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Some new fixed point theorems for α-Geraghty contraction type maps in metric spaces

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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 α-Geraghty 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., [210] 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, Amini-Harandi 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 ψ-Geraghty type contraction and supposedly improved and extended the results of Amini-Harandi and Emami [3]. Recently, Cho, Bae and Karapınar [13] defined the concept of α-Geraghty 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 [1323].

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 α-Geraghty 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:XX be a map and α:X×XR be a function. Then T is said to be α-admissible if α(x,y)1 implies α(Tx,Ty)1.

Definition 2 [16]

A map T:XX is said to be triangular α-admissible if:

(T1) T is α-admissible,

(T2) α(x,z)1 and α(z,y)1 imply α(x,y)1.

Let be the family of all functions β:[0,)[0,1) which satisfies the condition

lim n β( t n )=1implies lim n t n =0.

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 βF such that for all x,yX,

d(Tx,Ty)β ( d ( x , y ) ) d(x,y).

Then T has a unique fixed point x X and { T n x} converges to x for each xX.

Amini-Harandi and Emami [3] reconsidered Theorem 1 in the framework of partially ordered metric spaces.

Theorem 2 Let (X,,d) be a partially ordered complete metric space. Let f:XX be an increasing mapping such that there exists an element x 0 X with x 0 f x 0 . If there exists αF such that

d(fx,fy)α ( d ( x , y ) ) d(x,y),

for each x,yX with xy, then f has a fixed point provided that either f is continuous or X is such that if an increasing sequence { x n }x, then x n x for all n. Besides, if for each x,yX there exists zX 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 α:X×XR be a function. A map T:XX is called a generalized α-Geraghty contraction type map if there exists βF such that for all x,yX,

α(x,y)d(Tx,Ty)β ( M ( x , y ) ) M(x,y),

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, α:X×XR be a function, and let T:XX be a map. Suppose that the following conditions are satisfied:

  1. (1)

    T is a generalized α-Geraghty contraction type map;

  2. (2)

    T is triangular α-admissible;

  3. (3)

    there exists x 1 X such that α( x 1 ,T x 1 )1;

  4. (4)

    T is continuous.

Then T has a fixed point x X and { T n x 1 } converges to x .

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 α( x n , x n + 1 )1 for all n and x n xX as n, then there exists a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) ,x)1 for all k.

Adding the condition (H),

  1. (H)

    For all x,yFix(T), there exists zX such that α(x,z)1 and α(z,y)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 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:

  1. (iii)

    [16] For all xyX, there exists vX such that α(x,v)1 and α(v,y)1.

This condition implies α(x,y)1 for all xyX, 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 α-Geraghty contraction type map.

Definition 4 Let (X,d) be a complete metric space and let α:X×XR be a function. A map T:XX is called a generalized α-Geraghty contraction type map if there exists βF such that for all x,yX,

α(x,y)d(Tx,Ty)β ( M T ( x , y ) ) M T (x,y),

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:XX be a map and α:X×XR be a function. Then T is said to be α-orbital admissible if

(T3) α(x,Tx)1 implies α(Tx, T 2 x)1.

Definition 6 Let T:XX be a map and α:X×XR be a function. Then T is said to be triangular α-orbital admissible if T is α-orbital admissible and

(T4) α(x,y)1 and α(y,Ty)1 imply α(x,Ty)1.

Obviously, every α-admissible mapping is an α-orbital admissible mapping and every triangular α-admissible mapping is a triangular α-orbital admissible mapping. The following example shows that there exists a triangular α-admissible mapping which is not triangular α-admissible.

Example 7 Let X={0,1,2,3}, d:X×XR, d(x,y)=|xy|, T:XX such that T0=0, T1=2, T2=1, T3=3 and α:X×XR, α(x,y)=1 if (x,y){(0,1),(0,2),(1,1),(2,2),(1,2),(2,1),(1,3),(2,3)} and α(x,y)=0 otherwise. Since α(1,T1)=α(1,2)=1 and α(2,T2)=α(2,1)=1, T is α-orbital admissible. Since α(0,1)=α(1,2)=α(0,2)=1, α(0,2)=α(2,1)=α(0,1)=1, T is triangular α-orbital admissible. But α(0,1)=α(1,3)=1, α(0,3)=0, so T is not triangular α-admissible.

Lemma 8 Let T:XX be a triangular α-orbital admissible mapping. Assume that there exists x 1 X such that α( x 1 ,T x 1 )1. Define a sequence { x n } by x n + 1 =T x n . Then we have α( x n , x m )1 for all m,nN with n<m.

Proof Since T is α-orbital admissible and α( x 1 ,T x 1 )1, we deduce that α( x 2 , x 3 )=α(T x 1 ,T x 2 )1. By continuing this process, we get α( x n , x n + 1 )1 for all n1. Suppose that α( x n , x m )1 and prove that α( x n , x m + 1 )1, where m>n. Since T is triangular α-orbital admissible and α( x m , x m + 1 )1, we get that α( x n , x m + 1 )1. Hence, we have proved that α( x n , x m )1 for all m,nN with n<m. □

Theorem 4 Let (X,d) be a complete metric space, α:X×XR be a function, and let T:XX be a map. Suppose that the following conditions are satisfied:

  1. (1)

    T is a generalized α-Geraghty contraction type mapping;

  2. (2)

    T is a triangular α-orbital admissible mapping;

  3. (3)

    there exists x 1 X such that α( x 1 ,T x 1 )1;

  4. (4)

    T is continuous.

Then T has a fixed point x X and { T n x 1 } converges to x .

Proof Let x 1 X such that α( x 1 ,T x 1 )1. Define a sequence { x n } by x n + 1 =T x n for n1. If x n ( 0 ) = x n ( 0 ) + 1 for some n(0)1, then obviously T has a fixed point. Hence we suppose that x n x n + 1 for all n1. By Lemma 8 we have α( x n , x n + 1 )1 for all n1. Then we get, for all n1,

d( x n + 1 , x n + 2 )=d(T x n ,T x n + 1 )α( x n , x n + 1 )d(T x n ,T x n + 1 )β ( M T ( x n , x n + 1 ) ) M T ( x n , x n + 1 ),

where

M T ( x n , x n + 1 ) = max { d ( x n , x n + 1 ) , d ( x n , T x n ) , d ( x n + 1 , T x n + 1 ) , [ d ( x n , T x n + 1 ) + d ( x n + 1 , T x n ) ] / 2 } = max { d ( x n , x n + 1 ) , d ( x n + 1 , x n + 2 ) , d ( x n , x n + 2 ) / 2 } max { d ( x n , x n + 1 ) , d ( x n + 1 , x n + 2 ) , [ d ( x n , x n + 1 ) + d ( x n + 1 , x n + 2 ) ] / 2 } = max { d ( x n , x n + 1 ) , d ( x n + 1 , x n + 2 ) } .

Since β( M T ( x n , x n + 1 ))<1, the case d( x n , x n + 1 )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 r0 such that lim n d( x n , x n + 1 )=r. We will show that r=0. Suppose, to the contrary, that r>0. Then we have

d( x n + 1 , x n + 2 )/d( x n , x n + 1 )β ( M T ( x n , x n + 1 ) ) <1.

This implies that lim n β( M T ( x n , x n + 1 ))=1. Since βF, we get lim n M T ( x n , x n + 1 )=0, and then lim n 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 ϵ>0 such that, for all k1, there exists m(k)>n(k)>k with d( x n ( k ) , x m ( k ) )ϵ. Let m(k) be the smallest number satisfying the conditions above. Hence, we have d( x n ( k ) , x m ( k ) 1 )<ϵ. Therefore, we get

ϵd( x n ( k ) , x m ( k ) )d( x n ( k ) , x m ( k ) 1 )+d( x m ( k ) 1 , x m ( k ) )<ϵ+d( x m ( k ) 1 , x m ( k ) ).

Letting k, we have lim k d( x n ( k ) , x m ( k ) )=ϵ. Since

|d( x n ( k ) , x m ( k ) 1 )d( x n ( k ) , x m ( k ) )|d( x m ( k ) , x m ( k ) 1 ),

we get lim k d( x n ( k ) , x m ( k ) 1 )=ϵ. Similarly, we obtain

lim k d( x m ( k ) , x n ( k ) 1 )= lim k d( x m ( k ) 1 , x n ( k ) 1 )=ϵ.

By Lemma 8 we have α( x n ( k ) 1 , x m ( k ) 1 )1. Thus, we deduce that

d ( x n ( k ) , x m ( k ) ) = d ( T x n ( k ) 1 , T x m ( k ) 1 ) α ( x n ( k ) 1 , x m ( k ) 1 ) d ( T x n ( k ) 1 , T x m ( k ) 1 ) β ( M T ( x n ( k ) 1 , x m ( k ) 1 ) ) M T ( x n ( k ) 1 , x m ( k ) 1 ) ,

where

M T ( x n ( k ) 1 , x m ( k ) 1 ) = max { d ( x n ( k ) 1 , x m ( k ) 1 ) , d ( x n ( k ) 1 , T x n ( k ) 1 ) , d ( x m ( k ) 1 , T x m ( k ) 1 ) , [ d ( x n ( k ) 1 , T x m ( k ) 1 ) + d ( x m ( k ) 1 , T x n ( k ) 1 ) ] / 2 } = max { d ( x n ( k ) 1 , x m ( k ) 1 ) , d ( x n ( k ) 1 , x n ( k ) ) , d ( x m ( k ) 1 , x m ( k ) ) , [ d ( x n ( k ) 1 , x m ( k ) ) + d ( x m ( k ) 1 , x n ( k ) ) ] / 2 } .

Clearly, we deduce that

lim k M T ( x n ( k ) 1 , x m ( k ) 1 )=ϵ.

Hence, we have

d( x n ( k ) , x m ( k ) )/ M T ( x n ( k ) 1 , x m ( k ) 1 )β ( M T ( x n ( k ) 1 , x m ( k ) 1 ) ) .

Letting k, we conclude that lim k β( M T ( x n ( k ) 1 , x m ( k ) 1 ))=1, which yields that lim k M T ( x n ( k ) 1 , x m ( k ) 1 )=0. Hence, ϵ=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 = lim n x n X. By the continuity of T, we get lim n T x n =T x , and so x =T x , which means that x 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, α:X×XR be a function, and let T:XX be a map. Suppose that the following conditions are satisfied:

  1. (1)

    T is a generalized α-Geraghty contraction type mapping;

  2. (2)

    T is triangular α-orbital admissible mapping;

  3. (3)

    there exists x 1 X such that α( x 1 ,T x 1 )1;

  4. (4)

    if { x n } is a sequence in X such that α( x n , x n + 1 )1 for all n and x n xX as n, then there exists a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) ,x)1 for all k.

Then T has a fixed point x X and { T n x 1 } converges to x .

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 n1 converges to x X. By condition (4) of the hypothesis, we deduce that there exists a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) , x )1 for all k. Therefore, we have

d ( x n ( k ) + 1 , T x ) = d ( T x n ( k ) , T x ) α ( x n ( k ) , x ) d ( T x n ( k ) , T x ) β ( M T ( x n ( k ) , x ) ) M T ( x n ( k ) , x ) ,

where

M T ( x n ( k ) , x ) = max { d ( x n ( k ) , x ) , d ( x n ( k ) , T x n ( k ) ) , d ( x , T x ) , [ d ( x n ( k ) , T x ) + d ( x , T x n ( k ) ) ] / 2 } = max { d ( x n ( k ) , x ) , d ( x n ( k ) , x n ( k ) + 1 ) , d ( x , T x ) , [ d ( x n ( k ) , T x ) + d ( x , x n ( k ) + 1 ) ] / 2 } .

Now, we suppose that T x x , that is, d( x ,T x )>0. Letting k in the above equality, we get that

lim k M T ( x n ( k ) , x )=d( x ,T x ).

Since d( x n ( k ) + 1 ,T x )/ M T ( x n ( k ) , x )β( M T ( x n ( k ) , x )) for all k, letting k, we conclude that lim k β( M T ( x n ( k ) , x ))=1. This implies lim k M T ( x n ( k ) , x )=0. Hence, d( x ,T x )=0. This is a contradiction. Therefore, T x = x . □

For the uniqueness of a fixed point of a generalized α-Geraghty contraction type mapping, we consider the following hypothesis:

  1. (K)

    For all xyX, there exists vX such that α(x,v)1, α(y,v)1 and α(v,Tv)1.

Remark 9 Replacing condition (3) with condition (K) in the hypotheses of Theorem 4 or Theorem 5, we obtain that x is the unique fixed point of T. Suppose that x and y are two fixed points of T such that x y . Then, by (K), there exists vX such that α( x ,v)1, α( y ,v)1 and α(v,Tv)1. Since T is a triangular α-orbital admissible mapping, we get that α( x , T n v)1 and α( y , T n v)1 for all n1. Thus we have

d ( x , T n + 1 v ) α ( x , T n v ) d ( T x , T n + 1 v ) β ( M T ( x , T n v ) ) M T ( x , T n v )

for all n1, where

M T ( x , T n v ) = max { d ( x , T n v ) , d ( x , T x ) , d ( T n v , T n + 1 v ) , [ d ( x , T n + 1 v ) + d ( T x , T n v ) ] / 2 } = max { d ( x , T n v ) , d ( T n v , T n + 1 v ) , [ d ( x , T n + 1 v ) + d ( x , T n v ) ] / 2 } .

By Theorem 4 (resp. Theorem 5) we deduce that the sequence { T n v} converges to a fixed point z of T. Letting n in the above equality, we get lim n M T ( x , T n v)=d( x , z ). If we suppose x z , then we have d( x , T n + 1 v)/ M T ( x , T n v)β( M T ( x , T n v)), and letting n, we deduce that lim n β( M T ( x , T n v))=1. This implies lim n M T ( x , T n v)=0, so d( x , z )=0, which is a contradiction. Therefore, x = z . Similarly, we get y = z . Hence, x = y , which is a contradiction.

Example 10 Let X=[2,1]{0}[1,2], d:X×XR, d(x,y)=|xy|, T:XX such that Tx=x if x[2,1)(1,2], Tx=0 if x{1,0,1}, and α:X×XR, α(x,y)=1 if xy0 and α(x,y)=0 otherwise. Then T satisfies the conditions of Theorem 5. Clearly, X is a complete metric space. If α(x,Tx)1, then xTx0, so Tx=0, which implies α(Tx, T 2 x)1. Hence T is an α-orbital admissible mapping. For α(x,y)1 and α(y,Ty)1, we have Ty=0. Thus, xTy=0 and α(x,Ty)1. Therefore, T is a triangular α-orbital admissible mapping. Also, α(1,T1)1 and if { x n } is a sequence in X such that α( x n , x n + 1 )1 for all n and x n xX as n, then x=0 and α( x n ,x)1 for all n. For x,y[2,1), we have d(Tx,Ty)=|xy|1 and M T (x,y)2x2. Thus d(Tx,Ty) M T (x,y)/2. The case x,y(1,2] is similar. If x[2,1)(1,2] and y{1,0,1}, then d(Tx,Ty)=|x|, M T (x,y)2|x|, so d(Tx,Ty) M T (x,y)/2. For x,y{1,0,1}, we have d(Tx,Ty)=0 M T (x,y)/2. Therefore, taking β:[0,)[0,1), β(t)=1/2, we obtain that T is a generalized α-Geraghty contraction type mapping. Hence, T satisfies all the conditions of Theorem 5. However, since α(2,0)=α(0,2)=1, α(2,2)=0, T is not a triangular α-admissible mapping.

3 α-Orbital attractive mappings

Now we introduce a new concept.

Definition 11 Let T:XX be a map and α:X×XR be a function. Then T is said to be α-orbital attractive if

α(x,Tx)1impliesα(x,y)1 or α(y,Tx)1

for every yX.

Theorem 6 Let (X,d) be a complete metric space, α:X×XR be a function, and let T:XX be a map. Suppose that the following conditions are satisfied:

  1. (1)

    T is a generalized α-Geraghty contraction type mapping;

  2. (2)

    T is α-orbital admissible;

  3. (3)

    there exists x 1 X such that α( x 1 ,T x 1 )1;

  4. (4)

    T is α-orbital attractive.

Then T has a unique fixed point x X and { T n x 1 } converges to x .

Proof Let x 1 X such that α( x 1 ,T x 1 )1. Define a sequence { x n } by x n + 1 =T x n for n1. If x n ( 0 ) = x n ( 0 ) + 1 for some n(0)1, then obviously T has a fixed point. Hence we suppose that x n x n + 1 for all n1. Since T is α-orbital admissible, we have

α( x 1 , x 2 )=α( x 1 ,T x 1 )1impliesα(T x 1 ,T x 2 )=α( x 2 , x 3 )1.

Inductively, we get α( x n , x n + 1 )1 for all n1. Then we obtain

d( x n , x n + 1 )=d(T x n ,T x n + 1 )α( x n , x n + 1 )d(T x n ,T x n + 1 )β ( M T ( x n , x n + 1 ) ) M T ( x n , x n + 1 )

for all n1, where

M T ( x n , x n + 1 ) = max { d ( x n , x n + 1 ) , d ( x n , T x n ) , d ( x n + 1 , T x n + 1 ) , [ d ( x n , T x n + 1 ) + d ( x n + 1 , T x n ) ] / 2 } = max { d ( x n , x n + 1 ) , d ( x n + 1 , x n + 2 ) , d ( x n , x n + 2 ) / 2 } max { d ( x n , x n + 1 ) , d ( x n + 1 , x n + 2 ) , [ d ( x n , x n + 1 ) + d ( x n + 1 , x n + 2 ) ] / 2 } = max { d ( x n , x n + 1 ) , d ( x n + 1 , x n + 2 ) } .

If d( x n , x n + 1 )d( x n + 1 , x n + 2 ), we have

d( x n + 1 , x n + 2 )β ( d ( x n + 1 , x n + 2 ) ) d( x n + 1 , x n + 2 )<d( x n + 1 , x n + 2 ),

which is a contradiction. Thus, we get

d( x n + 1 , x n + 2 )β ( d ( x n , x n + 1 ) ) d( x n , x n + 1 )<d( x n , x n + 1 ).

Therefore, the sequence {d( x n , x n + 1 )} is positive and nonincreasing. Hence, there exists r0 such that lim n d( x n , x n + 1 )=r. We will show that r=0. Suppose, on the contrary, that r>0. Then we have

d( x n + 1 , x n + 2 )/d( x n , x n + 1 )β ( d ( x n , x n + 1 ) ) <1.

This implies that lim n β( x n , x n + 1 )=1. Since βF, we obtain that lim n 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 ϵ>0 such that, for all k1, there exists m(k)>n(k)>k with d( x n ( k ) , x m ( k ) )ϵ. Let m(k) be the smallest number satisfying the conditions above. Hence, we have d( x n ( k ) , x m ( k ) 1 )<ϵ. Therefore, we get

ϵd( x n ( k ) , x m ( k ) )d( x n ( k ) , x m ( k ) 1 )+d( x m ( k ) 1 , x m ( k ) )<ϵ+d( x m ( k ) 1 , x m ( k ) ).

Letting k, we have lim k d( x n ( k ) , x m ( k ) )=ϵ. Since

|d( x n ( k ) , x m ( k ) 1 )d( x n ( k ) , x m ( k ) )|d( x m ( k ) , x m ( k ) 1 ),

we get lim k d( x n ( k ) , x m ( k ) 1 )=ϵ. Similarly, we obtain

lim k d( x m ( k ) , x n ( k ) 1 )= lim k d( x m ( k ) 1 , x n ( k ) 1 )= lim k d( x m ( k ) 1 , x n ( k ) + 1 )=ϵ.

Since α( x n ( k ) 1 , x n ( k ) )1 and T is α-orbital attractive, we have

α( x n ( k ) 1 , x m ( k ) 1 )1orα( x m ( k ) 1 , x n ( k ) )1.

Hence, we get two cases as follows.

  1. (1)

    There exists an infinite subset I of N such that α( x n ( k ) 1 , x m ( k ) 1 )1 for every kI.

  2. (2)

    There exists an infinite subset J of N such that α( x m ( k ) 1 , x n ( k ) )1 for every kJ.

In the first case, we have

d ( x n ( k ) , x m ( k ) ) = d ( T x n ( k ) 1 , T x m ( k ) 1 ) α ( x n ( k ) 1 , x m ( k ) 1 ) d ( T x n ( k ) 1 , T x m ( k ) 1 ) β ( M T ( x n ( k ) 1 , x m ( k ) 1 ) ) M T ( x n ( k ) 1 , x m ( k ) 1 ) ,

where

M T ( x n ( k ) 1 , x m ( k ) 1 ) = max { d ( x n ( k ) 1 , x m ( k ) 1 ) , d ( x n ( k ) 1 , T x n ( k ) 1 ) , d ( x m ( k ) 1 , T x m ( k ) 1 ) , [ d ( x n ( k ) 1 , T x m ( k ) 1 ) + d ( x m ( k ) 1 , T x n ( k ) 1 ) ] / 2 } = max { d ( x n ( k ) 1 , x m ( k ) 1 ) , d ( x n ( k ) 1 , x n ( k ) ) , d ( x m ( k ) 1 , x m ( k ) ) , [ d ( x n ( k ) 1 , x m ( k ) ) + d ( x m ( k ) 1 , x n ( k ) ) ] / 2 } .

Letting k, kI, we conclude that

lim k , k I M T ( x n ( k ) 1 , x m ( k ) 1 )=ϵ,

and since

d( x n ( k ) , x m ( k ) )/ M T ( x n ( k ) 1 , x m ( k ) 1 )β ( M T ( x n ( k ) 1 , x m ( k ) 1 ) ) ,

we get that

lim k , k I β ( M T ( x n ( k ) 1 , x m ( k ) 1 ) ) =1.

Since βF, we obtain that

lim k , I M T ( x n ( k ) 1 , x m ( k ) 1 )=0,

which is a contradiction.

In the second case, we have

d ( x m ( k ) , x n ( k ) + 1 ) = d ( T x m ( k ) 1 , T x n ( k ) ) α ( x m ( k ) 1 , x n ( k ) ) d ( T x m ( k ) 1 , T x n ( k ) ) β ( M T ( x m ( k ) 1 , x n ( k ) ) ) M T ( x m ( k ) 1 , x n ( k ) ) ,

where

M T ( x m ( k ) 1 , x n ( k ) ) = max { d ( x m ( k ) 1 , x n ( k ) ) , d ( x m ( k ) 1 , T x m ( k ) 1 ) , d ( x n ( k ) , T x n ( k ) ) , [ d ( x m ( k ) 1 , T x n ( k ) ) + d ( x n ( k ) , T x m ( k ) 1 ) ] / 2 } = max { d ( x m ( k ) 1 , x n ( k ) ) , d ( x m ( k ) 1 , x m ( k ) ) , d ( x n ( k ) , x n ( k ) + 1 ) , [ d ( x m ( k ) 1 , x n ( k ) + 1 ) + d ( x m ( k ) , x n ( k ) ) ] / 2 } .

Letting k, kJ, we conclude that

lim k , k J M T ( x m ( k ) 1 , x n ( k ) )=ϵ,

and since

d( x m ( k ) , x n ( k ) + 1 )/ M T ( x m ( k ) 1 , x n ( k ) )β ( M T ( x m ( k ) 1 , x n ( k ) ) ) ,

we get that

lim k , k J β ( M T ( x m ( k ) 1 , x n ( k ) ) ) =1.

Since βF, we obtain that

lim k , k J M T ( x m ( k ) 1 , x n ( k ) )=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 = lim n x n X.

We claim that x =T x . Suppose, on the contrary, that x T x . Since T is α-orbital attractive, we have for every n1 that α( x n , x )1 or α( x , x n + 1 )1. Hence, there exists a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) , x )1 or α( x , x n ( k ) )1 for all k1. In the first case, we have

d( x n ( k ) + 1 ,Tx)α( x n ( k ) , x )d( x n ( k ) + 1 ,T x )β ( M T ( x n ( k ) , x ) ) M T ( x n ( k ) , x ),

for all k1, where

M T ( x n ( k ) , x ) = max { d ( x n ( k ) , x ) , d ( x n ( k ) , T x n ( k ) ) , d ( x , T x ) , [ d ( x n ( k ) , T x ) + d ( T x n ( k ) , x ) ] / 2 } .

Letting k, we conclude that

lim k M T ( x n ( k ) , x )=d( x ,T x ),

and since

d( x n ( k ) + 1 ,Tx)/ M T ( x n ( k ) , x )β ( M T ( x n ( k ) , x ) ) ,

we have

lim k β ( M T ( x n ( k ) , x ) ) =1.

Since βF, we obtain that

lim k M T ( x m ( k ) 1 , x n ( k ) )=0,

which is a contradiction. The second case is similar. Therefore, x =T x .

If y is another fixed point of T, from the hypothesis we deduce that α( x n , y )1 or α( y , x n + 1 )1. Hence, there exists a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) , y )1 or α( y , x n ( k ) )1 for all k1. In the first case, we have

d( x n ( k ) + 1 ,Ty)α( x n ( k ) , y )d( x n ( k ) + 1 ,T y )β ( M T ( x n ( k ) , y ) ) M T ( x n ( k ) , y ),

for all k1, where

M T ( x n ( k ) , y ) = max { d ( x n ( k ) , y ) , d ( x n ( k ) , T x n ( k ) ) , d ( y , T y ) , [ d ( x n ( k ) , T y ) + d ( T x n ( k ) , y ) ] / 2 } .

Letting k, we conclude that

lim k M T ( x n ( k ) , y )=d( x , y ),

and since

d( x n ( k ) + 1 ,Ty)/ M T ( x n ( k ) , y )β ( M T ( x n ( k ) , y ) ) ,

we obtain

lim k β ( M T ( x n ( k ) , y ) ) =1.

Since βF, we get that

lim k M T ( x n ( k ) , y )=0,

so d( x , y )=0. This is a contradiction. The second case is similar. □

Example 12 Let X={0,8,9,10}, d:X×XR, d(x,y)=|xy|, T:XX such that T0=T8=9, T9=T10=10, α:X×XR, α(x,y)=0 if (x,y){(8,9),(9,8)} and α(x,y)=1 otherwise. Obviously, T is α-orbital admissible and α-orbital attractive. Also, T is a generalized α-Geraghty contraction type mapping if we take β:[0,)[0,1), β(t)=1/2. Hence T satisfies all the conditions of Theorem 6. However, since α(8,0)=α(0,T0)=1, α(8,T0)=0, T is not triangular α-orbital admissible. Note that d(T8,T9)= M T (8,9)=1.

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Correspondence to Ovidiu Popescu.

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Keywords

  • Differential Geometry
  • Nonlinear Analysis
  • Fixed Point Theorem
  • Point Theory
  • Suitable Condition