Multivariate Multiplicative Functions of Uniform Random Vectors in Large Integer Domains

Abstract

For a wide class of sequences of integer domains ${\mathcal {D}}_n\subset {\mathbb {N}}^d$, $n\in {\mathbb {N}}$, we prove distributional limit theorems for $F(X_1^{(n)},\ldots ,X_d^{(n)})$, where F is a multivariate multiplicative function and $(X_1^{(n)},\ldots ,X_d^{(n)})$ is a random vector with uniform distribution on ${\mathcal {D}}_n$. As a corollary, we obtain limit theorems for the greatest common divisor and least common multiple of the random set $\{X_1^{(n)},\ldots ,X_d^{(n)}\}$. This generalizes previously known limit results for ${\mathcal {D}}_n$ being either a discrete cube or a discrete hyperbolic region.

Appendix A. On the Regular Growth Condition for Discrete Domains

The following definition can be found on p. 173 in [5].

Definition A.1

A sequence of finite sets ${\mathcal {D}}_n\subset {\mathbb {Z}}^d$ is said to be regularly growing to infinity if as $n\rightarrow \infty$,

$$\begin{aligned} \#{\mathcal {D}}_n\rightarrow \infty \quad \text {and}\quad \frac{\#({\mathcal {D}}^{1}_n\setminus {\mathcal {D}}_n)}{\#{\mathcal {D}}_n}\rightarrow 0, \end{aligned}$$

(29)

where for $A\subset {\mathbb {Z}}^d$ and $p\in {\mathbb {N}}$, we denote by

$$\begin{aligned} A^{p}:=\{x=(x_1,\ldots ,x_d)\in {\mathbb {Z}}^d:\textrm{dist}(x,A)\le p\}, \end{aligned}$$

and $\textrm{dist}$ is the supremum metric on ${\mathbb {Z}}^d$.

Proposition A.2

Assume that ${\mathcal {D}}_n\subset {\mathbb {Z}}^d$ is a sequence of finite sets and $\#{\mathcal {D}}_n\rightarrow \infty$ as $n\rightarrow \infty$. The following statements are equivalent:

• (i) Condition (7) holds for all $c\in {\mathbb {Z}}^d$.

• (ii) Condition (7) holds for $c=\pm e_k$, $k=1,\ldots ,d$.

• (iii) Condition (7) holds for $c=e_k$, $k=1,\ldots ,d$.

• (iv) The sequence ${\mathcal {D}}_n$ is regularly growing.

Proof

Condition (i) trivially implies condition (ii), and (ii) clearly implies (iii). The fact that (iiii)$\Longrightarrow$ (ii) follows from

$$\begin{aligned} \#(({\mathcal {D}}_n-e_k)\Delta {\mathcal {D}}_n)= & {} \#((({\mathcal {D}}_n-e_k)\Delta {\mathcal {D}}_n)+e_k)\\= & {} \#({\mathcal {D}}_n\Delta ({\mathcal {D}}_n+e_k))=\#(({\mathcal {D}}_n+e_k)\Delta {\mathcal {D}}_n). \end{aligned}$$

We now prove that (ii)$\Longrightarrow$(iv). Note that

$$\begin{aligned} {\mathcal {D}}^{1}_n=\bigcup _{k=1}^{d}({\mathcal {D}}_n\pm e_k). \end{aligned}$$

Thus,

$$\begin{aligned} \frac{\#({\mathcal {D}}^{1}_n\setminus {\mathcal {D}}_n)}{\#{\mathcal {D}}_n}\le \sum _{k=1}^{d}\frac{\#(({\mathcal {D}}_n\pm e_k)\setminus {\mathcal {D}}_n)}{\#{\mathcal {D}}_n}\le \sum _{k=1}^{d}\frac{\#(({\mathcal {D}}_n\pm e_k)\Delta {\mathcal {D}}_n)}{\#{\mathcal {D}}_n}. \end{aligned}$$

The right-hand side converges to 0, since by (7) every summand converges to 0.

We proceed to the proof of (iv)$\Longrightarrow$(i). Assume that (29) holds and fix $c\in {\mathbb {Z}}^d$. Using the inclusion $A{\setminus } B\subset (A{\setminus } C)\cup (C{\setminus } B)$ which holds for any sets A, B, C, we conclude that

$$\begin{aligned} ({\mathcal {D}}_n+c)\Delta {\mathcal {D}}_n\subset \bigcup _j \left( ({\mathcal {D}}_n+u_j)\setminus ({\mathcal {D}}_n+v_j)\right) , \end{aligned}$$

where the union is finite and for every index j, $u_j-v_j=\pm e_{k_j}$ for some $k_j\in \{1,\ldots ,d\}$. Since $\#{\mathcal {D}}_n=\#({\mathcal {D}}_n+x)$ for every $x\in {\mathbb {Z}}^d$, it suffices to check that, for every j,

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\#\left( ({\mathcal {D}}_n+u_j)\setminus ({\mathcal {D}}_n+v_j)\right) }{\#({\mathcal {D}}_n+v_j)}=0, \end{aligned}$$

but this follows from the inclusion $({\mathcal {D}}_n+u_j)=({\mathcal {D}}_n+v_j\pm e_{k_j})\subset ({\mathcal {D}}_n+v_j)^1$ and the fact that if (29) holds for a sequence ${\mathcal {D}}_n$, it also holds for the shifted sequence ${\mathcal {D}}_n+x$, for every fixed $x\in {\mathbb {Z}}^d$. $\square$

The following result is a combination of Proposition A.2 and Lemma 1.5 in [5]. In some cases, it is useful for checking (29).

Proposition A.3

Assume that $V_n$, $n\in {\mathbb {N}}$, is a sequence of bounded measurable subsets of ${\mathbb {R}}^d$ satisfying the so-called van Hove condition, meaning that for every $\varepsilon >0$

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\textrm{Vol}(\partial V_n\oplus B^d_{\varepsilon }(0))}{\textrm{Vol}(V_n)}=0, \end{aligned}$$

(30)

where $\partial V_n$ is the topological boundary of $V_n$. Then the sequence ${\mathcal {D}}_n:=V_n\cap {\mathbb {Z}}^d$ satisfies (29).

Our last auxiliary result provides sufficient conditions for (13). It has been used in the proof of Proposition 4.6.

Proposition A.4

Assume that there exist two sequences $(s_{1}(n),\ldots ,s_{d}(n))_{n\in {\mathbb {N}}}$ and $(c_{1}(n),\ldots ,c_{d}(n))_{n\in {\mathbb {N}}}$ of nonnegative integers such that the rectangle

satisfies

$$\begin{aligned} \#{\mathcal {D}}_n\subset \Pi _n\quad \text {and}\quad \overline{C}:=\sup _{n\in {\mathbb {N}}}\frac{\#\Pi _n}{\#{\mathcal {D}}_n}<\infty . \end{aligned}$$

(31)

Then (13) holds. More generally, if (13) holds with ${\mathcal {D}}_n$ replaced by some set $\Pi _n$ which satisfies (31), then (13) holds for ${\mathcal {D}}_n$.

Proof

Fix $i,j=1,\ldots ,d$, $i\ne j$. If (31) holds, then for all $n\in {\mathbb {N}}$ and all $a,b\in {\mathbb {N}}$ it holds

$$\begin{aligned} \frac{\#({\mathcal {D}}_n\cap {\mathbb {Z}}_{i}(a)\cap {\mathbb {Z}}_{j}(b))}{\#{\mathcal {D}}_n}\le \frac{\#(\Pi _n\cap {\mathbb {Z}}_{i}(a)\cap {\mathbb {Z}}_j(b))}{\#{\mathcal {D}}_n}\le \overline{C}\frac{\#(\Pi _n\cap {\mathbb {Z}}_{i}(a)\cap {\mathbb {Z}}_j(b))}{\#\Pi _n}. \end{aligned}$$

Since $i\ne j$, we obtain

$$\begin{aligned} \frac{\#(\Pi _n\cap {\mathbb {Z}}_{i}(a)\cap {\mathbb {Z}}_j(b))}{\#\Pi _n}\le \frac{1}{s_i(n)+1}\left\lfloor \frac{s_i(n)+1}{a}\right\rfloor \frac{1}{s_j(n)+1}\left\lfloor \frac{s_j(n)+1}{b}\right\rfloor \le \frac{1}{ab} \end{aligned}$$

and the desired estimate holds true with $K=\overline{C}$. $\square$