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Fixed point results for implicit contractions on spaces with two metrics
Journal of Inequalities and Applications volume 2014, Article number: 84 (2014)
Abstract
We establish new fixedpoint results involving implicit contractions on a metric space endowed with two metrics. The main results in this paper extend and generalize several existing fixedpoint theorems in the literature.
1 Introduction
Fixedpoint 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 fixedpoint 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 fixedpoint 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 fixedpoint results for implicit contractions involving αadmissible 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,+\mathrm{\infty})}^{6}\to \mathbb{R} satisfying the following conditions:

(i)
F is continuous;

(ii)
F is nondecreasing in the first variable;

(iii)
F is nonincreasing in the fifth variable;

(iv)
\mathrm{\exists}h\in (0,1)\mid F(u,v,v,u,u+v,0)\le 0\u27f9u\le hv.
Example Let F:{[0,+\mathrm{\infty})}^{6}\to \mathbb{R} be the function defined by
where q\in (0,1). We can check easily that F\in \mathcal{F}.
Let X be a nonempty set endowed with two metrics d and {d}^{\prime}. If {x}_{0}\in X and r>0, let
We denote by {\overline{B({x}_{0},r)}}^{{d}^{\prime}} the {d}^{\prime}closure of B({x}_{0},r).
Let T:{\overline{B({x}_{0},r)}}^{{d}^{\prime}}\to X and \alpha :X\times X\to [0,\mathrm{\infty}). We say that T is αadmissible (see [11]) if the following condition holds: for all x,y\in B({x}_{0},r), we have
We say that X satisfies the property (H) with respect to the metric d if the following condition holds:
If {lim}_{n\to \mathrm{\infty}}d({x}_{n},x)=0 for some x\in X and \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n, then there exist a positive integer κ and a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that \alpha ({x}_{n(k)},x)\ge 1 for all k\ge \kappa.
Our first result is the following.
Theorem 2.1 Let (X,{d}^{\prime}) be a complete metric space, d another metric on X, {x}_{0}\in X, r>0, T:{\overline{B({x}_{0},r)}}^{{d}^{\prime}}\to X, and \alpha :X\times X\to [0,\mathrm{\infty}). Suppose there exists F\in \mathcal{F} such that for x,y\in {\overline{B({x}_{0},r)}}^{{d}^{\prime}}, we have
In addition, assume the following properties hold:

(I)
d({x}_{0},T{x}_{0})<(1h)r and \alpha ({x}_{0},T{x}_{0})\ge 1;

(II)
T is αadmissible;

(III)
if d\ngeqq {d}^{\prime}, assume T is uniformly continuous from (B({x}_{0},r),d) into (X,{d}^{\prime});

(IV)
if d={d}^{\prime}, assume X satisfies the property (H) with respect to the metric d;

(V)
if d\ne {d}^{\prime}, assume T is continuous from ({\overline{B({x}_{0},r)}}^{{d}^{\prime}},{d}^{\prime}) into (X,{d}^{\prime}).
Then T has a fixed point.
Proof Let {x}_{1}=T{x}_{0}. From (I), we have
which implies that {x}_{1}\in B({x}_{0},r). Let {x}_{2}=T{x}_{1}. From (1), we have
From (I), we have
Since F is nondecreasing in the first variable (property (i)), we obtain
Since d({x}_{0},{x}_{2})\le d({x}_{0},{x}_{1})+d({x}_{1},{x}_{2}), using (iii), we obtain
which implies from (iv) that
Now, we have
This implies that {x}_{2}\in B({x}_{0},r). Again, let {x}_{3}=T{x}_{2}. Since T is αadmissible and \alpha ({x}_{0},{x}_{1})\ge 1, we have
Then, from (1), we obtain
Using (iii), we obtain
which implies from (iv) that
Now, we have
This implies that {x}_{3}\in B({x}_{0},r). Continuing this process, by induction, we can define the sequence \{{x}_{n}\} by
Such sequence satisfies the following property:
Since h\in (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}^{\prime}. If {d}^{\prime}\le d, the result follows immediately from (2). If d\ngeqq {d}^{\prime}, from (III), given \epsilon >0, there exists δ> such that
On the other hand, since \{{x}_{n}\} is Cauchy with respect to d, there exists a positive integer N such that
Using (3), we have
Thus we proved that \{{x}_{n}\} is Cauchy with respect to {d}^{\prime}.
Since (X,{d}^{\prime}) is complete, there exists z\in {\overline{B({x}_{0},r)}}^{{d}^{\prime}} such that
We shall prove that z is a fixed point of T. We consider two cases.
Case 1. If d={d}^{\prime}.
From (IV), there exist a positive integer κ and a subsequence \{{x}_{n(k)}\} of \{{x}_{n}\} such that
Using (1), for all k\ge \kappa, we obtain
Using (5) and condition (ii), for all k\ge \kappa, we obtain
Letting k\to \mathrm{\infty}, using (4) and the continuity of F, we obtain
which implies from (iv) that d(z,Tz)=0.
Case 2. If d\ne {d}^{\prime}.
In this case, using (V) and (4), we obtain
The uniqueness of the limit gives z=Tz. □
Taking d={d}^{\prime} in Theorem 2.1, we obtain the following result.
Theorem 2.2 Let (X,d) be a complete metric space, {x}_{0}\in X, r>0, T:{\overline{B({x}_{0},r)}}^{d}\to X, and \alpha :X\times X\to [0,\mathrm{\infty}). Suppose there exists F\in \mathcal{F} such that for x,y\in {\overline{B({x}_{0},r)}}^{d}, we have
In addition, assume the following properties hold:

(I)
d({x}_{0},T{x}_{0})<(1h)r and \alpha ({x}_{0},T{x}_{0})\ge 1;

(II)
T is αadmissible;

(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}^{\prime}) be a complete metric space, d another metric on X, T:X\to X, and \alpha :X\times X\to [0,\mathrm{\infty}). Suppose there exists F\in \mathcal{F} such that for x,y\in X, we have
In addition, assume the following properties hold:

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

(II)
T is αadmissible (x,y\in X, \alpha (x,y)\ge 1\u27f9\alpha (Tx,Ty)\ge 1);

(III)
if d\ngeqq {d}^{\prime}, assume T is uniformly continuous from (X,d) into (X,{d}^{\prime});

(IV)
if d={d}^{\prime}, assume X satisfies the property (H) with respect to the metric d;

(V)
if d\ne {d}^{\prime}, assume T is continuous from (X,{d}^{\prime}) into (X,{d}^{\prime}).
Then T has a fixed point.
Proof We take r>0 such that d({x}_{0},T{x}_{0})<(1h)r. From Theorem 2.1, T has a fixed point in {\overline{B({x}_{0},r)}}^{{d}^{\prime}}. □
Taking d={d}^{\prime} in Theorem 2.3, we obtain the following result.
Theorem 2.4 Let (X,d) be a complete metric space, T:X\to X, and \alpha :X\times X\to [0,\mathrm{\infty}). Suppose there exists F\in \mathcal{F} such that for x,y\in X, we have
In addition, assume the following properties hold:

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

(II)
T is αadmissible (x,y\in X, \alpha (x,y)\ge 1\u27f9\alpha (Tx,Ty)\ge 1);

(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 \alpha (x,y)=1
Taking \alpha (x,y):=1 for all x,y\in X, from Theorems 2.1, 2.2, 2.3, and 2.4, we obtain the following results that are generalizations of the fixedpoint results in [2, 3, 5, 8, 10, 12, 13].
Corollary 3.1 Let (X,{d}^{\prime}) be a complete metric space, d another metric on X, {x}_{0}\in X, r>0, and T:{\overline{B({x}_{0},r)}}^{{d}^{\prime}}\to X. Suppose there exists F\in \mathcal{F} such that for x,y\in {\overline{B({x}_{0},r)}}^{{d}^{\prime}}, we have
In addition, assume the following properties hold:

(I)
d({x}_{0},T{x}_{0})<(1h)r;

(II)
if d\ngeqq {d}^{\prime}, assume T is uniformly continuous from (B({x}_{0},r),d) into (X,{d}^{\prime});

(III)
if d\ne {d}^{\prime}, assume T is continuous from ({\overline{B({x}_{0},r)}}^{{d}^{\prime}},{d}^{\prime}) into (X,{d}^{\prime}).
Then T has a fixed point.
Corollary 3.2 Let (X,d) be a complete metric space, {x}_{0}\in X, r>0, and T:{\overline{B({x}_{0},r)}}^{d}\to X. Suppose there exists F\in \mathcal{F} such that for x,y\in {\overline{B({x}_{0},r)}}^{d}, we have
In addition, assume that d({x}_{0},T{x}_{0})<(1h)r. Then T has a fixed point.
Corollary 3.3 Let (X,{d}^{\prime}) be a complete metric space, d another metric on X, and T:X\to X. Suppose there exists F\in \mathcal{F} such that for x,y\in X, we have
In addition, assume the following properties hold:

(I)
if d\ngeqq {d}^{\prime}, assume T is uniformly continuous from (X,d) into (X,{d}^{\prime});

(II)
if d\ne {d}^{\prime}, assume T is continuous from (X,{d}^{\prime}) into (X,{d}^{\prime}).
Then T has a fixed point.
Corollary 3.4 Let (X,d) be a complete metric space and T:X\to X. Suppose there exists F\in \mathcal{F} such that for x,y\in X, we have
Then T has a fixed point.
Corollary 3.4 is an enriched version of Popa [8] that unifies the most important metrical fixedpoint 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
We say that (X,\u22b2) satisfies the property (H) with respect to the metric d if the following condition holds:
If {lim}_{n\to \mathrm{\infty}}d({x}_{n},x)=0 for some x\in X and {x}_{n}\u22b2{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)}\u22b2x for all k\ge \kappa.
From Theorems 2.1, 2.2, 2.3, and 2.4, we obtain the following results that are extensions and generalizations of the fixedpoint results in [14, 15].
At first, we denote by \tilde{\mathcal{F}} the set of functions F:{[0,+\mathrm{\infty})}^{6}\to \mathbb{R} satisfying the following conditions:

(j)
F\in \mathcal{F};
(jj) F(0,{t}_{2},{t}_{3},{t}_{4},{t}_{5},{t}_{6})\le 0 for all {t}_{i}\ge 0, i=2,\dots ,6.
We start with the following fixedpoint result.
Corollary 3.5 Let (X,{d}^{\prime}) be a complete metric space, d another metric on X, {x}_{0}\in X, r>0, and T:{\overline{B({x}_{0},r)}}^{{d}^{\prime}}\to X. Suppose there exists F\in \tilde{\mathcal{F}} such that for x,y\in {\overline{B({x}_{0},r)}}^{{d}^{\prime}} with x\u22b2y, we have
In addition, assume the following properties hold:

(I)
d({x}_{0},T{x}_{0})<(1h)r and {x}_{0}\u22b2T{x}_{0};

(II)
x,y\in {\overline{B({x}_{0},r)}}^{{d}^{\prime}}, x\u22b2y\u27f9Tx\u22b2Ty;

(III)
if d\ngeqq {d}^{\prime}, assume T is uniformly continuous from (B({x}_{0},r),d) into (X,{d}^{\prime});

(IV)
if d={d}^{\prime}, assume (X,\u22b2) satisfies the property (H) with respect to the metric d;

(V)
if d\ne {d}^{\prime}, assume T is continuous from ({\overline{B({x}_{0},r)}}^{{d}^{\prime}},{d}^{\prime}) into (X,{d}^{\prime}).
Then T has a fixed point.
Proof It follows from Theorem 2.1 by taking
□
Similarly, from Theorem 2.2, we obtain the following result.
Corollary 3.6 Let (X,d) be a complete metric space, {x}_{0}\in X, r>0, and T:{\overline{B({x}_{0},r)}}^{d}\to X. Suppose there exists F\in \tilde{\mathcal{F}} such that for x,y\in {\overline{B({x}_{0},r)}}^{d} with x\u22b2y, we have
In addition, assume the following properties hold:

(I)
d({x}_{0},T{x}_{0})<(1h)r and {x}_{0}\u22b2T{x}_{0};

(II)
x,y\in {\overline{B({x}_{0},r)}}^{{d}^{\prime}}, x\u22b2y\u27f9Tx\u22b2Ty;

(III)
(X,\u22b2) 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}^{\prime}) be a complete metric space, d another metric on X, and T:X\to X. Suppose there exists F\in \tilde{\mathcal{F}} such that for x,y\in X with x\u22b2y, we have
In addition, assume the following properties hold:

(I)
there exists {x}_{0}\in X such that {x}_{0}\u22b2T{x}_{0};

(II)
x,y\in X, x\u22b2y\u27f9Tx\u22b2Ty;

(III)
if d\ngeqq {d}^{\prime}, assume T is uniformly continuous from (X,d) into (X,{d}^{\prime});

(IV)
if d={d}^{\prime}, assume (X,\u22b2) satisfies the property (H) with respect to the metric d;

(V)
if d\ne {d}^{\prime}, assume T is continuous from (X,{d}^{\prime}) into (X,{d}^{\prime}).
Then T has a fixed point.
Finally, from Theorem 2.4, we obtain the following fixedpoint result.
Corollary 3.8 Let (X,d) be a complete metric space and T:X\to X. Suppose there exists F\in \tilde{\mathcal{F}} such that for x,y\in X with x\u22b2y, we have
In addition, assume the following properties hold:

(I)
there exists {x}_{0}\in X such that {x}_{0}\u22b2T{x}_{0};

(II)
x,y\in X, x\u22b2y\u27f9Tx\u22b2Ty;

(III)
(X,\u22b2) 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 fixedpoint 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\cup B\to A\cup B. Suppose there exists F\in \tilde{\mathcal{F}} such that for x\in A, y\in B, we have
In addition, assume that T(A)\subseteq B and T(B)\subseteq A.
Then T has a fixed point in A\cap B.
Proof Let X:=A\cup B. Clearly (since A and B are closed), (X,d) is a complete metric space. Define \alpha :X\times X\to [0,\mathrm{\infty}) by
Clearly (since F\in \tilde{\mathcal{F}}), for all x,y\in X, we have
Taking any point {x}_{0}\in A, since T(A)\subseteq B, we have T{x}_{0}\in B, which implies that \alpha ({x}_{0},T{x}_{0})\ge 1.
Now, let (x,y)\in X\times X such that \alpha (x,y)\ge 1. We have two cases.
Case 1. (x,y)\in A\times B.
Since T(A)\subseteq B and T(B)\subseteq A, we have (Tx,Ty)\in B\times A, which implies that \alpha (Tx,Ty)\ge 1.
Case 2. (x,y)\in B\times A.
In this case, we have (Tx,Ty)\in A\times B, which implies that \alpha (Tx,Ty)\ge 1.
Then T is αadmissible.
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\to \mathrm{\infty}}d({x}_{n},x)=0 for some x\in X, and \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n. From the definition of α, this implies that ({x}_{n},{x}_{n+1})\in (A\times B)\cup (B\times A) for all n. Since A and B are closed, we get x\in A\cap B. Then we have \alpha ({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\in A\cup B such that Tz=z. Since T(A)\subseteq B and T(B)\subseteq A, obviously, we have z\in A\cap 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/10.1186/1029242X201484
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DOI: https://doi.org/10.1186/1029242X201484
Keywords
 fixed point
 implicit contraction
 two metrics