# Fixed-Point Theorems forα-Admissible Mappings with w-Distance and Applications to Nonlinear Integral Equations

Authors: Fengrong Zhang; Haoyue Wang; Shuangqi Wu; Liangshi Zhao
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

Two fixed-point theorems forα-admissible mappings satisfying contractive inequality of integral type with w-distance in complete metric spaces are proved. Our results extend and improve a few existing results in the literature. As applications, we use the fixed-point theorems obtained in this paper to establish solvability of nonlinear integral equations. Examples are included.


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## Body

## 1. Introduction and Preliminaries

In 2002, Branciari [1] discussed the existence and uniqueness of fixed points for mappings satisfying contractive condition of integral type, which is a generalization of the Banach contraction principle in metric spaces.Theorem 1 (see [1]).
Letf be a mapping from a complete metric space X,d into itself satisfying(1)∫0dfx,fyϕtdt≤c∫0dx,yϕtdt,for all x,y∈X, where c∈0,1 is a constant and ϕ∈Φ2. Then, f has a unique fixed point a∈X and limn⟶∞fnx=a for each x∈X.In 2012, Samet et al. [2] introduced the concept of α-ψ-contractive type mappings and established some fixed-point theorems for these mappings in complete metric spaces.Definition 1 (see [2]).
LetX,d be a metric space, T:X⟶X and α:X×X⟶ℝ+ be two given mappings. Then, T is called an α-admissible mapping if(2)αx,y≥1⇒αTx,Ty≥1,∀x,y∈X.Theorem 2 (see [2]).
LetX,d be a complete metric space and T:X⟶X be an α-ψ-contractive mapping, that is,(3)αx,ydTx,Ty≤ψdx,y,∀x,y∈X,where ψ∈Φ1. Assume thata1 T is α-admissiblea2 there exists x0∈X such that αx0,Tx0≥1a3 T is continuous
Then,T has a fixed point, that is, there exists x∗∈X such that Tx∗=x∗.Theorem 3 (see [2]).
LetX,d be a complete metric space and T:X⟶X satisfy (3), a1, a2, and the followinga4 if xnn∈ℕ is a sequence in X such that αxn,xn+1≥1 for all n∈ℕ and xn⟶x∈X as n⟶∞, then αxn,x≥1 for all n∈ℕ.Then,T has a fixed point.In 1996, Kada et al. [3] introduced the notion of w-distance and elaborated with the help of examples, that the concept of w-distance is general than that of metric on a nonempty set. Several researchers [4–7] used this notion to obtain some fixed point results not only in usual metric spaces but also in other spaces.Definition 2 (see [3]).
LetX,d be a metric space. A function p:X×X⟶ℝ+ is called a w-distance in X if it satisfies the following:p1px,z≤px,y+py,z,∀x,y,z∈Xp2 for each x∈X, a mapping px,⋅:X⟶ℝ+ is lower semicontinuousp3 for any ε>0, there exists δ>0 such that pz,x≤δ and pz,y≤δ imply dx,y≤ε.
In 2016, Lakzian et al. [8] introduced the concept of generalized α−ψ−p-contractive mappings and proved the following fixed point results for such mappings, which generalize Theorems 2 and 3.Theorem 4 (see [8]).
Letp be a w-distance on a complete metric space X,d, and let f:X⟶X be an α−ψ−p-contractive mapping, that is,(4)αx,ypfx,fy≤ψpx,y,∀x,y∈X,where ψ∈Φ1. Assume thatb1 f is an α-admissible mappingb2 there exists a point x0∈X such that αx0,fx0≥1b3 either f is continuous or, for any sequence xnn∈ℕ in X if αxn,xn+1≥1 for all n∈ℕ and xn⟶x∈X as n⟶∞, then αxn,x≥1 for all n∈ℕ.Then, there exists a pointu∈X such that fu=u. Moreover, if αu,u≥1, then pu,u=0.Theorem 5 (see [8]).
Letp be a w-distance on a complete metric space X,d, and let f:X⟶X satisfy (4), b1, b2, and the followingb4 for every y∈X with y≠fy, infpx,y+px,fx:x∈X>0.
Then, there exists a pointu∈X such that fu=u. Moreover, if αu,u≥1, then pu,u=0.
Motivated by the results in [1–10], in this paper we prove the existence and uniqueness of fixed points for α-admissible mappings satisfying contractive inequality of integral type via w-distance in complete metric spaces, which are used to study solvability of nonlinear Fredholm and Volterra integral equations. Our results generalize Theorems 1–5 and two examples are given.
Throughout this paper, we denote byℕ the set of positive integers, ℕ0=0∪ℕ, ℝ=−∞,+∞, ℝ+=0,+∞, and(5)Φ1=ψψ:ℝ+⟶ℝ+satisfies thatψis nondecreasing and∑n=1∞ψnt<+∞,∀t>0;Φ2=ϕϕ:ℝ+⟶ℝ+is Lebesgue integrable,summable in each compact subset ofℝ+and∫0εϕtdt>0,∀ε>0;Φ3=ϕϕis inΦ2and∫0a+bϕtdt≤∫0aϕtdt+∫0bϕtdt,∀a,b∈ℝ+.Lemma 1 (see [10]).
Letϕ∈Φ2 and rnn∈ℕ be a nonnegative sequence. Then, limn⟶∞∫0rnϕtdt=0 if and only if limn⟶∞rn=0.


## 2. Fixed-Point Theorems

In this section, we prove the existence and uniqueness of fixed points forα-admissible mapping (6) with w-distance.Theorem 6.
Letp be a w-distance in a complete metric space X,d and let f:X⟶X satisfy that(6)αx,y∫0pfx,fyϕtdt≤ψ∫0px,yϕtdt,∀x,y∈X,where ψ,ϕ∈Φ1×Φ3. Assume thatc1 f is an α-admissible mappingc2 there exists a point x0∈X such that αx0,fx0≥1
if one of the following conditions holds:c3 f is continuousc4 for any sequence xnn∈ℕ in X if αxn,xn+1≥1 for all n∈ℕ and xn⟶x∈X as n⟶∞, then αxn,x≥1 for all n∈ℕ Then, 1 f has a fixed point u∈X and limn⟶∞fnx0=u. Moreover, pu,u=0 if αu,u≥1.
2 f has a unique fixed point u∈X if for any fixed points x,y∈X of f, there exists a point z∈X such that αz,x≥1 and αz,y≥1.Proof.
Firstly, we show (1). Define a sequencexnn∈ℕ in X by xn+1=fxn, ∀n∈ℕ0, where x0 satisfies c2. Assume that there exists some n0∈ℕ with xn0=xn0−1. Put u=xn0−1. Clearly, u is a fixed point of f and u=limn⟶∞fnx0.
Now, we assume thatxn≠xn−1 for all n∈ℕ. It follows from c1 and c2 that(7)αx0,x1=αx0,fx0≥1⇒αx1,x2=αfx0,fx1≥1.
It is easy to see that(8)αxn,xn+1≥1,∀n∈ℕ0.
In view of (6) and (8) and ψ,ϕ∈Φ1×Φ3, we conclude that(9)0≤∫0pxn,xn+1ϕtdt=∫0pfxn−1,fxnϕtdt≤αxn−1,xn∫0pfxn−1,fxnϕtdt≤ψ∫0pxn−1,xnϕtdt≤ψn∫0px0,x1ϕtdt⟶0asn⟶∞,which implies that(10)limn⟶∞∫0pxn,xn+1ϕtdt=0,
which together withψ,ϕ∈Φ1×Φ3 and Lemma 1 yields that(11)limn⟶∞pxn,xn+1=0.
Letε>0 and δ be defined by p3. Note that ∑n=1∞ψn∫0px0,x1ϕtdt<+∞, which means that there exists n0∈ℕ satisfying(12)∑k=n0∞ψk∫0px0,x1ϕtdt<∫0δϕtdt.
In the following, we claim thatxnn∈ℕ is a Cauchy sequence. Making use of (6), (9), (12), p1, and ψ,ϕ∈Φ1×Φ3, we infer that(13)∫0pxn,xmϕtdt≤∫0∑k=nm−1pxk,xk+1ϕtdt≤∑k=nm−1∫0pxk,xk+1ϕtdt≤∑k=nm−1ψk∫0px0,x1ϕtdt<∫0δϕtdt,∀m,n∈ℕwithm>n≥n0,which implies that(14)pxn,xm<δ,∀m,n∈ℕwithm>n≥n0.
It follows from (14) that(15)pxn0,xn<δ,pxn0,xm<δ,∀m,n∈ℕwithm>n≥n0,which together with p3 gives that(16)dxn,xm<ε,∀m,n∈ℕwithm>n≥n0,
That is,xnn∈ℕ is a Cauchy sequence. By completeness of X, there exists a point u∈X such that(17)limn⟶∞xn=u.
Assume thatc3 holds. Using (17) and c3, we gain that(18)u=limn⟶∞xn+1=limn⟶∞fxn=flimn⟶∞xn=fu.
Assume thatc4 holds. By virtue of (8), (17), and c4, we deduce that(19)αxn,u≥1,∀n∈ℕ.
Similar to the proofs of (12)–(14), we know that for each ε1>0 there exists n1∈ℕ satisfying(20)0≤pxn,xm<ε1,∀m,n∈ℕwithm>n≥n1,which together with p2 and (17) yields that(21)0≤pxn,u≤liminfm⟶∞pxn,xm≤ε1,∀n∈ℕwithn≥n1,that is,(22)limn⟶∞pxn,u=0.
In light of (6), (19), and ψ,ϕ∈Φ1×Φ3, we obtain that(23)0≤∫0pxn+1,fuϕtdt=∫0pfxn,fuϕtdt≤αxn,u∫0pfxn,fuϕtdt≤ψ∫0pxn,uϕtdt≤ψn+1∫0px0,uϕtdt⟶0asn⟶∞,which together with Lemma 1 implies that(24)limn⟶∞pxn+1,fu=0.
In view of (11), (24) and, p1, we infer that(25)0≤pxn,fu≤pxn,xn+1+pxn+1,fu⟶0asn⟶∞,that is,(26)limn⟶∞pxn,fu=0.
Letε2>0. It follows from p3 that there exists δ1>0 such that pu,v≤δ1 and pu,z≤δ1 imply dv,z<ε2. Combining (22) and (26), we know that there exists n2∈ℕ such that pxn,u≤δ1 and pxn,fu≤δ1 for all n≥n2. Hence, du,fu≤ε2. Letting ε2⟶0+, we have(27)u=fu.
Next, we show thatpu,u=0 if αu,u≥1. Suppose that pu,u>0. In view of (6) and ψ,ϕ∈Φ1×Φ3, we infer that(28)0<∫0pu,uϕtdt=∫0pfu,fuϕtdt≤αu,u∫0pfu,fuϕtdt≤ψ∫0pu,uϕtdt<∫0pu,uϕtdt,which is ridiculous. Hence, pu,u=0.
Secondly, we show (2). Suppose thatx and y are two fixed points of f in X. It follows that there exists a point z∈X satisfying(29)αz,x≥1,αz,y≥1.
In light of (29) and c1, we get that(30)αfnz,x≥1αfnz,y≥1,∀n∈ℕ,which together with (6) and ψ,ϕ∈Φ1×Φ3 imply that(31)0≤∫0pfn+1z,xϕtdt=∫0pffnz,fxϕtdt≤αfnz,x∫0pffnz,fxϕtdt≤ψ∫0pfnz,fxϕtdt≤ψn∫0pfz,xϕtdt⟶0asn⟶∞,which gives that(32)limn⟶∞∫0pfn+1z,xϕtdt=0.
Similarly, we conclude that(33)limn⟶∞∫0pfn+1z,yϕtdt=0.
Making use of (32), (33), and Lemma 1, we get that(34)limn⟶∞pfn+1z,x=0,limn⟶∞pfn+1z,y=0,which together with the proof of (27) yields similarly that x=y. That is, f has a unique fixed point in X. This completes the proof.Theorem 7.
Letp be a w-distance in a complete metric space X,d, and let f:X⟶X satisfy (6), c1, c2, and the followingc5infpx,y+px,fx:x∈X}>0 for each y∈X with y≠fy.
Then, (1) and (2) hold.Proof.
Following the proof of Theorem6, we deduce that xnn∈ℕ is a Cauchy sequence. By completeness of X,d, there exists a point u∈X that satisfies (17) holds. Suppose that u≠fu. In light of (11), (17), and c5, we conclude that(35)0<infpx,u+px,fx:x∈X≤infpxn,u+pxn,xn+1:n∈ℕ=0,which is impossible. Hence, u=fu. The rest of the proof is similar to that of Theorem 6 and is omitted. This completes the proof.Remark 1.
Theorem6 generalizes and improves Theorems 1–4. The following example manifests that Theorem 6 extends substantially Theorem 1.Example 1.
LetX=0,1/2∪2/3∪1 be endowed with the Euclidean metric d=⋅, p:X×X⟶ℝ+, ψ,ϕ:ℝ+⟶ℝ+, f:X⟶X, and α:X×X⟶ℝ+ be defined by(36)px,y=y,∀x,y∈X,ψt=45t,ϕt=1,∀t∈ℝ+,fx=23x2,∀x∈0,12,12,x=23,23,x=1,αx,y=1,ifx,y∈0,12∪23,0,otherwise.
It is clear thatp is a w-distance in X and ψ,ϕ∈Φ1×Φ3. Let x,y∈X. In order to verify (6), we have to consider the following cases.Case 1.
x
∈
0
,
1
/
2
∪
2
/
3 and y∈0,1/2. It follows that(37)αx,y∫0pfx,fyϕtdt=∫02/3y2dt=23y2≤45y=ψy=ψ∫0px,yϕtdt.Case 2.
x
∈
0
,
1
/
2
∪
2
/
3 and y=2/3. Note that(38)αx,y∫0pfx,fyϕtdt=∫01/2dt=12≤45⋅23=ψ23=ψ∫0px,yϕtdt.Case 3.
x
∉
0
,
1
/
2
∪
2
/
3 or y∉0,1/2∪2/3. It is clear that(39)αx,y∫0pfx,fyϕtdt=0≤ψ∫0px,yϕtdt.
That is, (6) holds. Let x,y∈X with αx,y≥1. It follows that x,y∈0,1/2∪2/3 and(40)fx=23x2,∀x∈0,12,12,x=23,fy=23y2,∀y∈0,12,12,y=23,which imply that fx∈0,1/2 and fy∈0,1/2, that is, αfx,fy=1. Hence, f is an α-admissible mapping.
Putx0=2/3∈X. It is clear that αx0,fx0=α2/3,1/2=1.
Finally, letxnn∈ℕ be a sequence in X such that αxn,xn+1≥1 for all n∈ℕ and xn⟶x∈X as n⟶∞. It is easy to see that xn∈0,1/2∪2/3 for all n∈ℕ. Since xnn∈ℕ is a sequence in the closed subset 0,1/2∪2/3 in the metric space X,d, it follows that the point x belongs to 0,1/2∪2/3. Therefore, αxn,x=1 for all n∈ℕ.
Hence, the conditions of Theorem6 are satisfied. It follows from Theorem 6 that f has a fixed point 0∈X.
However, we cannot invoke Theorem1 to show that the mapping f has a fixed point in X. Suppose that the conditions of Theorem 1 are satisfied. Clearly, y∗=2/3.(41)0<∫01/3ϕtdt=∫0df1/2,f2/3ϕtdt≤c∫0d1/2,2/3ϕtdt<∫01/6ϕtdt≤∫01/3ϕtdt,which is impossible. Thus, Theorem 1 is not applicable in proving the existence of fixed points for the mapping f in X.Remark 2.
Theorem7 generalizes Theorem 5. The example below is an application of Theorem 7.Example 2.
LetX=0∪1/n+12:n∈ℕ be endowed with the Euclidean metric d=⋅, p:X×X⟶ℝ+, ψ,ϕ:ℝ+⟶ℝ+, f:X⟶X, and α:X×X⟶ℝ+ be defined by(42)px,y=y2,∀x,y∈X,ψt=12t,ϕt=1,∀t∈ℝ+,fx=x2,∀x∈Xαx,y=1,ifx,y∈1n+12:n∈ℕ,0,otherwise.
It is clear thatp is a w-distance in X and ψ,ϕ∈Φ1×Φ3. Let x,y∈X. In order to show that f satisfies (6), we have to consider the following cases.Case 1.
x
,
y
∈
1
/
n
+
1
2
:
n
∈
ℕ. It follows that(43)αx,y∫0pfx,fyϕtdt=∫0y4dt=y4≤12y2=ψy2=ψ∫0px,yϕtdt.Case 2.
x
∉
1
/
n
+
1
2
:
n
∈
ℕ or y∉1/n+12:n∈ℕ. It is clear that(44)αx,y∫0pfx,fyϕtdt=0≤ψ∫0px,yϕtdt,that is, (6) holds. Let x,y∈X with αx,y≥1. It follows that x,y∈1/n+12:n∈ℕ and(45)fx=x2∈1n+12:n∈ℕ,fy=y2∈1n+12:n∈ℕ,which yield that αfx,fy=1. Hence, f is an α-admissible mapping. Put x0=1/4∈X. It follows that αx0,fx0=α1/4,1/16=1. Finally, we have(46)1n+12≠f1n+12,∀n∈ℕ,which implies that for each y∈1/n+12:n∈ℕ,(47)infpx,y+px,fx:x∈X=y2>0.
Hence, the conditions of Theorem7 are satisfied. It follows from Theorem 7 that f has a fixed point 0∈X.


## 3. Solvability of Nonlinear Integral Equations

The theory of nonlinear integral equations nowadays is a large topic which is found in many applications of various branches in mathematics. In this section, we prove the existence results of solutions for the following nonlinear Fredholm and Volterra integral equations, respectively, by using Theorem6:(48)xt=φt+∫abKt,s,xsds,∀t∈a,b,(49)xt=φt+∫atKt,s,xsds,∀t∈a,b,where a,b∈ℝ are constants with a<b, φ:a,b⟶ℝ, and K:a,b2×ℝ⟶ℝ are given functions.LetCa,b,ℝ denote the Banach space of all continuous functions x:a,b⟶ℝ with the norm x=supt∈a,bxt. Put X=Ca,b,ℝ and(50)dx,y=supt∈a,bxt−yt,∀x,y∈X.It is clearX,d is a complete metric space. Define mappings T and S as follows:(51)Txt=φt+∫abKt,s,xsds,∀t,x∈a,b×X,(52)Sxt=φt+∫atKt,s,xsds,∀t,x∈a,b×X.Theorem 8.
Letξ:ℝ×ℝ⟶ℝ, φ:a,b⟶ℝ, and K:a,b2×ℝ⟶ℝ satisfy the following:d1 φ and K are continuousd2 ξTxt,Tyt≥0 and ∀t,x,y∈a,b×X2 with ξxt,yt≥0d3 there exists x0∈X such that ξx0t,Tx0t≥0, ∀t∈a,bd4 if xnn∈ℕ is a sequence in X such that xn⟶x in X and ξxnt,xn+1t≥0, ∀n,t∈ℕ×a,b, then(53)ξxnt,xt≥0,∀n,t∈ℕ×a,b.d5 there exists ψ∈Φ1 with(54)Kt,s,ys≤ψsups∈a,bys−φtb−a,∀t,s,y∈a,b2×X.
Then, nonlinear Fredholm integral equation (48) has a solution in X.Proof.
Definep:X×X⟶ℝ+, ϕ:ℝ+⟶ℝ+, and α:X×X⟶ℝ+ by(55)px,y=supt∈a,byt,∀x,y∈X,ϕr=1,∀r∈ℝ+,(56)αx,y=1ifξxt,yt≥0,∀t∈a,b0otherwise,∀x,y∈X.
Clearly,p is a w-distance in X and c1,c2, and c4 follow from d2,d3, and d4, respectively. It is easy to verify that d1 and (51) ensure that, for each x∈X, Tx is a continuous function in a,b, which yields that T maps X into itself. Observe that (51) and d5 mean that(57)∫0Tytϕrdr=Tyt=φt+∫abKt,s,ysds≤φt+∫abKt,s,ysds≤φt+∫abψsups∈a,bys−φtb−ads=φt+1b−a∫abψpx,y−φtds=ψ∫0px,yϕrdr,∀t,x,y∈a,b×X2,which implies that(58)αx,y∫0pTx,Tyϕrdr=αx,y∫0supt∈a,bTytϕrdr≤ψ∫0px,yϕrdr,∀x,y∈X.
That is, (6) holds. Thus, Theorem 6 guarantees that T has a fixed point x∈X, which is a solution of nonlinear Fredholm integral equation (48) in X. This completes the proof.As in the proof of Theorem8, we get the following result and omit its proof.Theorem 9.
Letξ:ℝ×ℝ⟶ℝ, φ:a,b⟶ℝ, and K:a,b2×ℝ⟶ℝ satisfy d1,d4,d5, andd6 ξSxt,Syt≥0, ∀t,x,y∈a,b×X2 with ξxt,yt≥0;d7 there exists x0∈X such that ξx0t,Sx0t≥0, ∀t∈a,b. Then, nonlinear Volterra integral equation (49) has a solution in X.



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