# Fixed Point Theorems for Manageable Contractions with Application to Integral Equations

Authors: N. Hussain; I. Iqbal; Badriah A. S. Alamri; M. A. Kutbi
Journal: Journal of Function Spaces
 (2017)

---

## Abstract

In this paper we utilize the concept of manageable functions to define multivaluedα⁎-η⁎ manageable contractions and prove fixed point theorems for such contractions. As applications we deduce certain fixed point theorems which generalize and improve Nadler’s fixed point theorem, Mizoguchi-Takahashi’s fixed point theorem, and some other well-known results in the literature. Also, we give an illustrating example showing that our results are a proper generalization of Nadler’s theorem and provide an application to integral equations.


---

## Body

## 1. Introduction and Preliminaries

The Banach contraction principle [1] is an elementary result in metric fixed point theory. This golden principle has been broadened in several directions by different authors (see [1–18]). An interesting generalization is the elongation of the Banach contraction principle to multivalued maps, known as Nadler’s fixed point theorem [19] and Mizoguchi-Takahashi’s fixed point theorem [20]. In 2012, Samet et al. [18] defined α-ψ-contractive and α-admissible mappings and then Salimi et al. [17] generalized this idea by introducing function η and established fixed point theorems. Further Hasanzade Asl et al. [13] extended these notions to multivalued functions by introducing the concepts of α∗-ψ-contractive and α∗-admissible for multivalued mappings and proved some fixed point results.Hussain et al. [14] modified the notions of α∗-admissible as follows.Definition 1 (see [14]).
LetT:X→2X be a multifunction on a metric space (X,d) and α,η:X×X→R+ be two functions, where η is bounded; then T is an α∗-admissible mapping with respect to η if (1)αy,z≥ηy,zimpliesthatα∗Ty,Tz≥η∗Ty,Tz,y,z∈X,where(2)α∗A,B=infy∈A,z∈B⁡αy,z,η∗A,B=supy∈A,z∈B⁡ηy,z.Further, Ali et al. [3] generalized the results of Hussain et al. and introduced the following definition.Definition 2 (see [3]).
LetT:X→2X be a closed valued mapping on a metric space (X,d) and α,η:X×X→R+ be two functions. We say that T is generalized α∗-admissible mapping with respect to η if (3)αy,z≥ηy,zimpliesthatαu,v≥ηu,v,∀u∈Ty,v∈Tz.Very recently, Ali et al. [2] modified Definition 2 for the sequence of multivalued functions as follows.Definition 3 (see [2]).
LetTi:X→2Xi=1∞ be a sequence of closed valued maps on a metric space (X,d) and α,η:X×X→R+ be two functions; then the sequence Ti is α∗-admissible mapping with respect to η if (4)αy,z≥ηy,zimpliesthatαu,v≥ηu,v,∀u∈Tiy,v∈Tjz,for each i,j∈N. If, for all y,z∈X  α(y,z)=1, the sequence {Ti} is called α∗-subadmissible and, for η(y,z)=1, the sequence {Ti} is called α∗-admissible.Recently, Du and Khojasteh [10] initiated the concept of manageable functions and proved some fixed point theorems. In this paper, we introduce multivalued α∗-η∗ manageable contraction and prove certain fixed point results. We also prove common fixed point theorem for multivalued contraction. The investigated results of this paper conclude several existing fixed point results including Nadler’s theorem.Throughout this paper,CL(X) denotes the family of all nonempty closed subsets of a metric space (X,d). The Hausdorff metric H is defined on CL(X) by (5)HA,B=max⁡supx∈A⁡Dx,B,supy∈B⁡Dy,A,where D(x,B)=inf⁡d(x,y):y∈B. In the sequel, Fix{T} denotes the set of all fixed points of mapping T, Fix{Ti} denotes the set of all common fixed points of mappings Ti, Σ denotes the class of all functions σ:[0,+∞)→[0,1) fulfilling lim⁡supt→r+⁡σ(t)<1,  for all r∈[0,+∞), Φ denotes the set of all functions φ:[0,+∞)→[0,+∞) such that ∫0ϵ‍φ(t)dt exists and ∫0ϵ‍φ(t)dt>ϵ,   for each ϵ>0, Ψ denotes the class of all nondecreasing functions ψ:[0,+∞)→[0,+∞) such that ∑n=1∞‍ψn(t)<+∞   for all t>0, and Γ denotes the set of all L-functions. Recall that a function γ:[0,+∞)→[0,+∞) is said to be an L-function if γ(0)=0 and γ(t)>0  for all t>0 and for every t>0 there exists s>t such that γ(u)≤t  for u∈[t,s] [21].


## 2. Fixed Point and Common Fixed Point Results for Multivalued Contractions via Manageable Function

Consistent with Du and Khojasteh [10], we denote by Man(R)^ the set of all manageable functions ϑ:R×R→R fulfilling the following conditions:(ϑ1)
ϑt,s<s-t for all s,t>0;(ϑ2)
for any bounded sequencetn⊂0,+∞ and any nondecreasing sequence sn⊂(0,+∞), it holds that(6)limn→∞⁡sup⁡tn+ϑtn,snsn<1.Example 4 (see [10]).
Letr∈[0,1). Then ϑr:R×R→R defined by (7)ϑrt,s=rs-tis a manageable function.Example 5.
Letϑ:R×R→R defined by (8)ϑt,s=ψs-tift,s∈0,+∞×0,+∞,ft,sotherwise,where ψ∈Ψ and f:R×R→R is any function. Then ϑ(t,s)∈Man(R)^. Indeed, by using Lemma  1 of [12], we have, for any s,t>0,(9)ϑt,s=ψs-t<s-t,so, (ϑ1) holds. Let {tn}⊂(0,+∞) be a bounded sequence and let {sn}⊂(0,+∞) be a nonincreasing sequence. Then limn→∞sn=infn∈Nsn=a for some a∈[0,+∞); we get(10)limn→∞⁡sup⁡tn+ϑtn,snsn=limn→∞⁡sup⁡ψsnsn<limn→∞⁡snsn=1,so, (ϑ2) is also satisfied.Definition 6.
Let(X,d) be a metric space and T:X→2X be a closed valued mapping. Let α,η:X×X→R+ be two functions and ϑ∈Man(R)^. Then T is called a multivalued α∗-η∗-manageable contraction with respect to ϑ if for all y,z∈X(11)α∗Ty,Tz≥η∗Ty,TzimpliesϑHTy,Tz,dy,z≥0.Now we state and prove the main result of this section.Theorem 7.
Let(X,d) be a complete metric space and let T:X→2X be a closed valued map satisfying the following conditions:(1)
T is α∗-admissible map with respect to η;(2)
T is α∗-η∗ manageable contraction with respect to ϑ;(3)
there existsz0∈X and z1∈Tz0 such that α(z0,z1)≥η(z0,z1);(4)
for a sequence{zn}⊂X, limn→∞zn=x and α(zn,zn+1)≥η(zn,zn+1), for all n∈N, implies α(zn,x)≥η(zn,x) for all n∈N.
ThenFix{T}≠⌀.Proof.
Letz1∈Tz0 be such that α(z0,z1)≥η(z0,z1). Since T is α∗-admissible map with respect to η, then α∗(Tz0,Tz1)≥η∗(Tz0,Tz1). Therefore, from (11), we have (12)ϑHTz0,Tz1,dz0,z1≥0.If z1=z0, then z0∈FixT; also if z1∈Tz1, then z1∈FixT. So, we adopt that z0≠z1 and z1∉Tz1. Thus 0<dz1,Tz1≤HTz0,Tz1. Define λ:R×R→R by (13)λt,s=t+ϑt,ssift,s>00otherwise.By (ϑ1), we know that (14)0<λt,s<1,∀t,s>0.Also note that if ϑ(t,s)≥0, then (15)0<t≤sλt,s.So, from (12) and (14), we get (16)0<λHTz0,Tz1,dz0,z1<1.Let(17)ε1=1λHTz0,Tz1,dz0,z1-1dz1,Tz1.Since d(z1,Tz1)>0. So, by using (16), we get ε1>0 and (18)dz1,Tz1<dz1,Tz1+ε1=1λHTz0,Tz1,dz0,z1dz1,Tz1.This implies that there exists z2∈Tz1 such that (19)dz1,z2<1λHTz0,Tz1,dz0,z1dz1,Tz1.Note that z1≠z2 (since z1∉Tz1). Now if z2∈Tz2, then z2 is a fixed point of T. Otherwise, 0<d(z2,Tz2)<H(Tz1,Tz2). Also, since α∗(Tz0,Tz1)≥η∗(Tz0,Tz1), z1∈Tz0, and z2∈Tz1, then α(z1,z2)≥η(z1,z2). So, α∗(Tz1,Tz2)≥η∗(Tz1,Tz2), so, from (11), we get (20)ϑHTz1,Tz2,dz1,z2≥0.By taking(21)ε2=1λHTz1,Tz2,dz1,z2-1dz2,Tz2,there exists z3∈Tz2 with z3≠z2 such that(22)dz2,z3<1λHTz1,Tz2,dz1,z2dz2,Tz2.Hence, by induction, we form a sequence zn in X satisfying for each n∈N, zn∈Tzn-1, zn≠zn-1, zn∉Tzn, and α∗(zn-1,zn)≥η∗(zn-1,zn)(23)0<dxn,Txn≤HTzn-1,Tzn,(24)ϑHTzn-1,Tzn,dzn-1,zn≥0,(25)dzn,zn+1<1λHTzn-1,Tzn,dzn-1,zndzn,Tzn,by taking(26)εn=1λHTzn-1,Tzn,dzn-1,zn-1dzn,Tzn.By using (14), (15), (23), and (25), we get for each n∈N(27)dzn,Tzn≤dzn-1,znλHTzn-1,Tzn,dzn-1,zn≤dzn-1,zn.This implies that d(zn,Tzn)n∈N is a bounded sequence. By combining (25) and (27), for each n∈N, we get (28)dzn,zn+1<λHTzn-1,Tzn,dzn-1,zndzn-1,zn.Which means that {d(zn-1,zn)}n∈N is a monotonically decreasing sequence of nonnegative reals and so it must be convergent. So, let (29)limn→∞⁡dzn,zn+1=c≥0.From (ϑ2), we get(30)limn→∞⁡sup⁡λHTzn,Tzn,dzn-1,zn<1.Now, if, in (29), c>0, then, by taking limn→∞⁡sup in (28) and using (30), we have (31)c≤limn→∞⁡sup⁡λHTzn-1,Tzn,dzn-1,znc<c.This contradiction shows that c=0. Hence, (32)limn→∞⁡dzn,zn+1=0.Next, we prove that znn∈N is a Cauchy sequence in X. Let, for each n∈N, (33)σn=λHTzn-1,Tzn,dzn-1,zn.Then, from (16), we have σn∈(0,1). By (28), we obtain (34)dzn,zn+1<σndzn-1,zn.Equation (30) implies that limn→∞⁡σn<1, so there exists γ∈[0,1) and n0∈N, such that (35)σn≤γ,∀n∈N,n≥n0.For any n≥n0, since σn∈(0,1) for all n∈N and γ∈[0,1), (34) and (35) imply that (36)dzn,zn+1<σndzn-1,zn<σnσn-1dzn-2,zn-1<⋯<σnσn-1σn-2⋯σn0dz0,z1≤γn-n0+1dz0,z1.Put βn=γn-n0+1/1-γd(z0,z1), n∈N. For m,n∈N with m>n≥n0, we have from (36) that(37)dzn,zm≤∑j=nm-1dzj,zj+1<βn.Since γ∈[0,1), limn→∞⁡βn=0. Hence (38)limn→∞⁡sup⁡dzn,zm:m>n=0.This shows that zn is a Cauchy sequence in X. Completeness of X ensures the existence of z∈X such that zn→z as n→∞. Now, since α(zn,z)≥η(zn,z) for all n∈N, α∗(Tzn,Tz)≥η∗(Tzn,Tz), and so, from (11), we have (39)ϑHTzn,Tz,dzn,z≥0.Then, from (14) and (15), we have (40)HTzn,Tz≤λHTzn,Tz,dzn,zdzn,z<dzn,z.Since 0<d(z,Tz)≤H(Tzn,Tz)+d(zn,z), so, by using (40), we get (41)0<dz,Tz<2dzn,z.Letting limit n→∞ in the above inequality, we get (42)dz,Tz=0.Hence z∈Fix{T}.Example 8.
LetX=R with usual metric d. Then (X,d) is a complete metric space. Define T:X→CL(X), α,η:X×X→R+ and ϑ:R×R→R by (43)Tz=1,14zifz>1;0,z16ifz∈0,1;2,3otherwise,αy,z=2ifz,y∈0,1;12otherwise.η(y,z)=1, for all z,y∈X and ϑ(t,s)=as-t, where a∈1/16,1. Then ϑ is a manageable function. Indeed, for any s,t>0, we have (44)ϑt,s=as-t<s-t,so, (ϑ1) holds. Let tn⊂(0,+∞) be a bounded sequence and let sn⊂(0,+∞) be a nonincreasing sequence. Then (45)limn→∞⁡sup⁡tn+ϑtn,snsn=a<1,which means that (ϑ2) holds. Hence ϑ∈Man(R)^.
Sinceα(y,z)≥η(y,z) when z,y∈[0,1], this implies that (46)α∗Ty,Tz=infu∈Ty,v∈Tz⁡αu,v=2>1=supu∈Ty,v∈Tz⁡ηu,v=η∗Ty,Tz.Hence T is α∗-admissible mapping with respect to η.
Letα∗(Ty,Tz)≥η∗(Ty,Tz); then y,z∈[0,1]. This implies that(47)ϑHTy,Tz,dy,z=ϑy16-z16,y-z=ay-z-y16-z16=y-za-116≥0.Thus, all conditions of Theorem 7 are satisfied and 0 is a fixed point of T.
On the other hand, fory=-1 and z=0, we have (48)HTy,Tz=2>1=dy,z.This implies that T is not a multivalued contraction, so we cannot apply Nadler’s theorem [19] with this example.On bearingη(y,z)=1 in Theorem 7, we get the following corollary.Corollary 9.
Let(X,d) be a complete metric space and let T:X→2X be an α∗-admissible and closed valued map enjoying the following:(1)
α∗(Ty,Tz)≥1 implies ϑH(Ty,Ty),d(y,z)≥0;(2)
there existsz0∈X and z1∈Tz0 such that α(z0,z1)≥1;(3)
for a sequence{zn}⊂X, limn→∞⁡zn=x, and α(zn,zn+1)≥1, for all n∈N, one has α(zn,x)≥1 for all n∈N,for all y,z∈X and ϑ∈Man(R)^. Then Fix{T}≠⌀.By takingαy,z=1 in Theorem 7, we get the following corollary.Corollary 10.
Let(X,d) be a complete metric space and let T:X→2X be an η∗-admissible and closed valued map enjoying the following conditions:(1)
η∗(Ty,Tz)≤1 implies ϑH(Ty,Ty),d(y,z)≥0;(2)
there existsz0∈X and z1∈Tz0 such that η(z0,z1)≤1;(3)
for a sequence{zn}⊂X, limn→∞zn=x, and η(zn,zn+1)≤1, for all n∈N, one has η(zn,x)≤1 for all n∈N,for all y,z∈X and ϑ∈Man(R)^. Then Fix{T}≠⌀.Theorem 11.
Let(X,d) be a complete metric space and let the sequence {Ti:X→2X} of closed valued mappings enjoy the following with condition (4) of Theorem 7: (1)
Ti is α∗-admissible with respect to η;(2)
αy,z≥η(y,z) implies  ϑH(Tiy,Tjz),d(x,y)≥0;(3)
there existsz0∈X and yi∈Tiz0 for each i∈N such that α(z0,yi)≥ηz0,yi;for all y,z∈X, i,j∈N, and ϑ∈Man(R)^. Then Fix{Ti}≠⌀ for each i∈N.Proof.
Letz1∈T1z0 be such that α(z0,z1)≥η(z0,z1); then from (11) we have(49)ϑHT1z0,T2z1,dz0,z1≥0.If z1∈Tiz1, for each i∈N, then z1∈Fix{Ti}. Adopt that z1∉T2z1. Thus (50)0<dz1,T2z1≤HT1z0,T2z1.From (50), we get (51)0<λHT1z0,T2z1,dz0,z1<1,where λ is defined in (13). Let (52)ε1=αz0,z1ηz0,z1λHT1z0,T2z1,dz0,z1-1dz1,T2z1.Since α(z0,z1)≥η(z0,z1),  d(z1,T2z1)>0. So, by using (51), we get ε1>0 and (53)dz1,T2z1<dz1,T2z1+ε1=αz0,z1ηz0,z1λHT1z0,T2z1,dz0,z1dz1,T2z1.This implies that there exists z2∈T2z1 such that (54)dz1,z2<αz0,z1ηz0,z1λHT1z0,T2z1,dz0,z1dz1,T2z1.Note that z1≠z2 (since z1∉T2z1). Now if z2∈Tiz2, for each i∈N, then z2∈Fix{Ti}. Let z2∉T3z2; then (55)0<dz2,T3z2<HT2z1,T3z2.Since the sequence {Ti} is α∗-admissible with respect to η, so we have α(z1,z2)≥η(z1,z2) and, from condition (11), we get (56)ϑHT2z1,T3z2,dz1,z2≥0.By taking (57)ε2=αz1,z2ηz1,z2λHT2z1,T3z2,dz1,z2-1dz2,T3z2,there exists z3∈T3z2 with z3≠z2 such that (58)dz2,z3<αz1,z2ηz1,z2λHT2z1,T3z2,dz1,z2dz2,T3z2.Hence, by induction, we can establish a sequence {zn} in X satisfying for each n∈N, zn∈Tnzn-1, zn≠zn-1, zn∉Tizn for each n∈N, and α(zn-1,zn)≥η(zn-1,zn)(59)0<dzn,Tn+1zn≤HTnzn-1,Tn+1zn,(60)ϑHTnzn-1,Tn+1zn,dzn-1,zn≥0,(61)dzn,zn+1<αzn-1,znηzn-1,znλHTnzn-1,Tn+1zn,dzn-1,zn×dzn,Tn+1zn,by taking (62)εn=αzn-1,znηzn-1,znλHTnzn-1,Tn+1zn,dzn-1,zn-1dzn,Tn+1zn.By using (14), (15), (59), and (61), we get for each n∈N(63)dzn,Tn+1zn≤dzn-1,znλHTnzn-1,Tn+1zn,dzn-1,zn≤dzn-1,zn.Which means that d(zn,Tn+1zn)n∈N is a bounded sequence. By combining (61) and (63), for each n∈N, we get (64)dzn,zn+1<λHTnzn-1,Tn+1zn,dzn-1,zn×dzn-1,zn.Which means that {d(zn-1,zn)}n∈N is a monotonically decreasing sequence of nonnegative reals and so it must be convergent. So, let (65)limn→∞⁡dzn,zn+1=c≥0.From (ϑ2), we get(66)limn→∞⁡sup⁡λdzn,Tn+1zn,dzn-1,zn<1.Now, if, in (65), c>0, then, by taking limn→∞⁡sup in (64) and using (66), we have (67)c≤limn→∞⁡sup⁡λHTnzn-1,Tn+1zn,dzn-1,znc<c.This contradiction shows that c=0. Hence,(68)limn→∞⁡dzn,zn+1=0.Next, by following the same procedure as in proof of Theorem 7, we show that sequence znn∈N is a Cauchy sequence in X. Completeness of X guarantees that there is z∈X such that zn→z as n→∞. Now, since α(zn,z)≥η(zn,z) for all n∈N, and from condition (11), we have (69)ϑHTn+1zn,Tiz,dzn,z≥0.Then, from (14) and (15), we have (70)HTn+1zn,Tiz≤λHTn+1zn,Tiz,dzn,zdzn,z<dzn,z.Since 0<d(zn+1,Tiz)<H(Tn+1xn,Tiz), by using (70), we get (71)0<dzn+1,Tiz≤dzn,z.Letting limit n→∞ in the above inequality, we get (72)dz,Tiz=0.Hence z∈Fix{Tiz}.Let us takeTi=T for each i∈N; then Theorem 11 reduces to the following.Theorem 12.
Let(X,d) be a complete metric space and let the sequence T:X→2X be a closed valued mapping enjoying the following with conditions (3) and (4) of Theorem 7:(1)
T is generalized α∗-admissible with respect to η;(2)
α(y,z)≥η(y,z)impliesϑH(Ty,Tz),d(y,z)≥0;for all y,z∈X and ϑ∈Man(R)^. Then Fix{T}≠⌀.As an application of Theorems7 and 11, we can deduce the following result.Theorem 13.
Let(X,d) be a complete metric space and let T:X→2X be a closed valued mapping satisfying for all y,z∈X(73)ϑHTy,Tz,dy,z≥0,where ϑ∈Man(R)^. Then Fix{T}≠⌀.Proof.
Defineα,η:X×X→[0,∞) by (74)αy,z=dy,z,ηy,z=dy,z,for all y,z∈X. So that α(y,z)=η(y,z). This implies that α∗(Ty,Tz)=η∗(Ty,Tz) for all y,z∈X. That is, all the conditions of Theorem 7 hold true. Hence T has a fixed point.In the following corollaries, we obtain some known and some new results in literature via manageable functions.Corollary 14 (see [19]).
Let(X,d) be a complete metric space and let T:X→2X be a closed valued mapping satisfying for all y,z∈X(75)HTy,Tz≤kdy,z,where k∈(0,1). Then Fix{T}≠⌀.Proof.
DefineϑN:R×R→R by (76)ϑNt,s=ks-t∀t,s∈R×R,k∈0,1.Then ϑN∈Man(R)^, by Example 4. Therefore the result follows by taking ϑ=ϑN in Theorem 13.Corollary 15 (see [20]).
Let(X,d) be a complete metric space and let T:X→2X be a closed valued mapping satisfying for all y,z∈X(77)HTy,Tz≤σdy,zdy,z,where σ∈Σ. Then Fix{T}≠⌀.Proof.
DefineϑM:R×R→R by (78)ϑMt,s=sσs-tift,s∈0,+∞×0,+∞,ft,sotherwise,where σ∈Σ and f:R×R→R be any function. Then ϑM∈Man(R)^ (see Example C in [10]). Therefore the result follows by taking ϑ=ϑM in Theorem 13.Corollary 16.
Let(X,d) be a complete metric space and let T:X→2X be a closed valued mapping satisfying for all y,z∈X(79)∫0HTy,Tzφtdt≤αdy,z,where φ∈Φ. Then Fix{T}≠⌀.Proof.
DefineϑI:R×R→R by (80)ϑIt,s=s-∫0tφuduift,s∈0,+∞×0,+∞,ft,sotherwise,where φ∈Φ and f:R×R→R be any function. Then for s,t>0 we have (81)ϑIt,s=s-∫0tφudu<s-t.Now let {tn}⊂(0,+∞) be a bounded sequence and let {sn}⊂(0,+∞) be a nonincreasing sequence. Then limn→∞⁡sn=infn∈Nsn=a for some a∈[0,+∞) and (82)limn→∞⁡sup⁡tn+ϑtn,snsn=limn→∞⁡sup⁡tn+sn-∫0tφudusn<limn→∞⁡sup⁡tn+sn-tnsn=1.Hence ϑI∈Man(R)^. Therefore the result follows by taking ϑ=ϑI in Theorem 13.Corollary 17.
Let(X,d) be a complete metric space and let T:X→2X be a closed valued mapping satisfying for all y,z∈X(83)HTy,Tz<γdy,z,where γ∈Γ. Then Fix{T}≠⌀.Proof.
DefineϑL:R×R→R by (84)ϑLt,s=γs-tift,s∈0,+∞×0,+∞,ft,sotherwise,where γ∈Γ and f:R×R→R be any function. Then ϑL∈Man(R)^. Indeed, by using Lemma  10 of [22], we have for s,t>0(85)ϑLt,s=γs-t<s-t,so, (ϑ1) holds. Let {tn}⊂(0,+∞) be a bounded sequence and let {sn}⊂(0,+∞) be a nonincreasing sequence. Then limn→∞⁡sn=infn∈N⁡sn=a for some a∈[0,+∞) and (86)limn→∞⁡sup⁡tn+ϑtn,snsn=limn→∞⁡sup⁡γsnsn<limn→∞snsn=1.So, (ϑ2) is also satisfied. Therefore the result follows by taking ϑ=ϑL in Theorem 13.Remark 18.
SinceT is a Meir-Keeler multivalued mapping of a metric space (X,d) if and only if there exists (nondecreasing, right continuous) mapping γ∈Γ such that H(Ty,Tz)<γ(d(y,z)) (see [21], Theorem  2), therefore, from Corollary 17, we get the main result of [23].By takingϑ=ϑM in Theorem 12, as defined in Corollary 15, we get the following.Corollary 19 (see [3]).
Let(X,d) be a complete metric space and let T:X→2X be a closed valued mapping enjoying the following with conditions (3) and (4) of Theorem 7.(1)
T is generalized α∗-admissible with respect to η;(2)
α(y,z)≥η(y,z)  implies  H(Ty,Tz)≤σ(d(y,z))d(y,z),for all y,z∈X and σ∈Σ. Then Fix{T}≠⌀.On considering(87)ϑt,s=ψs-tift,s∈0,+∞×0,+∞;ft,sotherwise,where ψ∈Ψ and f:R×R→R is any function, in Theorems 7, 11, and 12, we get the following existing theorems, respectively.Corollary 20 (see [14]).
Let(X,d) be a complete metric space and let T:X→2X be a closed valued mapping fulfilling conditions (1), (3), and (4) with(88)α∗Ty,Tz≥η∗Ty,TzimpliesHTy,Tz≤ψdy,z,for all y,z∈X and ψ∈Ψ. Then FixT≠⌀.Corollary 21 (see [2]).
Let(X,d) be a complete metric space and let the sequence {Ti:X→2X}i=1∞ be α∗-admissible with respect to η fulfilling condition (4) of Theorem 7 and (3) of Theorem 11 with (89)αy,z≥ηy,zimpliesHTiy,Tjz≤ψdy,z,for all y,z∈X, i,j∈N and ψ∈Ψ. Then FixTi≠⌀.Corollary 22 (see [2]).
Let(X,d) be a complete metric space and let T:X→2X be a closed valued mapping and generalized α∗-admissible with respect to η fulfilling conditions (3) and (4) of Theorem 7 with (90)αy,z≥ηy,zimpliesHTy,Tz≤ψdy,z,for all y,z∈X and ψ∈Ψ. Then Fix{T}≠⌀.


## 3. Fixed Point Results in Partially Ordered Metric Spaces and an Application to Integral Equations

Let(X,d,⪯) be a partially ordered metric space. Recall that T:X→2X is monotone increasing if Ty⪯Tz for all y,z∈X, for which y⪯z (see [8]). There are many applications in differential and integral equations of monotone mappings in ordered metric spaces (see [15, 24–26] and references therein). In this section, from Theorems 7–13, we derive the following new results in partially ordered metric spaces and give an example to integral equations.Theorem 23.
Let(X,d,⪯) be a complete partially ordered metric space and let T:X→2X be a closed valued mapping satisfying the following assertions for all y,z∈X with y⪯z: (1)
T is monotone increasing;(2)
ϑH(Ty,Tz),d(y,z)≥0;(3)
there existsz0∈X and z1∈Tz0 such that z0⪯z1;(4)
for a sequence{zn}⊂X, limn→∞zn=z, and zn⪯zn+1 for all n∈N, one has zn⪯z for all n∈N.Then Fix{T}≠⌀.Proof.
Defineα,η:X×X→[0,∞) by (91)αy,z=1y⪯z0otherwiseηy,z=12y⪯z0otherwise.Then for y,z∈X with y⪯z, α(y,z)≥η(y,z) implies α∗(Ty,Tz)=1>1/2=η∗(Ty,Tz) and α∗(Ty,Tz)=η∗(Ty,Tz)=0 otherwise. Thus, all the conditions of Theorem 7 are satisfied and hence T has a fixed point.Theorem 24.
Let(X,d,⪯) be a complete partially ordered metric space and let the sequence {Ti:X→2X} of closed valued mappings enjoy the following assertions, for all y,z∈X with x⪯y and for each i,j∈N:(1)
ϑH(Tiy,Tjz),d(y,z)≥0;(2)
ify⪯z, then Tiy⪯z for each i∈N;(3)
there existsz0∈X and yi∈Tiz0 for each i∈N such that z0⪯yi;(4)
for a sequence{zn}⊂X, limn→∞zn=z, and zn⪯zn+1, for all n∈N, we have zn⪯z for all n∈N.Then FixTi≠⌀ for all i∈N.Proof.
By definingα,η:X×X→[0,∞) as in Theorem 23 and by using Theorem 11, we get the required result.Similar to the arguments of Theorems23 and 24, we conclude the following result and omit its proof.Theorem 25.
Let(X,d,⪯) be a complete partial ordered metric space and let T:X→2X be a closed valued mapping satisfying (73) of Theorem 13 for all x,y∈X with x⪯y. Then Fix{T}≠⌀.In case of single valued mapping, Theorems23–25 reduced to the following.Theorem 26.
Let(X,d,⪯) be a complete partially ordered metric space and let T:X→X be a self-map fulfilling the following assertions:(1)
T is monotone increasing;(2)
ϑd(Ty,Tz),d(y,z)≥0;(3)
there existsz0∈X and z1=Tz0 such that z0⪯z1;(4)
for a sequencezn⊂X, limn→∞zn=z, and zn⪯zn+1, for all n∈N, we have zn⪯z for all n∈N,for all y,z∈X with y⪯z and ϑ∈Man(R)^. Then FixT≠⌀.Theorem 27.
Let(X,d,⪯) be a complete partially ordered metric space and let the sequence {Ti:X→X} of self-mappings fulfill the following assertions: (1)
ϑd(Tiy,Tjz),d(y,z)≥0;(2)
Ti is nondecreasing for each i;(3)
there existsz0∈X and yi=Tiz0 for each i∈N such that z0⪯yi;(4)
for a sequencezn⊂X, limn→∞zn=z and zn⪯zn+1 for all n∈N, we have zn⪯z for all n∈N,for all y,z∈X with y⪯z, ϑ∈Man(R)^ and i∈N. Then FixTi≠⌀.Theorem 28.
Let(X,d,⪯) be a complete partial ordered metric space and let T:X→X be a self-map satisfying for all y,z∈X with y⪯z(92)ϑdTy,Tz,dy,z≥0,where ϑ∈Man(R)^. Then FixTi≠⌀.Now we give an application of our results and establish the existence of solution of the integral equation.(93)zr=∫bcBr,s,zsds+gr,t∈b,c.Let≪ be a partial order relation on Rn. Define T:X→X by (94)Tzr=∫bcBr,s,zsds+gr,r∈a,b.Theorem 29.
LetX=C([b,c],Rn) with the usual supremum norm. Suppose that (1)
B:[b,c]×[b,c]×Rn→Rn and g:Rn→Rn are continuous;(2)
there exists a continuous functionp:[b,c]×[b,c]→[b,c] such that (95)Br,s,u-Br,s,v≤pr,su-v,for eachr,s∈[b,c] and u,v∈Rn with u≪v;(3)
supr∈[b,c]∫bcp(r,s)ds=q≤1/4;(4)
there existsz0∈X and z1∈Tz0 such that z0⪯z1;(5)
for a sequencezn⊂X, limn→∞⁡zn=z, and ⪯zn+1, for all n∈N, one has zn⪯z for all n∈N.Then the integral equation (93) has a solution in X.Proof.
LetX=C([b,c],Rn) and z=maxr∈[b,c]zr, for z∈C([a,b]). Consider a partial order defined on X by(96)y,z∈Cb,c,Rn,y⪯ziffyr≪zr,forr∈b,c.
Then(X,·,⪯) is a complete partial ordered metric space and for any increasing sequence zn in X converging to z∈X, we have zn(r)≪z(r) for any r∈[b,c] (see [27]). By using (94) and conditions (2) and (3) and taking ϑr,s=1/2s-r for all y,z∈X with y⪯z, we obtain(97)Tyr-Tzr=∫bcBr,s,ysds-∫bcBr,s,zsds≤∫bcBr,s,ys-Br,s,zsds≤∫bcpr,sys-zsds≤14y-z.
This implies that(98)12y-z-Ty-Tz≥12y-z-14y-z=14y-z.
Soϑ(d(Ty,Tz),d(y,z))≥0 for all y,z∈X with y⪯z. Hence all the conditions of Theorem 26 are satisfied. Therefore T has a fixed point; consequently, integral equation (93) has a solution in X.



---

*Source: 7943896-2017-02-16.xml*