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\title{On direct and inverse problems related to some dilated sumsets}

\author{\firstname{Ramandeep} \lastname{Kaur}}
\address{Department of Mathematics, Akal University, Talwandi Sabo - 151302, India}
\email{ramandeepka71@gmail.com}

\author{\firstname{Sandeep} \lastname{Singh}\IsCorresp}
\address[1]{Department of Mathematics, Akal University, Talwandi Sabo - 151302, India}
\email{sandeepinsan86@gmail.com}

\thanks{The research of the second author is funded by NBHM (Sanction No. 02011/49/2023/R\&D-II/13983)}
\CDRGrant[NBHM]{02011/49/2023/R\&D-II/13983}

\begin{abstract}
Let $A$ be a nonempty finite set of integers. For a real number $m$, the set $m\cdot A=\{ma: a\in A\}$ denotes the set of $m$-dilates of $A$. In 2008, Bukh initiated an interesting problem of finding a lower bound for the sumset of dilated sets, i.e., a lower bound for $|\lambda_1\cdot A+\lambda_2\cdot A+\cdots+\lambda_h\cdot A|$, where $\lambda_1, \lambda_2, \dots, \lambda_h$ are integers and $A$ be a subset of integers. In particular, for nonempty finite subsets $A$ and $B$, the problem of dilates of $A$ and $B$ is defined as $A+k\cdot B=\{a+kb:a\in A$ and $b\in B\}$. In this article, we obtain the lower bound for the cardinality of $A+k\cdot B$ with $k\geq 3$ and describe sets for which equality holds. We also derive an extended inverse result with some conditions for the sumset $A+3\cdot B$.
\end{abstract}

\subjclass{11B13, 11B75}
\keywords{\kwd{Sum of dilates}
\kwd{direct and inverse problems}
\kwd{additive combinatorics}}


\dateposted{2024-02-02}
\begin{document}
\maketitle
\vspace{-0.5em}
\section{Introduction}

Let $A$ be a finite set of integers and $k$ be any integer. The $k$-dilation, $k \cdot A$ of $A$ is defined by $k\cdot A=\{ka: a\in A\}$. Classically, there are two types of problems of sumsets in additive number theory, called direct and inverse problems. In \emph{direct problems}, one starts with a set and tries to describe the size of sumsets (of any type) associated with given set, called direct problems. In case of \emph{inverse problems} one starts with the cardinality of a sumsets obtained from direct problem and tries to find the structure of set. More generally, in \emph{extended inverse problems} one tries to find the structure of a sumset by assuming some arbitrary cardinality of a sumset. The main aim is to find the lower bound for the cardinality of sets of type $\lambda_1\cdot A_1+\lambda_2\cdot A_2+\cdots+\lambda_h\cdot A_h$, where $\lambda_1\cdot A_1+\lambda_2\cdot A_2+\cdots+\lambda_h\cdot A_h= \{\lambda_1a_1+\lambda_2a_2+\cdots+\lambda_ha_h\;|\;a_i\in A_i\, and\; \lambda_i\in \mathbb{Z}, \; i=1,2,\dots,h\}$. In $2007,$ Bukh~\cite{bukh} gave an asymptotically sharp lower bound on the size of sumsets of the form $\lambda_1\cdot A+\lambda_2\cdot A+\cdots+\lambda_k\cdot A$, for arbitrary large integers $\lambda_1, \lambda_2, \dots, \lambda_k$ and integer set $A$. Bukh derived the lower bound for $\lambda_1\cdot A+\lambda_2\cdot A+\cdots+\lambda_k\cdot A$ with some error term $o(|A|)$. He proved that for every vector $\bar \lambda=(\lambda_1,\lambda_2,\dots,\lambda_k)\in \mathbb{Z}^{k}$ of coprime $k$-tuple, $|\lambda_1\cdot A+\lambda_2\cdot A+\cdots+\lambda_k\cdot A|\geq (|\lambda_1|+|\lambda_2|+\dots+|\lambda_k|)|A|-o(|A|)$ for a finite set $A\subset \mathbb{Z}$ with the error term $o(|A|)$ depending on $\bar \lambda$ only.

Confining ourselves to the sum of only two dilates, it is enough to consider only the sums $m\cdot A+ k\cdot B$, where $A$ and $B$ are non empty subsets of integers. When both $m$ and $k$ are equal to $1$, the sum $A+B$ is called the Minkowski sum of the sets $A$ and $B$. In a remarkable result by Nathanson~\cite{nathanson96}, it was proven that for non-empty subsets $A$ and $B$ of integers, $|A+B|\geq |A|+|B|-1$, and equality holds if and only if $A$ and $B$ are arithmetic progressions with the same common difference. Also, studying the dilated sumset $m\cdot A+ k\cdot B$, when $A=B$ presents an interesting problem. Researchers have dedicated considerable effort to investigate these sumset problems and have made significant advancements in this field of study. In $2010$, Cilleruelo et al. \cite{cilleruelo10} proved $|A+3\cdot A|\geq 4|A|-4$ and the equality holds only for $A=\{0,1,3\}$ or $A=\{0,1,4\}$ or $A=3\{0,1,\dots n\}\cup (3\{0,1,\dots,n\}+1)$ and all the affine transforms of these sets. In the same paper, they proposed the conjecture that $|A+k\cdot A|\geq (k+1)|A|-\bigl\lceil \frac{k^2+2k}{4}\bigr\rceil,$ where $A$ is any set of sufficiently large cardinality. This conjecture has been well studied in the past and is being studied presently. In $2009,$ Cilleruelo et al. \cite{cilleruelo9} confirmed the conjecture for a prime number $k$ such that $|A|\geq 3(k-1)^2(k-1)!$. In $2014,$ Du et al. \cite{du} verified the conjecture for $k$ to be prime power and product of two distinct primes such that $|A|\geq (k-1)^2k!$. Motivated by the work done on the cardinality of $A+k\cdot A,$ several authors proved various results on the cardinality of $m\cdot A+k\cdot A$. In $2011,$ Hamidoune and Rue~\cite{hamidoune} investigated the scenario, where $m$ is equal to $2$ and $k$ is an odd prime number. Consequently, they proved that for an odd prime $k$ and a finite set $A$ of integers with $|A|>8k^k$, $|2\cdot A+k\cdot A|\geq (k+2)|A|-k^2-k+2$. In $2013,$ Ljujic~\cite{ljujic} expanded upon this finding and derived the same limit for cases, where $k$ is a power of an odd prime or a product of two distinct odd primes. In $2013,$ Balog et al. \cite{balog} proved that $|p\cdot A+q\cdot A|\geq (p+q)|A|-(pq)^{(p+q-3)(p+q)+1},$ where $p<q$ are relatively primes and $A\subseteq \mathbb{Z}$. In $2020,$ Chahal and Pandey~\cite{chahal} handled the case for the cardinality of $3\cdot A+k\cdot A$, under some conditions on $A$ and also generalized this result for $q\cdot A+k\cdot A,$ where $q<k$ is an odd prime. In $2017,$ Freiman et al. \cite{freiman} proved that if $r\geq 3,$ then $|A+r\cdot A|\geq 4|A|-4$. For $r=2,$ they also obtained an extended inverse result, which states that if $|A+2\cdot A|<4|A|-4,$ then $A$ is a subset of arithmetic progression of length at most $2|A|-3$. In 2019, Bhanja et al. \cite{bhanja02} presented an alternative proof of the inequality $|A+r\cdot A|\geq 4|A|-4$ for $r\geq 3$. Additionally, they extended the inverse theorem to the cardinality of the sum of dilates $A+2\cdot B$, where $A$ and $B$ are subsets of the integers.

Let $A=\{a_0,a_1,\dots,a_{r-1}\}$ be a finite subset of integers such that $a_0<a_1<\dots<a_{r-1}$. Suppose $\ell^{*}(a_i)=a_{i}-a_{i-1}$ for all $i=1,2,\dots,r-1$. In Section~\ref{sec2}, we prove the following direct and inverse problem:

\begin{theo}\label{th4}
Let $k\geq 3$ be a positive integer and let $A$ and $B$ be nonempty finite subsets of integers with the properties such that $|A|\leq |B|$, $\ell^{*}(a_i)\leq k$, for all $1\leq i\leq r-1$ and $3\leq \ell^{*}(b_j)\leq k$ for all $1\leq j\leq l-1$. Then $|A+k\cdot B|\geq 3|A|+|B|-4$.

\smallskip\noindent
(Inverse Problem) Furthermore, if $|A+k\cdot B|=3|A|+|B|-4$, then $A$ and $B$ are arithmetic progressions.
\end{theo}

For any set $A$, we define $c_m(A)$ as the count of distinct classes of $A$ modulo $m$. In Section~\ref{sec3}, we obtain extended inverse problem for $|A+3\cdot B|$. More precisely, we prove the following theorem:

\begin{theo}\label{th5}
Let $A$, $B$ $\subseteq \mathbb{Z}$ be finite subsets such that $c_3(A)=t$ and $0\in A,\,B$ with properties
\begin{enumerate}
\item $d(A)=d(B)=1$
\item $\ell(A)\leq \ell(B)$
\item $h_A\leq h_B$.
\end{enumerate}
If $|A+3\cdot B|=|A|+t(|B|-1)+h\leq 2|A|+t(|B|-2)$ for some integer $h$, then both $A$ and $B$ are subsets of arithmetic progressions of length at most $|B|+h=|A+3\cdot B|-|A|-(t-1)|B|+t\leq |A|+|B|-3$.
\end{theo}

\pagebreak
By $d(A)$ we denote the greatest common divisor of $\{a_1-a_0,a_2-a_0,\dots,a_{r-1}-a_0\}$. Let $a_{i}^{\prime}= (a_i-a_0)/d(A)$ for $i= 1 $ to $ r-1$ and $\ell(A)=\max(A)-\min(A)=a_{r-1}-a_{0}$. The set $B=(a_{0}^{\prime},a_{1}^{\prime},\dots,a_{r-1}^{\prime})$ is called normal form of the set $A$. Clearly, $a_{0}^{\prime}<a_{1}^{\prime}<\dots<a_{r-1}^{\prime}$ and $d(B)=1$. Define $h_A=\ell(A)+1-|A|$ the number of holes in set $A$.

The following well known results of Liv~\cite{LS} and Stanchescu~\cite{S} are used frequently for proving our results.

\begin{theo}[{\cite{LS, S}}]\label{LSS}
Let $A$ and $B$ be finite subsets of $\mathbb{N}$ such that $0\in A\cap B$. Define
$\delta_{A,B}=\left\{
\begin{smallmatrix}
1 & \text{if }\ell(A)=\ell(B)\\
0 & \text{if }\ell(A)\neq \ell(B)
\end{smallmatrix}\right.$. Then the followings hold:
\begin{enumerate}
\item If $\ell(A)=\max(\ell(A),\ell(B))\geq |A|+|B|-1-\delta_{A,B}$ and $d(A)=1,$ then
\[
|A+B|\geq |A|+2|B|-2-\delta_{A,B}
\]
\item If $\max(\ell(A),\ell(B))\leq |A|+|B|-2-\delta_{A,B}$, then
\[
|A+B|\geq \max(\ell(A)+|B|,\ell(B)+|A|).
\]
\end{enumerate}
\end{theo}

\section{Proof of Theorem~\ref{th4}}\label{sec2}

\subsection*{Direct problem for \texorpdfstring{$|A+k\cdot B|$}{|A+k. B|}}

\begin{proof}
Let $A=\{a_0<a_1<\dots<a_{r-1}\}$ and $B=\{b_0<b_1<\dots<b_{l-1}\}$ be two finite sets of integers satisfying the given conditions. Consider the following sequence of distinct integers in the sumset $A+k\cdot B,$
\begin{align}
{\label {eq:1}}
a_0+k b_0&< a_1+kb_0<a_0+kb_1<a_1+kb_1<a_2+kb_1 \nonumber \\
&< a_1+kb_2< a_2+kb_2< a_3+kb_2< a_2+kb_3< a_3+kb_3 \nonumber \\
&< a_4+kb_3< a_3+kb_4< \cdots<a_{i-1}+kb_{i-1}<a_i+kb_{i-1}\nonumber\\
&<a_{i-1}+kb_i< a_{i}+kb_{i}< a_{i+1}+kb_i \nonumber \\
&< a_i+kb_{i+1}< a_{i+1}+kb_{i+1}<a_{i+2}+kb_{i+1}\nonumber\\
& <a_{i+1}+kb_{i+2}<\cdots <a_{r-2}+kb_{r-1}<a_{r-1}+kb_{r-1}\nonumber\\
&<a_{r-1}+kb_{r}<a_{r-1}+kb_{r+1}<\cdots<a_{r-1}+kb_{l-1}.
\end{align}
This list contains $3|A|-2+l-r=2|A|+l-2$ integers. To prove the result it remains to find $|A|-2$ more integers of $A+k\cdot B$. Take the following list of six consecutive integers of $ A+k\cdot B$ from~\eqref{eq:1} for every $1\leq i\leq r-2,$
\begin{equation}\label{eq:2}
a_{i-1}+kb_{i-1}<a_i+kb_{i-1}<a_{i-1}+kb_i<a_i+kb_i<a_{i+1}+kb_i<a_i+kb_{i+1}.
\end{equation}
We claim that for each list of integers of type~\eqref{eq:2}, there always exists an integer between $a_{i-1}+kb_{i-1}$ and $a_i+kb_{i+1}$ in the sumset $A+k\cdot B$, which is not in the list~\eqref{eq:1}. Let us verify our claim for every list of type~\eqref{eq:2}. Consider
\begin{align*}
&a_i+kb_{i-1}<a_{i+1}+kb_{i-1}<a_{i-1}+kb_i\\
\text{and}\quad
&a_{i+1}+kb_{i}<a_{i-1}+kb_{i+1}<a_{i}+kb_{i+1}.
\end{align*}
Clearly, $a_i+kb_{i-1}<a_{i+1}+kb_{i-1}$ and $a_{i-1}+kb_{i+1}<a_{i}+kb_{i+1}$ holds for every $i=1,2,\dots,r-2$. We need only to prove that $a_{i+1}+kb_{i-1}<a_{i-1}+kb_i$ and $a_{i+1}+kb_{i}<a_{i-1}+kb_{i+1}$.

On contrary suppose that $a_{i+1}+kb_{i-1}\geq a_{i-1}+kb_i$. It implies $a_{i+1}- a_{i-1}\geq k(b_i-b_{i-1})$, which is a contradiction. As maximum value of $a_{i+1}-a_{i-1}$ can be $2k,$ and $3\leq \ell^{*}(b_j)\leq k$ for all $1\leq j\leq l-1$. Hence $a_{i+1}+kb_{i-1}<a_{i-1}+kb_i$ and similarly $a_{i+1}+kb_{i}< a_{i-1}+kb_{i+1}$.

Next our aim is to show that for any two consecutive lists of six integers of the form~\eqref{eq:2}, we always have two distinct integers of $A+k\cdot B,$ that are not included in~\eqref{eq:1}. Let us consider two lists of six integers
\begin{align}\label{eq:3}
&a_{i-1}+kb_{i-1}<a_i+kb_{i-1}<a_{i-1}+kb_i<a_i+kb_i<a_{i+1}+kb_i<a_i+kb_{i+1}\\
\text{and}\quad
\label{eq:4}
&a_{i}+kb_{i}<a_{i+1}+kb_{i}<a_{i}+kb_{i+1}<a_{i+1}+kb_{i+1}<a_{i+2}+kb_{i+1}<a_{i+1}+kb_{i+2}.
\end{align}
Observe that, $x$ and $y$ in $A+k\cdot B$ such that $a_{i-1}+kb_{i-1}<x<a_i+kb_{i+1}$ and $a_{i}+kb_{i}<y<a_{i+1}+kb_{i+2},$ where $x$, $y$ not in lists~\eqref{eq:3}, \eqref{eq:4}. Our purpose is to show that either $x\neq y$ or there exists integer $z\neq x(=y)$ such that $z\in A+k\cdot B$ and lies between $a_{i-1}+kb_{i-1}$ and $a_{i+1}+kb_{i+2}$.

We claim that there exist two integers $x=a_{i-1}+kb_{i+1}$ and $y=a_{i+2}+kb_i$ satisfying $a_{i+1}+kb_i<a_{i-1}+kb_{i+1}$ and $a_{i+2}+kb_i<a_{i-1}+kb_{i+1}$.

For first identity, assume that $a_{i+1}+kb_i\geq a_{i-1}+kb_{i+1},$ which contradicts as $3\leq \ell^{*}(b_j)\leq k$ for all $1\leq j\leq l-1$. Similarly, if possible, let $a_{i+2}+kb_i\geq a_{i-1}+kb_{i+1}$, again a contradiction. Now, if $a_{i-1}+kb_{i+1}\neq a_{i+2}+kb_i$, then we get two distinct integers $x=a_{i-1}+kb_{i+1}$ and $y=a_{i+2}+kb_i,$ which are not in~\eqref{eq:3} and~\eqref{eq:4}. If $x=a_{i-1}+kb_{i+1}=a_{i+2}+kb_i=y$, we prove that in this case there also exist a new integer $z=a_i+kb_{i+2}$ in~\eqref{eq:4}, which is different from $x=y$. Clearly, $z>y=x$. We have to check that $a_i+kb_{i+2}>a_{i+2}+kb_{i+1}$. If possible, let
\begin{align*}
& a_i+kb_{i+2}\leq a_{i+2}+kb_{i+1}\\
& a_{i+2}-a_{i}\geq k(b_{i+2}-b_{i+1}).
\end{align*}
Since the maximum value of $a_{i+2}-a_{i}$ is $2k$. Therefore, our assumption is incorrect, leading to the confirmation and proof of our claim.

Thus in each case, we get two distinct elements of $A+k\cdot B,$ which are not in~\eqref{eq:3} and~\eqref{eq:4}. Hence, we get $|A|-2$ extra integers of $A+k\cdot B,$ which are not included in~\eqref{eq:1}. Consequently, $|A+k\cdot B|\geq 3|A|+|B|-4$.

\subsection*{Inverse Problem for \texorpdfstring{$|A+k\cdot B|$}{|A+k. B|}}

Let us begin with the case $|A|=|B|=r$ and assume that $A=\{a_0<a_1<\dots<a_{r-1}\}$ and $B=\{b_0<b_1<\dots<b_{r-1}\}$. The sumset $A+k\cdot B$ contains the following strictly increasing sequence of $3|A|-2$ integers
\begin{align}
a_0+k b_0&< a_1+kb_0<a_0+kb_1<a_1+kb_1<a_2+kb_1 \nonumber \\
&< a_1+kb_2< a_2+kb_2< a_3+kb_2< a_2+kb_3< a_3+kb_3 \nonumber \\
&< a_4+kb_3< a_3+kb_4< \cdots< a_{i}+kb_{i}< a_{i+1}+kb_i \nonumber \\
&< a_i+kb_{i+1}< a_{i+1}+kb_{i+1}<a_{i+2}+kb_{i+1}\nonumber\\
& <a_{i+1}+kb_{i+2}<\cdots <a_{r-2}+kb_{r-1}<a_{r-1}+kb_{r-1}.
\end{align}
Observe that the above sequence contains $|A|-2$ extra integers from the cardinality of $|A+k\cdot B|=4|A|-4$.

Since $a_{i-1}+k b_{i}<a_i+k b_i<a_i+kb_{i+1}$, $a_{i-1}+k b_{i}<a_{i-1}+k b_{i+1}<a_i+k b_{i+1}$ and also the cardinality of $|A+k\cdot B|=4|A|-4,$ it implies $a_i+k b_i=a_{i-1}+kb_{i+1},$ which gives $a_i-a_{i-1}=\linebreak k(b_{i+1}-b_i)$ for $i=1,2,\dots,r-2$. Similarly, from the inequalities $a_{i-1}+k b_{i-1}<a_{i-1}+k b_i<a_i+k b_{i}$ and $a_{i-1}+k b_{i-1}<a_{i}+k b_{i-1}<a_i+k b_{i}$, we have $a_{i-1}+k b_i=a_{i}+k b_{i-1}$. Thus $a_i-a_{i-1}=k(b_i-b_{i-1})$ for $i=1,2,\dots,r-2$. This completes the proof for the case $|A|=|B|$.

Further assume that $|A|<|B|$ and let $A=\{a_0<a_1<\dots<a_{r-1}\}$ and $B=\{b_0<b_1<\dots<b_{l-1}\}$. Suppose $0\leq m\leq l-r$. Let $B=B_0^{(m)}\cup B_1^{(m)}\cup B_2^{(m)}$, where $B_0^{(m)}=\{b_0,b_1,\dots,b_{m-1}\}, B_1^{(m)}=\{b_m,b_{m+1},\dots,b_{m+r-1}\}$, $B_2^{(m)}=\{b_{m+r}, b_{m+r+1},\dots,b_{l-1}\}$.

Therefore, $ A+k\cdot B\supseteq(a_0+k\cdot B_0^{(m)})\cup(A+k\cdot B_1^{(m)})\cup(a_{r-1}+k\cdot B_2^{(m)})$. It implies that $|a_0+k\cdot B_0^{(m)}|=m$, $|A+k\cdot B_1^{(m)}|\geq 4r-4$, $|a_{r-1}+k\cdot B_2^{(m)}|=l-m-r$. Thus
\begin{align*}
3r+l-4&=|A+k\cdot B|\\
&\geq |a_0+k\cdot B_0^{(m)}|+|A+k\cdot B_1^{(m)}|+|a_{r-1}+k\cdot B_2^{(m)}|\\
&\geq m+4r-4+l-m-r\\
&=3r+l-4.
\end{align*}
Hence the proof of the result.
\end{proof}


\section{Extended Inverse Problem for \texorpdfstring{$|A+3\cdot B|$}{|A+3.B|}}\label{sec3}

\begin{proof}
Let $A=\{a_0,a_1,\dots,a_{r-1}\}$ and $B=\{b_0,b_1,\dots,b_{l-1}\},$ where $a_0<a_1<\cdots<a_{r-1}$ and $b_0<b_1<\cdots<b_{l-1}$. Without loss of generality, we can assume that $a_0=0$ and $b_0=0$. Let $A_0, A_1$ and $A_2$ be three distinct congruence classes of $A,$ such that $A_0\subseteq 3\mathbb{Z}$, $A_1\subseteq 3\mathbb{Z}+1$ and $A_2\subseteq 3\mathbb{Z}+2$. We further assume that $|A_0|=m\geq 1$, $|A_1|=n\geq 0,$ $|A_2|=p\geq 0,$ and thus we have $r=m+n+p$.

\begin{proof}[Case 1. $|A_0|=m\geq 1$, $|A_1|=n\geq 1$, $|A_2|=p=0$ i.e. $c_3(A)=2$]
Assume that
\begin{align*}
A_0&=\{0=3x_0<3x_1<\cdots<3x_{m-1}\},
\\
A_0^{*}&=\frac{1}{3}\cdot A_0=\{0=x_0<x_1<\cdots<x_{m-1}\},
\\
A_1&=\{3y_0+1<3y_1+1<\cdots<3y_{n-1}+1\},
\\
A_1^{*}&=\frac{1}{3}\cdot (A_1-1)-y_0=\{0<y_1-y_0<y_2-y_0<\cdots<y_{n-1}-y_0\},
\end{align*}
Then $\ell(A_0^{*})=x_{m-1}<a_{r-1}=\ell(A)$ and $\ell(A_1^{*})=y_{n-1}-y_0<a_{r-1}=\ell(A)$. Now
\begin{align*}
|A+3\cdot B|&=|(A_0\cup A_1)+3\cdot B|\\
&=|A_0+3\cdot B|+|A_1+3\cdot B|\\
&=|3\cdot A_0^{*}+3\cdot B|+|3\cdot (A_1^{*}+y_0)+1+3\cdot B|\\
&=|A_0^{*}+B|+|A_1^{*}+B|.
\end{align*}
Further, We prove two inequalities in Claim~\ref{cla1} and Claim~\ref{cla2}.

\begin{enonce}{Claim}\label{cla1}
$\ell(B)\leq l+\max{(m,n)}-2\leq l+r-3$.
\end{enonce}

\begin{proof}[Proof of Claim~\ref{cla1}]
Since $\ell(B)\geq \ell(A)>\ell(A_0^{*})$ and $\ell(B)\geq \ell(A)>\ell(A_1^{*})$, therefore $\delta_{B,A_0^{*}}=\delta_{B,A_1^{*}}=0$.

Let's consider the case where $m\leq n$. Assuming Claim~\ref{cla1} is false, then $\ell(B)\geq l+n-1=|B+|A_1^{*}|-1\geq l+m-1=|B|+|A_0^{*}|-1$ and $d(B)=1$. Thus by Theorem~\ref{LSS}, $|A_0^{*}+B|\geq l+2|A_0^{*}|-2=l+2m-2$ and $|A_1^{*}+B|\geq l+2|A_1^{*}|-2=l+2n-2$

Hence $|A+3\cdot B|\geq 2l+2r-4,$ which contradicts to our hypothesis.

In the case where $n\leq m$, we can obtain the result by following the same approach as described earlier.

Thus $\ell(B)\leq l+\max{(m,n)}-2$. Since $r=m+n$ and $\max(m,n)\leq r-1$, therefore $\ell(B)\leq l+\max{(m,n)}-2\leq l+r-3$. This completes the proof of Claim~\ref{cla1}.
\end{proof}

\begin{enonce}{Claim}\label{cla2}
$|A+3\cdot B|\geq |A|+2(|B|-1)+h_{B}$.
\end{enonce}

\begin{proof}[Proof of Claim~\ref{cla2}]
Assume the case $m\leq n$. According to Claim~\ref{cla1}, it is evident that $\ell(B)\leq l+n-2$. Additionally, referring to Theorem~\ref{LSS}, we have $|A_1^{*}+B|\geq (n+l-1)+h_{B}$. Consequently,
\begin{align*}
|A+3\cdot B|&=|A_0^{*}+B|+|A_1^{*}+B|\\
&\geq (|A_0^{*}|+|B|-1)+(n+l-1)+h_{B}\\
&\geq (m+l-1)+(n+l-1)+h_{B}\\
&=2l+r-2+h_{B}.
\end{align*}
Similarly for the remaining case $(n\leq m)$, $|A+3\cdot B|\geq 2l+r-2+h_{B}$. Thus, we obtain that $h_B$ satisfies $0\leq h_B\leq |A+3\cdot B|-(2l+r-2)=h\leq r-3$. Therefore, $B\subseteq \{b_0,b_0+1,b_0+2,\dots,b_{l-1}\}\subseteq \{0,1,\dots,b_{l-1}\}$ and $B$ is an arithmetic progression of length at most $b_{l-1}+1=l+h_{B}\leq l+h\leq r+l-3$. As $\ell(A)\leq \ell(B),$ the set $A$ is also contained in A.P. of length at most $r+l-3$. The result can be easily verified for the case $|A_0|=m\geq 1$, $|A_1|=n=0$ and $|A_2|=p\geq 1$.
\end{proof}
\let\qed\relax
\end{proof}

\begin{proof}[Case 2. $|A_0|=m\geq 1$, $|A_1|=n\geq 1$, $|A_2|=p\geq 1$ i.e. $c_3(A)=3$]
Assume that
\begin{align*}
A_0&=\{0=3x_0<3x_1<\cdots<3x_{m-1}\},
\\
A_0^{*}&=\frac{1}{3}\cdot A_0=\{0=x_0<x_1<\cdots<x_{m-1}\},
\\
A_1&=\{3y_0+1<3y_1+1<\cdots<3y_{n-1}+1\},
\\
A_1^{*}&=\frac{1}{3}\cdot (A_1-1)-y_0=\{0<y_1-y_0<y_2-y_0<\cdots<y_{n-1}-y_0\},
\\
A_2&=\{3z_0+2<3z_1+2<\cdots<3z_{p-1}+2\},
\\
A_2^{*}&=\frac{1}{3}\cdot (A_2-2)-z_0=\{0<z_1-z_0<z_2-z_0<\cdots<z_{p-1}-z_0\}.
\end{align*}
Then $\ell(A_0^{*})=x_{m-1}<a_{r-1}=\ell(A)$, $\ell(A_1^{*})=y_{n-1}-y_0<a_{r-1}=\ell(A)$ and $\ell(A_2^{*})=z_{p-1}-z_0<a_{r-1}=\ell(A)$. Now
\begin{align*}
|A+3\cdot B|&=|(A_0\cup A_1\cup A_2)+3\cdot B|\\
&=|A_0+3\cdot B|+|A_1+3\cdot B|+|A_2+3\cdot B|\\
&=|3\cdot A_0^{*}+3\cdot B|+|3\cdot (A_1^{*}+y_0)+1+3\cdot B|+|3\cdot (A_2^{*}+z_0)+2+3\cdot B|\\
&=|A_0^{*}+B|+|A_1^{*}+B|+|A_2^{*}+B|.
\end{align*}
Furthermore, we establish two inequalities in claim~1 and claim~2.

\begin{enonce}{Claim}\label{cla3}
$\ell(B)\leq l+\max{(m,n,p)}-2\leq l+r-3$.
\end{enonce}

\begin{proof}[Proof of Claim~\ref{cla3}]
Since $\ell(B)\geq \ell(A)>\ell(A_0^{*})$, $\ell(B)\geq \ell(A)>\ell(A_1^{*})$ and $\ell(B)\geq \ell(A)>\ell(A_2^{*}),$ therefore $\delta_{B,A_0^{*}}=\delta_{B,A_1^{*}}=\delta_{B,A_2^{*}}=0$.

Let's start by considering the case where $m\leq n\leq p$. Assuming that Claim~\ref{cla3} is false, we can deduce that $\ell(B)\geq l+p-1=|B|+|A_2^{}|-1\geq l+n-1=|B|+|A_1^{}|-1\geq l+m-1=|B|+|A_0^{*}|-1$, while $d(B)=1$. Thus by Theorem~\ref{LSS}
\begin{multline}
|A_0^{*}+B|\geq l+2|A_0^{*}|-2=l+2m-2,\quad |A_1^{*}+B|\geq l+2|A_1^{*}|-2=l+2n-2\\ \text{and}\quad |A_2^{*}+B|\geq l+2|A_2^{*}|-2=l+2p-2.
\end{multline}
Hence $|A+3\cdot B|\geq 3l+2r-6,$ which contradicts our hypothesis.

For all the remaining cases we obtain the result by proceeding like above. Thus $\ell(B)\leq\linebreak l+\max{(m,n,p)}-2$. Since $r=m+n+p$ and $\max(m,n,p)\leq r-1,$ therefore, $\ell(B)\leq l+\linebreak \max{(m,n,p)}-2\leq l+r-3$. This completes the proof of Claim~\ref{cla3}.
\end{proof}

\begin{enonce}{Claim}\label{cla4}
$|A+3\cdot B|\geq |A|+3(|B|-1)+h_{B}$.
\end{enonce}

\begin{proof}[Proof of Claim~\ref{cla4}]
Assume the case $m\leq n\leq p$. By Claim~\ref{cla3}, observe that $\ell(B)\leq l+p-2$.

Also the By Theorem~1.3, $|A_2^{*}+B|\geq (p+l-1)+h_{B}$ and thus
\begin{align*}
|A+3\cdot B|&=|A_0^{*}+B|+|A_1^{*}+B|+|A_2^{*}+B|\\
&\geq (|A_0^{*}|+|B|-1)+(|A_1^{*}|+|B|-1)+|A_2^{*}+B|\\
&\geq (m+l-1)+(n+l-1)+(p+l-1)+h_{B}\\
&=3l+r-3+h_{B}.
\end{align*}
Similarly for all the remaining cases $|A+3\cdot B|\geq 3l+r-3+h_{B}$. Therefore, we can deduce that $h_B$ satisfies the inequality $0\leq h_B\leq |A+3\cdot B|-(3l+r-3)=h\leq r-3$. Consequently, it follows that $B\subseteq \{b_0,b_0+1,b_0+2,\dots,b_{l-1}\}\subseteq \{0,1,\dots,b_{l-1}\}$ and $B$ forms an arithmetic progression with a length of at most $b_{l-1}+1=l+h_{B}\leq l+h\leq r+l-3$. Since $\ell(A)\leq \ell(B)$, the set $A$ is also contained within an arithmetic progression with a length of at most $r+l-3$. This establishes the desired result. By combining both cases, we obtain the overall result.
\end{proof}
\let\qed\relax
\end{proof}
\let\qed\relax
\end{proof}

\subsection*{Acknowledgments}
%The authors are thankful to Dr MSG for their guidance.
The authors are thankful to Dr MSG for their guidance.
% and the reseach of second author is supported by NBHM.

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