ADMISSIBLE DIRICHLET SERIES

STANLEY BURRIS AND KAREN YEATS

Abstract. We propose a definition of admissible Dirichlet series as the analog of Hayman’s [6] 1956 definition of admissible power series.

1. Introduction

In 1956 Hayman defined admissible functions—they are analytic in a neighborhood of 0 and one can use the saddle point method to estimate the coefficients of the power series expansion of such functions. They include the functions e^{z} and

exp \frac{1}{1 − z}, are closed under product (of series with the same radius of convergence)

and under exponentiation.

In this paper a notion of admissibility for functions that have Dirichlet series expansions is proposed. We believe that this is a viable analog of Hayman’s definition because (1) this notion of admissible generalizes the conditions of Tenenbaum in [9], (2) there is a fundamental theorem (Theorem 7) that is the analog of Hayman’s fundamental theorem, and (3) a product of admissible Dirichlet series (with the same abscissa of convergence) is again admissible.

2. Definition of Admissible

Theorem 1. Suppose the function F(s) (A1) has a Dirchlet series expansion F(s) = ∑_{n≥1} f(n)n^{−s}, where the coefficients

f(n) are nonnegative real, with f(1) > 0, (A2) has abscissa of (absolute) convergence α ∈ [0, ∞), and (A3) F(s) has no zeros in its halfplane of convergence. Then there exists a Dirichlet series H(s) with real coefficients such that F(s) = e^{H(s)} for σ > α where s = σ + it.

Proof. This is a slight specialization of Theorem 11.14 in Apostol [1]. □

Date: October 10, 2002.

2000 Mathematics Subject Classification. Primary 41A60; Secondary 03C13, 11N45, 11N80, 11U99 .

Key words and phrases. Dirichlet series, saddlepoint method, asymptotic estimates, admissible functions.

Research of the first author was supported by a grant from NSERC.

Research of the second author was supported by a NSERC Undergraduate Student Research Award.

This paper benefited from discussions with R. Warlimont after he had received our preliminary version [3]. He suggested the use of an integral condition in the definition of admissible—the definition continued to evolve after this suggestion. Also he found that exp (ζ(s)^{k}) are examples of admissible functions.

Let ℝ be the set of real numbers. Assuming that F(s) satisfies (A1)–(A3) and F(s) = e^{H(s)}, we will need the basic facts about a Taylor series expansion with remainder of H(σ + it) about t = 0. With

(1)  a(s) := H′(s) b(s) := H′′(s) c(s) := H′′′(s)

we have, for σ > α and t ∈ ℝ,

(2)  H(σ + it) = H(σ) + ia(σ)t − \frac{b(σ)}{2}t^{2} + R(σ + it)

with the remainder term given by

(3)  R(σ + it) = −\frac{i}{2} ∫_{0}^{t} c(σ + iv)(t − v)^{2}dv.

Thus we can write

(4)  \frac{F(σ + it)}{F(σ)} = exp (ia(σ)t − \frac{b(σ)}{2}t^{2} + R(σ + it))

(5)  \frac{|F(σ + it)|}{F(σ)} = exp ( − \frac{b(σ)}{2}t^{2}) · | exp (R(σ + it))|.

Definition 2. Suppose F(s) satisfies (A1)–(A3), with F(s) = e^{H(s)}. Let a(s) and b(s) be the first two derivatives of H(s) as in (1), let R(σ + it) be the remainder term of the Taylor expansion for H(σ +it) as in (2), and suppose there is a function δ : (α, β) → (0, 1), for some β > α, such that as σ → α+ (A4) δ(σ) → 0 (A5) σ^{2}b(σ) → ∞ (A6) b(σ) · exp ( − b(σ)δ(σ)^{2}) → 0 (A7) R(σ + it) → 0 uniformly for |t| ≤ δ(σ)

(A8) \frac{σ\sqrt{b(σ)}}{F(σ)} ∫_{|t|≥δ(σ)} |F(σ + it)| \frac{dt}{σ^{2} + t^{2}} → 0 .

Then we say F(s) is admissible, as witnessed by δ(σ).

Remark 3. Except for §5 we can replace (A6) and (A8) by the following, giving a more general notion of admissible: (A6-) b(σ)δ(σ)^{2} → ∞

(A8-) \frac{σ\sqrt{b(σ)}}{F(σ)} ∫_{|t|≥δ(σ)} F(σ + it)x^{it} \frac{dt}{(σ + it)(σ + 1 + it)} → 0

uniformly for x > 0 The full strength of (A6) and (A8) are used to prove the product theorem in §5.

3. Asymptotic Estimates and Regular Variation

In this section we assume that F(s) is admissible, witnessed by δ(σ). (A5) implies

(6)  b(σ) → ∞  as σ → α+,

so there is a β > α such that

(7)  b(σ) > 0  for σ ∈ (α, β).

We will consistently use β as a number in (α, ∞) such that b(σ) > 0 on (α, β), keeping the original requirement that δ(σ) be defined on (α, β). From (A6) and (6) we have

(8)  b(σ)δ(σ)^{2} → ∞  as σ → α+.

Definition 4. The partial sums of the coefficients of F(s) and its integral are denoted as follows:

F (x) := ∑_{n≤x} f(n)

F̂(x) := ∫_{1}^{x} F (u)du.

Hayman [6] makes a direct application of Cauchy’s integral formula to express the coefficients of a power series. Tenenbaum [9] makes a direct application of Perron’s integral formula to express F (x). The next lemma, where the Perron formula is used to express F̂(x), is used to derive a formula that leads to the verification of regular variation at infinity for F (x). ^{1}

Lemma 5. For x > 0 and c > α,

F̂(x) = \frac{1}{2πi} ∫_{c−i∞}^{c+i∞} F(s) \frac{x^{s+1}}{s(s + 1)} ds .

Proof. See [7] or Lemma 11.22 in [2]. □

An elementary estimate will also be needed.

Lemma 6. For h, λ > 0 and κ ∈ ℝ,

∫_{−h}^{h} e^{iκ−λu^{2}}du = \sqrt{\frac{π}{λ}} e^{−κ^{2}/4λ}(1 + ε(h, κ, λ)),

where

|ε(h, κ, λ)| < \frac{2}{h\sqrt{λ}} .

The following gives the fundamental formula for F̂(x). It is this form, rather than the asymptotics that can be obtained by specializing σ to be the saddle point σ_{x}, that leads to a verification of regular variation at infinity.

Theorem 7. For x > 0 and σ > α

F̂(x) = \frac{x^{σ+1} F(σ)}{σ(σ + 1)\sqrt{2πb(σ)}}( exp (\frac{−(a(σ) + log x)^{2}}{2b(σ)} ) + R(x, σ))

where

R(x, σ) → 0 as σ → α+, uniformly for x > 0.

by S(z).

1Oppenheim ([7], [8]) appears to state this lemma for a particular choice of F(s), the zeta function connected with ‘Factorisatio Numerorum’, but it is a general result. On pages 210–211 of [2], one of the sources cited for Lemma 5, the several occurrences of exp(S(z)) need to be replaced

4 STANLEY BURRIS AND KAREN YEATS

Proof. For x > 0 and σ > α

F̂(x) = \frac{1}{2πi} ∫_{σ−i∞}^{σ+i∞} F(s) \frac{x^{s+1}}{s(s + 1)} ds  by Lemma 5
= \frac{x^{σ+1}}{2π} ∫_{−∞}^{∞} \frac{F(σ + it)x^{it}}{(σ + it)(σ + 1 + it)} dt
= \frac{x^{σ+1}}{2π} (J_{1}(σ, x) + J_{2}(σ, x))

where

J_{1}(σ, x) = ∫_{−δ(σ)}^{δ(σ)} \frac{F(σ + it)x^{it}}{(σ + it)(σ + 1 + it)} dt
J_{2}(σ, x) = ∫_{|t|≥δ(σ)} \frac{F(σ + it)x^{it}}{(σ + it)(σ + 1 + it)} dt.

Since

|J_{2}(σ, x)| ≤ ∫_{|t|≥δ(σ)} |F(σ + it)|\frac{dt}{σ^{2} + t^{2}}

by (A8) we immediately have

J_{2}(σ, x) = \frac{\sqrt{2π} F(σ)}{σ(σ + 1)\sqrt{b(σ)}} o(1)

as σ → α+, uniformly for x > 0.

Let us collect some simple facts before estimating J_{1}(σ, x). We easily have

(9)  \frac{σ + 1}{σ + 1 + it} = 1 + o(1)

as σ → α+, uniformly for |t| ≤ δ(σ), since

\frac{σ + 1}{σ + 1 + it} = 1 − \frac{it}{σ + 1 + it}

and

|\frac{it}{σ + 1 + it}| ≤ \frac{δ(σ)}{σ + 1} = o(1)  by (A4).

Also

(10) |\frac{σ}{σ + it} − 1| = |\frac{it}{σ + it}| ≤ \frac{|t|}{σ}.

Let

a(σ, x) = a(σ) + log x.

ADMISSIBLE DIRICHLET SERIES 5

Then by (4), (A7), (9) and (10), for σ ∈ (α, β) and x > 0

J_{1}(σ, x) = ∫_{|t|≤δ(σ)} \frac{F(σ + it)x^{it}}{(σ + it)(σ + it + 1)}dt
= \frac{F(σ)}{σ(σ + 1)} ∫_{|t|≤δ(σ)} exp (ia(σ, x)t − \frac{b(σ)}{2}t^{2} + R(σ + it))

· \frac{σ(σ + 1)}{(σ + it)(σ + 1 + it)}dt

= \frac{F(σ)}{σ(σ + 1)} ∫_{|t|≤δ(σ)} exp (ia(σ, x)t − \frac{b(σ)}{2}t^{2})(1 + o(1) + O(\frac{|t|}{σ}))dt
= \underbrace{\frac{F(σ)}{σ(σ + 1)} ∫_{|t|≤δ(σ)} exp (ia(σ, x)t − \frac{b(σ)}{2}t^{2})(1 + o(1))dt}_{J_{11}(σ,x)}

+ \underbrace{\frac{F(σ)}{σ(σ + 1)} ∫_{|t|≤δ(σ)} exp (ia(σ, x)t − \frac{b(σ)}{2}t^{2})O(\frac{|t|}{σ})dt}_{J_{12}(σ,x)} .

For J_{12}(σ, x) we have, for σ ∈ (α, β) and x > 0,

|J_{12}(σ, x)| = O(\frac{F(σ)}{σ(σ + 1)} ∫_{|t|≤δ(σ)} exp ( − \frac{b(σ)}{2}t^{2})\frac{|t|}{σ}dt)

= O(\frac{F(σ)}{σ^{2}(σ + 1)} ∫_{0}^{∞} exp ( − \frac{b(σ)}{2}t^{2})tdt)

= O(\frac{F(σ)}{σ(σ + 1)\sqrt{b(σ)}}(\frac{1}{σ\sqrt{b(σ)}})).

Thus by (A5)

J_{12}(σ, x) = \frac{\sqrt{2π} F(σ)}{σ(σ + 1)\sqrt{b(σ)}} o(1)

as σ → α+, uniformly for x > 0.

From Lemma 6 we have, for σ ∈ (α, β) and x > 0,

J_{11}(σ, x) = \frac{F(σ)}{σ(σ + 1)} ∫_{|t|≤δ(σ)} exp (ia(σ, x)t − \frac{b(σ)}{2}t^{2})(1 + o(1))dt
= \frac{F(σ)}{σ(σ + 1)}\sqrt{\frac{π}{b(σ)/2}}( exp (\frac{−a(σ, x)^{2}}{2b(σ)} )

+ ε(δ(σ), a(σ, x), b(σ)/2) + o(1))

= \frac{\sqrt{2π} F(σ)}{σ(σ + 1)\sqrt{b(σ)}} ( exp (\frac{−(a(σ) + log(x))^{2}}{2b(σ)} ) + o(1))

as σ → α+, uniformly for x > 0 since by Lemma 6 and (8)

|ε(δ(σ), a(σ, x), b(σ)/2)| < \frac{2}{δ(σ)\sqrt{b(σ)/2}} = o(1).

Combining these results we have

J(σ, x) := J_{1}(σ, x) + J_{2}(σ, x)
= \frac{\sqrt{2π} F(σ)}{σ(σ + 1)\sqrt{b(σ)}} ( exp (\frac{−(a(σ) + log(x))^{2}}{2b(σ)} ) + o(1))

as σ → α+, uniformly for x > 0, and the proof of the theorem is completed by observing that

F̂(x) = \frac{x^{σ+1}}{2π} J(σ, x).

□

Corollary 8. a(σ) is strictly increasing on (α, β) and as σ → α+ (a) a(σ) → −∞

(b) \frac{a(σ)^{2}}{b(σ)} → ∞

(c) (σ − α) · a(σ) → −∞.

Proof. We know that a′(σ) = b(σ) > 0 on (α, β), so a(σ) is strictly increasing on (α, β). Now F̂(1) = 0, so by Theorem 7 with x = 1 we have for σ ∈ (α, β)

0 = \frac{F(σ)}{σ(σ + 1)\sqrt{2πb(σ)}}( exp (\frac{−a(σ)^{2}}{2b(σ)} ) + R(1, σ))

so

0 = exp (\frac{−a(σ)^{2}}{2b(σ)} ) + R(1, σ).

Therefore

exp (\frac{−a(σ)^{2}}{2b(σ)} ) → 0 as σ → α+,

and thus (b) holds. Since a(σ) decreases on (α, β) as σ → α+ and b(σ) → ∞ by (6) we see that (a) follows from (b).

To prove (c) we first use (a) to choose a γ ∈ (α, β) so that a(σ) < 0 for σ ∈ (α, γ). Then for α < σ_{1} < σ < γ we have, by the mean value theorem,

\frac{1}{a(σ)} − \frac{1}{a(σ_{1})} = −\frac{a′(ξ)}{a(ξ)^{2}} (σ − σ_{1}) for some ξ ∈ (σ_{1}, σ)
= −\frac{b(ξ)}{a(ξ)^{2}} (σ − σ_{1})
= o(σ − σ_{1}) as σ → α+
= o(σ − α) as σ → α+.

Letting σ_{1} → α+

\frac{1}{a(σ)} = o(σ − α) as σ → α+

so (σ − α) · a(σ) → −∞ as σ → α+.

□

In view of Corollary 8(a), from now on we will assume that β was chosen small enough that a(σ) < 0 for σ ∈ (α, β).

Notice that a(σ) → −∞ as σ → α+ implies that for x sufficiently large the equation a(σ) + log x = 0 has, by the continuity of a(σ), a solution. In particular since a′(σ) is positive on (α, β), for

x ≥ x_{0} := exp ( − a(β)) + 1

one has a unique solution in (α, β).

Definition 9. For x ≥ x_{0} (as just described) let σ_{x} be the unique solution for σ ∈ (α, β) to the equation

(11)  a(σ) + log x = 0.

The function σ_{x} is strictly decreasing on [x_{0}, ∞) and

(12)  σ_{x} → α+ as x → ∞

since lim_{x→∞} a(σ_{x}) = − lim_{x→∞} log x = −∞.

Also note that if one puts σ = σ_{x} (where x ≥ x_{0}) in the expression for F̂(x) in Theorem 7 then it simplifies to

F̂(x) = \frac{x^{σ_{x}+1} F(σ_{x})}{σ_{x}(σ_{x} + 1)\sqrt{2πb(σ_{x})}}(1 + R(x, σ_{x})),

where R(x, σ_{x}) → 0 as x → ∞. So we have the following.

Corollary 10.

F̂(x) ∼ \frac{x^{σ_{x}+1} F(σ_{x})}{σ_{x}(σ_{x} + 1)\sqrt{2πb(σ_{x})}} as x → ∞.

The choice of σ = σ_{x} is what is commonly meant by ‘finding the saddlepoint’, and the resulting formula for F̂(x) is the result of ‘applying the saddlepoint method’. In reality the value s = σ_{x} is usually only near a saddle point of the integrand of the integral in Lemma 5, that is, a point s where the derivative of the integrand vanishes. By choosing the line of integration of this integral to pass through (a point near) the saddle point one hopes to concentrate the value of the integral in a small neighborhood of the real axis. Indeed, that is what happens for admissible functions. In the proof of Theorem 7, the value of F̂(x) is concentrated in the integral J_{1}(x) when σ = σ_{x} as x → ∞, leading to Corollary 10 above.

Corollary 11. As σ → α+

(a) \frac{F(σ)}{σ\sqrt{b(σ)}} → ∞ and

(b) F(σ) → ∞.

Proof. Note that F̂(2) > 0 (since f(1) > 0). Then for σ ∈ (α, β), by Theorem 7

(13)  F̂(2) = \frac{2^{σ+1} F(σ)}{σ(σ + 1)\sqrt{2πb(σ)}}( exp (\frac{−(a(σ) + log 2)^{2}}{2b(σ)} ) + R(2, σ)).

By Corollary 8(a) there is a γ ∈ (α, β) such that a(σ) is negative on (α, γ), and thus nonzero. For σ ∈ (α, γ) we then have

\frac{(a(σ) + log 2)^{2}}{b(σ)} = \frac{a(σ)^{2}}{b(σ)} (1 + \frac{2 log 2}{a(σ)} + \frac{(log 2)^{2}}{a(σ)^{2}} ).

By Corollary 8 the right hand side of this equation goes to ∞ as σ → α+, so

exp (\frac{−(a(σ) + log 2)^{2}}{2b(σ)} ) → 0 as σ → α+.

From Theorem 7 we know R(2, σ) → 0 as σ → α+; and clearly

\frac{2^{σ+1}}{(σ + 1)\sqrt{2π}} → \frac{2^{α+1}}{(α + 1)\sqrt{2π}} < ∞ as σ → α+.

The left side of (13) is a positive constant, so it follows that part (a) of this Corollary

must hold: \frac{F(σ)}{σ\sqrt{b(σ)}} → ∞ as σ → α+. Then part (a) and (A5) give F(σ) → ∞ as

σ → α+, which is part (b). □

The next corollary shows that as σ → α+ we have F(σ) growing much faster than any power of a(σ) or b(σ). This leads in turn to the fact that F(σ) grows much faster than any power of σ − α. Consequently F(s) cannot have a pole at α.

Corollary 12.

(a) For all ε > 0,

a(σ) = o(F(σ)^{ε})  and  b(σ) = o(F(σ)^{ε}) as σ → α+.

(b) For all r ∈ ℝ,

(σ − α)^{r}F(σ) → ∞ as σ → α+.

Proof. We break the proof of (a) into two claims.

Claim 1 : For all ε > 0 and all γ ∈ (α, β) there is a σ ∈ (α, γ) such that

\frac{|a(σ)|}{F(σ)^{ε}} < 1.

Assume not. Then we can choose ε > 0 and γ ∈ (α, β) such that for all σ ∈ (α, γ)

(14)  \frac{|a(σ)|}{F(σ)^{ε}} ≥ 1.

Then for α < σ_{1} < σ < γ by the mean value theorem

\frac{1}{F(σ)^{ε}} − \frac{1}{F(σ_{1})^{ε}} = \frac{−εF′(ξ)}{F(ξ)^{1+ε}}(σ − σ_{1}) for some ξ ∈ (σ_{1}, σ)
= −ε\frac{a(ξ)}{F(ξ)^{ε}} (σ − σ_{1})
= ε\frac{|a(ξ)|}{F(ξ)^{ε}} (σ − σ_{1})
≥ ε(σ − σ_{1}) by (14).

Letting σ_{1} → α+ gives 1/F(σ)^{ε} ≥ ε(σ − α); so

(15)  F(σ) ≤ (\frac{1}{ε(σ − α)})^{1/ε} for σ ∈ (α, γ).

We can also assume that γ ∈ (α, β) is such that (σ−α)|a(σ)| > 2/ε for σ ∈ (α, γ) by Corollary 8(c). So for α < σ < σ_{2} < γ

−\frac{F′(u)}{F(u)} = −a(u) > \frac{2}{ε(u − α)} for u ∈ [σ, σ_{2}]

which implies that

− ∫_{σ}^{σ_{2}} \frac{F′(u)}{F(u)}du > ∫_{σ}^{σ_{2}} \frac{2}{ε(u − α)}du,

that is,

−( log F(σ_{2}) − log F(σ)) > \frac{2}{ε} log (\frac{σ_{2} − α}{σ − α}).

From this inequality and (15)

log F(σ_{2}) + \frac{2}{ε} log (\frac{σ_{2} − α}{σ − α}) < log F(σ) ≤ \frac{1}{ε}( log \frac{1}{ε} + log \frac{1}{σ − α}).

Thus

\frac{1}{ε} log \frac{1}{σ − α} > log F(σ_{2}) + \frac{2}{ε} log(σ_{2} − α) − \frac{1}{ε} log \frac{1}{ε} + \frac{2}{ε} log \frac{1}{σ − α}
= C + \frac{2}{ε} log \frac{1}{σ − α}

where C is independent of σ. Hence

1 > \frac{Cε}{log (1/(σ − α))} + 2 → 2 as σ → α+,

which is a contradiction, proving Claim 1.

Claim 2 : For all ε > 0 there is a γ ∈ (α, β) such that

(16)  \frac{|a(σ)|}{F(σ)^{ε}} < 1 for σ ∈ (α, γ].

Let ε > 0 be given. From Claim 1 and Corollary 8(b) we know that there exists a γ ∈ (α, β) such that

(17)  \frac{|a(γ)|}{F(γ)^{ε}} < 1 and \frac{b(σ)}{a(σ)^{2}} < ε for σ ∈ (α, γ].

We will show this γ is such that (16) holds. Otherwise there is a σ ∈ (α, γ) such that |a(σ)|/F(σ)^{ε} ≥ 1. By the intermediate value theorem there must be a σ ∈ (α, γ) such that |a(σ)|/F(σ)^{ε} = 1. Letting σ_{1} be the largest such σ in (α, γ) we have

\frac{|a(σ_{1})|}{F(σ_{1})^{ε}} = 1 and \frac{|a(σ)|}{F(σ)^{ε}} < 1 for σ ∈ (σ_{1}, γ].

Equivalently

(18)  |a(σ_{1})| − F(σ_{1})^{ε} = 0 and |a(σ)| − F(σ)^{ε} < 0 for σ ∈ (σ_{1}, γ].

As |a(σ)| = −a(σ) on (α, β), from (18) we have

\frac{d}{dσ} ( − a(σ) − F(σ)^{ε})|_{σ=σ_{1}} ≤ 0.

Hence

0 ≤ a′(σ_{1}) + εF′(σ_{1})F(σ_{1})^{ε−1}
= b(σ_{1}) + εa(σ_{1})F(σ_{1})^{ε}
= b(σ_{1}) − εa(σ_{1})^{2}.

By (17) b(σ_{1}) < εa(σ_{1})^{2}. This is a contradiction, proving Claim 2.

From Claim 2 we immediately have a(σ) = O(F(σ)^{ε/2}), and thus by Corollary 11(b) a(σ) = o(F(σ)^{ε}). Then from Corollary 8(b)

b(σ) = o(a(σ)^{2}) = o(F(σ)^{ε}) as σ → α+.

This finishes the proof of (a).

Part (b) is now a trivial consequence of part (a) and Corollary 8(c).

□

Remark 13. Corollary 12(b) readily shows many Dirichlet series satisfying (A1)– (A3) are not admissible.

(a) ζ(s)^{k}, k = 1, 2, …, is not admissible as it has a pole at its abscissa α = 1.

(b) The zeta function

∏_{j=1}^{k} (1 − n_{j}^{−s})^{−m_{j}}

of a finitely generated multiplicative number system is not admissible as it has a pole at its abscissa α = 0.

Corollary 14. The function F̂(x) grows much faster than x^{α+1}, namely

lim_{x→∞} \frac{F̂(x)}{x^{α+1}} = ∞.

Proof. This is clear from Corollary 10, Corollary 11(a) and (12). □

Definition 15. For α ∈ ℝ, a real-valued function g(x) that is eventually defined on the reals and eventually positive is said to have regular variation at infinity with index α, written simply as g(x) ∈ RV_{α}, if for any y > 0

(19)  lim_{x→∞} \frac{g(xy)}{g(x)} = y^{α}.

Corollary 16. F̂(x) ∈ RV_{α+1}.

Proof. We assume x, y > 0. In the expressions for F̂(xy) and F̂(x) given by Theorem 7 let σ = σ_{x} (for x sufficiently large) and divide to obtain

(20)  \frac{F̂(xy)}{F̂(x)} = (y^{σ_{x}+1})\frac{exp ( − (log y)^{2}/(2b(σ_{x}))) + R(xy, σ_{x})}{1 + R(x, σ_{x})}
→ y^{α+1} as x → ∞

since both R(x, σ_{x}) → 0 and R(xy, σ_{x}) → 0 as x → ∞ by Theorem 7 and (12); and since b(σ_{x}) → ∞ as x → ∞ by (6) and (12). □

Lemma 17. Let G(s) = ∑_{n≥1} g(n)/n^{s} be a Dirichlet series with nonnegative real

coefficients and abscissa α ≥ 0, and let G(x) = ∑_{1≤n≤x} g(n), Ĝ(x) = ∫_{1}^{x} G(u)du.

If Ĝ(x) ∈ RV_{α+1} then

(a) G(x) ∈ RV_{α}, and

(b) G(x) ∼ \frac{α + 1}{x} Ĝ(x).

Proof. This is an immediate consequence of Lemma 11.21 from [2]. □

Corollary 18. F (x) ∈ RV_{α} and

F (x) ∼ \frac{α + 1}{x} F̂(x) ∼ \frac{x^{σ_{x}} F(σ_{x})}{σ_{x}\sqrt{2πb(σ_{x})}}

as x → ∞.

Proof. By Corollary 10, Corollary 16, Lemma 17 and (12). □

Corollary 19. The function F (x) grows much faster than x^{α}, namely

lim_{x→∞} \frac{F (x)}{x^{α}} = ∞.

Proof. By Corollary 11(a), Corollary 18 and (12). □

From this Corollary it is immediate that ζ(s) is not admissible (a fact already noted in Remark 13).

4. Tenenbaum’s Condtions

A version of admissibility conditions for Dirichlet series due to Tenenbaum [9], 1988, is given in the following.^{2}

Definition 20. Suppose F(s) satisfies conditions (A1)–(A3) and there is a function T : (α, β) → (0, ∞), for some β > α, such that as σ → α+ (T1) σ^{2}b(σ) → ∞

(T2) \frac{b(σ)^{3}}{c(σ)^{2}} → ∞

(T3) T (σ) \frac{|F(σ + it)|}{F(σ)} ≤ 1, for σ ∈ (α, β) and for δ(σ) ≤ |t| ≤ T (σ),

where δ(σ) = |b(σ)c(σ)|^{−1/5}

(T4) \frac{\sqrt{b(σ)}}{T (σ)} → 0

(T5) |c(σ + it)| ≤ |c(σ)| for σ ∈ (α, β) and t ∈ ℝ (T6) lim inf_{σ→α+} |c(σ)| > 0.

Then we say that F(s) is T-admissible, as witnessed by T (σ).

Tenenbaum uses T (σ) = σb(σ)/ε(σ) where ε(σ) → 0 as σ → α+. This choice of T (σ) makes condition (T4) unnecessary. Furthermore he gives an error term that is important to his applications in number theory, especially to the function ψ(x, y).

Theorem 21. If F(s) is T-admissible then it is admissible.

Proof. Let F(s) = exp (H(s)) be a T-admissible Dirichlet series as witnessed by T (σ) : (α, β) → (0, ∞). (T1) shows that

(21)  b(σ) → ∞ as σ → α+,

so we can assume that b(σ) is positive on (α, β). From (21) and (T4) it is clear that

T (σ) → ∞ as σ → α+.

2Tenenbaum actually uses T (σ) = σb(σ)/ε, which makes our (T4) unnecessary, and he uses the saddlepoint σ_{x} instead of σ in (T1)–(T3), which he labels as (H2)–(H4).

By (21) and (T6) one has

δ(σ) → 0 as σ → α+,

so (A4) holds. As (T1) is (A5) we only need to verify that (A6)–(A8) hold. For (A7) we have for σ ∈ (α, β) and |t| ≤ δ(σ)

|R(σ + it)| ≤ |c(σ)t^{3}|  by (3)
≤ |c(σ) · δ(σ)^{3}|
= |c(σ)| · |b(σ)c(σ)|^{−3/5}
= (\frac{c(σ)^{2}}{b(σ)^{3}} )^{1/5}
→ 0  as σ → α+ by (T2).

For (A6) we have from (T3) for σ ∈ (α, β)

T (σ) \frac{|F(σ + iδ(σ))|}{F(σ)} ≤ 1.

Multiplying this by (T4) gives

\sqrt{b(σ)} \frac{|F(σ + iδ(σ))|}{F(σ)} → 0,

which, in view of (5) and the fact that (A7) holds, gives (A6).

Finally (A8) is verified as follows, where σ ∈ (α, β):

σ\sqrt{b(σ)} ∫_{|t|≥δ(σ)} \frac{|F(σ + it)|}{F(σ)} \frac{dt}{σ^{2} + t^{2}}

≤ σ\sqrt{b(σ)} ∫_{δ(σ)≤|t|≤T (σ)} \frac{|F(σ + it)|}{F(σ)} \frac{dt}{σ^{2} + t^{2}} + 2σ\sqrt{b(σ)} ∫_{T (σ)}^{∞} \frac{dt}{t^{2}}

≤ \frac{\sqrt{b(σ)}}{T (σ)}(σ ∫_{δ(σ)≤|t|≤T (σ)} \frac{dt}{σ^{2} + t^{2}} ) + \frac{2σ\sqrt{b(σ)}}{T (σ)}  by (T3)
= o(1)  by (T4).
□

The conditions of Tenenbaum have proved to be very practical, giving the asymptotics for many naturally occurring examples of Dirichlet series to which the saddlepoint method applies.

Example 22. The function

F(s) := e^{ζ(s)}

is readily proved to be T-admissible, witnessed by T (σ) = b(σ), after noting

• ζ(s) = \frac{1}{s − 1} + g(s), where g(s) is holomorphic

• there is a constant C > 0 such that for σ ∈ [1, 2] and |t| ≥ 1 we have

|ζ(σ + it)| ≤ C log |t|. From the T-admissibility of exp (ζ(s)) one easily has the T-admissibility of

F_{λ}(s) := exp (ζ(s − λ)) = exp ( ∑_{n=1}^{∞} n^{λ} · n^{−s})

for λ ≥ 0.^{3}

Example 23. Tenenbaum studies the counting functions ψ(x, y) for the zeta functions

ζ(s, y) := ∏_{p≤y} (1 − p^{−s})^{−1}.

As noted in Remark 13, the functions ζ(s, y) are not admissible. These functions satisfy all the conditions for being T-admissible except (T3), and for y in a suitable range (depending on x) they satisfy (T3) provided σ = σ_{x}. This leads to asymptotics for ψ(x, y) as x and y tend to infinity with y suitably constrained.

Example 24. The function

F_{k}(s) := exp (\frac{1}{1 − k^{−s}} )

is admissible for k = 2, …, but not T-admissible. It is clear that each F_{k}(s) satisfies (A1)–(A3), and has a Dirichlet series expansion with abscissa of convergence α = 0. To see that F_{k}(s) is not T-admissible note that

\frac{|F_{k}(σ + it)|}{F_{k}(σ)}

is, for each σ > 0, positive and periodic as a function of t, and thus does not uniformly go to 0 on [δ(σ), T (σ)] as σ → α+. Consequently F_{k}(s) does not satisfy condition (T3).

To show that F_{k}(s) is admissible let

δ_{k}(σ) = (k^{σ} − 1)^{7/5}
T_{k}(σ) = (k^{σ} − 1)^{−3}.

One has

a_{k}(σ) = −\frac{k^{σ}}{(k^{σ} − 1)^{2}} log k

b_{k}(σ) = \frac{k^{2σ} + k^{σ}}{(k^{σ} − 1)^{3}} ( log k)^{2}

c_{k}(σ) = −\frac{k^{3σ} + 4k^{2σ} + k^{σ}}{(k^{σ} − 1)^{4}} ( log k)^{3}.

Verifying (A4)–(A6) is routine. For (A7) we proceed as in the proof of Theorem 21, namely for |t| ≤ δ(σ) one has

|R(σ + it)| ≤ |c(σ)δ(σ)^{3}| → 0 σ → α+.

3The asymptotics for F_{λ}(x) are also analyzed in §11.5 of [2] by the saddlepoint method, after changing the path of the Perron integral. (See Footnote 1 for errata to §11.5.)

R. Warlimont [12] first pointed out the example of exp(ζ(s)) to us. Later he found related examples of admissible functions, such as exp (ζ(s)^{k}), that subsequently turned out to be Tadmissible as well.

This leaves (A8), which is usually the challenging part of the verification of admissibility. First note that

\frac{σ\sqrt{b_{k}(σ)}}{F_{k}(σ)} ∫_{|t|≥T_{k}(σ)} |F_{k}(σ + it)| \frac{dt}{σ^{2} + t^{2}} ≤ σ\sqrt{b_{k}(σ)} ∫_{|t|≥T_{k}(σ)} \frac{dt}{t^{2}}

= \frac{σ\sqrt{b_{k}(σ)}}{T_{k}(σ)} → 0 as σ → α+.

Thus we only need to show that

\frac{σ\sqrt{b_{k}(σ)}}{F_{k}(σ)} ∫_{δ(σ)≤|t|≤T_{k}(σ)} |F_{k}(σ + it)| \frac{dt}{σ^{2} + t^{2}} → 0 as σ → α+.

Substituting τ = t log k, we need to show

\frac{σ\sqrt{b_{k}(σ)}}{F_{k}(σ)} ∫_{δ(σ) log k≤|τ|≤T_{k}(σ) log k} |F_{k}(σ + i\frac{τ}{log k} )| \frac{(log k)dτ}{(σ log k)^{2} + τ^{2}} → 0

as u → 1+. Letting u = k^{σ} it suffices to show

∫_{(u−1)^{7/5} log k}^{(u−1)^{−3} log k} \frac{log u}{(u − 1)^{3/2}} exp (\frac{2( cos(τ) − 1)}{(u − 1)(u^{2} − 2u cos(τ) + 1)})\frac{dτ}{(log u)^{2} + τ^{2}} → 0

as u → 1+.

One can do this by noting that as u → 1+ the integrand rapidly and uniformly approaches 0 outside neighborhoods of radius (u−1)^{7/5} about the points τ = 2mπ, indeed much faster than (u − 1)^{3}. Thus it suffices to show that

(22)  ∫_{U} \frac{log u}{(u − 1)^{3/2}} exp (\frac{2( cos(τ) − 1)}{(u − 1)(u^{2} − 2u cos(τ) + 1)})\frac{dτ}{(log u)^{2} + τ^{2}} → 0

as u → 1+, where U is the union of the intervals

[2mπ − (u − 1)^{7/5}, (2mπ + (u − 1)^{7/5}]

about the points 2mπ, m ≥ 1, such that 2mπ −(u−1)^{7/5} < (u−1)^{−3}. The integral in (22) is bounded by

(23)  2ζ(2) ∫_{0}^{(u−1)^{7/5}} \frac{log u}{(u − 1)^{3/2}} exp (\frac{2( cos(τ) − 1)}{(u − 1)(u^{2} − 2u cos(τ) + 1)})dτ.

Let J(u, τ) be the integrand in (23). Then

∫_{0}^{(u−1)^{7/5}} J(u, τ)dτ = ∫_{0}^{(u−1)^{3/2}} J(u, τ)dτ + ∫_{(u−1)^{3/2}}^{(u−1)^{7/5}} J(u, τ)dτ
≤ J(u, 0) · (u − 1)^{3/2} + J(u, (u − 1)^{3/2}) · (u − 1)^{7/5}
→ 0  as u → 1+.

This proves F_{k}(s) is admissible, and thus the class of admissible functions is wider than the class of T-admissible functions.

5. Closure under Product

The goal of this section is to prove that the product of two admissible functions F_{1}(s) and F_{2}(s) with the same abscissa of convergence is again admissible.

Theorem 25. Suppose F_{1}(s) and F_{2}(s) are admissible with the same abscissa of convergence α. Then F_{1}(s) · F_{2}(s) is admissible.

Proof. We assume F_{j}(s), δ_{j}(σ), b_{j}(s) satisfy (A1)–(A8) for j = 1, 2, and we assume β_{j} > α chosen such that b_{j}(σ) > 0 for σ ∈ (α, β_{j}).

Let

F(s) := F_{1}(s) · F_{2}(s)
β := min(β_{1}, β_{2})
δ(σ) := min (δ_{1}(σ), δ_{2}(σ)) for σ ∈ (α, β).

We have

F_{j}(s) = e^{H_{j}(s)}  (j = 1, 2)
F(s) = e^{H(s)}
H(s) = H_{1}(s) + H_{2}(s)
a(s) = a_{1}(s) + a_{2}(s)
b(s) = b_{1}(s) + b_{2}(s)
R(s) = R_{1}(s) + R_{2}(s).

It is easy to check that (A1)–(A5) hold for F(s).

Next,

R(σ + it) = R_{1}(σ + it) + R_{2}(σ + it)
→ 0  uniformly for |t| ≤ δ(σ)

since each of the R_{j} satisfy (A7) and since δ(σ) ≤ δ_{j}(σ) for j = 1, 2. So (A7) also holds for F.

To prove (A6) and (A8) for F we first observe that for σ > α, for t ∈ ℝ and for j = 1, 2

(24)  |F_{j}(σ + it)| ≤ F_{j}(σ),

and for σ ∈ (α, β) and j = 1, 2

(25)  b_{j}(σ) > 0
(26)  b(σ) = b_{1}(σ) + b_{2}(σ) ≤ 2 max (b_{1}(σ), b_{2}(σ)).

From (24) we have for σ > α, for t ∈ ℝ and for j = 1, 2

(27)  \frac{|F(σ + it)|}{F(σ)} = \frac{|F_{1}(σ + it)|}{F_{1}(σ)} · \frac{|F_{2}(σ + it)|}{F_{2}(σ)} ≤ \frac{|F_{j}(σ + it)|}{F_{j}(σ)} .

Choose γ_{1} ∈ (α, β) such that for σ ∈ (α, γ_{1}) and j = 1, 2

(28)  b_{j}(σ)δ_{j}(σ)^{2} > 1.

This is possible by (8).

Now suppose that we are given ε ∈ (0, 1).

16 STANLEY BURRIS AND KAREN YEATS

Choose γ_{2} ∈ (α, γ_{1}) such that for σ ∈ (α, γ_{2}) and j = 1, 2 b_{j}(σ) · exp ( − b_{j}(σ)δ_{j}(σ)^{2}) < ε^{2} (29)

σ\sqrt{b_{j}(σ)} ∫_{|t|≥δ_{j}(σ)} \frac{|F_{j}(σ + it)|}{F_{j}(σ)} \frac{dt}{σ^{2} + t^{2}} < ε. (30)

We can do this because the F_{j} satisfy (A6) and (A8).

Choose γ ∈ (α, γ_{2}) such that for σ ∈ (α, γ) and j = 1, 2

(31) \frac{|F_{j}(σ + it)|}{F_{j}(σ)} < 2 exp ( − b_{j}(σ)t^{2}/2) for |t| ≤ δ_{j}(σ).

In view of (5) we can do this because the F_{j} satisfy (A7). Claim: For σ ∈ (α, γ) b(σ) · exp ( − b(σ)δ(σ)^{2}) < 2ε^{2} (32)

σ\sqrt{b(σ)} ∫_{|t|≥δ(σ)} \frac{|F(σ + it)|}{F(σ)} \frac{dt}{σ^{2} + t^{2}} < 12ε. (33)

This will prove that (A6) and (A8) hold for F.

We start by fixing σ ∈ (α, γ).

\underline{Case (i):} δ_{2}(σ) ≤ δ_{1}(σ). Then δ(σ) = δ_{2}(σ).

\underline{Subcase (ia):} b_{1}(σ) ≤ b_{2}(σ).

Then by (26) and (29)

b(σ) · exp ( − b(σ)δ(σ)^{2}) < 2b_{2}(σ) · exp ( − b_{2}(σ)δ_{2}(σ)^{2})
< 2ε^{2}.

Also by (26), (27) for j = 2 and (30) for j = 2 we have

σ\sqrt{b(σ)} ∫_{|t|≥δ(σ)} \frac{|F(σ + it)|}{F(σ)} \frac{dt}{σ^{2} + t^{2}}

≤ σ\sqrt{2b_{2}(σ)} ∫_{|t|≥δ_{2}(σ_{2})} \frac{|F_{2}(σ + it)|}{F_{2}(σ)} \frac{dt}{σ^{2} + t^{2}}
< \sqrt{2}ε < 12ε.

\underline{Subcase (ib):} b_{2}(σ) ≤ b_{1}(σ).

By (29) for j = 2

\sqrt{b_{2}(σ)} exp ( − b_{2}(σ)δ_{2}(σ)^{2}/2) < ε,

so

(34)  \sqrt{b_{1}(σ)} exp ( − b_{1}(σ)δ_{2}(σ)^{2}/2) < ε

since

\sqrt{x} exp ( − xδ_{2}(σ)^{2}/2)

ADMISSIBLE DIRICHLET SERIES 17

is decreasing for x > 1/δ_{2}(σ)^{2}, and since by Subcase (ib) and (28) for j = 2

b_{1}(σ) ≥ b_{2}(σ) > \frac{1}{δ_{2}(σ)^{2}} .

Then by (26) and (34)

b(σ) · exp ( − b(σ)δ(σ)^{2}) < 2b_{1}(σ) · exp ( − b_{1}(σ)δ_{2}(σ)^{2})
< 2ε^{2}.

Also from (34) we have

\sqrt{b_{1}(σ)} exp ( − b_{1}(σ)t^{2}/2) < ε for δ_{2}(σ) ≤ |t|. Combined with (31) for j = 1 this gives

(35) \sqrt{b_{1}(σ)} \frac{|F_{1}(σ + it)|}{F_{1}(σ)} < 2ε for δ_{2}(σ) ≤ |t| ≤ δ_{1}(σ).

By (26) and (27) for j = 1

σ\sqrt{b(σ)} ∫_{|t|≥δ(σ)} \frac{|F(σ + it)|}{F(σ)} \frac{dt}{σ^{2} + t^{2}}

≤ σ\sqrt{2b_{1}(σ)} ∫_{|t|≥δ_{2}(σ)} \frac{|F_{1}(σ + it)|}{F_{1}(σ)} \frac{dt}{σ^{2} + t^{2}}
= \sqrt{2}(J_{1}(σ) + J_{2}(σ)),

where

J_{1}(σ) = σ\sqrt{b_{1}(σ)} ∫_{δ_{2}(σ)≤|t|≤δ_{1}(σ)} \frac{|F_{1}(σ + it)|}{F_{1}(σ)} \frac{dt}{σ^{2} + t^{2}}
J_{2}(σ) = σ\sqrt{b_{1}(σ)} ∫_{|t|≥δ_{1}(σ)} \frac{|F_{1}(σ + it)|}{F_{1}(σ)} \frac{dt}{σ^{2} + t^{2}} .

By (35)

J_{1}(σ) ≤ 2εσ ∫_{δ_{2}(σ)≤|t|≤δ_{1}(σ)} \frac{dt}{σ^{2} + t^{2}} < 2πε,

and by (30)

J_{2}(σ) < ε.

Thus

σ\sqrt{b(σ)} ∫_{|t|≥δ(σ)} \frac{|F(σ + it)|}{F(σ)} \frac{dt}{σ^{2} + t^{2}} < \sqrt{2}(2π + 1)ε < 12ε,

and the claim is proved in Case (i). Case (ii), where δ_{1}(σ) ≤ δ_{2}(σ), is handled likewise. So (A6) and (A8) hold for

F, and the theorem is proved.

□

6. Open questions

Problem 1. Is the sum of two admissible functions also admissible?

Problem 2. Is the product of any two admissible functions also admissible?

Problem 3. Given two admissible functions F_{j}(x) = exp (H_{j}(s)) is the function exp (H_{1}(s) · H_{2}(s)) admissible?

Problem 4. If F(s) is admissible, does it follow that e^{F(s)} is also admissible?

We suspect, by analogy with Hayman’s work, that this is true.

Problem 5. Can the notion of admissible be extended to include ζ(s)?

References

[1] Tom M. Apostol. Introduction to Analytic Number Theory. Springer-Verlag, New York, 1976.

[2] Stanley N. Burris. Number Theoretic Density and Logical Limit Laws. Mathematical Surveys and Monographs, 86. American Mathematical Society, Providence, RI, 2001.

[3] Stanley Burris and Karen Yeats, Toward a multiplicative version of Hayman admissibility: Progress report (Preprint Sept. 2001).

[4] Kevin J. Compton, A logical approach to asymptotic combinatorics. II. Monadic second-order properties. J. Combin. Theory Ser. A 50 (1989), 110–131.

[5] J.L. Geluk and L. de Haan, Regular variation, extensions and Tauberian theorems. CWI Tract 40, Centre for Mathematics and Computer Science, 1987.

[6] W. K. Hayman. A generalization of Stirling’s formula. Journal für die reine und angewandte Mathematik 196 (1956), 67-95.

[7] A. Oppenheim, On an arithmetic function. J. London Math. Soc. 1 (1926), 105–211.

[8] A. Oppenheim, On an arithmetic function (II ). J. London Math. Soc. 2 (1927), 123–130.

[9] Gérald Tenenbaum, La methode du col en thèorie analytique de nombres. Séminaire de Théorie des Nombres, Paris 1986–1987, 411–441, Prog. Math. 75, Birkhäuser Boston, 1988.

[10] Gérald Tenenbaum, Integration complexe: la methode du col en thèorie analytique de nombres. Journées Ètat de la Recherche, 7–9 décembre 2000, Université Bourdeaux 1.

[11] E.C. Titchmarsh, The Theory of the Riemann Zeta-function. 2nd Edition, 1981. Oxford University Press.

[12] R. Warlimont, Private Communication.

Dept. of Pure Mathematics, University of Waterloo, Waterloo, Ont., Canada N2L 3G1

   E-mail address: snburris@thoralf.uwaterloo.ca

Dept. of Pure Mathematics, University of Waterloo, Waterloo, Ont., Canada N2L 3G1

   E-mail address: kayeats@uwaterloo.ca