Higher Order of Convergence with Multivalued Contraction Mappings

Abstract

In this paper, we establish some theorems of fixed point on multivalued mappings satisfying contraction mapping by using gauge function. Furthermore, we use - and -order of convergence. Our main results extend many previous existing results in the literature. Consequently, to substantiate the validity of proposed method, we give its application in integral inclusion.

1. Introduction

A well-known mathematician, Banach, gave a main source of the existence of fixed points. An iterative method was used in the Banach contraction principle which converges to the fixed point linearly. In order to obtain higher order of convergence, Proinov [1] generalized Banach contraction theorem by generalizing the contractive condition which involves a gauge function of order . Later, Kiran and Kamran [2] extended his work and generalized using multivalued maps from a complete metric space into the space of all nonempty proximinal closed spaces. In this context, we study the multivalued contraction mapping involved in metric space from [3–7]. And some details for multivalued mapping and their fixed points are included in [1, 2, 8–15]. Recently, Petrusel investigated the local fixed point results for graphic contractions and multivalued locally contractive operators and proposed their application in optimization theory in [16, 17]. Sow proposed the strong convergence of a modified Mann algorithm for multivalued quasi-nonexpansive mappings and monotone mappings with an application in [18, 19]. (Proinov, General local convergence theory for a class of iterative processes and its applications to Newton’s method. Journal of Complexity, 25, 1, 38–62). We extend some results of Proinov to the case of multivalued maps from a metric space into the space of all bounded closed subspace of . We recall some notations which we used in this paper. Let be a metric space and be a subset of ; distance from every to subset is defined as there is a element such that . Now, distance from the point to set is defined as and , where . We denote as the class of all nonempty closed and bounded subsets of . Let be the generalized Hausdorff metric on generated by metric asfor each . The fixed point of is a point if . Throughout this paper, denotes an interval on containing 0, this interval is represented as or , and for polynomial, we used . We use the order of convergence as -order and -order with for iterative process:and use gauge function on , that is, . In this paper, we present some theorems of fixed point on multivalued mappings satisfying contraction mapping by using gauge function. Also, we use - and -order of convergence. In the last section, we applied our proposed method in integral theory.

2. Preliminaries

In this section, we take some results and definitions from [1, 19].

Definition 1. A gauge mapping with order which is greater than 1 defines with on if it fulfills the following conditions:(1) for each (2) for each The first condition of the above definition can easily show that it is equivalent to the condition and is nondecreasing on . Fixed point theory in metric space is full of fixed point theorems in different classes of contraction, which can be obtained by different properties of gauge function .

Example 1. where and where . Then,(1), where (2) because

Lemma 1. A mapping of order on fulfills the gauge condition and there is another mapping on which is nondecreasing and nonnegative such that

Now, the properties which the mapping has are

Now, the gauge mapping which is defined on is written as

For -order of convergence with at least order , sequence converges to . If a constant exists which is greater then zero such that for sufficiently large , then there are the different cases which depend on ; but, in that case, and . Now, the type of convergence linearly and quadratically depends on the order , respectively, 1 and 2.

Assume a sequence whose converges with order with constant and . Let us take a real number sequence whose order of convergence is and that converge to zero with such that .

3. Main Result

Definition 2. Let be a metric space and define a mapping such that be a Hausdorff metric defined on ; then,where the gauge mapping is .

Lemma 2. Let be a nonempty and arbitrary set and define a mapping such that a mapping which has the starting condition of with gauge mapping of order is on the interval which is . Now, for , each starting point and each . We have

Proof. Here, and is the starting point of . As , using Definition 2 and Lemma 2, we obtainThus, we obtainNow proveAs is nondecreasing on , we havewhere is thw gauge function and . Then,

Theorem 1. Assume a metric space and define a mapping . The initial condition of is and gauge mapping fulfills the condition for all . Let be in . Assume thatwhere is a nondecreasing mapping on such that . Then, has a unique fixed point at which we define . Furthermore, for every starting point of , Picard iteration remains in and converges -linearly to with error which we can estimate aswhere .

Proof. Substitute and obtainwhere .
Thus,Now, we prove that is unique. Suppose that contrary is another fixed point such that and . Then,This implies thatwhich is a contradiction such thatgiven is the initial point of Picard iteration. We obtainas . Then, we obtain . Now, following from the above inequality, we obtainand this implies thatgiven . Soand . Therefore,This proof is complete and it is necessary to note that .

Remark 1. Let be a number in . It is easily represented that following the assumption of theorem, the important point is fulfilled iff the gauge mapping has as the fixed point. Now, take as the root of mapping which can be selected from the theorem. So the fixed point of is 0.

Remark 2. Without using the measurement of the error, the above result will be true. If gamma is used as one of the nondecreasing mappings on , then gamma is right continuous.

Theorem 2. Assume a metric space and define a mapping . The initial condition of is and gauge mapping fulfills the condition for all . Let be in . Assume thatwhere is a nondecreasing mapping on such that is a gauge mapping which is strict of order on . Furthermore, for , it implies that is a nondecreasing nonnegative function on fulfilling

Then, is fixed point of which is unique with . Furthermore, for every starting point of , the following conditions are fulfilled:(1)The iterative sequence is in the set and it will converge to . If , then the sequence converges with -order .(2)For each , the error estimate is where , , and is a nonnegative mapping of such that for each .(3)If there is another mapping which is continuous at , then . Hence, we use the sequence which is iterative and converges to with the order - with aswhere is and a mapping which is real, is continuous at 0, and for all .

Proof. We have already proved that fixed point is unique. Now, we have to prove -order of convergence. Given that for all , we select a arbitrary number which is nonnegative and represented as as follows:Now, by result, we observe thatand . Then, Here, . According to abovementioned lemma, the function satisfiesNow, we shall prove the next part by induction. Substitute and obtain . Assume that it fulfills for . Now, we prove that for ,Now, we have to prove that .
We already know that

Corollary 1. Let be a metric space and define an operator and . Assume thatwhere a gauge mapping which is strict with order is . Then, the fixed point which is unique is contains the set . Furthermore, if , then for every point , the following conditions are fulfilled:(1)The sequence remains in and converge to with - order (2)Now, we have to estimate for all as where and is a nondecreasing nonnegative mapping on satisfying (3) we have to estimate

Proof. As we know that , now substitute and obtainwhere and it is unique. Picard sequence converges with -order asNow, we prove the other part by induction. Suppose ; then, equality holds . Suppose that it is true for ,For ,

Theorem 3. Assume a metric space and define a mapping . The initial condition of is and gauge mapping fulfills the condition for all . Let be in . Assume thatwhere is a nondecreasing mapping on such that is a gauge mapping which is strict of order on . Furthermore, for , this implies that :

Now, for the fixed point of the mapping , contains the set . Furthermore, for the starting point of , the following statements hold:(1)Picard sequence remains in and converges to with order of convergence -order . It converges with -order given that , where is contentious at .(2)For each , we estimate thatwhere and is nondecreasing and nonnegative on satisfying .

Proof. First, we prove that is a fixed point which is unique:Now, substituting above, thenThis implies that . Now, for uniqueness, suppose to be another fixed point and , i.e.,which implies thatwhich is a contradiction so . Now, the first part of the theorem is already proved in the previous theorem. Now, we have the second part:We prove it by mathematical induction. Take ; then, equality holds, i.e.,Assume that it is true for . Now, we prove that for ,We prove this by mathematical induction; now, for ,Assume that it is true for ; now, we prove that for ,and also note thatNow,Hence,

4. Application

We apply the proposed multivalued contraction mappings in integral theory.

Theorem 4. Consider the integralwhere is a compact subset of real line , and satisfies the following conditions:(1) is continuous on the rectangle , where (2) is continuous and bounded as for all :where with and . Then, the interval has the solution of integral.

Proof. Let denote those spaces which contain all those mappings which are continuous. And define metric space as . Consider to be a closed subspace of the space which is defined as . Define asFinding the solution of integral (56) becomes equivalent to the fixed point problem of (58). We represent that is well defined and is defined for every , is a compact subset of for each . For and by definition of , we obtain . Thus, . Since is continuous on , way integral also exists. So is defined for each . Now, we prove for each is a compact subset of . Let . Then, for some andwhich prove that ; then, since was arbitrary and so for each . Now, for compactness, first we prove that is compact. Consider a sequence and for some . Since is compact, then there exists subsequence of such that . Let ; then, we obtainand we note that , where andwhere inf apply on and :Now, is a nondecreasing function such thatFurthermore, note thatNow,and we haveThus, we obtainNow, interchanging and , we obtainThus,

Data Availability

No data were used to support this study.

Conflicts of Interest

All authors declare that there are no conflicts of interest.

Acknowledgments

This research project received a grant from Anhui Jianzhu University. The authors acknowledge the technical and financial support.