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Modified αψcontractive mappings with applications
Fixed Point Theory and Applications volume 2013, Article number: 151 (2013)
Abstract
The aim of this work is to modify the notions of αadmissible and αψcontractive mappings and establish new fixed point theorems for such mappings in complete metric spaces. Presented theorems provide main results of Karapinar and Samet (Abstr. Appl. Anal. 2012:793486, 2012) and Samet et al. (Nonlinear Anal. 75:21542165, 2012) as direct corollaries. Moreover, some examples and applications to integral equations are given here to illustrate the usability of the obtained results.
MSC:46N40, 47H10, 54H25, 46T99.
1 Introduction and preliminaries
Metric fixed point theory has many applications in functional analysis. The contractive conditions on underlying functions play an important role for finding solutions of metric fixed point problems. The Banach contraction principle is a remarkable result in metric fixed point theory. Over the years, it has been generalized in different directions by several mathematicians (see [1–25]). In 2012, Samet et al. [24] introduced the concepts of αψcontractive and αadmissible mappings and established various fixed point theorems for such mappings in complete metric spaces. Afterwards Karapinar and Samet [19] generalized these notions to obtain fixed point results. The aim of this paper is to modify further the notions of αψcontractive and αadmissible mappings and establish fixed point theorems for such mappings in complete metric spaces. Our results are proper generalizations of the recent results in [19, 24]. Moreover, some examples and applications to integral equations are given here to illustrate the usability of the obtained results.
Denote with Ψ the family of nondecreasing functions \psi :[0,+\mathrm{\infty})\to [0,+\mathrm{\infty}) such that {\sum}_{n=1}^{\mathrm{\infty}}{\psi}^{n}(t)<+\mathrm{\infty} for all t>0, where {\psi}^{n} is the n th iterate of ψ.
The following lemma is obvious.
Lemma 1.1 If \psi \in \mathrm{\Psi}, then \psi (t)<t for all t>0.
Definition 1.1 [24]
Let T be a selfmapping on a metric space (X,d) and let \alpha :X\times X\to [0,+\mathrm{\infty}) be a function. We say that T is an αadmissible mapping if
Definition 1.2 [24]
Let T be a selfmapping on a metric space (X,d). We say that T is an αψcontractive mapping if there exist two functions \alpha :X\times X\to [0,+\mathrm{\infty}) and \psi \in \mathrm{\Psi} such that
for all x,y\in X.
For the examples of αadmissible and αψcontractive mappings, see [19, 24] and the examples in the next section.
2 Main results
We first modify the concept of αadmissible mapping.
Definition 2.1 Let T be a selfmapping on a metric space (X,d) and let \alpha ,\eta :X\times X\to [0,+\mathrm{\infty}) be two functions. We say that T is an αadmissible mapping with respect to η if
Note that if we take \eta (x,y)=1, then this definition reduces to Definition 1.1. Also, if we take \alpha (x,y)=1, then we say that T is an ηsubadmissible mapping.
Our first result is the following.
Theorem 2.1 Let (X,d) be a complete metric space and let T be an αadmissible mapping with respect to η. Assume that
where \psi \in \mathrm{\Psi} and
Also, suppose that the following assertions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge \eta ({x}_{0},T{x}_{0});

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge \eta ({x}_{n},{x}_{n+1}) for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha ({x}_{n},x)\ge \eta ({x}_{n},x) for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
Proof Let {x}_{0}\in X be such that \alpha ({x}_{0},T{x}_{0})\ge \eta ({x}_{0},T{x}_{0}). Define a sequence \{{x}_{n}\} in X by {x}_{n}={T}^{n}{x}_{0}=T{x}_{n1} for all n\in \mathbb{N}. If {x}_{n+1}={x}_{n} for some n\in \mathbb{N}, then x={x}_{n} is a fixed point for T and the result is proved. Hence, we suppose that {x}_{n+1}\ne {x}_{n} for all n\in \mathbb{N}. Since T is a generalized αadmissible mapping with respect to η and \alpha ({x}_{0},T{x}_{0})\ge \eta ({x}_{0},T{x}_{0}), we deduce that \alpha ({x}_{1},{x}_{2})=\alpha (T{x}_{0},{T}^{2}{x}_{0})\ge \eta (T{x}_{0},{T}^{2}{x}_{0})=\eta ({x}_{1},{x}_{2}). Continuing this process, we get \alpha ({x}_{n},{x}_{n+1})\ge \eta ({x}_{n},{x}_{n+1}) for all n\in \mathbb{N}\cup \{0\}. Now, by (2.1) with x={x}_{n1}, y={x}_{n}, we get
On the other hand,
which implies
Now, if max\{d({x}_{n1},{x}_{n}),d({x}_{n},{x}_{n+1})\}=d({x}_{n},{x}_{n+1}) for some n\in \mathbb{N}, then
which is a contradiction. Hence, for all n\in \mathbb{N}, we have
By induction, we have
Fix \u03f5>0, there exists N\in \mathbb{N} such that
Let m,n\in \mathbb{N} with m>n\ge N. Then, by the triangular inequality, we get
Consequently, {lim}_{m,n,\to +\mathrm{\infty}}d({x}_{n},{x}_{m})=0. Hence \{{x}_{n}\} is a Cauchy sequence. Since X is complete, there is z\in X such that {x}_{n}\to z as n\to \mathrm{\infty}. Now, if we suppose that T is continuous, then we have
So, z is a fixed point of T. On the other hand, since
for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to z as n\to \mathrm{\infty}, we get
for all n\in \mathbb{N}\cup \{0\}. Then from (2.1) we have
where
Since M({x}_{n},z)>0, then
By taking limit as n\to \mathrm{\infty} in the above inequality, we have
which implies d(z,Tz)=0, i.e., z=Tz. □
By taking \eta (x,y)=1 in Theorem 2.1, we have the following result.
Corollary 2.1 Let (X,d) be a complete metric space and let T be an αadmissible mapping. Assume that for \psi \in \mathrm{\Psi},
Also, suppose that the following assertions hold:

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

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
By taking \alpha (x,y)=1 in Theorem 2.1, we have the following corollary.
Corollary 2.2 Let (X,d) be a complete metric space and let T be an ηsubadmissible mapping. Assume that for \psi \in \mathrm{\Psi},
Also, suppose that the following assertions hold:

(i)
there exists {x}_{0}\in X such that \eta ({x}_{0},T{x}_{0})\le 1;

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \eta ({x}_{n},{x}_{n+1})\le 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \eta ({x}_{n},x)\le 1 for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
Clearly, Corollary 2.1 implies the following results.
Corollary 2.3 (Theorem 2.1 and Theorem 2.2 of [24])
Let (X,d) be a complete metric space and let T be an αadmissible mapping. Assume that for \psi \in \mathrm{\Psi},
holds for all x,y\in X. Also, suppose that the following assertions hold:

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

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
Corollary 2.4 (Theorem 2.3 and Theorem 2.4 of [19])
Let (X,d) be a complete metric space and let T be an αadmissible mapping. Assume that for \psi \in \mathrm{\Psi},
where
Also, suppose that the following assertions hold:

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

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
Example 2.1 Let X=[0,\mathrm{\infty}) be endowed with the usual metric d(x,y)=xy for all x,y\in X and let T:X\to X be defined by
Define also \alpha :X\times X\to [0,+\mathrm{\infty}) and \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by
We prove that Corollary 2.1 can be applied to T. But Theorem 2.2 of [24] and Theorem 2.4 of [19] cannot be applied to T.
Clearly, (X,d) is a complete metric space. We show that T is an αadmissible mapping. Let x,y\in X, if \alpha (x,y)\ge 1, then x,y\in [0,1]. On the other hand, for all x\in [0,1] we have Tx\le 1. It follows that \alpha (Tx,Ty)\ge 1. Hence, the assertion holds. In reason of the above arguments, \alpha (0,T0)\ge 1.
Now, if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, then \{{x}_{n}\}\subset [0,1] and hence x\in [0,1]. This implies that \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}.
Let \alpha (x,y)\ge 1. Then x,y\in [0,1]. We get
That is,
All of the conditions of Corollary 2.1 hold. Hence, T has a fixed point. Let x=0 and y=1, then
That is, Theorem 2.2 of [24] cannot be applied to T.
Also, by a similar method, we can show that Theorem 2.4 of [19] cannot be applied to T.
By the following simple example, we show that our results improve the results of Samet et al. [24] and the results of Karapinar and Samet [19].
Example 2.2 Let X=[0,\mathrm{\infty}) be endowed with the usual metric d(x,y)=xy for all x,y\in X and let T:X\to X be defined by Tx=\frac{1}{4}x. Also, define \alpha :{X}^{2}\to [0,\mathrm{\infty}) by \alpha (x,y)=3 and \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by \psi (t)=\frac{1}{2}t.
Clearly, T is an αadmissible mapping. Also, \alpha (x,y)=3\ge 1 for all x,y\in X. Hence,
Then the conditions of Corollary 2.1 hold and T has a fixed point. But if we choose x=4 and y=8, then
That is, Theorem 2.2 of [24] cannot be applied to T. Similarly, we can show that Theorem 2.4 of [19] cannot be applied to T. Further notice that the Banach contraction principle holds for this example.
Example 2.3 Let X=[0,\mathrm{\infty}) be endowed with the usual metric d(x,y)=xy for all x,y\in X and let T:X\to X be defined by
Define also \alpha ,\eta :X\times X\to [0,+\mathrm{\infty}) and \psi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by
We prove that Corollary 2.2 can be applied to T. But the Banach contraction principle cannot be applied to T.
Clearly, (X,d) is a complete metric space. We show that T is an ηsubadmissible mapping. Let x,y\in X, if \eta (x,y)\le 1, then x,y\in [0,1]. On the other hand, for all x\in [0,1], we have Tx\le 1. It follows that \eta (Tx,Ty)\le 1. Also, \eta (0,T0)\le 1.
Now, if \{{x}_{n}\} is a sequence in X such that \eta ({x}_{n},{x}_{n+1})\le 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, then \{{x}_{n}\}\subset [0,1] and hence x\in [0,1]. This implies that \eta ({x}_{n},x)\le 1 for all n\in \mathbb{N}.
Let \eta (x,y)\le 1. Then x,y\in [0,1]. We get
That is,
Then the conditions of Corollary 2.2 hold. Hence, T has a fixed point. Let x=2, y=3 and r\in [0,1). Then
That is, the Banach contraction principle cannot be applied to T.
From our results, we can deduce the following corollaries.
Corollary 2.5 Let (X,d) be a complete metric space and let T be an αadmissible mapping. Assume that
holds for all x,y\in X, where \psi \in \mathrm{\Psi} and \ell >0. Also, suppose that the following assertions hold:

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

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
Proof Let \alpha (x,y)\ge 1. Then by (2.2) we have
Then d(Tx,Ty)\le \psi (d(x,y)). Hence, the conditions of Corollary 2.1 hold and f has a fixed point. □
Similarly, we have the following corollary.
Corollary 2.6 Let (X,d) be a complete metric space and let T be an αadmissible mapping. Assume that
hold for all x,y\in X, where \psi \in \mathrm{\Psi} and \ell >0. Also, suppose that the following assertions hold:

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

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
Notice that the main theorem of Dutta and Choudhury [9] remains true if ϕ is lower semicontinuous instead of continuous (see, e.g., [1, 8]).
We assume that
and
where \psi (t)=\phi (t)=0 if and only if t=0.
Theorem 2.2 Let (X,d) be a complete metric space and let T be an αadmissible mapping with respect to η. Assume that for \psi \in {\mathrm{\Psi}}_{1} and \phi \in \mathrm{\Phi},
Also, suppose that the following assertions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge \eta ({x}_{0},T{x}_{0});

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge \eta ({x}_{n},{x}_{n+1}) for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha (x,Tx)\ge \eta (x,Tx) for all n\in \mathbb{N}\cup \{0\}.
Then T has a fixed point.
Proof Let {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge \eta ({x}_{0},T{x}_{0}). Define a sequence \{{x}_{n}\} in X by {x}_{n}={T}^{n}{x}_{0}=T{x}_{n1} for all n\in \mathbb{N}. If {x}_{n+1}={x}_{n} for some n\in \mathbb{N}, then x={x}_{n} is a fixed point for T and the result is proved. We suppose that {x}_{n+1}\ne {x}_{n} for all n\in \mathbb{N}. Since T is an αadmissible mapping with respect to η and \alpha ({x}_{0},T{x}_{0})\ge \eta ({x}_{0},T{x}_{0}), we deduce that \alpha ({x}_{1},{x}_{2})=\alpha (T{x}_{0},{T}^{2}{x}_{0})\ge \eta (T{x}_{0},{T}^{2}{x}_{0})=\eta ({x}_{1},{x}_{2}). By continuing this process, we get \alpha ({x}_{n},{x}_{n+1})\ge \eta ({x}_{n},{x}_{n+1}) for all n\in \mathbb{N}\cup \{0\}. Clearly,
Now, by (2.4) with x={x}_{n1}, y={x}_{n}, we have
which implies
Since ψ is increasing, we get
for all n\in \mathbb{N}. That is, \{{d}_{n}:=d({x}_{n},{x}_{n+1})\} is a nonincreasing sequence of positive real numbers. Then there exists r\ge 0 such that {lim}_{n\to \mathrm{\infty}}{d}_{n}=r. We shall show that r=0. By taking the limit infimum as n\to \mathrm{\infty} in (2.5), we have
Hence \varphi (r)=0. That is, r=0. Then
Suppose, to the contrary, that \{{x}_{n}\} is not a Cauchy sequence. Then there is \epsilon >0 and sequences \{m(k)\} and \{n(k)\} such that for all positive integers k,
Now, for all k\in \mathbb{N}, we have
Taking limit as k\to +\mathrm{\infty} in the above inequality and using (2.6), we get
Since
and
then by taking the limit as k\to +\mathrm{\infty} in the above inequality, and by using (2.6) and (2.7), we deduce that
On the other hand,
Then, by (2.4) with x={x}_{n(k)} and y={x}_{m(k)}, we get
By taking limit as k\to \mathrm{\infty} in the above inequality and applying (2.7) and (2.8), we obtain
That is, \epsilon =0, which is a contradiction. Hence \{{x}_{n}\} is a Cauchy sequence. Since X is complete, then there is z\in X such that {x}_{n}\to z. First we assume that T is continuous. Then we deduce
So, z is a fixed point of T. On the other hand, since
for all n\in \mathbb{N}\cup 0 and {x}_{n}\to z as n\to \mathrm{\infty}, so
which implies
Now, by (2.4) we get
Passing limit inf as n\to \mathrm{\infty} in the above inequality, we have
That is, z=Tz. □
By taking \eta (x,y)=1 in Theorem 2.2, we deduce the following corollary.
Corollary 2.7 Let (X,d) be a complete metric space and let T be an αadmissible mapping. Assume that for \psi \in {\mathrm{\Psi}}_{1} and \phi \in \mathrm{\Phi},
Also, suppose that the following assertions hold:

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

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha (x,Tx)\ge 1.
Then T has a fixed point.
By taking \alpha (x,y)=1 in Theorem 2.2, we deduce the following corollary.
Corollary 2.8 Let (X,d) be a complete metric space and let T be an ηsubadmissible mapping. Assume that for \psi \in {\mathrm{\Psi}}_{1} and \phi \in \mathrm{\Phi},
Also, suppose that the following assertions hold:

(i)
there exists {x}_{0}\in X such that \eta ({x}_{0},T{x}_{0})\le 1;

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \eta ({x}_{n},{x}_{n+1})\le 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \eta (x,Tx)\le 1.
Then T has a fixed point.
Example 2.4 Let X=[0,\mathrm{\infty}) be endowed with the usual metric
for all x,y\in X, and let T:X\to X be defined by
Define also \alpha :X\times X\to [0,+\mathrm{\infty}) and \psi ,\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by
We prove that Corollary 2.7 can be applied to T, but the main theorem in [9] cannot be applied to T.
By a similar proof to that of Example 2.1, we show that T is an αadmissible mapping. Assume that \alpha (x,Tx)\alpha (y,Ty)\ge 1. Now, if x\notin [0,1], then \alpha (x,Tx)=\frac{1}{2} and so \alpha (x,Tx)\alpha (y,Ty)<1, which is contradiction. If y\notin [0,1]. Similarly, \alpha (x,Tx)\alpha (y,Ty)<1, which is contradiction. Hence, \alpha (x,Tx)\alpha (y,Ty)\ge 1 implies x,y\in [0,1]. Therefore, we get
That is,
The conditions of Corollary 2.7 are satisfied. Hence, T has a fixed point. Let x=2 and y=3, then
That is, the main theorem in [9] cannot be applied to T.
Example 2.5 Let X=[0,\mathrm{\infty}) be endowed with the usual metric d(x,y)=xy for all x,y\in X, and let T:X\to X be defined by
Define also \eta :X\times X\to [0,+\mathrm{\infty}) and \psi ,\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by
We prove that Corollary 2.8 can be applied to T, but the main theorem in [9] cannot be applied to T.
By a similar proof to that of Example 2.3, we can show that T is an ηsubadmissible mapping.
Assume that \eta (x,Tx)\eta (y,Ty)\le 1. Now, if x\notin [0,1], then \eta (x,Tx)\eta (y,Ty)>1, which is a contradiction. Similarly, y\notin [0,1] is a contradiction. Hence, \eta (x,Tx)\eta (y,Ty)\le 1 implies x,y\in [0,1]. We get
That is,
Then the conditions of Corollary 2.8 hold and T has a fixed point. Let x=2, y=3. Then T2=0 and T3=1, which implies
That is, the main theorem in [9] cannot be applied to T.
In 1984 Khan et al. [20] proved the following theorem.
Theorem 2.3 Let (X,d) be a complete metric space and let T be a selfmapping on X. Assume that
where \psi \in {\mathrm{\Psi}}_{1} and 0<c<1. Then T has a unique fixed point.
Theorem 2.4 Let (X,d) be a complete metric space and let T be a generalized αadmissible mapping with respect to η. Assume that
where \psi \in {\mathrm{\Psi}}_{1} and 0<c<1. Also, suppose that the following assertions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},{x}_{0})\ge \eta ({x}_{0},{x}_{0});

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n})\ge \eta ({x}_{n},{x}_{n}) for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha (x,x)\ge \eta (x,x).
Then T has a fixed point.
Proof Let {x}_{0}\in X such that \alpha ({x}_{0},{x}_{0})\ge \eta ({x}_{0},{x}_{0}). Define a sequence \{{x}_{n}\} in X by {x}_{n}={T}^{n}{x}_{0}=T{x}_{n1} for all n\in \mathbb{N}. If {x}_{n+1}={x}_{n} for some n\in \mathbb{N}, then x={x}_{n} is a fixed point for T and the result is proved. Hence, we suppose that {x}_{n+1}\ne {x}_{n} for all n\in \mathbb{N}. Since T is a generalized αadmissible mapping with respect to η and \alpha ({x}_{0},{x}_{0})\ge \eta ({x}_{0},{x}_{0}), we deduce that \alpha ({x}_{1},{x}_{1})=\alpha (T{x}_{0},T{x}_{0})\ge \eta (T{x}_{0},T{x}_{0})=\eta ({x}_{1},{x}_{1}). By continuing this process, we get \alpha ({x}_{n},{x}_{n})\ge \eta ({x}_{n},{x}_{n}) for all n\in \mathbb{N}\cup \{0\}. Clearly,
Now, by (2.9) with x={x}_{n1}, y={x}_{n}, we have
Since, ψ is increasing, we get
for all n\in \mathbb{N}. That is, \{{d}_{n}:=d({x}_{n},{x}_{n+1})\} is a nonincreasing sequence of positive real numbers. Then there exists r\ge 0 such that {lim}_{n\to \mathrm{\infty}}{d}_{n}=r. We shall show that r=0. By taking the limit as n\to \mathrm{\infty} in (2.10), we have
which implies \psi (r)=0, i.e., r=0. Then
Suppose, to the contrary, that \{{x}_{n}\} is not a Cauchy sequence. Proceeding as in the proof of Theorem 2.2, there exists \u03f5>0 such that for all k\in \mathbb{N} there exist n(k),m(k)\in \mathbb{N} with m(k)>n(k)\ge k such that
and
Clearly,
Then, by (2.9) with x={x}_{n(k)} and y={x}_{m(k)}, we get
Taking limit as k\to \mathrm{\infty} in the above inequality and applying (2.12) and (2.13), we get
and so \u03f5=0, which is a contradiction. Hence \{{x}_{n}\} is a Cauchy sequence. Since X is complete, then there is z\in X such that {x}_{n}\to z. First, we assume that T is continuous. Then, we deduce
So, z is a fixed point of T. On the other hand, since
for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to z as n\to \mathrm{\infty}, we get
which implies
Then by (2.12) we deduce
Taking limit as n\to \mathrm{\infty} in the above inequality, we have
and then z=Tz. □
Corollary 2.9 Let (X,d) be a complete metric space and let T be an αadmissible mapping. Assume that
where \psi \in {\mathrm{\Psi}}_{1} and 0<c<1. Also, suppose that the following assertions hold:

(i)
there exists {x}_{0}\in X such that \alpha ({x}_{0},{x}_{0})\ge 1;

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \alpha ({x}_{n},{x}_{n})\ge 1 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \alpha (x,x)\ge 1.
Then T has a fixed point.
Corollary 2.10 Let (X,d) be a complete metric space and let T be a generalized αadmissible mapping with respect to η. Assume that
where \psi \in {\mathrm{\Psi}}_{1} and 0<c<1. Also, suppose that the following assertions hold:

(i)
there exists {x}_{0}\in X such that \eta ({x}_{0},{x}_{0})\le 1;

(ii)
either T is continuous or for any sequence \{{x}_{n}\} in X with \eta ({x}_{n},{x}_{n})\le 1 for all n\in \mathbb{N}\cup 0 and {x}_{n}\to x as n\to +\mathrm{\infty}, we have \eta (x,x)\le 1.
Then T has a fixed point.
Example 2.6 Let X=[0,\mathrm{\infty}) be endowed with the usual metric
for all x,y\in X, and let T:X\to X be defined by
Define also \alpha :X\times X\to [0,+\mathrm{\infty}) and \psi ,\phi :[0,\mathrm{\infty})\to [0,\mathrm{\infty}) by
We prove that Corollary 2.9 can be applied to T. But Theorem 2.3 cannot be applied to T.
By a similar proof to that of Example 2.1, we show that T is an αadmissible mapping. Assume that \alpha (x,x)\alpha (y,y)\ge 1. Now, if x\notin [0,1], then \alpha (x,x)=\frac{1}{2} and so \alpha (x,x)\alpha (y,y)<1, which is contradiction. If y\notin [0,1]. Similarly, \alpha (x,x)\alpha (y,y)<1, which is contradiction. Hence, \alpha (x,x)\alpha (y,y)\ge 1 implies x,y\in [0,1]. Therefore, we get
That is,
Then the conditions of Corollary 2.9 hold. Hence, T has a fixed point. Let x=2 and y=3, then T2=8 and T3=18, and hence
That is, Theorem 2.3 cannot be applied to T.
3 Application to the existence of solutions of integral equations
Integral equations like (3.1) were studied in many papers (see [2, 11] and references therein). In this section, we look for a nonnegative solution to (3.1) in X=C([0,T],\mathbb{R}). Let X=C([0,T],\mathbb{R}) be the set of real continuous functions defined on [0,T] and let d:X\times X\to {\mathbb{R}}_{+} be defined by
for all x,y\in X. Then (X,d) is a complete metric space.
Consider the integral equation
and let F:X\to X defined by
We assume that

(A)
f:[0,T]\times \mathbb{R}\to \mathbb{R} is continuous;

(B)
p:[0,T]\to \mathbb{R} is continuous;

(C)
S:[0,T]\times \mathbb{R}\to [0,+\mathrm{\infty}) is continuous;

(D)
there exist \psi \in \mathrm{\Psi} and \theta :X\times X\to \mathbb{R} such that if \theta (x,y)\ge 0 for x,y\in X, then for every s\in [0,T] we have
\begin{array}{rcl}0& \le & f(s,x(s))f(s,y(s))\\ \le & \psi (max\{x(s)y(s),\frac{1}{2}[x(s)F(x(s))+y(s)F(y(s))],\\ \frac{1}{2}[x(s)F(y(s))+y(s)F(x(s))]\left\}\right);\end{array} 
(F)
there exists {x}_{0}\in X such that \theta ({x}_{0},F({x}_{0}))\ge 0;

(G)
if \theta (x,y)\ge 0, x,y\in X, then \theta (Fx,Fy)\ge 0;

(H)
if \{{x}_{n}\} is a sequence in X such that \theta ({x}_{n},{x}_{n+1})\ge 0 for all n\in \mathbb{N}\cup \{0\} and {x}_{n}\to x as n\to +\mathrm{\infty}, then \theta ({x}_{n},x)\ge 0 for all n\in \mathbb{N}\cup \{0\};

(J)
{\int}_{0}^{T}S(t,s)\phantom{\rule{0.2em}{0ex}}ds\le 1 for all t\in [0,T] and s\in \mathbb{R}.
Theorem 3.1 Under assumptions (A)(J), the integral equation (3.1) has a solution in X=C([0,T],\mathbb{R}).
Proof Consider the mapping F:X\to X defined by (3.2).
By the condition (D), we deduce
Then
Now, define \alpha :X\times X\to [0,+\mathrm{\infty}) by
That is, \alpha (x,y)\ge 1 implies
All of the hypotheses of Corollary 2.1 are satisfied, and hence the mapping F has a fixed point that is a solution in X=C([0,T],\mathbb{R}) of the integral equation (3.1). □
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Acknowledgements
The authors thank the referees for their valuable comments and suggestions. This research was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. Therefore, the second and third authors acknowledge with thanks DSR, KAU for financial support. The first author is thankful for support of Astara Branch, Islamic Azad University, during this research.
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Salimi, P., Latif, A. & Hussain, N. Modified αψcontractive mappings with applications. Fixed Point Theory Appl 2013, 151 (2013). https://doi.org/10.1186/168718122013151
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DOI: https://doi.org/10.1186/168718122013151