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αadmissible mappings and related fixed point theorems
Journal of Inequalities and Applications volume 2013, Article number: 114 (2013)
Abstract
In this paper, we prove the existence and uniqueness of a fixed point for certain αadmissible contraction mappings. Our results generalize and extend some wellknown results on the topic in the literature. We consider some examples to illustrate the usability of our results.
MSC:46N40, 47H10, 54H25, 46T99.
1 Introduction
Fixed point theory is one of the outstanding subfields of nonlinear functional analysis. It has been used in the research areas of mathematics and nonlinear sciences (see, e.g., [1–8]). In 1922 Banach [10] proved that in a complete metric space every contraction has a unique fixed point. In the proof of this theorem, he not only showed the existence and uniqueness of a fixed point, but also provided a method (generally, iterative) for constructing the fixed point. This property of the Banach theorem differentiates it from other fixed point theorems. Therefore, the Banach fixed point theorem has attracted great attention of authors since then (see, e.g., [11–48]). On the other hand, the fixed point technique suggested by Banach attracted many researchers to solve various concrete problems.
2 Main results
In an attempt to generalize the Banach contraction principle, many researchers extended the following result in certain directions.
Theorem 1 (See, e.g., [9, 37, 38])
Let (X,d) be a complete metric space and f:X\to X be a mapping. Assume that there exists a function \beta :[0,\mathrm{\infty})\to [0,1] such that, for any bounded sequence \{{t}_{n}\} of positive reals, \beta ({t}_{n})\to 1 implies {t}_{n}\to 0 and
for all x,y\in X. Then f has a unique fixed point.
Definition 2 (See, e.g., [40]) Let f:X\to X and \alpha :X\times X\to {\mathbb{R}}_{+}. We say that f is an αadmissible mapping if
Example 3 (cf. [40]) Let X=\mathbb{R}. Define f:X\to X and \alpha :X\times X\to [0,\mathrm{\infty}) by
Then f is αadmissible.
Our first result is the following.
Theorem 4 Let (X,d) be a complete metric space and f:X\to X be an αadmissible mapping. Assume that there exists a function \beta :[0,\mathrm{\infty})\to [0,1] such that, for any bounded sequence \{{t}_{n}\} of positive reals, \beta ({t}_{n})\to 1 implies {t}_{n}\to 0 and
for all x,y\in X where \ell \ge 1. Suppose that either

(a)
f is continuous, or

(b)
if \{{x}_{n}\} is a sequence in X such that {x}_{n}\to x, \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n, then \alpha (x,fx)\ge 1.
If there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1, then f has a fixed point.
Proof Let {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1. Define a sequence \{{x}_{n}\} in X by {x}_{n}={f}^{n}{x}_{0}=f{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 f and the result is proved. Hence, we suppose that {x}_{n+1}\ne {x}_{n} for all n\in \mathbb{N}. Since f is an αadmissible mapping and \alpha ({x}_{0},f{x}_{0})\ge 1, we deduce that \alpha ({x}_{1},{x}_{2})=\alpha (f{x}_{0},{f}^{2}{x}_{0})\ge 1. By continuing this process, we get \alpha ({x}_{n},f{x}_{n})\ge 1 for all n\in \mathbb{N}\cup \{0\}. By the inequality (2.1), we have
then
which implies d({x}_{n},{x}_{n+1})\le d({x}_{n1},{x}_{n}). It follows that the sequence \{d({x}_{n},{x}_{n+1})\} is decreasing. Thus, there exists d\in {\mathbb{R}}_{+} such that {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=d. We will prove that d=0. From (2.2) we have
which implies {lim}_{n\to \mathrm{\infty}}\beta (d({x}_{n1},{x}_{n}))=1. Using the property of the function β, we conclude that
Next, we will prove that \{{x}_{n}\} is a Cauchy sequence. 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, we have
By the triangle inequality, we derive that
. Taking the limit as k\to +\mathrm{\infty} in the above inequality and using (2.3), we get
Again, by the triangle inequality, we find that
and
Taking the limit as k\to +\mathrm{\infty}, together with (2.3) and (2.4), we deduce that
From (2.1), (2.4) and (2.5) we have
Hence,
Letting k\to \mathrm{\infty} in the above inequality, we get
That is, {lim}_{k\to \mathrm{\infty}}d({x}_{n(k)},{x}_{m(k)})=0<\epsilon, 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 suppose that f is continuous. Since f is continuous, then we have
So, z is a fixed point of f. Next, we suppose that (b) holds. Then \alpha (z,fz)\ge 1. Now, by (2.1) we have
That is, d(fz,{x}_{n+1})\le \beta (d(z,{x}_{n}))d(z,{x}_{n}), and so we get
Letting n\to \mathrm{\infty} in the above inequality, we get d(fz,z)=0, that is, z=fz. □
Example 5 Let X=[0,\mathrm{\infty}) be endowed with the usual metric d(x,y)=xy for all x,y\in X and f:X\to X be defined by
Define also \alpha :X\times X\to [0,+\mathrm{\infty}) and \beta :[0,\mathrm{\infty})\to [0,1] by
We prove that Theorem 4 can be applied to f, but Theorem 1 cannot be applied to f.
Clearly, (X,d) is a complete metric space. We show that f 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 fx\le 1. It follows that \alpha (fx,fy)\ge 1. Thus the assertion holds. In reason of the above arguments, \alpha (0,f0)\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,fx)\ge 1.
Let x,y\in [0,1] and y\ge x. We get
Otherwise, \alpha (x,fx)\alpha (y,fy)=0 and so
then the condition of Theorem 4 holds. Hence, f has a fixed point. Let x=2 and y=3. Then
that is, the contractive condition of Theorem 1 does not hold for this example.
Theorem 6 Let (X,d) be a complete metric space and f:X\to X be an αadmissible mapping. Assume that there exists a function \beta :[0,\mathrm{\infty})\to [0,1] such that, for any bounded sequence \{{t}_{n}\} of positive reals, \beta ({t}_{n})\to 1 implies {t}_{n}\to 0 and
for all x,y\in X. Suppose that either

(a)
f is continuous, or

(b)
if \{{x}_{n}\} is a sequence in X such that {x}_{n}\to x, \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n, then \alpha (x,fx)\ge 1.
If there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1, then f has a fixed point.
Proof Let {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1. Define a sequence \{{x}_{n}\} in X by {x}_{n}={f}^{n}{x}_{0}=f{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 f and the result is proved. Hence, we suppose that {x}_{n+1}\ne {x}_{n} for all n\in \mathbb{N}. As in Theorem 4, we conclude that \alpha ({x}_{n},f{x}_{n})\ge 1 for all n\in \mathbb{N}\cup \{0\}. Due to (2.6) we have
which yields that
So, we conclude that d({x}_{n},{x}_{n+1})\le d({x}_{n1},{x}_{n}). It follows that the sequence {d}_{n}:=d({x}_{n},{x}_{n+1}) is decreasing. Thus, there exists d\in {\mathbb{R}}_{+} such that {d}_{n}\to d as n\to \mathrm{\infty}. We claim that d=0. Suppose, to the contrary, that d>0. Considering (2.7), we obtain
which implies {lim}_{n\to \mathrm{\infty}}\beta (d({x}_{n1},{x}_{n}))=1. Hence, d={lim}_{n\to \mathrm{\infty}}{d}_{n}={lim}_{n\to \mathrm{\infty}}d({x}_{n1},{x}_{n})=0, which is a contradiction. Hence, we derive that
We prove that \{{x}_{n}\} is a Cauchy sequence. 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,
Following the related lines in the proof of Theorem 4, we get
and
Now, from (2.6), (2.8) and (2.9), we have
Hence,
By taking limit as k\to \mathrm{\infty}, we get
That is, {lim}_{k\to \mathrm{\infty}}d({x}_{n(k)},{x}_{m(k)})=0<\epsilon, 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 of all, we suppose that f is continuous. We obtain that
due to the continuity of f. Thus, we derived that z is a fixed point of f.
Next, we suppose that (b) holds. Then, \alpha (z,fz)\ge 1. Now, by (2.6) we have
That is, d(fz,{x}_{n+1})\le \beta (d(z,{x}_{n}))d(z,{x}_{n}), and so we get
By taking the limit as n\to \mathrm{\infty}, we get d(fz,z)=0, i.e., z=fz. □
Example 7 Let X=[0,\mathrm{\infty}) be endowed with the usual metric d(x,y)=xy for all x,y\in X and f:X\to X be defined by
Define also \alpha :X\times X\to [0,+\mathrm{\infty}) and \beta :[0,\mathrm{\infty})\to [0,1] by
We prove that Theorem 6 can be applied to f, but Theorem 1 cannot be applied to f.
By a similar method to that in the proof of Example 5, we can show that f is an αadmissible mapping and \alpha ({x}_{n},f{x}_{n})\ge 1, {x}_{n}\to x as n\to +\mathrm{\infty} implies that \alpha (x,fx)\ge 1. Clearly, \alpha (0,f0)\ge 1.
Let x,y\in [0,1]. Then
Otherwise, \alpha (x,fx)\alpha (y,fy)=0, and so
then the contractive condition of Theorem 6 holds and f has a fixed point. Let x=2 and y=4; then
That is, the contractive condition of Theorem 1 does not hold for this example.
Theorem 8 Let (X,d) be a complete metric space and f:X\to X be an αadmissible mapping. Assume that there exists a function \beta :[0,\mathrm{\infty})\to [0,1] such that, for any bounded sequence \{{t}_{n}\} of positive reals, \beta ({t}_{n})\to 1 implies {t}_{n}\to 0 and
for all x,y\in X. Suppose that either

(a)
f is continuous, or

(b)
if \{{x}_{n}\} is a sequence in X such that {x}_{n}\to x, \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n, then \alpha (x,fx)\ge 1.
If there exists {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1, then f has a fixed point.
Proof Let {x}_{0}\in X such that \alpha ({x}_{0},f{x}_{0})\ge 1. Define a sequence \{{x}_{n}\} in X by {x}_{n}={f}^{n}{x}_{0}=f{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 f and the result is proved. Hence, we suppose that {x}_{n+1}\ne {x}_{n} for all n\in \mathbb{N}. As in Theorem 4, we conclude that \alpha ({x}_{n},f{x}_{n})\ge 1 for all n\in \mathbb{N}\cup \{0\}. Now, by (2.10) we have
then
It yields that d({x}_{n},{x}_{n+1})\le d({x}_{n1},{x}_{n}). It follows that the sequence \{d({x}_{n},{x}_{n+1})\} is decreasing. Consequently, there exists d\in {\mathbb{R}}_{+} such that d({x}_{n},{x}_{n+1})\to d as n\to \mathrm{\infty}. Regarding (2.11), we observe that
Thus, we find that {lim}_{n\to \mathrm{\infty}}\beta (d({x}_{n1},{x}_{n}))=1 by the property of the function β. Hence,
Next, we will show that the sequence \{{x}_{n}\} is Cauchy. 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,
Again, by following the lines of the proof of Theorem 4, we derive that
and
Combining (2.10), (2.12) and (2.13), we have
Hence,
By taking limit as k\to \mathrm{\infty}, we get
That is, {lim}_{k\to \mathrm{\infty}}d({x}_{n(k)},{x}_{m(k)})=0. Hence \{{x}_{n}\} is a Cauchy sequence. Since X is complete, then there is z\in X such that {x}_{n}\to z.
First, suppose that f is continuous. Since f is continuous, then we have
So, z is a fixed point of f.
We suppose that (b) holds. Then \alpha (z,fz)\ge 1. Now, by (2.10) we have
That is, d(fz,{x}_{n+1})\le \beta (d(z,{x}_{n}))d(z,{x}_{n}), and so we get
Letting n\to \mathrm{\infty} in the above inequality, we get d(fz,z)=0, i.e., z=fz. □
Example 9 Let X=[0,\mathrm{\infty}) be endowed with the usual metric d(x,y)=xy for all x,y\in X and f:X\to X be defined by
Define also \alpha :X\times X\to [0,+\mathrm{\infty}) and \beta :[0,\mathrm{\infty})\to [0,1] by
We prove that Theorem 8 can be applied to f (here, a fixed point is u=\sqrt{5}2), but Theorem 1 cannot be applied to f.
By a similar method to that in the proof of Example 5, we can show that f is an αadmissible mapping and \alpha ({x}_{n},f{x}_{n})\ge 1, {x}_{n}\to x as n\to +\mathrm{\infty} implies that \alpha (x,fx)\ge 1. Clearly, \alpha (0,f0)\ge 1.
Let x,y\in [0,1]. Then
Otherwise, \alpha (x,fx)\alpha (y,fy)=0, and so
then the conditions of Theorem 8 hold and f has a fixed point. Let x=3 and y=4; then
That is, the contractive condition of Theorem 1 does not hold for this example.
Theorem 10 Assume that all the hypotheses of Theorems 4, 6 and 8 hold. Adding the following condition:

(c)
if x=fx then \alpha (x,fx)\ge 1,
we obtain the uniqueness of the fixed point of f.
Proof Suppose that z and {z}^{\ast} are two fixed points of f such that z\ne {z}^{\ast}. Then \alpha (z,fz)\ge 1 and \alpha ({z}^{\ast},f{z}^{\ast})\ge 1.
For Theorem 4 we have
For Theorem 6 we have
For Theorem 8 we have
Hence, all the three inequalities separately imply that \beta (d(z,{z}^{\ast}))=1. Thus d(z,{z}^{\ast})=0, i.e., z={z}^{\ast} as required. □
Remark 11 By utilizing the technique of Samet et al. [40], we can obtain corresponding coupled fixed point results from our Theorems 4, 6 and 8.
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Acknowledgements
This research was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah. The first author acknowledges with thanks DSR, KAU for financial support. The 3rd author is thankful for support of Astara Branch, Islamic Azad University, during this research.
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Hussain, N., Karapınar, E., Salimi, P. et al. αadmissible mappings and related fixed point theorems. J Inequal Appl 2013, 114 (2013). https://doi.org/10.1186/1029242X2013114
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DOI: https://doi.org/10.1186/1029242X2013114