The first part of the following definition was introduced in [17].
Definition 1 Let f,g:X\to X be selfmappings of a set X and \alpha :X\times X\to [0,\mathrm{\infty}) be a mapping, then the mapping f is called αadmissible if
x,y\in X,\phantom{\rule{1em}{0ex}}\alpha (x,y)\ge 1\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\alpha (fx,fy)\ge 1
and the pair (f,g) is called αadmissible if
x,y\in X,\phantom{\rule{1em}{0ex}}\alpha (x,y)\ge 1\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\alpha (fx,gy)\ge 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\alpha (gx,fy)\ge 1.
Example 2 Let X=\mathbb{R} and
\alpha (x,y)=\{\begin{array}{cc}1,\hfill & x,y\in [0,1],\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}
Then the pair ({x}^{1/2},{x}^{1/3}) is αadmissible but the pair ({x}^{1/2},x+1) is not αadmissible.
Definition 3 Let (X,d) be a metric space and f:X\to X be a selfmapping, \alpha :X\times X\to [0,\mathrm{\infty}) be a mapping. Then f is called MeirKeeler αcontractive if, given an \u03f5>0, there exists a \delta >0 such that
\u03f5\le d(x,y)<\u03f5+\delta \phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}\alpha (x,y)d(fx,fy)<\u03f5.
Definition 4 Let (X,d) be a metric space and f:X\to X be a selfmapping, \alpha :X\times X\to [0,\mathrm{\infty}) be a mapping. Then f is called generalized MeirKeeler αcontractive if, given an \u03f5>0, there exists a \delta >0 such that
\u03f5\le {M}_{f}(x,y)<\u03f5+\delta \phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}\alpha (x,y)d(fx,fy)<\u03f5,
where
{M}_{f}(x,y)=max\{d(x,y),d(x,fx),d(y,fy),\frac{d(x,fy)+d(y,fx)}{2}\}.
Definition 5 Let (X,d) be a metric space and f,g:X\to X be selfmappings, \alpha :X\times X\to [0,\mathrm{\infty}) be a mapping. Then the pair (f,g) is called generalized MeirKeeler αcontractive if, given an \u03f5>0, there exists a \delta >0 such that
\u03f5\le {M}_{(f,g)}(x,y)<\u03f5+\delta \phantom{\rule{1em}{0ex}}\text{implies}\phantom{\rule{1em}{0ex}}\alpha (x,y)d(fx,gy)<\u03f5,
(4)
where
{M}_{(f,g)}(x,y)=max\{d(x,y),d(x,fx),d(y,gy),\frac{d(x,gy)+d(y,fx)}{2}\}.
We write {M}_{f}(x,y)={M}_{(f,f)}(x,y).
Clearly, f is generalized MeirKeeler αcontractive if and only if (f,f) is generalized MeirKeeler αcontractive.
Definition 6 Let X be any set, {x}_{0}\in X and f, g be selfmaps of X. Define {x}_{2n+1}=f{x}_{2n} and {x}_{2n+2}=g{x}_{2n}, n=0,1,2,\dots . Then \{{x}_{n}\} is called the (f,g)orbit of {x}_{0}. If d is a metric on X, then (X,d) is called (f,g)orbitally complete if every Cauchy sequence in the (f,g)orbit of {x}_{0} is convergent and the map f or g is called orbitally continuous if it is continuous on the orbit.
The proof of the following lemma is immediate.
Lemma 7 Let f,g:X\to X be selfmappings of a set X, \alpha :X\times X\to [0,\mathrm{\infty}) be a mapping and \{{x}_{n}\} be the (f,g)orbit of {x}_{0} with \alpha ({x}_{0},f{x}_{0})\ge 1. If the pair (f,g) is αadmissible, then \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n=0,1,2,\dots .
Theorem 8 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:

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

2.
the pair (f,g) is generalized MeirKeeler αcontractive.
Then the sequence {d}_{n}=d({x}_{n},{x}_{n+1}) is monotone decreasing. If, moreover, we assume that

3.
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 and that f and g are continuous on the (f,g)orbit of {x}_{0}.
Then either (1) f or g has a fixed point in the (f,g)orbit \{{x}_{n}\} of {x}_{0} or (2) f and g have a common fixed point p and lim{x}_{n}=p. If, moreover, we assume that the following condition (H) holds: If \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x implies \alpha ({x}_{n},x)\ge 1 for all n, then uniqueness of the fixed point is obtained.
Proof Define {d}_{n}=d({x}_{n},{x}_{n+1}) for n=0,1,2,\dots . If {d}_{n}=0 for some even integer n, then f has a fixed point. If {d}_{n}=0 for some odd integer n, then g has a fixed point. Hence, we may assume that {d}_{n}\ne 0 for each n. The fact that the pair (f,g) is generalized MeirKeeler αcontractive implies that
\alpha (x,y)d(fx,gy)<{M}_{f}(x,y)\phantom{\rule{1em}{0ex}}\text{for each}x,y\in X,x,y\ne 0.
(5)
Note that assumption (3) implies that \alpha ({x}_{0},f{x}_{0})\ge 1. Hence, since (f,g) is αadmissible, then Lemma 7 implies that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n=0,1,2,\dots and hence by (5), we have
\begin{array}{rcl}{d}_{2n}& =& d(f{x}_{2n},g{x}_{2n1})\\ \le & \alpha ({x}_{2n},{x}_{2n1})d(f{x}_{2n},g{x}_{2n1})\\ <& max\{{d}_{2n1},\frac{d({x}_{2n1},{x}_{2n+1})}{2}\}\\ \le & max\{{d}_{2n1},\frac{{d}_{2n1}+{d}_{2n}}{2}\},\end{array}
(6)
whence {d}_{2n}<{d}_{2n1}. □
Similarly, it can be shown that {d}_{2n+1}<{d}_{2n}. Thus, \{{d}_{n}\} is monotone decreasing in n and converges to a limit, say ϱ.
Suppose \varrho >0. Then, for each \delta >0, there exists a positive integer N=N(\delta ) such that \varrho \le {d}_{N}=d({x}_{N},{x}_{N+1})<\varrho +\delta, where N can be chosen even. Thus, from assumption (1) and Lemma 7, we have {d}_{N+1}\le \alpha ({x}_{N},{x}_{N+1})d(f{x}_{N},g{x}_{N+1})<\varrho, a contradiction. Therefore, \varrho =0. To show that \{{x}_{n}\} is Cauchy, we assume the contrary. Thus, there exists an {\u03f5}^{\mathrm{\prime}}>0 such that for each integer N, there exist integers m>n>N such that d({x}_{m},{x}_{n})\ge {\u03f5}^{\mathrm{\prime}}. Define ϵ by {\u03f5}^{\mathrm{\prime}}=2\u03f5. Choose a number δ, 0<\delta <\u03f5, for which (4) is satisfied. Since \varrho =0, there exists an integer N=N(\delta ) such that {d}_{i}<\frac{\delta}{6} for i\ge N. With this choice of N, pick integers m>n>N such that
d({x}_{m},{x}_{n})\ge 2\u03f5>\delta +\u03f5,
(7)
in which it is clear that mn>6. Otherwise, d({x}_{m},{x}_{n})\le {\sum}_{i=0}^{5}{d}_{i+n}<\delta <\delta +\u03f5, contradicting (7). Without loss of generality, we may assume that n is even since from (7) it follows that d({x}_{m},{x}_{n+1})>\u03f5+\frac{\delta}{3}. From (7) there exists the smallest odd integer j>n such that
d({x}_{n},{x}_{j})\ge \u03f5+\frac{\delta}{3}.
(8)
Hence, d({x}_{n},{x}_{j2})<\u03f5+\frac{\delta}{3}, and so d({x}_{n},{x}_{j})\le d({x}_{n},{x}_{j2})+{d}_{j1}+{d}_{j}<\u03f5+\frac{\delta}{3}+2(\frac{\delta}{6})=\u03f5+\frac{2\delta}{3}. Therefore, we have
\begin{array}{rcl}\u03f5& <& d({x}_{n},{x}_{j})\le {M}_{(f,g)}({x}_{n},{x}_{j})\\ \le & max\{d({x}_{n},{x}_{j}),\frac{d({x}_{n},{x}_{j+1})+d({x}_{j},{x}_{n+1})}{2}\}\\ \le & \frac{d({x}_{n},{x}_{j})+{d}_{j}+d({x}_{j},{x}_{n})+{d}_{n}}{2}\\ \le & d({x}_{n},{x}_{j})+\frac{\delta}{6}\le \u03f5+\delta ,\end{array}
so that, by (7) and assumption (3), d({x}_{n+1},{x}_{j+1})\le \alpha ({x}_{n},{x}_{j})d({x}_{n+1},{x}_{j+1})<\u03f5. Then we have
\begin{array}{rcl}d({x}_{n},{x}_{j})& \le & {d}_{n}+d({x}_{n+1},{x}_{j+1})+{d}_{j}\\ <& \frac{\delta}{6}+\u03f5+\frac{\delta}{6}=\u03f5+\frac{\delta}{3}.\end{array}
This contradicts the choice of j in (8). Therefore, \{{x}_{n}\} is Cauchy.
Since X is (f,g)orbitally complete, \{{x}_{n}\} converges to some point p\in X. Since f and g are orbitally continuous, then p is a common fixed point of f and g. To prove uniqueness, assume p is the common fixed point obtained as {x}_{n}\to p and q is another common fixed point. Then (5) and the condition (H) yield
\begin{array}{rcl}d(p,q)& =& d(fp,q)\le d(fp,g{x}_{n})+d(g{x}_{n},q)\\ \le & \alpha ({x}_{n},p)d(fp,g{x}_{n})+d(g{x}_{n},q)\\ <& {M}_{(f,g)}({x}_{n},p)+d(g{x}_{n},q).\end{array}
If we let n\to \mathrm{\infty}, then we reach d(p,q)<d(p,q), which implies that p=q.
Corollary 9 Let (X,d) be an forbitally complete metric space, where f is a selfmapping of X. Also, let \alpha :X\times X\to [0,\mathrm{\infty}) be a mapping. Assume the following:

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

2.
f is generalized MeirKeeler αcontractive.
Then the sequence {d}_{n}=d({x}_{n},{x}_{n+1}) is monotone decreasing. If, moreover, we assume that

3.
on the forbit of {x}_{0}, we have \alpha ({x}_{n},{x}_{j})\ge 1 for all n even and j>n odd.
Then either (1) f has a fixed point in the forbit \{{x}_{n}\} of {x}_{0} or (2) f has a fixed point p and lim{x}_{n}=p. If, moreover, we assume that the following condition (H) holds: If \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x, then \alpha ({x}_{n},x)\ge 1 for all n, then uniqueness of the fixed point is obtained.
Since generalized MeirKeeler αcontractions are MeirKeeler αcontractions, then Corollary 9 is valid also for MeirKeeler αcontractions. In the following example, the existence and uniqueness of the fixed point cannot be proved in the category of MeirKeeler contractions, but can be proved by means of Corollary 9.
Example 10 Let X=[0,2] with the absolute value metric d(x,y)=xy. Define f:X\to X by
f(x)=\{\begin{array}{cc}0,\hfill & x=\frac{1}{4},\hfill \\ 1,\hfill & x\in [0,\frac{1}{2})\{\frac{1}{4}\},\hfill \\ \frac{3}{2},\hfill & x\in [\frac{1}{2},2].\hfill \end{array}
Then, for \u03f5=\frac{1}{2}, x=\frac{1}{4} and any \delta >0, we have \frac{1}{2}\le \frac{1}{4}y<\delta +\frac{1}{2} implies y\in [\frac{1}{2},2] and hence d(fx,fy)=d(0,\frac{3}{2})=\frac{3}{2}>\u03f5. Hence, f is not a MeirKeeler contraction. However, f is a MeirKeeler αcontraction, where
\alpha (x,y)=\{\begin{array}{cc}1,\hfill & x,y\in [\frac{1}{2},2],\hfill \\ 0,\hfill & \text{otherwise}.\hfill \end{array}
Indeed, for 0<\u03f5<1 (the case \u03f5\ge 1 is trivial, since fxfy\le 1), let \delta =(1\u03f5), then \u03f5\le \alpha (x,y)d(x,y)<\delta +\u03f5=1 implies that x,y\in [\frac{1}{2},2] and hence d(fx,fy)=\frac{3}{2}\frac{3}{2}=0<\u03f5. Also, notice that f is continuous on the orbit of {x}_{0}=1 and that \alpha ({x}_{n},{x}_{j})\ge 1 for all n, j. Clearly, p=\frac{3}{2} is the unique fixed point.
Remark 11 Note that the admissibility condition (1) in Theorem 8 is not enough to proceed to guarantee the existence of the fixed point. However, such an admissibility condition was used in obtaining the main result in Theorem 2.2 of [17].