# Fixed Points for Multivalued Suzuki Type (θ,R)-Contraction Mapping with Applications

Authors: Mujahid Abbas; Hira Iqbal; Adrian Petrusel
Journal: Journal of Function Spaces
 (2019)

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## Abstract

In this paper, we will introduce the concept of Suzuki type multivalued(θ,R)-contraction and we will prove some fixed point results in the setting of a metric space equipped with a binary relation. Our results generalize and extend various comparable results in the existing literature. Examples are provided to support the results proved here. As an application of our results, we obtain a homotopy result, proving the existence of a solution for a second-order differential equation and for a first-order fractional differential equation.


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## Body

## 1. Introduction and Preliminaries

Let(X,d) be a metric space and T:X→X be a mapping on X. An element x∈X is called a fixed point of T if it remains invariant under the action of T; that is, x=Tx. A mapping T on a metric space X is said to be a Banach contraction if (1)dTx,Ty≤kdx,yholds for all x,y∈X, where 0≤k<1. A Banach contraction mapping defined on a complete metric space has a unique fixed point. This result is known as Banach contraction principle. Several authors have extended and generalized Banach contraction principle in different directions.Jleli and Samet [1] suggested a modification in the contraction condition and introduced a ϑ-contraction mapping. Consistent with [1], the following notations, definitions, and results will be needed in the sequel.Suppose that(2)Ω=ϑ:0,∞→1,∞satisfyϑ1,ϑ2andϑ3where(ϑ1) ϑ is nondecreasing;(ϑ2) for each sequence {tn}⊆(0,∞), limn→∞ϑ(tn)=1 if and only if limn→∞tn=0+;(ϑ3) there exists r∈(0,1) and l∈(0,∞] such that limt→0+ϑ(t)-1/tr=l.Example 1.
Defineηi:(0,∞)→(1,∞) for i=1,2,3 by (3)η1t=etη2t=5tandη3t=etet.Then, η1,η2,η3∈Ω.Let(X,d) be a metric space and ϑ∈Ω. A mapping T:X→X is called a ϑ-contraction if for any x,y∈X, we have (4)ϑdTx,Ty≤ϑdx,yαwhenever d(Tx,Ty)>0 and 0≤α<1. Jleli and Samet [1] proved the following fixed point theorem in the framework of a generalized metric space in the sense of Branciar; i.e., the triangle inequality is replaced by the inequality d(x,y)≤d(x,u)+d(u,v)+d(v,y), for all pairwise distinct points x,y,u,v∈X.Theorem 2.
Let(X,d) be a complete generalized metric space in the sense of Branciari and T:X→X be a ϑ-contraction. Then T has a unique fixed point.Hussain et al. [2] considered the following class of mappings: (5)Θ=θ:0,∞→1,∞:θsatisfiesθ1-θ5where(θ1) θ is nondecreasing;(θ2) θ(t)=1 if and only if t=0;(θ3) for each sequence {tn}⊂(0,∞), limn→∞θ(tn)=1 if and only if limn→∞tn=0+;(θ4) there exists r∈(0,1) and l∈(0,∞] such that limt→0+θ(t)-1/tr=l;(θ5) θ(t1+t2)≤θ(t1)θ(t2).Example 3.
Letθ1,θ2:[0,∞)→[1,∞) be defined by θ1(t)=et and θ2(t)=5t, respectively, then θ1 and θ2∈Θ.Hussain et al. [2] proved the following result.Theorem 4.
Let(X,d) be a complete metric space and T:X→X be a continuous mapping. Suppose there exist θ∈Θ and α,β,γ,δ∈R+ with 0≤α+β+γ+2δ<1, such that for any x,y∈X, we have (6)θdTx,Ty≤θdx,yαθdx,Txβθdy,Tyγθdx,Ty+dy,Txδ.Then T has a unique fixed point.For other results in this direction, we refer to [3, 4] and references mentioned therein.Let(X,d) be a metric space and CB(X) (respectively K(X)) be the family of all nonempty closed and bounded (nonempty compact, respectively) subsets of X. For A,B∈CB(X) and x∈X, define (7)dx,A≔infa∈Adx,a,HA,B≔maxsupa∈Ada,B,supb∈Bdb,A.H is the Hausdorff-Pompeiu metric on CB(X) (or on K(X)) induced by d. Let T:X→CB(X) be a given multivalued mapping. An element x∈X is called a fixed point of T if x∈Tx. A mapping T:X→CB(X) is said to be a Nadler contraction if there exists k∈(0,1) such that (8)HTx,Ty≤kdx,y,for any x,y∈X. Nadler [5] obtained the following multivalued version of Banach contraction principle.Theorem 5.
Let(X,d) be a complete metric space and T:X→CB(X) be a Nadler contraction. Then T has at least one fixed point.Later on, many researchers have obtained fixed point results for multivalued mappings satisfying generalized contraction type conditions. For example, recently, Hançer et al. [6] proved the following fixed point result for multivalued θ-contractions.Theorem 6.
Let(X,d) be a complete metric space and T:X→K(X) be a multivalued mapping. Suppose that there exist θ∈Θ and 0≤α<1 such that (9)θHTx,Ty≤θdx,yα,for any x,y∈X, provided that H(Tx,Ty)>0. Then T has at least one fixed point.Durmaz [7] introduced a new type of generalized multivalued θ-contraction and proved some interesting fixed point results (see also [8]). Kikkawa and Suzuki [9] refined Nadler’ result by proving the following theorem.Theorem 7.
Letβ:[0,1)→(1/2,1] be defined as β(b)=1/1+b. Let (X,d) be a complete metric space and T:X→CB(X). Assume there exists b∈(0,1) such that (10)x,y∈Xwithβbdx,Tx≤dx,yimpliesHTx,Ty≤bdx,y.Then T has at least one fixed point.We denoteR+≔[0,∞) and define the following class of mappings, which was considered in [10].(11)Φ=φ:R+×R+→Randφs,t≤12s-t.Example 8.
Letφ1:R+×R+→R be defined by φ(s,t)=v(s)-u(t) where v,s:R+→R+ are given by v(s)=s/2 and u(t)=t. Obviously φ1∈Φ.Example 9.
Letφ2:R+×R+→R be defined by φ(s,t)=s/2-v(s,t)/u(s,t)t where v,s:R+×R+→R+ are defined by v(s,t)=st and u(s,t)=st+t for all s,t>0. Note that φ2∈Φ.Many results, dealing with existence of fixed points of mappings satisfying certain contraction type conditions in the framework of complete metric spaces endowed with a partial ordering, have appeared in the last decade. Ran and Reurings [11] proved an analogue of Banach’s fixed point theorem in a metric space endowed with partial ordering and gave an application of their results to solve matrix equations. Alam and Imdad [12] proved another variant of Banach’s fixed point theorem in a metric space equipped with a binary relation which generalized many comparable results, including Ran and Reuring’s result in [11]. Senapati and Dey [13] proved Banach’s fixed point theorem in metric spaces equipped with an arbitrary binary relation using ω-distance. They employed their results to prove the existence of solutions of nonlinear fractional differential equations and fractional thermostat model involving the Caputo fractional derivative. A very nice Ph.D. thesis was written on the same subject; see Dobrican [14].Let us first recall the following definitions.Definition 10.
LetX be a nonempty set and R be a binary relation defined on X×X. Then, x is R-related to y if and only if (x,y)∈R.We denoteN0≔0,1,2,…, and N≔1,2,…,.Definition 11.
LetX be a nonempty set and R a binary relation on X. A sequence {xn}⊂X is called R-preserving if (12)xn,xn+1∈R∀n∈N0.Definition 12.
Let(X,d) be a metric space. A binary relation R defined on X is called d-self closed if whenever {xn} is an R-preserving sequence and xn converges to x, then there exists a subsequence {xnk} of {xn} with either (xnk,x)∈R or (x,xnk)∈R for all k∈N0.Definition 13.
Let(X,d) be a metric space and R a binary relation defined on X. A mapping T:X→P(X) is R-closed if for any x,y∈X, (x,y)∈R implies that (u,v)∈R for any u∈Tx and v∈Ty.IfT:X→P(X) is a multivalued map, then we set (13)XT;R=x∈X:x,y∈Rfor  somey∈Tx.In particular, if T is single-valued, then we denote (14)X′T;R=x∈X:x,Tx∈R.Motivated by the results in [2, 10, 12], we introduce the concept of a Suzuki type multivalued (θ,R) -contraction mapping and present some fixed point results in metric spaces equipped with a binary relation. Our results extend and generalize several results given in [2, 15–19]. We also provide applications of our results to homotopy theory proving the existence of a solution of second-order differential equations and first-order fractional differential equations.


## 2. Multivalued Suzuki Type(θ,R)-Contraction

In this section, we obtain a fixed point result for multivalued Suzuki type(θ,R)-contraction in a metric space equipped with a binary relation R.Throughout this paperθ∈Θ satisfies the following additional property: (15)θ6:θinf⁡A=infa∈A⁡θawhere A⊂[0,∞).We will denote(16)Θ∗=θ:0,∞→1,∞:θsatisfiesθ1-θ6We start with the following definition.Definition 14.
Let(X,d) be a metric space and R a binary relation on X. Assume that φ∈Φ and θ∈Θ∗. A mapping T:X→P(X) is a multivalued Suzuki type (θ,R)-contraction if for any x,y∈X with (x,y)∈R(17)φdx,Tx,dx,y<0⟹θHTx,Ty≤θdx,yαθdx,Txβθdy,Tyγθdy,Tx+dx,Tyδ,where α,β,γ,δ∈R+ with 0≤α+β+γ+2δ<1.Our first main result is the following.Theorem 15.
Let(X,d) be a complete metric space, R a binary relation on X, and T:X→CB(X) a multivalued Suzuki type (θ,R)-contraction. Suppose that following conditions hold: (1)
X(T;R) is nonempty,(2)
T is R-closed,(3)
R is d-self closed or T has closed graph.
ThenT has at least one fixed point.Proof.
SinceX(T;R) is nonempty, if we choose x0∈X(T;R), then there exists some x1∈Tx0 such that (x0,x1)∈R. If x0=x1, the result follows. Assume that x0≠x1. As (x0,x1)∈R and (18)φdx0,Tx0,dx0,x1≤12dx0,Tx0-dx0,x1<dx0,Tx0-dx0,x1≤dx0,x1-dx0,x1=0,we have (19)θdx1,Tx1≤θHTx0,Tx1≤θdx0,x1αθdx0,Tx0βθdx1,Tx1γθdx1,Tx0+dx0,Tx1δ.Now d(x1,Tx0)=0 and d(x0,Tx0)≤d(x0,x1) imply that (20)θdx1,Tx1≤θdx0,x1α+βθdx1,Tx1γθ(dx0,Tx1δ.Then by d(x0,Tx1)≤d(x0,x1)+d(x1,Tx1) and (θ5), it follows that (21)θdx0,Tx1≤θdx0,x1θdx1,Tx1and hence (22)θdx1,Tx1≤θdx0,x1α+β+δθdx1,Tx1γ+δ.Hence, we obtain (23)θdx1,Tx11-γ-δ≤θdx0,x1α+β+δand, in conclusion, we get that (24)θdx1,Tx1≤θdx0,x1α+β+δ/1-γ-δ.By (θ6) we have (25)θdx1,Tx1=infy∈Tx1θdx1,y.Thus, (26)infy∈Tx1θdx1,y≤θdx0,x1α+β+δ/1-γ-δ.We can choose x2∈Tx1 such that(27)θdx1,x2≤θdx0,x1α+β+δ/1-γ-δ.As (x0,x1)∈R, x1∈Tx0, x2∈Tx1 and T is R- closed, we have that (x1,x2)∈R. If x1=x2, the result follows. Assume that x1≠x2. Also, (28)φdx1,Tx1,dx1,x2≤12dx1,Tx1-dx1,x2<dx1,Tx1-dx1,x2<0.Hence, (29)θdx2,Tx2≤θHTx1,Tx2≤θdx1,x2αθdx1,Tx1βθdx2,Tx2γθdx2,Tx1+dx1,Tx2δ.By d(x1,Tx2)≤d(x1,x2)+d(x2,Tx2) and (θ5), we have (30)θdx1,Tx2≤θdx1,x2θdx2,Tx2.As d(x2,Tx1)=0 and d(x1,Tx1)≤d(x1,x2), we obtain that (31)θdx2,Tx21-γ-δ≤θdx1,x2α+β+δ.Hence, (32)θdx2,Tx2≤θdx1,x2α+β+δ/1-γ-δ.By (θ6), we have (33)θdx2,Tx2=infy∈Tx2θdx2,y.Hence, (34)infy∈Tx2θdx2,y≤θdx1,x2α+β+δ/1-γ-δWe can choose x3∈Tx2 such that (35)θdx2,x3≤θdx1,x2α+β+δ/1-γ-δ.By (27), we get (36)θdx2,x3≤θdx0,x1α+β+δ/1-γ-δ2.Since (x1,x2)∈R, and T is R-closed, we have (x2,x3)∈R.
Continuing this way, we can obtain a sequence{xn} such that xn+1∈Txn and {xn} is R-preserving. Obviously, we have d(xn,Txn)≤d(xn,xn+1), for all natural numbers n≥0. Hence,(37)θdxn,xn+1≤θdxn-1,xnα+β+δ/1-γ-δ≤…≤θdx0,x1α+β+δ/1-γ-δn.Letting n→∞, we have (38)limn→∞θdxn,xn+1=1.It follows from (θ3) that (39)limn→∞dxn,xn+1=0.Now, we show that {xn} is a Cauchy sequence. If we set tn=d(xn,xn+1), then from (37), we obtain(40)1<θtn≤θtn-1α+β+δ/1-γ-δ≤⋯≤θt1α+β+δ/1-γ-δn.Further, from (θ4), there exist r∈(0,1) and l∈(0,∞] such that (41)limn→∞θtn-1tnr=l.Suppose l<∞. Let B=l/2>0. Then there exists n0∈N such that for all n≥n0, we obtain that (42)θtn-1tnr-l≤B.Hence, for all n≥n0, we have (43)θtn-1tnr-l≥-B.This implies that (44)ntnr≤Anθtn-1∀n≥n0and  forA=1B.If l=∞, then for B>0 there exists n0 such that for all n≥n0, we have (45)θtn-1tnr≥Bwhich implies that (46)ntnr≤Anθtn-1∀n≥n0,whereA=1B.Hence, for each case, we obtain that (47)ntnr≤Anθtn-1∀n≥n0.Thus, using (40) we have (48)ntnr≤Anθt1α+β+δ/1-γ-δn-1∀n≥n0.Therefore, (49)limn→∞ntnr=0.So, there exists n1∈N such that for all n≥n1, we have 0<n(tn)r<1 which implies that tn<(1/n)1/r. Let m>n>n1∈N. Then (50)dxn,xm≤dxn,xn+1+dxn+1,xn+2+⋯+dxm-1,xm=tn+tn+1+⋯+tm-1≤∑i=n∞ti<∑i=n∞1i1/r.By the convergence of the series ∑i=1∞(1/i)1/r we get d(xn,xm)→0 as m,n→∞. Hence, {xn} is Cauchy. Since (X,d) is complete, there exists x∗ in X such that xn→x∗. We show that T has a fixed point. Assume on the contrary that T does not have a fixed point. Then, d(xn,Txn)>0 for all natural numbers n≥0. As (xn,xn+1)∈R, we have (51)12dxn,Txn<dxn,Txn≤dxn,xn+1,which implies that (52)φdxn,Txn,dxn,xn+1<0.Furthermore, (53)dxn+1,Txn+1≤HTxn,Txn+1gives that(54)θdxn+1,Txn+1≤θHTxn,Txn+1≤θdxn,xn+1α·θdxn,Txnβ·θdxn+1,Txn+1γ·θdxn+1,Txn+dxn,Txn+1δ.Thus (55)θdxn+1,Txn+11-γ-δ≤θdxn,xn+1α+β+δ,which further implies that(56)θdxn+1,Txn+1≤θdxn,xn+1α+β+δ/1-γ-δ<θdxn,xn+1.Hence(57)dxn+1,Txn+1<dxn,xn+1,∀n∈N0.If R is d-self closed, there exists a subsequence {xnk} of {xn} such that either (xnk,x∗)∈R or (x∗,xnk)∈R. Assume that (xnk,x∗)∈R. If φ(d(xnk,Txnk),d(xnk,x∗))≥0, then we have (58)12dxnk,Txnk≥dxnk,x∗.From (57), we obtain that (59)dxnk,xnk+1≤dxnk,x∗+dx∗,xnk+1≤12dxnk,Txnk+12dxnk+1,Txnk+1<12dxnk,xnk+1+12dxnk,xnk+1=dxnk,xnk+1,a contradiction. Hence, φ(d(xnk,Txnk+1),d(xnk,x∗))<0 for all k∈N0.
By our assumptionx∗∉Tx∗. Thus (60)dx∗,Tx∗≤dx∗,xnk+dxnk,Tx∗≤dx∗,xnk+HTxnk,Tx∗,which implies that(61)θdx∗,Tx∗≤θdx∗,xnkθHTxnk,Tx∗.Consequently, (62)θdx∗,xnkθHTxnk,Tx∗≤θdx∗,xnkθdxnk,x∗αθdxnk,Txnkβθdx∗,Tx∗γθdx∗,Txnk+dxnk,Tx∗δ.Also, (63)dxnk,Tx∗≤dxnk,x∗+dx∗,Tx∗.From (61), it follows that(64)θdx∗,Tx∗≤θdx∗,xnk1+α+δθdx∗,Tx∗δ+γθdxnk,xnk+1βθdx∗,xnk+1δ.Letting k→∞ in (64) we obtain that (65)θdx∗,Tx∗≤θdx∗,Tx∗δ+γ<θdx∗,Tx∗,a contradiction. Hence, x∗∈Tx∗.
IfT has closed graph, since xn+1∈Txn for each n∈N0 and limn→∞xn=x∗, we get that x∗∈Tx∗.If we takeβ,γ,δ=0 in Theorem 15, we obtain a Suzuki type generalization of the result in [6] in the framework of a complete metric space equipped with a binary relation R.Corollary 16.
Let(X,d) be a complete metric space, R a binary relation on X and T:X→CB(X). Assume that φ∈Φ and θ∈Θ∗. Suppose that there exists 0≤α<1 such that for any x,y∈X with (x,y)∈R, (66)φdx,Tx,dx,y<0implies that (67)θHTx,Ty≤θdx,yα.If conditions (1)-(3) in Theorem 15 are satisfied, then T has a fixed point.If we takeθ(t)=etn, Theorem 15, then we have the following multivalued extension of Cirić result in [16].Corollary 17.
Let(X,d) be a complete metric space, R a binary relation on X, and T:X→CB(X). Suppose that φ∈Φ and there exist α,β,γ,δ∈R+ with 0≤α+β+γ+2δ<1, such that for any x,y∈X with (x,y)∈R, (68)φdx,Tx,dx,y<0implies that (69)HTx,Ty≤αdx,y+βdx,Tx+γdy,Ty+δdx,Ty+dy,Tx.Assume that conditions (1)-(3) in Theorem 15 are satisfied. Then T has a fixed point.Remark 18.
Note that the conclusion of Corollary17 can be written as (70)HTx,Ty≤α2dx,y+β2dx,Tx+γ2dy,Ty+δ2dx,Ty+dTx,y+2αβdx,ydx,Tx+2αγdx,ydy,Ty+2αδdx,ydx,Ty+dy,Tx+2βγdx,Txdy,Ty+2βδdy,Tydx,Ty+dy,Tx+2γδdy,Tydx,Ty+dy,TxNotice that if we takeα,β,γ=0 in Theorem 15, using Remark 18, we obtain the following multivalued Suzuki type generalization of Chatterjea’s result in [15].Corollary 19.
Let(X,d) be a complete metric space, R a binary relation on X, and T:X→CB(X). Suppose that φ∈Φ and there exists δ∈[0,1/2) such that for any x,y∈X with (x,y)∈R, (71)φdx,Tx,dx,y<0implies that (72)HTx,Ty≤δ2dx,Ty+dy,TxAssume that conditions (1)-(3) in Theorem 15 are satisfied, then T has a fixed point.If we takeα=δ=0 in Theorem 15, using Remark 18, we obtain the following multivalued Kannan type result in [17].Corollary 20.
Let(X,d) be a complete metric space, R a binary relation on X, and T:X→CB(X). Suppose that φ∈Φ and there exist β,γ∈R+ with 0≤β+γ<1 such that for any x,y∈X with (x,y)∈R, (73)φdx,Tx,dx,y<0implies that (74)HTx,Ty≤β2dx,Tx+γ2dy,Ty+2βγdx,Txdy,Ty.Assume that conditions (1)-(3) in Theorem 15 are satisfied. Then T has a fixed point.Takingδ=0 in Theorem 15, we have a multivalued extension and generalization of Reich’s result in [20].Corollary 21.
Let(X,d) be a complete metric space, R a binary relation on X, and T:X→CB(X). Suppose that φ∈Φ and there exist α,β,γ∈R+ with 0≤α+β+γ<1, such that for any x,y∈X with (x,y)∈R, the following implication is true (75)φdx,Tx,dx,y<0implying that (76)HTx,Ty≤α2dx,y+β2dx,Tx+γ2dy,Ty+2αβdx,ydx,Tx+2αγdx,ydy,Ty2βγdx,Txdy,Ty.Assume that conditions (1)-(3) in Theorem 15 are satisfied. Then T has a fixed point.Similarly, if we takeθ(t)=etn, then we obtain the following corollary.Corollary 22.
Let(X,d) be a complete metric space, R a binary relation on X, and T:X→CB(X). Suppose that φ∈Φ and there exists α,β,γ,δ∈R+ with 0≤α+β+γ+2δ<1, such that for any x,y∈X with (x,y)∈R, (77)φdx,Tx,dx,y<0implies that (78)HTx,Tyn≤αdx,yn+βdx,Txn+γdy,Tyn+δdx,Ty+dy,Txn.Assume that conditions (1)-(3) in Theorem 15 are satisfied. Then T has a fixed point.We now give an example of a multivalued Suzuki type(θ,R)-contraction which is neither a multivalued Banach contraction nor a multivalued (θ,R)-contraction.Example 23.
LetX=x1,x2,x3. Define the binary relation on X as follows: (79)R=x1,x1,x1,x2,x2,x1,x2,x3,x2,x2,x3,x2,x3,x3Let d:X×X→R+ be defined by(80)dx1,x2=4,dx2,x3=1,dx1,x3=3anddx,y=dy,x,dx,x=0,forx,y∈X.Define the mapping T:X→CB(X) by (81)Tx=x1,x2,wheneverx∈x1,x2x1,wheneverx=x3.Clearly, T is R-closed and X(T;R) is nonempty. Indeed, if x=x1, Tx1=x1,x2, then (x1,x2)∈R. Take an R-preserving sequence {xn} such that {xn} converges to x and (xn,xn+1)∈R for all n∈N0. Then, (xn,xn+1)∈{(x1,x1),(x1,x2),(x2,x1),(x2,x3), (x2,x2),(x3,x2),(x3,x3)} for all n∈N0. Thus, {xn}⊂{x1,x2} or {x2,x3}. Since both {x1,x2} and {x2,x3} are closed, either (xn,x)∈R or (x,xn)∈R.
Letφ(r,s)=r/2-s and θ(t)=et. We consider following cases: (1)
If(x,y)=(x3,x2), then 1/2d(x3,Tx3)-d(x3,x2)=3/2-1>0.(2)
If(x,y)=(x2,x3), then clearly 1/2d(x2,Tx2)-d(x2,x3)<0. For α,β,γ=0 and δ=0.9 we have (82)θHTx2,Tx3≤θdx2,x3αθdx2,Tx2βθdx3,Tx3γ(θdx2,Tx3+dx3,Tx2δwhich implies thate4<(e5)0.9.
All other cases are trivial. Thus all the conditions of Theorem15 are satisfied. Moreover, T has two fixed points x1 and x2.
Note that, for(x3,x2)∈R, (83)HTx3,Tx2=4>k1=kdx3,x2for any 0<k<1. Thus T is not a Nadler’s contraction. Also, (84)θHTx3,Tx2=e4>e1k=θdx3,x2kfor any 0<k<1. Thus, T is not multivalued (θ,R)-contraction.
IfT:X→X is a single-valued map, then we have the following result.Theorem 24.
Let(X,d) be a complete metric space, R a binary relation on X, and T:X→X. Suppose that φ∈Φ, θ∈Θ∗ and there exist α,β,γ,δ∈R+ with 0≤α+β+γ+2δ<1, such that for any x,y∈X with (x,y)∈R, (85)φdx,Tx,dx,y<0implies that (86)θdTx,Ty≤θdx,yαθdx,Txβθdy,Tyγθdx,Ty+dy,Txδ.
In addition, assume that the following conditions also hold:(1)
X′(T;R) is nonempty,(2)
ifx,y∈X such that (x,y)∈R, then (Tx,Ty)∈R,(3)
R is d-self closed or T has closed graph.
ThenT has at least one fixed point.Remark 25.
The above theorem generalizes various existing results in literature. If in Theorem24 we take φ(s,t)=1/2s-t, then we have the following: (1)
We obtain a Suzuki type generalization ofθ-contraction result in [2] in the setup of metric spaces endowed with binary relation. If θ(t)=et, the above theorem also generalizes the result in [16].(2)
Ifθ(t)=et, β,γ,δ=0, and α<1, we obtain a Suzuki type result in the setting of metric spaces endowed with binary relation (see [19]).(3)
Ifθ(t)=et, α,β,γ=0, and δ<1/2, we obtain a Suzuki type version of Chatterjea result [15] in the setup of metric spaces endowed with binary relation.(4)
Ifθ(t)=et, α,δ=0, and β+γ<1, we obtain a Suzuki type result for generalized Kannan mappings [17] in the setup of metric spaces endowed with binary relation.Example 26.
LetX=[0,3]∪[5,9] be the usual metric space with binary relation defined as follows: (87)R=1,1,1,3,3,1,3,3,1,5,5,5,5,9,9,5,9,9.Define the mapping T:X→X by (88)Tx=2x+3,whenever1≤x≤39,whenever5≤x≤9.Let φ(s,t)=1/2s-t and θ(t)=5t. Note that T is not a (θ,R)-contraction because (89)θdT1,T3=θd1,5=54>θd1,5k=54k.Consider the following cases: (1)
If(x,y)∈1,1,1,3,3,1,3,3, then 1/2d(x,Tx)-d(x,y)<0 does not hold.(2)
If(x,y)∈5,5,5,9,9,5,9,9, then clearly T is a Suzuki type (θ,R)-contraction.(3)
If(x,y)=(1,5), then we have (90)12d1,T1-d1,5=12d1,5-d1,5<0Forα,β,γ=0.1 and δ=0.4, we obtain that (91)θdT1,T5<θd1,5αθd1,T1βθd5,T5γθd1,T5+d5,T1δwhich implies that(92)54<540.1540.1540.1580.4.Clearly, T is R-closed and X′(T;R) is nonempty, since (5,9)∈R. Take an R-preserving sequence {xn} such that {xn} converges to x and we have (xn,xn+1)∈R for all n∈N0. Note that (xn,xn+1)∉1,5. Then, (xn,xn+1)∈1,1,1,3,3,1,3,3,5,5,5,9,9,5,9,9 for n∈N0. Thus, {xn} is a subset of {1,3}, {3,5}, or {5,9}. Since all of these sets are closed, either (xn,x)∈R or (x,xn)∈R. Thus all the conditions of Theorem 24 are satisfied. Moreover, x=9 is a fixed point of T.


## 3. Application to Homotopy Results

In this section, as an application of our above fixed point result for Suzuki type(θ,R)-contractions, we obtain a homotopy result for this class of multivalued mappings. For the beginning, we give a local fixed point result for multivalued Suzuki type (θ,R)-contraction. Let B(x0,r) be an open ball and B(x0,r)¯ the closure of B(x0,r).Theorem 27.
Let(X,d) be a complete metric space, R a binary relation on X, x0∈X and r>0. Suppose that T:B(x0,r)¯→CB(X) be a multivalued Suzuki type (θ,R)-contraction with θ(d(x0,Tx0))<(θ(r))1-k where 0<k<1. Assume that the following conditions are also satisfied: (1)
x0∈X(T;R),(2)
T is R-closed,(3)
T has closed graph or R is d-self closed.
ThenT has a fixed point in B(x0,r)¯.Proof.
Sincex0∈X(T;R), there exists x1∈Tx0 such that (x0,x1)∈R. If x0=x1, the result follows. Assume that x0≠x1. Let 0<s<r be such that B(x0,s)⊂B(x0,r)¯ and θ(d(x0,Tx0))<(θ(s))1-k<(θ(r))1-k. Clearly, d(x1,Tx1)≤H(Tx0,Tx1). Now (93)φdx0,Tx0,dx0,x1≤12dx0,Tx0-dx0,x1<dx0,Tx0-dx0,x1<0implies that (94)θdx1,Tx1≤θHTx0,Tx1≤θdx0,x1αθdx0,Tx0βθdx1,Tx1γθdx0,Tx1+dx1,Tx0δ.Thus, we have (95)θdx1,Tx1≤θdx0,x1α+β+δθdTx1,x1γ+δ.Hence (96)θdx1,Tx11-γ-δ≤θdx0,x1α+β+δ,which implies that (97)θdx1,Tx1≤θdx0,x1α+β+δ/1-γ-δ.By the condition (θ6), we get (98)infy∈Tx1θdx1,y≤θdx0,x1α+β+δ/1-γ-δ. We may now choose x2∈Tx1 such that (x1,x2)∈R and (99)θdx1,x2≤θdx0,x1kwhere k≔α+β+δ/1-γ-δ<1. Furthermore,(100)θdx1,x2≤θdx0,x1k<θs1-kk.Note that (101)θdx0,x2≤θdx0,x1θdx1,x2<θs1-kθsk1-k<θs1-k2<θs.Thus x2∈B(x0,s) as d(x0,x2)<s. Continuing this way, we obtain a sequence {xn} with following properties: (1)
xn∈B(x0,s),(2)
(xn,xn+1)∈R,(3)
xn+1∈Txn∀n∈N0,(4)
θ(d(xn,xn+1))<(θ(s))kn(1-k).
By similar arguments to those in the proof of Theorem15, we obtain that {xn} is a Cauchy sequence which converges to u∈B(x0,r)¯ and u∈Tu.Now we present the following homotopy result.Theorem 28.
Let(X,d) be a complete metric space, U an open set of X, R a d-self closed binary relation on X and Λ:U¯×[0,1]→P(X) be a multivalued mapping with closed values. Suppose that for each t∈[0,1], Λ(·,t):U¯→CB(X) satisfies the conditions (1)-(2) in Theorem 27. Assume that the following conditions are also satisfied: (1)
x∉Λ(x,t), for each x∈∂U and each t∈[0,1];(2)
Λ(·,t):U¯→CB(X) is a multivalued Suzuki type (θ,R)-contraction with closed graph, for each t∈[0,1];(3)
there exists a continuous increasing functionη:[0,1]→R such that (102)θ(HΛx,t,Λx,s≤θηt-ηsfor allt,s∈[0,1] and each x∈U¯.ThenΛ(·,0) has a fixed point if and only if Λ(·,1) has a fixed point.Proof.
Suppose thatΛ(·,0) has a fixed point z, then by (1) z∈U. Define (103)Q=(x,t∈U×0,1:x∈Λx,tAs (z,0)∈Q, so Q≠∅. Define partial order on Q by (104)x,t≼y,sif  and  only  ift≤sand (105)θdx,y≤θηt-ηs2/1-k,wherek=α+β+δ1-γ-δ.Let A be a totally ordered set in Q and t∗=supt:x,t∈A. Consider a sequence {(xn,tn)} in A such that (xn,tn)≼(xn+1,tn+1) and tn→t∗ as n→∞. Then, we have(106)θdxm,xn≤(θηtm-ηtn)2/1-k.As (107)limm,n→∞ηtm-ηtn=0,by (θ3), we have (108)limm,n→∞θηtm-ηtn=1.Thus (106) gives that (109)1≤limm,n→∞θdxm,xn≤1and (110)limm,n→∞dxm,xn=0.Hence {xn} is Cauchy sequence which converges to an element x∗∈U¯. As xn∈Λ(xn,tn) for n∈N and Λ is closed, so x∗∈Λ(x∗,t∗). Also, from (1) we have x∗∈U. Hence, (x∗,t∗)∈Q. Since A is totally ordered, therefore (x,t)≤(x∗,t∗) for each (x,t)∈A. That is, (x∗,t∗) is an upper bound of A. By Zorn’s Lemma, the set Q admits a maximal element (x0,t0)∈Q. We claim that t0=1. Assume, on the contrary, that t0<1. Choose r>0 and t∈(t0,1] such that B(x0,r)⊂U and (111)θr=θ(ηt-ηt02/1-k.Note that (112)θdx0,Λx0,t≤θdx0,Λx0,t0+HΛx0,t0,Λx0,t≤θηt-ηt0=θr1-k/2<θr1-k.Thus Λ(·,t):B(x0,r)¯→CB(X) satisfies all assumptions of Theorem 27. Hence, for all t∈[0,1], there exists x∈B(x0,r)¯ such that x∈Λ(x,t). Hence (x,t)∈Q. Now, d(x0,x)<r implies that (113)θdx0,x<θr=θ(ηt-ηt02/1-k,which further implies that (x0,t0)<(x,t), a contradiction to the fact that (x0,t0) is a maximal element.
Conversely, ifΛ(·,1) has a fixed point, then by similar arguments to those given before, we obtain that Λ(.,0) has a fixed point.


## 4. Existence of a Solution of Second-Order Differential Equation

In this section, we study the existence of solution of a two-point boundary value problem associated with a second-order differential equation. LetX=C[0,1] be the space of all continuous functions defined on [0,1]. The metric on X is given by(114)dx,y=x-y∞=maxt∈0,1xt-yt.Define the binary relation on X by (115)x,y∈Rif  and  only  ifx≤yi.e.,xt≤yt,t∈0,1.Note that the space X=(C[0,1],d) is complete metric space. We consider the following two-point boundary value problem:(116)-x′′t=ft,xt,t∈0,1x0=x1=0,where f:[0,1]×R→R is a continuous function. Then, the problem (116) can be written in the following integral form:(117)xt=∫01Gt,sfs,xsds,where the associated Green function is (118)Gt,s=t1-s;0≤t≤s≤1s1-t;0≤s≤t≤1,see [21] for details.Theorem 29.
Assume that the following conditions are satisfied:(1)
f:[0,1]×R→R is continuous,(2)
f(s,.):R→R is increasing for all s∈[0,1],(3)
there existsτ∈[1,∞) such that the following condition holds for all x,y∈X, with x≤y, (119)fs,x-fs,y≤8e-τxs-ys,∀s∈0,1,(4)
there existsx0∈X such that for all t∈[0,1], we have (120)x0t≤∫01Gt,sfs,x0sds.Then the problem (116) has a solution in X.Proof.
If we define the mappingT:X→X by (121)Txt=∫01Gt,sfs,xsds,then our problem can be written as a fixed point equation x=Tx.
Obviously,T is continuous. As f(s,.) is increasing, for any x,y∈X with x(t)≤y(t) for all t∈[0,1], we obtain that (122)∫01Gt,sfs,xsds≤∫01Gt,sfs,ysds,i.e.,Txt≤Tyt.Thus, T is R-closed. If x,y∈X such that x(t)≤y(t), then we have(123)Txt-Tyt=∫01Gt,sfs,xsds-∫01Gt,sfs,ysds≤∫01Gt,sfs,xs-fs,ysds≤∫018e-τxs-ysGt,sds≤8e-τdx,y∫01Gt,sds.As ∫01G(t,s)ds=t/2-t2/2, we get that supt∈[0,1](∫01G(t,s)ds)=1/8. The inequality (123) becomes (124)Txt-Tyt≤8e-τdx,y18≤e-τdx,y.Hence, we have (125)maxt∈0,1Txt-Tyt≤e-τdx,y,which implies that (126)dTx,Ty≤e-τdx,y.Taking square root on both sides and passing through exponential function, we obtain that (127)edTx,Ty≤ee-τdx,y=edx,ye-τ,where e-τ<1 as τ≥1. Hence, (128)edTx,Ty≤edx,yαwithα=e-τ∈0,1.Thus, (129)θdTx,Ty≤θdx,yα,where θ(t)=et and (x,y)∈R.
As above inequality is true for anyx,y∈X with x(t)≤y(t), so is for any φ∈Φ, with φ(d(x,Tx),d(x,y))<0. Thus (130)θdTx,Ty≤θdx,yα.By Theorem 24 we get that (116) has a solution in X.


## 5. Existence of a Solution of Fractional Boundary Value Problem

In this section, we investigate the existence of solutions of a nonlinear fractional differential equation. Let the spaceX and the metric d be defined as in the above section.Consider the following fractional differential equation(131)​CDβxt=ft,xt;0<t<1,1<β≤2,with boundary conditions(132)x0=0,Ix1=x′0.Here CDβ stands for the Caputo fractional derivative of order β, defined by (133)​CDβft=1Γn-β∫0tt-sn-β-1fnsds,(where we consider n-1<β<n and n=[β]+1) and Iβf denotes the Riemann-Liouville fractional integral of order β of a continuous function f, given by (134)Iβft=1Γβ∫0tt-sβ-1fsds,withβ>0.Senapati and Dey ([13]) showed that the problem (131)+(132) can be written in the following integral form:(135)xt=1Γβ∫0tt-sβ-1fs,xsds+2tΓβ∫01∫0ss-rβ-1fr,xrdrds.Theorem 30.
Suppose that following conditions are satisfied:(1)
f:[0,1]×R→R is a continuous function,(2)
f(s,·):R→R is an increasing function,(3)
for everyx,y∈X with x≤y, the following condition holds: (136)fs,x-fs,y≤e-τΓβ+14xs-ys,∀s∈0,1,whereτ∈[1,∞),(4)
there existsx0∈X such that, for all t∈[0,1], we have (137)x0t≤1Γβ∫0tt-sβ-1fs,x0sds+2tΓβ∫01∫0ss-rβ-1fr,x0rdrds.Then, (131)+(132) has at least one solution in C[0,1].Proof.
Define the mappingT:X→X by (138)Txt=1Γβ∫0tt-sβ-1fs,xsds+2tΓβ∫01∫0ss-rβ-1fr,xrdrds.Then (135) can be written as a fixed point equation for T; i.e., x=Tx. Consider on X≔C[0,1] the binary relation defined as follows(139)xRyif  and  only  ifxt≤yt,∀t∈0,1.By the given assumption (4), X′(T;R) is nonempty. If x,y∈X are such that x(t)≤y(t) for every t, then by assumption that f(s,·) is increasing we have (140)1Γβ∫0tt-sβ-1fs,xsds+2tΓβ∫01∫0ss-rβ-1fr,xrdrds≤1Γβ∫0tt-sβ-1fs,ysds+2tΓβ∫01∫0ss-rβ-1fr,yrdrds,which implies that T(x)(t)≤T(y)(t). Therefore, T is R-closed. Nieto and López [22] have shown that if there exists a sequence {xn} in X such that xn(t)≤xn+1(t) and xn converges to some x, then xn(t)≤x(t). Hence, R is d-self closed. If x,y∈X such that x(t)≤y(t), then (141)Txt-Tyt=1Γβ∫0tt-sβ-1fs,xsds-1Γβ∫0tt-sβ-1fs,ysds+2tΓβ∫01∫0ss-rβ-1fr,xrdrds-2tΓβ∫01∫0ss-rβ-1fr,yrdrds≤1Γβ∫0tt-sβ-1fs,xsds-fs,ysds+2Γβ∫01∫0ss-rβ-1fr,xr-fr,yrdrds≤e-τΓβ+14Γβ∫0tt-sβ-1xs-ysds+2e-τΓβ+14Γβ∫01∫0ss-rβ-1xr-yrdrds≤e-τΓβ+14Γβdx,y∫0tt-sβ-1ds+2e-τΓβ+14Γβdx,y∫01∫0ss-rβ-1drds≤e-τΓβΓβ+14ΓβΓβ+1dx,y+2e-τBβ+1,1ΓβΓβ+14ΓβΓβ+1dx,y≤e-τ4dx,y+e-τ2dx,y,where B is the beta function. From the above inequality, we obtain that (142)dTx,Ty≤e-τdx,y.Taking square root on both sides and passing through exponential function, we have (143)edTx,Ty≤ee-τdx,y,that is, (144)edTx,Ty≤edx,yα,where α=e-τ<1. Since the above inequality holds for any x,y∈X such that x(t)≤y(t) so is true for any φ∈Φ, with (145)φdx,Tx,dx,y<0.Hence we have (146)θdTx,Ty≤θdx,yα.Thus T is a Suzuki type (θ,R)-contractive mapping. Since all the conditions of Theorem 24 are satisfied, the problem (131) has at least one solution.



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