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Some new fixed point theorems for αGeraghty contraction type maps in metric spaces
Fixed Point Theory and Applications volume 2014, Article number: 190 (2014)
Abstract
We generalize the results obtained in Cho et al. (Fixed Point Theory Appl. 2013:329, 2013) and give other conditions to prove the existence and uniqueness of a fixed point of αGeraghty contraction type maps in the context of a complete metric space.
1 Introduction and preliminaries
The Banach contraction principle [1] is one of the earliest and most important results in fixed point theory. Because of its application in many disciplines such as chemistry, physics, biology, computer science and many branches of mathematics, a lot of authors have improved, generalized and extended this classical result in nonlinear analysis; see, e.g., [2–10] and the references therein.
One of the interesting results was given by Geraghty [6] in the setting of complete metric spaces by considering an auxiliary function. Later, AminiHarandi and Emami [3] characterized the result of Geraghty in the context of a partially ordered complete metric space, and Caballero et al. [11] discussed the existence of a best proximity point of Geraghty contraction. Gordji et al. [12] defined the notion of ψGeraghty type contraction and supposedly improved and extended the results of AminiHarandi and Emami [3]. Recently, Cho, Bae and Karapınar [13] defined the concept of αGeraghty contraction type maps in the setting of a metric space and proved the existence and uniqueness of a fixed point of such maps in the context of a complete metric space. Very recently, Karapınar and Samet [14] proved that the results of Gordji et al. [12] and all results inspired by the paper of Gordji et al. [12] are equivalent to existing results in the literature. For other results related to Geraghty contractions, see [13–23].
In this paper, we generalize the results obtained in [13] and give other conditions to prove the existence and uniqueness of a fixed point of αGeraghty contraction type maps in the context of a complete metric space. Now, we remind some basic definitions and remarkable results on the topic in the literature.
Definition 1 [24]
Let T:X\to X be a map and \alpha :X\times X\to R be a function. Then T is said to be αadmissible if \alpha (x,y)\ge 1 implies \alpha (Tx,Ty)\ge 1.
Definition 2 [16]
A map T:X\to X is said to be triangular αadmissible if:
(T1) T is αadmissible,
(T2) \alpha (x,z)\ge 1 and \alpha (z,y)\ge 1 imply \alpha (x,y)\ge 1.
Let ℱ be the family of all functions \beta :[0,\mathrm{\infty})\to [0,1) which satisfies the condition
By using such maps, Geraghty [6] proved the following result.
Theorem 1 Let (X,d) be a complete metric space and let T be a mapping on X. Suppose that there exists \beta \in \mathcal{F} such that for all x,y\in X,
Then T has a unique fixed point {x}_{\ast}\in X and \{{T}^{n}x\} converges to {x}_{\ast} for each x\in X.
AminiHarandi and Emami [3] reconsidered Theorem 1 in the framework of partially ordered metric spaces.
Theorem 2 Let (X,\u2aaf,d) be a partially ordered complete metric space. Let f:X\to X be an increasing mapping such that there exists an element {x}_{0}\in X with {x}_{0}\u2aaff{x}_{0}. If there exists \alpha \in \mathcal{F} such that
for each x,y\in X with x\u2ab0y, then f has a fixed point provided that either f is continuous or X is such that if an increasing sequence \{{x}_{n}\}\to x, then {x}_{n}\u2aafx for all n. Besides, if for each x,y\in X there exists z\in X which is comparable to x and y, then f has a unique fixed point.
Definition 3 [24]
Let (X,d) be a metric space and \alpha :X\times X\to R be a function. A map T:X\to X is called a generalized αGeraghty contraction type map if there exists \beta \in \mathcal{F} such that for all x,y\in X,
where M(x,y)=max\{d(x,y),d(x,Tx),d(y,Ty)\}.
Cho et al. [13] proved the following interesting result.
Theorem 3 Let (X,d) be a complete metric space, \alpha :X\times X\to R be a function, and let T:X\to X be a map. Suppose that the following conditions are satisfied:

(1)
T is a generalized αGeraghty contraction type map;

(2)
T is triangular αadmissible;

(3)
there exists {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1;

(4)
T is continuous.
Then T has a fixed point {x}_{\ast}\in X and \{{T}^{n}{x}_{1}\} converges to {x}_{\ast}.
The continuity of the mapping T can be replaced by a suitable condition (4′) (see Theorem 2.2 [13]):
(4′) 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\in X as n\to \mathrm{\infty}, then there exists a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that \alpha ({x}_{n(k)},x)\ge 1 for all k.
Adding the condition (H),

(H)
For all x,y\in Fix(T), there exists z\in X such that \alpha (x,z)\ge 1 and \alpha (z,y)\ge 1, where Fix(T) denotes the set of fixed points of T,
to the hypotheses of Theorem 3, Cho et al. [13] obtained that {x}_{\ast} is the unique fixed point of T (see Theorem 2.3 [13]). However, we think that this condition is not appropriate. To verify such a condition assumes to know Fix(T) and then the uniqueness of a fixed point is trivial. Also, we think that the following condition is not more appropriate:

(iii)
[16] For all x\ne y\in X, there exists v\in X such that \alpha (x,v)\ge 1 and \alpha (v,y)\ge 1.
This condition implies \alpha (x,y)\ge 1 for all x\ne y\in X, and then the utility of α is 0. For more details, see Theorem 19 [16].
2 Fixed point theorems
In this section, we prove that the results of Cho et al. [13] are still available if we replace condition (2) of Theorem 3 with a weaker condition, and if we extend the notion of generalized αGeraghty contraction type map.
Definition 4 Let (X,d) be a complete metric space and let \alpha :X\times X\to R be a function. A map T:X\to X is called a generalized αGeraghty contraction type map if there exists \beta \in \mathcal{F} such that for all x,y\in X,
where {M}_{T}(x,y)=max\{d(x,y),d(x,Tx),d(y,Ty),[d(x,Ty)+d(y,Tx)]/2\}.
Now, we introduce two new concepts.
Definition 5 Let T:X\to X be a map and \alpha :X\times X\to R be a function. Then T is said to be αorbital admissible if
(T3) \alpha (x,Tx)\ge 1 implies \alpha (Tx,{T}^{2}x)\ge 1.
Definition 6 Let T:X\to X be a map and \alpha :X\times X\to R be a function. Then T is said to be triangular αorbital admissible if T is αorbital admissible and
(T4) \alpha (x,y)\ge 1 and \alpha (y,Ty)\ge 1 imply \alpha (x,Ty)\ge 1.
Obviously, every αadmissible mapping is an αorbital admissible mapping and every triangular αadmissible mapping is a triangular αorbital admissible mapping. The following example shows that there exists a triangular αadmissible mapping which is not triangular αadmissible.
Example 7 Let X=\{0,1,2,3\}, d:X\times X\to R, d(x,y)=xy, T:X\to X such that T0=0, T1=2, T2=1, T3=3 and \alpha :X\times X\to R, \alpha (x,y)=1 if (x,y)\in \{(0,1),(0,2),(1,1),(2,2),(1,2),(2,1),(1,3),(2,3)\} and \alpha (x,y)=0 otherwise. Since \alpha (1,T1)=\alpha (1,2)=1 and \alpha (2,T2)=\alpha (2,1)=1, T is αorbital admissible. Since \alpha (0,1)=\alpha (1,2)=\alpha (0,2)=1, \alpha (0,2)=\alpha (2,1)=\alpha (0,1)=1, T is triangular αorbital admissible. But \alpha (0,1)=\alpha (1,3)=1, \alpha (0,3)=0, so T is not triangular αadmissible.
Lemma 8 Let T:X\to X be a triangular αorbital admissible mapping. Assume that there exists {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1. Define a sequence \{{x}_{n}\} by {x}_{n+1}=T{x}_{n}. Then we have \alpha ({x}_{n},{x}_{m})\ge 1 for all m,n\in N with n<m.
Proof Since T is αorbital admissible and \alpha ({x}_{1},T{x}_{1})\ge 1, we deduce that \alpha ({x}_{2},{x}_{3})=\alpha (T{x}_{1},T{x}_{2})\ge 1. By continuing this process, we get \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\ge 1. Suppose that \alpha ({x}_{n},{x}_{m})\ge 1 and prove that \alpha ({x}_{n},{x}_{m+1})\ge 1, where m>n. Since T is triangular αorbital admissible and \alpha ({x}_{m},{x}_{m+1})\ge 1, we get that \alpha ({x}_{n},{x}_{m+1})\ge 1. Hence, we have proved that \alpha ({x}_{n},{x}_{m})\ge 1 for all m,n\in N with n<m. □
Theorem 4 Let (X,d) be a complete metric space, \alpha :X\times X\to R be a function, and let T:X\to X be a map. Suppose that the following conditions are satisfied:

(1)
T is a generalized αGeraghty contraction type mapping;

(2)
T is a triangular αorbital admissible mapping;

(3)
there exists {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1;

(4)
T is continuous.
Then T has a fixed point {x}_{\ast}\in X and \{{T}^{n}{x}_{1}\} converges to {x}_{\ast}.
Proof Let {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1. Define a sequence \{{x}_{n}\} by {x}_{n+1}=T{x}_{n} for n\ge 1. If {x}_{n(0)}={x}_{n(0)+1} for some n(0)\ge 1, then obviously T has a fixed point. Hence we suppose that {x}_{n}\ne {x}_{n+1} for all n\ge 1. By Lemma 8 we have \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\ge 1. Then we get, for all n\ge 1,
where
Since \beta ({M}_{T}({x}_{n},{x}_{n+1}))<1, the case d({x}_{n},{x}_{n+1})\le d({x}_{n+1},{x}_{n+2}) is impossible, so we have d({x}_{n},{x}_{n+1})>d({x}_{n+1},{x}_{n+2}). Thus, the sequence \{d({x}_{n},{x}_{n+1})\} is positive and decreasing. Therefore, there exists r\ge 0 such that {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=r. We will show that r=0. Suppose, to the contrary, that r>0. Then we have
This implies that {lim}_{n\to \mathrm{\infty}}\beta ({M}_{T}({x}_{n},{x}_{n+1}))=1. Since \beta \in \mathcal{F}, we get {lim}_{n\to \mathrm{\infty}}{M}_{T}({x}_{n},{x}_{n+1})=0, and then {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=0. This is a contradiction.
Next, we shall show that \{{x}_{n}\} is a Cauchy sequence. Suppose, to the contrary, that \{{x}_{n}\} is not a Cauchy sequence. Then there exists \u03f5>0 such that, for all k\ge 1, there exists m(k)>n(k)>k with d({x}_{n(k)},{x}_{m(k)})\ge \u03f5. Let m(k) be the smallest number satisfying the conditions above. Hence, we have d({x}_{n(k)},{x}_{m(k)1})<\u03f5. Therefore, we get
Letting k\to \mathrm{\infty}, we have {lim}_{k\to \mathrm{\infty}}d({x}_{n(k)},{x}_{m(k)})=\u03f5. Since
we get {lim}_{k\to \mathrm{\infty}}d({x}_{n(k)},{x}_{m(k)1})=\u03f5. Similarly, we obtain
By Lemma 8 we have \alpha ({x}_{n(k)1},{x}_{m(k)1})\ge 1. Thus, we deduce that
where
Clearly, we deduce that
Hence, we have
Letting k\to \mathrm{\infty}, we conclude that {lim}_{k\to \mathrm{\infty}}\beta ({M}_{T}({x}_{n(k)1},{x}_{m(k)1}))=1, which yields that {lim}_{k\to \mathrm{\infty}}{M}_{T}({x}_{n(k)1},{x}_{m(k)1})=0. Hence, \u03f5=0, which is a contradiction. Thus, we get that \{{x}_{n}\} is a Cauchy sequence. Since X is a complete metric space, it follows that there exists {x}_{\ast}={lim}_{n\to \mathrm{\infty}}{x}_{n}\in X. By the continuity of T, we get {lim}_{n\to \mathrm{\infty}}T{x}_{n}=T{x}_{\ast}, and so {x}_{\ast}=T{x}_{\ast}, which means that {x}_{\ast} is a fixed point of T. □
Like in [13] we can replace the continuity of the operator T by a suitable condition.
Theorem 5 Let (X,d) be a complete metric space, \alpha :X\times X\to R be a function, and let T:X\to X be a map. Suppose that the following conditions are satisfied:

(1)
T is a generalized αGeraghty contraction type mapping;

(2)
T is triangular αorbital admissible mapping;

(3)
there exists {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1;

(4)
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\in X as n\to \mathrm{\infty}, then there exists a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that \alpha ({x}_{n(k)},x)\ge 1 for all k.
Then T has a fixed point {x}_{\ast}\in X and \{{T}^{n}{x}_{1}\} converges to {x}_{\ast}.
Proof Following the lines in the proof of Theorem 4, we have that the sequence \{{x}_{n}\} defined by {x}_{n+1}=T{x}_{n} for all n\ge 1 converges to {x}_{\ast}\in X. By condition (4) of the hypothesis, we deduce that there exists a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that \alpha ({x}_{n(k)},{x}_{\ast})\ge 1 for all k. Therefore, we have
where
Now, we suppose that T{x}_{\ast}\ne {x}_{\ast}, that is, d({x}_{\ast},T{x}_{\ast})>0. Letting k\to \mathrm{\infty} in the above equality, we get that
Since d({x}_{n(k)+1},T{x}_{\ast})/{M}_{T}({x}_{n(k)},{x}_{\ast})\le \beta ({M}_{T}({x}_{n(k)},{x}_{\ast})) for all k, letting k\to \mathrm{\infty}, we conclude that {lim}_{k\to \mathrm{\infty}}\beta ({M}_{T}({x}_{n(k)},{x}_{\ast}))=1. This implies {lim}_{k\to \mathrm{\infty}}{M}_{T}({x}_{n(k)},{x}_{\ast})=0. Hence, d({x}_{\ast},T{x}_{\ast})=0. This is a contradiction. Therefore, T{x}_{\ast}={x}_{\ast}. □
For the uniqueness of a fixed point of a generalized αGeraghty contraction type mapping, we consider the following hypothesis:

(K)
For all x\ne y\in X, there exists v\in X such that \alpha (x,v)\ge 1, \alpha (y,v)\ge 1 and \alpha (v,Tv)\ge 1.
Remark 9 Replacing condition (3) with condition (K) in the hypotheses of Theorem 4 or Theorem 5, we obtain that {x}_{\ast} is the unique fixed point of T. Suppose that {x}_{\ast} and {y}_{\ast} are two fixed points of T such that {x}_{\ast}\ne {y}_{\ast}. Then, by (K), there exists v\in X such that \alpha ({x}_{\ast},v)\ge 1, \alpha ({y}_{\ast},v)\ge 1 and \alpha (v,Tv)\ge 1. Since T is a triangular αorbital admissible mapping, we get that \alpha ({x}_{\ast},{T}^{n}v)\ge 1 and \alpha ({y}_{\ast},{T}^{n}v)\ge 1 for all n\ge 1. Thus we have
for all n\ge 1, where
By Theorem 4 (resp. Theorem 5) we deduce that the sequence \{{T}^{n}v\} converges to a fixed point {z}_{\ast} of T. Letting n\to \mathrm{\infty} in the above equality, we get {lim}_{n\to \mathrm{\infty}}{M}_{T}({x}_{\ast},{T}^{n}v)=d({x}_{\ast},{z}_{\ast}). If we suppose {x}_{\ast}\ne {z}_{\ast}, then we have d({x}_{\ast},{T}^{n+1}v)/{M}_{T}({x}_{\ast},{T}^{n}v)\le \beta ({M}_{T}({x}_{\ast},{T}^{n}v)), and letting n\to \mathrm{\infty}, we deduce that {lim}_{n\to \mathrm{\infty}}\beta ({M}_{T}({x}_{\ast},{T}^{n}v))=1. This implies {lim}_{n\to \mathrm{\infty}}{M}_{T}({x}_{\ast},{T}^{n}v)=0, so d({x}_{\ast},{z}_{\ast})=0, which is a contradiction. Therefore, {x}_{\ast}={z}_{\ast}. Similarly, we get {y}_{\ast}={z}_{\ast}. Hence, {x}_{\ast}={y}_{\ast}, which is a contradiction.
Example 10 Let X=[2,1]\cup \{0\}\cup [1,2], d:X\times X\to R, d(x,y)=xy, T:X\to X such that Tx=x if x\in [2,1)\cup (1,2], Tx=0 if x\in \{1,0,1\}, and \alpha :X\times X\to R, \alpha (x,y)=1 if xy\ge 0 and \alpha (x,y)=0 otherwise. Then T satisfies the conditions of Theorem 5. Clearly, X is a complete metric space. If \alpha (x,Tx)\ge 1, then xTx\ge 0, so Tx=0, which implies \alpha (Tx,{T}^{2}x)\ge 1. Hence T is an αorbital admissible mapping. For \alpha (x,y)\ge 1 and \alpha (y,Ty)\ge 1, we have Ty=0. Thus, xTy=0 and \alpha (x,Ty)\ge 1. Therefore, T is a triangular αorbital admissible mapping. Also, \alpha (1,T1)\ge 1 and 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\in X as n\to \mathrm{\infty}, then x=0 and \alpha ({x}_{n},x)\ge 1 for all n. For x,y\in [2,1), we have d(Tx,Ty)=xy\le 1 and {M}_{T}(x,y)\ge 2x\ge 2. Thus d(Tx,Ty)\le {M}_{T}(x,y)/2. The case x,y\in (1,2] is similar. If x\in [2,1)\cup (1,2] and y\in \{1,0,1\}, then d(Tx,Ty)=x, {M}_{T}(x,y)\ge 2x, so d(Tx,Ty)\le {M}_{T}(x,y)/2. For x,y\in \{1,0,1\}, we have d(Tx,Ty)=0\le {M}_{T}(x,y)/2. Therefore, taking \beta :[0,\mathrm{\infty})\to [0,1), \beta (t)=1/2, we obtain that T is a generalized αGeraghty contraction type mapping. Hence, T satisfies all the conditions of Theorem 5. However, since \alpha (2,0)=\alpha (0,2)=1, \alpha (2,2)=0, T is not a triangular αadmissible mapping.
3 αOrbital attractive mappings
Now we introduce a new concept.
Definition 11 Let T:X\to X be a map and \alpha :X\times X\to R be a function. Then T is said to be αorbital attractive if
for every y\in X.
Theorem 6 Let (X,d) be a complete metric space, \alpha :X\times X\to R be a function, and let T:X\to X be a map. Suppose that the following conditions are satisfied:

(1)
T is a generalized αGeraghty contraction type mapping;

(2)
T is αorbital admissible;

(3)
there exists {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1;

(4)
T is αorbital attractive.
Then T has a unique fixed point {x}_{\ast}\in X and \{{T}^{n}{x}_{1}\} converges to {x}_{\ast}.
Proof Let {x}_{1}\in X such that \alpha ({x}_{1},T{x}_{1})\ge 1. Define a sequence \{{x}_{n}\} by {x}_{n+1}=T{x}_{n} for n\ge 1. If {x}_{n(0)}={x}_{n(0)+1} for some n(0)\ge 1, then obviously T has a fixed point. Hence we suppose that {x}_{n}\ne {x}_{n+1} for all n\ge 1. Since T is αorbital admissible, we have
Inductively, we get \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\ge 1. Then we obtain
for all n\ge 1, where
If d({x}_{n},{x}_{n+1})\le d({x}_{n+1},{x}_{n+2}), we have
which is a contradiction. Thus, we get
Therefore, the sequence \{d({x}_{n},{x}_{n+1})\} is positive and nonincreasing. Hence, there exists r\ge 0 such that {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=r. We will show that r=0. Suppose, on the contrary, that r>0. Then we have
This implies that {lim}_{n\to \mathrm{\infty}}\beta ({x}_{n},{x}_{n+1})=1. Since \beta \in \mathcal{F}, we obtain that {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=0, which is a contradiction. Hence r=0.
Now, we shall show that \{{x}_{n}\} is a Cauchy sequence. Suppose, on the contrary, that \{{x}_{n}\} is not a Cauchy sequence. Then there exists \u03f5>0 such that, for all k\ge 1, there exists m(k)>n(k)>k with d({x}_{n(k)},{x}_{m(k)})\ge \u03f5. Let m(k) be the smallest number satisfying the conditions above. Hence, we have d({x}_{n(k)},{x}_{m(k)1})<\u03f5. Therefore, we get
Letting k\to \mathrm{\infty}, we have {lim}_{k\to \mathrm{\infty}}d({x}_{n(k)},{x}_{m(k)})=\u03f5. Since
we get {lim}_{k\to \mathrm{\infty}}d({x}_{n(k)},{x}_{m(k)1})=\u03f5. Similarly, we obtain
Since \alpha ({x}_{n(k)1},{x}_{n(k)})\ge 1 and T is αorbital attractive, we have
Hence, we get two cases as follows.

(1)
There exists an infinite subset I of N such that \alpha ({x}_{n(k)1},{x}_{m(k)1})\ge 1 for every k\in I.

(2)
There exists an infinite subset J of N such that \alpha ({x}_{m(k)1},{x}_{n(k)})\ge 1 for every k\in J.
In the first case, we have
where
Letting k\to \mathrm{\infty}, k\in I, we conclude that
and since
we get that
Since \beta \in \mathcal{F}, we obtain that
which is a contradiction.
In the second case, we have
where
Letting k\to \mathrm{\infty}, k\in J, we conclude that
and since
we get that
Since \beta \in \mathcal{F}, we obtain that
which is a contradiction. Thus, we get that \{{x}_{n}\} is a Cauchy sequence. Since X is a complete metric space, it follows that there exists {x}_{\ast}={lim}_{n\to \mathrm{\infty}}{x}_{n}\in X.
We claim that {x}_{\ast}=T{x}_{\ast}. Suppose, on the contrary, that {x}_{\ast}\ne T{x}_{\ast}. Since T is αorbital attractive, we have for every n\ge 1 that \alpha ({x}_{n},{x}_{\ast})\ge 1 or \alpha ({x}_{\ast},{x}_{n+1})\ge 1. Hence, there exists a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that \alpha ({x}_{n(k)},{x}_{\ast})\ge 1 or \alpha ({x}_{\ast},{x}_{n(k)})\ge 1 for all k\ge 1. In the first case, we have
for all k\ge 1, where
Letting k\to \mathrm{\infty}, we conclude that
and since
we have
Since \beta \in \mathcal{F}, we obtain that
which is a contradiction. The second case is similar. Therefore, {x}_{\ast}=T{x}_{\ast}.
If {y}_{\ast} is another fixed point of T, from the hypothesis we deduce that \alpha ({x}_{n},{y}_{\ast})\ge 1 or \alpha ({y}_{\ast},{x}_{n+1})\ge 1. Hence, there exists a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that \alpha ({x}_{n(k)},{y}_{\ast})\ge 1 or \alpha ({y}_{\ast},{x}_{n(k)})\ge 1 for all k\ge 1. In the first case, we have
for all k\ge 1, where
Letting k\to \mathrm{\infty}, we conclude that
and since
we obtain
Since \beta \in \mathcal{F}, we get that
so d({x}_{\ast},{y}_{\ast})=0. This is a contradiction. The second case is similar. □
Example 12 Let X=\{0,8,9,10\}, d:X\times X\to R, d(x,y)=xy, T:X\to X such that T0=T8=9, T9=T10=10, \alpha :X\times X\to R, \alpha (x,y)=0 if (x,y)\in \{(8,9),(9,8)\} and \alpha (x,y)=1 otherwise. Obviously, T is αorbital admissible and αorbital attractive. Also, T is a generalized αGeraghty contraction type mapping if we take \beta :[0,\mathrm{\infty})\to [0,1), \beta (t)=1/2. Hence T satisfies all the conditions of Theorem 6. However, since \alpha (8,0)=\alpha (0,T0)=1, \alpha (8,T0)=0, T is not triangular αorbital admissible. Note that d(T8,T9)={M}_{T}(8,9)=1.
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Popescu, O. Some new fixed point theorems for αGeraghty contraction type maps in metric spaces. Fixed Point Theory Appl 2014, 190 (2014). https://doi.org/10.1186/168718122014190
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DOI: https://doi.org/10.1186/168718122014190