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On ({\alpha}^{\ast},\psi )contractive multivalued mappings
Fixed Point Theory and Applications volume 2013, Article number: 137 (2013)
Abstract
In this paper, we generalize the contractive condition for multivalued mappings given by Asl, Rezapour and Shahzad in 2012. We establish some fixed point theorems for multivalued mappings from a complete metric space to the space of closed or bounded subsets of the metric space satisfying generalized ({\alpha}^{\ast},\psi )contractive condition.
MSC:47H10, 54H25.
1 Introduction
Samet et al. [1] introduced the notion of αψcontractive selfmappings of a metric space. Recently, Asl et al. [2] introduced the notion of {\alpha}^{\ast}ψcontractive mappings to extend the notion αψcontractive mappings. In this paper, we generalize the notion of {\alpha}^{\ast}ψcontractive mappings and prove some fixed point theorems for such mappings.
Let Ψ be a 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 each t>0, where {\psi}^{n} is the n th iterate of ψ. It is known that for each \psi \in \mathrm{\Psi}, we have \psi (t)<t for all t>0 and \psi (0)=0 for t=0 [1]. Let (X,d) be a metric space. A mapping G:X\to X is called αψcontractive if there exist two functions \alpha :X\times X\to [0,\mathrm{\infty}) and \psi \in \mathrm{\Psi} such that \alpha (x,y)d(Gx,Gy)\le \psi (d(x,y)) for each x,y\in X. A mapping G:X\to X is called αadmissible [1] if \alpha (x,y)\ge 1\Rightarrow \alpha (Gx,Gy)\ge 1. We denote by N(X) the space of all nonempty subsets of X, by B(X) the space of all nonempty bounded subsets of X and by \mathit{CL}(X) the space of all nonempty closed subsets of X. For A\in N(X) and x\in X, d(x,A)=inf\{d(x,a):a\in A\}. For every A,B\in B(X), \delta (A,B)=sup\{d(a,b):a\in A,b\in B\}. When A=\{x\}, we denote \delta (A,B) by \delta (x,B). For every A,B\in \mathit{CL}(X), let
Such a map H is called generalized Hausdorff metric induced by d. Let (X,\u2aaf,d) be an ordered metric space and A,B\subseteq X. We say that A{\prec}_{r}B if for each a\in A and b\in B, we have a\u2aafb. We give a few definitions and the result due to Asl et al. [2] for convenience.
Definition 1.1 [2]
Let (X,d) be a metric space and let \alpha :X\times X\to [0,\mathrm{\infty}) be a mapping. A mapping G:X\to \mathit{CL}(X) is {\alpha}^{\ast}admissible if \alpha (x,y)\ge 1\Rightarrow {\alpha}^{\ast}(Gx,Gy)\ge 1, where {\alpha}^{\ast}(Gx,Gy)=inf\{\alpha (a,b):a\in Gx,b\in Gy\}.
Definition 1.2 [2]
Let (X,d) be a metric space. A mapping G:X\to \mathit{CL}(X) is called {\alpha}^{\ast}ψcontractive 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.
Theorem 1.3 [2]
Let (X,d) be a complete metric space, let \alpha :X\times X\to [0,\mathrm{\infty}) be a function, let \psi \in \mathrm{\Psi} be a strictly increasing map and T be a closedvalued, {\alpha}^{\ast}admissible and {\alpha}^{\ast}ψcontractive multifunction on X. Suppose that there exist {x}_{0}\in X and {x}_{1}\in G{x}_{0} such that \alpha ({x}_{0},{x}_{1})\ge 1. Assume that if \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n and {x}_{n}\to x, then \alpha ({x}_{n},x)\ge 1 for all n. Then G has a fixed point.
2 Main results
We begin this section by introducing the following definition.
Definition 2.1 Let (X,d) be a metric space and let G:X\to \mathit{CL}(X) be a mapping. We say that G is generalized ({\alpha}^{\ast},\psi )contractive if there exists \psi \in \mathrm{\Psi} such that
for each x\in X and y\in Gx, where {\alpha}^{\ast}(Gx,Gy)=inf\{\alpha (a,b):a\in Gx,b\in Gy\}.
Note that an {\alpha}^{\ast}ψcontractive mapping is generalized ({\alpha}^{\ast},\psi )contractive. In case when \psi \in \mathrm{\Psi} is strictly increasing, generalized ({\alpha}^{\ast},\psi )contractive is called strictly generalized ({\alpha}^{\ast},\psi )contractive. The following lemma is inspired by [[3], Lemma 2.2].
Lemma 2.2 Let (X,d) be a metric space and B\in \mathit{CL}(X). Then, for each x\in X with d(x,B)>0 and q>1, there exists an element b\in B such that
Proof It is given that d(x,B)>0. Choose
Then, by using the definition of d(x,B), it follows that there exists b\in B such that
□
Lemma 2.3 Let (X,d) be a metric space and G:X\to \mathit{CL}(X). Assume that there exists a sequence \{{x}_{n}\} in X such that {lim}_{n\to \mathrm{\infty}}d({x}_{n},G{x}_{n})=0 and {x}_{n}\to x\in X. Then x is a fixed point of G if and only if the function f(\xi )=d(\xi ,G\xi ) is lower semicontinuous at x.
Proof Suppose f(\xi )=d(\xi ,G\xi ) is lower semicontinuous at x, then
By the closedness of G it follows that x\in Gx. Conversely, suppose that x is a fixed point of G, then f(x)=0\le {lim\hspace{0.17em}inf}_{n}f({x}_{n}). □
Theorem 2.4 Let (X,d) be a complete metric space and let G:X\to \mathit{CL}(X) be an {\alpha}^{\ast}admissible strictly generalized ({\alpha}^{\ast},\psi )contractive mapping. Assume that there exist {x}_{0}\in X and {x}_{1}\in G{x}_{0} such that \alpha ({x}_{0},{x}_{1})\ge 1. Then x is a fixed point of G if and only if f(\xi )=d(\xi ,G\xi ) is lower semicontinuous at x.
Proof By the hypothesis, there exist {x}_{0}\in X and {x}_{1}\in G{x}_{0} such that \alpha ({x}_{0},{x}_{1})\ge 1. If {x}_{0}={x}_{1}, then we have nothing to prove. Let {x}_{0}\ne {x}_{1}. If {x}_{1}\in G{x}_{1}, then {x}_{1} is a fixed point. Let {x}_{1}\notin G{x}_{1}. Since G is {\alpha}^{\ast}admissible, so {\alpha}^{\ast}(G{x}_{0},G{x}_{1})\ge 1, we have
For given q>1 by Lemma 2.2, there exists {x}_{2}\in G{x}_{1} such that
It follows from (2.3), (2.4) and (2.1) that
It is clear that {x}_{1}\ne {x}_{2} and \alpha ({x}_{1},{x}_{2})\ge 1. Thus {\alpha}^{\ast}(G{x}_{1},G{x}_{2})\ge 1. Since ψ is strictly increasing, by (2.5), we have
Put {q}_{1}=\frac{\psi (q\psi (d({x}_{0},{x}_{1})))}{\psi (d({x}_{1},{x}_{2}))}, then {q}_{1}>1. If {x}_{2}\in G{x}_{2}, then {x}_{2} is a fixed point. Let {x}_{2}\notin G{x}_{2}, then by Lemma 2.2, there exists {x}_{3}\in G{x}_{2} such that
It is clear that {x}_{2}\ne {x}_{3}, \alpha ({x}_{2},{x}_{3})\ge 1 and \psi (d({x}_{2},{x}_{3}))<{\psi}^{2}(q\psi (d({x}_{0},{x}_{1}))). Now put {q}_{2}=\frac{{\psi}^{2}(q\psi (d({x}_{0},{x}_{1})))}{\psi (d({x}_{2},{x}_{3}))}. Then {q}_{2}>1. If {x}_{3}\in G{x}_{3}, then {x}_{3} is a fixed point. Let {x}_{3}\notin G{x}_{3}. Then by Lemma 2.2 there exists {x}_{4}\in G{x}_{3} such that
By continuing the same process, we get a sequence \{{x}_{n}\} in X such that {x}_{n+1}\in G{x}_{n}. Also, {x}_{n}\ne {x}_{n+1}, \alpha ({x}_{n},{x}_{n+1})\ge 1 and 0<d({x}_{n},{x}_{n+1})<{\psi}^{n1}(q\psi (d({x}_{0},{x}_{1}))) or
For each m>n, we have
Since \psi \in \mathrm{\Psi}, it follows that \{{x}_{n}\} is a Cauchy sequence in X. Thus there is x\in X such that {x}_{n}\to x. Letting n\to \mathrm{\infty} in (2.6), we have
The rest of the proof follows from Lemma 2.3. □
Example 2.5 Let X=\mathbb{R} be endowed with the usual metric d. Define G:X\to \mathit{CL}(X) and \alpha :X\times X\to [0,\mathrm{\infty}) by
and
Let \psi (t)=\frac{t}{2} for all t\ge 0. For each x\in X and y\in Gx, we have
Hence G is a strictly generalized ({\alpha}^{\ast},\psi )contractive mapping. Clearly, G is {\alpha}^{\ast}admissible. Also, we have {x}_{0}=1 and {x}_{1}=1\in G{x}_{0} such that \alpha ({x}_{0},{x}_{1})=1. Therefore, all conditions of Theorem 2.4 are satisfied and G has infinitely many fixed points. Note that Theorem 1.3 in Section 1 is not applicable here. For example, take x=1 and y=1.
Corollary 2.6 Let (X,\u2aaf,d) be a complete ordered metric space, \psi \in \mathrm{\Psi} be a strictly increasing map and G:X\to \mathit{CL}(X) be a mapping such that for each x\in X and y\in Gx with x\u2aafy, we have
Also, assume that

(i)
there exist {x}_{0}\in X and {x}_{1}\in G{x}_{0} such that {x}_{0}\u2aaf{x}_{1},

(ii)
if x\u2aafy, then Gx{\prec}_{r}Gy.
Then x is a fixed point of G if and only if f(\xi )=d(\xi ,G\xi ) is lower semicontinuous at x.
Proof Define \alpha :X\times X\to [0,\mathrm{\infty}) by
By using condition (i) and the definition of α, we have \alpha ({x}_{0},{x}_{1})=1. Also, from condition (ii), we have x\u2aafy implies Gx{\prec}_{r}Gy; by using the definitions of α and {\prec}_{r}, we have \alpha (x,y)=1 implies {\alpha}^{\ast}(Gx,Gy)=1. Moreover, it is easy to check that G is a strictly generalized ({\alpha}^{\ast},\psi )contractive mapping. Therefore, by Theorem 2.4, x is a fixed point of G if and only if f(\xi )=d(\xi ,G\xi ) is lower semicontinuous at x. □
Definition 2.7 Let (X,d) be a metric space and G:X\to B(X) be a mapping. We say that G is a generalized ({\alpha}^{\ast},\psi ,\delta )contractive mapping if there exists \psi \in \mathrm{\Psi} such that
for each x\in X and y\in Gx, where {\alpha}^{\ast}(Gx,Gy)=inf\{\alpha (a,b):a\in Gx,b\in Gy\}.
Lemma 2.8 Let (X,d) be a metric space and G:X\to B(X). Assume that there exists a sequence \{{x}_{n}\} in X such that {lim}_{n\to \mathrm{\infty}}\delta ({x}_{n},G{x}_{n})=0 and {x}_{n}\to x\in X. Then \{x\}=Gx if and only if the function f(\xi )=\delta (\xi ,G\xi ) is lower semicontinuous at x.
Proof Suppose that f(\xi )=\delta (\xi ,G\xi ) is lower semicontinuous at x, then
Hence, \{x\}=Gx because \delta (A,B)=0 implies A=B=\{a\}. Conversely, suppose that \{x\}=Gx. Then f(x)=0\le {lim\hspace{0.17em}inf}_{n}f({x}_{n}). □
Theorem 2.9 Let (X,d) be a complete metric space and let G:X\to B(X) be an {\alpha}^{\ast}admissible generalized ({\alpha}^{\ast},\psi ,\delta )contractive mapping. Assume that there exist {x}_{0}\in X and {x}_{1}\in G{x}_{0} such that \alpha ({x}_{0},{x}_{1})\ge 1. Then there exists x\in X such that \{x\}=Gx if and only if f(\xi )=\delta (\xi ,G\xi ) is lower semicontinuous at x.
Proof By the hypothesis of the theorem, there exist {x}_{0}\in X and {x}_{1}\in G{x}_{0} such that \alpha ({x}_{0},{x}_{1})\ge 1. Assume that {x}_{0}\ne {x}_{1}, for otherwise, {x}_{0} is a fixed point. Let {x}_{1}\notin G{x}_{1}. As G is {\alpha}^{\ast}admissible, we have {\alpha}^{\ast}(G{x}_{0},G{x}_{1})\ge 1. Then
Since G{x}_{1}\ne \mathrm{\varnothing}, there is {x}_{2}\in G{x}_{1}. Then
From (2.12) and (2.13), we have
Since ψ is nondecreasing, we have
As {x}_{2}\in G{x}_{1}, we have \alpha ({x}_{1},{x}_{2})\ge 1. Since G{x}_{2}\ne \mathrm{\varnothing}, there is {x}_{3}\in G{x}_{2}. Assume that {x}_{2}\ne {x}_{3}, for otherwise, {x}_{2} is a fixed point of G. Then
Since ψ is nondecreasing, we have
By continuing in this way, we get a sequence \{{x}_{n}\} in X such that {x}_{n+1}\in G{x}_{n} and {x}_{n}\ne {x}_{n+1} for n=0,1,2,3,\dots . Further we have
For each m>n, we have
Since \psi \in \mathrm{\Psi}, it follows that \{{x}_{n}\} is a Cauchy sequence in X. As X is complete, there exists x\in X such that {x}_{n}\to x. Letting n\to \mathrm{\infty} in (2.18), we have
The rest of the proof follows from Lemma 2.8. □
Example 2.10 Let X=\{0,2,4,6,8,10,\dots \} be endowed with the usual metric d. Define G:X\to B(X) and \alpha :X\times X\to [0,\mathrm{\infty}) by
and
Let \psi (t)=\frac{t}{2} for all t\ge 0. For each x\in X and y\in Gx, we have
Hence G is a generalized ({\alpha}^{\ast},\psi ,\delta )contractive mapping. Clearly, G is {\alpha}^{\ast}admissible. Also, we have {x}_{0}=0\in X and {x}_{1}=0\in G0 such that \alpha ({x}_{0},{x}_{1})=1. Therefore, all conditions of Theorem 2.9 are satisfied and G has infinitely many fixed points.
Corollary 2.11 Let (X,\u2aaf,d) be a complete ordered metric space, \psi \in \mathrm{\Psi} and G:X\to B(X) be a mapping such that for each x\in X and y\in Gx with x\u2aafy, we have
Also, assume that

(i)
there exists {x}_{0}\in X such that \{{x}_{0}\}{\prec}_{1}G{x}_{0}, i.e., there exists {x}_{1}\in G{x}_{0} such that {x}_{0}\u2aaf{x}_{1},

(ii)
if x\u2aafy, then Gx{\prec}_{r}Gy.
Then there exists x\in X such that \{x\}=Gx if and only if f(\xi )=\delta (\xi ,G\xi ) is lower semicontinuous at x.
Proof Define \alpha :X\times X\to [0,\mathrm{\infty}) by
By using condition (i) and the definition of α, we have \alpha ({x}_{0},{x}_{1})=1. Also, from condition (ii), we have x\u2aafy implies Gx{\prec}_{r}Gy, by using the definitions of α and {\prec}_{r}, we have \alpha (x,y)=1 implies {\alpha}^{\ast}(Gx,Gy)=1. Moreover, it is easy to check that G is a generalized ({\alpha}^{\ast},\psi ,\delta )contractive mapping. Therefore, by Theorem 2.9, there exists x\in X such that \{x\}=Gx if and only if f(\xi )=\delta (\xi ,G\xi ) is lower semicontinuous at x. □
References
Samet B, Vetro C, Vetro P: Fixed point theorems for α  ψ contractive type mappings. Nonlinear Anal. 2012, 75: 2154–2165. 10.1016/j.na.2011.10.014
Asl JH, Rezapour S, Shahzad N: On fixed points of α  ψ contractive multifunctions. Fixed Point Theory Appl. 2012., 2012: Article ID 212. doi:10.1186/1687–1812–2012–212
Kamran T: MizoguchiTakahashi’s type fixed point theorem. Comput. Math. Appl. 2009, 57: 507–511.
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Authors are grateful to referees for their suggestions and careful reading.
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Ali, M.U., Kamran, T. On ({\alpha}^{\ast},\psi )contractive multivalued mappings. Fixed Point Theory Appl 2013, 137 (2013). https://doi.org/10.1186/168718122013137
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DOI: https://doi.org/10.1186/168718122013137
Keywords
 αadmissible
 {\alpha}^{\ast}ψcontractive mapping
 generalized ({\alpha}^{\ast},\psi )contractive mapping
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