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 $F:{[0,+\mathrm{\infty })}^6\to \mathbb{R}$ satisfying the following conditions:

• (i) F is continuous;

• (ii) F is non-decreasing in the first variable;

• (iii) F is non-increasing in the fifth variable;

• (iv) $\mathrm{\exists }h\in (0,1)\mid F(u,v,v,u,u+v,0)\le 0⟹u\le hv$.

Example Let $F:{[0,+\mathrm{\infty })}^6\to \mathbb{R}$ be the function defined by

$$F(t_1,t_2,t_3,t_4,t_5,t_6):=t_1-qmax\{t_2,t_3,t_4,\frac{t_5+t_6}{2}\},$$

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

$$B(x_0,r):=\{x\in X:d(x_0,x)<r\}.$$

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 α-admissible (see [11]) if the following condition holds: for all $x,y\in B(x_0,r)$, we have

$$\alpha (x,y)\ge 1⟹\alpha (Tx,Ty)\ge 1.$$

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

$$F(\alpha (x,y)d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

(1)

In addition, assume the following properties hold:

• (I) $d(x_0,T{x}_0)<(1-h)r$ and $\alpha (x_0,T{x}_0)\ge 1$;

• (II) T is α-admissible;

• (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

$$d(x_0,x_1)=d(x_0,T{x}_0)\le (1-h)r<r,$$

which implies that $x_1\in B(x_0,r)$. Let $x_2=T{x}_1$. From (1), we have

$$F(\alpha (x_0,x_1)d(T{x}_0,T{x}_1),d(x_0,x_1),d(x_0,x_1),d(x_1,x_2),d(x_0,x_2),0)\le 0.$$

From (I), we have

$$d(T{x}_0,T{x}_1)\le \alpha (x_0,x_1)d(T{x}_0,T{x}_1).$$

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

$$F(d(x_1,x_2),d(x_0,x_1),d(x_0,x_1),d(x_1,x_2),d(x_0,x_2),0)\le 0.$$

Since $d(x_0,x_2)\le d(x_0,x_1)+d(x_1,x_2)$, using (iii), we obtain

$$F(d(x_1,x_2),d(x_0,x_1),d(x_0,x_1),d(x_1,x_2),d(x_0,x_1)+d(x_1,x_2),0)\le 0,$$

which implies from (iv) that

$$d(x_1,x_2)\le hd(x_0,x_1)\le h(1-h)r<r.$$

Now, we have

$$d(x_0,x_2)\le d(x_0,x_1)+hd(x_0,x_1)=(1+h)d(x_0,x_1)\le (1+h)(1-h)r<r.$$

This implies that $x_2\in B(x_0,r)$. Again, let $x_3=T{x}_2$. Since T is α-admissible and $\alpha (x_0,x_1)\ge 1$, we have

$$d(x_2,x_3)\le \alpha (x_1,x_2)d(T{x}_1,T{x}_2).$$

Then, from (1), we obtain

$$F(d(x_2,x_3),d(x_1,x_2),d(x_1,x_2),d(x_2,x_3),d(x_1,x_3),0)\le 0.$$

Using (iii), we obtain

$$F(d(x_2,x_3),d(x_1,x_2),d(x_1,x_2),d(x_2,x_3),d(x_1,x_2)+d(x_2,x_3),0)\le 0,$$

which implies from (iv) that

$$d(x_2,x_3)\le hd(x_1,x_2)\le h^2(1-h)r<r.$$

Now, we have

$$d(x_0,x_3)\le d(x_0,x_2)+d(x_2,x_3)\le (1+h)(1-h)r+h^2(1-h)r=(1-h^3)r<r.$$

This implies that $x_3\in B(x_0,r)$. Continuing this process, by induction, we can define the sequence $\{x_n\}$ by

$$x_{n+1}=T{x}_n,\mathrm{\forall }n\in \mathbb{N}.$$

Such sequence satisfies the following property:

$$x_n\in B(x_0,r),\alpha (x_n,x_{n+1})\ge 1\text{and}d(x_n,x_{n+1})\le h^n(1-h)r,\mathrm{\forall }n\in \mathbb{N}.$$

(2)

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

$$x,y\in B(x_0,r),d(x,y)<\delta ⟹d^{\prime }(Tx,Ty)<\epsilon .$$

(3)

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

$$d(x_n,x_m)<\delta ,\mathrm{\forall }n,m\ge N.$$

Using (3), we have

$$d^{\prime }(x_{n+1},x_{m+1})<\epsilon ,\mathrm{\forall }n,m\ge N.$$

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

$$\underset{n\to \mathrm{\infty }}{lim}{d}^{\prime }(x_n,z)=0.$$

(4)

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

Case 1. If $d=d^{\prime }$.

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

$$\alpha (x_{n(k)},z)\ge 1,\mathrm{\forall }k\ge \kappa .$$

(5)

Using (1), for all $k\ge \kappa$, we obtain

$$\begin{array}{c}F(\alpha (x_{n(k)},z)d(T{x}_{n(k)},Tz),d(x_{n(k)},z),d(x_{n(k)},x_{n(k)+1}),d(z,Tz),d(x_{n(k)},Tz),d(z,x_{n(k)+1}))\hfill \\ \le 0.\hfill \end{array}$$

Using (5) and condition (ii), for all $k\ge \kappa$, we obtain

$$F(d(x_{n(k)+1},Tz),d(x_{n(k)},z),d(x_{n(k)},x_{n(k)+1}),d(z,Tz),d(x_{n(k)},Tz),d(z,x_{n(k)+1}))\le 0.$$

Letting $k\to \mathrm{\infty }$, using (4) and the continuity of F, we obtain

$$F(d(z,Tz),0,0,d(z,Tz),d(z,Tz),0)\le 0,$$

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

$$\underset{n\to \mathrm{\infty }}{lim}{d}^{\prime }(T{x}_n,Tz)=\underset{n\to \mathrm{\infty }}{lim}{d}^{\prime }(x_{n+1},Tz)=0.$$

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

$$F(\alpha (x,y)d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

In addition, assume the following properties hold:

• (I) $d(x_0,T{x}_0)<(1-h)r$ and $\alpha (x_0,T{x}_0)\ge 1$;

• (II) T is α-admissible;

• (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

$$F(\alpha (x,y)d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

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 α-admissible ($x,y\in X$, $\alpha (x,y)\ge 1⟹\alpha (Tx,Ty)\ge 1$);

• (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 $d(x_0,T{x}_0)<(1-h)r$. 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

$$F(\alpha (x,y)d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

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 α-admissible ($x,y\in X$, $\alpha (x,y)\ge 1⟹\alpha (Tx,Ty)\ge 1$);

• (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 fixed-point 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

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

In addition, assume the following properties hold:

• (I) $d(x_0,T{x}_0)<(1-h)r$;

• (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

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

In addition, assume that $d(x_0,T{x}_0)<(1-h)r$. Then T has a fixed point.

Corollary 3.3 Let $(X,d^{\prime })$ 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

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

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

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

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

$$(x,y)\in X\times X,x⊲y⟺x⪯y\text{or}y⪯x.$$

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 $x_n⊲x_{n+1}$ 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 $k\ge \kappa$.

From Theorems 2.1, 2.2, 2.3, and 2.4, we obtain the following results that are extensions and generalizations of the fixed-point 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 fixed-point 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 $x⊲y$, we have

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

In addition, assume the following properties hold:

• (I) $d(x_0,T{x}_0)<(1-h)r$ and $x_0⊲T{x}_0$;

• (II) $x,y\in {\overline{B(x_0,r)}}^{d^{\prime }}$, $x⊲y⟹Tx⊲Ty$;

• (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 It follows from Theorem 2.1 by taking

$$\alpha (x,y):=\{\begin{array}{ll}1& \text{if }x⊲y;\\ 0& \text{if }x⋪y.\end{array}$$

 □

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 $x⊲y$, we have

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

In addition, assume the following properties hold:

• (I) $d(x_0,T{x}_0)<(1-h)r$ and $x_0⊲T{x}_0$;

• (II) $x,y\in {\overline{B(x_0,r)}}^{d^{\prime }}$, $x⊲y⟹Tx⊲Ty$;

• (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 $(X,d^{\prime })$ 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 $x⊲y$, we have

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

In addition, assume the following properties hold:

• (I) there exists $x_0\in X$ such that $x_0⊲T{x}_0$;

• (II) $x,y\in X$, $x⊲y⟹Tx⊲Ty$;

• (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.

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

Corollary 3.8 Let $(X,d)$ be a complete metric space and $T:X\to X$. Suppose there exists $F\in \tilde{\mathcal{F}}$ such that for $x,y\in X$ with $x⊲y$, we have

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

In addition, assume the following properties hold:

• (I) there exists $x_0\in X$ such that $x_0⊲T{x}_0$;

• (II) $x,y\in X$, $x⊲y⟹Tx⊲Ty$;

• (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 $T:A\cup B\to A\cup B$. Suppose there exists $F\in \tilde{\mathcal{F}}$ such that for $x\in A$, $y\in B$, we have

$$F(d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

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

$$\alpha (x,y):=\{\begin{array}{ll}1& \text{if }(x,y)\in (A\times B)\cup (B\times A);\\ 0& \text{if }(x,y)\notin (A\times B)\cup (B\times A).\end{array}$$

Clearly (since $F\in \tilde{\mathcal{F}}$), for all $x,y\in X$, we have

$$F(\alpha (x,y)d(Tx,Ty),d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx))\le 0.$$

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 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 ${lim}_{n\to \mathrm{\infty }}d(x_n,x)=0$ 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.