1 Introduction

The concept of density in number theory has been extensively studied for many years, with various variants introduced over time, such as upper and lower density, natural density, analytic density, and Banach density. Among these, upper and lower Banach densities are commonly used in number theory. The upper Banach density measures the density of a subset of natural numbers within the set of all natural numbers. Furthermore, it is popular among mathematicians who work on combinatorial number theory problems using ergodic methods [3].

The upper Banach density has found applications in various areas of number theory, including prime numbers, additive number theory, and Diophantine equations (see [2, 3] or [5]). One notable application is in the study of Szemer’di’s theorem [9], which states that any set of integers with positive upper density contains arbitrarily long arithmetic progressions.

There are many interesting results about upper Banach density (see [1] or [3]) in combinatorial number theory and additive number theory. For instance, in [4], Grekos, Toma, and Tomanov’a presented and discussed three equivalent definitions of the so-called uniform or Banach density of a set of positive integers. Additionally, Leonetti and Tringali [7], introduced a new upper density for a set of positive integers. They demonstrated that the upper asymptotic, upper logarithmic, upper Banach, upper Buck, upper Pólya, and upper analytic densities, along with all upper \(\alpha \)-densities, are upper densities in the sense of their definition.

However, the exploration of upper Banach density in additive number theory remains a challenging domain. One of the major untouched problems in this area is finding the growth and structure of sums of sets of zero lower Banach density but positive upper Banach density.

Consider a subset S of \({\mathbb {N}}\). Define the upper density \({\overline{d}}(S)\) of S by

$$\begin{aligned} {\overline{d}}(S)=\limsup _{n\rightarrow \infty } \frac{\left| S\cap [0, n-1]\right| }{n}, \end{aligned}$$

where \(|\cdot |\) is the number of elements of a set. The upper Banach density of S, denoted as \(\textrm{BD}^*(S)\), is defined by

$$\begin{aligned} {\textrm{BD}}^*(S)=\limsup _{|I|\rightarrow \infty } \frac{\left| S\cap I\right| }{|I|}, \end{aligned}$$

where I ranges over finite continuous integer intervals of \({\mathbb {N}}\). That is, for some sequence of intervals \(\{I_k\}\), \(I_k=[a_k, b_k]\), \(b_k-a_k\rightarrow \infty \),

$$\begin{aligned} \frac{\left| S\cap I_k\right| }{|I_k|}\rightarrow {\textrm{BD}}^*(S), \end{aligned}$$

and for any other sequence with \(|I_k|\rightarrow \infty \),

$$\begin{aligned} \limsup _{k\rightarrow \infty }\frac{\left| S\cap I_k\right| }{|I_k|}\le {\textrm{BD}}^*(S). \end{aligned}$$

In [8], Li, Tu, and Ye presented an intriguing result (Theorem 1) concerning the upper Banach density of a subset S of \({\mathbb {N}}\), which sheds light on the connection between topological entropy and Banach mean sensitivity for a transitive system. They employed a proof technique for the aforementioned theorem, relying on topological dynamical systems and utilizing the methods of dynamical systems and the Furstenberg correspondence principle.

Theorem 1

If S has positive upper Banach density and \(W^*\) is an IP-set, then there are \(q, q_1, q_2\in S\) and \(l_1, l_2\in W^*\) with \(l_i=q_i-q\) for each \(i=1, 2\) and

$$\begin{aligned} {\textrm{BD}}^*\big ((S-l_1)\cap (S-l_2)\big )\ge \frac{1}{4}\left( {{\textrm{BD}}^*(S)}\right) ^4. \end{aligned}$$

In this paper, we present a combinatorial proof approach for the aforementioned theorem, which is elementary in nature and avoids the need for specialized knowledge and techniques in topological dynamical systems and ergodic theory. Part of our inspiration stems from [6], a work by Kerr and Li, wherein they established a connection between measure entropy tuples and combinatorial independence. Our main results are as follows.

Theorem 2

Let S be a subset of \({\mathbb {N}}\) satisfying \({\textrm{BD}}^*(S) > 0\). Suppose \(c>0\), and let \(W \subset {\mathbb {Z}}\) be a finite nonempty subset of \({\mathbb {Z}}\) with \(|W| > \frac{2(1+c)}{{\textrm{BD}}^*(S)}\). Then, there exist two distinct positive integers \(l_1\) and \(l_2\) in W such that:

$$\begin{aligned} {\textrm{BD}}^*\big ((S+l_1)\cap (S+l_2)\big )\ge \frac{c}{4(1+c)^2} \left( {{\textrm{BD}}^*(S)}\right) ^2. \end{aligned}$$

Taking \(c=1\), it follows that:

Corollary 1

Let S be a subset of \({\mathbb {N}}\) satisfying \({\textrm{BD}}^*(S) > 0\). Let \(W \subset {\mathbb {Z}}\) be an finite nonempty subset of \({\mathbb {Z}}\) with \(|W| > \frac{4}{{\textrm{BD}}^*(S)}\). Then, there exist two distinct positive integers \(l_1\) and \(l_2\) in W such that:

$$\begin{aligned} {\textrm{BD}}^*\big ((S+l_1)\cap (S+l_2)\big )\ge \frac{1}{16} \left( {{\textrm{BD}}^*(S)}\right) ^2. \end{aligned}$$

In Li, Tu, and Ye’s result (Theorem 1), the IP-set \(W^*\) is an infinite subset of \({\mathbb {N}}\). In this paper, we establish a finite IP-set revision of Theorem 1 using our result (Corollary 1).

Proposition 3

Let S be a subset of \({\mathbb {N}}\) satisfying \({\textrm{BD}}^*(S) > 0\), and let \(M>\frac{4}{{\textrm{BD}}^*(S)}\) be a positive integer. Suppose that \(\{p_1, p_2, \cdots , p_{2M}\}\) are 2M integers and satisfy \(\sum _{i=1}^{n} p_i<p_{n+1}\) for all \(1\le n\le 2M-1\). Let \(W^*\) be an IP-set generated by \(\{p_i\}_{i=1}^{2M}\), then there are \(q, q_1, q_2\in S\) and \(l_1, l_2\in W^*\) with \(l_i=q_i-q\) for each \(i=1, 2\) and

$$\begin{aligned} {\textrm{BD}}^*\big ((S-l_1)\cap (S-l_2)\big )\ge \frac{1}{16^3}\left( {{\textrm{BD}}^*(S)}\right) ^4. \end{aligned}$$

Moreover, employing analogous arguments as those applied to the previous theorem, we obtain:

Theorem 4

Let S be a subset of \({\mathbb {N}}\) satisfying \({\overline{d}}(S) > 0\). Suppose \(c>0\), and let \(W \subset {\mathbb {N}}\) be a finite nonempty subset of \({\mathbb {N}}\) with \(|W| > \frac{2(1+c)}{{\overline{d}}(S)}\). Then, there exist two distinct positive integers \(l_1\) and \(l_2\) in W such that:

$$\begin{aligned} {\overline{d}}\big ((S+l_1)\cap (S+l_2)\big )\ge \frac{c}{4(1+c)^2} \left( {\overline{d}}(S)\right) ^2. \end{aligned}$$

2 Combinational Proof of the Main Theorem

In the first part of this section, we will review the concept of an IP set.

Let \(F\subset {\mathbb {N}}\). We say that F is an IP-set if there is a subsequence \(\{p_i\}_{i\in {\mathbb {N}}}\) of \({\mathbb {N}}\) such that

$$\begin{aligned} \left\{ p_{i_1}+\cdots +p_{i_k} \, : \, k\in {\mathbb {N}}\right\} \subset F. \end{aligned}$$

The following example illustrates the necessity of the lower bound for the finite set W in Theorem 2.

Example 1

Consider a positive integer \(p\in {\mathbb {N}}\), and let \(S=\{p\, m \,:\, m\in {\mathbb {N}}\}\). It is evident that \({\textrm{BD}}^*(S)=1/p\). Now, suppose we define

$$\begin{aligned} W:=\{0, 1, \cdots , p-1\}. \end{aligned}$$

It is straightforward to observe that \(|W|=p\), indicating that \(|W|<\frac{2(1+c)}{{\textrm{BD}}^*(S)}\) for any \(c>0\). Simultaneously, for any distinct pair \(l_1, l_2\in W\), it is clear that \((W+l_1)\cap (W+l_2)=\emptyset \).

Proof of Theorem 2

Let \(q={\textrm{BD}}^*(S)>0\). According to the definition of upper Banach density, there exists a sequence of intervals \([M_k, N_k]\subset {\mathbb {N}}\) such that

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\left| S\cap [M_k, N_k]\right| }{N_k-M_k+1}=q\quad {\text{ and } }\quad \lim _{k\rightarrow \infty } (N_k-M_k)=+\infty . \end{aligned}$$
(1)

Since \(|W| > \frac{2(1+c)}{{\textrm{BD}}^*(S)}\), we can choose a subset \(W_1\subset W\) such that

$$\begin{aligned} |W_1|=\left\lceil \llceil \frac{2(1+c)}{{\textrm{BD}}^*(S)}\right\rceil \rrceil , \end{aligned}$$
(2)

where \(\lceil x\rceil \) denotes the smallest integer greater than x.

Denote by \(J_k=S\cap [M_k, N_k]\) and \(p_k=N_k-M_k+1\). Take

$$\begin{aligned} F_k=\bigcap _{s\in W_1} [M_k-s, N_k-s]\quad {\text{ and }}\quad M=\sum _{s\in W_1} |s|. \end{aligned}$$

It is evident that

$$\begin{aligned} \begin{aligned} \left| [M_k, N_k]\backslash F_k\right|&=\left| \bigcup _{s\in W_1} [M_k, N_k]\backslash [M_k-s, N_k-s]\right| \\&\le \sum _{s\in W_1}\left| [M_k, N_k]\backslash [M_k-s, N_k-s]\right| \\&\le \sum _{s\in W_1} |s|=M. \end{aligned} \end{aligned}$$

Thus, by (1), one has

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\left| [M_k, N_k]\backslash F_k\right| }{N_k-M_k+1}\le \lim _{k\rightarrow \infty }\frac{M}{p_k}=0. \end{aligned}$$
(3)

It is easy to see that

$$\begin{aligned} \left| J_k\cap F_k\right| =\left| J_k\right| - \left| J_k\backslash F_k\right| \ge \left| J_k\right| -\left| [M_k, N_k]\backslash F_k\right| . \end{aligned}$$

Therefore, it follows that

$$\begin{aligned} \frac{|J_k|}{p_k}-\frac{\left| [M_k, N_k]\backslash F_k\right| }{p_k}\le \frac{\left| J_k\cap F_k\right| }{p_k}\le \frac{\left| J_k\right| }{p_k}. \end{aligned}$$

Taking \(k\rightarrow \infty \) and combining with (1) and (3), we obtain

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{\left| J_k\cap F_k\right| }{p_k}=q. \end{aligned}$$

Let \(I_k=J_k\cap F_k\). Thus, there exists \(K\in {\mathbb {N}}\) such that for every \(k>K\), we have

$$\begin{aligned} |I_k|\ge \frac{q}{2}p_k. \end{aligned}$$
(4)

Now, consider \(k\in {\mathbb {N}}\) with \(k>K\).

Observation. \(F_k=\{t\in {\mathbb {Z}}\,: \, t+W_1\subset [M_k, N_k]\}\).

Let \(t\in {\mathbb {Z}}\) such that \(t+W_1\subset [M_k, N_k]\). Then we have

$$\begin{aligned} \begin{aligned} t+s\in [M_k, N_k]\ {\text{ for } \text{ all }}\ s\in W_1&\Leftrightarrow t\in [M_k-s, N_k-s]\ {\text{ for } \text{ all }}\ s\in W_1\\&\Leftrightarrow t\in \bigcap _{s\in W_1}[M_k-s, N_k-s]=F_k. \end{aligned} \end{aligned}$$

Hence, we have established the observation.

Note that \(I_k\subset F_k\). Based on the observation, we have

$$\begin{aligned} \bigcup _{s\in W_1} (I_k+s)\subset [M_k, N_k]. \end{aligned}$$
(5)

Let \(\kappa \) be the maximum value of \(|(I_k+r)\cap (I_k+s)|/p_k\) for all distinct \(r, s\in W_1\). For each \(s\in W_1\), we define the set

$$\begin{aligned} \Omega _s=\bigcup _{r\in W_1\backslash \{s\}}\big ((I_k+r)\cap (I_k+s)\big ) \end{aligned}$$

which has a cardinality of at most \(\kappa \cdot |W_1|\cdot p_k\).

Note that the sets \((I_k+s)\backslash \Omega _s\) for \(s\in W_1\) are pairwise disjoint. In the following, we will count the number \(\sum _{r\in W_1}|I_k+r|\) using two different approaches.

Consider \(n\in \bigcup _{s\in W_1} (I_k+s)\).

Case 1. The positive integer n belongs to only one \((I_k+s)\) for some \(s\in W_1\).

This implies that \(n\not \in (I_k+r)\) for all \(r\in W_1\backslash \{s\}\). According to the definition of \(\Omega _r\), it follows that \(n\not \in \Omega _r\) for all \(r\in W_1\). Therefore, we conclude that

$$\begin{aligned} n\in \bigcup _{r\in W_1} (I_k+r)\backslash \Omega _r. \end{aligned}$$

Case 2. The positive integer \(n\in (I_k+s_1)\cap (I_k+s_2)\cap \cdots \cap (I_k+s_l)\), where \(l\ge 2\), \(s_1, \cdots s_l\in W_1\), \(s_i\ne s_j\) \((i\ne j)\) and \(n\not \in (I_k+r)\) for all \(r\in W_1\backslash \{s_1, \cdots , s_l\}\).

In this situation, the positive integer n is counted exactly l times in the sum \(\sum _{r\in W_1} |I_k+r|\). Moreover, since \(l\ge 2\), it is evident that \(n\in \Omega _{s_i}\) for \(i=1, \cdots , l\), indicating that n is counted at least l times in the sum \(\sum _{r\in W_1} |\Omega _r|\).

Based on the above arguments, we have

$$\begin{aligned} \begin{aligned} \sum _{r\in W_1} \left| I_k+r\right|&\le \sum _{r\in W_1} \left| \Omega _r\right| +\left| \bigcup _{r\in W_1}(I_k+r)\backslash \Omega _r\right| \\&\le \sum _{r\in W_1} \left| \Omega _r\right| + \left| [M_k, N_k]\right| \quad \quad \quad \quad {\text{(by } }(2.5))\\&\le \kappa \cdot |W_1|^2\cdot p_k + p_k. \end{aligned} \end{aligned}$$

On the other hand, noting that \(|W_1|=\lceil 2(1+c)/{\textrm{BD}}^*(S)\rceil \ge 2(1+c)/q\) and (4), we have

$$\begin{aligned} \begin{aligned} \sum _{r\in W_1} \left| I_k+r\right|&= \sum _{r\in W_1} \left| I_k\right| =\left| I_k\right| \cdot |W_1|\\&\ge \frac{q |W_1|}{2} p_k\ge (1+c) p_k. \end{aligned} \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \kappa \ge \frac{c}{|W_1|^2}. \end{aligned}$$

It follows that we can find distinct \(l^k_1, l^k_2\in W_1\) such that

$$\begin{aligned} \left| (I_k+l^k_1)\cap (I_k+l^k_2)\right| \ge \frac{c\cdot p_k}{|W_1|^2}. \end{aligned}$$
(6)

Let \(\Delta (W_1)=\{(s, s)\,: \, s\in W_1\}\). The above arguments demonstrate that for all sufficiently large values of k, specifically greater than K as defined in (4), we can find a pair \((l^k_1, l^k_2)\in W_1\times W_1\backslash \Delta (W_1)\) satisfying (6).

By the pigeonhole principle, there exist integers \(K<k_1<k_2<\cdots<k_n<\cdots \) such that \(l^{k_n}_1\) is constant, denoted as \(l_1\), and \(l^{k_n}_2\) is also constant, denoted as \(l_2\), for all \(n\in {\mathbb {N}}\). The fact that \((l^{k_n}_1, l^{k_n}_2)\in W_1\times W_1\backslash \triangle (W_1)\) implies that \(l_1, l_2\in W_1\) and \(l_1\ne l_2\). Meanwhile, the inequality (6) implies that

$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{\left| \left( I_{k_n}+l_1\right) \cap \left( I_{k_n}+l_2\right) \right| }{p_{k_n}}\ge \frac{c}{|W_1|^2}. \end{aligned}$$

According to (5), we know that \(\left( I_{k_n}+l_1\right) \cap \left( I_{k_n}+l_2\right) \subset [M_k, N_k]\). Thus, we obtain

$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{\left| \left( S+l_1\right) \cap \left( S+l_2\right) \cap \left[ M_{k_n}, N_{k_n}\right] \right| }{N_{k_n}-M_{k_n}+1}\ge \frac{c}{|W_1|^2}. \end{aligned}$$

Combining with (2) and above inequality, we get

$$\begin{aligned} {\textrm{BD}}^*\left( \left( S+l_1\right) \cap \left( S+l_2\right) \right) \ge \frac{c}{\left\lceil \llceil \frac{2(1+c)}{{\textrm{BD}}^*(S)}\right\rceil \rrceil ^2} \ge \frac{c}{4(1+c)^2} \left( {{\textrm{BD}}^*(S)}\right) ^2. \end{aligned}$$

Hence, the proof of Theorem 2 is completed. \(\square \)

Proof of Proposition 3

By the assumption, we know that \(\sum _{i=1}^n p_i<p_{n+1}\) for \(1\le n\le 2M-1\). Denote by \(P=\{-\sum _{i=1}^n \, p_i:\, n=1, \cdots , M\}\).

For the set P, by Corollary 1, there exist \(n_1<n_2\) such that

$$\begin{aligned} {\textrm{BD}}^*\left( (S-\sum _{i=1}^{n_1} p_i)\cap (S-\sum _{i=1}^{n_2} p_i)\right) \ge \frac{1}{16}\left( {\textrm{BD}}^*(S)\right) ^2. \end{aligned}$$

Denote by \(S_1=(S-\sum _{i=1}^{n_1} p_i)\cap (S-\sum _{i=1}^{n_2} p_i)\). Let \(r_1=\sum _{i=n_1+1}^{n_2} p_i\) and \(S_2=S\cap (S-r_1)\). It is clear that \(S_2=S_1-\sum _{i=1}^{n_1} p_i\). Thus \(r_1\in W^*\) and \({\textrm{BD}}^*(S_2)= {\textrm{BD}}^*(S_1)>0\).

Note that \(r_1\le \sum _{i=1}^{M} p_i<p_{M+1}\). Let \(Q=\{-\sum _{i=M+1}^{M+n} \, p_i:\, n=1, \cdots , M\}\). With the same arguments, there is \(r_2\in W^*\) such that \(S_3=S_2\cap (S_2-r_2)\) satisfying

$$\begin{aligned} {\textrm{BD}}^*(S_3)\ge \frac{1}{16}\left( {\textrm{BD}}^*(S_2)\right) ^2\ge \frac{1}{16^3}\left( {\textrm{BD}}^*(S)\right) ^4. \end{aligned}$$
(7)

Define \(l_1=r_1\) and \(l_2=r_1+r_2\). It is easy to see that \(l_1, l_2\in W^*\). Let \(q\in S_3\). Then

$$\begin{aligned} \begin{aligned} q&+r_1\in (S_3+r_1)\subset S \\ q&+r_1+r_2\in (S_3+r_2)+r_1\subset (S_2+r_1)\subset S. \end{aligned} \end{aligned}$$

These facts imply that \(q\in (S-l_1)\cap (S-l_2)\), that is \(S_3\subseteq (S-l_1)\cap (S-l_2)\). Hence we get \({\textrm{BD}}^*((S-l_1)\cap (S-l_2))\ge {\textrm{BD}}^*(S_3)\), as desired. \(\square \)

We know that the lower bound of the finite set W in Theorem 2 is necessary. It would be intriguing to explore whether the following problem has a positive answer:

Problem 1

Is the lower bound \(\frac{2(1+c)}{{\textrm{BD}}^*(S)}\) of the finite set W in Theorem 2 the optimal lower bound?