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\begin{document}
 \title[A common fixed point theorem via two and three mapping]{A common fixed point theorem via two and three mapping in Banach algebra}
\author{Sanjeev Verma, Kuldip Raj and Sunil K. Sharma}%
\address {School of  Mathematics\newline
Shri Mata Vaishno Devi University\newline Katra-182320, J\&K, India}%
\email{vsanjev28@gmail.com}
\email{kuldipraj68@gmail.com}%
\email{sunilksharma42@gmail.com}%
\thanks{}%
\subjclass[2020]{47H09, 47H10, 26A25.}%
\keywords {Common fixed point theorem, Banach algebra, measure of non-compactness, Schaurder fixed point theorem, Darbo's fixed point theorem}%

%\date{February 8, 2006}%
%\dedicatory{}%
%\commby{}%
% ----------------------------------------------------------------
\begin{abstract}
In this paper an attempt is made to prove common fixed point theorem for two and three mappings in Banach algebra by using measure of non-compactness. We also make an effort to prove the common fixed point theorem for two and three commuting maps.
\end{abstract}
\maketitle
\section{Introduction and preliminaries}
Schauder [18] used Compactness in fixed point theory and G. Darbo in 1955 [15] used non compactness and non compact operators in fixed point theory. The foremost aim of study of non compact operators is to create a new family of operators which transform the bounded set to compact set. \\\\
{\bf Definition 1.1:}
	Let $M$ be a metric space and $N$ be its subset then Kuratowski's measure of non compactness is defined as
\begin{equation*}
\gamma(N) = \displaystyle\inf\Big\{\theta>0:N = \cup_{i=1}^n N_i\;\;\mbox{for some}\;\;N_i\;\;\mbox{with diam}(N_i)\leq\theta, 1\leq i\leq n < \infty\Big\}.
\end{equation*} Here diam $(N_i) = \sup\{d(a,b): a, b\in N_i, i = 1,2,\cdots, n\}$ (see [16]).\\
 {\bf Definition 1.2: }
 Let $S$ be a Banach space and $N$ be a non empty bounded subset of $N_S$, where $N_S$ denotes the family of non empty subsets of $S$. Then the mapping $\delta: N_S \rightarrow [0,\infty)$ is called Hausdorff measure of non compactness of $N\subseteq N_S$ and is defined as
 \begin{equation*}
 \delta(N) = \displaystyle\inf\Big\{\alpha>0:N\;\; \mbox{has a finite}\;\; \alpha-\mbox{net in S}\Big\}\;\; \mbox{(see [12])}.\end{equation*}
{\bf Definition 1.3: }
Let $(M,d)$ be a complete metric space and $N$ be a non empty bounded subset of $N_M$,  where $N_M$ denotes the family of non empty subsets of $M$. Then the mapping $\delta: N_M \rightarrow[o,\infty)$ is called Hausdorff measure of non compactness of $N\subseteq N_M$ and is defined as
\begin{equation*}
	\delta(N) = \displaystyle\inf\Big\{\alpha>0:N \subseteq\cup_{i=1}^n B(x_i,r_i), x_i\in M, r_i< \alpha\;\;i=1,2,3,....,n\Big\}\;\; \mbox{(see [12])}.\end{equation*}
%{\bf Definition 1.4: }
%Let $(M,d)$ be a complete metric space and $N$ be a non empty bounded subset of $N_M$ then the mapping $\epsilon: N_M \rightarrow [0,\infty)$ is called Istratescu measure of non compactness of $N \subseteq N_M$ s.t
%\begin{equation*}
%	\epsilon(N) = \displaystyle\inf\Big\{\beta>0:N\;\; \mbox{has no infinite}\;\; \beta-\mbox{ discrete subsets}\Big\}\;\; \mbox{(see [12])}.\end{equation*}
%\textbf{Theorem 1.5:} { Let $(M,d)$ be a complete metric space. If $\{\phi_n\}$ be a decreasing sequence of non-empty, closed and bounded subsets of $M$ s.t ${\displaystyle\lim_{n\rightarrow\infty} \gamma(\phi_n)=0}$ then the intersection ${\phi_{\infty}=\bigcap_{i=1}^\infty\phi_n} $	is non empty and compact subset of $M$ (see [12]).}\\\\
%{\bf Proposition 1.6:} {Let $ N, N_1$ and $N_2$ be non empty and bounded subsets of a complete metric space  $(M,d)$ then }\\
%	1.	$\gamma(N) = 0 \Longleftrightarrow \overline{N}$ is compact, $\overline{N}$ is the closure of $N$                        \\
%	2. $\gamma(N)=\gamma(\overline{N})$\\
 %   3. $N_1\subseteq {N_2}\rightarrow\gamma(N_1)\leq\gamma(N_2)$\\
  %  4.  $\gamma(N_1\bigcup{N_2})=max\{\gamma(N_1),\gamma(N_2)\}$\\
   % 5. $\gamma(N_1\bigcap{N_2})\leq\min\{\gamma(N_1),\gamma(N_2)\}$.\\
    %For proof we refer to $[14]$.\\
%{\bf Proposition 1.7:} {Let $ N, N_1$ and $N_2$ be non empty and bounded subsets of a complete metric space  $(M,d)$ then}\\
 % 	1. $\delta(N) = 0 \Longleftrightarrow \overline{N}\;\; \mbox{is compact},\;\;\overline{N}$ is the closure of $N$ \\
  %	2. $\delta(N)=\delta(\overline{N})$\\
  	%3. If $N_1\subseteq {N_2}\rightarrow\delta(N_1)\leq\delta(N_2)$\\
  	%4. $\delta(N_1\bigcup{N_2})= \max\{\delta(N_1),\delta(N_2)\}$\\
  	%5. $\delta(N_1\bigcap{N_2}) \leq\min\{\delta(N_1),\delta(N_2)\}$\\
%{\bf Proposition 1.8:} {Let $ N, N_1$ and $N_2$ be non empty and bounded subsets of a Banach space  $(S,\Vert.\Vert)$  over $F$ then}\\
 %1. $\delta(N_1+N_2)\leq \delta(N_1) +\delta (N_2)$ \\
 %2. $\delta(N+x)\ = \delta(N) ,\forall x\in S$ \\
 %3. $\delta(\lambda(N)=\vert\lambda\vert\delta(N),\forall \lambda\in F$ \\
 %4. $\delta(N)\ = \delta(Conv(N)),$ where $Conv(N)$ is the convex hull of $N$\\
 %{\bf Note:} Similar Properties hold for Istratescu measure of Non Compactness\\
\textbf{Theorem 1.4: Schauder's fixed point theorem:} {Let $S$ be a Banach space and $\aleph$ be a  non-empty bounded subsets of S then every continuous mapping $ P: \aleph\rightarrow\aleph$ has atleast one fixed point (see [12]).} We use abbreviation S. f. t for Schauder's fixed point theorem throughout the paper. \\
{\bf Definition 1.5:} A function $m*: {\mu_s} \longrightarrow \bf{R_+}$ (where $S $ is the Banach space and ${\mu_s}$ is the family of all non empty bounded subsets of $S$) is said to be measure of non compactness in the space $S$ if it satisfies the following conditions.\\
1. The family $\mbox{Ker}\; m*= \{E\in\mu_s : m^*(E)=0\}$ is non empty and $\mbox{Ker}\; m^*\subseteq\nu_s$,  where $\nu_s$ is the family of non empty relatively compact subsets of $S$.\\
2. $ E_1\subseteq {E_2}\Rightarrow {m^*(E_1)}\leq m^*(E_2)$\\
3. $ m^*(E)= m^*(\overline{E})$\\
4. $ m^*(E)=m^*(conv({E}))$\\
5. $ m^*\{\lambda E_1+(1-\lambda)(E_2)\}\leq\lambda m^*(E_1)+(1-\lambda)m^*(E_2), \forall\in [0,1]$\\
6. If $\{E_n\}$ is a sequence of closed sets from $\mu_s$ such that $E_{n+1}\subseteq E_n$ for $n=1,2,3,....$ and $\displaystyle\lim_{n\rightarrow\infty}m^*(E_n)=0$, then the intersection $E_{\infty}=\displaystyle\cap_{i=1}^\infty E_n$	is non empty (see [12]).\\
{\bf Definition 1.6:} Let $\aleph$ be a non empty bounded, closed and convex subset of a Banach space $S$. A self mapping $P: \aleph\rightarrow\aleph $ is said  to be a $m^*$ contractive mapping if there exists some constant $k \in (0,1)$ s.t $m^*(P(E))\leq k{m^*(E)},$ for non empty subset $E$  of $\aleph.$\\
\textbf{Theorem 1.7: Darbo's fixed point theorem:} { Let $S$ be a Banach space and $\aleph$ be a  non-empty, closed and bounded subsets of $S$ and let mapping $P: \aleph\rightarrow\aleph$ be continuous and if $P$ is a $m*$ contraction then $P$ has atleast one fixed point (see [5]).\\
{\bf Definition 1.8:} Let $T$ be a non empty set then $T$ is said to be an algebra if\\
1. $((T,+,.)$ is a vector space over $F$,\\
2. $ (T,+,\circ)$ is a ring  and\\
3. $(\alpha a)\circ b=\alpha(a\circ b)=a\circ(\alpha b),$ $\forall \alpha\in F $and $a,b\in T$ (see [13]).\\
{\bf Definition 1.9:} If $T$ is an algebra and {$\Vert.\Vert$} a norm on $T$ satisfying {$\Vert ab \Vert \leq \Vert a \Vert \Vert b \Vert$}, $\forall a,b\in T$  then the pair {$(T,\Vert.\Vert)$} is called normed algebra. A complete norm algebra $T$ is called Banach algebra (see [13]).\\For example, let $K$ be a compact Hausdorff space and  $T = C(K)$ then with respect to point wise multiplication of functions, $T$ is a commutative unital algebra with norm $\Vert f \Vert_\infty = \displaystyle \sup_{t\in K} |f(t)|$ is a Banach algebra. For more details about Banach algebra and measure of non compactness one may refer to ([1], [2], [3], [4], [6], [7], [9], [10], [11], [14], [17]) and references therein.\\
{\bf Note:} $m*$ is the measure of non-compactness satisfying the conditions of definition 1.5. Now if the Banach space $S$ has the structure of Banach algebra $T$  and for given subsets $E_1$, $E_2$ of $\mu_T$, let $E_1 E_2 =\{ab: a\in E_1, b\in E_2\}$ then  the following inequality holds
$$m^* (E_1 E_2) \leq (|| E_1 ||) m*(E_2)+ (|| E_2||)m*(E_1).$$ The foremost goal of this paper is to prove common fixed point theorem for two and three commuting mapping in Banach algebra by using measure of non compactness.	
\section{Common fixed point theorem for two continuous linear operators on Banach Algebra $T$ }
\textbf{Theorem 2.1} \it{Let $\aleph$ be a non empty bounded, closed and convex subset of the Banach algebra $T$ and operators $P_1, P_2, Q_1, Q_2$ transforms $\aleph$ into $T$ are continuous such that $ P_1(\aleph), P_2(\aleph), Q_1(\aleph),$ $Q_2(\aleph)$ are bounded. Suppose $P = P_1Q_1$ and $Q = P_2Q_2$ be the operators from $\aleph$ into $\aleph$ are continuous and $Q$ is linear such that $Q(P(E))\subseteq P(E)$, $E\subseteq \aleph$.
If the operators $P_1, P_2, Q_1,$ $Q_2$ on set $\aleph$ satisfies the following conditions
$m^*(P_1(E))\leq\xi_1(m^*(E)),$ $ m^*(Q_1(E))\leq \xi_2(m^*(E)),$ $m^*(P_2(E))\leq\xi'_{1}(m^*(E)),$
 $ m^*(Q_2(E)) \leq \xi'_{2} (m^*(E))$,
where $E$ is the non empty subset of $\aleph$, $m^*$ is an arbitrary measure of non compactness defined
on $\mu_T $ and $\xi_1, \xi_2, \xi'_1, \xi'_2$ are non decreasing functions from $R_+$ to  $R_+ $ such that $$\displaystyle\lim_{n\rightarrow\infty}\xi^n_{1}(x)=0, $$ $$\displaystyle\lim_{n\rightarrow\infty}\xi^n_{2}(x)=0,$$
$$\displaystyle\lim_{n\rightarrow\infty}\xi'^n_{1}(x)=0, $$ $$\displaystyle\lim_{n\rightarrow\infty}\xi'^n_{2}(x)=0$$
and $$\displaystyle\lim_{n\rightarrow\infty}(||P_1(\aleph)|| \xi_2 +||Q_1(\aleph)||\xi_1)^n(x)=0,$$
$$\displaystyle\lim_{n\rightarrow\infty}(||P_2(\aleph)|| \xi'_2 +||Q_2(\aleph)||\xi'_1)^n(x)=0,$$ for any $x\geq 0$,
then $P$ and  $Q$ have a common fixed point.}
\begin{proof} Let $E$ be a non empty subset of $\aleph$. Then in view of the assumption that $m^*$ is a measure of non compactness defined on $\mu_T$, we have \begin{eqnarray*}
m^*(P(E)) & = & m^*(P_1Q_1(E))\\ &\leq & m^*(P_1(E)Q_1(E))\\ &\leq & m^*(Q_1(E))||P_1(E)||+m^*(P_1(E))||Q_1(E)||\\ &\leq& m^*(Q_1(E))||P_1(\aleph)|| + m^*(P_1(E)||Q_1(\aleph))||\\ &\leq & \xi_2(m^*(E))||P_1(\aleph)|| + \xi_1(m^*(E))||Q_1(\aleph)||\\ & = & (\xi_2)||P_1(\aleph)||+ (\xi_1)||Q_1(\aleph)||)(m^*(E) \;\;\;\;\;\;\;\ (2. 1)\end{eqnarray*}
Let $\psi_1(x)= ((\xi_2)||P_1(\aleph)||+ (\xi_1)||Q_1(\aleph)||)(x). $ So from equation (2. 1), we have
$m^*(P(E))\leq \psi_1(m^*(E))$.
Similarly,
$m^*(Q(E))\leq \psi_2(m^*(E))$, where $\psi_2(x)= ((\xi'_2)||P_2(\aleph)||+ (\xi'_1)||Q_2(\aleph)||)(x). $ \\
Since $\xi_1,\xi_2,\xi'_1,\xi'_2$ are non decreasing functions from $R_+$to $ R_+ $ so $\psi_1$ and $\psi_2$ are also non decreasing thus because of given condition
$\displaystyle\lim_{n\rightarrow\infty} \psi^n_{1}(x) =0$ and $\displaystyle\lim_{n\rightarrow\infty} \psi^n_{2}(x) =0$.\\
Now we can define a sequence of subsets $\{\aleph_n\}$ of $T$ as $\aleph=\aleph_0 ,\aleph_n = convP\aleph_{n-1},  n\geq 1$.\\
Then $\aleph_n \subseteq \aleph_{n-1}$ and $Q(\aleph_n)\subset \aleph_n$ \;\;\;\;\;\;\; (2. 2)\\
Clearly $\aleph_1 \subset \aleph_{0} $ and $Q(\aleph_1)\subseteq Conv(QP(\aleph_{0}))\subseteq Conv(P(\aleph_{0}))=\aleph_1$. Therefore  $Q(\aleph_1)\subseteq \aleph_1$ and so equation (2. 2) holds for n=1. Suppose it is true for $n\geq 1$, then $(\aleph_{n+1})= Conv(P(\aleph_n))\subseteq Conv(P(\aleph_{n-1}))= \aleph_n ,$  as $\aleph_n \subseteq \aleph_{n-1}$, thus $\aleph_{n+1} \subseteq \aleph_n $
and $$Q(\aleph_{n+1})=Q(Conv(P(\aleph_n)))\subseteq Conv(QP(\aleph_{n}))\subseteq Conv P(\aleph_n)=\aleph_{n+1}.$$ Hence $Q(\aleph_{n+1}) \subseteq \aleph_{n+1}$ and this implies that $\aleph_{0}\supset \aleph_1 \supset \aleph_2 ...  \;\;\   .$\\
If there exists $n\geq 0$ such that $m^*(\aleph_n)=0$,  then $\aleph_n$ is relatively compact and since $P(\aleph_n)\subseteq Conv P(\aleph_n)=\aleph_{n+1}\subseteq \aleph_n,$ so by S. f. t $P$ has a fixed point .\\
 Now, we assume that $m^*(\aleph_n)\neq 0 , n\geq 0$. Then by assumption,
\begin{eqnarray*}
 m^*(\aleph_{n+1}) & = & m^*(Conv(P(\aleph_n)))\\ & = & m^*(P(\aleph_n))\\ &\leq& \psi_1 (m^*(\aleph_{n}))\\ & = & \psi_1(m^*(Conv P(\aleph_{n-1} )))\\ &\leq& \psi^2_{1} (m^*(\aleph_{n-1}))\\& & \cdots\cdots\cdots\\& \leq & \psi^n_{1}(m^*(\aleph_{0})).
 \end{eqnarray*}
Since $ \psi_1 :\mathbb{R_+} \rightarrow \mathbb{R_+}$ is a non decreasing and $\displaystyle \lim_{n\rightarrow\infty}\psi^n_{1} (x)=0$. If we consider $r=\displaystyle\lim_{n\rightarrow\infty} m^*(\aleph_{n+1}) \leq \displaystyle\lim_{n\rightarrow\infty} m^*\psi^n_{1}(m^*(\aleph_0))=\displaystyle\lim_{n\rightarrow\infty}\psi^n_{1}(x)=0,$ for $ x\geq 0, x=m^*(\aleph_0)$\\
then, $m^*(\aleph_n)\rightarrow 0$ as $ n\rightarrow \infty$. Since $\aleph_{n+1} \subseteq \aleph_n$ and $P(\aleph_{n}) \subseteq \aleph_n, n\geq 1$.\\
 Thus, by condition $(6)$ in the definition of $m^*$, we have $\displaystyle\aleph_{\infty} =\cap^{\infty}_{n=1}\aleph_{n}$ is non empty, closed and convex subset in $\aleph$. Moreover $\aleph_{\infty}$ is invariant under the operator $P$ and belongs to $\mbox{Ker}m^*$ is relatively compact. Hence, by S. f. t $P$ has a fixed point in $\aleph$ .\\
Now, Suppose $G_P =\{t\in \aleph : P(t)=t\}$. Then clearly $G_P$ is closed by continuity of $P$  and by assumption $Q$ is linear, we have $Q(G_P)\subseteq G_P $. So $Q(t)$ is a fixed point of $P$ for any $t\in G_P$ and $ m^*(G_P)=m^*(P(G_P))\leq \psi_1 (m^*(G_P))\leq m^*(G_P).$\\
Therefore, $m^*(G_P)=0 $ and $G_P$ is compact. Then by S. f. t, $Q$ has a fixed point and set $G_Q =\{t\in \aleph : Q(t) = t\}$ is closed by continuity of $Q$.\\
Also $Q(G_P) \subseteq G_P$ thus by S. f. t, $P(t)$ is a fixed point of $Q$, for all $t\in G_Q$.
Since $G_P \cap G_Q \subseteq G_P \subseteq \aleph $ is compact  subset  and $P,Q :G_P\cap G_Q \rightarrow G_P\cap G_Q$ are continuous self maps so by S. f. t   $P \;\&\; Q$ have a common fixed point in $\aleph$.
\end{proof}
In next theorem we prove common fixed point theorem for two continuous Commutative operators on Banach Algebra $T$.\\\\
\textbf{Theorem 2.2} \it{Let $\aleph$ be a non empty bounded, closed and convex subset of the Banach algebra $T$ and operators $P_1, P_2, Q_1, Q_2$ transforms $\aleph$ into $T$ are continuous and mutually commutative  s.t $ P_1(\aleph), P_2(\aleph), Q_1(\aleph),$ $Q_2(\aleph)$ are bounded. Suppose $P = P_1Q_1$ and $Q = P_2Q_2$ be the commutative operators from $\aleph$ into $\aleph$ are continuous. If the operators $P_1, P_2, Q_1,$ $Q_2$ on set $\aleph$ satisfies the following conditions	$m^*(P_1(E))\leq\xi_1(m^*(E)), m^*(Q_1(E))\leq \xi_2(m^*(E)),$ $m^*(P_2(E))\leq\xi'_{1}(m^*(E)),
	m^*(Q_2(E)) \leq \xi'_{2} (m^*(E))$,
	where $E$ is the non empty subset of $\aleph$ and $m^*$ is an arbitrary measure of non compactness defined
	on $\mu_T $. Also $\xi_1, \xi_2, \xi'_1, \xi'_2$ are non decreasing functions from $R_+$ to  $R_+ $ s.t $\displaystyle\lim_{n\rightarrow\infty}\xi^n_{1}(x)=0$, $\displaystyle\lim_{n\rightarrow\infty}\xi^n_{2}(x)=0$,
	$\displaystyle\lim_{n\rightarrow\infty}\xi'^n_{1}(x)=0$,
	$\displaystyle\lim_{n\rightarrow\infty}\xi'^n_{2}(x)=0$
	and $$\displaystyle\lim_{n\rightarrow\infty}(||P_1(\aleph)|| \xi_2 +||Q_1(\aleph)||\xi_1)^n(x)=0,$$
	$$\displaystyle\lim_{n\rightarrow\infty}(||P_2(\aleph)|| \xi'_2 +||Q_2(\aleph)||\xi'_1)^n(x)=0,$$ for any $x\geq 0$,
	then $P$ and  $Q$ have a common fixed point.}
\begin{proof}: Let $E$ be a non empty subset of $\aleph$ then by definition of $m^*$ on $\mu_T$, we have \begin{eqnarray*}
		m^*(P(E)) & = & m^*(P_1Q_1(E))\\ &\leq& m^*(P_1(E)Q_1(E))\\ &\leq & m^*(Q_1(E))||P_1(E)||+m^*(P_1(E))||Q_1(E)||\\ &\leq& m^*(Q_1(E))||P_1(\aleph)|| + m^*(P_1(E)||Q_1(\aleph))||\\ &\leq & \xi_2(m^*(E))||P_1(\aleph)|| + \xi_1(m^*(E))||Q_1(\aleph)||\\ & = & (\xi_2)||P_1(\aleph)||+ (\xi_1)||Q_1(\aleph)||)(m^*(E)  \;\;\;\;\     ........(2.3)\end{eqnarray*}
	Let $\psi_1(x)= ((\xi_2)||P_1(\aleph)||+ (\xi_1)||Q_1(\aleph)||)(x). $ So from equation (2.3), we have
	$m^*(P(E))\leq \psi_1(m^*(E))$. Similarly
	$m^*(Q(E))\leq \psi_2(m^*(E))$, where $\psi_2(x)= ((\xi'_2)||P_2(\aleph)||+ (\xi'_1)||Q_2(\aleph)||)(x). $
	Since $\xi_1, \xi_2, \xi'_1, \xi'_2$ are non decreasing functions from $R_+$ into $ R_+ $ so $\psi_1$ , $\psi_2$ are also non decreasing functions. Now because of given condition, we have
	$\displaystyle\lim_{n\rightarrow\infty} \psi^n_{1}(x) =0$ and $\displaystyle\lim_{n\rightarrow\infty} \psi^n_{2}(x) =0$.
	Now we can define a sequence of subsets $\{\aleph_n\}$ of $T$ as $\aleph=\aleph_0 ,\aleph_n = convP\aleph_{n-1},  n\geq 1$.
	Then $\aleph_n \subseteq \aleph_{n-1}$ and $Q(\aleph_n)\subset \aleph_n$ \;\;\;\;\;\;\ ...(2.4). Clearly $\aleph_1 \subset \aleph_{0} $ and $Q(\aleph_1)\subseteq Conv(QP(\aleph_{0}))=Conv(PQ(\aleph_{0}))\subseteq Conv.(P(\aleph_{0}))=\aleph_1$   (as $PQ=QP$).
	Therefore, $Q(\aleph_1)\subseteq \aleph_1$ and so (2.4) holds for n=1. Suppose it is true for $n\geq 1$ then $(\aleph_{n+1})= Conv(P(\aleph_n))\subseteq Conv(P(\aleph_{n-1}))= \aleph_n ,$    as $\aleph_n \subseteq \aleph_{n-1}$.
	Thus $\aleph_{n+1} \subseteq \aleph_n $
	and $Q(\aleph_{n+1})=Q(Conv(P(\aleph_n)))\subseteq Conv(QP(\aleph_{n}))\subseteq Conv P(\aleph_n)=\aleph_{n+1}$. Thus $Q(\aleph_{n+1}) \subseteq \aleph_{n+1}$ and this implies that $\aleph_{0}\supset \aleph_1 \supset \aleph_2 ...\;\;\;\;.$\\
	If there exists $n\geq 0$ s.t $m^*(\aleph_n)=0$  then $\aleph_n$ is relatively compact. Since $P(\aleph_n)\subseteq Conv P(\aleph_n)=\aleph_{n+1}\subseteq \aleph_n,$ so by S. f. t $P$ has a fixed point.
	Now, we assume that $m^*(\aleph_n)\neq 0 , n\geq 0$. Then by assumption, \begin{eqnarray*}m^*(\aleph_{n+1})&=& m^*(Conv(P(\aleph_n)))\\ &=& m^*(P(\aleph_n))\\ &\leq& \psi_1 (m^*(\aleph_{n}))\\ &=& \psi_1(m^*( Conv P(\aleph_{n-1} )))\\ &\leq& \psi^2_{1} (m^*(\aleph_{n-1}))\\ & &\cdots\cdots\\ &\leq & \psi^n_{1}(m^*(\aleph_{0})).\end{eqnarray*} Since $ \psi_1 :\mathbb{R_+} \rightarrow \mathbb{R_+}$ is a non decreasing and $\displaystyle \lim_{n\rightarrow\infty}\psi^n_{1} (x)=0$. If we consider $r=\displaystyle\lim_{n\rightarrow\infty} m^*(\aleph_{n+1}) \leq \displaystyle\lim_{n\rightarrow\infty} m^*\psi^n_{1}(m^*(\aleph_0))=\displaystyle\lim_{n\rightarrow\infty}\psi^n_{1}(x)=0,$ for $ x\geq 0 ,x=m^*(\aleph_0)$,
	then, $m^*(\aleph_n)\rightarrow 0$ as $ n\rightarrow \infty$. Since $\aleph_{n+1} \subseteq \aleph_n$ and $P(\aleph_{n}) \subseteq \aleph_n , n\geq 1$,
	so by condition $(6)$ in the definition of $m^*$, we have $\displaystyle\aleph_{\infty} =\cap^{\infty}_{n=1}\aleph_{n}$ is a non empty closed and convex subset in $\aleph$. Moreover $\aleph_{\infty}$ is invariant under the operator $P$ and belongs to $Kerm^*$ is relatively compact. So, by S. f. t $P$ has a fixed point in $\aleph$.\\
	Now, Suppose $G_P=\{t\in \aleph : P(t)=t\}$ then clearly $G_P$ is closed by continuity of $P$  and we have $Q(G_P)\subseteq G_P $ then $Q(t)$ is a fixed point of $P$ for any $t\in G_P$ and $ m^*(G_P)=m^*(P(G_P))\leq \psi_1 (m^*(G_P))\leq m^*(G_P).$\\
	Therefore, $m^*(G_P)=0 $ and $G_P$ is compact then by S. f. t $Q$ has a fixed point and set $G_Q =\{t\in \aleph : Q(t) = t\}$ is closed by continuity of $Q$. Also $Q(G_P) \subseteq G_P$ then by S. f. t, $P(t)$ is a fixed point of $Q$, for all $t\in G_Q$.
	Since $G_P \cap G_Q \subseteq G_P \subseteq \aleph $ is compact  subset  and $P,Q :G_P\cap G_Q \rightarrow G_P\cap G_Q$ are continuous self maps so by S. f. t   $P\;\&\;Q$ have a common fixed point in $\aleph$.
\end{proof}
\section{Common fixed point theorem for three continuous Commutative operators on Banach Algebra $T$}
\textbf{Theorem 3.1} \it{Let $\aleph$ be a non empty bounded, closed and convex subset of the Banach algebra $T$ and operators $P_1, P_2, P_3, Q_1, Q_2, Q_3$ transforms $\aleph$ into $T$ are continuous and mutually commutative  such that  $ P_1(\aleph), P_2(\aleph), P_3(\aleph), Q_1(\aleph),$ $Q_2(\aleph), Q_3(\aleph)$ are bounded. Suppose $P = P_1Q_1$,  $Q = P_2Q_2$ and $R = P_3 Q_3$ be three continuous and commuting mappings from $\aleph$ into $\aleph$ such that $Q$ and $R$ are linear and  $Q(P(E))\subseteq P(E)$, $R(P(E)) \subseteq P(E)$, $E\subseteq \aleph.$
If the operators $P_1, P_2, P_3, Q_1,$ $Q_2, Q_3$ on  $\aleph$ satisfies the following conditions
$$m^*(P_1(E))\leq\xi_1(m^*(E)),$$ $$ m^*(Q_1(E))\leq \xi_2(m^*(E)),$$ $$m^*(P_2(E))\leq\xi'_{1}(m^*(E)),$$ $$ m^*(Q_2(E)) \leq \xi'_{2} (m^*(E)),$$
$$ m^*(P_3(E))\leq\xi''_1(m^*(E)),$$  and $$ m^*(Q_3(E))\leq \xi''_2(m^*(E)), $$
where $E$ is the non empty subset of $\aleph$ and $m^*$ is an arbitrary measure of non compactness defined
	on $\mu_T $ and $\xi_1, \xi_2, \xi'_1, \xi'_2, \xi''_1, \xi''_2$ are non decreasing functions from $R_+$ to  $R_+ $ s.t $\displaystyle\lim_{n\rightarrow\infty}\xi^n_{1}(x)=0$, $\displaystyle\lim_{n\rightarrow\infty}\xi^n_{2}(x)=0$,
	$\displaystyle\lim_{n\rightarrow\infty}\xi'^n_{1}(x)=0$,
	$\displaystyle\lim_{n\rightarrow\infty}\xi'^n_{2}(x)=0$, $\displaystyle\lim_{n\rightarrow\infty}\xi''^n_{1}(x)=0$,
	$\displaystyle\lim_{n\rightarrow\infty}\xi''^n_{2}(x)=0$
	and $$\displaystyle\lim_{n\rightarrow\infty}(||P_1(\aleph)|| \xi_2 +||Q_1(\aleph)||\xi_1)^n(x)=0,$$
	 $$\displaystyle\lim_{n\rightarrow\infty}(||P_2(\aleph)|| \xi'_2 +||Q_2(\aleph)||\xi'_1)^n(x)=0,$$ $$\displaystyle\lim_{n\rightarrow\infty}(||P_3(\aleph)|| \xi''_2 +||Q_3(\aleph)||\xi''_1)^n(x) = 0,$$  for any $x\geq 0$,
	then we have\\
	1. $P$, $Q$ and $R$ have a common fixed point in $\aleph$.\\
2. If $Q(Conv(\aleph))\subseteq Conv(Q(\aleph)) $ then $PQ$, $Q$ and $R$ have a fixed point in $\aleph$.}
\begin{proof}: Let $E$ be a non empty subset of $\aleph$ and $m^*$ be a measure of non compactness defined on $\mu_T$, $T$ is a Banach algebra, then we have \begin{eqnarray*}
		m^*(P(E)) & = & m^*(P_1Q_1(E))\\ &\leq& m^*(P_1(E)Q_1(E))\\ &\leq & m^*(Q_1(E))||P_1(E)||+m^*(P_1(E))||Q_1(E)||\\ &\leq& m^*(Q_1(E))||P_1(\aleph)|| + m^*(P_1(E)||Q_1(\aleph))||\\ &\leq & \xi_2(m^*(E))||P_1(\aleph)|| + \xi_1(m^*(E))||Q_1(\aleph)||\\ & = & (\xi_2)||P_1(\aleph)||+ (\xi_1)||Q_1(\aleph)||)(m^*(E)       ...  \;\;\ .(3.1)\end{eqnarray*}
	Let $\psi_1(x)= ((\xi_2)||P_1(\aleph)||+ (\xi_1)||Q_1(\aleph)||)(x). $ So from equation (3.1), we have
	$m^*(P(E))\leq \psi_1(m^*(E)).$
	Similarly $m^*(Q(E))\leq \psi_2(m^*(E))$ and  $m^*(R(E))\leq \psi_3(m^*(E)),$ \\where $$\psi_2(x)= ((\xi'_2)||P_2(\aleph)||+ (\xi'_1)||Q_2(\aleph)||)(x)$$ and  $$\psi_3(x)= ((\xi''_2)||P_3(\aleph)||+ (\xi''_1)||Q_3(\aleph)||)(x). $$
	Since $\xi_1, \xi_2, \xi'_1, \xi'_2, \xi''_1, \xi''_2$ are non decreasing functions from $R_+$ to $ R_+ $ so $\psi_1$, $\psi_2$ and $\psi_3$ are also non decreasing because of given conditions, so we have
	$$\displaystyle\lim_{n\rightarrow\infty} \psi^n_{1}(x) =0, \displaystyle\lim_{n\rightarrow\infty} \psi^n_{2}(x) =0 \;\; \mbox{and}\;\; \displaystyle\lim_{n\rightarrow\infty} \psi^n_{3}(x) =0.$$
	Now we can define a sequence of subsets $\{\aleph_n\}$ of $T$ as $\aleph = \aleph_0,\; \aleph_n = convP\aleph_{n-1},  n\geq 1$,\\
	then $\aleph_n \subseteq \aleph_{n-1}$ and $Q(\aleph_n)\subset \aleph_n$  ........(3.2)\\
	Clearly $\aleph_1 \subset \aleph_{0} $ and $Q(\aleph_1)\subseteq Conv(QP(\aleph_{0}))=Conv(PQ(\aleph_{0}))\subseteq Conv(P(\aleph_{0}))=\aleph_1$   as $PQ=QP$. Therefore,  $Q(\aleph_1)\subseteq \aleph_1$ and so (3.2) holds for $n=1.$ Suppose it is true for $n\geq 1$, then $(\aleph_{n+1})= Conv(P(\aleph_n))\subseteq Conv(P(\aleph_{n-1}))= \aleph_n,$ because $\aleph_n \subseteq \aleph_{n-1}$.\\
	Thus $\aleph_{n+1} \subseteq \aleph_n $
	and $Q(\aleph_{n+1})=Q(Conv(P(\aleph_n)))\subseteq Conv(QP(\aleph_{n}))\subseteq Conv P(\aleph_n)=\aleph_{n+1}$.So $Q(\aleph_{n+1}) \subseteq \aleph_{n+1}$ and this implies that $\aleph_{0}\supset \aleph_1 \supset \aleph_2 ...   \;\;\ .$\\
	If there exists $n\geq 0$ s.t $m^*(\aleph_n)=0$  then $\aleph_n$ is relatively compact. Since $P(\aleph_n)\subseteq Conv P(\aleph_n)=\aleph_{n+1}\subseteq \aleph_n,$ so by S. f. t $P$ has a fixed point.\\
Now, we assume that $m^*(\aleph_n)\neq 0, n\geq 0$. Then by supposition
\begin{eqnarray*}
 m^*(\aleph_{n+1})& = & m^*(Conv(P(\aleph_n)))\\ & = & m^*(P(\aleph_n))\\ &\leq&  \psi_1 (m^*(\aleph_{n}))\\& = & \psi_1(m^*( Conv P(\aleph_{n-1} )))\\ &\leq& \psi^2_{1} (m^*(\aleph_{n-1}))\\& &\cdots\cdots \\&\leq& \psi^n_{1}(m^*(\aleph_{0})).
\end{eqnarray*}
Since  $ \psi_1 :\mathbb{R_+} \rightarrow \mathbb{R_+}$ is a non decreasing function and $\displaystyle \lim_{n\rightarrow\infty}\psi^n_{1} (x)=0$.
 If we consider $r=\displaystyle\lim_{n\rightarrow\infty} m^*(\aleph_{n+1}) \leq \displaystyle\lim_{n\rightarrow\infty} \psi^n_{1}(m^*(\aleph_0))=\displaystyle\lim_{n\rightarrow\infty}\psi^n_{1}(x)=0,$ for $ x\geq 0 \;\ \mbox{and} \;\; x=m^*(\aleph_0)$, then $m^*(\aleph_n)\rightarrow 0$ as $ n\rightarrow \infty$. Since $\aleph_{n+1} \subseteq \aleph_n$ and $P(\aleph_{n}) \subseteq \aleph_n, n\geq 1$, so by condition $(6)$ in the definition of $m^*$, we have $\displaystyle\aleph_{\infty} =\cap^{\infty}_{n=1}\aleph_{n}$ is a non empty closed and convex subset of $\aleph$. Moreover, $\aleph_{\infty}$ is invariant under the operator $P$ and belongs to Ker $m^*$ is relatively compact. Thus, by S. f. t $P$ has a fixed point in $\aleph$.	Similarly, we have
	$$Q(\aleph_{n}) \subseteq \aleph_n,$$ $$R(\aleph_{n}) \subseteq \aleph_n$$ and $$Q(\aleph_{\infty}) \subseteq \aleph_{\infty},$$ $$R(\aleph_{\infty}) \subseteq \aleph_{\infty}.$$
	Thus,  by S. f. t $Q$, $R$ have fixed point in $\aleph$.
Now consider $G_R=\{t\in \aleph : R(t)=t\}$. Then clearly $G_R$ is closed, convex and bounded by continuity of $R$ is a subset of $\aleph$ such that $Q(G_R)\subseteq G_R $ and $R(G_R)\subseteq G_R,$ we have  $$ m^*(G_R)=m^*(R(G_R))\leq \psi_3 (m^*(G_R))\leq m^*(G_R).$$
	Therefore, $m^*(G_R)=0 $ and $G_R$ is compact. Thus by S. f. t $Q$ has a fixed point in $G_R$ and so $Q$ and $R$ have a common fixed point.
	Next consider $G =\{t\in \aleph : Q(t)=R(t) = t\}$ is closed  and convex subset of $\aleph$ and $P(G)\subseteq G$, so $P$ have fixed point in $G$. Hence $P, Q, R$ have a common fixed point in $\aleph$.\\
	$(ii)$ Given $Q(Conv(\aleph))\subseteq Conv(Q(\aleph))$, Consider the sequence $\{\aleph_n\}$ such that $\aleph=\aleph_0, \aleph_n = conv(QP(\aleph_{n-1})),  n\geq 1$. We have clearly
$\aleph_{n+1}=(Conv(QP(\aleph_n)))\subseteq Conv(QP(\aleph_{n-1}))\subseteq Conv PQ(\aleph_{n-1})=\aleph_{n}$  and so $\aleph_n \subseteq \aleph_{n-1}$ ,$\forall n\geq 1$. Thus it follows that $$R(\aleph_n)\subseteq R(\aleph_{n-1}) ,\forall n\geq 1.$$
So $\{m^*(R(\aleph_n))\}$ is a positive non increasing sequence of real numbers such that $$\displaystyle\lim_{n\rightarrow \infty}m^*(R(\aleph_n))= 0.$$
Now choose $\aleph'_n =\overline{R(\aleph_n)}$ such that $m^*(\aleph'_n) = m^*(R(\aleph_n))=m^*(\overline{R(\aleph_n)})$  (by definition of $m^*$). So $\displaystyle\lim_{n\rightarrow \infty}m^*(\aleph'_n)=0$. Since $\{\aleph'_n\}$ is a nested sequence and ${\aleph'_{n+1} \subseteq \aleph'_n}, \forall n\geq 1,$ we have $\aleph'_{\infty}=\displaystyle\cap^{\infty}_{n=1}\aleph'_n$ is non empty and $m^*(\aleph'_{\infty})\leq m^*(\aleph'_n)$, $\forall n\geq 1$. Thus $\displaystyle\lim_{n\rightarrow \infty} m^*(\aleph'_{\infty})=0$ and $\overline{\aleph'_{\infty}} ={\aleph'_{\infty}}$ is compact and convex  as $R$ is continuous linear map. Now $(PQ=QP)(\aleph_n)\subseteq \aleph_n $ and $(QP) (\aleph_n)\subseteq Conv(QP)\aleph_n\subseteq Conv(QP)(\aleph_{n-1})=\aleph_n,       \;\;\ n=1,2,3...\;\; . $ Thus, $(QP)(\aleph_n)\subseteq \aleph_n$, $\forall n\in \mathbb{N}$ and $$(Q)(\aleph_n)\subseteq \aleph_n ,$$ $$(R)(\aleph_n)\subseteq \aleph_n ,$$  $$(QP)(\aleph'_n) = QP (\overline{R(\aleph_n))}\subseteq\overline{QP(R\aleph_n)} \subseteq \overline{R(QP(\aleph_n))}\subseteq \overline{R(\aleph_n)}=\aleph'_n.$$\\
 Hence, $$(QP)(\aleph'_n)\subseteq \aleph'_n(Q)(\aleph'_n)\subseteq \aleph'_n(R)(\aleph'_n)\subseteq \aleph'_{n}$$ and $$(QP)(\aleph'_{\infty})\subseteq \aleph'_{\infty}, (Q)(\aleph'_{\infty})\subseteq \aleph'_{\infty}, (R)(\aleph'_{\infty})\subseteq \aleph'_{\infty}.$$ So, by S. f. t $QP, Q, R$ have a fixed point in ${\aleph}$.
\end{proof}
\textbf{Theorem 3.2} \it{Let $\aleph$ be a non empty bounded, closed and convex subset of the Banach algebra $T$ and operators $P_1, P_2, P_3, Q_1, Q_2, Q_3$ transforms $\aleph$ into $T$ are continuous  such  that  $ P_1(\aleph), P_2(\aleph), P_3(\aleph), Q_1(\aleph),$ $Q_2(\aleph), Q_3(\aleph)$ are bounded. Suppose $P = P_1Q_1$,  $Q = P_2Q_2$ and $R = P_3 Q_3$ be three continuous mappings from $\aleph$ into $\aleph$ such that $Q$ and $R$ are linear with  $Q(P(E))\subseteq P(E)$ and $R(P(E)) \subseteq P(E)$, \mbox{for} $E\subseteq \aleph.$
If the operators $P_1, P_2, P_3, Q_1,$ $Q_2, Q_3$ on  $\aleph$ satisfies the following conditions
$$m^*(P_1(E))\leq\xi_1(m^*(E)),$$ $$ m^*(Q_1(E))\leq \xi_2(m^*(E)),$$ $$m^*(P_2(E))\leq\xi'_{1}(m^*(E)),$$ $$ m^*(Q_2(E)) \leq \xi'_{2} (m^*(E)),$$
$$ m^*(P_3(E))\leq\xi''_1(m^*(E)),$$  $$ m^*(Q_3(E))\leq \xi''_2(m^*(E)), $$
where $m^*$ is an arbitrary measure of non compactness defined
	on $\mu_T $ and $\xi_1, \xi_2, \xi'_1, \xi'_2, \xi''_1, \xi''_2$ are non decreasing functions from $R_+$ to  $R_+ $ s.t $\displaystyle\lim_{n\rightarrow\infty}\xi^n_{1}(x)=0$, $\displaystyle\lim_{n\rightarrow\infty}\xi^n_{2}(x)=0$,
	$\displaystyle\lim_{n\rightarrow\infty}\xi'^n_{1}(x)=0$,
	$\displaystyle\lim_{n\rightarrow\infty}\xi'^n_{2}(x)=0$, $\displaystyle\lim_{n\rightarrow\infty}\xi''^n_{1}(x)=0$,
	$\displaystyle\lim_{n\rightarrow\infty}\xi''^n_{2}(x)=0$
	and $$\displaystyle\lim_{n\rightarrow\infty}(||P_1(\aleph)|| \xi_2 +||Q_1(\aleph)||\xi_1)^n(x)=0,$$
	 $$\displaystyle\lim_{n\rightarrow\infty}(||P_2(\aleph)|| \xi'_2 +||Q_2(\aleph)||\xi'_1)^n(x)=0,$$ $$\displaystyle\lim_{n\rightarrow\infty}(||P_3(\aleph)|| \xi''_2 +||Q_3(\aleph)||\xi''_1)^n(x) = 0,$$  for any $x\geq 0$,
	then we have\\
	1. $P$, $Q$ and $R$ have a common fixed point in $\aleph$.\\
2. If $Q(Conv(\aleph))\subseteq Conv(Q(\aleph)) $ then $PQ$, $Q$ and $R$ have a fixed point in $\aleph$.}
\begin{proof}: The proof is straightforward, so we omit the details.\end{proof}


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