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Common fixedpoint results for generalized Berindetype contractions which involve altering distance functions
Fixed Point Theory and Applications volume 2014, Article number: 24 (2014)
Abstract
In this paper, the existence and the uniqueness of a common fixed point for two selfmappings satisfying some generalized Berindetype 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
Fixedpoint 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 fixedpoint 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 fixedpoint 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
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 $x,y\in X$.
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 selfmaps on a metric space $(X,d)$. A map f is called generalized almost gcontraction if there exist a constant $\lambda \in [0,1)$ and some $L\ge 0$ such that
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 selfmaps 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
for all $x,y\in X$.
A new category of contractive fixedpoint 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 Berindetype 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 selfmappings satisfying some generalized Berindetype contractions which involve an altering distance function with $\lambda \in [0,1]$. These results extend and generalize several wellknown 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 semicontinuous function with $\phi (t)=0$ if and only if $t=0$. Moreover, suppose that f and g are selfmaps satisfying the inequality
where
and
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
and
for all $n=0,1,2,\dots $ . Let
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 $y={x}_{n1}$ in (2.1). We get
where
and
i.e.,
Thus, we have
where
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 $d({x}_{n},{x}_{n+1})\le d({x}_{n1},{x}_{n})$. Arguing by contradiction, we assume $d({x}_{n},{x}_{n+1})>d({x}_{n1},{x}_{n})$. Therefore, we have three cases.
Case 1: $u({x}_{n},{x}_{n1})=d({x}_{n1},{x}_{n})$. Then
Since ψ is increasing, we have $d({x}_{n},{x}_{n+1})<\lambda d({x}_{n1},{x}_{n})$, which is a contradiction.
Case 2: $u({x}_{n},{x}_{n1})=d({x}_{n},{x}_{n+1})$. Then
Since ψ is increasing, we have $d({x}_{n},{x}_{n+1})<\lambda d({x}_{n},{x}_{n+1})$, which is impossible.
Case 3: $u({x}_{n},{x}_{n1})=\frac{1}{2}d({x}_{n1},{x}_{n+1})$. Then
Since ψ is increasing, we have
which leads to
but $d({x}_{n},{x}_{n+1})>d({x}_{n1},{x}_{n})$, therefore
which is impossible since $\lambda /(2\lambda )\le 1$. Hence, from the above we obtain $d({x}_{n},{x}_{n+1})\le d({x}_{n1},{x}_{n})$.
Similarly, we can prove that $d({x}_{n},{x}_{n+1})\le d({x}_{n1},{x}_{n})$ 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
Next, we want to show that $d=0$. We have two cases.
Case 1. When $u({x}_{n},{x}_{n1})\in \{d({x}_{n1},{x}_{n}),d({x}_{n},{x}_{n+1})\}$, as ψ is continuous and φ is lower semicontinuous and from (2.4) we get
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 $u({x}_{n},{x}_{n1})=\frac{1}{2}d({x}_{n1},{x}_{n+1})$, we suppose that $d\ne 0$; then
Now, we have two subcases.
Subcase 1. $\lambda <1$. Then as $n\to \mathrm{\infty}$ we obtain
which leads to a contradiction if $d\ne 0$.
Subcase 2. $\lambda =1$. Then
As $n\to \mathrm{\infty}$, we have
Since ψ is an increasing function, we get
By taking $n\to \mathrm{\infty}$ in $\psi ({d}_{n})\le \psi (\frac{\lambda}{2}d({x}_{n1},{x}_{n+1}))\phi (\frac{\lambda}{2}d({x}_{n1},{x}_{n+1}))$ and using (2.5), we have
i.e.
Hence, $\phi (d)=0$ and then $d=0$.
From the above we obtain $d=0$, i.e.,
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
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
Then we have
by taking $k\to \mathrm{\infty}$, we obtain
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
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
where
and
Let $n\to \mathrm{\infty}$, we get
where
If $d({x}^{\ast},f{x}^{\ast})\ne 0$, then
which is a contradiction. Hence, we obtain
Similarly, when we take $x={x}_{2n}$ and $y={x}^{\ast}$ in (2.1) for all n we get
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
where
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 semicontinuous function with $\phi (t)=0$ if and only if $t=0$. If f and g are selfmaps satisfying the inequality
where
and
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 semicontinuous function with $\phi (t)=0$ if and only if $t=0$. If f is a selfmap satisfying the inequality
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 fixedpoint result of Dutta and Choudhury [[20], Theorem 2.1] and Theorem 2.2 of Doric [18].
If we take $\psi (t)=t$ and $\phi (t)=(1k)t$ 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 selfmaps satisfying the inequality
where
and
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 selfmap satisfying the inequality
where
and
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 selfmaps satisfying the inequality
where
and
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 $d(x,y)={\parallel xy\parallel}_{\mathrm{\infty}}$ for every $x,y\in X$. Let $\phi :{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ be a function such that

φ is lower semicontinuous;

φ is increasing;

$\phi (t)=0\iff t=0$.
Theorem 3.1 Consider the Urysohn integral equations
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
Then there exists a solution to (3.1).
Proof Define $f,g:X\to X$ by $f(x)={\int}_{a}^{b}{K}_{1}(t,s,x(s))\phantom{\rule{0.2em}{0ex}}ds+h(t)$ and $g(x)={\int}_{a}^{b}{K}_{2}(t,s,x(s))\phantom{\rule{0.2em}{0ex}}ds+q(t)$. It is obvious that
Thus
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). □
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Acknowledgements
The authors would like to acknowledge the financial support received from Universiti Kebangsaan Malaysia under the research grant UKMDIP201231.
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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 fixedpoint results for generalized Berindetype contractions which involve altering distance functions. Fixed Point Theory Appl 2014, 24 (2014). https://doi.org/10.1186/16871812201424
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Keywords
 metric space
 common fixed point
 generalized Berindetype contractions
 Urysohn integral equations