\to\infty}\int_{-A}^{A}\frac{f(c+iy)}{(c+iy)^{1+k}}e^{t(c+iy)}dy\,,$

which is the equality we intended to prove. ∎

It is important for our later purposes that the preceding integral formula for
the summatory function of a given $\lambda$-Dirichlet series $D$ extends
considerably, whenever the limit function of $f$ extends holomorphically to
the full right half-plane – still satisfying a ’finite order type-condition’
(a result sketched in [13, p.51]).

######  Corollary 2.14.

For $k\geq 0$ let $D=\sum a_{n}e^{-\lambda_{n}s}$ be $(\lambda,k)$-Riesz
summable for some $s_{0}\in[\operatorname{Re}>0]$. Assume that the limit
function $f$ of $D$ extends holomorphically to $[\operatorname{Re}>0]$ such
that

$\forall\varepsilon,\delta>0\,\,\,\exists t_{0}>0\,\,\,\forall|t|\geq
t_{0},\sigma>\delta\colon|f(\sigma+it)|\leq\varepsilon|\sigma+it|^{1+k}.$

Then for each $x>0$ and $c>0$

$S_{x}^{\lambda,k}(D)(0)=\frac{\Gamma(1+k)}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{f(s)}{s^{1+k}}e^{xs}ds.$

###### Proof.

By Theorem 2.13 we know that that for $\sigma>\operatorname{Re}s_{0}$

$S_{x}^{\lambda,k}(D)(0)=\frac{\Gamma(1+k)}{2\pi
i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{f(s)}{s^{1+k}}e^{xs}ds.$

We claim that for each $x>0$ and $c>0$

(18)
$\int_{c-i\infty}^{c+i\infty}\frac{f(s)}{s^{1+k}}e^{xs}ds=\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{f(s)}{s^{1+k}}e^{xs}ds.$

Indeed, an application of Cauchy’s integral theorem shows that (18) follows,
once we prove that

(19) $\lim_{T\to\infty}\int_{c}^{\sigma}\frac{f(y\pm iT)}{(y\pm
iT)^{1+k}}e^{x(y\pm iT)}dy=0.$

Indeed, for $\varepsilon>0$ and $\delta=c$, choose $t_{0}>0$ according to the
assumption. Then for all $T>t_{0}$

$\displaystyle\Big{|}\int_{c}^{\sigma}\frac{f(y\pm iT)}{(y\pm
iT)^{1+k}}e^{x(y\pm
iT)}dy\Big{|}\leq\varepsilon\int_{c}^{\sigma}e^{xy}dy=\varepsilon(e^{x\sigma}-e^{xc})x^{-1},$

which implies (19). ∎

Finally, we show that the coefficients of a $\lambda$-Dirichlet series are
uniquely determined by the values of its limit function on some abscissa -
another crucial point for our coming purposes.

######  Corollary 2.15.

Let $D=\sum a_{n}e^{-\lambda_{n}s}$ be $(\lambda,k)$-Riesz summable at
$s_{0}=\sigma_{0}+i\tau_{0}$ for some $k\geq 0$. If the limit function $f$ of
$D$ vanishes on some vertical line $[\operatorname{Re}=\sigma]$, where
$\sigma>\sigma_{0}$, then $a_{n}=0$ for all $n$.

###### Proof.

Let first $\lambda_{1}<x<\lambda_{2}$. Then by Theorem 2.13 and our assumption

$a_{1}e^{-\lambda_{1}s_{0}}(1-\frac{\lambda_{1}}{x})^{k}=S_{x}^{\lambda,k}(D)(0)=\frac{\Gamma(1+k)}{2\pi
i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{f(s)e^{xs}}{s^{1+k}}ds=0,$

which implies $a_{1}=0$. Proceeding successively by considering a sequence
$(x_{n})$ with $\lambda_{n}<x_{n}<\lambda_{n+1}$, we obtain that $a_{n}=0$ for
all $n$. ∎

###  2.5. Bohr-Cahen formula

The following formula for the abscissa of convergence is again due to Hardy
and Riesz [13, Theorem 31].

######  Theorem 2.16.

Let $D=\sum a_{n}e^{-\lambda_{n}s}$ and $k\geq 0$. Then

$\sigma_{c}^{\lambda,k}(D)\leq\limsup_{x\to\infty}\frac{\log(|R_{x}^{\lambda,k}(D)(0)|)}{x},$

with equality whenever $\sigma_{c}^{\lambda,k}(D)$ is non-negative.

The proof will be given in Section 3.4, mainly as a consequence of Theorem
2.9. Note that

$\limsup_{x\to\infty}\frac{\log(|R_{x}^{\lambda,k}(D)(0)|)}{x}=\limsup_{x\to\infty}\frac{\log(|S_{x}^{\lambda,k}(D)(0)|)}{x}.$

Recall that Theorem 2.16 for the special case $k=0$ was already presented in
Corollary 2.5, and that we in Corollary 2.6 formulated a uniform counterpart.

Again a simple analysis shows that the proof of Theorem 2.16 also works for
vector-valued Dirichlet series. Hence we exactly as in Section 2.1 obtain the
following uniform variant of Theorem 2.16 (note that for $0<k<1$ a direct, but
technically more involved, argument was given in [20, Lemma 3.8]).

######  Corollary 2.17.

Let $D=\sum a_{n}e^{-\lambda_{n}s}$ and $k\geq 0$. Then

$\sigma_{u}^{\lambda,k}(D)\leq\limsup_{x\to\infty}\frac{\log(\sup_{t\in\mathbb{R}}|R_{x}^{\lambda,k}(D)(it)|)}{x},$

with equality whenever $\sigma_{u}^{\lambda,k}(D)$ is non-negative.

##  3\. Summatory functions

This section is devoted to the proof of Theorem 2.9 and Theorem 2.16, where a
couple of technical lemmas are needed to properly prepare the proofs.

###  3.1. Integral forms

Given a frequency $\lambda$ and a $\lambda$ Dirichlet series $D$, we in (15)
defined the summatory function $S^{\lambda,k}_{x}(D)$. We start with the
following integral form of this function, which simple as it is, seems to be
the seed for most of the coming arguments.

######  Lemma 3.1.

Let $D\in\mathcal{D}(\lambda)$ and $k>0$. Then for all $s,w\in\mathbb{C}$ and
all $x\geq 0$

(20)
$S^{\lambda,k}_{x}(D)(s+w)=-\int_{0}^{x}S^{\lambda,0}_{t}(D)(s)\Big{(}e^{-w\,\bullet}(x-\,\bullet)^{k}\Big{)}^{\prime}(t)dt\,.$

In particular, for $w=0$ we have

(21)
$S^{\lambda,k}_{x}(D)(s)=k\int_{0}^{x}S^{\lambda,0}_{t}(D)(s)(x-t)^{k-1}dt\,,$

and for $w=0$ and $k=1$

(22) $\frac{d}{dt}S^{\lambda,1}_{t}(D)(s)_{|t=x}=S^{\lambda,0}_{x}(D)(s)\,.$

###### Proof.

It suffices to do the proof for $s=0$. As in the proof of Lemma 2.3 we write
$S^{\lambda,k}_{x}(D)(w)$ as a Stietjes integral,

$S^{\lambda,k}_{x}(D)(w)=\sum_{\lambda_{n}<x}a_{n}e^{-\lambda_{n}w}(x-\lambda_{n})^{k}=\int_{0}^{x}e^{-wt}(x-t)^{k}dS^{\lambda,0}_{t}(D)(0)\,,$

and then the conclusion again follows by partial integration. The remaining
assertions are then obvious consequences. ∎

Recall that, the first statement of Theorem 2.9 claims $(\lambda,k)$-Riesz
summability of the Dirichlet series on the half plane
$[\operatorname{Re}>\operatorname{Re}\leavevmode\nobreak\ s_{0}]$, given
$(\lambda,k)$-Riesz summability at $s_{0}$. Rougly speaking, the strategy for
proving this claim is to substitute the expression $S^{\lambda,0}_{t}(s)$ of
(20) by $S^{\lambda,k}_{t}(s)$. In other terms, we want to increase the order
of the summatory function of $f$ on the right hand side of (20) from zero to
$k$. As the following lemma shows, partial integration allows to increase the
order to some $m\in\mathbb{N}$ with $m<k\leq m+1$. Then, given
$k\notin\mathbb{N}$, we utilize the subsequent Lemma 3.3 to drop the order to
$k$ as desired.

Again using partial integration, we start to elaborate the preceding lemma.

######  Lemma 3.2.

Let $D\in\mathcal{D}(\lambda)$, $k>0$ and $m\in\mathbb{N}_{0}$ such that
$m<k\leq m+1$. Then for all $s,w\in\mathbb{C}$ and all $x>0$

$\displaystyle S_{x}^{\lambda,k}(D)(s+w)$
$\displaystyle=-e^{-wx}S_{x}^{\lambda,k}(D)(s)+\frac{(-1)^{m}}{(m+1)!}\int_{0}^{x}S_{t}^{\lambda,m+1}(D)(s)\partial^{m}h_{w}(t)dt,$

where
$h_{w}(t)=\partial^{2}\big{(}(e^{-w\bullet}-e^{-wx})(x-\bullet)^{k}\big{)}(t)$.

###### Proof.

We start proving the claim for $m=0$ and $0<k\leq 1$. By Lemma 2.3 we have

$\displaystyle S^{\lambda,k}_{x}(D)(s+w)$
$\displaystyle=-\int_{0}^{x}S_{t}^{\lambda,0}(D)(s)(e^{-w\bullet}(x-\bullet)^{k})^{\prime}(t)dt$
$\displaystyle=-ke^{-wx}\int_{0}^{x}S_{t}^{\lambda,0}(D)(s)(x-t)^{k-1}dt-\int_{0}^{x}S_{t}^{\lambda,0}(D)(s)\big{(}(e^{-w\bullet}-e^{-wx})(x-\bullet)^{k}\big{)}^{\prime}(t)dt.$

By (21) we know that

$ke^{-wx}\int_{0}^{x}S_{t}^{\lambda,0}(D)(s)(x-t)^{k-1}dt=e^{-wx}S_{x}^{\lambda,k}(D)(s)\,,$

and hence by (22) and partial integration we eventually obtain

$S_{x}^{\lambda,k}(D)(s+w)=-e^{-wx}S_{x}^{\lambda,k}(D)(s)+\int_{0}^{x}S_{t}^{\lambda,1}(D)(s)\big{(}(e^{-w\bullet}-e^{-wx})(x-\bullet)^{k}\big{)}^{\prime\prime}(t)dt,$

which proves the claim for $0<k\leq 1$ with $m=0$. If $k>1$ and $m\leq k<m+1$,
then we take into account that

$\frac{d}{dt}S_{t}^{\lambda,q+1}(D)(s)=-(q+1)S_{t}^{\lambda,q}(D)(s),\leavevmode\nobreak\
\leavevmode\nobreak\ q\in\mathbb{N}_{0},$

and apply $m$-times partial integration, which leads to our claim. ∎

For the next lemma we need basics on the classical Gamma function

$\Gamma(z)=\int_{0}^{\infty}x^{z-1}e^{-x}dx,\,\,\,\leavevmode\nobreak\
\operatorname{Re}\leavevmode\nobreak\ z>0,$

as well as the Beta function

$B(p,q)=\int_{0}^{1}y^{p-1}(1-y)^{q-1}dy,\,\,\,\leavevmode\nobreak\
p,q\in[\operatorname{Re}>0].$

Therefore, we recall the important property that

(23) $B(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$

which shows that for $x>0$

(24)
$\int_{0}^{x}y^{p}(x-y)^{q-p-1}dy=x^{q}\frac{\Gamma(p+1)\Gamma(q-p)}{\Gamma(q+1)}.$

Indeed, by substitution we have

$\displaystyle\int_{0}^{x}y^{p}(x-y)^{q-p-1}dy$
$\displaystyle=x^{q}\int_{0}^{x}x^{-1}(\frac{y}{x})^{p}(1-\frac{y}{x})^{q-p-1}dy$
$\displaystyle=x^{q}\int_{0}^{1}\alpha^{p}(1-\alpha)^{q-p-1}d\alpha=x^{q}\frac{\Gamma(p+1)\Gamma(q-p)}{\Gamma(q+1)}.$

Moreover, recall that for every $z\in[\operatorname{Re}>0]$

(25) $\Gamma(1+z)=\Gamma(z)z.$

The next lemma presents a device for passing to lower orders of the summatory
function. Observe that a combination of Lemma 3.2 and the following, reveals a
reformulation of Lemma 3.2 entirely written with respect to the order $k$.
This is one of the key steps in the proof of Theorem 2.9.

######  Lemma 3.3.

Given $D\in\mathcal{D}(\lambda)$ and $k,\mu>0$, for all $x>0$

$S_{x}^{\lambda,k+\mu}(D)(0)=\frac{\Gamma(k+\mu+1)}{\Gamma(k+1)\Gamma(\mu)}\int_{0}^{x}S_{u}^{k,\lambda}(D)(0)(x-u)^{\mu-1}du\,.$

This formula is taken from [13, Lemma 6], and following the idea of this
manuscript we for the sake of completeness repeat its proof.

###### Proof.

We start with the integral on the left side, and use (21), Fubini’s theorem
and (24) to obtain

$\displaystyle\int_{0}^{x}S_{u}^{\lambda,k}(D)(0)(x-u)^{\mu-1}du$
$\displaystyle=\int_{0}^{x}k\int_{t}^{x}S_{t}^{\lambda,0}(D)(0)(u-t)^{k-1}(x-u)^{\mu-1}dudt$
$\displaystyle=k\int_{0}^{x}S_{t}^{\lambda,0}(D)(0)\int_{t}^{x}(u-t)^{k-1}(x-u)^{\mu-1}dudt$
$\displaystyle=\frac{\Gamma(k)\Gamma(\mu)}{\Gamma(k+\mu)}k\int_{0}^{x}S_{t}^{\lambda,0}(D)(0)(x-t)^{\mu+k-1}dt=\frac{\Gamma(k)\Gamma(\mu)}{\Gamma(k+\mu)}\frac{k}{\mu+k}S_{t}^{\lambda,\mu+k}(D)(0)\,,$

the conclusion. ∎

Recall that Theorem 2.9 in (17) provides us with an integral representation
for $(\lambda,k)$-Riesz limits. We need another important tool to prove it.

######  Lemma 3.4.

Let $D\in\mathcal{D}(\lambda)$ and $0<p<q$. Then for all $w,s\in\mathbb{C}$
and all $x>0$

$\int_{0}^{\infty}S_{x}^{\lambda,q}(D)(s)w^{q+1}e^{-wx}dx=\frac{\Gamma(q+1)}{\Gamma(p+1)}\int_{0}^{\infty}S_{u}^{\lambda,p}(D)(s)w^{p+1}e^{-wu}du\,.$

###### Proof.

Again we assume that $s=0$. We first use Lemma 3.3 (applied to $k=p$ and
$\mu=q-p$) to show that

$\displaystyle\int_{0}^{\infty}S_{x}^{\lambda,q}(D)(0)e^{-wx}dx$
$\displaystyle=\frac{\Gamma(q+1)}{\Gamma(p+1)\Gamma(q-p)}\int_{0}^{\infty}\int_{0}^{x}S_{u}^{\lambda,p}(D)(0)(x-u)^{q-p-1}du\,\,e^{-wx}dx$
$\displaystyle=\frac{\Gamma(q+1)}{\Gamma(p+1)\Gamma(q-p)}\int_{0}^{\infty}S_{u}^{\lambda,p}(D)(0)\int_{u}^{\infty}(x-u)^{q-p-1}e^{-wx}dx\,\,du\,.$

Now we claim that

$\int_{u}^{\infty}(x-u)^{q-p-1}e^{-wx}dx=w^{p-q}e^{-wu}\Gamma(q-p).$

Indeed, since both sides define holomorphic functions in $w$, it suffices to
check for all $w=\sigma>0$. In this case we obtain by substitution that

$\displaystyle\int_{u}^{\infty}(x-u)^{q-p-1}e^{-wx}dx=\int_{0}^{\infty}\alpha^{q-p-1}e^{-\sigma(\alpha+u)}d\alpha$
$\displaystyle=e^{-\sigma
u}\sigma^{p-q}\int_{0}^{\infty}\beta^{q-p-1}e^{-\beta}d\beta=e^{-\sigma
u}\sigma^{p-q}\Gamma(q-p).$

Altogether

$w^{q+1}\int_{0}^{\infty}S_{x}^{\lambda,q}(D)(0)e^{-wx}dx=w^{p+1}\frac{\Gamma(q+1)}{\Gamma(p+1)}\int_{0}^{\infty}S_{u}^{\lambda,q}(D)(0)e^{-wu}du\,.\qed$

###  3.2. Estimates

We prove two estimates for Riesz means, which seem to be of independent
interest. Among others, they play an important role in the proof of (the
forthcoming) Theorem 2.16.

######  Lemma 3.5.

Given $q>p\geq 0$, $s\in\mathbb{C}$ and $D\in\mathcal{D}(\lambda)$, we for all
$x>0$ have

$|R_{x}^{\lambda,q}(D)(s)|\leq\sup_{0<y<x}|R_{y}^{\lambda,p}(D)(s)|.$

###### Proof.

From Lemma 3.3 (with $k=p$ and $\mu=q-p$) we get

$R_{x}^{\lambda,q}(D)(s)=x^{-q}\frac{\Gamma(q+1)}{\Gamma(1+p)\Gamma(q-p)}\int_{0}^{x}R^{\lambda,p}_{y}(D)(s)y^{p}(x-y)^{q-p-1}dy\,,$

and then we conclude from (24) that

$|R_{x}^{\lambda,q}(D)(s)|\leq\sup_{0<y<x}|R_{x}^{\lambda,p}(D)(s)|.\qed$

######  Lemma 3.6.

Let $D\in\mathcal{D}(\lambda)$ and $k>0$. Then for
$s_{0}=\sigma+it,s\in\mathbb{C}$ and $x>0$

$|R^{\lambda,k}_{x}(D)(s)|\leq C(s_{0},k)e^{\sigma
x}\sup_{0<y<x}|R^{\lambda,k}_{y}(D)(s+s_{0})|\,.$

###### Proof.

Using translation, we may assume without loss of generality that $s=0$. Then
by Lemma 3.2 the choice $s=s_{0}$ and $w=-s_{0}$ leads to

$\displaystyle R_{x}^{\lambda,k}(D)(0)$
$\displaystyle=R_{x}^{\lambda,k}(D)(s_{0}-s_{0})$
$\displaystyle=e^{s_{0}x}R_{x}^{\lambda,k}(D)(s_{0})+\frac{(-1)^{m}}{(m+1)!}x^{-k}\int_{0}^{x}S^{\lambda,m+1}_{t}(D)(s_{0})\partial^{m}h_{-s_{0}}(t)dt.$

By Lemma 3.5 we have

$|R^{\lambda,m+1}_{t}(s_{0})|\leq\sup_{0<y<t}|R^{\lambda,k}_{y}(s_{0})|\,,$

and so

$\displaystyle|x^{-k}\int_{0}^{x}S^{\lambda,m+1}_{t}(s_{0})\partial^{m}h_{-s_{0}}(t)dt|\leq\sup_{0<y<x}|R^{\lambda,k}_{y}(s_{0})|x^{-k}\int_{0}^{x}t^{m+1}\partial^{m}h_{-s_{0}}(t)dt,$

where

$\displaystyle\partial^{m}h_{-\sigma}(t)=s_{0}^{m+2}e^{s_{0}t}(x-t)^{k}$
$\displaystyle+\sum_{j=1}^{m+1}\sigma^{j}e^{s_{0}t}c_{k,j}(x-\cdot)^{k-(m+2-j)}(t)$
$\displaystyle+(e^{s_{0}t}-e^{s_{0}x})c_{k,m+1}(x-t)^{k-(m+2)}.$

We now take into account that

$|e^{s_{0}t}-e^{s_{0}x}|\leq|e^{\sigma t}-e^{\sigma
x}|\frac{|s_{0}|}{\sigma}\leq(x-t)\sigma e^{\sigma
t}\frac{|s_{0}|}{\sigma}=(x-t)e^{\sigma t}|s_{0}|\,,$

and hence

$\displaystyle|x^{-k}\int_{0}^{x}t^{m+1}\partial^{m}h_{-s_{0}}(t)dt|\leq
C(s_{0},k)e^{\sigma
x}x^{-k}\sum_{j=1}^{m+2}\int_{0}^{x}t^{m+1}(x-t)^{k-(m+2-j)}dt$
$\displaystyle\leq C(s_{0},k)e^{\sigma
x}\sum_{j=1}^{m+2}x^{m+2+j}\int_{0}^{x}t^{m+1}(1-\frac{t}{x})^{k-(m+2-j)}dt\,.$

Finally, we calculate

$\displaystyle\int_{0}^{x}t^{m+1}(1-\frac{t}{x})^{k-(m+2-j)}dt$
$\displaystyle=x^{m+2}\int_{0}^{x}(\frac{t}{x})^{m+1}(1-\frac{t}{x})^{k-(m+2-j)}x^{-1}dt$
$\displaystyle=x^{m+2}\int_{0}^{1}y^{m+1}(1-y)^{k-(m+2-j)}dy\,,$

and by (24) with $x=1$

$\displaystyle\int_{0}^{1}y^{m+1}(1-y)^{k-(m+2-j)}dy=\frac{\Gamma(m+2)\Gamma(k-(m+2-j)+1)}{\Gamma(k+j+1)}.$

This finishes the proof. ∎

###  3.3. Two crucial lemmas

The following two lemmas finish our preparation of the proofs of Theorem 2.9
and Theorem 2.16 (the only proofs we’ve left out so far). We believe that they
in a sense form the ’technical heart’ of much of the theory of Riesz summation
as created in [13]. In fact, they will be used at various places of this work
(see e.g. the proof of one of our main contributions, Theorem 4.7).

Basically, we execute the indicated strategy for proving Theorem 2.9 from the
previous section, by carefully calculating the function $h_{\omega}$ from
Lemma 3.2.

######  Lemma 3.7.

Let $D\in\mathcal{D}(\lambda)$, $k>0$ and $m\in\mathbb{N}_{0}$ such that
$m<k\leq m+1$. Then there is a constant $C=C(m,k)$ such that for all
$w=\sigma+i\tau\in[\operatorname{Re}>0]$, all $s\in\mathbb{C}$ and all $x>0$

$\displaystyle\Big{|}\Gamma(m+2)^{-1}$
$\displaystyle\int_{0}^{x}S_{t}^{\lambda,m+1}(D)(s)w^{m+2}(x-t)^{k}e^{-wt}dt-
S_{x}^{\lambda,k}(D)(s+w)\Big{|}$ $\displaystyle\leq e^{-\sigma
x}|S_{x}^{\lambda,k}(D)(s)|+C\sum_{j=1}^{m+1}|w|^{j}\int_{0}^{x}|S_{y}^{\lambda,k}(D)(s)|e^{-\sigma
y}(x-y)^{j-1}dy.$

In particular, there is a constant $L=L(m,k)$ such that for all
$w=\sigma+it\in[\operatorname{Re}\leavevmode\nobreak\ >\leavevmode\nobreak\
0]$ with $|\arg(w)|\leq\gamma$ and all $x>0$

$\displaystyle\Big{|}\Gamma(m+2)^{-1}\int_{0}^{x}S_{t}^{\lambda,m+1}(D)(0)w^{m+2}(x-t)^{k}e^{-wt}dt-
S_{x}^{(\lambda,k)}(D)(w)\Big{|}$ $\displaystyle\leq e^{-\sigma
x}|S_{x}^{\lambda,k}(D)(0)|+L\sum_{j=1}^{m+1}|\sec(\gamma)|^{j}\int_{0}^{x}|S_{y}^{\lambda,k}(D)(0)|y^{-j}(x-y)^{j-1}dy.$

###### Proof.

Let $k>0$ and $m\in\mathbb{N}_{0}$ such that $m<k\leq m+1$. Then by Lemma 3.2
for every $s,w\in\mathbb{C}$

(26)
$S_{x}^{\lambda,k}(D)(s+w)=-e^{-wx}S_{x}^{\lambda,k}(D)(s)+\frac{(-1)^{m}}{(m+1)!}\int_{0}^{x}S_{t}^{\lambda,m+1}(D)(s)\partial^{m}h_{w}(t)dt,$

where
$h_{w}(t)=\partial^{2}\big{(}(e^{-w\cdot}-e^{-wx})(x-\cdot)^{k}\big{)}(t)$. We
calculate

$\displaystyle\partial^{m}h_{w}(t)=$ $\displaystyle(-w)^{m+2}e^{-wt}(x-t)^{k}$
$\displaystyle+\sum_{j=1}^{m+1}(-w)^{j}e^{-wt}c_{k,j}(x-t)^{k-(m+2-j)}$
$\displaystyle+(e^{-wt}-e^{-wx})c_{k,m+1}(x-t)^{k-(m+2)}=:(g_{1}+g_{2}+g_{3})(t).$

The aim is to replace in (26) the summatory function of order $m+1$ by the
summatory function of order $k$. If $k=m+1$, then we are fine – but if
$k<m+1$, then we need Lemma 3.3 (with the choices $\mu=m-k+1$ und $\kappa=k$):

$S_{t}^{m+1,\lambda}(D)(s)=\frac{\Gamma(2+m)}{\Gamma(1+k)\Gamma(1+m-k)}\int_{0}^{t}S_{y}^{\lambda,k}(D)(s)(t-y)^{m-k}dy.$

For the rest of the proof we assume that $k<m+1$; in fact, it is easier to
handle the case $k=m+1$ following basically the same lines. We write

$C(m,k)=\frac{\Gamma(2+m)}{\Gamma(1+k)\Gamma(1+m-k)}\,,$

apply Fubini’s theorem and obtain for $i=2,3$

$\displaystyle C(m,k)^{-1}$
$\displaystyle\int_{0}^{x}S_{t}^{\lambda,m+1}(D)(s)g_{i}(t)dt=\int_{0}^{x}\int_{0}^{t}S_{y}^{\lambda,k}(D)(s)(t-y)^{m-k}dy\leavevmode\nobreak\
g_{i}(t)dt$
$\displaystyle=\int_{0}^{\infty}S_{y}^{\lambda,k}(D)(s)\int_{0}^{x}\chi_{(0,t)}(y)(t-y)^{m-k}g_{i}(t)\leavevmode\nobreak\
dt\leavevmode\nobreak\ dy$
$\displaystyle=\int_{0}^{x}S_{y}^{\lambda,k}(D)(s)\int_{y}^{x}(t-y)^{m-k}g_{i}(t)\leavevmode\nobreak\
dt\leavevmode\nobreak\ dy.$

Consequently, using (26), we eventually obtain

$\displaystyle
S_{x}^{\lambda,k}(D)(s+w)-\Gamma(m+2)^{-1}\int_{0}^{x}S_{t}^{\lambda,m+1}(D)(s)w^{m+2}e^{-wt}(x-t)^{k}dt$
$\displaystyle=-e^{-wx}S_{x}^{\lambda,k}(f)(s)$
$\displaystyle\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\
+(-1)^{m}C(m,k)\int_{0}^{x}S_{y}^{\lambda,k}(D)(s)\int_{y}^{x}(t-y)^{m-k}(g_{2}+g_{3})(t)\leavevmode\nobreak\
dt\leavevmode\nobreak\ dy.$

Then, in view of our claim, it remains to control the second summand. To do
this we for arbitrary $c,d\in\mathbb{R}$ compute

$\displaystyle\int_{y}^{x}e^{-wt}(t-y)^{c}(x-t)^{d}dt$
$\displaystyle=\int_{0}^{x-y}e^{-w(y+\alpha)}\alpha^{c}(x-y-\alpha)^{d}d\alpha$
$\displaystyle=(x-y)^{c+d+1}\int_{0}^{1}e^{-w(y+\beta(x-y))}\beta^{c}(1-\beta)^{d}d\beta$
$\displaystyle=(x-y)^{c+d+1}e^{-wy}\int_{0}^{1}e^{-w\beta(x-y)}\beta^{c}(1-\beta)^{d}d\beta.$

Then for $i=2$ we obtain with $w=\sigma+i\tau\in[\operatorname{Re}>0]$ (see
also (24) with $x=1$) we get

$\displaystyle|\int_{y}^{x}(t-y)^{m-k}g_{2}(t)\leavevmode\nobreak\ dt|$
$\displaystyle=|\sum_{j=1}^{m+1}(-1)^{j}w^{j}c_{k,j}(x-y)^{j-1}e^{-wy}\int_{0}^{1}e^{-w\beta(x-y)}\beta^{m-k}(1-\beta)^{k-(m+2-j)}d\beta|$
$\displaystyle\leq C_{1}(m,k)\sum_{j=1}^{m+1}|w|^{j}(x-y)^{j-1}e^{-\sigma
y}\int_{0}^{1}\beta^{m-k}(1-\beta)^{k-(m+2-j)}d\beta$
$\displaystyle=C_{1}(m,k)\frac{\Gamma(1+m-k)\Gamma(k-m-1+j)}{\Gamma(j)}\sum_{j=1}^{m+1}|w|^{j}(x-y)^{j-1}e^{-\sigma
y}\,.$

Finally, we claim that

$\displaystyle|\int_{y}^{x}(t-y)^{m-k}g_{3}(t)\leavevmode\nobreak\ dt|\leq
C(m,k)|w|e^{-\sigma y}\,.$

Indeed, by the mean value theorem we for $0\leq t\leq x$ and
$w\in[\operatorname{Re}>0]$ have

$|e^{-wt}-e^{-wx}|\leq|e^{-\sigma t}-e^{-\sigma
x}|\frac{|w|}{\sigma}\leq(x-t)\sigma e^{-\sigma
t}\frac{|w|}{\sigma}=(x-t)e^{-\sigma t}|w|\,.$

Then (again using (24) with $x=1$)

$\displaystyle|\int_{y}^{x}(t-y)^{m-k}g_{3}(t)\leavevmode\nobreak\ dt|$
$\displaystyle=|c_{k,m+1}\int_{y}^{x}(t-y)^{m-k}(e^{-wt}-e^{-wx})(x-t)^{k-(m+2)}dt|$
$\displaystyle\leq c_{k,m+1}\int_{y}^{x}(t-y)^{m-k}|e^{-\sigma t}-e^{-\sigma
x}|\frac{|w|}{\sigma}(x-t)^{k-(m+2)}dt$ $\displaystyle\leq e^{-\sigma
y}|w|c_{k,m+1}\int_{y}^{x}(t-y)^{m-k}(x-t)^{k-m-1}dt$
$\displaystyle=e^{-\sigma
y}|w|c_{k,m+1}(x-y)^{m-k+(k-m-1)+1}\int_{0}^{1}\beta^{m-k}(1-\beta)^{k-m-1}d\beta$
$\displaystyle=C_{2}(m,k)\frac{\Gamma(1+m-k)\Gamma(k-m)}{\Gamma(1)}e^{-\sigma
y}|w|.$

This completes the argument for the first part. For the second, take $s=0$,
and observe that $e^{\sigma y}\leq\frac{(\sigma y)^{j}}{j!}$ and
$\frac{|w|}{\sigma}\leq\sec(\gamma)$ . ∎

Roughly speaking, the integral representation (17) (from Theorem 2.9) for
$(\lambda,k)$-Riesz limits is determined by the first summand of the left-hand
side of the first inequality of Lemma 3.7, tending $x\to\infty$. The next
lemma proves existence of this limit – given a growth condition on the
summatory function.

######  Lemma 3.8.

Given $D\in\mathcal{D}(\lambda)$, let $s\in\mathbb{C}$ and $k\geq 0$, $q>0$.
Then for all $w\in\mathbb{C}$ and $\varepsilon>0$ such that
$0<\varepsilon<\operatorname{Re}w$ and

$|S_{t}^{\lambda,q}(D)(s)|\leq C(q,\varepsilon)t^{q}e^{\varepsilon
t},\,\,t>0\,,$

we have

(27)
$\lim_{x\to\infty}\int_{0}^{x}S_{t}^{\lambda,q}(D)(s)\Big{(}1-\frac{t}{x}\Big{)}^{k}w^{q+1}e^{-wt}dt=\int_{0}^{\infty}S_{y}^{\lambda,q}(D)(s)w^{q+1}e^{-wy}dy\,.$

Moreover, the convergence in (27) is uniform on each cone
$|\arg(w-s)|\leq\gamma<\frac{\pi}{2}$, whenever

$|S_{t}^{\lambda,q}(D)(s)|\leq C(q)t^{q}\,,\,\,t>0\,.$

###### Proof.

We assume without loss of generality that $s=0$. In order to prove (27), we
fix $w$ and $\varepsilon$, and put $\sigma=\operatorname{Re}w-\varepsilon$.
Then for all $x,t$

$\displaystyle\Big{|}S_{t}^{\lambda,q}(D)(0)\Big{(}1-\frac{t}{x}\Big{)}^{k}w^{q+1}e^{-wt}\chi_{[0,x]}(t)\Big{|}\leq
C(q,\varepsilon)|w|^{q+1}e^{-\sigma t}t^{q}\,,$

and hence (27) is an immediate consequence of the dominated convergence
theorem. The proof of the ’moreover’ is similar: For each $w$ with
$|\arg(w)|\leq\gamma<\frac{\pi}{2}$ we abbreviate
$\sigma_{w}=\operatorname{Re}w$. Then for all $x,t,w$

$\displaystyle\Big{|}S_{t}^{\lambda,q}(D)(0)\Big{(}1-\frac{t}{x}\Big{)}^{k}w^{q+1}e^{-wt}\chi_{[0,x]}(t)\Big{|}\leq
C(q)\sec{(\gamma)}t^{q}\sigma_{w}^{q+1}e^{-\sigma_{w}t}\,,$

and hence by the substitution $u=\sigma_{w}t$ we get for all $x,u,w$

$\displaystyle\Big{|}S_{\frac{u}{\sigma_{w}}}^{\lambda,q}(D)(0)\Big{(}1-\frac{\frac{u}{\sigma_{w}}}{x}\Big{)}^{k}w^{q+1}e^{-w\frac{u}{\sigma_{w}}}\chi_{[0,x]}(\frac{u}{\sigma_{w}})\Big{|}\frac{1}{\sigma_{w}}\leq
C(q)\sec{(\gamma)}u^{q}e^{-u}\,.$

Hence again by the dominated convergence theorem (uniform variant)

$\int_{0}^{x\sigma_{w}}S_{\frac{u}{\sigma_{w}}}^{\lambda,q}(D)(0)\Big{(}1-\frac{\frac{u}{\sigma_{w}}}{x}\Big{)}^{k}w^{q+1}e^{-w\frac{u}{\sigma_{w}}}\frac{1}{\sigma_{w}}du\to\int_{0}^{\infty}S_{\frac{u}{\sigma_{w}}}^{\lambda,q}(D)(0)w^{q+1}e^{-w\frac{u}{\sigma_{w}}}\frac{1}{\sigma_{w}}du$

uniformly in $w$ as $x\to\infty$. Substituting back, finishes the proof. ∎

###  3.4. Proofs of Theorem 2.9 and Theorem 2.16

###### Proof of Theorem 2.9.

Let $k>0$ and $m\in\mathbb{N}_{0}$ such that $m<k\leq m+1$. If $D$ is
$(\lambda,k)$-summable at $s_{0}=\sigma_{0}+i\tau_{0}$, then for some constant
$C=C(s_{0},k)$ we by Lemma 3.6 have that for all $x>0$

(28) $|R_{x}^{\lambda,k}(D)(0)|\leq Ce^{\sigma_{0}x}.$

We fix $s=\sigma+i\tau\in[\operatorname{Re}>\sigma_{0}]$, and deduce from
Lemma 3.8 and Lemma 3.4 that

$\displaystyle\lim_{x\to\infty}\frac{x^{-k}}{\Gamma(m+2)}\int_{0}^{x}S_{t}^{\lambda,m+1}(D)(0)s^{m+2}(x-t)^{k}e^{-st}dt$
$\displaystyle=s^{m+2}s^{k-(m+1)}\Gamma(m+2)^{-1}\frac{\Gamma(m+2)}{\Gamma(k+1)}\int_{0}^{\infty}S_{y}^{\lambda,k}(D)(0)e^{-st}dy$
$\displaystyle=\Gamma(1+k)^{-1}s^{k+1}\int_{0}^{\infty}S_{y}^{\lambda,k}(D)(0)e^{-st}dy.$

According to Lemma 3.7 we have to show that

$e^{-\sigma
x}|R_{x}^{(\lambda,k)}(D)(0)|+C\sum_{j=1}^{m+1}x^{-k}|s|^{j}\int_{0}^{x}|S_{y}^{\lambda,k}(D)(0)|e^{-\sigma
y}(x-y)^{j-1}dy$

vanishes as $x\to\infty$. Observe that (28) implies the claim for the first
summand. Writing $\varepsilon=\sigma-\sigma_{0}>0$ we use again (28) to see
that for $j=1,\ldots,m+1$

$\displaystyle x^{-k}\int_{0}^{x}|S_{y}^{\lambda,k}(D)(0)|e^{-\sigma
y}(x-y)^{j-1}dy$ $\displaystyle\leq Cx^{-k}\int_{0}^{x}e^{-\varepsilon
y}y^{k}(x-y)^{j-1}dy$ $\displaystyle\leq
Cx^{m-k}\int_{0}^{\infty}e^{-\varepsilon y}y^{k}dy,$

which vanishes as $x\to\infty$, since $k>m$. It remains to prove the ’moreover
part’. We may assume without loss of generality that $\lambda_{1}=0$,
$s_{0}=0$ and $\lim_{x\to\infty}R_{x}^{\lambda,k}(D)(0)=0$; if this particular
case is settled and $s_{0}$ is arbitrary, then we consider the modified
Dirichlet series $\sum b_{n}e^{-s_{0}\lambda_{n}}e^{-s\lambda_{n}}$ with
$b_{1}=a_{1}-\lim_{x\to\infty}R_{x}^{\lambda,k}(D)(s_{0})$ and $b_{n}=a_{n}$,
$n\geq 2$, which then converges uniformly on each cone
$|\arg(s)|\leq\gamma<\frac{\pi}{2}$, and so implies the conclusion. So let
$s_{0}=0$ and according to Lemma 3.7 and Lemma 3.8, we have to check that
uniformly for all $s\in[\operatorname{Re}>0]$ with
$\arg(s)\leq\gamma<\frac{\pi}{2}$

(29)
$\lim_{x\to\infty}\sum_{j=1}^{m+1}|\sec(\gamma)|^{j}x^{-k}\int_{0}^{x}|S^{\lambda,k}_{y}(D)(0)|y^{-j}(x-y)^{j-1}dy=0.$

Therefore, choose $\varepsilon>0$ and $x_{0}>1$ such that
$|R_{x}^{\lambda,k}(D)(0)|\leq\varepsilon$ for all $x>x_{0}$. We split the
integral

$\displaystyle x^{-k}\int_{0}^{x}|S^{\lambda,k}_{y}(D)(0)|y^{-j}(x-y)^{j-1}dy$
$\displaystyle=x^{-k}\int_{0}^{x_{0}}|S^{\lambda,k}_{y}(D)(0)|y^{-j}(x-y)^{j-1}dy+x^{-k}\int_{x_{0}}^{x}|S^{\lambda,k}_{y}(D)(0)|y^{-j}(x-y)^{j-1}dy.$

Then note that for $1\leq j\leq m+1$ and $x\geq 2x_{0}$

$\displaystyle
x^{-k}\int_{0}^{x_{0}}|S^{\lambda,k}_{y}(D)(0)|y^{-j}(x-y)^{j-1}dy\leq
C(D,x_{0})x^{-k}\int_{0}^{x_{0}}y^{k-j}(x-y)^{j-1}dy$ $\displaystyle\leq
C(D,x_{0})x^{m-k}\int_{0}^{x_{0}}y^{k-j}dy\leq
C(D,x_{0})x^{m-k}x_{0}^{k+1-j},$

which vanishes as $x\to\infty$, since $k>m$, and moreover for $x\geq 2x_{0}$

$\displaystyle
x^{-k}\int_{x_{0}}^{x}|S_{y}^{\lambda,k}(D)(0)|y^{-j}(x-y)^{j-1}dy\leq
x^{-k}\varepsilon\int_{x_{0}}^{x}y^{k-j}(x-y)^{j-1}dy$
$\displaystyle\leq\varepsilon
x^{-k}x^{k-j}\int_{x_{0}}^{x}(x-y)^{j-1}dy=\varepsilon
x^{-j}j^{-1}(x-x_{0})^{j}\leq\varepsilon(1-\frac{x_{0}}{x})^{j}\leq\varepsilon.$

This finishes the proof of (29). ∎

###### Proof of Theorem 2.16.

Assume that
$\sigma_{0}>\limsup_{x\to\infty}x^{-1}\log(|R_{x}^{\lambda,k}(D)(0)|)=L.$ Then
(28) is satisfied and from the previous proof of Theorem 2.9 we see that $D$
is $(\lambda,k)$-summable on $[\operatorname{Re}>\sigma_{0}]$. Now assume that
$\sigma^{\lambda,k}_{c}(D)\geq 0$ and let $\varepsilon>0$. We define
$\sigma_{0}=\sigma^{\lambda,k}_{c}(D)+\varepsilon$. Then $D$ is
$(\lambda,k)$-summable at $\sigma_{0}$ and so by Lemma 3.6 we have

$|R_{x}^{\lambda,k}(D)(0)|\leq
C(\sigma_{0})e^{\sigma_{0}x},\leavevmode\nobreak\ x>0,$

and so $L\leq\sigma_{0}=\sigma^{\lambda,k}_{c}(D)+\varepsilon$ for all
$\varepsilon>0$. Hence $L=\sigma^{\lambda,k}_{c}(D)$, whenever
$\sigma_{c}^{\lambda,k}(D)\geq 0.$ ∎

##  4\. A new scale of Banach spaces

We start repeating some of our basic definitions from the introduction. Given
a frequency $\lambda=(\lambda_{n})$, we call a $\lambda$-Dirichlet series
$D=\sum a_{n}e^{-\lambda_{n}s}$ a $\lambda$-Riesz germ of the holomorphic
function $f:\leavevmode\nobreak\ [\operatorname{Re}>0]\to\mathbb{C}$, whenever
$\sigma_{c}^{\lambda,m}(D)<\infty$ for some $m\geq 0$ and $f$ is the
holomorphic extension of the limit function of $D$ to all of
$[\operatorname{Re}>0]$.

Two remarks are in order.

######  Remark 4.1.

Let $f:\leavevmode\nobreak\ [\operatorname{Re}>0]\to\mathbb{C}$ be a
holomorphic function generated by the $\lambda$-Riesz germ $D=\sum
a_{n}e^{-\lambda_{n}s}$. Then $D$ by Corollary 2.15 is unique, and as a
consequence we may assign to every such $f$ the unique sequence
$(a_{n}(f))_{n}=(a_{n})_{n}$, which we (like for Dirichlet series) again call
the ’sequence of Bohr coefficients of $f$’.

######  Remark 4.2.

Let $f:\leavevmode\nobreak\ [\operatorname{Re}>0]\to\mathbb{C}$ be
holomorphic. Then $f$ by Theorem 2.8 has a $\lambda$-Riesz germ if and only if
$f$ on some half-plane and for some $m$ equals the limit function of some
$\lambda$-Dirichlet series under $(e^{\lambda},m)$-Riesz summation. This is an
immediate consequence of Theorem 2.8.

Moreover, given a holomorphic function $f:\leavevmode\nobreak\
[\operatorname{Re}>0]\to\mathbb{C}$ generated by the $\lambda$-Riesz germ $D$,
the $x$th Riesz mean of order $k\geq 0$ of $f$ in $s\in\mathbb{C}$ is given by

$R_{x}^{\lambda,k}(f)(s)=\sum_{\lambda<x}a_{n}(f)e^{-\lambda_{n}s}\Big{(}1-\frac{\lambda_{n}}{x}\Big{)}^{k}\,,$

and its summatory function in $s\in\mathbb{C}$ by

$S_{x}^{\lambda,k}(f)(s)=\sum_{\lambda<x}a_{n}(f)e^{-\lambda_{n}s}(x-\lambda_{n})^{k}\,,\,\,x>0\,.$

As explained above the spaces

$H_{\infty,\ell}^{\lambda}[\operatorname{Re}\leavevmode\nobreak\
>\leavevmode\nobreak\ 0],\,\ell\geq 0,$

accumulate all holomorphic functions $f:\leavevmode\nobreak\
[\operatorname{Re}>0]\to\mathbb{C}$, which are generated by a $\lambda$-Riesz
germ and satisfy the growth condition

(30)
$\|f\|_{\infty,\ell}=\sup_{\operatorname{Re}s>0}\frac{|f(s)|}{(1+|s|)^{\ell}}<\infty\,.$

For each such $f$

$\sup_{\operatorname{Re}s>0}\frac{|f(s)|}{(1+|s|)^{\ell}}\leq\sup_{\operatorname{Re}s>0}\frac{|f(s)|}{|(1+s)^{\ell}|}\leq
2^{\ell}\sup_{\operatorname{Re}s>0}\frac{|f(s)|}{(1+|s|)^{\ell}}\,,$

since $|1+s|\leq 1+|s|\leq 2|1+s|$ for each $s\in[\operatorname{Re}>0]$.
Obviously, the pairs

$(H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0],\|\cdot\|_{\infty,\ell})$

form an increasing scale of normed spaces. That all these spaces in fact are
complete, so Banach spaces, is a non-trivial fact which is given in Theorem
4.16.

Finally, we remark that a holomorphic function satisfying the growth condition
(30), by Remark 4.2 belongs to
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ if and only if there is a
$\lambda$-Dirichlet series $D=\sum a_{n}e^{-\lambda_{n}s}$ such that for some
$m,\sigma\geq 0$ and for every $s\in[\operatorname{Re}>0]$

(31)
$f(s)=\lim_{x\to\infty}R_{x}^{e^{\lambda},k}(D)=\lim_{x\to\infty}\sum_{\lambda_{n}<x}a_{n}e^{-\lambda_{n}s}(1-e^{\lambda_{n}-x})^{k}.$

###  4.1. Ordinary and power case

We start considering the power case $\lambda=(n)$. As already announced in the
introduction the scale
$H_{\infty,\ell}^{(n)}[\operatorname{Re}>0]\,,\,\ell\geq 0,$ collapses to the
case $\ell=0$.

######  Proposition 4.3.

For every $\ell\geq 0$

$H_{\infty}^{(n)}[\operatorname{Re}>0]=H_{\infty,\ell}^{(n)}[\operatorname{Re}>0]\,.$

###### Proof.

Take $f\in H_{\infty,\ell}^{(n)}[\operatorname{Re}>0]$, and let $D=\sum
a_{n}e^{-ns}$ be its associated Dirichlet series. From Corollary 4.9 (proved
below) and (the easy part ${}^{\prime}\geq^{\prime}$ of) Theorem 2.8 we know
that $D$ is $((e^{n}),\ell+1)$-Riesz summmable on $[\operatorname{Re}>0]$ with
limit $f$. Hence Remark 2.7, (iii) (see also [9, Proposition 1.2, (3)]) shows
that $D$ converges on $[\operatorname{Re}>0]$ to $f$. Then it is well-known
that $D$ converges uniformly on $[\operatorname{Re}>\varepsilon]$ for every
$\varepsilon>0$ (indeed, for the frequency $\lambda=(n)$ the abscissas of
convergence and absolutely convergence coincide), so in particular $f$ is
bounded on $[\operatorname{Re}>1]$. Moreover,
$f(\sigma+\boldsymbol{\cdot}):\mathbb{R}\to\mathbb{C}$ is $2\pi$-periodic for
every $\sigma>0$, which implies that

$\displaystyle\sup_{\begin{subarray}{c}0<\sigma<2\\\
t\in\mathbb{R}\end{subarray}}|f(\sigma+it)|$
$\displaystyle=\sup_{\begin{subarray}{c}0<\sigma<2\\\
t\in[0,2\pi]\end{subarray}}|f(\sigma+it)|$
$\displaystyle\leq\|f\|_{\infty,\ell}\sup_{\begin{subarray}{c}0<\sigma<2\\\
t\in[0,2\pi]\end{subarray}}(1+|\sigma+it|)^{\ell}\leq\|f\|_{\infty,\ell}(1+\sqrt{4+4\pi^{2}})^{\ell}\,.$

All in all we conclude that $f\in H_{\infty}^{(n)}[\operatorname{Re}>0]$ . ∎

Next we consider the frequency $\lambda=(\log n)$, which generates ordinary
Dirichlet series $\sum a_{n}n^{-s}$. Looking at (31) (with $x=\log y$), we see
that a holomorphic function $f\colon[\operatorname{Re}>0]\to\mathbb{C}$
belongs to the Banach space $\mathcal{H}^{(\log
n)}_{\infty,1}[\operatorname{Re}>0]$ if only if (4) holds, and there exists an
ordinary Dirichlet series $D=\sum a_{n}n^{-s}$ such that for some
$m,\sigma\geq 0$

$f(s)=\lim_{y\to\infty}\sum_{n<y}a_{n}n^{-s}(1-\frac{n}{y})^{m},\leavevmode\nobreak\
\leavevmode\nobreak\ s\in[\operatorname{Re}>\sigma].$

Actually, we will see in Corollary 4.25 that Theorem 4.10 in the special case
$\lambda=(\log n)$ is valid for $k=\ell$, and therefore in the particular case
$\ell=1$

$f(s)=\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\sum_{k=1}^{n}a_{k}k^{-s},\leavevmode\nobreak\
\leavevmode\nobreak\ s\in[\operatorname{Re}>0],$

so that eventually we arrive at the following result.

######  Proposition 4.4.

Let $f:[\operatorname{Re}>0]\to\mathbb{C}$ be holomorphic. Then $f\in
H_{\infty,1}^{(\log n)}$ if and only if (4) holds, and there exists an
ordinary Dirichlet series which on $[\operatorname{Re}>0]$ is Cesàro summable
with limit $f$.

Note that the space $H^{(\log n)}_{\infty,\ell}[\operatorname{Re}>0]$ is
strictly larger than $H^{(\log n)}_{\infty}[\operatorname{Re}>0]$. To see an
example, we again look at the eta-function $\eta$ (defined in the
introduction), for which

$\ell>\dfrac{1}{2}\,\,\,\,\,\Rightarrow\,\,\,\,\,\eta\in H^{(\log
n)}_{\infty,\ell}[\operatorname{Re}>0]\,\,\,\,\,\Rightarrow\,\,\,\,\,\ell\geq\dfrac{1}{2}\,.$

In particular, $\eta\notin\mathcal{H}_{\infty}((\log n))$. Alternatively, this
may also be seen as a consequence of Bohr’s inequality (see [21, Theorem
4.4.1, p. 116]): Assuming $\eta\in\mathcal{H}_{\infty}((\log n))$ implies that
$N=\sum_{n=1}^{N}|(-1)^{p_{n}}|\leq\|D\|_{\infty}<\infty$ for all $N$, which
clearly is a contradiction.

This example in particular shows that, although we in general have

$\mathcal{H}^{\lambda}_{\infty,\ell_{1}}[\operatorname{Re}>0]\subset\mathcal{H}^{\lambda}_{\infty,\ell_{2}}[\operatorname{Re}>0]\,\,\,\text{
for all $0\leq\ell_{1}<\ell_{2}$ }\,,$

this inclusion for certain $\lambda$’s may be strict whereas for other
$\lambda$’s it actually may be an equality (of sets).

###  4.2. Perron formula – a variant

We now prove an important variant of Perron’s formula from Theorem 2.13 for
the new scale of Banach spaces.

######  Theorem 4.5.

Let $f\in H^{\lambda}_{\infty,\ell}[\operatorname{Re}>0]$ and $k>\ell\geq 0$.
Then for all $s_{0}\in[\operatorname{Re}\geq 0]$, $x>0$ and $c>0$

$S_{x}^{\lambda,k}(f)(s_{0})=\frac{\Gamma(1+k)}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{f(s+s_{0})}{s^{1+k}}e^{xs}ds.$

###### Proof.

After translation we may assume that $s_{0}=0.$ By assumption the Dirichlet
series $D=\sum a_{n}e^{-\lambda_{n}s}$ is $(\lambda,m)$-Riesz summable at some
$s_{1}\in[\operatorname{Re}>0]$ for some $m\geq 0$. Hence $D$ is
$(\lambda,m+k)$-Riesz summable at $s_{1}$ and for every $\varepsilon,\sigma>0$
and $|t|\geq t_{0}>0$

$\frac{|f(\sigma+it)|}{|\sigma+it|^{1+m+k}}\leq\|f\|_{\infty,\ell}|\sigma+it|^{-(1+m+k-\ell)}\leq
t_{0}^{-(1+m+k-\ell)}\|f\|_{\infty,\ell}\leq\varepsilon$

for $t_{0}$ large enough. Corollary 2.14 then implies that for all $c,x>0$

$\sum_{\lambda_{n}<x}a_{n}(x-\lambda_{n})^{m+k}=\frac{\Gamma(1+m+k)}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{f(s)e^{xs}}{s^{1+m+k}}ds.$

Now we differentiate $m$-times and obtain

$\frac{d^{m}}{dx}\int_{c-i\infty}^{c+i\infty}\frac{f(s)e^{xs}}{s^{1+k+m}}ds=\int_{c-i\infty}^{c+i\infty}\frac{f(s)e^{xs}}{s^{1+k}}ds$

and

$\frac{d^{m}}{dx}\sum_{\lambda_{n}<x}a_{n}(x-\lambda_{n})^{m+k}=\prod_{j=0}^{m-1}(m+k-j)\sum_{\lambda_{n}<x}a_{n}(x-\lambda_{n})^{k}.$

Note that the differentiation under the integral is legit, since for every
$j=0,\ldots,m-1$

$\int_{c-i\infty}^{c+i\infty}|\frac{f(s)e^{xs}}{s^{1+k+j}}|ds\leq
e^{xc}\|f\|_{\infty,\ell}\int_{c-i\infty}^{c+i\infty}|s|^{-(1+k+j-\ell)}ds<\infty.$

Additionally by the functional equation of the Gamma function we have

$\Gamma(1+k)\Gamma(1+m+k)=\prod_{j=0}^{m-1}(m+k-j),$

so that altogether we obtain

$S_{x}^{\lambda,k}(f)(0)=\sum_{\lambda_{n}<x}a_{n}(x-\lambda_{n})^{k}=\frac{\Gamma(1+k)}{2\pi
i}\int_{c-i\infty}^{c+i\infty}\frac{f(s)e^{xs}}{s^{1+k}}ds.\qed$

###  4.3. A maximal inequality for Riesz means

The following maximal inequality is going to be our major tool to perform a
substantial structure theory of the spaces
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. It shows that the linear
projections

$R_{x}^{\lambda,k}:H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]\to
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]\,,\,\,\,\,f\to
R_{x}^{\lambda,k}f,\leavevmode\nobreak\ x>0,$

are uniformly bounded.

######  Theorem 4.6.

Let $k>\ell\geq 0$ and $\lambda=(\lambda_{n})$ a frequency. Then there is a
constant $C=C(k,\ell,\lambda)>0$ such that for all $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$

$\sup_{x>0}\|R_{x}^{\lambda,k}(f)\|_{\infty,\ell}\leq C\|f\|_{\infty,\ell}\,.$

In particular, for every $\sigma\geq 0$ there is a constant
$C=C(\ell,k,\lambda,\sigma)>0$ such that for every $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ and $t\in\mathbb{R}$

$\sup_{x>0}|R_{x}^{\lambda,k}(f)(\sigma+it)|\leq
C(1+|t|)^{\ell}\|f\|_{\infty,\ell}\,.$

###### Proof.

The first estimate is trivial, and for the second assume first that
$\lambda_{1}>0$. Let $s=\sigma+it\in[\operatorname{Re}>0]$ and
$y\geq\lambda_{1}$. Then by Theorem 4.5 with $c=y^{-1}$

$\displaystyle|R_{y}^{(\lambda,k)}(D)(s)|\leq\frac{e\Gamma(1+k)}{2\pi}y^{-k}\int_{\mathbb{R}}\frac{|f(s+\frac{1}{y}+ix)|}{|\frac{1}{y}+ix|^{1+k}}dx\,,$

and

$\displaystyle\|f\|_{\infty,\ell}^{-1}y^{-k}\int_{\mathbb{R}}\frac{|f(s+\frac{1}{y}+ix)|}{|\frac{1}{y}+ix|^{1+k}}dx\leq\int_{\mathbb{R}}\frac{y|1+s+\frac{1}{y}+ix|^{\ell}}{|1+ixy|^{1+k}}dx$
$\displaystyle\leq
2^{\max\\{0,\ell-1\\}}\int_{\mathbb{R}}\frac{y(|1+s|^{\ell}+|\frac{1}{y}+ix|^{\ell})}{|1+ixy|^{1+k}}dx$
$\displaystyle\leq
2^{\max\\{0,\ell-1\\}}\Big{(}(1+|s|)^{\ell}\int_{\mathbb{R}}\frac{y}{|1+ixy|^{1+k}}dx+y^{-\ell}\int_{\mathbb{R}}\frac{y}{|1+ixy|^{1+k-\ell}}dx\Big{)}$
$\displaystyle\leq
2^{\max\\{0,\ell-1\\}}\Big{(}(1+|s|)^{\ell}\int_{\mathbb{R}}\frac{1}{|1+iv|^{1+k}}dv+\lambda_{1}^{-\ell}\int_{\mathbb{R}}\frac{1}{|1+iv|^{1+k-\ell}}dv\Big{)}$
$\displaystyle\leq
2^{\max\\{0,\ell-1\\}}\Big{(}(1+|s|)^{\ell}\big{(}1+\frac{1}{k}\big{)}+\lambda_{1}^{-\ell}\big{(}1+\frac{1}{k-\ell}\big{)}\Big{)}$
$\displaystyle\leq
2^{\max\\{0,\ell-1\\}}\Big{(}\big{(}1+\frac{1}{k}\big{)}+\lambda_{1}^{-\ell}\big{(}1+\frac{1}{k-\ell}\big{)}\Big{)}(1+|s|)^{\ell}\,.$

Assume now that $\lambda_{1}=0$ and choose $0<y<\lambda_{2}$. Then by the
previous calculation

$|a_{0}|\leq\|f\|_{\infty,\ell}2^{\max\\{0,\ell-1\\}}\Big{(}\big{(}1+\frac{1}{k}\big{)}+y^{-\ell}\big{(}1+\frac{1}{k-\ell}\big{)}\Big{)}(1+|s|)^{\ell}.$

Tending $y\to\lambda_{2}$ we obtain

$|a_{0}|\leq C(\lambda,k,\ell)(1+|s|)^{\ell}\|f\|_{\infty,\ell}.$

The ’in particular’ part is then a straight forward consequence, since

$1+|s|\leq(1+\sigma)\big{(}1+\frac{|t|}{1+\sigma}\big{)}\leq(1+\sigma)(1+|t|)\,.\qed$

###  4.4. Uniform Riesz approximation

Next we give the main application of the preceding maximal theorem. We prove
that for $k>\ell$ all functions in a bounded subset of
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$, after translation by some
arbitrary $u>0$, are uniformly $(\lambda,k)$-Riesz approximable in the norm of
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$.

######  Theorem 4.7.

Let $\lambda=(\lambda_{n})$ be a frequency and $k>\ell\geq 0$. Then for every
bounded set $M\subset H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ and
every choice of $\varepsilon,u>0$ there is $x_{0}>0$ such that

$\sup_{x>x_{0}}\sup_{f\in
M}\|f(u+\boldsymbol{\cdot})-R_{x}^{\lambda,k}(f)(u+\boldsymbol{\cdot})\|_{\infty,\ell}\leq\varepsilon.$

###### Proof.

Let $m\in N_{0}$ such that $m<k\leq m+1$, and $C=\sup_{f\in
M}\|f\|_{\infty,\ell}<\infty$. Fix $f\in M$ and $u>0$. Then by Lemma 3.7 and
Theorem 4.6 we for every $s\in[\operatorname{Re}>0]$ have

$\displaystyle\Big{|}\frac{x^{-k}}{\Gamma(m+2)}\int_{0}^{x}\frac{S_{t}^{\lambda,m+1}(f)(s)}{(1+|s|)^{\ell}}u^{m+2}(x-t)^{k}e^{-ut}dt-\frac{R_{x}^{\lambda,k}(f)(s+u)}{(1+|s|)^{\ell}}\Big{|}$
$\displaystyle\leq
e^{-ux}\frac{|R_{x}^{\lambda,k}(f)(s)|}{(1+|s|)^{\ell}}+C_{1}(m,k)x^{-k}\int_{0}^{x}\frac{|S_{y}^{\lambda,k}(f)(s)|}{(1+|s|)^{\ell}}\sum_{j=1}^{m+1}u^{j}e^{-uy}(x-y)^{j-1}dy$
$\displaystyle\leq
e^{-ux}C+C_{2}(m,k,u)C\sum_{j=1}^{m+1}x^{-k}\int_{0}^{x}y^{k}(x-y)^{j-1}e^{-uy}dy,$

where

$\displaystyle
x^{-k}\int_{0}^{x}y^{k}(x-y)^{j-1}e^{-uy}dy=x^{j}\int_{0}^{x}x^{-1}(\frac{y}{x})^{k}(1-\frac{y}{x})^{j-1}e^{-uy}dy$
$\displaystyle=x^{j}e^{-ux}\int_{0}^{1}\alpha^{k}(1-\alpha)^{j-1}d\alpha=x^{j}e^{-ux}\frac{\Gamma(k+1)\Gamma(j)}{\Gamma(k+1-j)}.$

Hence for $x\geq 1$ and $f\in M$

$\displaystyle\sup_{s\in[\operatorname{Re}>0]}\Big{|}\frac{x^{-k}}{\Gamma(m+2)}\int_{0}^{x}\frac{S_{t}^{\lambda,m+1}(f)(s)}{(1+|s|)^{\ell}}u^{m+2}(x-t)^{k}e^{-ut}dt-\frac{R_{x}^{\lambda,k}(D)(s+u)}{(1+|s|)^{\ell}}\Big{|}$
$\displaystyle\leq Ce^{-ux}(1+C_{3}(m,k,u)x^{m+1}),$

which vanishes uniformly for all $f\in M$ as $x\to\infty$. Additionally,
following the proof of Lemma 3.8 and Lemma 3.4 the limit

$\displaystyle\lim_{x\to\infty}\frac{x^{-k}}{\Gamma(m+2)}\int_{0}^{x}$
$\displaystyle\frac{S_{t}^{\lambda,m+1}(f)(s)}{(1+|s|)^{\ell}}u^{m+2}(x-t)^{k}e^{-ut}dt$
$\displaystyle=\frac{1}{\Gamma(1+k)}\int_{0}^{\infty}\frac{S_{t}^{\lambda,k}(f)(s)}{(1+|s|)^{\ell}}u^{k+1}e^{-ut}dt=\frac{f(s+u)}{(1+|s|)^{\ell}},$

exists uniformly on $[\operatorname{Re}>0]$ and $f\in M$, where the last
equality holds by Theorem 2.9 (or more precisely Remark 2.11). This proves the
claim. ∎

The following corollary is an immediate consequence considering singleton
sets.

######  Corollary 4.8.

Let $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ and $k>\ell\geq 0$.
Then for every $u>0$

$\text{$\lim_{x\to\infty}R_{x}^{\lambda,k}(f)(u+\boldsymbol{\cdot})=f(u+\boldsymbol{\cdot})$
\,\,\,in $H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$}\,.$

In particular, for every $u>0$

$\lim_{x\to\infty}\sup_{s\in[\operatorname{Re}>u]}\big{|}\frac{f(s)-R_{x}^{\lambda,k}(f)(s)}{(1+s)^{\ell}}\big{|}=0.$

To check the ’in particular’ part fix $u>0$ and let $s=w+u$, where
$w\in[\operatorname{Re}>0]$. Then for large $x$ by the first part

$\displaystyle\frac{|f(s)-R_{x}^{\lambda,k}(f)(s)|}{|1+s|^{\ell}}$
$\displaystyle=\frac{|f(u+w)-R_{x}^{\lambda,k}(f)(u+w)|}{|1+w|^{\ell}}\frac{|1+w|^{\ell}}{|1+u+w|^{\ell}}\leq\varepsilon\frac{|1+u+w|^{\ell}}{|1+u+w|^{\ell}}=\varepsilon.$

###  4.5. Theorem 41 of M. Riesz revisted

Corollary 4.8 immediately implies the following reformulation of an important
result of M. Riesz published in [13, Theorem 41]. Actually, this result
motivated much of the content of this article.

######  Corollary 4.9.

If $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$, then for every
$k>\ell\geq 0$ and $s\in[\operatorname{Re}>0]$

$f(s)=\lim_{x\to\infty}R_{x}^{\lambda,k}(f)(s).$

Let us point out that the mentioned result of Riesz actually is formulated in
terms of $(e^{\lambda},k)$-Riesz summability (recall the definition of
$(\lambda,k)$-Riesz means of second kind from (13)). In its original form
given in [13] it reads as follows .

######  Theorem 4.10.

Let $f\colon[\operatorname{Re}>0]\to\mathbb{C}$ be holomorphic with a
$\lambda$-Riesz germ $D=\sum a_{n}e^{-\lambda_{n}s}$. Assume that there is
$\ell\geq 0$ such that

(32) $\forall\leavevmode\nobreak\ \varepsilon>0\leavevmode\nobreak\
\exists\leavevmode\nobreak\ C>0\colon|f(s)|\leq
C|s|^{\ell},\leavevmode\nobreak\ \leavevmode\nobreak\
s\in[\operatorname{Re}>\varepsilon].$

Then for every $k>\ell$ and $s\in[\operatorname{Re}>0]$

$f(s)=\lim_{x\to\infty}T_{x}^{\lambda,k}(D)(s)=\lim_{x\to\infty}\sum_{\lambda_{n}<x}a_{n}\big{(}1-e^{\lambda_{n}-x})^{k}e^{-\lambda_{n}s}.$

Let us explain why Theorem 4.10 and Corollary 4.9 in fact are equivalent.
Recall first, that by Theorem 2.8 every $\lambda$-Dirichlet series is
$(\lambda,k$)-Riesz summable on all half-planes
$[\operatorname{Re}>\sigma],\,\sigma>0$ if and only if this series is
$(e^{\lambda},k)$-Riesz summable on all such half-planes. So it is obvious
that Theorem 4.10 is applicable to every $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$; indeed, given
$\varepsilon>0$, we have

$\sup_{\operatorname{Re}s>\varepsilon}\frac{|f(s)|}{|s|^{\ell}}\leq\|f\|_{\infty,\ell}\sup_{\operatorname{Re}s>\varepsilon}\frac{(1+|s|)^{\ell}}{|s|^{\ell}}<\infty\,.$

Conversely, if a holomorphic function $f$ on $[\operatorname{Re}>0]$ has a
$\lambda$-Riesz germ and satisfies (32), then
$f_{\varepsilon}=f(\varepsilon+\cdot)\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ for every $\varepsilon>0$, and
consequently we may apply Corollary 4.9.

The decision on why we focus on $(\lambda,k)$-Riesz means instead of
$(e^{\lambda},k)$-Riesz means is based on the integral representation

$R_{x}^{\lambda,k}(f)(s_{0})=\frac{\Gamma(1+k)}{2\pi
i}x^{-k}\int_{c-i\infty}^{c+i\infty}f(s+s_{0})\frac{e^{xs}}{s^{1+k}}ds$

from Theorem 2.13 of the $x$th Riesz means of first kind of a Dirichlet series
$D=\sum a_{n}e^{-\lambda_{n}s}$ in terms of its limit function $f$ , whereas
by [13, Theorem 40, p. 51] for Riesz means of second kind we have

$T_{x}^{\lambda,k}(D)(s_{0})=\frac{\Gamma(1+k)}{2\pi
i}\int_{c-i\infty}^{c+i\infty}f(s+s_{0})\frac{\Gamma(s)e^{xs}}{\Gamma(1+k+s)}ds.$

In fact, we believe that the handling of the kernels $e^{xs}/s^{1+k}$ is more
simple than that of the kernels $\Gamma(s)e^{xs}/\Gamma(1+k+s)$.

###  4.6. The bounded case $\ell=0$

For $\ell=0$ and $f\in H_{\infty,0}^{\lambda}[\operatorname{Re}>0]$, Corollary
4.8 shows that for every $\sigma>0$

$f=\lim_{x\to\infty}R_{x}^{\lambda,k}(f)\,\,\,\,\text{ uniformly on
$[\operatorname{Re}>\sigma]$ }\,.$

Hence, as the uniform limit of uniformly almost periodic function on
$\mathbb{R}$ is again uniformly almost periodic, the function $t\mapsto
f(\sigma+it)$ is uniformly almost periodic for every $\sigma>0$, this has the
following consequence.

######  Corollary 4.11.

For every frequency $\lambda$

$H_{\infty,0}^{\lambda}[\operatorname{Re}>0]=H_{\infty}^{\lambda}[\operatorname{Re}>0].$

###  4.7. Boundedness of coefficient functionals

We explained above that every function in
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]\to\mathbb{C}$ is uniquely
determined by its Bohr coefficients $a_{N}(f),N\leavevmode\nobreak\
\in\leavevmode\nobreak\ \mathbb{N}$ (Remark 4.1). The following result shows
that this assignment in fact is continuous.

######  Theorem 4.12.

Let $k,\ell\geq 0$ with $k>\ell$ and $\lambda=(\lambda_{n})$ a frequency. Then
for each $N\in\mathbb{N}$ the coefficient mapping

$\mathcal{C}_{N}\colon
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]\to\mathbb{C},\leavevmode\nobreak\
\leavevmode\nobreak\ f\mapsto a_{N}(f)$

is bounded. More precisely, with $\lambda_{-1}:=0$

$\|\mathcal{C}_{N}\|\leq
C(k,\ell,\lambda_{1})\left(\big{(}\frac{\lambda_{N+1}}{\lambda_{N+1}-\lambda_{N}}\big{)}^{k}+\big{(}\frac{\lambda_{N}}{\lambda_{N}-\lambda_{N-1}}\big{)}^{k}\right)\,,\leavevmode\nobreak\
\leavevmode\nobreak\ N\in\mathbb{N}.$

Since $a_{N}(f)=\sum_{n=1}^{N}a_{n}(f)-\sum_{n=1}^{N-1}a_{n}(f)$, the proof
follows from the following independently interesting lemma.

######  Lemma 4.13.

Let $k,\ell\geq 0$ with $k>\ell$. Then for every $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ and $M\in\mathbb{N}$

$|\sum_{n=1}^{M}a_{n}(f)|\leq
C(k,\ell,\lambda)\Big{(}\frac{\lambda_{M+1}}{\lambda_{M+1}-\lambda_{M}}\Big{)}^{k}\,\|f\|_{\infty,\ell}\,.$

###### Proof.

From [20, Lemma 3.5] we know that for every $M\in\mathbb{N}$

$|\sum_{n=1}^{M}a_{\lambda_{n}}(f)|\leq
3\big{(}\frac{\lambda_{M+1}}{\lambda_{M+1}-\lambda_{M}}\big{)}^{k}\sup_{\lambda_{1}<x<\lambda_{M+1}}|\sum_{\lambda_{n}<x}a_{\lambda_{n}}(f)(1-\frac{\lambda_{n}}{x})^{k}|.$

Then the conclusion follows from Theorem 4.6. ∎

###  4.8. A Montel theorem

The following Montel type theorem extends [11, Theorem 3.4] from
$H_{\infty}^{\lambda}[\operatorname{Re}>0]$ to
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0],\,\ell\geq 0$.

Recall from (5) that in the ordinary case $\lambda=(\log n)$ we have the
isometric, coefficient preserving identity
$H_{\infty}^{\lambda}[\operatorname{Re}>0]=\mathcal{H}_{\infty}(\lambda)$, so
Montel’s theorem holds in $\mathcal{H}_{\infty}(\lambda)$, i.e. for each
bounded sequence $(D_{N})$ in $\mathcal{H}_{\infty}(\log n)$ there is a
subsequence $(D_{N_{k}})$ and $D\in\mathcal{H}_{\infty}(\log n)$ such that for
every $\varepsilon>0$ the sequence of all translations
$D^{\varepsilon}_{N_{k}}=\sum a_{n}n^{-\varepsilon}n^{-s}$ converges to
$D\in\mathcal{H}_{\infty}(\log n)$.

This important special case was first proved by Bayart in [1], and within the
setting of Hardy space of ordinary and general Dirichlet series it was
substantially extended in [8], [9], and [11]. See also the discussion in
Section 4.13.

######  Theorem 4.14.

Let $(f_{N})$ be a bounded sequence in
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. Then there is a subsequence
$(f_{N_{j}})_{j}$ of $(f_{N})$ and $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ such that for every $u>0$

$\text{$f_{N_{j}}(u+\boldsymbol{\cdot})\to f(u+\boldsymbol{\cdot})$ \,\,\,\,in
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$}\,.$

In fact we are going to deduce this result from the following more general
version.

######  Theorem 4.15.

Let $(f_{N})$ be a bounded sequence in
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. Assume that for every $n$
the limit

$a_{n}=\lim_{N\to\infty}a_{n}(f_{N})$

exists. Then there exists $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ such that $a_{n}(f)=a_{n}$ for
all $n$, and for every $u>0$

$\text{$f_{N}(u+\boldsymbol{\cdot})\to f(u+\boldsymbol{\cdot})$ \,\,\,in
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$}\,.$

###### Proof.

We define formally the Dirichlet series $E=\sum a_{n}e^{-\lambda_{n}s}$. Then
Theorem 4.6 implies that

$|R_{x}^{\lambda,k}(E)(0)|=\lim_{N\to\infty}|R_{x}^{\lambda,k}(f_{N})(0)|\leq
C\sup_{N\in\mathbb{N}}\|f_{N}\|_{\infty,\ell}.$

Hence, applying Theorem 2.16, we see that $E$ is $(\lambda,k)$-Riesz summable
on $[\operatorname{Re}>0]$. Moreover, denoting by $f$ its limit function, we
by Theorem 4.6 have

$\frac{|f(s)|}{(1+|s|)^{\ell}}=\lim_{x\to\infty}\lim_{N\to\infty}\frac{|R_{x}^{\lambda,k}(f_{N})(s)|}{(1+|s|)^{\ell}}\leq
C\sup_{N\in\mathbb{N}}\|f_{N}\|_{\infty,\ell}\,,$

and so $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. It remains to
show that $f_{N}(u+\boldsymbol{\cdot})\to f(u+\boldsymbol{\cdot})$ in
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. Therefore fix $u>0$ and
applying Theorem 4.7 to the bounded set $M=\\{f_{N}\mid
N\in\mathbb{N}\\}\cup\\{f\\}$, we for every $\varepsilon>0$ obtain some $x>0$
such that

$\displaystyle\|f(u+\cdot)-f_{N}(u+\cdot)\|_{\infty,\ell}\leq\|f(u+\cdot)-R_{x}^{\lambda,k}(f)(u+\cdot)\|_{\infty,\ell}$
$\displaystyle+\|R_{x}^{\lambda,k}(f-f_{N})(u+\cdot)\|_{\infty,\ell}+\|f(u+\cdot)-R_{x}^{\lambda,k}(f_{N})(u+\cdot)\|_{\infty,\ell}$
$\displaystyle\leq
2\varepsilon+\|R_{x}^{\lambda,k}(f-f_{N})(u+\cdot)\|_{\infty,\ell}.$

Now tending $N\to\infty$ gives the claim. ∎

###### Proof of Theorem 4.14.

From Theorem 4.12 we know that $|a_{n}(g)|\leq C(n)\|g\|_{\infty,\ell}$ for
every $g\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. This allows to
apply a diagonal process to find a subsequence $(N_{j})$ such that the limit

$a_{n}=\lim_{j\to\infty}a_{n}(f_{N_{j}})$

exists for every $n$. Then Theorem 4.15 finishes the proof. ∎

###  4.9. Completeness

As we have already announced several times, all spaces
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0],\,\ell\geq 0$ are indeed
Banach spaces, which after all seems non-trivial.

######  Theorem 4.16.

For each $\ell\geq 0$

$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$

together with the norm $\|\cdot\|_{\infty,\ell}$ forms a Banach space.

###### Proof.

Given a Cauchy sequence $(f_{N})$, Theorem 4.14 proves the existence of a
subsequence $(N_{j})$ and $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$, such that for every $u>0$ we
have

$\lim_{j\to\infty}\|f_{N_{j}}(u+\cdot)-f(u+\cdot)\|_{\infty,\ell}=0,$

which in particular shows that $(f_{N_{j}})$ converges pointwise to $f$ on
$[\operatorname{Re}>0]$. Hence for large $N$ and $j=j(s)$ we obtain

$\displaystyle|\frac{f(s)-f_{N}(s)}{(1+s)^{\ell}}|\leq|\frac{f(s)-f_{N_{j}}(s)}{(1+s)^{\ell}}|+\|f_{N_{j}}-f_{N}\|_{\infty,\ell}\leq|f(s)-f_{N_{j}}(s)|+\varepsilon,$

and so $\|f-f_{N}\|_{\infty,\ell}\leq 2\varepsilon$ tending $j\to\infty$. ∎

###  4.10. Behavior far left

The following result shows that the norm of any $f\in
H^{\lambda}_{\infty,\ell}[\operatorname{Re}>0]$ determines the growth of $f$
on vertical abscissas close to the imaginary axis.

######  Theorem 4.17.

Let $f\in H^{\lambda}_{\infty,\ell}[\operatorname{Re}>0]$ with $\ell\geq 0$.
Then

$\|f\|_{\infty,\ell}=\lim_{\sigma\to
0}\sup_{t\in\mathbb{R}}\Big{|}\frac{f(\sigma+it)}{(1+(\sigma+it))^{\ell}}\Big{|}\,.$

The proof needs a well-known Phragmen-Lindelöf type lemma (see e.g. [3, p.
138]).

######  Lemma 4.18.

Let $g$ be a bounded function on the strip
$[\sigma_{1}\leq\operatorname{Re}\leq\sigma_{1}]$, which is holomorphic on
$[\sigma_{1}<\operatorname{Re}<\sigma_{1}]$. Then the function

$L_{g}(\sigma)=\sup_{t\in\mathbb{R}}\big{|}g(\sigma+it)\big{|}\,,\,\,\,\sigma\in[\sigma_{1},\sigma_{2}]\,.$

is logarithmically convex on $[\sigma_{1},\sigma_{2}]$ , i.e. $\log
L_{g}(\sigma)$ is convex on $[\sigma_{1},\sigma_{2}]$ .

###### Proof of Theorem 4.17.

Define the function

$g(s)=\frac{f(s+\varepsilon)}{(1+s)^{\ell}}\,,$

and for each $x$

$g_{x}(s)=\frac{\sum_{\lambda_{n}<x}a_{n}\big{(}1-\frac{\lambda_{n}}{x}\big{)}^{k}e^{-s\lambda_{n}}}{(1+s)^{\ell}}=\frac{R_{x}^{\lambda,k}(f)(s)}{(1+s)^{\ell}}\,,$

which are both obviously holomorphic and bounded on $[\operatorname{Re}>0]$.
Note now that $L_{g_{x}}(\sigma)$ for each $x>0$ is non-increasing for
$\sigma>0$; indeed, by Lemma 4.18 we know that $\log L_{g_{x}}(\sigma)$ is
convex and moreover $\log L_{g_{x}}(\infty)=-\infty$. But since Theorem 4.8
implies that for every $u>0$

$g_{x}\to g\,\,\,\text{ uniformly on $[\operatorname{Re}>u]$ as
$x\to\infty$}\,,$

we see that $L_{g}(\sigma)$ is non-increasing for $\sigma>0$. This gives

$\|f\|_{\infty,\ell}=\sup_{\sigma>0}\big{|}L_{g_{x}}(\sigma)\big{|}=\lim_{\sigma\to
0}\big{|}L_{g_{x}}(\sigma)\big{|}\,,$

which is exactly what we were looking for. ∎

###  4.11. Behavior far right

In contrast to what we did in Section 4.10 we now prove in which sense the
growth of functions $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$
subsides on abscissas far right.

######  Theorem 4.19.

Let $\ell\geq 0$ and $\lambda=(\lambda_{n})$ a frequency with $\lambda_{1}>0$.
Then for every $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ we have

$\lim_{\sigma\to\infty}\sup_{s\in[\operatorname{Re}>\sigma]}\frac{|f(s)|}{(1+|s|)^{\ell}}=\lim_{\sigma\to\infty}\sup_{\begin{subarray}{c}\delta\geq\sigma\\\
t\in\mathbb{R}\end{subarray}}\frac{|f(\sigma+it)|}{(1+|t|)^{\ell}}=0.$

###### Proof.

Observe, that it suffices to check the equality for the second limit. Moreover
recall that by Corollary 4.9 the Dirichlet series $D=\sum
a_{n}e^{-\lambda_{n}s}$ associated to $f$ is $(\lambda,k)$-Riesz summable on
$[\operatorname{Re}>0]$ for every $k>\ell$. Hence, fixing $k>\ell$,
$t\in\mathbb{R}$ and $\sigma>0$, we apply Theorem 2.9 to the translation $\sum
a_{n}e^{-\lambda_{n}it}e^{-\lambda_{n}s}$ and obtain for $\delta>\sigma$

$f(\delta+it)=\frac{\delta^{1+k}}{\Gamma(1+k)}\int_{\lambda_{1}}^{\infty}e^{-\delta
y}S_{y}^{\lambda,k}(D)(it)dy.$

Hence with Theorem 4.6

$\displaystyle\Gamma(1+k)|f(\delta+it)|\leq\delta^{1+k}\int_{\lambda_{1}}^{\infty}e^{-\delta
y}y^{k}\Big{|}R_{y}^{(\lambda,k)}(D)(it)\Big{|}dy$
$\displaystyle\leq\|f\|_{\infty,\ell}(1+|t|)^{\ell}\delta^{1+k}\int_{\lambda_{1}}^{\infty}e^{-\delta
y}y^{k}dy\leq\|f\|_{\infty,\ell}(1+|t|)^{\ell}\delta\int_{\lambda_{1}}^{\infty}e^{-\delta
y}(\delta y)^{k}dy$
$\displaystyle\leq\|f\|_{\infty,\ell}(1+|t|)^{\ell}\int_{\delta\lambda_{1}}^{\infty}e^{-y}y^{k}dy\leq\|f\|_{\infty,\ell}(1+|t|)^{\ell}\int_{\sigma\lambda_{1}}^{\infty}e^{-y}y^{k}dy\,,$

and so

$\lim_{\sigma\to\infty}\,\,\sup_{\begin{subarray}{c}\delta\geq\sigma\\\
t\in\mathbb{R}\end{subarray}}\frac{|f(\delta+it)|}{(1+|t|)^{\ell}}\leq\frac{\|f\|_{\infty,\ell}}{\Gamma(1+k)}\lim_{\sigma\to\infty}\,\,\int_{\sigma\lambda_{1}}^{\infty}e^{-y}y^{k}dy=0\,.\qed$

With a similar technique we obtain the following corollary – needed for the
proof of Theorem 4.21 below.

######  Corollary 4.20.

Let $\lambda_{1}=0$, and $D=\sum a_{n}e^{-\lambda_{n}s}$ with limit function
$f$ and $\sigma_{c}^{\lambda,m}(D)<\infty$ for some $m\geq 0$. Then for every
$\gamma>0$

$\lim_{\sigma\to\infty}\sup_{\begin{subarray}{c}s\in[\operatorname{Re}\geq\sigma]\\\
|Im(s)|\leq\gamma\end{subarray}}|f(s)-a_{1}|=0.$

###### Proof.

We may assume that $a_{1}=0$. Fix some $\sigma_{0}>\sigma_{c}^{\lambda,m}(D)$
and $\gamma>0$. Then by Theorem 2.9 the $(\lambda,m)$-Riesz means of $D$
converge uniformly on $[\sigma_{0},+\infty]\leavevmode\nobreak\
+\leavevmode\nobreak\ i[-\gamma,\gamma]$. Hence,

$\sup_{\sigma>\sigma_{0}}\sup_{|t|\leq\gamma}|R_{y}^{\lambda,m}(D)(\sigma+it)|=C(\sigma_{0})<\infty.$

Now, we follow the calculation of the proof of Theorem 4.19. After translation
about $2\sigma_{0}+it$ and replacing the expression
$\|f\|_{\infty,\ell}(1+|t|)^{\ell}$ by $C(\sigma_{0})$, we obtain

$\lim_{\sigma\to\infty}\sup_{\begin{subarray}{c}s\in[\operatorname{Re}\geq\sigma]\\\
|Im(s)|\leq\gamma\end{subarray}}|f(s)|\leq\frac{C(\sigma_{0})}{\Gamma(1+m)}\lim_{\sigma\to\infty}\int_{\lambda_{2}(\sigma-2\sigma_{0})}^{\infty}e^{-y}y^{m}dy=0.\qed$

###  4.12. Finite order

In (2) we defined what it means that a holomorphic function
$f:[\operatorname{Re}>0]\to\mathbb{C}$ has finite uniform order $\nu_{f}$ (see
also [13, p.14]) The following result shows that our new scale of Banach
spaces is intimately linked with this notion.

######  Theorem 4.21.

Let $f\colon[\operatorname{Re}>0]\to\mathbb{C}$ be a holomorphic function
which has a Riesz germ.

  * (1)

If $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$, then $f$ has finite
uniform order $\nu_{f}\leq\ell$ on $[\operatorname{Re}>0]$.

  * (2)

Conversely, if $f$ has finite uniform order $\nu_{f}$ on
$[\operatorname{Re}>0]$, then

$f_{\mu}=f(\mu+\cdot)\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$

for all $\mu>0$ and $\ell>\nu_{f}$.

The proof relies on Corollary 4.20 and the following lemma, which covers the
first part of Theorem 4.21.

######  Lemma 4.22.

Let $\ell\geq 0$ and $\lambda=(\lambda_{n})$ an arbitrary frequency. Then for
every $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ we have

$\sup_{\begin{subarray}{c}\sigma>0\\\
t\in\mathbb{R}\end{subarray}}\frac{|f(\sigma+it)|}{(1+|t|)^{\ell}}<\infty\,.$

###### Proof.

Since $|a_{1}|\leq C\|f\|_{\infty,\ell}$ by Theorem 4.12, we may assume that
$\lambda_{1}>0$. Note first that by Theorem 4.19 there is some $\sigma_{0}>0$
such that

$\sup_{\begin{subarray}{c}\sigma>\sigma_{0}\\\
t\in\mathbb{R}\end{subarray}}\frac{|f(\sigma+it)|}{(1+|t|)^{\ell}}\leq 1\,.$

On the other hand, for $\sigma\leq\sigma_{0}$ and $t\in\mathbb{R}$ we have
that

$\frac{|f(\sigma+it)|}{(1+|t|)^{\ell}}=\frac{|f(\sigma+it)|}{(1+|\sigma+it|)^{\ell}}\frac{(1+|\sigma+it|)^{\ell}}{(1+|t|)^{\ell}}\leq\frac{|f(\sigma+it)|}{(1+|\sigma+it|)^{\ell}}(1+\sigma_{0})^{\ell}\,.\qed$

###### Proof of Theorem 4.21.

(1) Let $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. Then by Lemma
4.22 we know that

$\exists\leavevmode\nobreak\ R>0\leavevmode\nobreak\
\forall\leavevmode\nobreak\ \sigma>0,\leavevmode\nobreak\
t\in\mathbb{R}\colon|f(\sigma+it)|\leq R(1+|t|)^{\ell}.$

Then $f$ satisfies (2) with $t_{0}=1$ and $C=R2^{\ell}$, since
$(1+|t|)^{\ell}\leq 2^{\ell}|t|^{\ell}$ for all $|t|\geq 1$.

(2) Let $f$ satisfy (2) for some $\ell>\nu_{f}$, i.e. there are $C,t_{0}>0$
such that for all $\sigma>0$ and all $|t|>|t_{0}|$

$|f(\sigma+it)|\leq C|t|^{\ell}\leq C(1+|t|)^{\ell}\,.$

Fix some $\mu>0$, and assume without loss of generality that $\lambda_{1}=0$.
Then we know from Corollary 4.20 (with $\gamma=t_{0}$) that there is some
$\sigma_{0}>\mu$ with $|f(\sigma+it)|\leq C$ for all $\sigma>\sigma_{0}$ and
$|t|\leq t_{0}$. Since the continuous function $f_{\mu}$ is bounded on the
rectangle $[\mu,\sigma_{0}]\times[-t_{0},t_{0}]$, we have that

$\displaystyle\|f_{\mu}\|_{\infty,\ell}\leq\sup_{\begin{subarray}{c}\sigma>0\\\
|t|\geq
t_{0}\end{subarray}}\frac{|f(s)|}{|(1+s)^{\ell}|}+\sup_{\begin{subarray}{c}\sigma_{0}>\sigma>\mu\\\
|t|\leq
t_{0}\end{subarray}}\frac{|f(s)|}{|(1+s)^{\ell}|}+\sup_{\begin{subarray}{c}\sigma>\sigma_{0}\\\
|t|\leq t_{0}\end{subarray}}\frac{|f(s)|}{|(1+s)^{\ell}|},$

which then is finite. ∎

###  4.13. Equivalence

Recall that $\lambda$-Dirichlet series $D=\sum a_{n}e^{-\lambda_{n}s}$
converge on half-planes in the complex plane, where they define holomorphic
functions, and in general the largest possible half-planes of convergence,
uniform convergence, or absolute convergence differ.

In the early days of the theory, a very prominent research problem was to
characterize the class of those frequencies $\lambda$ for which boundedness of
the limit function of $D$ on $[\operatorname{Re}>0]$ implies uniform
convergence of $D$ on every smaller half plane
$[\operatorname{Re}>\varepsilon]$. In other words, if
$\mathcal{D}_{\infty}(\lambda)$ stands for all $\lambda$-Dirichlet series
$D=\sum a_{n}e^{-\lambda_{n}s}$ that on $[\operatorname{Re}>0]$ converges to a
bounded (holomorphic) function, then the question is whether all these series
even converges uniformly on all smaller half-planes
$[\operatorname{Re}>\varepsilon]$ .

Bohr was the first who in [5] managed to isolate a prominent class of such
$\lambda$’s, namely those for which there is some $\beta>0$ such that

(BC) $\frac{1}{\lambda_{n+1}-\lambda_{n}}=O(e^{\beta\lambda_{n}})\,;$

slightly more precise, we in this case say that $\lambda$ satisfies (BC) with
constant $\beta$. We say that ’Bohr’s theorem holds for $\lambda$’, whenever
every $D\in\mathcal{D}_{\infty}(\lambda)$ converges uniformly on
$[\operatorname{Re}>\varepsilon]$ for every $\varepsilon>0$. Later this result
was extended by Landau in [18] assuming the less restrictive assumption

(LC) $\forall\leavevmode\nobreak\ \delta>0\leavevmode\nobreak\
\exists\leavevmode\nobreak\ C>0\leavevmode\nobreak\
\forall\leavevmode\nobreak\ n\colon\leavevmode\nobreak\ \leavevmode\nobreak\
\lambda_{n+1}-\lambda_{n}\geq Ce^{-e^{\delta\lambda_{n}}}.$

Very recently, Bayart in [2] added a further interesting condition, which
provides a nontrivial extension of (LC): A frequency $\lambda$ satisfies
$(NC)$ if

(NC) $\forall\leavevmode\nobreak\ \delta>0\leavevmode\nobreak\
\exists\leavevmode\nobreak\ C>0\leavevmode\nobreak\
\forall\leavevmode\nobreak\ m>n\colon\leavevmode\nobreak\ \leavevmode\nobreak\
\log\bigg{(}\frac{\lambda_{m}+\lambda_{n}}{\lambda_{m}-\lambda_{n}}\bigg{)}+(m-n)\leq
Ce^{\delta\lambda_{n}}\,.$

Examples show that any concrete $\lambda$ may or may not satisfy Bohr’s
theorem, and we refer to [2] and [20] for detailed information on all this.

In this context, an important recent achievement is that the property
’$\lambda$ satisfies Bohr’s theorem’ in fact is equivalent to a number of
facts which seem absolutely unavoidable for a reasonable abstract theory of
general Dirichlet series. The following so-called ’equivalence theorem’ is
taken from [6, Theorem 4.6]) (in fact, the main part of this result was proved
earlier in [9, Theorem 2.16] and [10, Theorem 5.1]).

######  Theorem 4.23.

Let $\lambda$ be a frequency. Then the following are equivalent:

  * (1)

$\lambda$ satisfies Bohr’s theorem.

  * (2)

$\mathcal{D}_{\infty}(\lambda)$ is a Banach space.

  * (3)

$\mathcal{D}_{\infty}(\lambda)=\mathcal{H}_{\infty}(\lambda)$ isometrically
and coefficient preserving.

  * (4)

$\mathcal{D}_{\infty}(\lambda)=H_{\infty}^{\lambda}[\operatorname{Re}>0]$
isometrically and coefficient preserving.

  * (5)

$\lambda$ satisfies Montel’s theorem in $\mathcal{D}_{\infty}(\lambda)$ .

Let us take some time to explain this result in more detail. Note first that
$\mathcal{D}_{\infty}(\lambda)$ endowed with the sup norm
$\|D\|_{\infty}=\sup_{\operatorname{Re}>0}|f(s)|$, where $f$ denotes the limit
function of $D$, forms a normed space. But unfortunately this space in general
is not complete, and so the theorem ensures that this only holds true if
$\lambda$ satisfies Bohr’s theorem – in particular under each of the
conditions of Bayart, Bohr, or Landau.

To see a non-trivial example, $(\sqrt{\log n})$ doesn’t satisfy $(BC)$ but
$(LC)$. Hence, Bohr’s theorem holds for $(\sqrt{\log n})$, and so
$\mathcal{D}_{\infty}((\sqrt{\log n}))$ forms a Banach space, a concrete fact
unknown so far.

An important step is to understand that the Banach space
$H_{\infty}^{\lambda}[\operatorname{Re}>0]$ always contains the (in general
non-complete) space $\mathcal{D}_{\infty}(\lambda)$ isometrically – see [20,
Proposition 3.4]. Hence both spaces in general differ, but in view of the
theorem they coincide iff Bohr’s theorem holds true for $\lambda$.

Moreover, as already mentioned in Section 4.8, Bayart proved a ’Montel type
theorem’ for $\mathcal{D}_{\infty}((\log n))=\mathcal{H}_{\infty}((\log n))$,
which is very useful at many occasions. So the equivalence theorem shows that
we here may replace the frequency $(\log n)$ by a given frequency $\lambda$ if
and only if $\lambda$ satisfies Bohr’s theorem.

The idea now is to extend parts of Theorem 4.23 to the new setting. Given a
frequency $\lambda$ and $\ell\geq 0$, let $\mathcal{D}_{\infty,\ell}(\lambda)$
denote the space of all $\lambda$-Dirichlet series $D=\sum
a_{n}e^{-\lambda_{n}s}$ that are $(\lambda,\ell)$-Riesz summable on all of
$[\operatorname{Re}>0]$ and have limit functions $f\in
H_{\infty}^{\lambda}[\operatorname{Re}>0]$. Together with the norm
$\|D\|_{\infty,\ell}=\|f\|_{\infty,\ell}$, this leads to another scale of
normed spaces of $\lambda$-Dirichlet series.

Recall again from Corollary 4.9 that the $\lambda$-Riesz germ of each $f\in
H_{\infty}^{\lambda}[\operatorname{Re}>0]$ is $(\lambda,k)$-Riesz summable for
each $k>\ell$, and so one of the questions in the following will be under with
additional assumptions on $\lambda$ this even holds for $k=\ell$.

As mentioned above, $\mathcal{D}_{\infty}(\lambda)$ is always an isometric
subspace of $H_{\infty}^{\lambda}[\operatorname{Re}>0]$, hence in the
particular case $\ell=0$ we by definition have the isometric equality

(33) $\mathcal{D}_{\infty,0}(\lambda)=\mathcal{D}_{\infty}(\lambda)\,,$

and hence Theorem 4.23 settles the case $\ell=0$. For $\ell>0$ the following
result partly serves as a sort of substitute of Theorem 4.23.

######  Theorem 4.24.

Let $\lambda$ be a frequency and $\ell\geq 0$. Then the following are
equivalent:

  * (1)

$\mathcal{D}_{\infty,\ell}(\lambda)$ satisfies Montel’s theorem.

  * (2)

$\mathcal{D}_{\infty,\ell}(\lambda)$ is a Banach space.

  * (3)

The $\lambda$-Riesz germ of every $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ $(\lambda,\ell)$-Riesz
summable on $[\operatorname{Re}>0]$.

  * (4)

$\mathcal{D}_{\infty,\ell}(\lambda)=H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$
holds isometrically and coefficient preserving.

###### Proof.

If $\mathcal{D}_{\infty,\ell}(\lambda)$ satisfies Montel’s theorem, then the
same proof as in Theorem 4.16 shows that $\mathcal{D}_{\infty,\ell}(\lambda)$
is complete. Assume that $\mathcal{D}_{\infty,\ell}(\lambda)$ is complete, and
let $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ with the associated
$\lambda$-Riesz germ $D$. By Theorem 4.8 the Riesz means of $f$, after
translation, converge to $f$ in
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. Since all Riesz means
$R_{x}^{\lambda,k}(D)=R_{x}^{\lambda,k}(f)\in\mathcal{D}_{\infty,\ell}(\lambda)$,
we obtain that for every $u>0$ the translation $D_{u}=\sum
a_{n}e^{-\lambda_{n}u}e^{-\lambda_{n}s}\in\mathcal{D}_{\infty,\ell}(\lambda)$,
which eventually shows that $D\in\mathcal{D}_{\infty,\ell}(\lambda).$ Hence
$\mathcal{D}_{\infty,\ell}(\lambda)=H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$.
Finally, if the $\lambda$-Riesz germ of every function in
$H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ converges on
$[\operatorname{Re}>0]$, then
$\mathcal{D}_{\infty,\ell}(\lambda)=H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$
holds isometrically and coefficient preserving. Conversely, if
$\mathcal{D}_{\infty,\ell}(\lambda)=H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$
holds true, then trivially by Theorem 4.14 Montel’s theorem holds in
$\mathcal{D}_{\infty,\ell}(\lambda)$. ∎

For concrete frequencies $\lambda$, in particular $\lambda=(\log n)$, the
following result is our main application of Theorem 4.24.

######  Theorem 4.25.

Let $\ell>0$. If (BC) holds for $\lambda$, then the $\lambda$-Riesz germ $D$
of every $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ is
$(\lambda,\ell)$-Riesz summable on $[\operatorname{Re}>0]$. In particular, all
equivalent statements from Theorem 4.24 hold.

It seems very interesting to decide whether or not the preceding result even
holds under the conditions (LC) or more generally (NC).

Divided into three lemmas of independent interest, the proof of Theorem 4.25
is given at the very end of this section. The first lemma gives a necessary
condition under which $H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ and
$\mathcal{D}_{\infty,\ell}(\lambda)$ coincide, and is based on a Phragmen-
Lindelöf type argument again borrowed from [13].

######  Lemma 4.26.

Let $\ell>0$ and $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. Assume
that there exist $\sigma>0$ and $0\leq\ell^{\prime}<\ell$ such that
$f_{\sigma}=f(\sigma+\cdot)\in
H_{\infty,\ell^{\prime}}^{\lambda}[\operatorname{Re}>0]$. Then the
$\lambda$-Riesz germ $D$ of $f$ is $(\lambda,\ell)$-summable on
$[\operatorname{Re}>0]$, i.e. $D\in\mathcal{D}_{\infty,\ell}(\lambda)$.

###### Proof.

Fix $\varepsilon>0$. Since $f\in
H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$, $f$ by Theorem 4.21 has
finite uniform order $\leq\ell$ on $[\operatorname{Re}>0]$ and finite uniform
order $\leq\ell^{\prime}$ on $[\operatorname{Re}>\sigma$] (in the sense of
(2)). Hence by [13, Theorem 14] there are $C,t_{0}>0$ such that for every
$\varepsilon\leq\beta\leq\sigma+1$ and every $|t|\geq t_{0}$

$|f(\beta+it)|\leq C|t|^{k(\beta)}\,,$

where

$\kappa(\beta)=\frac{\ell(\sigma+1-\beta)+\ell^{\prime}(\beta-\varepsilon)}{\sigma+1-\varepsilon}.$

Since $\kappa^{\prime}(\beta)=\ell^{\prime}-\ell<0$, we obtain that the order
of $f$ on $[\operatorname{Re}=\beta]$ is strictly smaller than $\ell$.
Moreover, Theorem 4.17 shows that $f(\beta+\cdot)\in
H_{\infty,\kappa(\beta)}^{\lambda}[\operatorname{Re}>0]$ for every
$\varepsilon\leq\beta\leq\sigma+1$. Since $\kappa(\beta)<\ell$ for every
$\varepsilon<\beta\leq\sigma+1$, we by Corollary 4.9 obtain
$\sigma_{c}^{\lambda,\ell}(D)\leq\varepsilon$ for every $\varepsilon>0$. ∎

The next tool, based on the preceding one, is a reformulation of [13, Theorem
44]. For the sake of completeness we include an argument within our new
setting.

######  Lemma 4.27.

Let $\ell>0$, and $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$. Then
the $\lambda$-Riesz germ $D$ of $f$ is $(\lambda,\ell)$-Riesz summable on
$[\operatorname{Re}>0]$, i.e. $D\in\mathcal{D}_{\infty,\ell}(\lambda)$,
provided $\sigma_{a}(D)<\infty$.

###### Proof.

Since $\sigma_{a}(D)<\infty$, we have that

$f(s)=\sum_{n=1}^{\infty}a_{n}(f)e^{-\lambda_{n}s},\leavevmode\nobreak\
\leavevmode\nobreak\ s\in[\operatorname{Re}>\sigma_{a}(D)]\,,$

and so we see that $f$ is bounded on $[\operatorname{Re}>\sigma_{a}(D)+1]$.
Hence Lemma 4.26 with $\sigma=\sigma_{a}(D)+1$ and $\ell^{\prime}=0$ implies
$D\in\mathcal{D}_{\infty,\ell}(\lambda)$. ∎

For the last lemma we need further notation and information. Given a frequency
$\lambda$, define

$L(\lambda):=\sup_{D\in\mathcal{D}(\lambda)}\sigma_{a}(D)-\sigma_{c}(D)\,,$

where we recall that $\sigma_{c}(D)$ and $\sigma_{a}(D)$ define the abscissas
of convergence and absolute convergence. Then straightforward arguments show
that

(34) $L(\lambda)=\sigma_{c}\left(\sum
e^{-\lambda_{n}s}\right)=\sigma_{a}\left(\sum e^{-\lambda_{n}s}\right)\,.$

With a less obvious arguments, Bohr in [4, §3, Hilfssatz 4] showed that his
condition (BC) implies $L(\lambda)<\leavevmode\nobreak\ \infty$. In passing we
remark (although this is not needed for what’s coming) that by [4, §3,
Hilfssatz 4]

$L(\lambda)=\limsup_{N\to\infty}\frac{\log(N)}{\lambda_{N}}\,.$

It is evident that $L((n))=0$ and $L((\log n))=1$.

######  Lemma 4.28.

Let $\ell>0$, and $f\in H_{\infty,\ell}^{\lambda}[\operatorname{Re}>0]$ with
its associated $\lambda$-Riesz germ $D$. Then $\sigma_{a}(D)\leq
L(\lambda)+\beta\ell<\infty$, whenever $\lambda$ satisfies (BC) with exponent
$\beta$.

To see an example, we note that for $\lambda=(n)$ we get the conclusion
$\sigma_{a}(D)\leq\leavevmode\nobreak\ 0$, whereas for $\lambda=(\log n)$ we
obtain that $\sigma_{a}(D)\leq 1+\ell$.

###### Proof.

Condition (BC) for $\lambda$ with exponent $\beta$ means that there is some
$C>0$ such that for all $n$

$\frac{1}{\lambda_{n+1}-\lambda_{n+1}}\leq Ce^{\lambda_{n}\beta}\,.$

Adding more elements to $\lambda$, we assume without loss of generality that
$\lambda_{n+1}-\lambda_{n}\leq 1$ for all $n$ (see the first part of the proof
of [20, Theorem 4.2] for details). We now claim that the Bohr coefficients
$(a_{n}(D))=(a_{n}(f))$ of $D$ for every $\delta>0$ and $k>\ell$ satisfy

(35) $\displaystyle A=\sup_{n}|a_{n}(f)e^{-(\beta
k+\delta)\lambda_{n}}|<\infty\,.$

Indeed, given $\delta>0$ and $k>\ell$, by Theorem 4.12 we conclude that for
every $n$

$\displaystyle|a_{n}(f)|$ $\displaystyle\leq
C(k,\ell,\lambda_{1})\Big{(}\big{(}\frac{\lambda_{n+1}}{\lambda_{n+1}-\lambda_{n}}\big{)}^{k}+\big{(}\frac{\lambda_{n}}{\lambda_{n}-\lambda_{n-1}}\big{)}^{k}\Big{)}\|f\|_{\infty,\ell}$
$\displaystyle\leq 2C(k,\ell,\lambda_{1})\Big{(}\lambda_{n+1}^{k}e^{\beta
k\lambda_{n}}+\lambda_{n}^{k}e^{\beta
k\lambda_{n-1}}\Big{)}\,\,\|f\|_{\infty,\ell}$ $\displaystyle\leq
2C(k,\ell,\lambda_{1})\Big{(}(\lambda_{n+1})^{k}e^{\beta
k\lambda_{n}}\Big{)}\,\,\|f\|_{\infty,\ell}\,.$

Hence, choosing $D(\delta,k,\lambda)>0$ such that $(1+\lambda_{n})^{k}\leq
D(\delta,k,\lambda)e^{\delta\lambda_{n}}$ for all $n$, we finally obtain (35),
namely that for all $n$

$|a_{n}(f)|\leq 2C(k,\ell,\lambda_{1})D(\delta,k,\lambda)e^{(\delta+\beta
k)\lambda_{n}}\|f\|_{\infty,\ell}\,.$

Then by (4.13) and by the fact that under Bohr’s condition
$L(\lambda)<\infty$, we see that for every $\varepsilon>0$

$\displaystyle\sum|a_{n}|e^{-((\beta
k+\delta)+(L(\lambda)+\varepsilon))\lambda_{n}}$
$\displaystyle=\sum|a_{n}e^{-(\beta
k+\delta)\lambda_{n}}|e^{-(L(\lambda)+\varepsilon)\lambda_{n}}$
$\displaystyle\leq A\sum e^{-(L(\lambda)+\varepsilon)\lambda_{n}}<\infty\,.$

Consequently, $\sigma_{a}(D)\leq(\beta k+\delta)+(L(\lambda)+\varepsilon)$,
which is the conclusion (whenever $k\to\ell$, $\delta\to 0$, and
$\varepsilon\to 0$). ∎

Applying first Lemma 4.28 and then second Lemma 4.27, we finally obtain as
desired Theorem 4.25.

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