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# Common fixed-point results for generalized Berinde-type contractions which involve altering distance functions

## Abstract

In this paper, the existence and the uniqueness of a common fixed point for two self-mappings satisfying some generalized Berinde-type contractions which involve an altering distance function with $λ∈[0,1]$ is established. Our theorems extend, unify, and generalize several existing results in the literature. An application of an integral equation is presented.

MSC:54H25, 54C60, 54E50.

## 1 Introduction and preliminaries

Fixed-point theory is one of the most fruitful and effective tools in mathematics which has enormous applications within as well as outside mathematics. In 1922, Banach established the famous fixed-point theorem which is called the Banach contraction principle. This principle is a forceful tool in nonlinear analysis. It has many applications in solving nonlinear equations.

Later, Berinde  studied many interesting fixed-point theorems for many kinds of contraction mappings. In  and , he defined the almost contraction map as follows.

Definition 1.1 Let $(X,d)$ be a metric space. A map $f:X→X$ is called an almost contraction if there exist a constant $λ∈[0,1)$ and some $L≥0$ such that

$d(fx,fy)≤λd(x,y)+Ld(y,fx)$

for all $x,y∈X$.

Let $(X,d)$ be a metric space. A map $f:X→X$ is said to satisfy ‘condition (B)’ if there exist a constant $λ∈[0,1)$ and some $L≥0$ such that for all $x,y∈X$.

$d(fx,fy)≤λd(x,y)+Lmin { d ( x , f x ) , d ( y , f y ) , d ( x , f y ) , d ( y , f x ) } .$

Recently, Babu et al.  considered the class of mappings that satisfy ‘condition (B)’.

More recently, Abbas and Ilic in  introduced the following definition.

Definition 1.2 Let f and g be two self-maps on a metric space $(X,d)$. A map f is called generalized almost g-contraction if there exist a constant $λ∈[0,1)$ and some $L≥0$ such that

$d(fx,fy)≤λM(x,y)+Lmin { d ( g x , f x ) , d ( g y , f y ) , d ( g x , f y ) , d ( g y , f x ) }$

for all $x,y∈X$, where $M(x,y)=max{d(gx,gy),d(gx,fx),d(fy,gy), 1 2 (d(fx,gy)+d(gx,fy))}$.

If $g= I X$, $I X$ is the identity map on X, then they note that f satisfies ‘generalized condition (B)’.

Furthermore, Ćirić et al.  introduced the concept of the almost generalized contractive condition as follows.

Definition 1.3 Let f and g be two self-maps on a metric space $(X,d)$. They are said to satisfy almost generalized contractive condition if there exist a constant $λ∈[0,1)$ and some $L≥0$ such that

$d ( f x , g y ) ≤ λ max { d ( x , y ) , d ( x , f x ) , d ( y , g y ) , 1 2 ( d ( f x , y ) + d ( x , g y ) ) } + L min { d ( x , f x ) , d ( y , g y ) , d ( x , g y ) , d ( y , f x ) }$

for all $x,y∈X$.

A new category of contractive fixed-point problems was addressed by Khan et al. . In their study they introduced the notion of an altering distance function, which is a control function that alters the distance between two points in a metric space.

Definition 1.4 The function $ψ:[0,∞)→[0,∞)$ is called an altering distance function if the following properties are satisfied:

1. (i)

ψ is continuous and nondecreasing;

2. (ii)

$ψ(t)=0⇔t=0$.

In the literature, there has been extensive research on common fixed points by using Berinde-type contractions, see [10, 11], by using an altering distance function, see  and by using many other kinds of methods, see .

The aim of this work is to prove that there is a unique common fixed point for two self-mappings satisfying some generalized Berinde-type contractions which involve an altering distance function with $λ∈[0,1]$. These results extend and generalize several well-known compatible recent and classical results in the literature. As an application, the existence of a solution for the Urysohn integral equation is presented.

## 2 Main results

Theorem 2.1 Let $(X,d)$ be a complete metric space. Suppose $ψ:[0,∞)→[0,∞)$ is an altering distance function and $φ:[0,∞)→[0,∞)$ is a lower semi-continuous function with $φ(t)=0$ if and only if $t=0$. Moreover, suppose that f and g are self-maps satisfying the inequality

$ψ ( d ( f x , g y ) ) ≤ψ ( λ u ( x , y ) ) −φ ( λ u ( x , y ) ) +LN(x,y),$
(2.1)

where

$u(x,y)∈ { d ( x , y ) , d ( x , f x ) , d ( y , g y ) , 1 2 ( d ( f x , y ) + d ( x , g y ) ) }$

and

$N(x,y)=min { d ( x , y ) , d ( x , f x ) , d ( y , g y ) , d ( f x , y ) , d ( x , g y ) } ,$

with $L≥0$ and $0≤λ≤1$. Then f and g have a unique common fixed point.

Proof We prove the theorem in several steps.

Step 1. Let $x 0 ∈X$ be an arbitrary point.

Taking $x 1 =f x 0$ and $x 2 =g x 1$, then let $x 3 =f x 2$ and $x 4 =g x 3$. Continue in this way, we can choose a sequence ${ x n }$ in X so that

$x 2 n + 1 =f x 2 n$

and

$x 2 n + 2 =g x 2 n + 1$

for all $n=0,1,2,…$ . Let

$d n =d( x n , x n + 1 ).$
(2.2)

We will prove that ${ d n }$ is a decreasing sequence which converges to 0.

If n is an even number, put $x= x n$ and $y= x n − 1$ in (2.1). We get

$ψ ( d ( x n + 1 , x n ) ) = ψ ( d ( f x n , g x n − 1 ) ) ≤ ψ ( λ u ( x n , x n − 1 ) ) − φ ( λ u ( x n , x n − 1 ) ) + L N ( x n , x n − 1 ) ,$
(2.3)

where

$u( x n , x n − 1 )∈ { d ( x n , x n − 1 ) , d ( x n , f x n ) , d ( x n − 1 , g x n − 1 ) , 1 2 ( d ( f x n , x n − 1 ) + d ( x n , g x n − 1 ) ) }$

and

$N( x n , x n − 1 )=min { d ( x n , x n − 1 ) , d ( x n , f x n ) , d ( x n − 1 , g x n − 1 ) , d ( f x n , x n − 1 ) , d ( x n , g x n − 1 ) } ,$

i.e.,

$N( x n , x n − 1 )=0.$

Thus, we have

$ψ ( d ( x n , x n + 1 ) ) ≤ψ ( λ u ( x n , x n − 1 ) ) −φ ( λ u ( x n , x n − 1 ) ) ,$

where

$u( x n , x n − 1 )∈ { d ( x n − 1 , x n ) , d ( x n , x n + 1 ) , 1 2 d ( x n − 1 , x n + 1 ) } .$

If $x n = x n + 1$ for some $n≥0$, then the proof will be finished. Therefore, we suppose that $x n ≠ x n + 1$ for all $n≥0$. Now, we shall show that $d( x n , x n + 1 )≤d( x n − 1 , x n )$. Arguing by contradiction, we assume $d( x n , x n + 1 )>d( x n − 1 , x n )$. Therefore, we have three cases.

Case 1: $u( x n , x n − 1 )=d( x n − 1 , x n )$. Then

$ψ ( d ( x n , x n + 1 ) ) ≤ψ ( λ d ( x n − 1 , x n ) ) −φ ( λ d ( x n − 1 , x n ) ) <ψ ( λ d ( x n − 1 , x n ) ) .$

Since ψ is increasing, we have $d( x n , x n + 1 )<λd( x n − 1 , x n )$, which is a contradiction.

Case 2: $u( x n , x n − 1 )=d( x n , x n + 1 )$. Then

$ψ ( d ( x n , x n + 1 ) ) ≤ψ ( λ d ( x n , x n + 1 ) ) −φ ( λ d ( x n , x n + 1 ) ) <ψ ( λ d ( x n , x n + 1 ) ) .$

Since ψ is increasing, we have $d( x n , x n + 1 )<λd( x n , x n + 1 )$, which is impossible.

Case 3: $u( x n , x n − 1 )= 1 2 d( x n − 1 , x n + 1 )$. Then

$ψ ( d ( x n , x n + 1 ) ) ≤ψ ( λ 2 d ( x n − 1 , x n + 1 ) ) −φ ( λ 2 d ( x n − 1 , x n + 1 ) ) ≤ψ ( λ 2 d ( x n − 1 , x n + 1 ) ) .$

Since ψ is increasing, we have

$d( x n , x n + 1 )≤ λ 2 d( x n − 1 , x n + 1 )≤ λ 2 ( d ( x n − 1 , x n ) + d ( x n , x n + 1 ) ) ,$

$d( x n , x n + 1 )≤ λ 2 − λ d( x n − 1 , x n ),$

but $d( x n , x n + 1 )>d( x n − 1 , x n )$, therefore

$d( x n , x n + 1 )< λ 2 − λ d( x n , x n + 1 ),$

which is impossible since $λ/(2−λ)≤1$. Hence, from the above we obtain $d( x n , x n + 1 )≤d( x n − 1 , x n )$.

Similarly, we can prove that $d( x n , x n + 1 )≤d( x n − 1 , x n )$ also in the case when n is an odd number.

Therefore, we find that ${ d n }$ is a decreasing sequence and bounded below. Thus, ${ d n }$ is convergent. Let

(2.4)

Next, we want to show that $d=0$. We have two cases.

Case 1. When $u( x n , x n − 1 )∈{d( x n − 1 , x n ),d( x n , x n + 1 )}$, as ψ is continuous and φ is lower semi-continuous and from (2.4) we get

$ψ(d)≤ψ(λd)−φ(λd).$

If $λ=0$, then we have $ψ(d)≤0⇒d=0$. If $λ≠0$, then we have $φ(λd)≤ψ(λd)−ψ(d)≤0$. Thus $φ(λd)=0$, which implies $d=0$.

Case 2. When $u( x n , x n − 1 )= 1 2 d( x n − 1 , x n + 1 )$, we suppose that $d≠0$; then

$ψ ( d n ) ≤ ψ ( λ 2 d ( x n − 1 , x n + 1 ) ) − φ ( λ 2 d ( x n − 1 , x n + 1 ) ) ≤ ψ ( λ 2 d ( x n − 1 , x n + 1 ) ) ≤ ψ ( λ 2 ( d ( x n − 1 , x n ) + d ( x n , x n + 1 ) ) ) .$

Now, we have two subcases.

Subcase 1. $λ<1$. Then as $n→∞$ we obtain

$ψ(d)≤ψ(λd),$

which leads to a contradiction if $d≠0$.

Subcase 2. $λ=1$. Then

$ψ( d n )≤ψ ( 1 2 d ( x n − 1 , x n + 1 ) ) ≤ψ ( 1 2 ( d ( x n − 1 , x n ) + d ( x n , x n + 1 ) ) ) .$

As $n→∞$, we have

$ψ(d)≤ψ ( 1 2 lim n → ∞ d ( x n − 1 , x n + 1 ) ) ≤ψ(d).$

Since ψ is an increasing function, we get

$lim n → ∞ d( x n − 1 , x n + 1 )=2d.$
(2.5)

By taking $n→∞$ in $ψ( d n )≤ψ( λ 2 d( x n − 1 , x n + 1 ))−φ( λ 2 d( x n − 1 , x n + 1 ))$ and using (2.5), we have

$ψ(d)≤ψ ( 1 2 ( 2 d ) ) −φ ( 1 2 ( 2 d ) ) ,$

i.e.

$φ(d)≤0.$

Hence, $φ(d)=0$ and then $d=0$.

From the above we obtain $d=0$, i.e.,

(2.6)

Step 2. We prove that ${ x n }$ is a Cauchy sequence. Suppose that ${ x n }$ is not a Cauchy sequence, then there exists $ε>0$ for which we can find subsequences ${ x n ( k ) }$ and ${ x m ( k ) }$ of ${ x n }$ with $n(k)>m(k)≥k$ such that

$d( x n ( k ) , x m ( k ) )≥ε/k.$
(2.7)

Furthermore, corresponding to $m(k)$, we can choose $n(k)$ in such a way that it is the smallest integer with $n(k)>m(k)$ and satisfying (2.7). Then

$d( x n ( k ) − 1 , x m ( k ) )<ε/k.$

Then we have

$ε / k ≤ d ( x n ( k ) , x m ( k ) ) ≤ d ( x n ( k ) , x n ( k ) − 1 ) + d ( x n ( k ) − 1 , x m ( k ) ) ≤ d ( x n ( k ) , x n ( k ) − 1 ) + ε / k ,$

by taking $k→∞$, we obtain

$lim k → ∞ d( x n ( k ) , x m ( k ) )=0,$

which is a contradiction. So, ${ x n }$ is a Cauchy sequence in a complete metric space X and hence it is convergent in X. Let

$lim n → ∞ x n = x ∗ .$

Step 3. Let us now prove that $x ∗$ is a common fixed point of f and g. Put $x= x ∗$ and $y= x 2 n + 1$ in (2.1) for all n, and we obtain

$ψ ( d ( f x ∗ , g x 2 n + 1 ) ) ≤ψ ( λ u ( x ∗ , x 2 n + 1 ) ) −φ ( λ u ( x ∗ , x 2 n + 1 ) ) +LN ( x ∗ , x 2 n + 1 ) ,$

where

$u ( x ∗ , x 2 n + 1 ) ∈ { d ( x ∗ , x 2 n + 1 ) , d ( x ∗ , f x ∗ ) , d ( x 2 n + 1 , g x 2 n + 1 ) , 1 2 ( d ( f x ∗ , x 2 n + 1 ) + d ( x ∗ , g x 2 n + 1 ) ) }$

and

$N ( x ∗ , x 2 n + 1 ) =min { d ( x ∗ , x 2 n + 1 ) , d ( x ∗ , f x ∗ ) , d ( x 2 n + 1 , g x 2 n + 1 ) , d ( f x ∗ , x 2 n + 1 ) , d ( x ∗ , g x 2 n + 1 ) } .$

Let $n→∞$, we get

$ψ ( d ( f x ∗ , x ∗ ) ) ≤ψ ( λ lim n → ∞ u ( x ∗ , x 2 n + 1 ) ) −φ ( λ lim n → ∞ u ( x ∗ , x 2 n + 1 ) ) ,$

where

$lim n → ∞ u ( x ∗ , x 2 n + 1 ) ∈ { 0 , d ( x ∗ , f x ∗ ) , 1 2 d ( x ∗ , f x ∗ ) } .$

If $d( x ∗ ,f x ∗ )≠0$, then

$ψ ( d ( x ∗ , f x ∗ ) ) <ψ ( λ d ( x ∗ , f x ∗ ) ) orψ ( d ( x ∗ , f x ∗ ) ) <ψ ( λ 2 d ( x ∗ , f x ∗ ) ) ,$

which is a contradiction. Hence, we obtain

$d ( x ∗ , f x ∗ ) =0or x ∗ =f x ∗ .$
(2.8)

Similarly, when we take $x= x 2 n$ and $y= x ∗$ in (2.1) for all n we get

$x ∗ =g x ∗ .$
(2.9)

Equations (2.8) and (2.9) show that $x ∗$ is a common fixed point of f and g.

Step 4. Let us now show the uniqueness. Let $y ∗$ be another common fixed point of f and g. Then from (2.1) we have

$ψ ( d ( f x ∗ , g y ∗ ) ) =ψ ( d ( x ∗ , y ∗ ) ) ≤ψ ( λ u ( x ∗ , y ∗ ) ) −φ ( λ u ( x ∗ , y ∗ ) ) +LN ( x ∗ , y ∗ ) ,$

where

$u ( x ∗ , y ∗ ) ∈ { 0 , d ( x ∗ , y ∗ ) } andN ( x ∗ , y ∗ ) =0.$

Then we obtain $x ∗ = y ∗$. □

Corollary 2.2 Let $(X,d)$ be a complete metric space. Suppose $ψ:[0,∞)→[0,∞)$ is an altering distance function and $φ:[0,∞)→[0,∞)$ is a lower semi-continuous function with $φ(t)=0$ if and only if $t=0$. If f and g are self-maps satisfying the inequality

$ψ ( d ( f x , g y ) ) ≤ψ ( λ M ( x , y ) ) −φ ( λ M ( x , y ) ) +LN(x,y),$

where

$M(x,y)=max { d ( x , y ) , d ( x , f x ) , d ( y , g y ) , 1 2 ( d ( f x , y ) + d ( x , g y ) ) }$

and

$N(x,y)=min { d ( x , y ) , d ( x , f x ) , d ( y , g y ) , d ( f x , y ) , d ( x , g y ) } ,$

with $L≥0$ and $0≤λ≤1$, then f has a unique fixed point.

Proof Since $M(x,y)∈{d(x,y),d(x,fx),d(y,gy), 1 2 (d(fx,y)+d(x,gy))}$, the result follows from Theorem 2.1. □

Remark 2.3 In Corollary 2.2:

1. (i)

If $g=f$ and $λ=1$, we obtain a metric version of Theorem 12 of Aydi et al. .

2. (ii)

If $L=0$ and $λ=1$, we get Theorem 2.1 of Doric .

3. (iii)

If $ψ(t)=t$, $L=0$ and $λ=1$, we get Theorem 2.1 of Zhang and Song .

By taking $f=g$ and $L=0$ in Corollary 2.2, we obtain the following result.

Corollary 2.4 Let $(X,d)$ be a complete metric space. Suppose $ψ:[0,∞)→[0,∞)$ is an altering distance function and $φ:[0,∞)→[0,∞)$ is a lower semi-continuous function with $φ(t)=0$ if and only if $t=0$. If f is a self-map satisfying the inequality

$ψ ( d ( f x , f y ) ) ≤ψ ( λ u ( x , y ) ) −φ ( λ u ( x , y ) ) ,$
(2.10)

where $u(x,y)∈{d(x,y),d(x,fx),d(y,fy), 1 2 (d(fx,y)+d(x,fy))}$, $0≤λ≤1$, then f has a unique fixed point.

Remark 2.5 Corollary 2.4 extends the main fixed-point result of Dutta and Choudhury [, Theorem 2.1] and Theorem 2.2 of Doric .

If we take $ψ(t)=t$ and $φ(t)=(1−k)t$ with $k<1$ in Theorem 2.1, we have the following corollary.

Corollary 2.6 Let $(X,d)$ be a complete metric space. If f and g are self-maps satisfying the inequality

$d(fx,gy)≤αu(x,y)+LN(x,y),$
(2.11)

where

$u(x,y)∈ { d ( x , y ) , d ( x , f x ) , d ( y , g y ) , 1 2 ( d ( f x , y ) + d ( x , g y ) ) }$

and

$N(x,y)=min { d ( x , y ) , d ( x , f x ) , d ( y , g y ) , d ( f x , y ) , d ( x , g y ) } ,$

with $L≥0$ and $0≤α<1$, then f and g have a unique common fixed point.

If we take $f=g$ in Corollary 2.6 we obtain the following result.

Corollary 2.7 Let $(X,d)$ be a complete metric space. If f is a self-map satisfying the inequality

$d(fx,fy)≤αu(x,y)+LN(x,y),$

where

$u(x,y)∈ { d ( x , y ) , d ( x , f x ) , d ( y , f y ) , 1 2 ( d ( f x , y ) + d ( x , f y ) ) }$

and

$N(x,y)=min { d ( x , y ) , d ( x , f x ) , d ( y , f y ) , d ( f x , y ) , d ( x , f y ) } ,$

with $L≥0$ and $0≤α<1$, then f has a unique fixed point.

By the aid of Lemma 2.1 of , we have the following result as a consequence of Corollary 2.7.

Corollary 2.8 Let $(X,d)$ be a complete metric space. If f and g are self-maps satisfying the inequality

$d(fx,fy)≤αu(gx,gy)+LN(gx,gy),$

where

$u(gx,gy)∈ { d ( g x , g y ) , d ( g x , f x ) , d ( g y , f y ) , 1 2 ( d ( f x , g y ) + d ( g x , f y ) ) }$

and

$N(gx,gy)=min { d ( g x , g y ) , d ( g x , f x ) , d ( g y , f y ) , d ( f x , g y ) , d ( g x , f y ) } ,$

with $L≥0$ and $0≤α<1$, then f and g have a unique common fixed point.

Remark 2.9 Corollary 2.8 extends the results of Abbas et al. [, Theorem 2.1] and Jleli et al. [, Corollary 3.2].

## 3 Applications: existence of a common solution to Urysohn integral equations

Throughout this section we take $X=C([a,b],R)$ (the set of continuous functions defined in $I=[a,b]$). We define the metric $d:X×X→R$ by $d(x,y)= ∥ x − y ∥ ∞$ for every $x,y∈X$. Let $φ: R + → R +$ be a function such that

• φ is lower semi-continuous;

• φ is increasing;

• $φ(t)=0⇔t=0$.

Theorem 3.1 Consider the Urysohn integral equations

$x ( t ) = ∫ a b K 1 ( t , s , x ( s ) ) d s + h ( t ) , x ( t ) = ∫ a b K 2 ( t , s , x ( s ) ) d s + q ( t ) ,$
(3.1)

where $t∈I⊂R$ and $x,h,q∈X$. Assume that, for $K 1 , K 2 :I×I×R→R$, we have

$| ∫ a b K 1 ( t , s , x ( s ) ) ds− ∫ a b K 2 ( t , s , x ( s ) ) ds+h(t)−q(t)|≤|x(t)−y(t)|−φ ( sup t ∈ I | x ( t ) − y ( t ) | ) .$

Then there exists a solution to (3.1).

Proof Define $f,g:X→X$ by $f(x)= ∫ a b K 1 (t,s,x(s))ds+h(t)$ and $g(x)= ∫ a b K 2 (t,s,x(s))ds+q(t)$. It is obvious that

$∥ f − g ∥ ∞ ≤ ∥ x − y ∥ ∞ −φ ( ∥ x − y ∥ ∞ ) .$

Thus

$d(fx,gy)≤d(x,y)−φ ( d ( x , y ) )$

for all $x,y∈X$. Now, all the assumptions of Theorem 2.1 are satisfied with $ψ(t)=t$, for all $t∈ R +$, $u(x,y)=d(x,y)$, and $L=0$. Therefore, f and g have a common fixed point, that is, a solution to the integral equation (3.1). □

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## Acknowledgements

The authors would like to acknowledge the financial support received from Universiti Kebangsaan Malaysia under the research grant UKM-DIP-2012-31.

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### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All three authors contributed equally and significantly in writing this paper. All three authors read and approved the final manuscript.

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Shaddad, F., Md Noorani, M.S. & Alsulami, S.M. Common fixed-point results for generalized Berinde-type contractions which involve altering distance functions. Fixed Point Theory Appl 2014, 24 (2014). https://doi.org/10.1186/1687-1812-2014-24 