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Some new generalizations of MizoguchiTakahashi type fixed point theorem
Journal of Inequalities and Applications volume 2013, Article number: 493 (2013)
Abstract
In the light of the paper of Hasanzade Asl et al. (Fixed Point Theory Appl. 2012:212, 2012, doi:10.1186/168718122012212), we obtain a fixed point theorem for multivalued mappings on a complete metric space. Our result is a generalized version of some results in the literature, including the famous result of MizoguchiTakahashi (J. Math. Anal. Appl. 141:177188, 1989). Also, we give some examples to illustrate our result.
MSC:54H25, 47H10.
1 Introduction and preliminaries
Let (X,d) be a metric space, and let \mathit{CB}(X) denote the class of all nonempty, closed and bounded subsets of X. It is well known that H:\mathit{CB}(X)\times \mathit{CB}(X)\to \mathbb{R} defined by
is a metric on \mathit{CB}(X), which is called a Hausdorff metric, where d(x,B)=inf\{d(x,y):y\in B\}. Let T:X\to \mathit{CB}(X) be a map, then T is called a multivalued contraction if for all x,y\in X, there exists \lambda \in [0,1) such that
In 1969, Nadler [1] proved a fundamental fixed point theorem for multivalued maps: Every multivalued contraction on a complete metric space has a fixed point.
Then, a lot of generalizations of the result of Nadler have been given (see, for example, [2–5]). One of the most important generalizations of it was given by Mizoguchi and Takahashi [6]. We can find both a simple proof of MizoguchiTakahashi fixed point theorem and an example showing that it is a real generalization of Nadler’s result in [7]. We can also find some important results about this direction in [8–12].
Definition 1 [2]
A function k:[0,\mathrm{\infty})\to [0,1) is said to be an \mathcal{MT}function if it satisfies lim{sup}_{s\to {t}^{+}}k(s)<1 for all t\in [0,\mathrm{\infty}) (MizoguchiTakahashi’s condition).
Lemma 1 [9]
Let k:[0,\mathrm{\infty})\to [0,1) be an \mathcal{MT}function, then the function h:[0,\mathrm{\infty})\to [0,1) defined as h(t)=\frac{1+k(t)}{2} is also an \mathcal{MT}function.
Lemma 2 [9]
k:[0,\mathrm{\infty})\to [0,1) is an \mathcal{MT}function if and only if for each t\in [0,\mathrm{\infty}), there exist {r}_{t}\in [0,1) and {\epsilon}_{t}>0 such that k(s)\le {r}_{t} for all s\in [t,t+{\epsilon}_{t}).
Theorem 1 [6]
Let (X,d) be a complete metric space, and let T:X\to \mathit{CB}(X) be a multivalued map. Assume
for all x,y\in X, where k is an \mathcal{MT}function. Then T has a fixed point.
Recently, Samet et al. [13] introduced the notion of αψcontractive mappings and gave some fixed point results for such mappings. Their results are closely related to some ordered fixed point results. Then, using their idea, some authors presented fixed point results for single and multivalued mappings (see, for example, [13–17]). First, we recall these results. Denote by 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.
Definition 2 [13]
Let (X,d) be a metric space, T be a selfmap on X,\psi \in \mathrm{\Psi} and \alpha :X\times X\to [0,\mathrm{\infty}) be a function. Then T is called αψcontractive whenever
for all x,y\in X.
Note that every Banach contraction mapping is an αψcontractive mapping with \alpha (x,y)=1 and \psi (t)=\lambda t for some \lambda \in [0,1).
Definition 3 [13]
T is called αadmissible whenever \alpha (x,y)\ge 1 implies \alpha (Tx,Ty)\ge 1.
There exist some examples for αadmissible mappings in [13]. For convenience, we mention in here one of them. Let X=[0,\mathrm{\infty}). Define T:X\to X and \alpha :X\times X\to [0,\mathrm{\infty}) by Tx=\sqrt{x} for all x\in X and \alpha (x,y)={e}^{xy} for x\ge y and \alpha (x,y)=0 for x<y. Then T is αadmissible.
Definition 4 [14]
α is said to have (B) property whenever \{{x}_{n}\} is a sequence in X such that \alpha ({x}_{n},{x}_{n+1})\ge 1 for all n\in \mathbb{N} and {x}_{n}\to x, then \alpha ({x}_{n},x)\ge 1 for all n\in \mathbb{N}.
Theorem 2 (Theorem 2.1 of [13])
Let (X,d) be a complete metric space and T:X\to X be an αadmissible and αψcontractive mapping. If there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1 and T is continuous, then T has a fixed point.
Remark 1 If we assume that α has (B) property instead of the continuity of T, then again T has a fixed point (Theorem 2.2 of [13]). If for each x,y\in X there exists z\in X such that \alpha (x,z)\ge 1 and \alpha (y,z)\ge 1, then X is said to have (H) property. Therefore, if X has (H) property in Theorem 2.1 and Theorem 2.2 in [13], then the fixed point of T is unique (Theorem 2.3 of [13]).
Then some generalizations of αψcontractive mappings are given as follows.
Definition 5 [14]
T is called a Ćirić type αψgeneralized contractive mapping whenever
for all x,y\in X, where
Note that every Ćirić type generalized contraction mapping is a Ćirić type αψgeneralized contractive mapping with \alpha (x,y)=1 and \psi (t)=\lambda t for some \lambda \in [0,1).
Theorem 3 (Theorem 2.3 of [14])
Let (X,d) be a complete metric space and T:X\to X be an αadmissible and Ćirić type αψgeneralized contractive mapping. If there exists {x}_{0}\in X such that \alpha ({x}_{0},T{x}_{0})\ge 1 and T is continuous or α has (B) property, then T has a fixed point. If X has (H) property, then the fixed point of T is unique.
We can find some fixed point results for singlevalued mappings in these directions in [15, 17]. Now we recall some multivalued case.
Let (X,d) be a metric space and T:X\to \mathit{CB}(X) be a multivalued mapping. Then T is called multivalued αψcontractive whenever
for all x,y\in X and T is called multivalued {\alpha}_{\ast}ψcontractive whenever
where {\alpha}_{\ast}(Tx,Ty)=inf\{\alpha (a,b):a\in Tx,b\in Ty\}. Similarly, if we replace d(x,y) with m(x,y), we can obtain Ćirić type multivalued αψgeneralized contractive and Ćirić type multivalued {\alpha}_{\ast}ψgeneralized contractive mappings on X.
Let (X,d) be a metric space and T:X\to \mathit{CB}(X) be a multivalued mapping.

(a)
T is said to be αadmissible whenever for each x\in X and y\in Tx with \alpha (x,y)\ge 1 implies \alpha (y,z)\ge 1 for all z\in Ty.

(b)
T is said to be {\alpha}_{\ast}admissible whenever for each x\in X and y\in Tx with \alpha (x,y)\ge 1 implies {\alpha}_{\ast}(Tx,Ty)\ge 1.
Remark 2 It is clear that {\alpha}_{\ast}admissible maps are also αadmissible, but the converse may not be true as shown in the following example.
Example 1 Let X=[1,1] and \alpha :X\times X\to [0,\mathrm{\infty}) be defined by \alpha (x,x)=0 and \alpha (x,y)=1 for x\ne y. Define T:X\to \mathit{CB}(X) by
Let x=1 and y=0\in Tx=\{0,1\}, then \alpha (x,y)\ge 1, but {\alpha}_{\ast}(Tx,Ty)={\alpha}_{\ast}(\{0,1\},\{1\})=0. Thus T is not {\alpha}_{\ast}admissible. Now we show that T is αadmissible with the following cases:
Case 1. If x=0, then y=1 and \alpha (x,y)\ge 1. Also, \alpha (y,z)\ge 1 since z=1\in Ty=\{1\}.
Case 2. If x=1, then y\in \{0,1\} and \alpha (x,y)\ge 1. Also, \alpha (y,z)\ge 1 for all z\in Ty.
Case 3. If x\notin \{1,0\}, then y=x and \alpha (x,y)\ge 1. Also, \alpha (y,z)\ge 1 since z=x\in Ty=\{x\}.
The purpose of this work is to present some generalizations of MizoguchiTakahashi’s fixed point theorem using this new idea.
2 Main results
Theorem 4 Let (X,d) be a complete metric space, and let T:X\to \mathit{CB}(X) be an αadmissible multivalued mapping such that
for all x,y\in X, where k is an \mathcal{MT}function. Suppose that there exist {x}_{0}\in X and {x}_{1}\in T{x}_{0} such that \alpha ({x}_{0},{x}_{1})\ge 1. If T is continuous or α has (B) property, then T has a fixed point.
Proof Define h(t)=\frac{1+k(t)}{2}, then from Lemma 1, h:[0,\mathrm{\infty})\to [0,1) is an \mathcal{MT}function. Let {x}_{0} and {x}_{1} be as mentioned in the hypothesis. If {x}_{0}={x}_{1}, then {x}_{0} is a fixed point of T. Assume {x}_{0}\ne {x}_{1}, then \frac{1k(d({x}_{0},{x}_{1}))}{2}d({x}_{0},{x}_{1})>0. Therefore there exists {x}_{2}\in T{x}_{1} such that
Since T is αadmissible, {x}_{1}\in T{x}_{0} and \alpha ({x}_{0},{x}_{1})\ge 1, then \alpha ({x}_{1},u)\ge 1 for all u\in T{x}_{1}. Thus \alpha ({x}_{1},{x}_{2})\ge 1 since {x}_{2}\in T{x}_{1}. If {x}_{1}={x}_{2}, then {x}_{1} is a fixed point of T. Assume {x}_{1}\ne {x}_{2}, then \frac{1k(d({x}_{1},{x}_{2}))}{2}d({x}_{1},{x}_{2})>0. Therefore there exists {x}_{3}\in T{x}_{2} such that
Again, since T is αadmissible, then \alpha ({x}_{2},{x}_{3})\ge 1. In this way, we can construct a sequence \{{x}_{n}\} in X such that {x}_{n+1}\in T{x}_{n}, \alpha ({x}_{n},{x}_{n+1})\ge 1 and
for all n\in \mathbb{N}. Since h(t)<1 for all t\in [0,\mathrm{\infty}), then \{d({x}_{n},{x}_{n+1})\} is a nonincreasing sequence in [0,\mathrm{\infty}) and so there exists \lambda \ge 0 such that {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=\lambda. Now since h is an \mathcal{MT}function, then lim{sup}_{s\to {\lambda}^{+}}h(s)<1 and h(\lambda )<1. Therefore from Lemma 2 there exist r\in [0,1) and \epsilon >0 such that h(s)\le r for all s\in [\lambda ,\lambda +\epsilon ). Since {lim}_{n\to \mathrm{\infty}}d({x}_{n},{x}_{n+1})=\lambda, then there exists {n}_{0}\in \mathbb{N} such that \lambda \le d({x}_{n},{x}_{n+1})<\lambda +\epsilon for all n\ge {n}_{0} and so
for all n\ge {n}_{0}. Thus, we have
and so \{{x}_{n}\} is a Cauchy sequence. Since X is complete, there exists z\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=z.
If T is continuous, then from the inequality d({x}_{n+1},Tz)\le H(T{x}_{n},Tz), we have d(z,Tz)=0 and so z\in Tz.
Now assume that α has (B) property. Then \alpha ({x}_{n},z)\ge 1 for all n\in \mathbb{N}. Therefore
and, taking limit n\to \mathrm{\infty}, we have d(z,Tz)=0 and so z\in Tz. □
Although {\alpha}_{\ast}admissibility implies αadmissibility of T, we will give the following theorem. However, the contractive condition is slightly different from (2.1).
Theorem 5 Let (X,d) be a complete metric space, and let T:X\to \mathit{CB}(X) be an {\alpha}_{\ast}admissible multivalued mapping such that
for all x,y\in X, where k is an \mathcal{MT}function. Suppose that there exist {x}_{0}\in X and {x}_{1}\in T{x}_{0} such that \alpha ({x}_{0},{x}_{1})\ge 1. If T is continuous or α has (B) property, then T has a fixed point.
Proof Define h(t)=\frac{1+k(t)}{2}, then from Lemma 1, h:[0,\mathrm{\infty})\to [0,1) is an \mathcal{MT}function. Let {x}_{0} and {x}_{1} be as mentioned in the hypothesis. If {x}_{0}\in T{x}_{0}, then {x}_{0} is a fixed point of T. Let {x}_{0}\notin T{x}_{0}. Since {x}_{0}\ne {x}_{1}, then \frac{1k(d({x}_{0},{x}_{1}))}{2}d({x}_{0},{x}_{1})>0. If {x}_{1}\in T{x}_{1}, {x}_{1} is a fixed point of T. Let {x}_{1}\notin T{x}_{1}. Also, since T is {\alpha}_{\ast}admissible, {\alpha}_{\ast}(T{x}_{0},T{x}_{1})\ge 1. Therefore, there exists {x}_{2}\in T{x}_{1} such that
Since \alpha ({x}_{1},{x}_{2})\ge {\alpha}_{\ast}(T{x}_{0},T{x}_{1})\ge 1, then {\alpha}_{\ast}(T{x}_{1},T{x}_{2})\ge 1. Therefore there exists {x}_{3}\in T{x}_{2} such that
Again, if {x}_{2}\in T{x}_{2}, {x}_{2} is a fixed point of T. Let {x}_{2}\notin T{x}_{2}. Since \alpha ({x}_{2},{x}_{3})\ge {\alpha}_{\ast}(T{x}_{1},T{x}_{2})\ge 1, then {\alpha}_{\ast}(T{x}_{2},T{x}_{3})\ge 1. In this way, we can construct a sequence \{{x}_{n}\} in X such that {x}_{n+1}\in T{x}_{n}, \alpha ({x}_{n},{x}_{n+1})\ge 1 and
for all n\in \mathbb{N}. As in the proof of Theorem 4, we can show that \{{x}_{n}\} is a Cauchy sequence in X. Since X is complete, there exists z\in X such that {lim}_{n\to \mathrm{\infty}}{x}_{n}=z.
If T is continuous, then from the inequality d({x}_{n+1},Tz)\le H(T{x}_{n},Tz), we have d(z,Tz)=0 and so z\in Tz.
Now assume that α has (B) property. Then \alpha ({x}_{n},z)\ge 1 for all n\in \mathbb{N}. Since T is {\alpha}_{\ast}admissible, {\alpha}_{\ast}(T{x}_{n},Tz)\ge 1. Therefore
and, taking limit n\to \mathrm{\infty}, we have d(z,Tz)=0 and so z\in Tz. □
Now we give an example to illustrate our main theorems. Note that Theorem 1 cannot be applied to this example.
Example 2 Let X=[1,1] and d(x,y)=xy. Define T:X\to \mathit{CB}(X) by
and \alpha :X\times X\to [0,\mathrm{\infty}) by
Then T is {\alpha}_{\ast}admissible and
for all x,y\in X, where k is any \mathcal{MT}function. Indeed, first we show that T is {\alpha}_{\ast}admissible. If \alpha (x,y)\ge 1, then x,y\in [\frac{1}{2},\frac{1}{2}] and hence
Therefore T is {\alpha}_{\ast}admissible.
Now we consider the following cases:
Case 1. Let x,y\in X with \{x,y\}\cap \{[1,\frac{3}{4})\cup (\frac{3}{4},1]\}\ne \mathrm{\varnothing}, then {\alpha}_{\ast}(Tx,Ty)=0. Thus (2.2) is satisfied.
Case 2. Let x,y\in X with x,y\in [\frac{3}{4},\frac{3}{4}], then
and so again (2.2) is satisfied.
Now, if x,y\in (\frac{3}{4},1] with x\ne y, we have
Therefore there is no \mathcal{MT}function satisfying (1.1).
Remark 3 If we take \alpha :X\times X\to [0,\mathrm{\infty}) by \alpha (x,y)=1, then any multivalued mappings T:X\to \mathit{CB}(X) are αadmissible as well as {\alpha}_{\ast}admissible. Therefore, MizoguchiTakahashi’s fixed point theorem is a special case of Theorem 4 and Theorem 5.
We can obtain some ordered fixed point results from our theorems as follows. First we recall some ordered notions. Let X be a nonempty set and ⪯ be a partial order on X.
Definition 8 [18]
Let A, B be two nonempty subsets of X, the relations between A and B are defined as follows:
(r_{1}) If for every a\in A there exists b\in B such that a\u2aafb, then A{\prec}_{1}B.
(r_{2}) If for every b\in B there exists a\in A such that a\u2aafb, then A{\prec}_{2}B.
(r_{3}) If A{\prec}_{1}B and A{\prec}_{2}B, then A\prec B.
Remark 4 [18]
≺_{1} and ≺_{2} are different relations between A and B. For example, let X=\mathbb{R}, A=[\frac{1}{2},1], B=[0,1], ⪯ be the usual order on X, then A{\prec}_{1}B but A{\nprec}_{2}B; if A=[0,1], B=[0,\frac{1}{2}], then A{\prec}_{2}B while A{\nprec}_{1}B.
Remark 5 [18]
≺_{1}, ≺_{2} and ≺ are reflexive and transitive, but are not antisymmetric. For instance, let X=\mathbb{R}, A=[0,3], B=[0,1]\cup [2,3], ⪯ be the usual order on X, then A\prec B and B\prec A, but A\ne B. Hence, they are not partial orders.
Corollary 1 Let (X,\u2aaf) be a partially ordered set and suppose that there exists a metric d in X such that (X,d) is a complete metric space. Let T:X\to \mathit{CB}(X) be a multivalued mapping such that
for all x,y\in X with x\u2aafy, where k is an \mathcal{MT}function. Suppose that there exists {x}_{0}\in X such that \{{x}_{0}\}{\prec}_{1}T{x}_{0}. Assume that for each x\in X and y\in Tx with x\u2aafy, we have y\u2aafz for all z\in Ty. If T is continuous or X satisfies the following condition:
then T has a fixed point.
Proof Define the mapping \alpha :X\times X\to [0,\mathrm{\infty}) by
Then we have
for all x,y\in X. Also, since \{{x}_{0}\}{\prec}_{1}T{x}_{0}, then there exists {x}_{1}\in T{x}_{0} such that {x}_{0}\u2aaf{x}_{1} and so \alpha ({x}_{0},{x}_{1})\ge 1. Now let x\in X and y\in Tx with \alpha (x,y)\ge 1, then x\u2aafy and so, by the hypotheses, we have y\u2aafz for all z\in Ty. Therefore, \alpha (y,z)\ge 1 for all z\in Ty. This shows that T is αadmissible. Finally, if T is continuous or X satisfies (2.3), then T is continuous or α has (B) property. Therefore, from Theorem 4, T has a fixed point. □
Remark 6 We can give a similar corollary using ≺_{2} instead of ≺_{1}.
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Mınak, G., Altun, I. Some new generalizations of MizoguchiTakahashi type fixed point theorem. J Inequal Appl 2013, 493 (2013). https://doi.org/10.1186/1029242X2013493
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DOI: https://doi.org/10.1186/1029242X2013493