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Common fixed points of generalized MeirKeeler αcontractions
Fixed Point Theory and Applicationsvolume 2013, Article number: 260 (2013)
Abstract
Motivated by Abdeljawad (Fixed Point Theory Appl. 2013:19, 2013), we establish some common fixed point theorems for three and four selfmappings satisfying generalized MeirKeeler αcontraction in metric spaces. As a consequence, the results of Rao and Rao (Indian J. Pure Appl. Math. 16(1):12491262, 1985), Jungck (Int. J. Math. Math. Sci. 9(4):771779, 1986), and Abdeljawad itself are generalized, extended and improved. Sufficient examples are given to support our main results.
MSC:47H10, 54H25.
1 Introduction and preliminaries
The MeirKeeler contractive condition [1] is one of the interesting aspects to study metrical fixed point theory, that is, for given $\u03f5>0$, there exists a $\delta >0$ such that
This contraction has further been generalized and studied by various authors (see [2–15]). Very recently, Abdeljawad [16] (see also [17]) established some fixed point results for αcontractivetype maps (due to Samet et al. [18]) to MeirKeeler versions for single and a pair of maps. In this article, we prove some common fixed point theorems for three and four selfmappings satisfying generalized MeirKeeler αcontractions. Thus, we provide an affirmative answer to the question of Abdeljawad (see [16], Remark 17).
Let us recall some definitions, which we will use in our main results.
Let $f,g:X\to X$ be selfmappings of a set X, and let $\alpha :X\times X\to [0,\mathrm{\infty})$ be a mapping, then the mapping f is called αadmissible if
and the pair $(f,g)$ is called αadmissible if
Let f and g ($f\ne g$) be two selfmappings defined on a metric space $(X,d)$, then f is called gabsorbing if there exists some real number $R>0$ such that $d(gx,gfx)\le Rd(fx,gx)$ for all x in X. Analogously, g will be called fabsorbing if there exists some real number $R>0$ such that $d(fx,fgx)\le Rd(fx,gx)$ for all x in X. The pair of selfmaps $(f,g)$ will be called absorbing if it is both gabsorbing as well as fabsorbing. In particular, if we take g to be the identity map on X, then f is trivially Iabsorbing. Similarly, I is also fabsorbing in respect to f.
Definition 1.3 (cf. [21])
Two selfmappings f and g of a metric space $(X,d)$ are called reciprocally continuous if and only if $fg{x}_{n}\to ft$ and $gf{x}_{n}\to gt$ whenever $\{{x}_{n}\}$ is a sequence in X such that ${lim}_{n\to \mathrm{\infty}}f{x}_{n}={lim}_{n\to \mathrm{\infty}}g{x}_{n}=t$ for some $t\in X$.
2 Main results
We begin with the following definitions.
Definition 2.1 Let $f,g,T:X\to X$ be three selfmappings of a nonempty set X, and let $\alpha :T(X)\times T(X)\to [0,\mathrm{\infty})$ be a mapping, then the pair $(f,g)$ is called αadmissible with respect to T (in short, $(f,g)$ is ${\alpha}_{T}$admissible) if for all $x,y\in X$,
Definition 2.2 Let $f,g,S,T:X\to X$ be four selfmappings of a nonempty set X, and let $\alpha :S(X)\cup T(X)\times S(X)\cup T(X)\to [0,\mathrm{\infty})$ be a mapping, then the pair $(f,g)$ is called αadmissible with respect to S and T (in short, $(f,g)$ is ${\alpha}_{S,T}$admissible) if for all $x,y\in X$,
Clearly, if $S=T=I$ (identity map), then the definitions above imply Definition 1.1.
In order to extend and improve the result contained in [16] for three selfmappings, we now introduce the concept of generalized MeirKeeler ${\alpha}_{T}$contractive mappings as follows.
Definition 2.3 Let $(X,d)$ be a metric space, and $f,g,T:X\to X$ are selfmappings. Then we say that the pair $(f,g)$ is a generalized MeirKeeler ${\alpha}_{T}$contractive pair of type ${m}_{3}$ (${M}_{3}$, respectively) if given an $\u03f5>0$, there exists a $\delta >0$ such that
where
and
Definition 2.4 Let f, g, and T be three selfmappings on a metric space $(X,d)$ such that $f(X)\cup g(X)\subseteq T(X)$. If for a point ${x}_{0}\in X$, there exists a sequence $\{{x}_{n}\}$ such that $T{x}_{2n+1}=f{x}_{2n}$, $T{x}_{2n+2}=g{x}_{2n+1}$, $n=0,1,2,\dots $ , then $\mathcal{O}(f,g,T,{x}_{0})=\{T{x}_{n}:n=1,2,\dots \}$ is called the orbit for $(f,g,T)$ at ${x}_{0}$. The space $(X,d)$ is called $(f,g,T)$orbitally complete at ${x}_{0}$ iff every Cauchy sequence in $\mathcal{O}(f,g,T,{x}_{0})$ converges to a point in X. X is called $(f,g,T)$orbitally complete if it is so at every $x\in X$.
Our first result is the following.
Theorem 2.1 Let $(X,d)$ be an $(f,g,T)$orbitally complete metric space. Suppose that $(f,g)$ is generalized MeirKeeler ${\alpha}_{T}$contractive pair of type ${m}_{3}$ and satisfies the following conditions:

(i)
$(f,g)$ is ${\alpha}_{T}$admissible;

(ii)
there exists ${x}_{0}\in X$ such that $\alpha (T{x}_{0},f{x}_{0})\ge 1$;

(iii)
on the $(f,g,T)$orbit of ${x}_{0}$, we have $\alpha (T{x}_{n},T{x}_{j})\ge 1$ for all n even and $j>n$ odd.
Then $\{T{x}_{n}\}$ is a Cauchy sequence. Moreover, if

(iv)
$\alpha (T{x}_{n},T{x}_{n+1})\ge 1$ for all n, and $T{x}_{n}\to x$ implies that $\alpha (T{x}_{n},Tx)\ge 1$ for all n;

(v)
one of the pairs $(f,T)$ and $(g,T)$ is absorbing as well as reciprocal continuous.
Then f, g, and T have a common fixed point.
Proof Let ${x}_{0}\in X$ such that $\alpha (T{x}_{0},f{x}_{0})\ge 1$. Define the sequences $\{{x}_{n}\}$ and $\{T{x}_{n}\}$ in X given by the rule
Since $(f,g)$ is ${\alpha}_{T}$admissible, we have
which gives
Again by (i), we have
which gives
Inductively, we have
The fact that $(f,g)$ is generalized MeirKeeler ${\alpha}_{T}$contractive implies that
Now, to obtain a common fixed point of f, g, and T, we take the following steps.
Step 1: We show that there exists a point $z\in X$ such that $T{x}_{n}\to z$ as $n\to \mathrm{\infty}$. For this, first, we claim that $\{T{x}_{n}\}$ is a Cauchy sequence. Two cases arise: either $T{x}_{n}=T{x}_{n+1}$ for some n or $T{x}_{n}\ne T{x}_{n+1}$ for each n.
Case I: Suppose that $T{x}_{n}=T{x}_{n+1}$ for some n. We first assume that n is even, i.e., $T{x}_{2m}=T{x}_{2m+1}$ but $T{x}_{2m+1}\ne T{x}_{2m+2}$, then by (6),
which is a contradiction. Hence $T{x}_{2m+1}=T{x}_{2m+2}$. By proceeding in this way, we obtain $T{x}_{2m+k}=T{x}_{2m}$ for all $k\in \mathcal{N}$. Similar is the case when n is odd. Thus, we conclude that $\{T{x}_{n}\}$ is a Cauchy sequence.
Case II: Suppose that $T{x}_{n}\ne T{x}_{n+1}$ for all integers n. Applying (6), we have
Similarly, it can be shown that
Thus, $\{d(T{x}_{n},T{x}_{n+1})\}$ is strictly decreasing sequence of positive real numbers, and, therefore, converges to a limit $r\ge 0$. If possible, suppose that $r>0$. Then given $\delta >0$, there exists a positive integer $N=N(\delta )$ such that
where $d(T{x}_{2n},T{x}_{2n+1})\le m({x}_{2n},{x}_{2n+1})$. So by Eqs. (5) and (6), we have
that is, $d(T{x}_{2n+1},T{x}_{2n+2})<r$, which is a contradiction. Hence
We now show that $\{T{x}_{n}\}$ is a Cauchy sequence.
Suppose that it is not. Then there exists an $\u03f5>0$ such that for each positive integer m, n with $m>n>N$, we have $d(T{x}_{m},T{x}_{n})\ge 2\u03f5$. Choose a number δ, $0<\delta <\u03f5$ for which contractive condition (4) is satisfied. Since $d(T{x}_{n},T{x}_{n+1})\to 0$, there exists integer $N=N(\delta )$ such that $d(T{x}_{i},T{x}_{i+1})<\frac{\delta}{6}$ for all $i\ge N$. With this choice of N, pick m, n with $m>n>N$ such that
in which it is clear that $mn>6$. Otherwise, we have
which contradicts (9). Also from (9), it follows that
Without loss of generality, we may assume that n is even. Suppose that
then
which is a contradiction to (9). So we have
Similarly, suppose that
then
which is a contradiction to (9). So we have
Thus, there exists the smallest odd integer $j>n$ such that
and hence,
Now,
Thus, there exists an odd integer $j\in (n,m)$ such that
Since we have
So, using (4) and assumption (iii), we get
that is, $d(T{x}_{n+1},T{x}_{j+1})<\u03f5$. But then
which contradicts (11). Therefore, $\{T{x}_{n}\}$ is a Cauchy sequence. Since X is $(f,g,T)$orbitally complete, so there exists a point $z\in X$ such that $T{x}_{n}\to z$ as $n\to \mathrm{\infty}$. Consequently, $f{x}_{2n}\to z$ and $g{x}_{2n+1}\to z$.
Step 2: We show that z is common fixed point of $(f,g,T)$. In view of assumption (v), without loss of generality, let the pair $(f,T)$ be absorbing and reciprocal continuous. Then the reciprocal continuity of f and T implies that
Since T is fabsorbing, so there exists an $R>0$ such that
Letting $n\to \mathrm{\infty}$, we get $fT{x}_{2n}\to z$. Similarly, since f is Tabsorbing, so we have
letting $n\to \mathrm{\infty}$, we get $Tf{x}_{2n}\to z$. By the uniqueness of the limit, we have $z=fz=Tz$.
Now, suppose that $z\ne gz$, then by assumption (iv) and Eq. (6), we have
Letting $n\to \mathrm{\infty}$, we get $d(z,gz)\le \frac{1}{2}d(z,gz)$, which implies that $z=gz$. Thus, z is a common fixed point of f, g, and T. This completes the proof of the theorem. □
By putting $f=g$ and $T=I$ (identity map) in Theorem 2.1, we get the following result as a corollary.
Corollary 2.1 Let $(X,d)$ be an forbitally complete metric space, where f is a selfmapping on X. Also, let $\alpha :X\times X\to [0,\mathrm{\infty})$ be a mapping. Assume the following:

(i)
f is αadmissible;

(ii)
there exists an ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$;

(iii)
for given $\u03f5>0$, there exists a $\delta >0$ such that
$$\begin{array}{r}\u03f5\le {m}_{1}(x,y)<\u03f5+\delta \phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}\alpha (x,y)d(fx,fy)<\u03f5,\\ \mathit{\text{where}}{m}_{1}(x,y)=max\{d(x,y),\frac{1}{2}[d(x,fx)+d(y,fy)],\frac{1}{2}[d(x,fy)+d(y,fx)]\};\end{array}$$ 
(iv)
on the forbit of ${x}_{0}$, we have $\alpha ({x}_{n},{x}_{j})\ge 1$ for all n even and $j>n$ odd.
Then, f has a fixed point in the forbit $\{{x}_{n}\}$ of ${x}_{0}$, or f has a fixed point z and ${lim}_{n\to \mathrm{\infty}}{x}_{n}=z$.
Example 2.1 Let $X=[0,2]$ be endowed with the standard metric $d(x,y)=xy$ for all $x,y\in X$. Define $f:X\to X$ by
Then f is not a MeirKeeler contraction. To see this consider $\u03f5=\frac{1}{2}$, $x=\frac{1}{4}$, and $y=\frac{3}{4}$, then for any $\delta >0$, we have $\u03f5\le {m}_{1}(x,y)<\u03f5+\delta $, but $d(fx,fy)=d(0,\frac{3}{2})=\frac{3}{2}>\u03f5$. However, f is a generalized MeirKeeler αcontraction, where $\alpha :X\times X\to [0,\mathrm{\infty})$ is defined by
Clearly, f has two fixed points, namely $x=0$ and $x=\frac{3}{2}$. Notice that $\alpha (\frac{3}{2},0)=0<1$.
For the uniqueness of the fixed point of a generalized MeirKeeler αcontractive mapping, we will consider the following hypothesis.

(H)
For all fixed points x and y of $(f,g,T)$, we have $\alpha (Tx,Ty)\ge 1$.
Theorem 2.2 Adding condition (H) to the hypotheses of Theorem 2.1 (resp., Corollary 2.1), we obtain the uniqueness of the common fixed point of f, g, and T.
Proof Let z be the common fixed point obtained as $T{x}_{n}\to z$ and u is another common fixed point. Then, (6) and condition (H) yield to
Thus, we reach $d(z,u)<d(z,u)$, and hence $z=u$. □
The following example illustrates Theorem 2.2.
Example 2.2 Let $X=[2,20]$ and d be the usual metric on X. Define $f,g,T:X\to X$ as follows:
In this example the mappings f, g, and T do not satisfy the general MeirKeeler contractive condition. To see this, consider $\u03f5=\frac{3}{4}$, $x=3$ and $y\in [2,3)$, then for any $\delta >0$, we have $\u03f5\le m(x,y)<\u03f5+\delta $, but $d(fx,gy)=d(3,2)=1>\u03f5$. However, f, g, and T satisfy the generalized MeirKeeler αcontractive condition (4) with the mapping $\alpha :T(X)\times T(X)\to [0,\mathrm{\infty})$ defined by
Also, all the hypotheses of Theorem 2.1 with condition (H) are satisfied, and clearly $x=3$ is our unique common fixed point. Indeed, hypothesis (ii) is satisfied with ${x}_{0}=3$, and here $T{x}_{n}=3$ is a sequence, for which hypotheses (iii) and (iv) are satisfied. Also in view of the sequence ${x}_{n}=3$, here both pairs $(f,T)$ and $(g,T)$ are reciprocal continuous as well as absorbing. Notice that $x=3$ is the point of discontinuity of the mappings g and T.
Theorem 2.3 The conclusion of Theorem 2.1 remains true if the assumption (v) of Theorem 2.1 is replaced by one of the following conditions:

(a)
$d(gx,Ty)\le max\{d(y,gx),d(y,Tx)\}$ for all $x,y\in X$ with righthand side positive.

(b)
$d(fx,Ty)\le max\{d(y,Tx),d(y,fx)\}$ for all $x,y\in X$ with righthand side positive.
Proof In view of Theorem 2.1, we have that $\{T{x}_{n}\}$ is a Cauchy sequence, and $T{x}_{n}\to z\in X$ as $n\to \mathrm{\infty}$, and, consequently, $f{x}_{2n}$ and $g{x}_{2n+1}$ also converge to z as $n\to \mathrm{\infty}$.
Clearly, $T{x}_{n}\ne z$ for infinitely many n. We can as well assume that $T{x}_{n}\ne z$ for all n.
If (a) holds, then
Letting $n\to \mathrm{\infty}$, we get $d(z,Tz)\le 0$, i.e., $Tz=z$. If (b) holds, then also $Tz=z$.
Now, suppose that $z\ne gz$. Since $T{x}_{2n}\ne T{x}_{2n+1}$, so by assumption (iv) and Eq. (6), we have
letting $n\to \mathrm{\infty}$, we get $d(z,gz)\le \frac{1}{2}d(z,gz)$, which implies that $z=gz$.
Now, let $fz\ne z=Tz$, then again by the process above, we have
letting $n\to \mathrm{\infty}$, we get $d(fz,z)\le \frac{1}{2}d(z,fz)$, which implies that $fz=z$. Thus, z is the common fixed point of f, g, and T. □
The following example demonstrates Theorem 2.3.
Example 2.3 Let $X=[0,1]$ and d be the usual metric on X. Define $f,g,T:X\to X$ as follows:
Here the mappings f, g, and T satisfy all the conditions of Theorem 2.3 with the mapping $\alpha :T(X)\times T(X)\to [0,\mathrm{\infty})$ defined by
Clearly, none of the pairs $(f,T)$ and $(g,T)$ are reciprocal continuous. To see this consider the sequence ${x}_{n}=\frac{1}{2}+\frac{1}{n}$, then ${lim}_{n\to \mathrm{\infty}}f{x}_{n}={lim}_{n\to \mathrm{\infty}}T{x}_{n}=\frac{1}{4}$, but ${lim}_{n\to \mathrm{\infty}}fT{x}_{n}={lim}_{n\to \mathrm{\infty}}f(\frac{1}{4}+\frac{1}{2n})=\frac{1}{20}\ne 0=f(\frac{1}{4})$. Therefore, $(f,T)$ is not reciprocal continuous. To see that $(g,T)$ is not reciprocal continuous, one can consider the sequence ${y}_{n}=\frac{1}{4}+\frac{1}{n}$. Here, the involved mappings satisfy condition (a) of Theorem 2.3, and they have the unique common fixed $x=0$.
Remark 2.1 Theorem 2.3 generalizes and extends Theorem 1.2 of Rao and Rao [22].
Theorem 2.4 Theorem 2.1 remains true if we replace ${m}_{3}(x,y)$ by ${M}_{3}(x,y)$ and condition (iv) by the following (iv′):
(iv′) $\alpha (T{x}_{n},T{x}_{n+1})\ge 1$ for all n and $T{x}_{n}\to x$ implies that $\alpha (T{x}_{n},Tx)\ge K$ for all n, where $K>1$.
Proof The proof of $z=fz=Tz$ follows from Theorem 2.1. Now, suppose that $z\ne gz$, then by the help of condition (iv’), we have
By letting $n\to \mathrm{\infty}$, we conclude that $d(z,gz)\le {K}^{1}d(z,gz)<d(z,gz)$, and hence $z=gz$. Thus, z is a common fixed point of f, g, and T. □
Example 2.2 above also satisfies Theorem 2.4.
Remark 2.2 Theorem 2.4 generalizes and extends Theorem 1.3 of Rao and Rao [22].
By taking $T=I$ (identity map) in Theorem 2.4, we derive the following result as a corollary.
Corollary 2.2 Let $(X,d)$ be an $(f,g)$orbitally complete metric space, where f, g are selfmappings of X. Also, let $\alpha :X\times X\to [0,\mathrm{\infty})$ be a mapping. Assume the following:

(i)
$(f,g)$ is αadmissible, and there exists an ${x}_{0}\in X$ such that $\alpha ({x}_{0},f{x}_{0})\ge 1$;

(ii)
for given $\u03f5>0$, there exists a $\delta >0$ such that
$$\u03f5\le M(x,y)<\u03f5+\delta \phantom{\rule{1em}{0ex}}\mathit{\text{implies that}}\phantom{\rule{1em}{0ex}}\alpha (x,y)d(fx,gy)<\u03f5,$$
where

(iii)
on the $(f,g)$orbit of ${x}_{0}$, we have $\alpha ({x}_{n},{x}_{j})\ge 1$ for all n even and $j>n$ odd;

(iv)
$\alpha ({x}_{n},{x}_{n+1})\ge 1$ for n, and ${x}_{n}\to x$ implies that $\alpha ({x}_{n},x)\ge K$ for all n, where $K>1$.
Then, the pair $(f,g)$ has a common fixed point provided it is absorbing as well as reciprocal continuous.
Remark 2.3 Corollary 2.2 improves Theorem 8 contained in [16].
The next result is a common fixed point theorem for four selfmappings.
Theorem 2.5 Let f, g, S, and T be four selfmappings on a complete metric space $(X,d)$ such that $f(X)\subseteq T(X)$ and $g(X)\subseteq S(X)$, and they satisfy the following conditions:

(i)
the pair $(f,g)$ is ${\alpha}_{S,T}$admissible;

(ii)
there exists a point ${x}_{0}\in X$ such that $\alpha (S{x}_{0},f{x}_{0})\ge 1$;

(iii)
for given $\u03f5>0$, there exists a $\delta >0$ such that
$$\u03f5\le {m}_{4}(x,y)<\u03f5+\delta \phantom{\rule{1em}{0ex}}\u27f9\phantom{\rule{1em}{0ex}}\alpha (Sx,Ty)d(fx,gy)<\u03f5,$$(12)
where

(iv)
there exists a sequence $\{{x}_{n}\}$ in X such that $\alpha (S{x}_{n},T{x}_{j})\ge 1$ for all n even and $j>n$ odd;
Then f, g, S, and T have a common fixed point provided both the pair $(f,S)$ and $(g,T)$ are absorbing as well as reciprocal continuous.
Proof Let ${x}_{0}\in X$ such that $\alpha (S{x}_{0},f{x}_{0})\ge 1$. Define sequences $\{{x}_{n}\}$ and $\{{y}_{n}\}$ in X as
This can be done since $f(X)\subseteq T(X)$ and $g(X)\subseteq S(X)$.
Since $(f,g)$ is ${\alpha}_{S,T}$admissible, we have
which gives
Again by (i), we have
which gives
Inductively, we obtain
that is, $\alpha (S{x}_{n+1},T{x}_{n+2})\ge 1$, when n is odd and $\alpha (T{x}_{n+1},S{x}_{n+2})\ge 1$ when n is even.
By assumption (iii), we have
Now, we claim that $\{{y}_{n}\}$ is a Cauchy sequence.
Case I: If ${y}_{n}={y}_{n+1}$ for some n. We first assume that n is odd, i.e., ${y}_{2m+1}={y}_{2m+2}$ and suppose that ${y}_{2m+2}\ne {y}_{2m+3}$, then by applying (13) and (14), we get
a contradiction. Hence ${y}_{2m+2}={y}_{2m+3}$. By proceeding in this manner, we obtain ${y}_{2m+k}={y}_{2m+1}$ for all $k\ge 1$. Similarly, when we assume n as even, then we obtain ${y}_{2m+k}={y}_{2m}$ for all $k\ge 1$, and so $\{{y}_{n}\}$ is a Cauchy sequence.
Case II: If ${y}_{n}\ne {y}_{n+1}$ for each n. Applying (13) and (14), we get
Similarly, we obtain $d({y}_{2n1},{y}_{2n})<d({y}_{2n2},{y}_{2n1})$. Thus, $\{d({y}_{n},{y}_{n+1})\}$ is a strictly decreasing sequence of positive numbers, and, therefore, tends to a limit $r\ge 0$. If possible, suppose that $r>0$. Then given $\delta >0$, there exists a positive integer N such that for each $n\ge N$, we have
where $d(S{x}_{2n+2},T{x}_{2n+1})\le {m}_{4}({x}_{2n+2},{x}_{2n+1})$. Then by applying (14), we have
that is, $d({y}_{2n+2},{y}_{2n+1})<r$, which is a contradiction, and hence,
Now, we show that $\{{y}_{n}\}$ is a Cauchy sequence. Suppose that it is not, then there exists an $\u03f5>0$ such that for each integer N, there exist integers $m>n>N$ such that $d({y}_{m},{y}_{n})\ge 2\u03f5$. Choose a number δ, $0<\delta <\u03f5$, for which contractive condition (12) is satisfied. By virtue of (16), there exists an integer N such that $d({y}_{i},{y}_{i+1})<\frac{\delta}{6}$ for all $i\ge N$. With this choice of N, pick integers $m>n>N$ such that
in which it is clear that $mn>6$. Also from (17), it follows that $d({y}_{m},{y}_{n+1})>\u03f5+\frac{\delta}{3}$.
If not, then
which is a contradiction. Without loss of generality, we can assume that n is even. From (17), there exists the smallest odd integer $j>n$ such that
and hence $d({y}_{n},{y}_{j2})<\u03f5+\frac{\delta}{3}$. So we have
Thus, there exists an odd integer $j\in (n,m)$ such that
Therefore, we have
so that by (12) and assumption (iv), we get
i.e., $d({y}_{n+1},{y}_{j+1})<\u03f5$. But then
which contradicts (19). Therefore, $\{{y}_{n}\}$ is a Cauchy sequence. By the completeness of X, there exists a $z\in X$ such that ${y}_{n}\to z$ as $n\to \mathrm{\infty}$ and, consequentially, $f{x}_{2n}$, $T{x}_{2n+1}$, $g{x}_{2n+1}$ and $S{x}_{2n+2}\to z$ as $n\to \mathrm{\infty}$.
Since the pair $(f,S)$ is reciprocal continuous and absorbing, so by reciprocal continuity, we have $fS{x}_{2n}\to fz$ and $Sf{x}_{2n}\to Sz$ as $n\to \mathrm{\infty}$. By absorbing property, there is an $R>0$ such that $d(f{x}_{2n},fS{x}_{2n})\le Rd(f{x}_{2n},S{x}_{2n})$ and $d(S{x}_{2n},Sf{x}_{2n})\le Rd(f{x}_{2n},S{x}_{2n})$, which letting $n\to \mathrm{\infty}$ gives $fS{x}_{2n}\to z$ and $Sf{x}_{2n}\to z$. Thus, we have $z=fz=Sz$. Similarly, the absorbing and reciprocal continuity of the pair $(g,T)$ provides us $z=gz=Tz$. Thus, z is a common fixed point of f, g, S, and T. □
Theorem 2.6 Adding the condition (H2): For all common fixed points x and y of f, g, S, and T, $\alpha (Sx,Ty)\ge 1$, to the hypotheses of Theorem 2.5, the uniqueness of the fixed point is obtained.
Remark 2.4 Theorem 2.6 generalizes, extends and improves the results of Jungck (Theorem 3.1, [8]), Cho et al. (Theorem 3.2, [4]) and Rao and Rao [22].
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Acknowledgements
The authors would like to express their thanks to Prof. M. A Khamsi and the referees for their helpful comments and suggestions. The first author is thankful to S. V. National Institute of Technology, Surat, India for awarding Senior Research Fellow. The third author gratefully acknowledges the support from the CSIR, Govt. of India, Grant No.25(0215)/13/EMRII.
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Keywords
 common fixed points
 MeirKeeler contraction
 generalized MeirKeeler αcontraction
 αadmissible
 reciprocally continuous
 absorbing maps