1. Introduction

For nonempty sets $A,B$ of integers, define

\begin{equation*}A+A=\{a+a':a,a'\in A\}\hskip3mm\textrm{and}\hskip3mm A-A=\{a-a':a,a'\in A\}.\end{equation*}

Let $\mathbb{Z}$ be the set of integers and $\mathbb{N}$ the set of positive integers. For any integer n, let $r_{A}(n)$ be the number of representations of n in the form $n=a+a'$, where $a\leqslant a'$ and $a, a'\in A$, and $d_{A}(n)$ be the number of representations of n in the form $n=a-a'$, where $a, a'\in A$. Clearly, $d_{A}(-n)=d_{A}(n)$ for any positive integer n. Let $|A|$ be the cardinality of the set A and $\max A$ be the maximal element in A. For a real number x, denote $|x|$ by the absolute value of x, $\lfloor x\rfloor$ by the largest integer no larger than x, $A+x=\{a+x:a\in A\}$, and $A(-x,x)$ by the number of elements $a\in A$ with $-x\leqslant a\leqslant x$.

The famous Erdős-Turán Conjecture [Reference Erdős and Turán3] states that if A is a set of positive integers such that $r_A(n)\geqslant 1$ for all sufficiently large n, then

\begin{equation*}\limsup_{n\rightarrow\infty}r_A(n)=\infty.\end{equation*}

In 2004, Nešetřil and Serra [Reference Nešetřil and Serra6] initially introduced the notation of “bounded” property. For a positive integer n, denote the binary support of n by the subset S(n) of nonnegative integers consisting of the exponents in the binary expansion of n, i.e., $n=\sum_{i\in S(n)} 2^i$, and $S(-n)=-S(n)$. Define $S(0)=\emptyset$. A set A of integers is called bounded if there is a function $f:\mathbb{N}\bigcup\{0\}\rightarrow \mathbb{N}\bigcup\{0\}$ such that $f(0)=0$ and for each $n\in A+A$ there exists a pair $x,y\in A$ with

\begin{equation*}n=x+y,\hskip6mm|S(x)\bigcup S(y)|\leqslant f(|S(n)|).\end{equation*}

Obviously, if A is a set of positive integers and the binary expansion of each element in A has no two consecutive 1’s, then A is a bounded set with $f(n)=n$. Nešetřil and Serra [Reference Nešetřil and Serra6] confirmed the Erdős-Turán conjecture for a class of “bounded” bases.

For a set A of integers, A is a basis for $\mathbb{Z}$ if $r_A(n)\geqslant 1$ for all integers n and a unique representation basis for $\mathbb{Z}$ if $r_A(n)=1$ for all integers n. For the unique representation basis for $\mathbb{Z}$, by considering the bounded property, Nešetřil and Serra [Reference Nešetřil and Serra6] also obtained the following result:

Theorem B. ([Reference Nathanson5, Theorem 2])

There is a unique representation basis A for $\mathbb{Z}$ such that

\begin{eqnarray*}\frac{2\log x}{\log 5}+2(1-\frac{\log 3}{\log 5})\leqslant A(-x,x)\leqslant \frac{2\log x}{\log 3}+2 \end{eqnarray*}

for all $x\geqslant 1$.

Afterwards, Xiong and Tang [Reference Xiong and Tang7] extended Theorem B by considering the structure of difference, and constructed a unique representation basis A of integers such that $d_A(n)=1$ for all positive integers n and

\begin{equation*}\frac{4(\log x-\log 2)}{\log 15}-1 \lt A(-x,x) \lt \frac{4(\log x-\log 2)}{\log 3}+7\end{equation*}

for all x > 1.

In this paper, based on the above results, we incorporate the bounded property and prove that:

Theorem 1.1. There exists a bounded basis A of integers such that $r_A(n)=1$ for all integers n and $d_A(n)=1$ for all positive integers n, and

\begin{equation*}A(-x,x) \gt \frac{4}{\log 5}\log\log x+c \hskip2mm \text{for all} \hskip2mm x\geqslant 1,\end{equation*}

where c is an absolute constant.

On the other hand, similar to [Reference Nathanson5] and [Reference Xiong and Tang7], we also obtain the following result:

Theorem 1.2. Let f(x) be a function such that $\lim_{x\rightarrow\infty}f(x)=\infty$. Then there exists a bounded basis A of integers such that $r_A(n)=1$ for all integers n and $d_A(n)=1$ for all positive integers n, and

\begin{equation*}A(-x,x)\leqslant f(x) \hskip2mm\textrm{for all sufficiently large}\hskip2mm x.\end{equation*}

Furthermore, noting that if $r_{A}(n)=2$ for infinitely many integers, then $d_A(n)\geqslant 2$ for infinitely many integers n, Cilleruelo and Nathanson [Reference Cilleruelo and Nathanson2] posed the following problem:

Cilleruelo-Nathanson Problem. Give general conditions for functions f ₁ and f ₂ to assure that there exists a set A such that $d_A(n)\equiv f_1(n)$ and $r_{A}(n)\equiv f_2(n)$. Is the condition $\liminf_{u\rightarrow \infty}f_1(u)\geqslant 2$ and $\liminf_{|u|\rightarrow \infty}f_2(u)\geqslant 2$ sufficient?

In 2011, Y.G. Chen and the author [Reference Chen and Fang1] answered this problem affirmatively. In this paper, we also consider the bounded property and obtain that:

2. Proof of Theorem 1.1 and Theorem 1.2

The main idea is from [Reference Nathanson5]-[Reference Xiong and Tang7]. During the induction process, we focus on the choice of critical values. Denote $\sigma(n)$ by

\begin{equation*}S(\sigma(n))=\{i\in S(n): i-1\not\in S(n)\} \hskip2mm \textrm{for positive integer} \hskip2mm n\end{equation*}

and

\begin{equation*}S(\sigma(n))=\{i\in S(n): i+1\not\in S(n)\} \hskip2mm \textrm{for negative integer} \hskip2mm n.\end{equation*}

It easily follows from the definition of $\sigma(n)$ that $|S(n+\sigma(n))|=|S(\sigma(n))|\leqslant |S(n)|$, $S(\sigma(n))$ and $S(n+\sigma(n))$ has no two consecutive integers.

Lemma 2.1. ([Reference Fang4, Lemma 2.1])

Let $x,y,z$ be integers with yz > 0 such that

1. (i) $|S(|y|)|\leqslant |S(|z|)|$;

2. (ii) a > b for any $a\in S(|z|)$ and $b\in S(|x|)\bigcup S(|y|)$;

3. (iii) each of S(x), S(y) and S(z) has no two consecutive integers.

Then

\begin{equation*}|S(x)\bigcup S(y+z)|\leqslant 4|S(x+y+z)|.\end{equation*}

Proof of Theorem 1.1. We will construct finite sets of integers $A_1\subseteq A_2\subseteq\cdots\subseteq A_k\subseteq \cdots$ such that for any positive integer k, we have:

1. (i) $|A_k|=4k+3$;

2. (ii) $r_{A_k}(n)\leqslant 1$ for all $n\in \mathbb{Z}$ and $d_{A_k}(n)\leqslant 1$ for all $n\in \mathbb{N}$;

3. (iii) $r_{A_{k}}(n)=1$ for all $n\in \mathbb{Z}$ with $|n|\leqslant \lfloor\frac{k}{2}\rfloor$ and $d_{A_k}(n)=1$ for all $n\in \mathbb{N}$ with $1\leqslant n\leqslant k$;

4. (iv) $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in A_k$;

5. (v) the binary support of each element in A_k has no two consecutive integers;

6. (vi) $d_{k} \lt 172d_{k-1}^5$, where $d_k=\max\{|a|:a\in A_k\}$ and $d_0=1$.

Let $A_1=\{-32,-10,0,9,33,128,129\}$. Then $d_1=129$, $r_{A_1}(0)=1$, $r_{A_1}(1)=r_{A_1}(-1)=d_{A_1}(1)=1$, $r_{A_1}(n)\leqslant 1$ for all $n\in \mathbb{Z}$ and $d_{A_1}(n)\leqslant 1$ for all $n\in \mathbb{N}$, $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in A_1$, and the binary support of each element in A ₁ has no two consecutive integers. Thus, (i)-(vi) hold for k = 1.

Assume that we have already obtained a set A_k of integers satisfying (i)-(vi) for some positive integer k. By the definition of d_k we know that $A_k\subseteq [-d_k,d_k]$. Since $r_{A_k}(0)\leqslant 1$, we have $d_k\in A_k$ and $-d_k\not\in A_k$, or $-d_k\in A_k$ and $d_k\not\in A_k$. Thus, $A_k+A_k\subseteq [-2d_k+2,2d_k]$ or $[-2d_k,2d_k-2]$. In any case, we have $A_k-A_k\subseteq [-2d_k+1,2d_k-1]$. Write

\begin{equation*} u_k=\min\{|n|:n\not\in A_k+A_k\},\hskip4mm v_k=\min\{n \gt 0:n\not\in A_k-A_k\}.\end{equation*}

It follows that

\begin{equation*} 2\leqslant u_k\leqslant 2d_k-1,\hskip3mm 2\leqslant v_k\leqslant 2d_k.\end{equation*}

Let

(2.1)\begin{eqnarray} a_k=\lceil\max\{\log_2\left(\frac{3d_k+1-\sigma(u_k)}{4^{|S(u_k)|-1}}\right), \log_2d_k+1, \max S(u_k)+3\}\rceil, \end{eqnarray}

where $\lceil x \rceil$ is the least integer no less than x. Then

\begin{equation*} \sigma(u_k)+2^{a_k}4^{|S(u_k)|-1}\geqslant 3d_k+1.\end{equation*}

Take

(2.2)\begin{eqnarray} x_k=u_k+\sigma(u_k)+2^{a_k}+2^{a_k+2}+2^{a_k+4}+\cdots+2^{a_k+2(|S(u_k)|-1)}. \end{eqnarray}

Thus, $x_k-u_k\geqslant 3d_k+1$. Furthermore,

(2.3)\begin{eqnarray} x_k=u_k+\sigma(u_k)+2^{a_k}\frac{4^{|S(u_k)|}-1}{3} \lt 4d_k+\frac{1}{3}4^{|S(u_k)|}2^{a_k}, \end{eqnarray}

where the last inequality is based on the facts that $\sigma(u_k)\leqslant u_k$ and $u_k\leqslant 2d_k-1$. If $a_k=\lceil \log_2\left(\frac{3d_k+1-\sigma(u_k)}{4^{|S(u_k)|-1}}\right)\rceil$, then

\begin{equation*}4d_k+\frac{1}{3}4^{|S(u_k)|}2^{a_k} \lt 4d_k+\frac{1}{3}4^{|S(u_k)|}(2\cdot\frac{3d_k}{4^{|S(u_k)|-1}})=12d_k.\end{equation*}

If $a_k=\lceil \log_2d_k+1\rceil$, then by $|S(u_k)|\leqslant \max S(u_k)+1$ and $2^{\max S(u_k)}\leqslant u_k$ we know that

\begin{eqnarray*}4d_k+\frac{1}{3}4^{|S(u_k)|}2^{a_k} & \lt &4d_k+\frac{1}{3}4^{|S(u_k)|}(2\cdot2^{\log_2d_k+1})\\ &\leqslant &4d_k+\frac{4}{3}d_k4^{\max S(u_k)+1}\leqslant 4d_k+\frac{16}{3}d_ku_k^2\\ &\leqslant &4d_k+\frac{64}{3}d_k^3 \lt 22d_k^3.\end{eqnarray*}

If $a_k=\max S(u_k)+3$, then

\begin{eqnarray*}4d_k+\frac{1}{3}4^{|S(u_k)|}2^{a_k} &=&4d_k+\frac{8}{3}4^{|S(u_k)|}2^{\max S(u_k)}\leqslant 4d_k+\frac{32}{3}2^{3\max S(u_k)}\\ &\leqslant &4d_k+\frac{32}{3}u_k^3\leqslant 86d_k^3.\end{eqnarray*}

In any case,

(2.4)\begin{eqnarray} 4d_k+\frac{1}{3}4^{|S(u_k)|}2^{a_k}\leqslant 86d_k^3. \end{eqnarray}

It infers from (2.1) and (2.3) that

\begin{equation*} x_k \lt 86d_k^3.\end{equation*}

Let

(2.5)\begin{eqnarray} b_k=\lceil\max\{\log_2\left(\frac{3x_k+2u_k-\sigma(v_k)}{4^{|S(v_k)|-1}}\right),\max S(v_k)+3, a_k+2|S(u_k)|-1\}\rceil. \end{eqnarray}

Then

\begin{equation*} \sigma(v_k)+2^{b_k}4^{|S(v_k)|-1}\geqslant3x_k+2u_k.\end{equation*}

Take

(2.6)\begin{eqnarray} y_k=\sigma(v_k)+2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots+2^{b_k+2(|S(v_k)|-1)}. \end{eqnarray}

Thus, $y_k\geqslant3x_k+2u_k$. Furthermore,

(2.7)\begin{eqnarray} y_k+v_k=v_k+\sigma(v_k)+2^{b_k}\frac{4^{|S(v_k)|}-1}{3} \lt 4d_k+\frac{1}{3}4^{|S(v_k)|}2^{b_k}, \end{eqnarray}

where the last inequality is based on the facts that $\sigma(v_k)\leqslant v_k$ and $v_k\leqslant 2d_k$. If $b_k=\lceil \log_2\left(\frac{3x_k+2u_k-\sigma(v_k)}{4^{|S(v_k)|-1}}\right)\rceil$, then by $x_k \lt 86 d_k^3$ we know that

\begin{equation*}4d_k+\frac{1}{3}4^{|S(v_k)|}2^{b_k} \lt 4d_k+\frac{1}{3}4^{|S(v_k)|}(2\cdot\frac{3x_k+2u_k}{4^{|S(v_k)|-1}}) =4d_k+\frac{8}{3}(3x_k+2u_k) \lt 689d_k^3.\end{equation*}

If $b_k=\max S(v_k)+3$, then we could deduce from $|S(v_k)|\leqslant \max S(v_k)+1$ and $2^{\max S(v_k)}\leqslant v_k$ that

\begin{eqnarray*}4d_k+\frac{1}{3}4^{|S(v_k)|}2^{b_k} &=&4d_k+\frac{8}{3}4^{|S(v_k)|}2^{\max S(v_k)}\leqslant 4d_k+\frac{32}{3}2^{3\max S(v_k)}\\ &\leqslant &4d_k+\frac{32}{3}v_k^3\leqslant 86d_k^3.\end{eqnarray*}

If $b_k=a_k+2|S(u_k)|-1$, then it infers from (2.4) that

\begin{eqnarray*}4d_k+\frac{1}{3}4^{|S(v_k)|}2^{b_k} &=&4d_k+\frac{1}{3}4^{|S(u_k)|}2^{a_k}\cdot\frac{1}{2}4^{|S(v_k)|}\leqslant 86d_k^3\cdot\frac{1}{2}4^{|S(v_k)|}\\ &\leqslant& 86d_k^3\cdot\frac{1}{2}(2d_k)^2=172d_k^5. \end{eqnarray*}

In any case,

\begin{eqnarray*}4d_k+\frac{1}{3}4^{|S(v_k)|}2^{b_k}\leqslant 172d_k^5.\end{eqnarray*}

It infers from (2.5) and (2.7) that

\begin{equation*} y_k+v_k \lt 172d_k^5.\end{equation*}

To sum up,

(2.8)\begin{equation}3d_k \lt x_k-u_k \lt x_k \lt y_k \lt y_k+v_k \lt 172d_k^5. \end{equation}

Now we divide into the following two cases according to $u_k\not\in A_k+A_k$ or $u_k\in A_k+A_k$.

Case 1. $u_k\not\in A_k+A_k$.

Let

\begin{equation*}B_{k+1}=A_k\bigcup\{x_k,-x_k+u_k\} \hskip3mm \textrm{and} \hskip2mm A_{k+1}=B_{k+1}\bigcup\{y_k,y_k+v_k\}.\end{equation*}

It follows from $x_k \gt x_k-u_k \gt 3d_k$, $3x_k+2u_k\leqslant y_k \lt y_k+v_k$ and the definitions of d_k, x_k, y_k, we know that $r_{A_{k+1}}(u_k)=d_{A_{k+1}}(v_k)=1$, $r_{A_{k+1}}(n)\leqslant 1$ for all $n\in \mathbb{Z}$ and $d_{A_{k+1}}(n)\leqslant 1$ for all $n\in \mathbb{N}$. Thus, (ii) holds. By $a_k\geqslant\max S(u_k)+3$, $b_k\geqslant\max S(v_k)+3$ and the definition of $A_{k+1}$ we know that (i) and (v) hold. We will prove that $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in A_{k+1}$. If x = y, then

\begin{equation*}|S(x)\bigcup S(y)|=|S(x)|=|S(2x)|=|S(x+y)|\leqslant 4|S(x+y)|.\end{equation*}

So we only need to consider x ≠ y.

Firstly, we will prove that $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in B_{k+1}$ with x ≠ y. Noting that

\begin{eqnarray*}|S(x_k)|&\leqslant& |S(u_k+\sigma(u_k))|+\left|S\left(2^{a_k}+2^{a_k+2} +2^{a_k+4}+\cdots+2^{a_k+2(|S(u_k)|-1)}\right)\right|\\ &=&|S(\sigma(u_k))|+|S(u_k)|\leqslant 2|S(u_k)|\end{eqnarray*}

and

\begin{eqnarray*}|S(-x_k+u_k)|&=&\left|S\left(\sigma(u_k)+2^{a_k}+2^{a_k+2} +2^{a_k+4}+\cdots+2^{a_k+2(|S(u_k)|-1)}\right)\right|\\ &\leqslant& |S(\sigma(u_k))|+|S(u_k)|\leqslant 2|S(u_k)|,\end{eqnarray*}

we have

\begin{equation*}|S(x_k)\bigcup S(-x_k+u_k)|\leqslant 4|S(u_k)|=4|S(x_k+(-x_k+u_k))|.\end{equation*}

Let $x\in A_k$. By (2.1) we have $a_k\geqslant \log_2 d_k+1$, then $a_k \gt \max S(|x|)$. Also by (2.1), we know that $a_k\geqslant\max S(u_k)+3$. Taking $y_1=u_k+\sigma(u_k)$ and $z_1=2^{a_k}+2^{a_k+2}+2^{a_k+4}+\cdots+2^{a_k+2(|S(u_k)|-1)}$ in Lemma 2.1, we have

\begin{eqnarray*}|S(x)\bigcup S(x_k)|=|S(x)\bigcup S(y_1+z_1)|\leqslant 4|S(x+y_1+z_1)|=4|S(x+x_k)|.\end{eqnarray*}

Taking $y_2=-\sigma(u_k)$ and $z_2=-(2^{a_k}+2^{a_k+2}+2^{a_k+4}+\cdots+2^{a_k+2(|S(u_k)|-1)})$ in Lemma 2.1, we have

\begin{eqnarray*}|S(x)\bigcup S(-x_k+u_k)|=|S(x)\bigcup S(y_2+z_2)|\leqslant 4|S(x+y_2+z_2)|=4|S(x-x_k+u_k)|.\end{eqnarray*}

Thus, $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in B_{k+1}$ with x ≠ y.

Now, we will prove that $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in A_{k+1}$ with x ≠ y. It follows from (2.6) that

\begin{eqnarray*}|S(y_k)|&=&\left|S\left(\sigma(v_k)+2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots +2^{b_k+2(|S(v_k)|-1)}\right)\right|\\ &\leqslant& |S(\sigma(v_k))|+\left|S\left(2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots +2^{b_k+2(|S(v_k)|-1)}\right)\right|\\ &=&|S(\sigma(v_k))|+|S(v_k)|\leqslant 2|S(v_k)|\end{eqnarray*}

and

\begin{eqnarray*}|S(y_k+v_k)|&=&\left|S\left(v_k+\sigma(v_k)+2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots +2^{b_k+2(|S(v_k)|-1)}\right)\right|\\ &\leqslant&|S(v_k+\sigma(v_k))|+|S(v_k)|\leqslant 2|S(v_k)|,\end{eqnarray*}

we have

\begin{equation*}|S(y_k)\bigcup S(y_k+v_k)|\leqslant 4|S(v_k)|.\end{equation*}

By $b_k\geqslant\max S(v_k)+3$ and

\begin{eqnarray*}S(2y_k+v_k)=S(2\sigma(v_k)+v_k+2\left(2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots +2^{b_k+2(|S(v_k)|-1)}\right)),\end{eqnarray*}

we know that

\begin{eqnarray*}|S(2y_k+v_k)|&=&|S(2\sigma(v_k)+v_k)|+\left|S\left(2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots +2^{b_k+2(|S(v_k)|-1)}\right)\right|\\ &\geqslant& \left|S\left(2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots +2^{b_k+2(|S(v_k)|-1)}\right)\right|=|S(v_k)|.\end{eqnarray*}

Thus,

\begin{equation*}|S(y_k)\bigcup S(y_k+v_k)|\leqslant 4|S(2y_k+v_k)|.\end{equation*}

Let $x\in B_{k+1}$. By (2.5) we have $b_k\geqslant a_k+2|S(u_k)|-1$, namely, $b_k \gt a_k+2(|S(u_k)|-1)$. Then $b_k \gt \max S(|x|)$. Also by (2.5), we know that $b_k\geqslant\max S(v_k)+3$. Taking $y_1=\sigma(v_k)$ and $z_1=2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots +2^{b_k+2(|S(v_k)|-1)}$ in Lemma 2.1, we have

\begin{eqnarray*}|S(x)\bigcup S(y_k)|=|S(x)\bigcup S(y_1+z_1)|\leqslant 4|S(x+y_1+z_1)|=4|S(x+y_k)|.\end{eqnarray*}

Taking $y_2=\sigma(v_k)+v_k$ and $z_2=2^{b_k}+2^{b_k+2}+2^{b_k+4}+\cdots +2^{b_k+2(|S(v_k)|-1)}$ in Lemma 2.1, we have

\begin{eqnarray*}|S(x)\bigcup S(y_k+v_k)|=|S(x)\bigcup S(y_2+z_2)|\leqslant 4|S(x+y_2+z_2)|=4|S(x+y_k+v_k)|.\end{eqnarray*}

To sum up, $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in A_{k+1}$.

Case 2. $u_k\in A_k+A_k$. Then $-u_k\not\in A_k+A_k$.

Let

\begin{equation*}B_{k+1}=\{x_k,-x_k-u_k\} \hskip3mm \textrm{and} \hskip3mm A_{k+1}=\{y_k,y_k+v_k\},\end{equation*}

where x_k and y_k are defined in (2.2) and (2.6). Similar to Case 1, we know that $A_{k+1}$ satisfies (i)-(ii), (iv)-(v) and $r_{A_{k+1}}(-u_k)=d_{A_{k+1}}(v_k)=1$.

In both cases, it follows from (2.8) and the construction of $A_{k+1}$ that $d_{k+1} \lt 172d_k^5$. Thus, (vi) holds.

Now we will prove that (iii) holds. (The proof of (iii) is the same as in [Reference Xiong and Tang7, Theorem 1.1], we also give the details for the sake of completeness). If $u_k\not\in A_k+A_k$, then by Case 1 we know that $u_k\in A_{k+1}+A_{k+1}$, thus, $u_{k+2}\geqslant u_{k+1} \gt u_k$ if $-u_k\in A_{k+1}+A_{k+1}$. Otherwise, if $-u_k\not\in A_{k+1}+A_{k+1}$, then $u_{k+1}=u_k\in A_{k+1}+A_{k+1}$ and $-u_{k+1}\in A_{k+2}+A_{k+2}$ by Case 2, thus, $u_{k+2} \gt u_{k+1}=u_k$. If $u_k\in A_k+A_k$, then by Case 2 we know that $-u_k\in A_{k+1}+A_{k+1}$. It follows from $u_k\in A_k+A_k\subseteq A_{k+1}+A_{k+1}$ that $u_{k+2}\geqslant u_{k+1} \gt u_k$. In both cases, $u_{k+2} \gt u_k$. It follows from $u_2\geqslant 2$ that $u_{2k}\geqslant u_2+k-1\geqslant k+1$. Thus, for any positive integer k we have

\begin{equation*}\{-k,\cdots,-1,0,1,\cdots,k\}\subseteq A_{2k}+A_{2k}.\end{equation*}

Similarly, $v_k \lt v_{k+1}$. It infers from $v_1\geqslant 2$ that $v_k\geqslant k+1$. Thus, for any positive integer k we have

\begin{equation*}\{-k,\cdots,-1,0,1,\cdots,k\}\subseteq A_{k}-A_{k}.\end{equation*}

Namely, $r_{A_{k+1}}(n)=1$ for all $n\in \mathbb{Z}$ with $|n|\leqslant \lfloor\frac{k+1}{2}\rfloor$ and $d_{A_{k+1}}(n)=1$ for all $n\in \mathbb{N}$ with $1\leqslant n\leqslant k+1$. Thus, (iii) holds.

Let

\begin{equation*}A=\bigcup_{k=1}^{\infty} A_k.\end{equation*}

Then $r_A(n)=1$ for all $n\in \mathbb{Z}$ and $d_A(n)=1$ for all $n\in \mathbb{N}$, $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in A$. Furthermore, we could deduce from (vi) and $d_0=1$ that

\begin{eqnarray*} d_k\leqslant c_1^{5^k}, \hskip3mm \textrm{where} \hskip3mm c_1=\sqrt[4]{172}. \end{eqnarray*}

For sufficiently large x, there exists a positive integer k such that $d_k\leqslant x \lt d_{k+1}$. It follows from $4k+3\leqslant A(-x,x)\leqslant 4k+7$ that

\begin{equation*}A(-x,x) \gt \frac{4}{\log 5}\log\log x+c, \hskip2mm\textrm{where}\hskip2mm c \hskip2mm\textrm{is an absolute constant}.\end{equation*}

This completes the proof of Theorem 1.1. $\Box$

Proof of Theorem 1.2. In the proof of Theorem 1.1, the only constraint on the choice of x_k (resp. y_k) is the size of the value a_k (resp. b_k). The following proof is similar to [Reference Nathanson5, Theorem 1] and [Reference Xiong and Tang7, Theorem 1.1]. We apply the method of Theorem 1.1 by replacing a_k with $s_k(\geqslant a_k)$. Namely, take

(2.9)\begin{equation}x_k=u_k+\sigma(u_k)+2^{s_k}+2^{s_k+2} +2^{s_k+4}+\cdots+2^{s_k+2(|S(u_k)|-1)}.\end{equation}

Given a function f(x) tending to infinity, we shall take induction on k to construct a non-decreasing sequence of integers $\{h_k\}_{k=1}^\infty$ such that $A(-x,x)\leqslant f(x)$ for all integers x with $h_1\leqslant x\leqslant d_k$. Firstly, choose $h_1\geqslant d_1$ so that $f(x)\geqslant 11$ for $x\geqslant h_1$. Then

\begin{equation*} A(-x,x)\leqslant 11\leqslant f(x) \hskip2mm\textrm{for}\hskip2mm h_1\leqslant x\leqslant d_2.\end{equation*}

Suppose that for some integer $k\geqslant 2$, we have already selected an integer $h_{k-1}\geqslant d_{k-1}$ such that

\begin{equation*} f(x)\geqslant 4k+3 \hskip2mm\textrm{for}\hskip2mm x\geqslant h_{k-1}, \hskip3mm A(-x,x)\leqslant f(x) \hskip2mm\textrm{for}\hskip2mm h_1\leqslant x\leqslant d_k.\end{equation*}

Noting that f(x) tends to infinity, there exist positive integers h_k and $s_{k+1}$ with $h_k\geqslant d_k$ and $h_k \lt x_{k+1}-u_{k+1}$ (taking large $s_{k+1}$ in (2.9)) such that $f(x)\geqslant 4k+7$ for $x\geqslant h_{k}$. It follows that

\begin{equation*} A(-x,x)\leqslant 4k+7\leqslant f(x) \hskip2mm\textrm{for}\hskip2mm h_k\leqslant x\leqslant d_{k+1}.\end{equation*}

For $d_k\leqslant x\leqslant h_k$, we could deduce from the construction of $A_{k+1}\setminus A_k$ and the fact $h_k \lt x_{k+1}-u_{k+1}$ that

\begin{equation*} A(-x,x)=A_k(-x,x)=4k+3\leqslant f(x)\hskip2mm\textrm{for}\hskip2mm d_k\leqslant x\leqslant h_k.\end{equation*}

To sum up,

\begin{equation*} A(-x,x)\leqslant f(x) \hskip2mm\textrm{for}\hskip2mm d_k\leqslant x\leqslant d_{k+1}.\end{equation*}

By the induction hypothesis we know that $A(-x,x)\leqslant f(x)$ for $h_1\leqslant x\leqslant d_{k+1}$. It follows that

\begin{equation*}A(-x,x)\leqslant f(x) \hskip3mm\textrm{for all}\hskip3mm x\geqslant h_1. \end{equation*}

This completes the proof of Theorem 1.2.

3. Proof of Theorem 1.3

To give the proof of Theorem 1.3, we need the following preliminary lemmas. The idea is from [Reference Chen and Fang1, Theorem 1.2], [Reference Fang4, Theorem 1.1] and [Reference Nešetřil and Serra6, Theorem 5].

Lemma 3.1. Let $f_1:\mathbb{N}\rightarrow \mathbb{N}$ and $f_2:\mathbb{Z}\rightarrow \mathbb{N}$ be two functions such that

(3.1)\begin{eqnarray} \liminf_{u\rightarrow \infty}f_1(u)\geqslant 2 \hskip4mm \textrm{and} \hskip4mm \liminf_{|u|\rightarrow \infty}f_2(u)\geqslant 2. \end{eqnarray}

Let $B\subseteq \mathbb{Z}$ be a finite set with $|B|\geqslant 2$ such that:

1. (i) $d_B(n)\leqslant f_1(n)$ for all $n\in \mathbb{N}$ and $r_B(n)\leqslant f_2(n)$ for all $n\in \mathbb{Z}$;

2. (ii) $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in B$;

3. (iii) the binary support of each element in B has no two consecutive integers.

   If k is a positive integer with $d_B(k) \lt f_1(k)$, then there exists a finite set D with $B\subseteq D\subseteq \mathbb{Z}$ such that:

4. (iv) $d_D(k)=d_B(k)+1$;

5. (v) $d_D(n)\leqslant f_1(n)$ for all $n\in \mathbb{N}$ and $r_D(n)\leqslant f_2(n)$ for all $n\in \mathbb{Z}$;

6. (vi) $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in D$;

7. (vii) the binary support of each element in D has no two consecutive integers.

Proof. Let $B=\{b_1,b_2,\cdots,b_s\}$, where $b_1 \lt b_2 \lt \cdots \lt b_s$. Let $m=2\max_{1\leqslant j\leqslant s}|b_j|+k$. By (3.1), we could choose a subset U_k of positive integers such that:

1. (1) $|U_k|=|S(k)|$;

2. (2) $\min U_k \gt k+3\max ||B||$, where $||B||=\{|b|:b\in B\}$;

3. (3) U_k has no two consecutive integers;

4. (4) $f_1(n)\geqslant 2$ and $f_2(n)\geqslant 2$ for all integers $n\in [b-m,b+m]\bigcap \mathbb{Z}$, where $b=\sigma(k)+\sum_{i\in U_k} 2^i$.

Let

\begin{equation*}D=B\bigcup \{b,b+k\}.\end{equation*}

Then

\begin{equation*}D+D=\{2b,2b+k,2b+2k\}\bigcup\{B+B\}\bigcup\{B+b\}\bigcup\{B+b+k\}\end{equation*}

and

\begin{equation*}D-D=\pm\{\{k\}\bigcup\{B-B\}\bigcup\{B-b\}\bigcup\{B-b-k\}\}.\end{equation*}

We could deduce from (i)-(iii), the definition of D and the fact $b \gt 3\max ||B||+k$ that $d_D(k)=d_B(k)+1$, and the binary support of each element in D has no two consecutive integers. Furthermore, $r_D(2b)=r_D(2b+k)=r_D(2b+2k)=1$. It also follows from

\begin{eqnarray*}&&b-m \lt b+b_1 \lt b+b_2 \lt \cdots \lt b+b_s \lt b+m,\\ &&b-m \lt b+k+b_1 \lt b+k+b_2 \lt \cdots \lt b+k+b_s \lt b+m\end{eqnarray*}

and (4) that $r_D(n)\leqslant 2\leqslant f_2(n)$ for each $n\in \{B+b\}\bigcup\{B+b+k\}$. Noting that the sets $\{2b,2b+k,2b+2k\}$, B + B, B + b and $B+b+k$ are pairwise disjoint, we know that $r_D(n)\leqslant f_2(n)$ for all integers n. Similarly, by

\begin{eqnarray*}&&b-m \lt b-b_s \lt b-b_{s-1} \lt \cdots \lt b-b_1 \lt b+m,\\ &&b-m \lt b+k-b_s \lt b+k-b_{s-1} \lt \cdots \lt b+k-b_1 \lt b+m,\end{eqnarray*}

and (4) we have $d_D(n)\leqslant 2\leqslant f_1(n)$ for each $n\in \{B-b\}\bigcup\{B-b-k\}$. Noting that the sets B − B, B − b and $B-b-k$ are pairwise disjoint, we know that $d_D(n)\leqslant f_1(n)$ for all positive integers n. By the same proof as in Theorem 1.1, we know that $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in D$.

This completes the proof of Lemma 3.1.

Proof. Let $B=\{b_1,b_2,\cdots,b_s\}$, where $b_1 \lt b_2 \lt \cdots \lt b_s$. Let $m=2\max_{1\leqslant j\leqslant s}|b_j|+|k|$. By (3.1), we could choose a subset U_k of positive integers such that:

1. (1) $|U_k|=|S(k)|$;

2. (2) $\min U_k \gt |k|+3\max ||B||$;

3. (3) U_k has no two consecutive integers;

4. (4) $f_1(n)\geqslant 2$ and $f_2(n)\geqslant 2$ for all integers $n\in [b-m,b+m]\bigcap \mathbb{Z}$, where

   \begin{eqnarray*}b= \begin{cases} k+\sigma(k)+\sum_{i\in U_k} 2^i, &\textrm{if}\hskip2mm k \gt 0, \cr \sum_{i\in U_k} 2^i, & \textrm{if}\hskip2mm k=0, \cr k+\sigma(k)+\sum_{i\in U_k} 2^{-i}, & \textrm{if}\hskip2mm k \lt 0.\end{cases} \end{eqnarray*}

Let

\begin{equation*}D=B\bigcup \{b,-b+k\}.\end{equation*}

Then

\begin{equation*}D+D=\{k,2b,-2b+2k\}\bigcup(B+B)\bigcup(B+b)\bigcup(B-b+k)\end{equation*}

and

\begin{equation*}D-D=\pm\{\{2b-k\}\bigcup(B-B)\bigcup(B-b)\bigcup(B+b-k)\}.\end{equation*}

We could deduce from (i)-(iii), the definition of D and the fact $b \gt 3\max ||B||+k$ that $r_D(k)=r_B(k)+1$, and the binary support of each element in D has no two consecutive integers. Furthermore, $r_D(2b)=r_D(-2b+2k)=1$. It also follows from

\begin{eqnarray*}&&b-m \lt b+b_1 \lt b+b_2 \lt \cdots \lt b+b_s \lt b+m,\\ &&b-m \lt -b+k+b_1 \lt -b+k+b_2 \lt \cdots \lt -b+k+b_s \lt b+m\end{eqnarray*}

and (4) that $r_D(n)\leqslant 2\leqslant f_2(n)$ for each $n\in \{B+b\}\bigcup\{B-b+k\}$. Noting that the sets $\{2b,-2b+2k\}$, B + B, B + b and $B-b+k$ are pairwise disjoint, we know that $r_D(n)\leqslant f_2(n)$ for all integers n. Similarly, by

\begin{eqnarray*}&&-2b-k \lt b-m \lt b-b_s \lt b-b_{s-1} \lt \cdots \lt b-b_1 \lt b+m \lt 2b+k,\\ &&b-m \lt b-k+b_1 \lt b-k+b_2 \lt \cdots \lt b-k+b_s \lt b+m\end{eqnarray*}

and (4) we have $d_D(n)\leqslant 2\leqslant f_1(n)$ for each $n\in \{B-b\}\bigcup\{B+b-k\}$. Noting that the sets $\{2b-k\}$, B − B, B − b and $B+b-k$ are pairwise disjoint, we know that $d_D(n)\leqslant f_1(n)$ for all positive integers n. By the same proof as in Theorem 1.1, we know that $|S(x)\bigcup S(y)|\leqslant 4|S(x+y)|$ for $x,y\in D$.

This completes the proof of Lemma 3.2.

Proof of Theorem 1.3. Theorem 1.3 follows from Lemma 3.1 and Lemma 3.2. The proof is similar to Theorem 1.1, we omit the detail here.