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Fixed point results for implicit contractions on spaces with two metrics

Abstract

We establish new fixed-point results involving implicit contractions on a metric space endowed with two metrics. The main results in this paper extend and generalize several existing fixed-point theorems in the literature.

1 Introduction

Fixed-point theory is a major branch of nonlinear analysis because of its wide applicability. The existence problem of fixed points of mappings satisfying a given metrical contractive condition has attracted many researchers in past few decades. The Banach contraction principle [1] is one of the most important theorems in this direction. Many generalizations of this famous principle exist in the literature, see, for examples, [2–6] and references therein. On the other hand, several classical fixed-point theorems have been unified by considering general contractive conditions expressed by an implicit condition, see for examples, Turinici [7], Popa [8, 9], Berinde [10], and references therein.

This paper presents fixed-point theorems for implicit contractions on a metric space endowed with two metrics. This paper will be divided into two main sections. Section 2 presents local and global fixed-point results for implicit contractions involving α-admissible mappings, a recent concept introduced in [11]. Section 3 presents some interesting consequences that can be obtained from the results established in the previous section.

2 Main results

Let ℱ be the set of functions F: [ 0 , + ∞ ) 6 →R satisfying the following conditions:

  1. (i)

    F is continuous;

  2. (ii)

    F is non-decreasing in the first variable;

  3. (iii)

    F is non-increasing in the fifth variable;

  4. (iv)

    ∃h∈(0,1)∣F(u,v,v,u,u+v,0)≤0⟹u≤hv.

Example Let F: [ 0 , + ∞ ) 6 →R be the function defined by

F( t 1 , t 2 , t 3 , t 4 , t 5 , t 6 ):= t 1 −qmax { t 2 , t 3 , t 4 , t 5 + t 6 2 } ,

where q∈(0,1). We can check easily that F∈F.

Let X be a nonempty set endowed with two metrics d and d ′ . If x 0 ∈X and r>0, let

B( x 0 ,r):= { x ∈ X : d ( x 0 , x ) < r } .

We denote by B ( x 0 , r ) ¯ d ′ the d ′ -closure of B( x 0 ,r).

Let T: B ( x 0 , r ) ¯ d ′ →X and α:X×X→[0,∞). We say that T is α-admissible (see [11]) if the following condition holds: for all x,y∈B( x 0 ,r), we have

α(x,y)≥1⟹α(Tx,Ty)≥1.

We say that X satisfies the property (H) with respect to the metric d if the following condition holds:

If lim n → ∞ d( x n ,x)=0 for some x∈X and α( x n , x n + 1 )≥1 for all n, then there exist a positive integer κ and a subsequence { x n ( k ) } of { x n } such that α( x n ( k ) ,x)≥1 for all k≥κ.

Our first result is the following.

Theorem 2.1 Let (X, d ′ ) be a complete metric space, d another metric on X, x 0 ∈X, r>0, T: B ( x 0 , r ) ¯ d ′ →X, and α:X×X→[0,∞). Suppose there exists F∈F such that for x,y∈ B ( x 0 , r ) ¯ d ′ , we have

F ( α ( x , y ) d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.
(1)

In addition, assume the following properties hold:

  1. (I)

    d( x 0 ,T x 0 )<(1−h)r and α( x 0 ,T x 0 )≥1;

  2. (II)

    T is α-admissible;

  3. (III)

    if d≱ d ′ , assume T is uniformly continuous from (B( x 0 ,r),d) into (X, d ′ );

  4. (IV)

    if d= d ′ , assume X satisfies the property (H) with respect to the metric d;

  5. (V)

    if d≠ d ′ , assume T is continuous from ( B ( x 0 , r ) ¯ d ′ , d ′ ) into (X, d ′ ).

Then T has a fixed point.

Proof Let x 1 =T x 0 . From (I), we have

d( x 0 , x 1 )=d( x 0 ,T x 0 )≤(1−h)r<r,

which implies that x 1 ∈B( x 0 ,r). Let x 2 =T x 1 . From (1), we have

F ( α ( x 0 , x 1 ) d ( T x 0 , T x 1 ) , d ( x 0 , x 1 ) , d ( x 0 , x 1 ) , d ( x 1 , x 2 ) , d ( x 0 , x 2 ) , 0 ) ≤0.

From (I), we have

d(T x 0 ,T x 1 )≤α( x 0 , x 1 )d(T x 0 ,T x 1 ).

Since F is non-decreasing in the first variable (property (i)), we obtain

F ( d ( x 1 , x 2 ) , d ( x 0 , x 1 ) , d ( x 0 , x 1 ) , d ( x 1 , x 2 ) , d ( x 0 , x 2 ) , 0 ) ≤0.

Since d( x 0 , x 2 )≤d( x 0 , x 1 )+d( x 1 , x 2 ), using (iii), we obtain

F ( d ( x 1 , x 2 ) , d ( x 0 , x 1 ) , d ( x 0 , x 1 ) , d ( x 1 , x 2 ) , d ( x 0 , x 1 ) + d ( x 1 , x 2 ) , 0 ) ≤0,

which implies from (iv) that

d( x 1 , x 2 )≤hd( x 0 , x 1 )≤h(1−h)r<r.

Now, we have

d( x 0 , x 2 )≤d( x 0 , x 1 )+hd( x 0 , x 1 )=(1+h)d( x 0 , x 1 )≤(1+h)(1−h)r<r.

This implies that x 2 ∈B( x 0 ,r). Again, let x 3 =T x 2 . Since T is α-admissible and α( x 0 , x 1 )≥1, we have

d( x 2 , x 3 )≤α( x 1 , x 2 )d(T x 1 ,T x 2 ).

Then, from (1), we obtain

F ( d ( x 2 , x 3 ) , d ( x 1 , x 2 ) , d ( x 1 , x 2 ) , d ( x 2 , x 3 ) , d ( x 1 , x 3 ) , 0 ) ≤0.

Using (iii), we obtain

F ( d ( x 2 , x 3 ) , d ( x 1 , x 2 ) , d ( x 1 , x 2 ) , d ( x 2 , x 3 ) , d ( x 1 , x 2 ) + d ( x 2 , x 3 ) , 0 ) ≤0,

which implies from (iv) that

d( x 2 , x 3 )≤hd( x 1 , x 2 )≤ h 2 (1−h)r<r.

Now, we have

d( x 0 , x 3 )≤d( x 0 , x 2 )+d( x 2 , x 3 )≤(1+h)(1−h)r+ h 2 (1−h)r= ( 1 − h 3 ) r<r.

This implies that x 3 ∈B( x 0 ,r). Continuing this process, by induction, we can define the sequence { x n } by

x n + 1 =T x n ,∀n∈N.

Such sequence satisfies the following property:

x n ∈B( x 0 ,r),α( x n , x n + 1 )≥1andd( x n , x n + 1 )≤ h n (1−h)r,∀n∈N.
(2)

Since h∈(0,1), it follows from (2) that { x n } is a Cauchy sequence with respect to the metric d.

Now, we shall prove that { x n } is also a Cauchy sequence with respect to d ′ . If d ′ ≤d, the result follows immediately from (2). If d≱ d ′ , from (III), given ε>0, there exists δ> such that

x,y∈B( x 0 ,r),d(x,y)<δ⟹ d ′ (Tx,Ty)<ε.
(3)

On the other hand, since { x n } is Cauchy with respect to d, there exists a positive integer N such that

d( x n , x m )<δ,∀n,m≥N.

Using (3), we have

d ′ ( x n + 1 , x m + 1 )<ε,∀n,m≥N.

Thus we proved that { x n } is Cauchy with respect to d ′ .

Since (X, d ′ ) is complete, there exists z∈ B ( x 0 , r ) ¯ d ′ such that

lim n → ∞ d ′ ( x n ,z)=0.
(4)

We shall prove that z is a fixed point of T. We consider two cases.

Case 1. If d= d ′ .

From (IV), there exist a positive integer κ and a subsequence { x n ( k ) } of { x n } such that

α( x n ( k ) ,z)≥1,∀k≥κ.
(5)

Using (1), for all k≥κ, we obtain

F ( α ( x n ( k ) , z ) d ( T x n ( k ) , T z ) , d ( x n ( k ) , z ) , d ( x n ( k ) , x n ( k ) + 1 ) , d ( z , T z ) , d ( x n ( k ) , T z ) , d ( z , x n ( k ) + 1 ) ) ≤ 0 .

Using (5) and condition (ii), for all k≥κ, we obtain

F ( d ( x n ( k ) + 1 , T z ) , d ( x n ( k ) , z ) , d ( x n ( k ) , x n ( k ) + 1 ) , d ( z , T z ) , d ( x n ( k ) , T z ) , d ( z , x n ( k ) + 1 ) ) ≤0.

Letting k→∞, using (4) and the continuity of F, we obtain

F ( d ( z , T z ) , 0 , 0 , d ( z , T z ) , d ( z , T z ) , 0 ) ≤0,

which implies from (iv) that d(z,Tz)=0.

Case 2. If d≠ d ′ .

In this case, using (V) and (4), we obtain

lim n → ∞ d ′ (T x n ,Tz)= lim n → ∞ d ′ ( x n + 1 ,Tz)=0.

The uniqueness of the limit gives z=Tz. □

Taking d= d ′ in Theorem 2.1, we obtain the following result.

Theorem 2.2 Let (X,d) be a complete metric space, x 0 ∈X, r>0, T: B ( x 0 , r ) ¯ d →X, and α:X×X→[0,∞). Suppose there exists F∈F such that for x,y∈ B ( x 0 , r ) ¯ d , we have

F ( α ( x , y ) d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    d( x 0 ,T x 0 )<(1−h)r and α( x 0 ,T x 0 )≥1;

  2. (II)

    T is α-admissible;

  3. (III)

    X satisfies the property (H) with respect to the metric d.

Then T has a fixed point.

From Theorem 2.1, we can deduce the following global result.

Theorem 2.3 Let (X, d ′ ) be a complete metric space, d another metric on X, T:X→X, and α:X×X→[0,∞). Suppose there exists F∈F such that for x,y∈X, we have

F ( α ( x , y ) d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    there exists x 0 ∈X such that α( x 0 ,T x 0 )≥1;

  2. (II)

    T is α-admissible (x,y∈X, α(x,y)≥1⟹α(Tx,Ty)≥1);

  3. (III)

    if d≱ d ′ , assume T is uniformly continuous from (X,d) into (X, d ′ );

  4. (IV)

    if d= d ′ , assume X satisfies the property (H) with respect to the metric d;

  5. (V)

    if d≠ d ′ , assume T is continuous from (X, d ′ ) into (X, d ′ ).

Then T has a fixed point.

Proof We take r>0 such that d( x 0 ,T x 0 )<(1−h)r. From Theorem 2.1, T has a fixed point in B ( x 0 , r ) ¯ d ′ . □

Taking d= d ′ in Theorem 2.3, we obtain the following result.

Theorem 2.4 Let (X,d) be a complete metric space, T:X→X, and α:X×X→[0,∞). Suppose there exists F∈F such that for x,y∈X, we have

F ( α ( x , y ) d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    there exists x 0 ∈X such that α( x 0 ,T x 0 )≥1;

  2. (II)

    T is α-admissible (x,y∈X, α(x,y)≥1⟹α(Tx,Ty)≥1);

  3. (III)

    X satisfies the property (H) with respect to the metric d.

Then T has a fixed point.

3 Consequences

We present here some interesting consequences that can be obtained from our main results.

3.1 The case α(x,y)=1

Taking α(x,y):=1 for all x,y∈X, from Theorems 2.1, 2.2, 2.3, and 2.4, we obtain the following results that are generalizations of the fixed-point results in [2, 3, 5, 8, 10, 12, 13].

Corollary 3.1 Let (X, d ′ ) be a complete metric space, d another metric on X, x 0 ∈X, r>0, and T: B ( x 0 , r ) ¯ d ′ →X. Suppose there exists F∈F such that for x,y∈ B ( x 0 , r ) ¯ d ′ , we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    d( x 0 ,T x 0 )<(1−h)r;

  2. (II)

    if d≱ d ′ , assume T is uniformly continuous from (B( x 0 ,r),d) into (X, d ′ );

  3. (III)

    if d≠ d ′ , assume T is continuous from ( B ( x 0 , r ) ¯ d ′ , d ′ ) into (X, d ′ ).

Then T has a fixed point.

Corollary 3.2 Let (X,d) be a complete metric space, x 0 ∈X, r>0, and T: B ( x 0 , r ) ¯ d →X. Suppose there exists F∈F such that for x,y∈ B ( x 0 , r ) ¯ d , we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume that d( x 0 ,T x 0 )<(1−h)r. Then T has a fixed point.

Corollary 3.3 Let (X, d ′ ) be a complete metric space, d another metric on X, and T:X→X. Suppose there exists F∈F such that for x,y∈X, we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    if d≱ d ′ , assume T is uniformly continuous from (X,d) into (X, d ′ );

  2. (II)

    if d≠ d ′ , assume T is continuous from (X, d ′ ) into (X, d ′ ).

Then T has a fixed point.

Corollary 3.4 Let (X,d) be a complete metric space and T:X→X. Suppose there exists F∈F such that for x,y∈X, we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

Then T has a fixed point.

Corollary 3.4 is an enriched version of Popa [8] that unifies the most important metrical fixed-point theorems for contractive mappings in Rhoades’ classification [6].

3.2 The case of a partial ordered set

Let ⪯ be a partial order on X. Let ⊲ be the binary relation on X defined by

(x,y)∈X×X,x⊲y⟺x⪯yory⪯x.

We say that (X,⊲) satisfies the property (H) with respect to the metric d if the following condition holds:

If lim n → ∞ d( x n ,x)=0 for some x∈X and x n ⊲ x n + 1 for all n, then there exist a positive integer κ and a subsequence { x n ( k ) } of { x n } such that x n ( k ) ⊲x for all k≥κ.

From Theorems 2.1, 2.2, 2.3, and 2.4, we obtain the following results that are extensions and generalizations of the fixed-point results in [14, 15].

At first, we denote by F ˜ the set of functions F: [ 0 , + ∞ ) 6 →R satisfying the following conditions:

  1. (j)

    F∈F;

(jj) F(0, t 2 , t 3 , t 4 , t 5 , t 6 )≤0 for all t i ≥0, i=2,…,6.

We start with the following fixed-point result.

Corollary 3.5 Let (X, d ′ ) be a complete metric space, d another metric on X, x 0 ∈X, r>0, and T: B ( x 0 , r ) ¯ d ′ →X. Suppose there exists F∈ F ˜ such that for x,y∈ B ( x 0 , r ) ¯ d ′ with x⊲y, we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    d( x 0 ,T x 0 )<(1−h)r and x 0 ⊲T x 0 ;

  2. (II)

    x,y∈ B ( x 0 , r ) ¯ d ′ , x⊲y⟹Tx⊲Ty;

  3. (III)

    if d≱ d ′ , assume T is uniformly continuous from (B( x 0 ,r),d) into (X, d ′ );

  4. (IV)

    if d= d ′ , assume (X,⊲) satisfies the property (H) with respect to the metric d;

  5. (V)

    if d≠ d ′ , assume T is continuous from ( B ( x 0 , r ) ¯ d ′ , d ′ ) into (X, d ′ ).

Then T has a fixed point.

Proof It follows from Theorem 2.1 by taking

α(x,y):= { 1 if  x ⊲ y ; 0 if  x ⋪ y .

 □

Similarly, from Theorem 2.2, we obtain the following result.

Corollary 3.6 Let (X,d) be a complete metric space, x 0 ∈X, r>0, and T: B ( x 0 , r ) ¯ d →X. Suppose there exists F∈ F ˜ such that for x,y∈ B ( x 0 , r ) ¯ d with x⊲y, we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    d( x 0 ,T x 0 )<(1−h)r and x 0 ⊲T x 0 ;

  2. (II)

    x,y∈ B ( x 0 , r ) ¯ d ′ , x⊲y⟹Tx⊲Ty;

  3. (III)

    (X,⊲) satisfies the property (H) with respect to the metric d;

Then T has a fixed point.

From Theorem 2.3, we obtain the following global result.

Corollary 3.7 Let (X, d ′ ) be a complete metric space, d another metric on X, and T:X→X. Suppose there exists F∈ F ˜ such that for x,y∈X with x⊲y, we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    there exists x 0 ∈X such that x 0 ⊲T x 0 ;

  2. (II)

    x,y∈X, x⊲y⟹Tx⊲Ty;

  3. (III)

    if d≱ d ′ , assume T is uniformly continuous from (X,d) into (X, d ′ );

  4. (IV)

    if d= d ′ , assume (X,⊲) satisfies the property (H) with respect to the metric d;

  5. (V)

    if d≠ d ′ , assume T is continuous from (X, d ′ ) into (X, d ′ ).

Then T has a fixed point.

Finally, from Theorem 2.4, we obtain the following fixed-point result.

Corollary 3.8 Let (X,d) be a complete metric space and T:X→X. Suppose there exists F∈ F ˜ such that for x,y∈X with x⊲y, we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume the following properties hold:

  1. (I)

    there exists x 0 ∈X such that x 0 ⊲T x 0 ;

  2. (II)

    x,y∈X, x⊲y⟹Tx⊲Ty;

  3. (III)

    (X,⊲) satisfies the property (H) with respect to the metric d.

Then T has a fixed point.

3.3 The case of cyclic mappings

From Theorem 2.4, we obtain the following fixed-point result that is a generalization of Theorem 1.1 in [16].

Corollary 3.9 Let (Y,d) be a complete metric space, {A,B} a pair of nonempty closed subsets of Y, and T:A∪B→A∪B. Suppose there exists F∈ F ˜ such that for x∈A, y∈B, we have

F ( d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

In addition, assume that T(A)⊆B and T(B)⊆A.

Then T has a fixed point in A∩B.

Proof Let X:=A∪B. Clearly (since A and B are closed), (X,d) is a complete metric space. Define α:X×X→[0,∞) by

α(x,y):= { 1 if  ( x , y ) ∈ ( A × B ) ∪ ( B × A ) ; 0 if  ( x , y ) ∉ ( A × B ) ∪ ( B × A ) .

Clearly (since F∈ F ˜ ), for all x,y∈X, we have

F ( α ( x , y ) d ( T x , T y ) , d ( x , y ) , d ( x , T x ) , d ( y , T y ) , d ( x , T y ) , d ( y , T x ) ) ≤0.

Taking any point x 0 ∈A, since T(A)⊆B, we have T x 0 ∈B, which implies that α( x 0 ,T x 0 )≥1.

Now, let (x,y)∈X×X such that α(x,y)≥1. We have two cases.

Case 1. (x,y)∈A×B.

Since T(A)⊆B and T(B)⊆A, we have (Tx,Ty)∈B×A, which implies that α(Tx,Ty)≥1.

Case 2. (x,y)∈B×A.

In this case, we have (Tx,Ty)∈A×B, which implies that α(Tx,Ty)≥1.

Then T is α-admissible.

Finally, we shall prove that X satisfies the property (H) with respect to the metric d.

Let { x n } be a sequence in X such that lim n → ∞ d( x n ,x)=0 for some x∈X, and α( x n , x n + 1 )≥1 for all n. From the definition of α, this implies that ( x n , x n + 1 )∈(A×B)∪(B×A) for all n. Since A and B are closed, we get x∈A∩B. Then we have α( x n ,x)=1 for all n. Thus, we proved that X satisfies the property (H) with respect to the metric d.

Now, from Theorem 2.4, T has a fixed point in X, that is, there exists z∈A∪B such that Tz=z. Since T(A)⊆B and T(B)⊆A, obviously, we have z∈A∩B. □

Author’s contributions

The author read and approved the final manuscript.

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Acknowledgements

This project was supported by King Saud University, Deanship of Scientific Research, College of Science Research Center.

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Samet, B. Fixed point results for implicit contractions on spaces with two metrics. J Inequal Appl 2014, 84 (2014). https://doi.org/10.1186/1029-242X-2014-84

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