DIRICHLET DENSITY EXTENDS ASYMPTOTIC DENSITY IN MULTIPLICATIVE SYSTEMS

JASON P. BELL AND STANLEY N. BURRIS

Abstract. Dirichlet density extends global asymptotic density in an additive number system A whose generating series A(x) diverges at its radius of convergence (see [1], p. 50). This note shows that the parallel result holds for a multiplicative number system A, namely if the abcissa of convergence α of the generating series A(x) is finite and A(α) = ∞ then Dirichlet density extends global asymptotic density. Using this result it is proved that global asymptotic density Δ satisifies

Δ(P_{1}^{≥m_{1}} ⋯ P_{k}^{≥m_{k}} A) = Δ(P_{1}^{≥m_{1}}A) ⋯ Δ(P_{k}^{≥m_{k}} A),

where the P_{i} are pairwise disjoint sets of indecomposables, provided partition sets of A have global asymptotic density. This answers a question in [2].

1. Introduction

The relevant notation from [1] is presented in this section. Given a Dirichlet series

R(x) = ∑_{n≥1} r(n)n^{−x},

the global count function for R(x) is

R(x) = ∑_{n≤x} r(n). (1.1)

A multiplicative number system

A = (A, P, ·, 1, ∥ ∥)

consists of a countable free monoid (A, ·, 1), freely generated by its set P of indecomposable elements, with a multiplicative norm ∥ ∥ from A to the positive integers, that is,

∥a · b∥ = ∥a∥ · ∥b∥
∥a∥ = 1 if and only if a = 1,

with the property that only finitely many elements of A can have the same norm. α is the abscissa of convergence of A(x).

For B ⊆ A the generating series B(x) for B is the Dirichlet series

B(x) = ∑_{n≥1} b(n)n^{−x}

1991 Mathematics Subject Classification. Primary 60F20; Secondary 05A16. The authors would like to thank NSERC for support of this research.

where b(n) is the number of elements in B of norm n. The (global) asymptotic density Δ(B) is defined by

Δ(B) = lim_{n→∞} \frac{B(n)}{A(n)},

providing this limit exists. The Dirichlet density ∂(B) of B is given by

∂(B) = lim_{x→α+} \frac{B(x)}{A(x)},

provided the limit exists.^{∗} Let

B^{+}(x) = ∑_{n≥1} B(n)n^{−x},

and let A^{+}(x) have abcissa of convergence α^{+}. A modified Dirichlet density of B is given by the following limit when it exists:

∂^{+}(B) = lim_{x→α^{+}+} \frac{B^{+}(x)}{A^{+}(x)}.

Lemma 1 (Proposition 9.11 in [1]). If α^{+} is finite and A^{+}(α^{+}) = ∞ then

Δ ⊆ ∂^{+},

that is, for any B ⊆ A, if Δ(B) exists then so does ∂^{+}(B), and they are equal.

2. Dirichlet Density Extends Global Asymptotic Density

Let A be a multiplicative number system. This section shows that α < ∞ and A(α) = ∞ imply ∂^{+} = ∂, and thus Δ ⊆ ∂.

Lemma 2. α^{+} = α + 1.

Proof. Let A^{+}(n) = A(1) + ⋯ + A(n). Then, by Proposition 7.3 (b) of [1],

α^{+} = lim sup_{n→∞} \frac{log A^{+}(n)}{log n}.

Since

\frac{n − 1}{2}A(n/2) ≤ ∑_{n/2≤m≤n} A(m) ≤ A^{+}(n) ≤ nA(n)

it follows that

\frac{log A(n/2) + log(n − 1) − 1}{log n} ≤ \frac{log A^{+}(n)}{log n} ≤ \frac{log A(n) + log n}{log n}.

Taking the limsup as n → ∞ gives

α + 1 ≤ α^{+} ≤ α + 1.

Thus α is finite iff α^{+} is finite. From now on it is assumed that α and α^{+} are finite.

Lemma 3. For x > α and B ⊆ A, lim_{n→∞} \frac{B(n)}{n^{x}} = 0.

∗The notation x → α+ means that x approaches α from the right.

DIRICHLET DENSITY EXTENDS ASYMPTOTIC DENSITY 3

Proof. Let α_{B} be the abcissa of convergence of B(x). Then

lim sup_{n→∞} \frac{log B(n)}{log n} = \begin{cases}0 & if B is finite \\ α_{B} & else. \end{cases}

Since both 0 and α_{B} are at most α and α < x it follows, for n sufficiently large, that

0 ≤ B(n) < n^{(α+x)/2},

and thus eventually

0 ≤ \frac{B(n)}{n^{x}} < n^{(α−x)/2}.

Now note that the right side goes to 0 as n → ∞ since α − x < 0. Lemma 4. For x > α and B ⊆ A,

B(x) = x ∫_{1}^{∞} \frac{B(t)}{t^{x+1}}dt.

Proof. Use summation by parts on B(x) = ∑_{n≥1} b(n)n^{−x} (see [3], Theorem 421 on p. 346) and Lemma 3 above. Lemma 5. For x > α and B ⊆ A,

|B^{+}(x + 1) − \frac{1}{x}B(x)| ≤ B^{+}(x + 2).

Proof.

|B^{+}(x + 1) − \frac{1}{x}B(x)| = | ∑_{n=1}^{∞} \frac{B(n)}{n^{x+1}} − ∫_{1}^{∞} \frac{B(t)}{t^{x+1}}dt|
≤ ∑_{n=1}^{∞} B(n)|\frac{1}{n^{x+1}} − ∫_{n}^{n+1} \frac{dt}{t^{x+1}} |
≤ ∑_{n=1}^{∞} B(n)|\frac{1}{n^{x+1}} − \frac{1}{(n + 1)^{x+1}} |
≤ (x + 1) ∑_{n=1}^{∞} B(n)\frac{1}{n^{x+2}}
= (x + 1)B^{+}(x + 2).

Lemma 6. If A(α) = ∞ then ∂ = ∂^{+}. Proof. Given B ⊆ A let

f(x) = A^{+}(x + 1) − \frac{1}{x}A(x)

g(x) = B^{+}(x + 1) − \frac{1}{x}B(x).

Then, by Lemma 5, both f(x) and g(x) are bounded for x ∈ (α, α + 1), and one has

\frac{B^{+}(x + 1)}{A^{+}(x + 1)} = \frac{B(x) + xg(x)}{A(x) + xf(x)}.

Thus, in view of Lemma 2, ∂^{+}(B) exists iff ∂(B) exists, and if so they are equal.

Lemma 6 answers the question posed at the bottom of p. 162 of [1].

Lemma 7. A(α) = ∞ iff A^{+}(α^{+}) = ∞.

Proof. If α = 0 then

A(α) = ∑_{n} a(n) = ∞
A^{+}(α^{+}) = ∑_{n} A(n)/n = ∞.

Now suppose α > 0. For x > α Lemma 5 gives

A^{+}(x + 1) ≤ \frac{1}{x}A(x) + (x + 1)A^{+}(x + 2),

so, taking the limsup as x → α+ gives, in view of Lemma 2,

A^{+}(α^{+}) ≤ \frac{1}{α}A(α) + (α + 1)A^{+}(α^{+} + 1).

As A^{+}(α^{+} + 1) < ∞ it follows that

A^{+}(α^{+}) = ∞ implies A(α) = ∞.

Likewise

\frac{1}{x}A(x) ≤ A^{+}(x + 1) + (x + 1)A^{+}(x + 2)

gives

A(α) = ∞ implies A^{+}(α^{+}) = ∞.

Theorem 1. If A(α) = ∞ then Δ ⊆ ∂, that is, Dirichlet density extends global asymptotic density.

Proof. This follows from Lemma 1, Lemma 6 and Lemma 7.

The parallel result for additive number systems was proved in Proposition 3.13 of [1].

3. An Application to Density of Partition Sets

Let A be a multiplicative number system. For B_{1} and B_{2} subsets of A the product B_{1}B_{2} is defined in the obvious way:

B_{1}B_{2} = {b_{1}b_{2} : b_{i} ∈ B_{i}}.

Given a nonempty subset Q of the indecomposables P and a positive integer m let

Q^{≥m}

be the set of all elements of A that can be written as a product involving at least m factors from Q, and only factors from Q. Then let

Q^{≥0} = {1} ∪ Q^{≥1}.

If P_{1}, …, P_{k} are nonempty and pairwise disjoint subsets of P, and if m_{1}, …, m_{k} are nonnegative integers, then

P_{1}^{≥m_{1}} ⋯ P_{k}^{≥m_{k}}A (3.1)

is the set of all elements of A that can be written as a product involving at least m_{i} factors from P_{i}, for 1 ≤ i ≤ k. If the P_{i} also cover P then one can omit the factor A, that is,

P_{1}^{≥m_{1}} ⋯ P_{k}^{≥m_{k}}A = P_{1}^{≥m_{1}} ⋯ P_{k}^{≥m_{k}}.

Sets of this form are special cases of partition sets defined in [1]. If the P_{i} do not cover P then one has

P_{1}^{≥m_{1}} ⋯ P_{k}^{≥m_{k}}A = P_{1}^{≥m_{1}} ⋯ P_{k}^{≥m_{k}}(P ∖ ⋃_{i=1}^{k} P_{i})^{≥0},

and thus P_{1}^{≥m_{1}} ⋯ P_{k}^{≥m_{k}}A is again a partition set.^{∗} For Q ⊆ P define

∂_{Q}(Q^{≥m}) = lim_{x→α+} \frac{Q^{≥m}(x)}{Q^{≥0}(x)}.

Theorem 2. Suppose A is such that all sets of the form (3.1) have global asymptotic density, and A(α) = ∞. Then

Δ(P_{1}^{≥m_{1}} ⋯ P_{k}^{≥m_{k}}A) = Δ(P_{1}^{≥m_{1}}A) ⋯ Δ(P_{k}^{≥m_{k}}A),

where the P_{i} are nonempty pairwise disjoint sets of indecomposables.

Proof. There are two cases to consider, depending on whether or not the P_{i} cover P. We will give the details for the case that they do not, the other case being a minor variation. Let P_{k+1} = P ∖ (P_{1} ∪ ⋯ ∪ P_{k}), a nonempty subset of P, and let m_{k+1} = 0. Then

Δ( ∏_{i=1}^{k} P_{i}^{≥m_{i}}A) = Δ( ∏_{i=1}^{k+1} P_{i}^{≥m_{i}})

= ∂( ∏_{i=1}^{k+1} P_{i}^{≥m_{i}})  by Theorem 1

= ∏_{i=1}^{k+1} ∂_{P_{i}}(P_{i}^{≥m_{i}})  by Theorem 9.37, p. 172 of [1]

= ∏_{i=1}^{k+1} lim_{x→α+} \frac{P_{i}^{≥m_{i}}(x)}{P_{i}^{≥0}(x)}

= ∏_{i=1}^{k+1} lim_{x→α+} \frac{P_{i}^{≥m_{i}}(x) · (P \ P_{i})^{≥0}(x)}{P_{i}^{≥0}(x) · (P \ P_{i})^{≥0}(x)}

that α < ∞.

∗It is easy to show that all partition sets of A have global asymptotic density iff all sets of the form (3.1) have global asymptotic density. By a result of Warlimont, these conditions guarantee

= ∏_{i=1}^{k+1} lim_{x→α+} \frac{(P_{i}^{≥m_{i}}(P \ P_{i})^{≥0})(x)}{A(x)}  by Cor. 9.31, p. 170 of [1]

= ∏_{i=1}^{k+1} lim_{x→α+} \frac{(P_{i}^{≥m_{i}}A)(x)}{A(x)}

= ∏_{i=1}^{k} lim_{x→α+} \frac{(P_{i}^{≥m_{i}}A)(x)}{A(x)}

= ∏_{i=1}^{k} ∂(P_{i}^{≥m_{i}}A)

= ∏_{i=1}^{k} Δ(P_{i}^{≥m_{i}}A)  by Theorem 1.

This answers Question 4, p. 493, from [1] in the affirmative.^{∗} One can also write the conclusion of this theorem as

Δ( ⋂_{i=1}^{k} P_{i}^{≥m_{i}}A) = Δ( ∏_{i=1}^{k} P_{i}^{≥m_{i}}A) = ∏_{i=1}^{k} Δ(P_{i}^{≥m_{i}}A),

showing that sets of the form P_{i}^{≥m_{i}}A behave like independent events, provided the P_{i} are disjoint.

From this result it easily follows, for a multiplicative number system satisfying the hypotheses of Theorem 2, that the values of Δ(B) for partition sets B are completely determined by the values of Δ(Q^{≥m}A), where Q ⊆ P and m is a nonnegative integer.

Problem 1. If A is a multiplicative number system such that each set of the form Q^{≥m}A has global asymptotic density, where Q ⊆ P, does it follow that that all sets of the form (3.1) have global asymptotic density?

An affirmative answer would imply that all partition sets of A have global asymptotic density.

References

[1] Stanley N. Burris, Number Theoretic Density and Logical Limit Laws. Math. Surveys and Monographs 86, Amer. Math. Soc., 2001.

[2] Stanley Burris and András Sárközy, Fine spectra and limit laws I: First-order laws. Can. J. Math. 49 (1997), 468 - 498.

[3] G.H. Hardy and E.M. Wright, An Introduction to the Theory of Numbers, Oxford Press, 5th ed., 1979.

δ( ∑_{i=1}^{k}(≥ m_{i})P_{i} + A) = ∏_{i=1}^{k} δ((≥ m_{i})P_{i} + A).

∗A similar theorem holds for additive number systems for which partition sets have asymptotic density, namely

Department of Mathematics, University of California San Diego, La Jolla, CA 92093-0112. USA

E-mail address: jbell@math.ucsd.edu

Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario N2L 3G1 Canada

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