Abstract
In this manuscript, we define a new contraction mapping, Pata-Ćirić type contraction at a point, that merges distinct contractions defined by Seghal, Pata and Ćirić. We proved that in a complete space, each Pata-Ćirić type contraction at a point possesses a fixed point. We express an example to illustrate the observed result.
1. Introduction and Preliminaries
In recent decades, the fixed point theory has not only been one of the most interesting research topics in nonlinear functional analysis, but also one of the most dynamic and productive research areas. Although the first published work on this subject was conducted a century ago, the fixed point theory continues to be the center of attention. The reason why the fixed point theory is still the attraction center can be explained as follows: different problems in various disciplines can be turned into fixed point problems. Mostly, solving the fixed problem is easier than the original representation. Note that in the metric fixed point theory, not only the existence (mostly, also uniqueness) of the fixed point (which corresponds to the solution) is guaranteed, but also, it is shown "how to reach" the mentioned fixed point and hence the solution.
Before giving the main results of the manuscript, we shall introduce and fix the notations that we deal with it. Throughout this article, we use the letter S to denote a non-empty set. We presume that d forms a metric over S. Thereafter, the pair denotes metric space. Unless otherwise stated in the particular conditions, the pair expresses that the corresponding metric space is complete. The expressions stand for the set of all “real numbers”, “positive real numbers”, and “non-negative real numbers”. In addition, the letters and are restricted to indicate the set of all positive integers and all non-negative integers, respectively.
In the following theorem, Bryant [1] proved the analog of Banach’s fixed point theorem [2] for not the mapping itself but for its iterated form of the given mapping.
Theorem 1
([1]).A self-mapping T onadmits a unique fixed point, if there existandso that
for all.
Inspired from the idea of Bryant [1], we give the following example to illustrate the necessity of the result of Bryant [1]
Example 1.
Letbe defined by
The inequality (1) holds for. Despite T not being continuous,
Theorem 2
([3]).A continuous self-mapping T onadmits a unique fixed point, if it satisfies the condition: there exists a constantsuch that, for each, there is a positive integersuch that
for all.
Immediately after the publication of Seghal’s article, Guseman [4] showed that the theorem of continuity was unnecessary, see also [5,6].
Throughout this work, we note as the set of all auxiliary functions which are increasing, continuous at zero, and . Furthermore, for an arbitrary point in a , we consider the function
that will be called “the zero of S”.
Theorem 3
([7]).A self-mapping T, defined on, admits a fixed point, if for everyand allthe inequality below holds:
whereand,andare fixed constants.
Definition 1
([8]).A self-mapping T, defined on, is called Pata type contraction at a point if for everyand for anythere exists a positive integersuch that
for allwhereand, , are fixed constants.
Theorem 4
([8]).Suppose that a self-mapping T onis a Pata type contraction at a point. Then, T admits a unique fixed point.
2. Main Results
Definition 2.
A self-mapping T, defined on, is called Pata-Ćirić type contraction at a point if for everyand for eachthere exists a positive integersuch that the following inequality is fulfilled:
for all, whereand,andare fixed constants, with
Theorem 5.
Every Pata-Ćirić type contraction at a point overadmits a unique fixed point. Moreover,asfor each. In addition, there isso thatis continuous at
Proof.
Take an arbitrary point . We presume that , because, on the contrary, forms a fixed point for T and it terminates the proof.
Starting with such a point , we construct a sequence like this:
where, . Iteratively, we find
and thereby we have
for any By (5), for we have
Let such that
Then
and further
where is chosen so that
Continuing in this way, we deduce that
for
Let q be an arbitrary but fixed positive integer and k a positive integer, depending on and q. (We can suppose that and also ). We denote this as
Let such that with , and , , be fixed constants. Using the triangle inequality, by (5) we have
where
Our next purpose is to prove that the sequence is bounded. Taking into account that ), the above inequality becomes
and therefore,
We assert that the sequence is bounded. To use the method of Reductio ad Absurdum, we suppose that the sequence is unbounded. Attendantly, we can find a sub-sequence such that For this reason, there is number such that for each . Thus,
and letting we have
Since as , taking into account the properties of the function , we obtain
which is a contradiction. Therefore, our assumption is false, so that the set is bounded and varying we get that the set is bounded. Thus, taking into account (7), there exists a positive number K such that
for In order to prove that the sequence is Cauchy, let . Denoting by , we have
where in is chosen such that
Let . We have
where Continuing in this way, we found that
and then
Dividing by , the previous inequality gives
Thereof, the sequence is Cauchy on On account of completeness, there is so that
The first goal is to indicate that . Otherwise, it turns to . Accordingly, we have
By the triangle inequality, we have , and letting in the above inequality we have
but this implies that
Taking , since and using the property of function we have that
which is a contradiction. Therefore, , that is u is a fixed point of
Letting and keeping in mind the property of , we have that is, the fixed point of is unique.
Finally, because
taking into account this uniqueness, we have
We are now concerned with the second part of the theorem. Let u be a fix point of T in S. For we define
for and reasoning by contradiction, we will proof that for Presuming that there is such that , since we have,
Since, by the triangle inequality,
and
we get
and then
Keeping, the continuity of the function in mind, together with the property , we find that , letting in the above inequality. It contradicts our assumption We have proved that the sequence is non-increasing, so that there exists such that For all we have
Thus,
or, letting
By repeating the above arguments, when we get Since were fixed, we find that
For the last part, let be an arbitrary sequence, with . We shall indicate that by the method of Reductio ad Absurdum. Suppose, on the contrary, that there is so that . By (5), we have
Therefore, as we have
Furthermore,
and letting , we get which is a contradiction. This means that and since , we conclude that is continuous at ☐
Example 2.
On the setwe define the distance d as follows:
| (,) | |||||
| 0 | 1 | 2 | 3 | 1 | |
| 1 | 0 | 1 | 2 | 2 | |
| 2 | 1 | 0 | 1 | 3 | |
| 3 | 2 | 1 | 0 | 4 | |
| 1 | 2 | 3 | 4 | 0 |
Let the mapping, where
It easy to see that the Theorem 3 it cannot be applied, because for, andwe have
On the other hand, since
choosingandwe have:
for every, so that the mapping T has a unique fixed point.
3. Application
We will consider in this section the problem of the existence and uniqueness of solutions for a second-order nonlinear integrodifferential equation, as an application of our main results.
Let be a Banach space endowed with the norm . We denote by E the Banach space of all continuous functions from in , with the norm
and by the complete metric space with
(Here, )
Moreover, we recall here that the set of bounded linear operators in is said to be strongly continuous cosine family if and only if:
- ;
- is strongly continuous in , for any fixed ;
- , for
Let the be the associated sine family of that cosine family, defined as follows
and two positive constants , such that
for all . Let , be continuous function and the nonlinear integro-differential equation
with the initial value
where are given. Here , defined by
is the infinitesimal generator of a cosine family .
Theorem 6.
The Equation (13) admits an unique mild solutiononprovided that the following hypothesis is satisfied:
- there exist a continuous functionsuch thatfor everyand a nonnegative constant Υwith, such thatandfor alland
Proof.
Considering the mapping be defined by
the Equation (13) becomes , that is, the fixed point of T is the mild solution for (13)–(14). In this case, we have
so that,
Similarly,
or, equivalent
Taking into account the conditions imposed in the hypothesis of the theorem regarding the constant , we can find such that . We have, then,
Therefore, considering , , , we can easily see that for z arbitrary in , there exists such that
that is, the mapping is a Pata-Ćirić type contraction at a point. Therefore, since all the assumption of Theorem 5 are fulfilled, the problem (13)–(14) admits an unique solution. (Moreover, we mention that in the case , the presumptions of the Theorem 5 are not satisfied.) ☐
4. Conclusions
In this paper, we combine and extend Pata type contractions and Ćirić type contraction at a point. With this work, we underline the contribution of V. Pata [7] at the fixed point theory by defining an auxiliary distance function where x is an arbitrary but fixed point. Indeed, in all metric fixed point theory proofs, we choose an arbitrary point and built a constructive (iterative) sequence based on the given mapping that is called Picard sequence:
For a self-mapping f on a metric space X and arbitrary point "x" (renamed as ""). Then, ,
Note that in the original proof of Banach (and many others in the consecutive papers on the metric fixed point theory) for any point “x”, this sequence converges to the fixed point of T. Here, V.Pata, suggest such auxiliary distance function by initiating from an arbitrary point “x”, construct a sequence to refine Banach’s fixed point theorem.
In this note, we employ the approach of Pata in a more general case to generalize and unify several existing results in the literature. Thus, several consequences of our results can be observed by using the examples that have been introduced in [8,9,10,11,12].
Author Contributions
Writing—review and editing, E.K., A.F. and V.R. All authors contributed equally and significantly in writing this article. All authors have read and agreed to the published version of the manuscript.
Funding
This research received no external funding
Conflicts of Interest
The authors declare no conflict of interest.
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