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 $\lambda \in [0,1]$ 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 [1–5] studied many interesting fixed-point theorems for many kinds of contraction mappings. In [3] and [4], he defined the almost contraction map as follows.

Definition 1.1 Let $(X,d)$ be a metric space. A map $f:X\to X$ is called an almost contraction if there exist a constant $\lambda \in [0,1)$ and some $L\ge 0$ such that

$$d(fx,fy)\le \lambda d(x,y)+Ld(y,fx)$$

for all $x,y\in X$.

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

$$d(fx,fy)\le \lambda d(x,y)+Lmin\{d(x,fx),d(y,fy),d(x,fy),d(y,fx)\}.$$

Recently, Babu et al. [6] considered the class of mappings that satisfy ‘condition (B)’.

More recently, Abbas and Ilic in [7] 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 $\lambda \in [0,1)$ and some $L\ge 0$ such that

$$d(fx,fy)\le \lambda M(x,y)+Lmin\{d(gx,fx),d(gy,fy),d(gx,fy),d(gy,fx)\}$$

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

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. [8] 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 $\lambda \in [0,1)$ and some $L\ge 0$ such that

$$\begin{array}{rcl}d(fx,gy)& \le & \lambda max\{d(x,y),d(x,fx),d(y,gy),\frac{1}{2}(d(fx,y)+d(x,gy))\}\\ +Lmin\{d(x,fx),d(y,gy),d(x,gy),d(y,fx)\}\end{array}$$

for all $x,y\in X$.

A new category of contractive fixed-point problems was addressed by Khan et al. [9]. 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 $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is called an altering distance function if the following properties are satisfied:

• (i) ψ is continuous and nondecreasing;

• (ii) $\psi (t)=0\iff t=0$.

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 [12–14] and by using many other kinds of methods, see [15–17].

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 $\lambda \in [0,1]$. 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 $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is an altering distance function and $\phi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is a lower semi-continuous function with $\phi (t)=0$ if and only if $t=0$. Moreover, suppose that f and g are self-maps satisfying the inequality

$$\psi (d(fx,gy))\le \psi (\lambda u(x,y))-\phi (\lambda u(x,y))+LN(x,y),$$

(2.1)

where

$$u(x,y)\in \{d(x,y),d(x,fx),d(y,gy),\frac{1}{2}(d(fx,y)+d(x,gy))\}$$

and

$$N(x,y)=min\{d(x,y),d(x,fx),d(y,gy),d(fx,y),d(x,gy)\},$$

with $L\ge 0$ and $0\le \lambda \le 1$. Then f and g have a unique common fixed point.

Proof We prove the theorem in several steps.

Step 1. Let $x_0\in X$ 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

$$x_{2n+1}=f{x}_{2n}$$

and

$$x_{2n+2}=g{x}_{2n+1}$$

for all $n=0,1,2,\dots$ . 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

$$\begin{array}{rcl}\psi (d(x_{n+1},x_n))& =& \psi (d(f{x}_n,g{x}_{n-1}))\\ \le & \psi (\lambda u(x_n,x_{n-1}))-\phi (\lambda u(x_n,x_{n-1}))+LN(x_n,x_{n-1}),\end{array}$$

(2.3)

where

$$u(x_n,x_{n-1})\in \{d(x_n,x_{n-1}),d(x_n,f{x}_n),d(x_{n-1},g{x}_{n-1}),\frac{1}{2}(d(f{x}_n,x_{n-1})+d(x_n,g{x}_{n-1}))\}$$

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

$$\psi (d(x_n,x_{n+1}))\le \psi (\lambda u(x_n,x_{n-1}))-\phi (\lambda u(x_n,x_{n-1})),$$

where

$$u(x_n,x_{n-1})\in \{d(x_{n-1},x_n),d(x_n,x_{n+1}),\frac{1}{2}d(x_{n-1},x_{n+1})\}.$$

If $x_n=x_{n+1}$ for some $n\ge 0$, then the proof will be finished. Therefore, we suppose that $x_n\ne x_{n+1}$ for all $n\ge 0$. Now, we shall show that $d(x_n,x_{n+1})\le d(x_{n-1},x_n)$. 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

$$\psi (d(x_n,x_{n+1}))\le \psi (\lambda d(x_{n-1},x_n))-\phi (\lambda d(x_{n-1},x_n))<\psi (\lambda d(x_{n-1},x_n)).$$

Since ψ is increasing, we have $d(x_n,x_{n+1})<\lambda d(x_{n-1},x_n)$, which is a contradiction.

Case 2: $u(x_n,x_{n-1})=d(x_n,x_{n+1})$. Then

$$\psi (d(x_n,x_{n+1}))\le \psi (\lambda d(x_n,x_{n+1}))-\phi (\lambda d(x_n,x_{n+1}))<\psi (\lambda d(x_n,x_{n+1})).$$

Since ψ is increasing, we have $d(x_n,x_{n+1})<\lambda d(x_n,x_{n+1})$, which is impossible.

Case 3: $u(x_n,x_{n-1})=\frac{1}{2}d(x_{n-1},x_{n+1})$. Then

$$\psi (d(x_n,x_{n+1}))\le \psi (\frac{\lambda }{2}d(x_{n-1},x_{n+1}))-\phi (\frac{\lambda }{2}d(x_{n-1},x_{n+1}))\le \psi (\frac{\lambda }{2}d(x_{n-1},x_{n+1})).$$

Since ψ is increasing, we have

$$d(x_n,x_{n+1})\le \frac{\lambda }{2}d(x_{n-1},x_{n+1})\le \frac{\lambda }{2}(d(x_{n-1},x_n)+d(x_n,x_{n+1})),$$

which leads to

$$d(x_n,x_{n+1})\le \frac{\lambda }{2-\lambda }d(x_{n-1},x_n),$$

but $d(x_n,x_{n+1})>d(x_{n-1},x_n)$, therefore

$$d(x_n,x_{n+1})<\frac{\lambda }{2-\lambda }d(x_n,x_{n+1}),$$

which is impossible since $\lambda /(2-\lambda )\le 1$. Hence, from the above we obtain $d(x_n,x_{n+1})\le d(x_{n-1},x_n)$.

Similarly, we can prove that $d(x_n,x_{n+1})\le d(x_{n-1},x_n)$ 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

$$d_n\to d\text{as }n\to \mathrm{\infty }.$$

(2.4)

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

Case 1. When $u(x_n,x_{n-1})\in \{d(x_{n-1},x_n),d(x_n,x_{n+1})\}$, as ψ is continuous and φ is lower semi-continuous and from (2.4) we get

$$\psi (d)\le \psi (\lambda d)-\phi (\lambda d).$$

If $\lambda =0$, then we have $\psi (d)\le 0\Rightarrow d=0$. If $\lambda \ne 0$, then we have $\phi (\lambda d)\le \psi (\lambda d)-\psi (d)\le 0$. Thus $\phi (\lambda d)=0$, which implies $d=0$.

Case 2. When $u(x_n,x_{n-1})=\frac{1}{2}d(x_{n-1},x_{n+1})$, we suppose that $d\ne 0$; then

$$\begin{array}{rcl}\psi (d_n)& \le & \psi (\frac{\lambda }{2}d(x_{n-1},x_{n+1}))-\phi (\frac{\lambda }{2}d(x_{n-1},x_{n+1}))\\ \le & \psi (\frac{\lambda }{2}d(x_{n-1},x_{n+1}))\\ \le & \psi \left(\frac{\lambda }{2}(d(x_{n-1},x_n)+d(x_n,x_{n+1}))\right).\end{array}$$

Now, we have two subcases.

Subcase 1. $\lambda <1$. Then as $n\to \mathrm{\infty }$ we obtain

$$\psi (d)\le \psi (\lambda d),$$

which leads to a contradiction if $d\ne 0$.

Subcase 2. $\lambda =1$. Then

$$\psi (d_n)\le \psi (\frac{1}{2}d(x_{n-1},x_{n+1}))\le \psi \left(\frac{1}{2}(d(x_{n-1},x_n)+d(x_n,x_{n+1}))\right).$$

As $n\to \mathrm{\infty }$, we have

$$\psi (d)\le \psi (\frac{1}{2}\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_{n+1}))\le \psi (d).$$

Since ψ is an increasing function, we get

$$\underset{n\to \mathrm{\infty }}{lim}d(x_{n-1},x_{n+1})=2d.$$

(2.5)

By taking $n\to \mathrm{\infty }$ in $\psi (d_n)\le \psi (\frac{\lambda }{2}d(x_{n-1},x_{n+1}))-\phi (\frac{\lambda }{2}d(x_{n-1},x_{n+1}))$ and using (2.5), we have

$$\psi (d)\le \psi (\frac{1}{2}(2d))-\phi (\frac{1}{2}(2d)),$$

i.e.

$$\phi (d)\le 0.$$

Hence, $\phi (d)=0$ and then $d=0$.

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

$$d_n=d(x_n,x_{n+1})\to 0\text{as }n\to \mathrm{\infty }.$$

(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 $\epsilon >0$ for which we can find subsequences $\{x_{n(k)}\}$ and $\{x_{m(k)}\}$ of $\{x_n\}$ with $n(k)>m(k)\ge k$ such that

$$d(x_{n(k)},x_{m(k)})\ge \epsilon /k.$$

(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

$$d(x_{n(k)-1},x_{m(k)})<\epsilon /k.$$

Then we have

$$\begin{array}{rcl}\epsilon /k& \le & d(x_{n(k)},x_{m(k)})\le d(x_{n(k)},x_{n(k)-1})+d(x_{n(k)-1},x_{m(k)})\\ \le & d(x_{n(k)},x_{n(k)-1})+\epsilon /k,\end{array}$$

by taking $k\to \mathrm{\infty }$, we obtain

$$\underset{k\to \mathrm{\infty }}{lim}d(x_{n(k)},x_{m(k)})=0,$$

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

$$\underset{n\to \mathrm{\infty }}{lim}{x}_n=x^{\ast }.$$

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

$$\psi \left(d(f{x}^{\ast },g{x}_{2n+1})\right)\le \psi \left(\lambda u(x^{\ast },x_{2n+1})\right)-\phi \left(\lambda u(x^{\ast },x_{2n+1})\right)+LN(x^{\ast },x_{2n+1}),$$

where

$$u(x^{\ast },x_{2n+1})\in \{d(x^{\ast },x_{2n+1}),d(x^{\ast },f{x}^{\ast }),d(x_{2n+1},g{x}_{2n+1}),\frac{1}{2}(d(f{x}^{\ast },x_{2n+1})+d(x^{\ast },g{x}_{2n+1}))\}$$

and

$$N(x^{\ast },x_{2n+1})=min\{d(x^{\ast },x_{2n+1}),d(x^{\ast },f{x}^{\ast }),d(x_{2n+1},g{x}_{2n+1}),d(f{x}^{\ast },x_{2n+1}),d(x^{\ast },g{x}_{2n+1})\}.$$

Let $n\to \mathrm{\infty }$, we get

$$\psi \left(d(f{x}^{\ast },x^{\ast })\right)\le \psi \left(\lambda \underset{n\to \mathrm{\infty }}{lim}u(x^{\ast },x_{2n+1})\right)-\phi \left(\lambda \underset{n\to \mathrm{\infty }}{lim}u(x^{\ast },x_{2n+1})\right),$$

where

$$\underset{n\to \mathrm{\infty }}{lim}u(x^{\ast },x_{2n+1})\in \{0,d(x^{\ast },f{x}^{\ast }),\frac{1}{2}d(x^{\ast },f{x}^{\ast })\}.$$

If $d(x^{\ast },f{x}^{\ast })\ne 0$, then

$$\psi \left(d(x^{\ast },f{x}^{\ast })\right)<\psi \left(\lambda d(x^{\ast },f{x}^{\ast })\right)\text{or}\psi \left(d(x^{\ast },f{x}^{\ast })\right)<\psi \left(\frac{\lambda }{2}d(x^{\ast },f{x}^{\ast })\right),$$

which is a contradiction. Hence, we obtain

$$d(x^{\ast },f{x}^{\ast })=0\text{or}{x}^{\ast }=f{x}^{\ast }.$$

(2.8)

Similarly, when we take $x=x_{2n}$ and $y=x^{\ast }$ in (2.1) for all n we get

$$x^{\ast }=g{x}^{\ast }.$$

(2.9)

Equations (2.8) and (2.9) show that $x^{\ast }$ is a common fixed point of f and g.

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

$$\psi \left(d(f{x}^{\ast },g{y}^{\ast })\right)=\psi \left(d(x^{\ast },y^{\ast })\right)\le \psi \left(\lambda u(x^{\ast },y^{\ast })\right)-\phi \left(\lambda u(x^{\ast },y^{\ast })\right)+LN(x^{\ast },y^{\ast }),$$

where

$$u(x^{\ast },y^{\ast })\in \{0,d(x^{\ast },y^{\ast })\}\text{and}N(x^{\ast },y^{\ast })=0.$$

Then we obtain $x^{\ast }=y^{\ast }$. □

Corollary 2.2 Let $(X,d)$ be a complete metric space. Suppose $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is an altering distance function and $\phi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is a lower semi-continuous function with $\phi (t)=0$ if and only if $t=0$. If f and g are self-maps satisfying the inequality

$$\psi (d(fx,gy))\le \psi (\lambda M(x,y))-\phi (\lambda M(x,y))+LN(x,y),$$

where

$$M(x,y)=max\{d(x,y),d(x,fx),d(y,gy),\frac{1}{2}(d(fx,y)+d(x,gy))\}$$

and

$$N(x,y)=min\{d(x,y),d(x,fx),d(y,gy),d(fx,y),d(x,gy)\},$$

with $L\ge 0$ and $0\le \lambda \le 1$, then f has a unique fixed point.

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

Remark 2.3 In Corollary 2.2:

• (i) If $g=f$ and $\lambda =1$, we obtain a metric version of Theorem 12 of Aydi et al. [10].

• (ii) If $L=0$ and $\lambda =1$, we get Theorem 2.1 of Doric [18].

• (iii) If $\psi (t)=t$, $L=0$ and $\lambda =1$, we get Theorem 2.1 of Zhang and Song [19].

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 $\psi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is an altering distance function and $\phi :[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is a lower semi-continuous function with $\phi (t)=0$ if and only if $t=0$. If f is a self-map satisfying the inequality

$$\psi (d(fx,fy))\le \psi (\lambda u(x,y))-\phi (\lambda u(x,y)),$$

(2.10)

where $u(x,y)\in \{d(x,y),d(x,fx),d(y,fy),\frac{1}{2}(d(fx,y)+d(x,fy))\}$, $0\le \lambda \le 1$, then f has a unique fixed point.

Remark 2.5 Corollary 2.4 extends the main fixed-point result of Dutta and Choudhury [[20], Theorem 2.1] and Theorem 2.2 of Doric [18].

If we take $\psi (t)=t$ and $\phi (t)=(1-k)t$ 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

$$d(fx,gy)\le \alpha u(x,y)+LN(x,y),$$

(2.11)

where

$$u(x,y)\in \{d(x,y),d(x,fx),d(y,gy),\frac{1}{2}(d(fx,y)+d(x,gy))\}$$

and

$$N(x,y)=min\{d(x,y),d(x,fx),d(y,gy),d(fx,y),d(x,gy)\},$$

with $L\ge 0$ and $0\le \alpha <1$, 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

$$d(fx,fy)\le \alpha u(x,y)+LN(x,y),$$

where

$$u(x,y)\in \{d(x,y),d(x,fx),d(y,fy),\frac{1}{2}(d(fx,y)+d(x,fy))\}$$

and

$$N(x,y)=min\{d(x,y),d(x,fx),d(y,fy),d(fx,y),d(x,fy)\},$$

with $L\ge 0$ and $0\le \alpha <1$, then f has a unique fixed point.

By the aid of Lemma 2.1 of [21], 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

$$d(fx,fy)\le \alpha u(gx,gy)+LN(gx,gy),$$

where

$$u(gx,gy)\in \{d(gx,gy),d(gx,fx),d(gy,fy),\frac{1}{2}(d(fx,gy)+d(gx,fy))\}$$

and

$$N(gx,gy)=min\{d(gx,gy),d(gx,fx),d(gy,fy),d(fx,gy),d(gx,fy)\},$$

with $L\ge 0$ and $0\le \alpha <1$, then f and g have a unique common fixed point.

Remark 2.9 Corollary 2.8 extends the results of Abbas et al. [[22], Theorem 2.1] and Jleli et al. [[23], Corollary 3.2].

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

Throughout this section we take $X=C([a,b],\mathbb{R})$ (the set of continuous functions defined in $I=[a,b]$). We define the metric $d:X\times X\to \mathbb{R}$ by $d(x,y)={\parallel x-y\parallel }_{\mathrm{\infty }}$ for every $x,y\in X$. Let $\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ be a function such that

• φ is lower semi-continuous;

• φ is increasing;

• $\phi (t)=0\iff t=0$.

Theorem 3.1 Consider the Urysohn integral equations

$$\begin{array}{r}x(t)={\int }_a^b{K}_1(t,s,x(s))ds+h(t),\\ x(t)={\int }_a^b{K}_2(t,s,x(s))ds+q(t),\end{array}$$

(3.1)

where $t\in I\subset \mathbb{R}$ and $x,h,q\in X$. Assume that, for $K_1,K_2:I\times I\times \mathbb{R}\to \mathbb{R}$, we have

$$|{\int }_a^b{K}_1(t,s,x(s))ds-{\int }_a^b{K}_2(t,s,x(s))ds+h(t)-q(t)|\le |x(t)-y(t)|-\phi \left(\underset{t\in I}{sup}\right|x(t)-y(t)\left|\right).$$

Then there exists a solution to (3.1).

Proof Define $f,g:X\to X$ by $f(x)={\int }_a^b{K}_1(t,s,x(s))ds+h(t)$ and $g(x)={\int }_a^b{K}_2(t,s,x(s))ds+q(t)$. It is obvious that

$${\parallel f-g\parallel }_{\mathrm{\infty }}\le {\parallel x-y\parallel }_{\mathrm{\infty }}-\phi \left({\parallel x-y\parallel }_{\mathrm{\infty }}\right).$$

Thus

$$d(fx,gy)\le d(x,y)-\phi (d(x,y))$$

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