Existence, uniqueness, Ulam–Hyers–Rassias stability, well-posedness and data dependence property related to a fixed point problem in γ-complete metric spaces with application to integral equations

In this paper, we consider a rational contraction on metric spaces and investigate the fixed point problem associated with it. We assume that the metric space is -complete, which is a concept introduced by Kutbi and Sintunavarat in the paper [11]. The uniqueness of the fixed point is obtained under additional conditions. Rational contractions were first introduced by Dass et al. [6] and have been considered in fixed point theory in recent works like [4, 8]. Our investigation of the different aspects of the fixed point problem is performed in a metric space without completeness property. In most of the works on similar problems, the results are obtained by employing metric completeness. Instead, we assume the weaker concept of -completeness. There is a flexibility in such assumption since the choice of can be different subject to certain restrictions. This is one of the main motivations behind our considerations of the problems discussed in this paper. We impose an admissibility condition on the concerned mapping. The assumption of the contractive inequality is restricted to certain pairs of points. These assumptions are in tune with certain recent trends appearing in metric fixed point theory. Further, there are scopes of extending our present results, which are discussed at the end of the paper.

We investigate Ulam–Hyers–Rassias stability of the fixed point problem. It is a general type of stability, which is considered in several areas of mathematics. Introduced by Ulam [25] through a mathematical question posed in 1940 and later elaborated by Hyers [9] and Rassias [18], such stabilities have a very large literature today [10, 16, 19].

Well-posedness and data dependence property associated with this problem are also investigated.

Finally, we have an application of our results to a problem of a nonlinear integral equation.

Definition 1. (See [1].) An element is called a fixed point of a mapping

Several sufficient conditions have been discussed for the existence of fixed points of , where has a metric defined on it. The study is a part of the subject domain known as metric fixed point theory. The subject is widely recognized to have been originated in the work of Banach in 1922 [1], which is known as the Banach’s contraction mapping principle and is instrumental to the proofs of many important results. In subsequent times, many metric fixed point results were proved and applied to different problem arising in mathematics. Today fixed point methods are recognized as strong mathematical methods. References [12, 13] describe this development to a considerable extent.

Definition 2. (See [22].) A function where is a nonempty set, has triangular property if for imply

Admissibility conditions have recently been used for obtaining fixed point results. Various admissibility criteria were introduced in the study of fixed points of mappings. We refer the reader to [21–23] for some details on admissibility conditions.

Definition 3. (See [22].) Let be a nonempty set, and .The mapping is called -admissible if implies that

Definition 4. Let be a nonempty set and A function is said to have -directed property if for every there existswith such that

Definition 5. (See [23].) A metric space is said to have regular property with respect to a mapping if for any sequence in with limit implies for all .

Example 1. Let be equipped with usual metric. Let and be respectively defined as follows:

Here (i) is a -admissible mapping; (ii) has triangular property; (iii) has regular property with respect to .

Recently, Kutbi and Sintunavarat coined the concept of -complete metric space in the paper [11].

Definition 6. Let be a metric space and .A Cauchy sequence in is called a-Cauchy sequence if for all .

Definition 7. (See [11].) A metric space is said to be -complete, where , if every -Cauchy sequence in . converges to a point in .

Remark 1. If is a complete metric space, then is also a -complete metric space for any but the converse is not true.

Example 2. Let be equipped with usual metric . Let be defined as

then is a-complete metric space. Here is not a complete metric space. Indeed, is a Cauchy sequence in such that for all , then [2019, 2020] for all .As [2019, 2020] is a closed subset of , it follows that there exists [2019, 2020] such that .

Definition 8. Let be a mapping and , where are two metric spaces. The mapping is said to be -continuous at if for any sequence in for all imply that

Remark 2. The continuity of a mapping implies its -continuity for any .In general, the converse is not true.

Problem P. Let be a mapping, where is a metric space. Consider the problem of finding a point satisfying .

Our paper is characterized by the following features.

1. 1. We consider rational contractions in our theorems.
2. 2. We prove our main result with a generalized notion of completeness assumption of the underlying space and without the continuity assumption on the mapping.
3. 3. We investigate Ulam–Hyers–Rassias stability of the fixed point problem.
4. 4. We investigate well-posedness of the problem.
5. 5. We investigate data dependence of fixed point set and solution of the integral equation.
6. 6. We apply our theorem to a problem of an integral equation.

In this section, we establish some fixed point results and illustrate them with examples. We discuss the uniqueness of the fixed point under some additional assumptions. We deduce a corollary of the main result.

Let be a metric space and be two mappings.

We designate the following properties by (A1), (A2) and (A3):

1. (A1) Z has regular property with respect to γ;

2. (A2) γ has triangular property;

3. (A3) γ has F -directed property.

Theorem 1. Let be a metric space and be a function such that is -complete. Let be a γ-admissible mapping and there exists (0, 1) such that for with

If there exists such that and property (A1) holds, then has a fixed point in .

Proof. Let be such that. We construct a sequence in such that

As and is -admissible, we have . Since is

Let

By (1)–(4) we have

Therefore,

Suppose that . From (4) and (5) we have

which is a contradiction. Therefore,, that is, is a monotone decreasing sequence of nonnegative real numbers. Then from (5) we have

By repeated application of (6) we have

With the help of (7), we have

which implies that is a -Cauchy sequence in . As is -complete, there exists such that

By (3), (8) and property (A1) we have for all . Using (2), we have

Taking limit as in (9) and using (8), we have

which implies that , that is,, that is, is a fixed point of .

Remark 3. By Remark 1, Theorem 1 is still valid if one considers to be a complete metric space instead of a-complete metric space.

We present the following illustrative example in support of Theorems 1.

Example 3. Using the metric space , mappings and as in Example 1, we see that is regular with respect to (see Example 1), that is, property (A1) holds, and is a-admissible mapping. Let

Let with Then [0, 1] and [0, 1/8]. Therefore, it is required to verify the inequality in Theorem 1 for [0, 1] and [0, 1/8]. Now, and

Hence, all the conditions of Theorem 1 are satisfied, and 0 is a fixed point of .

Note 1. Theorem 1 is still valid if one considers the -continuity of instead of taking property (A1). Then the portion of the proof just after (8) of Theorem 1 is replaced by the following portion:

Using the -continuity assumption of , we have

Hence, that is, is a fixed point of .

We present the following illustrative example in view of Note 1.

Example 4. Let be equipped with usual metric . Let and be respectively defined as follows:

then is a -complete metric space. Here is not a complete metric space. Let us consider the sequence , where Here as, and for all . But and hence, is not regular with respect to . Also, is -admissible. Here the function is not continuous but -continuous. Choose

Let with Then (0, 1). In view of the above and Note 1, it only remains to be verified that the inequality in Theorem 1 is valid for all (0, 1). Now,and

Here 0 is a fixed point of .

Remark 4. By Remark 2, Theorem 1 is still valid if one considers the continuity of instead of taking property (A1).

Theorem 2. In addition to the hypothesis of Theorem ., suppose that properties (A2) and (A3) hold. Then has a unique fixed point.

Proof. By Theorem 1 the set of fixed points of is nonempty. If possible, let and be two fixed points of . Thenand . Our aim is to show that . By property (A3) there exists with such that Put and let Then Similarly, as in the proof of Theorem 1, we define a sequence such that

As is a -admissible mapping, we have

Arguing similarly as in proof of Theorem 1, we prove that is a-Cauchy sequence in , and there exists such that

We claim that

As and , by property (A2) we have Therefore, our claim is true for . We assume that holds for some. By (10), Applying property (A2), we haveand this proves our claim.

By (1) and (12) we have, for all ,

Taking limit as in (13) and using (11), we have

which is a contradiction unless, that is, , that is,

Similarly, we can show that

From (14) and (15) we have . Therefore, fixed point of is unique.

We present some special cases illustrating the applicability of Theorem 1.

Remark 5. Choosing for all we have a corollary.

Corollary 1. Let be a complete metric space. Then has a unique fixed point if for some (0, 1) and for all one of the following inequalities holds:

(i)

(ii)

(iii)

(iv)

In [19], one can find the following definition as well as some related notions concerning the Ulam–Hyers stability, which is relevant to the present considerations. Let be a metric space and be a mapping. We say that the fixed point problem Ulam–Hyers stable if there is such that for with there exists satisfying

Definition 9. (See [24].) Problem P is called Ulam–Hyers stable if there exists a function which is monotone increasing and continuous at 0 with such that for each and for each solution of the inequality there exists a solution such that

Remark 6. If is defined as , where is a constant, then Definition 9 reduces to Definition in [10].

Let us consider the fixed point Problem and the following inequation:

In the next theorem, we take the following additional condition to assure the Ulam– Hyers stablity via-admissible mapping.

(A4) For any solution of Problem P and any solution of (16), one has

Theorem 3. In addition to the hypothesis of Theorem ., suppose that (A4) holds. Then the fixed point Problem . is Ulam–Hyers stable.

Proof. By Theorem 2 there exists unique such that So, is a solution of Problem P. Let be a solution of (16). Then. By property (A4) we have With the help of (1), we have

which implies that

Let be defined by

The function is monotone increasing, continuous, and By (17) we have

Therefore, the fixed point Problem P is Ulam–Hyers stable.

The notion of well-posedness of a fixed point problem has evoked much interest to several mathematicians (see, for example, [16, 17]). Let be a metric space and be a mapping. The fixed point problem of is said to be well-posed if has a unique fixed point and for any sequence in implies

Definition 10. (See [10].) Problem P is called well-posed if (i) has a unique fixed point , (ii) as whenever is a sequence in with

In the next theorem, we take the following condition to assure the well-posedness via - admissible mapping.

(A5) If is any solution of Problem P and is any sequence in for which

Theorem 4. In addition to the hypothesis of Theorem P, suppose that (A5) holds. Then the fixed point Problem . is well-posed.

Proof. By Theorem 2 there exists unique such that is a solution of Problem P. Let be a sequence in for whichAs (A5) holds, we have for all. By (1) we have

which implies that

Thus, and hence, the fixed point Problem P is well-posed.

In this section, we investigate the data dependence of fixed points.

Definition 11. Let be two mappings, where is a metric space such that for all, where is some positive number. Then the problem of data dependence is to estimate the distance between the fixed points of these two mappings.

Several research papers on data dependence have been published in the recent literature, some of which we mention in references [3, 5, 20].

Theorem 5. In addition to the hypothesis of Theorem ., suppose that be a mapping with nonempty fixed point set. If for each fixed point u of and there exists such that for all, then where s and t are fixed points of F and T, respectively.

Proof. By Theorem 2 there exists unique such that Suppose is a fixed point of . Take Then Let Then by definition of we have

Applying the assumption of the theorem, we have then by admissibility property of we haveInductively, arguing similarly as in the proof of Theorem 1, we have a sequence in such that

Arguing similarly as in proof of Theorem 1, we can prove that

• (7) is satisfied;

• is a -Cauchy sequence in the metric space , and there exists such that

• is a fixed point of , that is, as fixed point of is unique, we have Using triangular property, we have

Taking limit as in the above inequality and using (18), we have

We have already mentioned in introduction that fixed point theorems in metric spaces are widely investigated and have applications in differential and integral equations (see [2,15, 22]). In this section, we deal with a nonlinear integral equation. In the first part, we apply Theorems 1 and 2 to prove the existence and uniqueness of solution of Fredholm-type nonlinear integral equations. In the remaining part, we discuss three aspects of the same integral equation, namely, Ulam–Hyers stability, well-posedness and data dependence.

We consider the following Fredholm-type nonlinear integral equation:

where the unknown function takes real values.

The space of all real valued continuous functions on [a, b] endowed with the metric is complete. Let this metric space be endowed with a partial ordered relation defined as ifand only if for all

Problem I. To find out a solution of the Fredholm-type integral equation

under some appropriate conditions on g, h and K.

We take the following assumptions:

(I1) are continuous mappings

(I2) implies for all

(I3) for all with and for all

(I4) where

(I5) There exists such that

(I6) For every([a, b]), there exists such that and for all

Theorem 6. Let and let satisfy assumptions (I1).(I5). Then nonlinear integral equation (19) has a solution in

Proof. Define a mapping

Let([a, b]) and Then hence, by (I2) we have

which implies

Let and Then .Hence, by (13) we have for all

Let be defined by

Now,implies Thus, by (22) the contraction condition holds for all

By (21), for with we have It follows that for Hence, -admissible.

Suppose that is a convergent sequence in with limit and for all. Then for all and for all, which implies that for all and for all that is,for all . It follows that if is a convergent sequence in with limit and then for all . Therefore, has-regular property, that is, property (A1) holds.

By (I5) there exists such that

So,This implies that there existssuch that

being complete, is a -complete metric space (see Remark 1).

All the assumptions of Theorem 1 are satisfied. Therefore, has a fixed point, that is, the integral equation (19) has a solution in

Example 5. Consider the integral equation

Observe that this equation is a special case of (19) with

• are continuous mappings, and hence, assumption (I1)holds.

• Assumption (I2) holds. To check, let Then

since the function is increasing in [0, 1].

Assumption (I3) holds. To check, let Then for all[0,1], we have

• Assumption (I4) holds. To check, let [0, 1]. Then

where

• Assumption (I5) holds. To check, let for all [0, 1],Then such that for all, that is,

Therefore, all the assumptions of Theorem 6 are satisfied. Hence, integral equation (23) has a solution in([0, 1]).

Theorem 7. In addition to the hypothesis of Theorem 6, suppose that assumption (I6) holds. Then nonlinear integral equation (19) has a unique solution.

Proof. First, we show that has-triangular property. Let and and. By definition of we have and that is, and for all [a, b], which imply thatfor all [a, b], that is,, that is,Hence,has-triangular property. Therefore, property (A2) holds.

By assumption (I6), for, there exists such that and

Hence, for all[a, b], that is,Thus, for there exists with such that Therefore, has -directed property, that is, property (A3) holds.

All the assumptions of Theorem 2 are satisfied. Thus, by Theorems 2 and 6 . has a unique fixed point, that is, the nonlinear integral equation (19) has a unique solution in

Being motivated by Definition 9, we give definitions of Ulam–Hyers stability for the case of integral equation (19).

Definition 12. Problem I is called Ulam–Hyers stable if there exists a function, which is monotone increasing, continuous at 0 with such that for each and for each solution of the inequality

there exists a solution of the integral equation (19) such that

Let us consider the following the integral inequality:

In the following next theorem, we add a new condition to assure the Ulam–Hyers stability of the integral equation (19):

(I7) For any solution of (19) and any solution of (24), one has

Theorem 8. Let all the hypothesis of Theorem 7 hold. Then integral equation (19) has a unique solution. Also suppose that (I7) holds. Then Problem . is Ulam–Hyers stable, and for given and for any solution u. of (24), we have

where is a mapping given by φ(.) = (. + 2)t/(2(1 − .)) for all t ∈ [0, ∞) and |.(t, s)| ≤ m.

Proof. By Theorem 7 the integral equation (19) has a unique solution ... Hence, it is a unique fixed point of the function . : Z Z defined by (20). Let .. is a solution of the integral inequation (24), hence, .. is a solution of .(x, Fx) ≤ ., and by (I7), .. ... By the definition of γ, γ(.., u.) ≥ 1, that is, property (A4) holds. By application of Theorem 3 the fixed point problem . = Fx is Ulam–Hyers stable. Therefore, the solution of the integral equation (19) is Ulam–Hyers stable, and

where . : [0, ∞) → [0, ∞) is a mapping given by .(.) = (. + 2)t/(2(1 − .)) for all . ∈ [0, ∞).

Being motivated by Definition 10, we give definitions of well-posedness for the case of integral equation (19).

Definition 13. Problem I is called well-posed if (i) integral equation (19) has a unique solution .. in .([a, b]), (ii) x. x. in .([a, b]), whenever x. is a sequence in .([a, b]) satisfying

In the following theorem, we add a new condition to assure the well-posedness for integral equation (19).

(I8) If .. is a solution of the integral equation (19) and {x.} is any sequence in .([a, b]) such that sup then x. ≺ .. for all ..

Theorem 9. Let all the hypothesis of Theorem . hold. Then the integral equation (19) has a unique solution x.. Also suppose that (I8) holds. Then Problem . is well-posed.

Proof. By Theorem 7 the integral equation (19) has a unique solution ... Hence, it is a unique fixed point of the function . : . → . defined by (20). Let {x.} be a sequence in .([a, b]) such that sup

Then by assumption (I8) we have x. x. for all .. From definition of . we have .(x., x.) = 1 for all ., that is, property (A5) holds. By application of Theorem 4 the fixed point Problem P, that is, the problem . = Fx, is well-posed. Therefore, Problem I is well-posed.

Being motivated by Definition 11, we give definitions of data dependence for the case of integral equation.

Definition 14. Let .. .([a, b]) be the unique solution of the integral equation (19) and .. be the solution of the integral equation .(.) = .(.) + . ∫ b K .t, s). (s, x(.)) d. for all . ∈ [a, b], where . ∈ .[a, b] and .. : [a, b] ×R → [0, ∞), .. : [a, b] ×[a, b] → [0, ∞) are continuous mappings. The problem of data dependence is to find sup.∈[a,b] |..(.) − ..(.)|.

Theorem 10. Let all the hypothesis of Theorem . hold and x. be the unique solution of the integral equation (19). Also suppose that if x be any solution of the integral equation

where p ∈ .([a, b]) and h. : [a, b] × [a, b] → [0, ∞), K. : [a, b] × [a, b] → [0, ∞) are continuous mappings, then for all t ∈ [a, b],

Further suppose that there exist ν, η > . such that

And

Then

Proof. By Theorem 7 the integral equation (19) has a unique solution ... Let us define a map . : . → . by

Since . is a solution of (25), it is a fixed point of the mapping . defined by (26). By the assumption of the theorem we have .(.) ≤ . (.)(.), . ∈ [a, b], which implies that

which implies that sup.∈[a,b] |. (.)(.) − . (.)(.)| ≤ . for all . ∈ .([a, b]). So, .(Fx, Tx) ≤ . for all . ∈ .. Thus, all the hypothesis of Theorem 5 are met. Therefore, we have .(x, x.) ≤ M/(1 − .) = (. + λη(. − .)).(1 − .(. − .).).

The result of the Theorem 1 is also valid if we replace the constant . by a Mizoguchi– Takahashi function [7, 14]. Here we have not proceeded with it but this can be taken up in a future work. Also the corresponding problem with multivalued mappings and possible applications to integral inclusion problems is supposed to be of considerable interest. One reason for it is that, in general, the fixed point sets of multivalued mappings are mathematically complicated in their structures. This can also be taken up in future works.