# Fixed Point Results via Real-Valued Function Satisfying Integral Type Rational Contraction

Authors: Naveen Mani; Amit Sharma; Rahul Shukla
Journal: Abstract and Applied Analysis
 (2023)

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## Abstract

In this article, we mainly discuss the existence and uniqueness of fixed point satisfying integral type contractions in complete metric spaces via rational expression using real-valued functions. We improve and unify many widely known results from the literature. Among these, the work of Rakotch (1962), Branciari (2002), and Liu et al. (2013) is extended. Finally, we conclude with an example presented graphically in favour of our work.


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## Body

## 1. Introduction

We start this section by recalling the definition of Lebesgue-integrable function. NotifyL as a function defined as
(1)L=l:R+⟶R+,which is nonnegative, summable on each compact subset of R+, and such that for each ε>0,(2)∫0εlmdm>0.Branciari [1] in 2002 independently and essentially deduced the following result, as an extension of most famous problem of Banach in 1922.Theorem 1 [1].
LetP,d be a complete metric space, 0<s<1, and U:P⟶P is a map. If for each g,h∈P(3)∫0dUg,Uhlmdm≤s∫0dg,hlmdm,where l∈L. Then, z∈P is a unique fixed point of U.Rhoades [2] made a major extension in 2003 of Branciari [1] by proving a more genral result. His proof introduced a number of interesting ideas for other reseachers to study on integral type of contracions. Prior to Branciari [1] works, Kumar et al. [3] had been able to derive Jungck’s [4] fixed point result in sense of integral type contractions. Mocanu and Popa [5] proposed following lemmas that are useful for deriving our main theorem.Lemma 1 [5].
Letl∈L and rss∈N be a nonnegative sequence with lims⟶∞rs=c then
(4)lims⟶∞∫0rslmdm=0implieslims⟶∞rs=0.Lemma 2 [5].
Letl∈L and rss∈N be a nonnegative sequence with lims⟶∞rs=c then
(5)lims⟶∞∫0rslmdm=∫0clmdm.Further, Liu et al. [6] extended Branciari’s work by including real-valued function and improved the result of Rakotch [7].Theorem 2 [6].
LetU be a self-mapping on a complete metric space P,d satisfying
(6)∫0dUg,Uhlmdm≤δdg,h∫0dg,hlmdm,for each g,h∈P, δ:0,∞⟶0,1 s.t.
(7)lim supδs<1,∀s>0,and l∈L. Then Pz=z for all z∈U.In addition to previous findings, Gupta et al. [8] in 2012 proposed a work for 2 compatible self-maps and derived a result satisfying integral type contraction. In contrast to Rakotch’s result, further in 2013, Gupta and Mani [9] placed a rational contraction using real-valued function and established their theorem.Theorem 3 [9].
LetU be a self-map on a complete metric space P,d. If for each g,h∈P(8)∫0dUg,Uhlmdm≤γdg,h∫0wg,hlmdm,where
(9)wg,h=maxdg,Ug·dh,Uhdg,h,dg,h,
l∈L and a function γ:0,∞⟶0,1 with lims⟶nsupγs<1 for all n>0. Then, U has a unique fixed point in P.About the same time, Liu et al. [10] come with different approach and set up three distinct results for integral type contractions. These studies further give other aspects of integral contractions for researchers, in particular related problems on real-valued functions. Some motivated results on integral type contractions and in metric spaces are refer to see [11–17].This article is devoted to state the theorem containing real-valued function and to prove the theorem satisfying integral type rational contraction. Our finding extends and generalized some renowned result. An example with graphical representation has been given in favour of our work.


## 2. Fixed Point via Rational Contraction and by Using Real-Valued Function

Geraghty [18] defined the following class of test function which is more general than the Rakotch [7].Definition 1 [18].
DefineS=γγ:0,∞⟶0,1 satisfies the condition
(10)γti⟶1impliesti⟶0.Example 1.
Define the function(11)γt=logt2+1t,
Clearly,γti<1 and γti⟶1impliesti⟶0.Theorem 4.
LetU be a self-mapping on a complete metric space P,d and are such that for each g,h∈P(12)∫0dUg,Uhlmdm≤γdg,h∫0wg,hlmdm,where
(13)wg,h=maxdg,h,dg,Ugdh,Uh1+dg,h,dh,Uhdg,Uh1+dUg,Uh,l∈L and γ∈S. Then U has a unique fixed point.Proof.
Set initial approximationg0∈P as an any arbitrary point in P. In general, construct gi in P such that
(14)Ugi=gi+1.
First, we assert thatlimi⟶∞dgi,gi+1=0.
From Equation (12), ∀i≥0, we have
(15)∫0dgi,gi+1lmdm≤γdgi−1,gi∫0wgi−1,gilmdm,where
(16)wgi−1,gi=maxdgi−1,gi,dgi−1,Ugi−1dgi,Ugi1+dgi−1,gi,dgi,Ugi−1dgi−1,Ugi1+dUgi−1,Ugi,=maxdgi−1,gi,dgi−1,gidgi,gi+11+dgi−1,gi,dgi,gidgi−1,gi+11+dgi,gi+1,=maxdgi−1,gi,dgi,gi+1.
Now, ifdgi,gi+1>dgi−1,gi, then wgi−1,gi=dgi,gi+1. Hence, from Equation (15) and using the fact that γ∈S, we arrived at a contradiction. Therefore, dgi,gi+1<dgi−1,gi and so wgi−1,gi=dgi−1,gi.
Thus, Equation (15) implies that
(17)∫0dgi,gi+1lmdm≤γdgi−1,gi∫0dgi−1,gilmdm.
Sinceγ∈S, we have
(18)∫0dgi,gi+1lmdm≤∫0dgi−1,gilmdm.
Similarly,(19)∫0dgi−1,gilmdm≤∫0dgi−2,gi−1lmdm.
Thus, a monotone decreasing sequence∫0dgi,gi+1lmdm of nonnegative reals has obtained, and so there exists s≥0 such that
(20)limi⟶∞∫0dgi,gi+1lmdm=s.
Assume thats>0. Letting i⟶∞ in Equation (15) and using Equation (20), we get s≤s, as γ∈S, a contradiction, implies s=0 and hence
(21)limi⟶∞∫0dgi,gi+1lmdm=0.
Lemma1 implies
(22)limi⟶∞dgi,gi+1=0.
Next, we assert that sequencegi is Cauchy.
Assume that for anϵ>0, there exists subsequences gis and gws of gi with ws>is≥s,s>0 satisfying
(23)dgws,gis≥ϵanddgws−1,gis<ϵ,∀s≥0, consider
(24)∫0εlmdm≤∫0dgws,gislmdm≤γdgws−1,gis−1∫0wgws−1,gis−1lmdm,where
(25)wgws−1,gis−1=maxdgws−1,gis−1,dgws−1,gwsdgis−1,gis1+dgws−1,gis−1,dgis−1,gwsdgws−1,gis1+dgws,gis−1.
Triangle inequality implies that(26)dgws−1,gis−1=dgws−1,gis+dgis,gis−1.
From Equation (23), dgws−1,gis−1≤ϵ+dgis,gis−1.
On taking limits and using Equations (22), we get
(27)limi⟶∞dgws−1,gis−1=ϵ.
Again, we know(28)dgis−1,gws≤dgis−1,gws−1+dgws−1,gws.
Therefore, on takinglims⟶∞ and using (23) and (22), we get
(29)lims⟶∞dgis−1,gws=ϵ.
Hence, from Equation (25), on taking lims⟶∞ and using Equations (23), (27), and (29), we get
(30)lims⟶∞wgws−1,gis−1=ϵ.
Thus, on lettinglims⟶∞, Equation (24) implies that
(31)∫0εlmdm≤γϵ∫0ϵlmdm,where we arrived at a contradiction as γ∈S. Therefore, sequence gi is Cauchy. Call a limit v such that from (12)
(32)limi⟶∞gi+1=limi⟶∞Ugi=v.
Now, assert thatv is a fixed point of U.
Indeed, continuity ofU implies that
(33)v=limi⟶∞Ugi=Ulimi⟶∞gi=Uv.
Secondly, assumeU is not continuous and also let Uv≠v. Then, clearly dUv,v≠0. Assume that dUv,v>0. Therefore
(34)0<∫0dUv,vlmdm=limi⟶∞∫0dUv,gi+1lmdm≤limi⟶∞∫0wv,gilmdm,where
(35)wv,gi=maxdv,gi,dv,Uvdgi,Ugi1+dv,gi,dgi,Uvdv,Ugi1+dUv,Ugi.
Takelimi⟶∞, we obtain
(36)limi⟶∞wv,gi=maxdv,v,dv,Uvdv,v1+dv,Ugi,dv,Uvdv,v1+dUv,v,=0.
Hence from (34),
(37)0<∫0dUv,vlmdm=∫0dUv,Ugilmdm≤0,
Therefore,Uv=v.
For uniqueness, assume there exist a points≠v other than v s.t dUs,s=0. Consider,
(38)∫0ds,vlmdm=∫0dUs,Uvlmdm≤γds,v∫0ws,vlmdm,where
(39)ws,v=maxds,v,ds,Uvdv,Uv1+ds,v,dv,Usds,Uv1+dUs,Uv,=maxds,v,ds,sdv,v1+ds,v,dv,sds,v1+ds,v.
Sinceds,v/1+ds,v<1, therefore
(40)ws,v≤maxds,v,0,ds,v≤ds,v.
Using the fact thatγ∈S and from (38), we have
(41)∫0ds,vlmdm<∫0ds,vlmdm.
This impliess=v, and hence, fixed point of U is unique. This accomplished our proof.Theorem 5.
Let a self-mapU on a complete metric space P,d such that for each g,h∈P(42)∫0dUg,Uhlmdm≤δdg,h∫0dg,hlmdm+γdg,h∫0ig,hlmdm,where
(43)ig,h=maxdg,Ugdh,Uh1+dg,h,dh,Uhdg,Uh1+dUg,Uh.
l∈L and δ,γ∈S with δm+γm<1. Then, U has a unique fixed point.Proof.
Sinceδ,γ∈S with δm+γm<1. Let γm=maxδm,γm. Then from Equation (42), we have
(44)∫0dUg,Uhlmdm≤δdg,h∫0dg,hlmdm+γdg,h∫0ig,hlmdm<γdg,h∫0maxdg,h,ig,hlmdm≤γdg,h∫0wg,hlmdm,where
(45)wg,h=maxdg,h,dg,Ugdh,Uh1+dg,h,dh,Uhdg,Uh1+dUg,Uh.
Rest of the proof is on the same line of Theorem4.If we takelt=1, then we have the following two consequence results from our main theoremCorollary 1.
Let a self-mapU on a complete metric space P,d such that for each g,h∈P(46)dUg,Uh≤γdg,hwg,h,where
(47)wg,h=maxdg,h,dg,Ugdh,Uh1+dg,h,dh,Uhdg,Uh1+dUg,Uh,and γ∈S. Then, U has a unique fixed point.Corollary 2.
Let a self-mapU on a complete metric space P,d such that for each g,h∈P(48)dUg,Uh≤δdg,hdg,h+γdg,hig,h,where
(49)ig,h=maxdg,Ugdh,Uh1+dg,h,dh,Uhdg,Uh1+dUg,Uh,and δ,γ∈S with δt+γt<1. Then U has a unique fixed point.Example 2.
LetP=N. Define the metric dg,h=maxg,h, for all g,h∈P. Clearly, P,d is a metric space. Define a function U:P⟶P as Ug=g,∀g∈P. Also defined l∈L as lm=2m,∀m∈R+ and δ:R+⟶0,1 is defined by δm=9/10
Figure1 showing the plot of inequality (12) satisfying the Example 2. Thus, all the conditions of Theorem 4 are satisfied. Clearly, 1∈P is a fixed-point of U.Remark 1.
Theorem4 and Theorem 5 are unified and extended results of Liu et al. [10] and Branciari [1].Remark 2.
Corollary2 is an extension of the result of Rakotch [7] with more general test functions.Remark 3.
In Theorem5, on letting γt=0, we obtain the result of 1.2 (result of Liu et al. [10]).Remark 4.
If we takeγt=0 in Corollary 2, we deduc the result of Geraghty [18].Figure 1
Graphical representation of inequality satisfying Example2. Also showing the graph of curves y=x and y=x with fixed point x=1.


## 3. Conclusion

We conclude this note by mentioning that our proved result is a further extension of Branciari result into other settings. Some remarks and an example are given to justify that our results are extension and generalized version of some known results of literature.



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