# 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  is one of the most important theorems in this direction. Many generalizations of this famous principle exist in the literature, see, for examples,  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 , Popa [8, 9], Berinde , 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 . Section 3 presents some interesting consequences that can be obtained from the results established in the previous section.

## 2 Main results

Let be the set of functions $F:{\left[0,+\mathrm{\infty }\right)}^{6}\to \mathbb{R}$ 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)

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

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

$F\left({t}_{1},{t}_{2},{t}_{3},{t}_{4},{t}_{5},{t}_{6}\right):={t}_{1}-qmax\left\{{t}_{2},{t}_{3},{t}_{4},\frac{{t}_{5}+{t}_{6}}{2}\right\},$

where $q\in \left(0,1\right)$. 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\left({x}_{0},r\right):=\left\{x\in X:d\left({x}_{0},x\right)

We denote by ${\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}$ the ${d}^{\prime }$-closure of $B\left({x}_{0},r\right)$.

Let $T:{\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}\to X$ and $\alpha :X×X\to \left[0,\mathrm{\infty }\right)$. We say that T is α-admissible (see ) if the following condition holds: for all $x,y\in B\left({x}_{0},r\right)$, we have

$\alpha \left(x,y\right)\ge 1\phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}\alpha \left(Tx,Ty\right)\ge 1.$

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

If ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},x\right)=0$ for some $x\in X$ and $\alpha \left({x}_{n},{x}_{n+1}\right)\ge 1$ for all n, then there exist a positive integer κ and a subsequence $\left\{{x}_{n\left(k\right)}\right\}$ of $\left\{{x}_{n}\right\}$ such that $\alpha \left({x}_{n\left(k\right)},x\right)\ge 1$ for all $k\ge \kappa$.

Our first result is the following.

Theorem 2.1 Let $\left(X,{d}^{\prime }\right)$ be a complete metric space, d another metric on X, ${x}_{0}\in X$, $r>0$, $T:{\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}\to X$, and $\alpha :X×X\to \left[0,\mathrm{\infty }\right)$. Suppose there exists $F\in \mathcal{F}$ such that for $x,y\in {\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}$, we have

$F\left(\alpha \left(x,y\right)d\left(Tx,Ty\right),d\left(x,y\right),d\left(x,Tx\right),d\left(y,Ty\right),d\left(x,Ty\right),d\left(y,Tx\right)\right)\le 0.$
(1)

In addition, assume the following properties hold:

1. (I)

$d\left({x}_{0},T{x}_{0}\right)<\left(1-h\right)r$ and $\alpha \left({x}_{0},T{x}_{0}\right)\ge 1$;

2. (II)

3. (III)

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

4. (IV)

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

5. (V)

if $d\ne {d}^{\prime }$, assume T is continuous from $\left({\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }},{d}^{\prime }\right)$ into $\left(X,{d}^{\prime }\right)$.

Then T has a fixed point.

Proof Let ${x}_{1}=T{x}_{0}$. From (I), we have

$d\left({x}_{0},{x}_{1}\right)=d\left({x}_{0},T{x}_{0}\right)\le \left(1-h\right)r

which implies that ${x}_{1}\in B\left({x}_{0},r\right)$. Let ${x}_{2}=T{x}_{1}$. From (1), we have

$F\left(\alpha \left({x}_{0},{x}_{1}\right)d\left(T{x}_{0},T{x}_{1}\right),d\left({x}_{0},{x}_{1}\right),d\left({x}_{0},{x}_{1}\right),d\left({x}_{1},{x}_{2}\right),d\left({x}_{0},{x}_{2}\right),0\right)\le 0.$

From (I), we have

$d\left(T{x}_{0},T{x}_{1}\right)\le \alpha \left({x}_{0},{x}_{1}\right)d\left(T{x}_{0},T{x}_{1}\right).$

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

$F\left(d\left({x}_{1},{x}_{2}\right),d\left({x}_{0},{x}_{1}\right),d\left({x}_{0},{x}_{1}\right),d\left({x}_{1},{x}_{2}\right),d\left({x}_{0},{x}_{2}\right),0\right)\le 0.$

Since $d\left({x}_{0},{x}_{2}\right)\le d\left({x}_{0},{x}_{1}\right)+d\left({x}_{1},{x}_{2}\right)$, using (iii), we obtain

$F\left(d\left({x}_{1},{x}_{2}\right),d\left({x}_{0},{x}_{1}\right),d\left({x}_{0},{x}_{1}\right),d\left({x}_{1},{x}_{2}\right),d\left({x}_{0},{x}_{1}\right)+d\left({x}_{1},{x}_{2}\right),0\right)\le 0,$

which implies from (iv) that

$d\left({x}_{1},{x}_{2}\right)\le hd\left({x}_{0},{x}_{1}\right)\le h\left(1-h\right)r

Now, we have

$d\left({x}_{0},{x}_{2}\right)\le d\left({x}_{0},{x}_{1}\right)+hd\left({x}_{0},{x}_{1}\right)=\left(1+h\right)d\left({x}_{0},{x}_{1}\right)\le \left(1+h\right)\left(1-h\right)r

This implies that ${x}_{2}\in B\left({x}_{0},r\right)$. Again, let ${x}_{3}=T{x}_{2}$. Since T is α-admissible and $\alpha \left({x}_{0},{x}_{1}\right)\ge 1$, we have

$d\left({x}_{2},{x}_{3}\right)\le \alpha \left({x}_{1},{x}_{2}\right)d\left(T{x}_{1},T{x}_{2}\right).$

Then, from (1), we obtain

$F\left(d\left({x}_{2},{x}_{3}\right),d\left({x}_{1},{x}_{2}\right),d\left({x}_{1},{x}_{2}\right),d\left({x}_{2},{x}_{3}\right),d\left({x}_{1},{x}_{3}\right),0\right)\le 0.$

Using (iii), we obtain

$F\left(d\left({x}_{2},{x}_{3}\right),d\left({x}_{1},{x}_{2}\right),d\left({x}_{1},{x}_{2}\right),d\left({x}_{2},{x}_{3}\right),d\left({x}_{1},{x}_{2}\right)+d\left({x}_{2},{x}_{3}\right),0\right)\le 0,$

which implies from (iv) that

$d\left({x}_{2},{x}_{3}\right)\le hd\left({x}_{1},{x}_{2}\right)\le {h}^{2}\left(1-h\right)r

Now, we have

$d\left({x}_{0},{x}_{3}\right)\le d\left({x}_{0},{x}_{2}\right)+d\left({x}_{2},{x}_{3}\right)\le \left(1+h\right)\left(1-h\right)r+{h}^{2}\left(1-h\right)r=\left(1-{h}^{3}\right)r

This implies that ${x}_{3}\in B\left({x}_{0},r\right)$. Continuing this process, by induction, we can define the sequence $\left\{{x}_{n}\right\}$ by

${x}_{n+1}=T{x}_{n},\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.$

Such sequence satisfies the following property:

${x}_{n}\in B\left({x}_{0},r\right),\phantom{\rule{1em}{0ex}}\alpha \left({x}_{n},{x}_{n+1}\right)\ge 1\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}d\left({x}_{n},{x}_{n+1}\right)\le {h}^{n}\left(1-h\right)r,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n\in \mathbb{N}.$
(2)

Since $h\in \left(0,1\right)$, it follows from (2) that $\left\{{x}_{n}\right\}$ is a Cauchy sequence with respect to the metric d.

Now, we shall prove that $\left\{{x}_{n}\right\}$ 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

$x,y\in B\left({x}_{0},r\right),\phantom{\rule{1em}{0ex}}d\left(x,y\right)<\delta \phantom{\rule{1em}{0ex}}⟹\phantom{\rule{1em}{0ex}}{d}^{\prime }\left(Tx,Ty\right)<\epsilon .$
(3)

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

$d\left({x}_{n},{x}_{m}\right)<\delta ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n,m\ge N.$

Using (3), we have

${d}^{\prime }\left({x}_{n+1},{x}_{m+1}\right)<\epsilon ,\phantom{\rule{1em}{0ex}}\mathrm{\forall }n,m\ge N.$

Thus we proved that $\left\{{x}_{n}\right\}$ is Cauchy with respect to ${d}^{\prime }$.

Since $\left(X,{d}^{\prime }\right)$ is complete, there exists $z\in {\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}$ such that

$\underset{n\to \mathrm{\infty }}{lim}{d}^{\prime }\left({x}_{n},z\right)=0.$
(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 $\left\{{x}_{n\left(k\right)}\right\}$ of $\left\{{x}_{n}\right\}$ such that

$\alpha \left({x}_{n\left(k\right)},z\right)\ge 1,\phantom{\rule{1em}{0ex}}\mathrm{\forall }k\ge \kappa .$
(5)

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

$\begin{array}{c}F\left(\alpha \left({x}_{n\left(k\right)},z\right)d\left(T{x}_{n\left(k\right)},Tz\right),d\left({x}_{n\left(k\right)},z\right),d\left({x}_{n\left(k\right)},{x}_{n\left(k\right)+1}\right),d\left(z,Tz\right),d\left({x}_{n\left(k\right)},Tz\right),d\left(z,{x}_{n\left(k\right)+1}\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le 0.\hfill \end{array}$

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

$F\left(d\left({x}_{n\left(k\right)+1},Tz\right),d\left({x}_{n\left(k\right)},z\right),d\left({x}_{n\left(k\right)},{x}_{n\left(k\right)+1}\right),d\left(z,Tz\right),d\left({x}_{n\left(k\right)},Tz\right),d\left(z,{x}_{n\left(k\right)+1}\right)\right)\le 0.$

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

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

which implies from (iv) that $d\left(z,Tz\right)=0$.

Case 2. If $d\ne {d}^{\prime }$.

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

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

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 $\left(X,d\right)$ be a complete metric space, ${x}_{0}\in X$, $r>0$, $T:{\overline{B\left({x}_{0},r\right)}}^{d}\to X$, and $\alpha :X×X\to \left[0,\mathrm{\infty }\right)$. Suppose there exists $F\in \mathcal{F}$ such that for $x,y\in {\overline{B\left({x}_{0},r\right)}}^{d}$, we have

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

In addition, assume the following properties hold:

1. (I)

$d\left({x}_{0},T{x}_{0}\right)<\left(1-h\right)r$ and $\alpha \left({x}_{0},T{x}_{0}\right)\ge 1$;

2. (II)

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 $\left(X,{d}^{\prime }\right)$ be a complete metric space, d another metric on X, $T:X\to X$, and $\alpha :X×X\to \left[0,\mathrm{\infty }\right)$. Suppose there exists $F\in \mathcal{F}$ such that for $x,y\in X$, we have

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

In addition, assume the following properties hold:

1. (I)

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

2. (II)

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

3. (III)

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

4. (IV)

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

5. (V)

if $d\ne {d}^{\prime }$, assume T is continuous from $\left(X,{d}^{\prime }\right)$ into $\left(X,{d}^{\prime }\right)$.

Then T has a fixed point.

Proof We take $r>0$ such that $d\left({x}_{0},T{x}_{0}\right)<\left(1-h\right)r$. From Theorem 2.1, T has a fixed point in ${\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}$. □

Taking $d={d}^{\prime }$ in Theorem 2.3, we obtain the following result.

Theorem 2.4 Let $\left(X,d\right)$ be a complete metric space, $T:X\to X$, and $\alpha :X×X\to \left[0,\mathrm{\infty }\right)$. Suppose there exists $F\in \mathcal{F}$ such that for $x,y\in X$, we have

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

In addition, assume the following properties hold:

1. (I)

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

2. (II)

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

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 $\alpha \left(x,y\right)=1$

Taking $\alpha \left(x,y\right):=1$ 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 $\left(X,{d}^{\prime }\right)$ be a complete metric space, d another metric on X, ${x}_{0}\in X$, $r>0$, and $T:{\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}\to X$. Suppose there exists $F\in \mathcal{F}$ such that for $x,y\in {\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}$, we have

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

In addition, assume the following properties hold:

1. (I)

$d\left({x}_{0},T{x}_{0}\right)<\left(1-h\right)r$;

2. (II)

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

3. (III)

if $d\ne {d}^{\prime }$, assume T is continuous from $\left({\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }},{d}^{\prime }\right)$ into $\left(X,{d}^{\prime }\right)$.

Then T has a fixed point.

Corollary 3.2 Let $\left(X,d\right)$ be a complete metric space, ${x}_{0}\in X$, $r>0$, and $T:{\overline{B\left({x}_{0},r\right)}}^{d}\to X$. Suppose there exists $F\in \mathcal{F}$ such that for $x,y\in {\overline{B\left({x}_{0},r\right)}}^{d}$, we have

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

In addition, assume that $d\left({x}_{0},T{x}_{0}\right)<\left(1-h\right)r$. Then T has a fixed point.

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

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

In addition, assume the following properties hold:

1. (I)

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

2. (II)

if $d\ne {d}^{\prime }$, assume T is continuous from $\left(X,{d}^{\prime }\right)$ into $\left(X,{d}^{\prime }\right)$.

Then T has a fixed point.

Corollary 3.4 Let $\left(X,d\right)$ 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

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

Then T has a fixed point.

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

### 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

$\left(x,y\right)\in X×X,\phantom{\rule{1em}{0ex}}x⊲y\phantom{\rule{1em}{0ex}}⟺\phantom{\rule{1em}{0ex}}x⪯y\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}y⪯x.$

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

If ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},x\right)=0$ for some $x\in X$ and ${x}_{n}⊲{x}_{n+1}$ for all n, then there exist a positive integer κ and a subsequence $\left\{{x}_{n\left(k\right)}\right\}$ of $\left\{{x}_{n}\right\}$ such that ${x}_{n\left(k\right)}⊲x$ 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 $\stackrel{˜}{\mathcal{F}}$ the set of functions $F:{\left[0,+\mathrm{\infty }\right)}^{6}\to \mathbb{R}$ satisfying the following conditions:

1. (j)

$F\in \mathcal{F}$;

(jj) $F\left(0,{t}_{2},{t}_{3},{t}_{4},{t}_{5},{t}_{6}\right)\le 0$ for all ${t}_{i}\ge 0$, $i=2,\dots ,6$.

Corollary 3.5 Let $\left(X,{d}^{\prime }\right)$ be a complete metric space, d another metric on X, ${x}_{0}\in X$, $r>0$, and $T:{\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}\to X$. Suppose there exists $F\in \stackrel{˜}{\mathcal{F}}$ such that for $x,y\in {\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}$ with $x⊲y$, we have

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

In addition, assume the following properties hold:

1. (I)

$d\left({x}_{0},T{x}_{0}\right)<\left(1-h\right)r$ and ${x}_{0}⊲T{x}_{0}$;

2. (II)

$x,y\in {\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}$, $x⊲y⟹Tx⊲Ty$;

3. (III)

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

4. (IV)

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

5. (V)

if $d\ne {d}^{\prime }$, assume T is continuous from $\left({\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }},{d}^{\prime }\right)$ into $\left(X,{d}^{\prime }\right)$.

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 $\left(X,d\right)$ be a complete metric space, ${x}_{0}\in X$, $r>0$, and $T:{\overline{B\left({x}_{0},r\right)}}^{d}\to X$. Suppose there exists $F\in \stackrel{˜}{\mathcal{F}}$ such that for $x,y\in {\overline{B\left({x}_{0},r\right)}}^{d}$ with $x⊲y$, we have

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

In addition, assume the following properties hold:

1. (I)

$d\left({x}_{0},T{x}_{0}\right)<\left(1-h\right)r$ and ${x}_{0}⊲T{x}_{0}$;

2. (II)

$x,y\in {\overline{B\left({x}_{0},r\right)}}^{{d}^{\prime }}$, $x⊲y⟹Tx⊲Ty$;

3. (III)

$\left(X,⊲\right)$ 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 $\left(X,{d}^{\prime }\right)$ be a complete metric space, d another metric on X, and $T:X\to X$. Suppose there exists $F\in \stackrel{˜}{\mathcal{F}}$ such that for $x,y\in X$ with $x⊲y$, we have

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

In addition, assume the following properties hold:

1. (I)

there exists ${x}_{0}\in X$ such that ${x}_{0}⊲T{x}_{0}$;

2. (II)

$x,y\in X$, $x⊲y⟹Tx⊲Ty$;

3. (III)

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

4. (IV)

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

5. (V)

if $d\ne {d}^{\prime }$, assume T is continuous from $\left(X,{d}^{\prime }\right)$ into $\left(X,{d}^{\prime }\right)$.

Then T has a fixed point.

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

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

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

In addition, assume the following properties hold:

1. (I)

there exists ${x}_{0}\in X$ such that ${x}_{0}⊲T{x}_{0}$;

2. (II)

$x,y\in X$, $x⊲y⟹Tx⊲Ty$;

3. (III)

$\left(X,⊲\right)$ 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 .

Corollary 3.9 Let $\left(Y,d\right)$ be a complete metric space, $\left\{A,B\right\}$ a pair of nonempty closed subsets of Y, and $T:A\cup B\to A\cup B$. Suppose there exists $F\in \stackrel{˜}{\mathcal{F}}$ such that for $x\in A$, $y\in B$, we have

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

In addition, assume that $T\left(A\right)\subseteq B$ and $T\left(B\right)\subseteq A$.

Then T has a fixed point in $A\cap B$.

Proof Let $X:=A\cup B$. Clearly (since A and B are closed), $\left(X,d\right)$ is a complete metric space. Define $\alpha :X×X\to \left[0,\mathrm{\infty }\right)$ by

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

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

Taking any point ${x}_{0}\in A$, since $T\left(A\right)\subseteq B$, we have $T{x}_{0}\in B$, which implies that $\alpha \left({x}_{0},T{x}_{0}\right)\ge 1$.

Now, let $\left(x,y\right)\in X×X$ such that $\alpha \left(x,y\right)\ge 1$. We have two cases.

Case 1. $\left(x,y\right)\in A×B$.

Since $T\left(A\right)\subseteq B$ and $T\left(B\right)\subseteq A$, we have $\left(Tx,Ty\right)\in B×A$, which implies that $\alpha \left(Tx,Ty\right)\ge 1$.

Case 2. $\left(x,y\right)\in B×A$.

In this case, we have $\left(Tx,Ty\right)\in A×B$, which implies that $\alpha \left(Tx,Ty\right)\ge 1$.

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

Let $\left\{{x}_{n}\right\}$ be a sequence in X such that ${lim}_{n\to \mathrm{\infty }}d\left({x}_{n},x\right)=0$ for some $x\in X$, and $\alpha \left({x}_{n},{x}_{n+1}\right)\ge 1$ for all n. From the definition of α, this implies that $\left({x}_{n},{x}_{n+1}\right)\in \left(A×B\right)\cup \left(B×A\right)$ for all n. Since A and B are closed, we get $x\in A\cap B$. Then we have $\alpha \left({x}_{n},x\right)=1$ 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 $T\left(A\right)\subseteq B$ and $T\left(B\right)\subseteq A$, 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/10.1186/1029-242X-2014-84 