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	\title[Normal and differentiable periodicity of linear shift...... ]{\LARGE 
		N\huge\MakeLowercase {ormal and differentiable periodicity of linear shift operators under partial sharing}}
	
	\author[A. Banerjee, J. Banerjee]{Abhijit Banerjee $^1$, Jhilik Banerjee$^{*}$ $^2$ }
	
	\address{$^{1}$ Department of Mathematics, University of Kalyani, West Bengal 741235, India.}
	\email{abanerjee\_kal@yahoo.co.in}
	\address{$^{2}$ Department of Mathematics, University of Kalyani, West Bengal 741235, India.}
	\email{ jhilikbanerjee38@gmail.com}
	
	
	%\dedicatory{This paper is dedicated to Professor ABCD}
	
	\subjclass[2010]{30D35.}
	
	\keywords{Meromorphic functions; partially CM sharing;  partially IM sharing; linear shift operator, order.}
	
	\date{Received: xxxxxx; Revised: yyyyyy; Accepted: zzzzzz.
		\newline \indent $^{*}$ Corresponding author}
	
	\begin{abstract}
	
	In this paper, we investigate the periodicity of linear shift operators associated with meromorphic functions under the framework of partial sharing conditions. Furthermore, we introduce the novel concept of differentiable periodicity, which provides a robust framework for analyzing the behavior of these operators under partial CM or IM sharing. By establishing comprehensive criteria for differentiable periodicity, this work offers a new perspective in the theory of shift operators. The findings not only extend but also refine and improve prior results, such as those presented in the papers of W J. Chen, Z. G. Huang (2022) and D. C. Pramanik, A. Sarkar (2024).
	
		%The primary objective of this paper is to determine sufficient conditions under which a linear shift operator exhibits periodicity. The investigation is based on the partial value-sharing criterion of meromorphic functions, with multiplicities disregarded. Notably, this work represents the first comprehensive result establishing the periodicity of a linear shift operator in its entirety. The findings presented in this study significantly extend and refine earlier results in the field.     

	  
	\end{abstract} \maketitle
	
	\section{Introduction and background}
At the outset, we introduce the following key notations, which are essential for establishing a comprehensive framework to analyze meromorphic functions and their behavior concerning shared values or sets. 
	\par Let  $f(z)$ be a non-constant meromorphic function and $a \in \mathbb{C}$. Define the function $\nu_{f}^{a}: \mathbb{C}\rightarrow \mathbb{N}$ by 
	\[\nu^{a}_f(z)=\begin{cases}
		0 &\text{if }f(z) \not =a\\d & \text{if }f(z)=a \text{ with multiplicity d}, 
	\end{cases}\] and set $\nu^{\infty}_{f}(z)=\nu^{0}_{\frac{1}{f}}(z)$. Next, define the reduced counting function $\ol \nu_{f}^{a} : \mathbb{C} \rightarrow \mathbb{N}$ as: $$\ol \nu_{f}^{a}(z) =\text{min} \{ \nu_{f}^{a}(z) ,1\}$$ and similarly $\ol \nu^{\infty}_{f}(z)=\ol \nu^{0}_{\frac{1}{f}}(z)$. Next for $m \in \mathbb{N}$ and $a\in \mathbb{C} \cup \{\infty\}$, define the reduced multiplicity counting function $\nu^{\leq m}_{(f,a)}:\mathbb{C} \cup \{\infty\} \rightarrow \mathbb{N} $ by 
\[\nu^{\leq m}_{(f,a)}(z)=\begin{cases}
	0 &\text{if }\nu^{a}_f(z) > m\\\nu^{a}_f & \text{if }\nu^{a}_f(z)\leq m
\end{cases}\]  and $\ol \nu^{\leq m}_{(f,a)}(z)=\text{min}\left\{\nu^{\leq m}_{(f,a)}(z),1\right\}$. When $m=\infty$, we interpret $\nu^{\leq m}_{(f,a)}(z)$ as $\nu^{a}_f(z) $.\par For $S \subset \mathbb{C}\cup \{\infty\}$, define the set $E(S,f)$  as \beas E(S,f) = \bigcup_{a \in S}\{(z,\nu^{a}_f(z)): z \in \mathbb{C}\}.\eeas \par In an analogous way, replacing $\nu^{a}_f(z)$ with $\ol \nu^{a}_f(z)$ in $ E(S,f)$, define $\ol E(S,f)$ as: 
\beas \ol E(S,f) = \bigcup_{a \in S}\{(z,\ol\nu^{a}_f(z)): z \in \mathbb{C}\}.\eeas
Using these definitions, we describe various types of set sharing between meromorphic functions $f$ and $g$:\\ $\bullet$ If $E(S,f)=E(S,g)$,  we say that $f$ and $g$ share the set $S$ CM.\\ $\bullet$  If $\ol E(S,f)=\ol E(S,g)$,  we say that $f$ and $g$ share the set $S$ IM.
\\$\bullet$ If $E(S,f) \subseteq E(S,g)$,  we say that $f$ and $g$ partially share $S$ CM. \\$\bullet$ If $\ol E(S,f) \subseteq \ol E(S,g)$,  we say that $f$ and $g$ partially share $S$ IM.\\
When $S$ is a singleton, the above definition corresponds to value sharing.
			%For a non-constant meromorphic function $f(z)$ and $S \subset \mathbb{C}\cup \{\infty\}$, let $E_f (S) = \bigcup_{a\in S}\{(z,p) \in \mathbb{C}\times\mathbb{N}: f(z)=a \text{ with multiplicity $p$}\} $ ($\ol E_f(S)=\bigcup_{a\in S}\{(z,1) \in \mathbb{C}\times\mathbb{N}: f(z)=a$\}.
	%	If $E(S,f)=E(S,g)\; (\ol E(S,f)=\ol E(S,g))$ we say that $f$ and $g$ share the set $S$ CM (IM). When $S$ is a singleton set, the definition coincides with the traditional definition of CM (IM) value sharing. In this context, if $E(S,f) \subseteq E(S,g)\;( \ol E(S,f) \subseteq \ol E(S,g))$, we say that $f(z)$ and $g(z)$ partially share $S$ CM (IM) i.e., counting multiplicities (ignoring multiplicities). 
		%If $\ol E(S,f) \subseteq \ol E(S,g)$, we say that $f(z)$ and $g(z)$ partially shares $aS$ IM (ignoring multiplicities). 
	%	If $S=\{a\}$ i.e., is a singleton set, then we say $f$ and $g$ share a value $a$ CM (IM).
			%	At the outset, we assume that the readers are acquainted with the conventional notations of value distribution theory defined on whole complex plane $\mathbb{C}$ such as $N(r,f)$, $m(r,f)$ $T(r,f)$, $S(r,f)$, etc, as outlined in \cite{Hayman_1964}, \cite{Yang_Springer_1993}. So we refrain from providing detail explanations. \par In value distribution theory we generally concern about the distribution of the zeros of the function $f(z)-a$, where $a\in \mathbb{C}$ and form which the idea of sharing of values or sets comes under consideration. For the standard notations of set sharing, we refer the readers to make a glance over the relevant information provided in the second paragraph of \cite{Ban_Georgian_2008}, which automatically includes the definition of value sharing. Below we invoke it.\par 
		
		
		%Let $S$ be a set of distinct elements of $\mathbb{C}\cup\{\infty\}$. We denote by $E_{f}(S)=E(S,f)=\everymath{\displaystyle}\bigcup_{a\in S}\{z: f(z)-a=0\}$, where each zero is counted according to its multiplicity. If multiplicities are not taken into account, the set is denoted by $\ol E(S,f)$.
%	Let $a \in \mathbb{C}\cup\{\infty\}$. We denote by $E(a,f)=\{z: f(z)-a=0\}$, where each zero is counted according to its multiplicity. If multiplicities are not taken into account, the set is denoted by $\ol E(a,f)=\{z: f(z)-a=0\}$.  In this context, if $E(a,f) \subseteq E(a,g)$, we say that $f(z)$ and $g(z)$ partially share $a$ CM (counting multiplicities). If $\ol E(a,f) \subseteq \ol E(a,g)$, we say that $f(z)$ and $g(z)$ partially shares $a$ IM (ignoring multiplicities).
	\par For further information or a detailed explanation of these concepts, readers are encouraged to consult the original sources cited in the manuscript: \cite{Ban_Georgian_2008}. Additionally, we assume that readers are familiar with the standard notations of value distribution theory on the entire complex plane $\mathbb{C}$, including $N(r,f)$, $m(r,f)$, $T(r,f)$, $S(r,f)$ etc. as outlined in \cite{Hayman_1964} and \cite{Yang_Springer_1993}.
	% Now, we shall calculate the counting functions of zeros of $L_{j}f^{(l)}(z)-a$ with multiplicity of $\geq 2,$ which is denoted by $N_{2)} \left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)$ counted according to its multiplicities.
	 We use the notation $N_{(k}(r,a,f(z))$ to denote the counting function of $a$-points of $f(z)$ of multiplicity $\geq k$.
	For $a\in \mathbb{C}\cup\infty$, we also use the notation $\ol N_{*}(r,a;f,g)$ to denote the reduced counting function of common $a$-points of $f$ and $g$ of different multiplicities.
	 \par The order and  hyper-order of $f$ are defined respectively as follows:\beas \rho(f)=  \limsup_{r\longrightarrow \infty} \frac{\log T(r,f)}{\log r}, \;\;\;\;\; \rho_2(f)=\limsup_{r\longrightarrow \infty} \frac{\log \log T(r,f)}{\log r}.\eeas 
	\par Further throughout the paper, we will assume $c \in \mathbb{C} \setminus\{0\}$ and the shift operator of $f$ will be denoted by $f(z+c)$. \par
	We begin our discussion with a result established by Qi  \cite{Qi_Ann polon_2011}.
	\begin{theoA}\cite{Qi_Ann polon_2011}
		Let $f$ be a entire function of finite order, $c\in \mathbb{C}$ and $a_i$, $i=1,2$ be three distinct periodic functions with period $c$. If $f(z)$ and $f(z+c)$ share $a_1$, $a_2$ IM, then $f(z)=f(z+c)$.
	\end{theoA} 
	
	For meromorphic function some analogous results are as follows: 
	\begin{theoB}\cite{Heittokangas_Com. var_2011}
		Let $f$ be a meromorphic function of finite order and $c \in \mathbb{C}$. If $f(z)$ and $f(z+c)$ share three distinct periodic functions $a_i$, $i=1,2,3$ with period $c$ CM, then $f(z)=f(z+c)$.
	\end{theoB}

\begin{theoC}\cite{Heittokangas_JMAA_2009}
 Under the same condition of {\em Theorem B}, if $f(z)$ and $f(z+c)$ share $a_1$, $a_2$ CM and $a_3$ IM, then $f(z)=f(z+c)$.
\end{theoC}
In this respect, further continuations of research can be found in \cite{Chen+Xu_Houston_2017} and \cite{Li+Yi_Bull.Korean_2016}.
%\begin{theoD}
%	Let $f$ be a non-constant meromorphic function of $\rho_2(f)<1$ and let $a_1,a_2,a_3 \in \mathbb{C} \cup \{\infty\}$. If $f(z)$ and $f(z+c)$ share $a_1$, $a_2$ CM and $a_3$ IM, where $c \in \mathbb{C}\setminus\{0\}$, then $f(z) \equiv f(z+c)$.
%\end{theoD}
%\begin{theoE}
%	Let $f$ be a non-constant entire function of $\rho_2(f)<1$ and $a_1$, $a_2$ be two distinct finite values. If $f(z)$ and $f(z+c)$ share $a_1$, $a_2$ IM, where $c  \in \mathbb{C}\setminus\{0\} $,then $f(z) \equiv f(z+c).$
%\end{theoE}
 Recently, Chen et al. introduced new types of structures that had not been considered earlier. In their work, Chen and Huang \cite{Chen+Huang_Contemp_2022} presented results focusing on partial and normal sharing, as outlined below:
\begin{theoD}\cite{Chen+Huang_Contemp_2022}
	Let $f(z)$ be a meromorphic function of finite order, $a \in \mathbb{C}\setminus \{0\}$ and $l,s$ be integers with $0 \leq l<s$. If $f^{(l)}(z)$ and $f^{(s)}(z+c)$ share $a$ CM and satisfy $E(0,f^{(l)}(z))  \subset E(0, f^{(s)}(z+c) )$ and $E(\infty, f^{(s)}(z+c)) \subset E(\infty, f^{(l)}(z) )$, then $f^{(l)}(z) \equiv f^{(s)}(z+c) $.
\end{theoD}
\begin{theoE}\cite{Chen+Huang_Contemp_2022}
	Let $f(z)$ be a transcendental meromorphic function of finite order, $0 \leq l<s$. Suppose that, $f^{(l)}(z)$ and $f^{(s)}(z+c)$ share a finite value $a(\not = 0)$ IM and satisfy $E(0,f^{(l)}(z)) \subset E(0,f^{(s)}(z+c))$ and $E(\infty, f^{(s)}(z+c)) \subset E(\infty, f^{(l)}(z))$. If $N\left(r, \frac{1}{f^{(l)}(z)}\right)+\ol N\left(r, \frac{1}{f(z)}\right)=S(r,f)$, then $f^{(l)}(z)\equiv f^{(s)}(z+c) $.
\end{theoE} 
Very recently, in \cite{Pramanik+Sarkar_Math.Studii_2024} the authors have been able to successfully remove the conditions over counting function in Theorem E. The result is as follows:
\begin{theoF}\cite{Pramanik+Sarkar_Math.Studii_2024}
	Let $f(z)$ be a non-constant meromorphic function of finite order, $0\leq l<s$. If $f^{(l)}(z)$ and $f^{(s)}(z+c)$ share a finite value $a(\not = 0)$ IM and satisfy $E(0,f^{(l)}(z)) \subset E(0,f^{(s)}(z+c))$ and $E(\infty, f^{(s)}(z+c)) \subset E(\infty, f^{(l)}(z))$, then $f^{(l)}(z)\equiv f^{(s)}(z+c) $.
\end{theoF}

\section{Motivation, main results and relevant examples}

The following example shows that Theorem A can not be extended for meromorphic function under partial sharing.
\begin{exm}
	Let us assume two distinct complex numbers $a,b$, $c=\pi i$ and $f(z)=\frac{ae^z-b}{e^z-1}$. Then, $f(z+c)=\frac{ae^z+b}{e^z+1}$. It is easy to check, $ E(a,f(z)) = E(a, f(z+c))$ and $ E(b,f(z)) =  E(b, f(z+c))$, but $f(z) \not = f(z+c).$
\end{exm} 
We see that in the above example the poles are not shared. 
In light of the above example, an intriguing question arises:

\par Can the periodicity of a meromorphic function be achieved under partial sharing on $\mathbb{C}$? If so, what would be the nature of the pole sharing?

The purpose of this paper is to address these questions and contribute to this line of inquiry. Specifically, we aim to establish our results within a more generalized framework, namely the linear shift operator, which is defined as follows:

For a non-constant meromorphic function $f(z)$, a non negative integer $j$, $a_i \in \mathbb{C}$, $a_j\not=  0$ and $c \in \mathbb{C}$, define $ L_{j}f(z)$ as
\bea\label{e1.1} L_{j}f(z)=\sum_{i=0}^{j} a_{i}f(z+ic).\eea


The following theorem is one of our main results, where we investigate the periodicity of the linear shift operator based solely on partial IM sharing which was not addressed earlier. 
\begin{theo}\label{t1}
	Let $f(z)$ be a non-constant meromorphic functions with $\rho_2(f) <1$ and let $a$, $b \in \mathbb{C} $ be two distinct numbers. Let $\ol E(a,L_{j}f(z)) \subseteq \ol E(a, L_{j}f(z+c)) $, $\ol E(b,L_{j}f(z)) \subseteq \ol E(b, L_{j}f(z+c)) $, $E(\infty,L_{j}f(z+c)) \subseteq E(\infty, L_{j}f(z))$ with $\ol N(r,f) =S(r,f)$,  then $L_{j}f(z)=L_{j}f(z+c)$.
\end{theo} 
The following example shows that, by no means, the linear shift operator $L_{j}(z+c)$ in Theorem \ref{t1}, can not be replaced by an arbitrary linear shift operator.
\begin{exm}
	Consider the function $L_{0}f(z)=f(z)=\frac{e^{\frac{2\pi i z}{c}}-1}{e^{\frac{2\pi i z}{c}}+1},$ where $c \in \mathbb{C}\setminus\{0\}$. Choose a linear shift operator $\tilde{L}_{3}f(z)$ by \beas\tilde{L}_{3}f(z)&=&3f(z+3c)-f(z+2c)-2f(z+c)-f(z)\\&=& -\frac{e^{\frac{2\pi i z}{c}}-1}{e^{2\frac{\pi i z}{c}}+1}. \eeas Clearly, $E(1,L_{0}f(z))=E(1,\tilde{L}_{3}f(z))$, $E(0,L_{0}f(z))=E(0,\tilde{L}_{3}f(z))$ and $E(\infty,L_{0}f(z))\\=E(\infty,\tilde{L}_{3}f(z))$, but $f(z) \not = \tilde{L}_{3}f(z).$
\end{exm}
The next example shows that for entire functions, in Theorem \ref{t1}, $f(z+c)$ can not be replaced by $\Delta_{c}f(z)=f(z+c)-f(z)$.
\begin{exm}
	Take $L_{0}f(z)=f(z)=\left(\frac{e^{\frac{\pi i z}{c}}-1}{2}\right)^2+1$. Then $\Delta_{c}f(z)= e^{\frac{\pi i z}{c}}$. Here, $f$ and $\Delta_{c}f$ share $1$, $0$ IM, but $f \not = \Delta_{c}f$.
\end{exm}
Next, we define differentiable periodic function.
\begin{defi}
	Let $F(z)$ be a non-constant meromorphic function and $l$, $s$ two non-negative integers. $F(z)$ is said to be differentiably periodic if $F^{(l)}(z) \equiv F^{(s)}(z+c)$, where $0  \leq l <s$.\end{defi}

In light of Theorem D, it becomes of significant interest to investigate the differentiable periodic property of $L_{j}f(z)$ as defined in (\ref{e1.1}) as this perspective provides a deeper understanding of the behavior of the linear shift operator $L_{j}$ under certain sharing conditions. Moreover, by relaxing the condition from CM sharing of the non-zero value $a$ to partial CM sharing of $a$, it can be shown that a similar result to Theorem D holds. The following theorem formalizes this extension:
%In this context, we present the following theorem, which demonstrates that the sharing conditions outlined in Theorem \ref{e1.1} can be relaxed in expense of allowing partial CM sharing to partial IM sharing, thereby broadening the scope of the result.
\begin{theo}\label{t2}
	Let $f(z)$ be a non-constant meromorphic function with $\rho_2(f)<1$, $a \in \mathbb{C}\setminus\{0\}$ and $l$, $s$ be two integers such that $0\leq l<s$. If $ E(a,L_{j}f^{(l)}(z)) \subseteq  E(a,  L_{j}f^{(s)}(z+c))$, $E(0,L_{j}f^{(l)}(z)) \subseteq  E(0,  L_{j}f^{(s)}(z+c))$, $  E(\infty,  L_{j}f^{(s)}(z+c)) \subseteq E(\infty,L_{j}f^{(l)}(z)) $, then $L_{j}f^{(l)}(z) \equiv L_{j}f^{(s)}(z+c). $
\end{theo}
\begin{note}
	We can see that {\em Theorem \ref{t2}} is not only extending {\em Theorem D}, but also improving it by relaxing sharing condition.
\end{note}
The following example shows that, sharing of the non-zero element in Theorem \ref{t2} can not be dropped.
\begin{exm}
	Set $L_{0}f(z)=\sin z$, then $L_{0}f^{\prime}(z+\frac{\pi}{2})=-\sin z$. Clearly, $E(0, L_{0}f(z))=E(0,L_{0}f^{\prime}(z+\frac{\pi}{2}))$ and $E(\infty, L_{0}f(z))=E(\infty,L_{0}f^{\prime}(z+\frac{\pi}{2}))$, but $E(1, L_{0}f(z))\not \subset E(1,L_{0}f^{\prime}(z+\frac{\pi}{2}))$ and $L_{0}f(z) \not = L_{0}f^{\prime}(z+\frac{\pi}{2}).$
\end{exm}

In view of Theorem \ref{t1}, it is timely and significant to investigate the differentiable periodic property of $L_{j}f(z)$ under the framework of partial IM value sharing over $\mathbb{C}$.
\begin{theo}\label{t3}
Under the same situation of {\em Theorem \ref{t2}}, if $\ol E(a,L_{j}f^{(l)}(z)) \subseteq \ol E(a,  L_{j}f^{(s)}(z+c))$, $E(0,L_{j}f^{(l)}(z)) \subseteq  E(0,  L_{j}f^{(s)}(z+c))$ and $  E(\infty,  L_{j}f^{(s)}(z+c)) \subseteq E(\infty,L_{j}f^{(l)}(z)) $ and $\ol N\left(r, \frac{1}{L_{j}f^{(l)}(z)}\right)=S(r,f)$, then $L_{j}f^{(l)}(z) \equiv L_{j}f^{(s)}(z+c). $
\end{theo}
The following example shows that sharing structure of $\infty$ in Theorem \ref{t3} can not be reversed.
\begin{exm}
	Take $L_{0}f(z)=\frac{2e^{\frac{2\pi i z}{c}}}{e^{\frac{2\pi i z}{c}}-1},$ where $c \in \mathbb{C}\setminus\{0\}$. Then, $L_{0}f^{\prime}(z+c)=\frac{-4e^{\frac{2\pi i z}{c}}}{(e^{\frac{2\pi i z}{c}}-1)^2}$. Clearly, $\ol E(1, L_{0}f(z)) \subset\ol E(1,L_{0}f^{\prime}(z+c))$, $E(0, L_{0}f(z)) =E(0,L_{0}f^{\prime}(z+c))$ and $\ol N\left(r,\frac{1}{L_{0}f(z)}\right)=S(r,f)$, but $E(\infty, L_{0}f^{\prime}(z+c)) \not \subset E(\infty,\\ L_{0}f(z))$ and $L_{0}f(z) \not = L_{0}f^{\prime}(z+c)$.
	
\end{exm}
The next theorem extends and refines Theorems E and F to the case of differentiably periodic linear shift operators. In fact, to prove the theorem, we have adopted a different approach than the one used in \cite{Chen+Huang_Contemp_2022} and \cite{Pramanik+Sarkar_Math.Studii_2024}. 
\begin{theo}\label{t4}
	Under the same situation of {\em Theorem \ref{t2}}, if $\ol E(a,L_{j}f^{(l)}(z)) = \ol E(a,  L_{j}f^{(s)}(z+c))$, $E(0,L_{j}f^{(l)}(z)) \subseteq  E(0,  L_{j}f^{(s)}(z+c))$ and $  E(\infty,  L_{j}f^{(s)}(z+c)) \subseteq E(\infty,L_{j}f^{(l)}(z)) $, then $L_{j}f^{(l)}(z) \equiv L_{j}f^{(s)}(z+c). $
\end{theo}
The following two examples show that Theorem \ref{t1} cannot be directly extended to set sharing structure.

\begin{exm}
	Let $L_{0}f(z)= \frac{2e^{\frac{\pi i z}{c}}}{1+e^{\frac{2\pi i z}{c}}},$ $c \in \mathbb{C}\setminus \{0\}$. Then we see that for a non-zero constant $\alpha$, $E(\{\alpha,-\alpha\},L_{0}f(z))=E(\{\alpha,-\alpha\},L_{0}f(z+c))$, $E(\infty,L_{0}f(z))=E(\infty,L_{0}f(z+c))$
	 %$L_{0}f(z)$ and $L_{0}f(z+c)$ share $\{\alpha, -\alpha\} $ CM,
	  but $L_{0}f(z) \not = L_{0}f(z+c)$.
\end{exm}
\begin{exm}
Let $L_{0}f(z)= \frac{1+e^{\frac{\pi i z}{c}}}{1-e^{\frac{2\pi i z}{c}}},$ $c \in \mathbb{C}\setminus \{0\}$. Then we see that for a non-zero constant $\alpha$, $E\left(\{\alpha,\frac{1}{\alpha}\},L_{0}f(z)\right)=E\left(\{\alpha,\frac{1}{\alpha}\},L_{0}f(z+c)\right)$, $E(\infty,L_{0}f(z))=E(\infty,L_{0}f(z+c))$
%$L_{0}f(z)$ and $L_{0}f(z+c)$ share $\{\alpha, \frac{1}{\alpha}\} $ CM,
 but $L_{0}f(z) \not = L_{0}f(z+c)$.
\end{exm}
 In view of Example 2.6 and 2.7, the following question is inevitable:
\begin{ques}
	Can we can replace the value sharing condition with set sharing condition in the {\em Theorems \ref{t1}-\ref{t4}}?
\end{ques}
Below we are presenting the following chart, which provides an in-depth comparison of the related theorems in terms of sufficient conditions in a compact form. Throughout the chart, we assume that  $a(\not = 0)$, $b$ are two distinct complex numbers.  

\begin{center}
	\includegraphics[scale=.51]{Table_3}
\end{center}
\section{Lemmas}
In this section, we are going to present some lemmas, which will be needed to proceed further.
	\begin{lem}\label{l3.1}\cite{Halburd+Korhonen_Tohge_TAMS_2014}
		Let $f(z)$ be a meromorphic function with $\rho_2(f)<1$ and $c\in \mathbb{C}$. Then we have \beas m\left(r, \frac{f(z+c)}{f(z)}\right)+m\left(r \frac{f(z)}{f(z+c)}\right)=S(r,f), \eeas where $S(r,f)=o(T(r,f))$, for all $r$, outside of a possible exceptional set $S$ with finite linear measure. 
	\end{lem}
	\begin{lem}\label{l3.2}\cite{Chiang+Feng_Ramanujan_2008}
		Let $f(z)$ be a non-constant meromorphic function of $\rho_2(f)<1$ and $c \in \mathbb{C}$. Then {\small \beas N(r,f(z+c))=N(r,f)+S(r,f),\; N\left(r, \frac{1}{f(z+c)}\right)= N\left(r,\frac{1}{f(z)}\right)+S(r,f)\eeas} and {\small \beas \ol{N}(r,f(z+c))=\ol{N}(r,f)+S(r,f),\; \ol{N}\left(r, \frac{1}{f(z+c)}\right)= \ol{N}\left(r,\frac{1}{f(z)}\right)+S(r,f).\eeas} 
	\end{lem}
\begin{lem}\label{l3.3}\cite{Yi+Yang_2003_Sci.Press}
	Let $f(z)$ be a non-constant meromorphic function and let $d_j, (j = 1,2,\ldots, q)$
	be $q$ distinct complex numbers. Then we have \beas m\left(r,\sum_{i=1}^{q}\frac{1}{f-d_i}\right)=\sum_{i=1}^{q}m\left(r,\frac{1}{f-d_i}\right)+O(1). \eeas
\end{lem}
\begin{lem}\label{l3.3e}\cite{Ban+Roy_Contemp_2023}
Let $f(z)$ be a meromorphic function of $\rho_2(f) < 1$, then we have
\beas  T(r, f(z+c)) = T(r, f(z)) + S(r, f).\eeas
\end{lem}
%\begin{lem}\label{l3.3e}\cite{Halburd+Korhonen_Tohge_TAMS_2014}
% Let $T : [0,+\infty) \rightarrow [0,+\infty)$ be a non-decreasing continuous function and let $ s \in(0,+\infty) $. If $\rho_2(T)<1$ and $\delta \in (0,1-\rho_2)$, then \beas T(r+s)=T(r)+o\left(\frac{T(r)}{r^\delta}\right) ,\eeas  where $r$ runs to infinity outside of a set of finite logarithmic measures.
%\end{lem}
%\begin{lem}\label{l3.3}\cite{Chiang+Feng_Ramanujan_2008}
%Let $f(z)$ be a meromorphic function of finite order $\rho$ and $c$ be a non-zero complex constant. Then for each $\epsilon >0$, we have \beas T(r,f(z+c))=T(r,f(z))+O(r^{\rho-1+\epsilon})+O(\log r).\eeas
%\end{lem}
\begin{lem}\label{l3.4}
	Let $f(z)$ be a meromorphic function of $\rho_2(f)<1$, then $S(r, L_jf(z))$ and $S(r, L_jf(z+c))$ can be replaced by $S(r,f)$.
\end{lem}
\begin{proof}
	In view of {\it Lemmas \ref{l3.1}, \ref{l3.3e}}, we have {\small \beas T(r, L_{j}f(z))&=&m(r,L_{j}f(z))+N(r,L_{j}f(z))\\&\leq& m\left(r,\frac{L_{j}f(z)}{f}\right)+m(r,f(z))+N(r,L_{j}f(z))+S(r,f) \\&\leq& (j+2)T(r,f(z))+S(r,f)\eeas} and so the lemma follows. 
	Similar result holds for $S(r, L_jf(z+c))$.
	% $S(r,L_{j}f^{(l)}(z))$ and $S(r,L_{j}f^{(s)}(z+c))$. 
	\end{proof}
\begin{lem}\label{l3.4e}
	Let $f(z)$ be a non-constant meromorphic function with $\rho_2(f)<1$, $c \in \mathbb{C}\setminus\{0\}$ and $l$, $s \geq 0$. Then, $S(r,L_{j}f^{(l)}(z))$ and $S(r,L_{j}f^{(s)}(z+c))$ can be replaced by $S(r,f)$.
\end{lem}
\begin{proof}
	In view of {\it Lemmas \ref{l3.1}. \ref{l3.3e}} and logarithmic derivative theorem, we obtain
	{\small\beas T(r, L_{j}f^{(l)}(z))&=&m(r,L_{j}f^{(l)}(z))+N(r,L_{j}f^{(l)}(z))\\&\leq& m\left(r,\frac{L_{j}f^{(l)}(z)}{L_{j}f(z)}\right)+m\left(r,\frac{L_{j}f(z)}{f(z)}\right)+N(r,L_{j}f^{(l)}(z))+S(r,f)\\&\leq& (l+j+1)T(r,f(z))+S(r,f). \eeas} Hence, the lemma follows. Similar results holds for  $S(r,L_{j}f^{(s)}(z+c))$.
\end{proof}
\begin{lem}\label{l3.5}
	Let $f(z)$ be a non-constant meromorphic function  with $\rho_2(f)<1$. For any complex number $b$ we have, \beas m\left(r, \frac{ L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)-b}\right)\leq S(r,f),\;\;\text{where } 0 \leq l<s.\eeas 
\end{lem}
\begin{proof}
	Using logarithmic derivative theorem and {\it Lemmas \ref{l3.1}, \ref{l3.4e}}, we obtain \beas  m\left(r, \frac{ L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)-b}\right) &\leq &m\left(r, \frac{ L_{j}f^{(s)}(z+c)}{L_{j}f^{(s)}(z)}\right)+m\left(r, \frac{L_{j}f^{(s)}(z)}{L_{j}f^{(l)}(z)-b}\right)  \\&=& S(r,f). \eeas
\end{proof}
\begin{lem}\label{l3.6}
	Under the same situation of {\em Lemma \ref{l3.5}}, if $a \in \mathbb{C}\setminus\{0\}$ and $0\leq l< s$; then \beas &&m\left(r,\frac{1}{L_{j}f^{(l)}(z)}\right)+m\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right) \\&&\leq m(r,L_{j}f^{(l)}(z))+N(r, L_{j}f^{(s)}(z+c) )-N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f). \eeas
\end{lem}

\begin{proof}
	Set \beas F(z)=\frac{1}{L_{j}f^{(l)}(z)}+\frac{1}{L_{j}f^{(l)}(z)-a}.\eeas Further, using {\it Lemmas \ref{l3.4e}, \ref{l3.5}} we get \beas m(r, L_{j}f^{(s)}(z+c)F(z)) &\leq& m\left(r,\frac{ L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)}\right)+m\left(r,\frac{ L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)-a}\right)\\&\leq& S(r,f).\eeas Thus, \beas m(r, F(z)) &\leq& m(r, L_{j}f^{(s)}(z+c)F(z))+m\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right) \\&\leq&m\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f).\eeas Next, using {\it Lemma \ref{l3.3}}, the First Fundamental Theorem and the above inequality we obtain, \beas  T(r,  L_{j}f^{(s)}(z+c))&=&T\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+O(1)\\&=&N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+m\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f)\\&\geq &  m(r,F(z))+N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f)\\&=&m\left(r,\frac{1}{L_{j}f^{(l)}(z)}\right)+m\left(r,\frac{1}{L_{j}f^{(l)}(z)-a}\right)\\&&+N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f).\eeas Thus, from the above inequality using {\it Lemmas \ref{l3.1}, \ref{l3.4e}, \ref{l3.5}} we have \beas&& m\left(r,\frac{1}{L_{j}f^{(l)}(z)}\right)+m\left(r,\frac{1}{L_{j}f^{(l)}(z)-a}\right) \\&\leq&  T(r,  L_{j}f^{(s)}(z+c))-N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f)\\&=&m(r,  L_{j}f^{(s)}(z+c))+N(r,  L_{j}f^{(s)}(z+c))-N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f)\\&\leq&m\left(r,\frac{ L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)}\right)+m(r,L_{j}f^{(l)}(z))+N(r,  L_{j}f^{(s)}(z+c))\\&&-N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f)\\&\leq & m(r,L_{j}f^{(l)}(z))+N(r,  L_{j}f^{(s)}(z+c))-N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f).\eeas
\end{proof}
\begin{lem}\label{l3.9}
	Let $f(z)$ a meromorphic function of $\rho_2(f)<1$ and $a \in \mathbb{C}\setminus\{0\}$. If $  E(\infty,  L_{j}f^{(s)}(z+c)) \subseteq E(\infty,L_{j}f^{(l)}(z)) $ for $0\leq l<s$, then $\ol N(r,L_{j}f^{(l)}(z))=\ol N(r,L_{j}f^{(s)}(z+c))=S(r,f)$.
\end{lem}
\begin{proof}
	Since $  E(\infty,  L_{j}f^{(s)}(z+c)) \subseteq E(\infty,L_{j}f^{(l)}(z)) $, we deduce \beas N(r,L_{j}f^{(s)}(z+c)) \leq N(r,L_{j}f^{(l)}(z))+S(r,f). \eeas Using {\it Lemma \ref{l3.4e}} we get, \beas  N(r,L_{j}f^{(s)}(z+c))&=&N(r,L_{j}f^{(s)}(z))+S(r,f)\\&=&N(r,L_{j}f^{(l)}(z))+(s-l)\ol N(r,L_{j}f^{(l)}(z))+S(r,f)\\&\leq & N(r,L_{j}f^{(l)}(z))+S(r,f),\eeas i.e., $\ol N(r,L_{j}f^{(l)}(z))=\ol N(r,L_{j}f^{(s)}(z+c))=S(r,f)$.
\end{proof}
%\begin{lem}\label{l3.10}
%	Let $f(z)$ be a non-constant meromorphic function of $\rho_2(f)<1$. If $E(0,L_{j}f^{(l)}(z)) \subseteq  E(0,  L_{j}f^{(s)}(z+c))$ and $  E(\infty,  L_{j}f^{(s)}(z+c)) \subseteq E(\infty,L_{j}f^{(l)}(z)) $, then $L_jf(z)$ is transcendental.
%\end{lem}
%\begin{proof}
%	Clearly, $\rho_2(L_jf(z)) \leq \rho_2(f) <1$. If $\rho_2(L_jf(z))$ is of infinite order, then $L_jf(z)$ can not be a rational function. If $\rho_2(L_jf(z))$ is finite then  
%	proceeding as the same way of {\it Lemma 4 of \cite{Pramanik+Sarkar_Math.Studii_2024}} we can prove that $L_jf(z)$ is transcendental.
%\end{proof}
\section{Proofs of the Theorems}
	\begin{proof}[\underline{\em\bf{Proof of Theorem \ref{t1}}}]The proof of the theorem is inspired by the idea of Lin-Ishizaki \cite{Lin+Ishizaki_JMAA_2018}.\\ 
On the contrary, we assume that, $L_{j}f(z) \not \equiv L_{j}f(z+c)$. Take a function \bea\label{e4.1} \Phi(z)= \frac{L_{j}f^\prime(z)(L_{j}f(z)-L_{j}f(z+c))}{(L_{j}f(z)-a)(L_{j}f(z)-b)}. \eea
$\bullet$ Let $z_0$ be an $a$- point of $L_{j}f(z)$ of order $p$ and same of $L_{j}f(z+c)$ of order $q$, then $z_0$ is a zero of $\Phi(z)$ of order $(p-1)+min \{p,q\}-p$, i.e., $z_0$ is not a pole of $\Phi(z)$. \\ $\bullet$ Similarly, $b$-points of $L_{j}f(z)$ are not poles of $\Phi$.\\$\bullet$ Let $z_1$ be a pole of $L_{j}f(z)$ of order $r$. Then, a simple calculation yields that $z_1$ is a simple pole of $\Phi(z)$, regardless of whether it is a pole of $L_{j}f(z+c)$ or not.
Hence, by {\it Lemma \ref{l3.2}}, \beas N(r,\Phi)  \leq \ol N(r,L_{j}f(z))\leq (j+1)\ol N(r,f)=S(r,f).\eeas Further, using {\it Lemmas \ref{l3.1}, \ref{l3.4}} and logarithmic derivative theorem we obtain \beas m(r,\Phi) &\leq& m\left(r,\frac{L_{j}f^\prime(z)}{L_{j}f(z)-a}\right)+m\left(r, \frac{L_{j}f(z)-L_{j}f(z+c)}{L_{j}f(z)-b}\right) \\&\leq& m\left(r,1-\frac{L_{j}f(z+c)-b}{L_{j}f(z)-b}\right)=S(r,f). \eeas
Hence. \bea \label{e4.2}T(r,\Phi)=S(r,f). \eea Now, we denote by $ N_{0}\left(r,\frac{1}{L_{j}f^\prime(z)}\right) $ counting function of those zeros of $L_{j}f^\prime(z)$, which are not zeros of $(L_jf(z)-a)(L_jf(z)-b)$ and similar notation for $L_jf(z+c)$. Now, from (\ref{e4.1}) and (\ref{e4.2}) we have \bea\label{e4.3} N_{0}\left(r,\frac{1}{L_{j}f^\prime(z)}\right) \leq N\left(r, \frac{1}{\Phi}\right)=S(r,f) .\eea  Here, \beas \Phi(z+c)= \frac{L_{j}f^\prime(z+c)(L_{j}f(z+c)-L_{j}f(z+c+c))}{(L_{j}f(z+c)-a)(L_{j}f(z+c)-b)}. \eeas From the above using {\it Lemma \ref{l3.3e}}, we get \bea\label{e4.4} N_{0}\left(r,\frac{1}{L_jf^\prime (z+c)}\right) \leq N\left(r, \frac{1}{\Phi(z+c)}\right)\leq T(r, \Phi(z+c))=S(r,f) . \eea Let us consider a function \bea\label{e4.5} H(z)=\frac{L_{j}f^\prime(z) (L_jf(z+c)-a)(L_jf(z+c)-b)}{L_jf^\prime (z+c)(L_jf(z)-a)(L_jf(z)-b)}.\eea Let $z_0$ be an $a$-point ($b$-point) of $L_jf(z)$ of multiplicity $m$ and that of $L_jf(z+c)$ of multiplicity $n$. Then $z_0$ is a pole of $H(z)$ of order $(m+n-1-m-n+1)=0$, i.e., $z_0$ is not a pole of $H(z)$. Note that, when $m>1$, then zeros of $L_jf(z)$ are zeros of $L_jf^{\prime}(z)$ of order $(m-1)$.\\  Let $z_1$ be a zero of	$L_jf^\prime (z+c)$. Then, the following cases arise: \\ \underline{\em\bf{Case 1:}} Suppose, $z_1$ is a common zero of $(L_jf(z)-a)\; ((L_jf(z)-b))$ and $(L_jf(z+c)-a) ((L_jf(z+c)-b))$.  By the similar arguments that were used before, we can show that $z_1$ is not a pole of $H(z)$. \\ \underline{\em\bf{Case 2:}} $z_1$ is a zero of $(L_jf(z+c)-a)((L_jf(z+c)-b))$ but not of $(L_jf(z)-a)((L_jf(z)-b))$. In that case also, $z_1$ is not a pole of $H(z)$.\\\underline{\em\bf{Case 3:}} If $z_1$ is not zero of $(L_jf(z+c)-a)((L_jf(z+c)-b))$. Then by (\ref{e4.4}), we can say that $z_1$ is not a pole of $H(z)$.  Therefore, combining the above cases we obtain \beas N(r,H) = S(r,f).\eeas Now, using {\it Lemma \ref{l3.1}} and logarithmic derivative theorem we have \beas m(r,H(z)) &\leq & m\left(r, \frac{L_{j}f^\prime(z)}{L_jf^\prime (z+c)}\right)+m\left(r,\frac{L_jf(z+c)-a}{L_jf(z)-a}\right) \\&&+ m\left(r,\frac{L_jf(z+c)-b}{L_jf(z)-b}\right) \\&\leq& S(r,f).\eeas Hence, using the above facts we have \bea\label{e4.6} T(r,H)=S(r,f). \eea \\  We denote by $\ol N_{\times}\left(r, \frac{1}{(L_jf(z+c)-a)(L_jf(z+c)-b)}\right)$, the reduced counting function of zeros of $(L_jf(z+c)-a)\; (\text{resp.}\; (L_jf(z+c)-b))$ which are not the zeros of $(L_jf(z)-a) \;(\text{resp.}\;(L_jf(z)-b))$. By a simple calculation we have, \beas \ol N_{\times}\left(r, \frac{1}{(L_jf(z+c)-a)(L_jf(z+c)-b)}\right) \leq \ol N\left(r, \frac{1}{H(z)}\right) \leq T(r,H)=S(r,f).\eeas Next (\ref{e4.1}) can be rewritten as \bea\label{e4.7} \frac{H(z)L_jf^\prime (z+c)}{(L_jf(z+c)-a)(L_jf(z+c)-b)} = \frac{L_{j}f^\prime(z)}{(L_jf(z)-a)(L_jf(z)-b)}. \eea  Henceforth by $\ol N_{(m,n)}(r,p;L_jf(z),L_jf(z+c))=\ol N_{(m,n)}(r,p),$  $p\in \{a,b\}$; we mean the reduced counting function of those common zeros of $(L_jf(z)-p)$ with multiplicity $m$ and of $(L_jf(z+c)-p)$ with multiplicity $n$.\\ \underline{\em\bf{Claim:}} $\ol N_{(m,n)}(r,p)=S(r,f)$, $p\in \{a,b\}$.
\par Without loss of generality take $p=a$. Suppose on the contrary, there exist $m$, $n$ such that $\ol N_{(m,n)}(r,p)\not=S(r,f)$. Let $z_2 \in \mathbb{C}$ such that $L_jf(z_2)=L_jf(z_2+c)=a$. Then (\ref{e4.7}) gives us, \beas H(z_2)=\frac{m}{n}. \eeas \\ \underline{\em\bf{Case 1:}} $m=n$. Then, $H(z_2)=1$. If $H(z) \not \equiv 1$, then using (\ref{e4.6}) we have \beas \ol N_{(m,n)}(r,a) \leq N\left(r,\frac{1}{H(z)-1}\right)=S(r,f),\eeas which is a contradiction. Hence, $H(z)\equiv 1.$ Therefore, (\ref{e4.7}) can be written as \beas  \frac{L_jf^\prime (z+c)}{(L_jf(z+c)-a)(L_jf(z+c)-b)} = \frac{L_{j}f^\prime(z)}{(L_jf(z)-a)(L_jf(z)-b)}, \eeas which implies that $L_jf(z)$ and $L_jf(z+c)$ share $a$, $b$ and $\infty$ CM and using Theorem B we have $L_jf(z)\equiv L_jf(z+c)$, a contradiction.
\\ \underline{\em\bf{Case 2:}} Suppose that $m \not = n,$  a contradiction. So, there exist $z_3 \in \mathbb{C}$ such that $H(z_3)=\frac{m}{n}$. Using similar arguments as in {\it Case 1}, we obtain $H(z)\equiv\frac{m}{n}$. Then (\ref{e4.7}) can be written as \beas\frac{mL_jf^\prime (z+c)}{(L_jf(z+c)-a)(L_jf(z+c)-b)} = \frac{nL_{j}f^\prime(z)}{(L_jf(z)-a)(L_jf(z)-b)}. \eeas Integrating both sides of the above equation we get \beas \left(\frac{L_jf(z+c)-a}{L_jf(z+c)-b}\right)^m = A \left(\frac{L_jf(z)-a}{L_jf(z)-b}\right)^n,\eeas where $A (\not = 0) \in \mathbb{C}$. Therefore,  by Valiron-Mohon'ko theorem and {\it Lemma \ref{l3.3e}}, $m=n$. \par Hence, the claim is true and $\ol N_{(m,n)}(r,p)=S(r,f),$ for $p \in \{a,b\}$. 
Now, \beas\ol N\left(r,\frac{1}{L_jf(z)-a}\right)&=&\sum_{m,n}\ol N_{(m,n)}(r,a)\\&=& \sum_{m\leq 2,n}\ol N_{(m,n)}(r,a)+\sum_{m \geq 3,n}\ol N_{(m,n)}(r,a)\\&\leq& \frac{1}{3}\sum_{m\geq 3,n}\frac{1}{m}N_{(m}\left(r,a;L_jf(z)\right)\\&  \leq & \frac{1}{3}T(r,L_jf(z))+S(r,f).\eeas Similarly, \beas \ol N\left(r,\frac{1}{L_jf(z)-b}\right) \leq \frac{1}{3}T(r,L_{j}f(z))+S(r,f).  \eeas
  Now, by the Second Fundamental Theorem we get \beas T(r,L_jf(z))& \leq &\ol N(r,L_jf(z))+\ol N\left(r,\frac{1}{L_jf(z)-a}\right)+\ol N\left(r,\frac{1}{L_jf(z)-b}\right)\\&&+S(r,f) \\&\leq& \frac{2}{3}T(r,L_jf(z))+(j+1)\ol N(r,f)+S(r,f),\eeas  which is not possible. Hence, $L_jf(z) \equiv L_jf(z+c)$. 
\end{proof}
\begin{proof}[\underline{\em\bf{Proof of Theorem \ref{t2}}}]
	Set \bea\label{e3.8} \Psi(z) \equiv \frac{ L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)}.\eea Since, $E(0,L_{j}f^{(l)}(z)) \subseteq E(0,  L_{j}f^{(s)}(z+c))$ and $ E(\infty,  L_{j}f^{(s)}(z+c))\subseteq E(\infty,L_{j}f^{(l)}(z))  $, $\Psi(z)$ is an entire function. Hence, \beas N(r,\Psi(z))=0. \eeas Further, by {\it Lemma \ref{l3.1}, \ref{l3.4e}} and logarithmic derivative theorem,  we know that \beas m(r,\Psi(z)) \leq m\left(r,  \frac{ L_{j}f^{(s)}(z+c)}{L_{j}f^{(s)}(z)} \right) + m\left(r,  \frac{L_{j}f^{(s)}(z)}{L_{j}f^{(l)}(z)} \right)=S(r,f). \eeas Therefore, \beas T(r,\Psi)=S(r,f).\eeas If possible, let $L_{j}f^{(l)}(z) \not \equiv  L_{j}f^{(s)}(z+c).$ Hence, $\Psi(z) \not = 1.$ 
	\\ \underline{\em\bf{Case 1:}} Let $\Psi(z)$ is constant and $z_0$ be an $a$-point of $L_{j}f^{(l)}(z)$. Since,  $ E(a,L_{j}f^{(l)}(z))\\ \subseteq  E(a,  L_{j}f^{(s)}(z+c))$, we can deduce that $\Psi(z_0)=1$, which implies that $\Psi(z)=1,$ which is not possible. Therefore, $a$ is an e.v.p. of $L_{j}f^{(l)}(z)$.
\\ \underline{\em\bf{Case 2:}} $\Psi(z)$ is not constant. If $z_0$ is $a$-point of $L_{j}f^{(l)}(z)$ and $ L_{j}f^{(s)}(z+c)$, by the same argument used as in {\it Case 1}, $\Psi(z_0)=1$. Then by the First Fundamental Theorem, we obtain, \beas  \ol N \left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right) \leq N\left(r, \frac{1}{\Psi(z)-1}\right)\leq T(r,\Psi)+O(1)=S(r,f).\eeas From (\ref{e3.8}), we have \bea\label{e3.9} \frac{\Psi^{\prime}(z)}{\Psi(z)}=\frac{L_{j}f^{(s+1)}(z+c)}{ L_{j}f^{(s)}(z+c)}-\frac{L_{j}f^{(l+1)}(z)}{L_{j}f^{(l)}(z)}. \eea Let $z_0$ be an $a$-point of $L_{j}f^{(l)}(z)$ of multiplicity $p \geq 2$. Since, $ E(a,L_{j}f^{(l)}(z)) \subseteq  E(a,  L_{j}f^{(s)}(z+c))$, $z_0$ is also an $a$-point of $ L_{j}f^{(s)}(z+c)$ with multiplicity $p \geq 2$. Then, from {(\ref{e3.9})}, it follows that $z_0$ is a zero of $\frac{\Psi^{\prime}(z)}{\Psi(z)}$ of multiplicity $p-1$. Hence, \beas N_{(2} \left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)& \leq& 2N\left(r,\frac{1}{\frac{\Psi^{\prime}(z)}{\Psi(z)}}\right) \\& \leq & 2T\left(r,\frac{\Psi^{\prime}(z)}{\Psi(z)}\right) +S(r,f) \\&\leq& 2N\left(r,\frac{\Psi^{\prime}(z)}{\Psi(z)}\right)+S(r,f)=S(r,f). \eeas  Combining the above cases we can say, \bea\label{e3.10} N\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)=S(r,f).\eea Since,  $E(0,L_{j}f^{(l)}(z)) \subseteq E(0,  L_{j}f^{(s)}(z+c))$ and $  E(\infty,  L_{j}f^{(s)}(z+c)) \subseteq E(\infty,L_{j}f^{(l)}(z))$, we deduce that \beas N\left(r,\frac{1}{L_{j}f^{(l)}(z)}\right)-N\left(r,\frac{1}{ L_{j}f^{(s)}(z+c)}\right)\leq 0\eeas and \beas N(r, L_{j}f^{(s)}(z+c))-N(r,L_{j}f^{(l)}(z)) \leq 0.\eeas 
%By {\it Lemma \ref{l3.6}}, we obtain \beas &&m\left(r,\frac{1}{L_{j}f^{(l)}(z)}\right)+m\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right) \\&&\leq m(r,L_{j}f^{(l)}(z))+N(r, L_{j}f^{(s)}(z+c) )-N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f). \eeas
 Adding $N\left(r,\frac{1}{L_{j}f^{(l)}(z)}\right)+N\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)$ with the both sides of the inequality of {\it Lemma \ref{l3.6}} and applying (\ref{e3.10}) and the First Fundamental Theorem we obtain, \beas 2T(r,L_{j}f^{(l)}(z)) &\leq & N\left(r,\frac{1}{L_{j}f^{(l)}(z)}\right)+N\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)\\&&+m(r,L_{j}f^{(l)}(z))+N(r, L_{j}f^{(s)}(z+c) )\\&&-N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)+S(r,f). \eeas i.e., {\small\beas T(r,L_{j}f^{(l)}(z)) &\leq& N\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)+\left(N\left(r,\frac{1}{L_{j}f^{(l)}(z)}\right)-N\left(r, \frac{1}{ L_{j}f^{(s)}(z+c)}\right)\right)\\&&+(N(r, L_{j}f^{(s)}(z+c)))-N(r,L_{j}f^{(l)}(z))+S(r,f) \\&=&S(r,f),\eeas} which is not possible. Therefore, $L_{j}f^{(l)}(z)  \equiv  L_{j}f^{(s)}(z+c).$ 
\end{proof}

\begin{proof}[\underline{\em\bf{Proof of Theorem \ref{t3}}}]
	Let us consider the following auxiliary function \bea\label{e3.11} U(z)= \frac{L_{j}f^{(l+1)}(z)(L_{j}f^{(l)}(z)-L_{j}f^{(s)}(z+c))}{L_{j}f^{(l)}(z)(L_{j}f^{(l)}(z)-a)}.\eea First assume that $U \not \equiv 0.$ \par Let $z_0$ be $a$-point of $L_{j}f^{(l)}(z)$ of multiplicity $p$ and that of $L_{j}f^{(s)}(z+c)$ of multiplicity $q$. Therefore, $z_0$ is a zero of $U$ of multiplicity $(p-1)+\min\{p,q\}-p \geq 0$. Hence, $z_0$ is not a pole of $U$. Further, as $E(0,L_{j}f^{(l)}(z)) \subseteq  E(0,  L_{j}f^{(s)}(z+c))$, if $z_1$ is a zero of $L_{j}f^{(l)}(z)$ of multiplicity $m$, it is a zero of $L_{j}f^{(s)}(z+c)$ of multiplicity same. Therefore, $z_1$ is a zero of $U$ of multiplicity $(m-1)+m-m=m-1 \geq 0.$ Hence, $z_1$ is not a pole of $U$. Next, let $z_2$ be a pole of $L_{j}f^{(l)}(z)$ of multiplicity $n$. By a simple calculation we can say that $z_2$ is a simple pole of $U$. Using {\it Lemma \ref{l3.9}} we have, \beas N(r,U) \leq \ol N (r,L_{j}f^{(l)}(z) )=S(r,f). \eeas Hence, \beas N(r,U)=S(r,f). \eeas Now, using {\it Lemmas \ref{l3.4e}, \ref{l3.5}} and logarithmic derivative theorem we obtain \beas m(r,U) &\leq& m\left(r,\frac{L_{j}f^{(l+1)}(z)}{L_{j}f^{(l)}(z)-a}\right)+m\left(r,1-\frac{L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)} \right) \\&=&S(r,f).\eeas Therefore, \bea\label{e3.12} T(r,U)=S(r,f).\eea Further, using {\it Lemmas \ref{l3.4e}, \ref{l3.5}} and logarithmic derivative theorem we have \beas&& m\left(r, \frac{1}{L_{j}f^{(l)}(z)}\right) \leq m\left(r, \frac{U}{L_{j}f^{(l)}(z)}\right)+m\left(r,\frac{1}{U}\right)+S(r,f)\\&\leq& m\left(r,\frac{L_{j}f^{(l+1)}(z)}{L_{j}f^{(l)}(z)(L_{j}f^{(l)}(z)-a)}\right)+m\left(r, 1-\frac{L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)}\right) +S(r,f)\\&=&S(r,f).\eeas Clearly from (\ref{e3.11}) and (\ref{e3.12}) using the First Fundamental Theorem we get \beas N\left(r, \frac{1}{L_{j}f^{(l)}(z)}\right)- \ol N\left(r, \frac{1}{L_{j}f^{(l)}(z)}\right) \leq N\left(r, \frac{1}{U}\right) \leq T(r,U)=S(r,f).\eeas By the hypothesis of Theorem \ref{t3} we get\beas N\left(r, \frac{1}{L_{j}f^{(l)}(z)}\right)=S(r,f).\eeas Hence, \beas T\left(r, \frac{1}{L_{j}f^{(l)}(z)}\right)=S(r,f) ,\eeas which is a contradiction. Therefore, $U \equiv 0$,
i.e., $L_{j}f^{(l)}(z) \equiv L_{j}f^{(s)}(z+c).$	\end{proof}
\begin{proof}[\underline{\em\bf{Proof of Theorem \ref{t4}}}]
	%Since, $E(0,L_{j}f^{(l)}(z)) \subseteq  E(0,  L_{j}f^{(s)}(z+c))$ and $  E(\infty,\\  L_{j}f^{(s)}(z+c)) \subseteq E(\infty,L_{j}f^{(l)}(z)) $, so \bea\label{e3.13}\frac{L_{j}f^{(s)}(z+c)}{L_{j}f^{(l)}(z)}=\gamma(z),\eea where $\gamma(z)$ is a meromorphic function such that $N(r,\gamma(z))=S(r,f).$ Using {\it Lemma \ref{l3.4e}, \ref{l3.5}} we get \beas m(r,\gamma(z))=S(r,f),\eeas which implies that \bea \label{e3.14}T(r,\gamma(z))=S(r,f).\eea
	
	Clearly, $\rho_2(L_jf(z)) \leq \rho_2(f) <1$. 
	%If $\rho_2(L_jf(z))$ is of infinite order, then $L_jf(z)$ can not be a rational function.
	% If $\rho_2(L_jf(z))$ is finite then  
		Proceeding as the same way of {\it Lemma 4 of \cite{Pramanik+Sarkar_Math.Studii_2024}} we can prove that $L_jf(z)$ is transcendental.
	Suppose that, $L_{j}f^{(l)}(z) \not \equiv L_{j}f^{(s)}(z+c),$ i.e., $\Psi (z) \not \equiv 1$, since otherwise we are done. Next considering the auxiliary function (\ref{e3.8}) and using the same arguments as used in the proof of Theorem 
	\ref{t2}, we have \beas T(r, \Psi)=S(r,f).\eeas
	  Let $z_0$ be a common $a$-point of $L_{j}f^{(l)}(z)$ and $L_{j}f^{(s)}(z+c)$. Then, $\Psi(z_0)=1.$\par  Therefore, using the First Fundamental Theorem we have {\small\bea \label{e4.13}\nonumber \ol N\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)&=&\ol N\left(r,\frac{1}{L_{j}f^{(s)}(z+c)-a}\right) \leq N\left(r, \frac{1}{\Psi(z)-1}\right)\\&\leq& T(r,\Psi(z))+S(r,f)=S(r,f).\eea }
	  
	  
	  
	  
%	Set \bea\label{e3.16} F(z)=\frac{L_{j}f^{(l+1)}(z)}{L_{j}f^{(l)}(z)(L_{j}f^{(l)}(z)-a)}-\frac{L_{j}f^{(s+1)}(z+c)}{L_{j}f^{(s)}(z+c)(L_{j}f^{(s)}(z+c)-a)}.\eea From (\ref{e3.16}),  applying {\it Lemma \ref{l3.4e}} and the logarithmic derivative theorem we get \beas m(r, F(z)) &\leq & m\left(r, \frac{L_{j}f^{(l+1)}(z)}{L_{j}f^{(l)}(z)}\right)+ m\left(r, \frac{L_{j}f^{(l+1)}(z)}{L_{j}f^{(l)}(z)-a}\right) \\&&+m\left(r, \frac{L_{j}f^{(s+1)}(z+c)}{L_{j}f^{(s)}(z+c)}\right)+m\left(r, \frac{L_{j}f^{(s+1)}(z+c)}{L_{j}f^{(s)}(z+c)-a}\right)\\&=& S(r,f).\eeas Further, using (\ref{e3.15}), from (\ref{e3.16}) we have, \beas N(r, F(z)) &\leq &\ol N_{*} (r,a;L_{j}f^{(l)}(z),L_{j}f^{(s)}(z+c) )+\ol N_{*} (r,0;L_{j}f^{(l)}(z),L_{j}f^{(s)}(z+c) )\\&\leq& \ol N\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)+\ol N\left(r, \frac{1}{ \Psi(z)}\right)+S(r,f)\\&=& S(r,f).  \eeas Hence, \bea \label{e3.17} T(r,F(z))=S(r,f).\eea\\{\bf{\underline{Case 1:}}} Suppose $F \equiv 0$. Integrating (\ref{e3.16}) we have, \bea\label{e3.18} \frac{L_{j}f^{(l)}(z)-a}{L_{j}f^{(l)}(z)} \equiv A \frac{L_{j}f^{(s)}(z+c)-a}{L_{j}f^{(s)}(z+c)},\eea where $A$ is a non-zero complex number. If $A=1$, we have $L_{j}f^{(l)}(z) \equiv L_{j}f^{(s)}(z+c),$ which is a contradiction. If $A \not = 1,$ from (\ref{e3.18}), a simple calculation yields, \beas L_{j}f^{(l)}(z)+\frac{a}{A-1} \equiv \frac{a}{A-1}\left\{\frac{L_{j}f^{(l)}(z)-a}{L_{j}f^{(s)}(z+c)-a}\right\}. \eeas  Clearly, $-\frac{a}{A-1}\not = 0,a.$ Using (\ref{e3.15}), from the above equation we get, \beas \ol N\left(r, \frac{1}{L_{j}f^{(l)}(z)+\frac{a}{A-1}}\right)=S(r,f). \eeas Using the Second Fundamental Theorem, {\it Lemma \ref{l3.9}} and (\ref{e3.15}) we have, {\small\beas &&T(r,L_{j}f^{(l)}(z))\\& \leq & \ol N\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)+\ol N\left(r, \frac{1}{L_{j}f^{(l)}(z)+\frac{a}{A-1}}\right)+\ol N(r,L_{j}f^{(l)}(z))+S(r,f)\\&=&S(r,f),\eeas} a contradiction. \\{\bf{\underline{Case 2:}}} Next, suppose that $F\not \equiv 0.$
	
	
	
	 From (\ref{e3.8}), we have \beas (L_{j}f^{(s)}(z+c)-a) = \Psi(z)\left( L_{j}f^{(l)}(z)-\frac{a}{\Psi(z)}\right). \eeas
	% we have \beas L_{j}f^{(l)}(z)-a=\frac{1}{\Psi(z)}(L_{j}f^{(s)}(z+c)-a\Psi(z)). \eeas
	 
	 Therefore, by (\ref{e4.13}) we have\beas \ol N\left(r,\frac{1}{L_{j}f^{(l)}(z)-\frac{a}{\Psi(z)}}\right)
	 \leq \ol N\left(r,\frac{1}{L_{j}f^{(s)}(z+c)-a}\right)=S(r,f).\eeas
	 
	
	%Since, $\Psi(z) \not \equiv 1$, there exists a complex number $z_0$ such that $\Psi(z_0)=d \not= 1.$ 
	%\\First suppose that, $a$ is not a Picard's exceptional value of $L_{j}f^{(l)}(z)$. Then from (\ref{e4.17}), we know there exist $z_1$ such that it is an a-point as well as $a\Psi(z_1)$ point of $L_{j}f^{(l)}(z)$.
	% Clearly, from (\ref{e4.17}), zeros of $\left(\Psi(z)-d\right)$ are zeros of $\left(L_{j}f^{(s)}(z+c)-ad\right)$. Hence, we get \beas \ol N\left(r,\frac{1}{L_{j}f^{(s)}(z+c)-ad}\right) =S(r,f).\eeas
	 
	 
	 By {\it Lemma \ref{l3.3e}}, we know $L_jf(z)$ is transcendental and so we have $\Psi(z)$ is small function with respect to $L_jf(z)$, by the Second Fundamental Theorem for small function, (\ref{e4.13}) and the above inequality we have, \beas T(r,L_{j}f^{(l)}(z)) &\leq & \ol N(r, L_{j}f^{(l)}(z))+\ol N\left(r, \frac{1}{L_{j}f^{(l)}(z)-a}\right)\\&&+\ol N\left(r,\frac{1}{L_{j}f^{(l)}(z)-\frac{a}{\Psi(z)}}\right)+S(r,f)\\&=&S(r,f), \eeas which is contradiction. Hence, $L_{j}f^{(l)}(z) \equiv L_{j}f^{(s)}(z+c).$
	\end{proof}
\section{Acknowledgment}
The authors wish to thank the reviewer for his/her valuable suggestions towards the improvement of the paper. 

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	\end{thebibliography}
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