# Fixed Points for Contractive Mappings of Integral Type Involvingω-Distance and Applications

Authors: Haiyan Gao; Na Liu; Liangshi Zhao
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we useω-distance to prove the existence, uniqueness, and iterative approximations of fixed points for a few contractive mappings of integral type in complete metric spaces. The proved results are used to investigate the solvability of certain nonlinear integral equations. Four examples are given.


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## Body

## 1. Introduction

The researchers [1–17] attained various generalizations of the well-known Banach contraction principle. In 2001, Rhoades [15] certified several fixed point results for the weakly contractive mappings. Branciari [1] introduced the notion of integral type contraction and established a nice fixed point result for the mapping. Many authors investigated the existence of fixed points for a lot of contractive mappings of integral type, for example, see [11–14]. Particularly, Liu et al. [14] obtained several fixed point theorems for contractive mappings of integral type in complete metric spaces.Kada et al. [8] introduced the concept of ω-distance in metric spaces and proved a few fixed point theorems for some contractive mappings by using ω-distance. It is clear that the results in [8] extended the Caristi’s fixed point theorem, Ekeland’s ε-variational’s principle, and the nonconvex minimization theorem. The researchers in [3, 5–7, 9, 10, 17] got several fixed point results for certain contractive mappings with respect to ω-distance.In this paper, we prove the existence, uniqueness, and iterative approximations of fixed points for several kinds of mappings, which satisfy some contractive conditions of integral type with respect toω-distance in complete metric spaces. We also construct four illustrative examples and give applications of the obtained results in nonlinear Fredholm and Volterra integral equations, respectively. Our results generalize or differ from the corresponding fixed point theorems in [1, 14, 15].


## 2. Preliminaries

Letℕ denotes the set of all positive integers, R=−∞,+∞, R+=0,+∞, N0=N∪0 and(1)Φ1=w∣w:R+→R+isLebesgueintegrable,summableoneachcompactsubsetofR+and∫0εwrdr>0foreachε>0;Φ2=w∣w∈Φ1andsatisfiesthat∫0uwrdr<∫0vwrdrforeachu,v∈R+withu<v;Φ3=w∣w:R+→R+isnondecreasing,continuous,w0=0,limt→+∞wt=+∞andwt>0foreacht>0;Φ4=w∣w:R+→R+islowersemi−continuous,w0=0andwt>0foreacht>0.Recall that a self-mappingf in a metric space X is called orbitally continuous if limn→∞fnx=u implies limn→∞fn+1x=fu for each fnxn∈N0⊆X and u∈X.


## 3. Fixed Point Results with respect toω-Distance

In this section, usingω-distance, we give four fixed point theorems for the contractive mappings (2), (38), (82), and (114) below.Theorem 1.
Letp be a ω-distance in a complete metric space X,d and let f:X→X satisfy that(2)∫0pfx,fywrdr≤∫0px,ywrdr−∫0ψpx,ywrdr,∀x,y∈X.Here, w,ψ∈Φ1×Φ4. Then, f possesses a unique fixed point u∈X such that pu,u=0,(3)limn→∞pfnx0,u=0andlimn→∞fnx0=uforeachx0∈X.Proof.
Firstly, we claim the existence of fixed points off in X. Put x0∈X and xn=fnx0 for each n∈N0. Now, we need to think over two situations as follows:Case 2.
xn0=xn0−1 for some n0∈N. Clearly, xn0−1 is a fixed point of f and limn→∞fnx0=xn0−1. Suppose that pxn0−1,xn0−1>0. Making use of (2) and w,ψ∈Φ1×Φ4, we obtain that(4)∫0pxn0−1,xn0−1wrdr=∫0pfxn0−1,fxn0−1wrdr≤∫0pxn0−1,xn0−1wrdr−∫0ψpxn0−1,xn0−1wrdr<∫0pxn0−1,xn0−1wrdr,which is ridiculous. Hence, pxn0−1,xn0−1=0, which means that(5)limn→∞pfnx0,xn0−1=pxn0−1,xn0−1=0;Case 3.
xn≠xn−1 for all n∈N. Suppose that(6)pxn0−1,xn0=0forsomen0∈N.
(2), (6), and w,ψ∈Φ1×Φ4 ensure that(7)0≤∫0pxn0,xn0+1wrdr=∫0pfxn0−1,fxn0wrdr≤∫0pxn0−1,xn0wrdr−∫0ψpxn0−1,xn0wrdr=0,that is,(8)∫0pxn0,xn0+1wrdr=0.The above equation andw∈Φ1 give that(9)pxn0,xn0+1=0.Combining (6), (9), and p1, we know that(10)0≤pxn0−1,xn0+1≤pxn0−1,xn0+pxn0,xn0+1=0,that is,(11)pxn0−1,xn0+1=0.Because of Lemma 1 in [8], (6), and (11), we deduce that xn0=xn0+1, which is contradictive, and, hence,(12)pxn−1,xn>0,∀n∈N.By means ofw,ψ∈Φ1×Φ4, (2) and (12), we have(13)∫0pxn,xn+1wrdr=∫0pfxn−1,fxnwrdr≤∫0pxn−1,xnwrdr−∫0ψpxn−1,xnwrdr<∫0pxn−1,xnwrdr,∀n∈N,which together with (12) and w∈Φ1 ensures that(14)0<pxn,xn+1<pxn−1,xn,∀n∈N.We see from (14) that pxn,xn+1n∈N0 is a positive and strictly decreasing sequence. It follows that(15)limn→∞pxn,xn+1=cforsomec≥0.Now, we claim thatc=0. Otherwise, c>0. In view of Lemma 2.1 in [12], (2), (15), and w,ψ∈Φ1×Φ4, we see that(16)∫0cwrdr=limsupn→∞∫0pxn,xn+1wrdr=limsupn→∞∫0pfxn−1,fxnwrdr≤limsupn→∞∫0pxn−1,xnwrdr−∫0ψpxn−1,xnwrdr≤limsupn→∞∫0pxn−1,xnwrdr−liminfn→∞∫0ψpxn−1,xnwrdr≤∫0cwrdr−∫0ψcwrdr<∫0cwrdr.It is ridiculous. Therefore,c=0. Consequently,(17)limn→∞pxn,xn+1=0.In the same way, we have(18)limn→∞pxn+1,xn=0.Now, we proceed to show that(19)limn,m→∞pxn,xm=0.Suppose that there exists a real numberε>0 such that for every k∈N, there exist mk,nk∈N satisfying(20)pxnk,xmk>ε,mk>nk>k,∀k∈N.For eachk∈N, mk denotes the least integer exceeding nk and satisfying (20). Obviously,(21)pxnk,xmk>εandpxnk,xmk−1≤ε,∀k∈N.On account ofp1 and (21), we obtain that(22)ε<pxnk,xmk≤pxnk,xnk−1+pxnk−1,xmk−1+pxmk−1,xmk≤pxnk,xnk−1+pxnk−1,xnk+pxnk,xmk−1+pxmk−1,xmk≤pxnk,xnk−1+pxnk−1,xnk+ε+pxmk−1,xmk,∀k∈N.Lettingk tend to infinity in (22) and taking advantage of (17), (18), and (21), we have(23)limk→∞pxnk,xmk=limk→∞pxnk−1,xmk−1=ε.In light of Lemma 2.1 in [12], (2), (23), and w,ψ∈Φ1×Φ4, we deduct that(24)∫0εwrdr=limsupk→∞∫0pxnk,xmkwrdr=limsupk→∞∫0pfxnk−1,fxmk−1wrdr≤limsupk→∞∫0pxnk−1,xmk−1wrdr−∫0ψpxnk−1,xmk−1wrdr≤limsupk→∞∫0pxnk−1,xmk−1wrdr−liminfk→∞∫0ψpxnk−1,xmk−1wrdr≤∫0εwrdr−∫0ψεwrdr<∫0εwrdr.It is ridiculous. Of course, (19) is true.Assume thatε>0 and δ denotes the real number appearing in (3) of [8]. By means of (19), we infer that there is N∈N satisfying(25)pxN,xn<δ,pxN,xm<δ,∀n,m>N,which ensures that(26)dxn,xm≤ε,∀n,m>N.Soxnn∈N0 is a Cauchy sequence. Completeness of X means that(27)limn→∞xn=uforsomeu∈X.According to (19), we are aware of the fact that for any ε>0 there is M∈N with(28)pxn,xm<ε,∀m>n>M,which together with (27) gives that(29)0≤pxn,u≤liminfm→∞pxn,xm≤ε,∀n>M.It follows that(30)limn→∞pxn,u=0.Taking account of (2), (30), w,ψ∈Φ1×Φ4, and Lemma 2.1 in [12], we infer that(31)0≤limsupn→∞∫0pfxn,fuwrdr≤limsupn→∞∫0pxn,uwrdr−∫0ψpxn,uwrdr≤limsupn→∞∫0pxn,uwrdr−liminfn→∞∫0ψpxn,uwrdr≤0−∫0ψ0wrdr=0.It follows that(32)limn→∞∫0pfxn,fuwrdr=0.Lemma 2.2 in [12] and the above equation give that(33)limn→∞pfxn,fu=limn→∞pxn+1,fu=0.We get from (1) in [8] and (17) that(34)0≤pxn,fu≤pxn,xn+1+pxn+1,fu→0asn→∞.Clearly,(35)limn→∞pxn,fu=0.Applying (30), (35), and Lemma 1 in [8], we gain that u=fu.Secondly, we prove thatpu,u=0. Suppose that pu,u>0. In view of w,ψ∈Φ1×Φ4 and (2), we receive that(36)0<∫0pu,uwrdr=∫0pfu,fuwrdr≤∫0pu,uwrdr−∫0ψpu,uwrdr<∫0pu,uwrdr,which is not possible. Hence, pu,u=0.Thirdly, we assert the uniqueness of fixed points off in X. Assume that f possesses two fixed points u,v∈X. We know, analogous to the proof of (36), that pu,u=pv,v=0. Assume that pu,v>0. Due to w,ψ∈Φ1×Φ4 and (2), we deduce that(37)0<∫0pu,vwrdr=∫0pfu,fvwrdr≤∫0pu,vwrdr−∫0ψpu,vwrdr<∫0pu,vwrdr,which is ridiculous. Consequently, pu,v=0. Using Lemma 1 in [8] and pu,u=0, we obtain that u=v.Theorem 4.
Letp be a ω-distance in a complete metric space X,d and let f:X→X satisfy that(38)∫0pfx,fywrdr≤∫0px,ywrdr−∫0ψpx,fywrdr,∀x,y∈X.Here, w,ψ∈Φ2×Φ4. Then, f possesses a unique fixed point u∈X such that pu,u=0,(39)limn→∞pfnx0,u=0andlimn→∞fnx0=uforeachx0∈X.Proof.
Firstly, we show thatf possesses fixed points in X. Let x0∈X and xn=fnx0 for each n∈N0. Now, we divide the proof into two steps.Step 5.
Putxn0=xn0−1 for some n0∈N. In addition, xn0−1 is a fixed point of f and limn→∞fnx0=xn0−1. Suppose that pxn0−1,xn0−1>0. Owing to (38) and w,ψ∈Φ2×Φ4, we acquire that(40)∫0pxn0−1,xn0−1wrdr=∫0pfxn0−1,fxn0−1wrdr≤∫0pxn0−1,xn0−1wrdr−∫0ψpxn0−1,fxn0−1wrdr=∫0pxn0−1,xn0−1wrdr−∫0ψpxn0−1,xn0−1wrdr<∫0pxn0−1,xn0−1wrdr,which is ridiculous. Hence, pxn0−1,xn0−1=0, which means that(41)limn→∞pfnx0,xn0−1=pxn0−1,xn0−1=0;Step 6.
xn≠xn−1 for all n∈N. Assume that(42)pxn0,xn0−1=0forsomen0∈N.Usingw,ψ∈Φ2×Φ4, (38), and (42), we conclude that(43)0≤∫0pxn0+1,xn0wrdr=∫0pfxn0,fxn0−1wrdr≤∫0pxn0,xn0−1wrdr−∫0ψpxn0,fxn0−1wrdr=∫0pxn0,xn0−1wrdr−∫0ψpxn0,xn0wrdr≤0.It becomes that(44)∫0pxn0+1,xn0wrdr=0.Thus,w∈Φ2 and the above equation guarantee that(45)pxn0+1,xn0=0.As a result of (42), (45), and (1) in [8], we deduce that(46)0≤pxn0+1,xn0−1≤pxn0+1,xn0+pxn0,xn0−1=0,in other words,(47)pxn0+1,xn0−1=0.From (45), (47), and Lemma 1 in [8], we obtain that xn0=xn0−1, which is impossible, and, hence,(48)pxn,xn−1>0,∀n∈N.Suppose that there existsq∈N with(49)pxq+1,xq>pxq,xq−1.In light of (38), (48), (49), and w,ψ∈Φ2×Φ4, we infer that(50)0<∫0pxq,xq−1wrdr<∫0pxq+1,xqwrdr=∫0pfxq,fxq−1wrdr≤∫0pxq,xq−1wrdr−∫0ψpxq,fxq−1wrdr=∫0pxq,xq−1wrdr−∫0ψpxq,xqwrdr≤∫0pxq,xq−1wrdr.It is ridiculous. By means of (48), we get that(51)0<pxn+1,xn≤pxn,xn−1,∀n∈N.Thus, (51) means that the sequence pxn+1,xnn∈N0 is both positive and decreasing. Consequently,(52)limn→∞pxn+1,xn=vforsomev≥0.Assume that(53)pxj+2,xj>pxj+1,xj−1forsomej∈N.In terms of (38), (53), and w,ψ∈Φ2×Φ4, we attain that(54)0≤∫0pxj+1,xj−1wrdr<∫0pxj+2,xjwrdr=∫0pfxj+1,fxj−1wrdr≤∫0pxj+1,xj−1wrdr−∫0ψpxj+1,fxj−1wrdr=∫0pxj+1,xj−1wrdr−∫0ψpxj+1,xjwrdr≤∫0pxj+1,xj−1wrdr,which is ridiculous. Hence,(55)0≤pxn+2,xn≤pxn+1,xn−1,∀n∈N.We get from (55) that the sequence pxn+2,xnn∈N0 is both nonnegative and nonincreasing. Thus,(56)limn→∞pxn+2,xn=bforsomeb≥0.Assume thatv>0. In view of (38), (52), (56), w,ψ∈Φ2×Φ4, and Lemma 2.1 in [12], we gain that(57)0≤∫0bwrdr=limsupn→∞∫0pxn+2,xnwrdr=limsupn→∞∫0pfxn+1,fxn−1wrdr≤limsupn→∞∫0pxn+1,xn−1wrdr−∫0ψpxn+1,fxn−1wrdr=limsupn→∞∫0pxn+1,xn−1wrdr−∫0ψpxn+1,xnwrdr≤limsupn→∞∫0pxn+1,xn−1wrdr−liminfn→∞∫0ψpxn+1,xnwrdr≤∫0bwrdr−∫0ψvwrdr<∫0bwrdr.It is impossible. Thus, (18) is true. Suppose that (6) holds. Taking advantage of (6), (38), and w,ψ∈Φ2×Φ4, we have(58)0≤∫0pxn0,xn0+1wrdr=∫0pfxn0−1,fxn0wrdr≤∫0pxn0−1,xn0wrdr−∫0ψpxn0−1,fxn0wrdr=∫0pxn0−1,xn0wrdr−∫0ψpxn0−1,xn0+1wrdr≤0.It follows that(59)∫0pxn0,xn0+1wrdr=0.Thus, the above equation andw∈Φ2 give (9). Using (6), (9), and (1) in [8], we conclude that(60)0≤pxn0−1,xn0+1≤pxn0−1,xn0+pxn0,xn0+1=0,that is, (11) holds. By virtue of (6), (11), and Lemma 1 in [8], we obtain that xn0=xn0+1, which is impossible. As a result, (12) holds. Assume that there exists q∈N with(61)pxq,xq+1>pxq−1,xq.We know from (12), (38), (61), and w,ψ∈Φ2×Φ4 that(62)0<∫0pxq−1,xqwrdr<∫0pxq,xq+1wrdr=∫0pfxq−1,fxqwrdr≤∫0pxq−1,xqwrdr−∫0ψpxq−1,fxqwrdr=∫0pxq−1,xqwrdr−∫0ψpxq−1,xq+1wrdr≤∫0pxq−1,xqwrdr,which is ridiculous. By means of (12), we obtain that(63)0<pxn,xn+1≤pxn−1,xn,∀n∈N.With the help of (63), there is a real number c≥0 satisfying (15). Assume that c>0. Let limsupn→∞pxn,xn=w. Clearly, there exists a subsequence xnkk∈N of xnn∈N0 with(64)limk→∞pxnk,xnk=w.Note that (38) and w,ψ∈Φ2×Φ4 infer that(65)0≤∫0pxn+1,xnwrdr=∫0pfxn,fxn−1wrdr≤∫0pxn,xn−1wrdr−∫0ψpxn,fxn−1wrdr=∫0pxn,xn−1wrdr−∫0ψpxn,xnwrdr≤∫0pxn,xn−1wrdr,∀n∈N.Lettingn→∞ in (65) and utilizing (18), w∈Φ2 and Lemma 2.2 in [12], we make a conclusion that(66)limn→∞∫0pxn,xn−1wrdr−∫0ψpxn,xnwrdr=0.It follows that(67)limn→∞∫0ψpxn,xnwrdr=limn→∞∫0pxn,xn−1wrdr−limn→∞∫0pxn,xn−1wrdr−∫0ψpxn,xnwrdr=0.In view of (18), (38), (64), (65), and (67), w,ψ∈Φ2×Φ4 and Lemma 2.2 in [12], we deduct that(68)0=limsupk→∞∫0pxnk+1,xnkwrdr=limsupk→∞∫0pfxnk,fxnk−1wrdr≤limsupk→∞∫0pxnk,xnk−1wrdr−∫0ψpxnk,fxnk−1wrdr=limsupk→∞∫0pxnk,xnk−1wrdr−∫0ψpxnk,xnkwrdr≤limsupk→∞∫0pxnk,xnk−1wrdr−liminfk→∞∫0ψpxnk,xnkwrdr≤0−∫0ψwwrdr,which together with w,ψ∈Φ2×Φ4 yields that w=0. It is obvious that(69)limn→∞pxn,xn=0.Using Lemma 2.2 in [12], (15), (38), (69), and w,ψ∈Φ2×Φ4, we find that(70)0=limsupn→∞∫0pxn+1,xn+1wrdr=limsupn→∞∫0pfxn,fxnwrdr≤limsupn→∞∫0pxn,xnwrdr−∫0ψpxn,fxnwrdr=limsupn→∞∫0pxn,xnwrdr−∫0ψpxn,xn+1wrdr≤limsupn→∞∫0pxn,xnwrdr−liminfn→∞∫0ψpxn,xn+1wrdr≤0−∫0ψcwrdr,which together with w,ψ∈Φ2×Φ4 implies that c=0. Thus, (17) holds.Now, we assert that (19) holds. Otherwise, there is a real number ε>0 such that for arbitrary k∈N, there are mk,nk∈N with (20) and (21). On account of (1) in [8] and (3.12), we gain that(71)ε<pxnk,xmk≤pxnk,xnk−1+pxnk−1,xmk≤pxnk,xnk−1+pxnk−1,xmk−1+pxmk−1,xmk≤pxnk,xnk−1+pxnk−1,xnk+pxnk,xmk−1+pxmk−1,xmk≤pxnk,xnk−1+pxnk−1,xnk+ε+pxmk−1,xmk,∀k∈N.Lettingk→∞ in (71) and making use of (17), (18), and (21), we require that(72)limk→∞pxnk,xmk=limk→∞pxnk−1,xmk−1=limk→∞pxnk−1,xmk=ε.Taking notice of (38), (72), w,ψ∈Φ2×Φ4, and Lemma 2.1 in [12], we receive that(73)∫0εwrdr=limsupk→∞∫0pxnk,xmkwrdr=limsupk→∞∫0pfxnk−1,fxmk−1wrdr≤limsupk→∞∫0pxnk−1,xmk−1wrdr−∫0ψpxnk−1,fxmk−1wrdr=limsupk→∞∫0pxnk−1,xmk−1wrdr−∫0ψpxnk−1,xmkwrdr≤limsupk→∞∫0pxnk−1,xmk−1wrdr−liminfk→∞∫0ψpxnk−1,xmkwrdr≤∫0εwrdr−∫0ψεwrdr<∫0εwrdr.It is ridiculous. That is, (19) is true.We deduce, similar to the proof of Theorem1, that (27) holds. It follows from (19) that for every real number ε>0 there is M∈N with(74)pxn,xm<ε,∀m>n>M,which together with (2) in [8] and (27) gives that(75)0≤pxn,u≤liminfm→∞pxn,xm≤ε,∀n>M,that is, (30) holds. In terms of (30), (38), w,ψ∈Φ2×Φ4, and Lemma 2.1 in [12], we get that(76)0≤limsupn→∞∫0pfxn,fuwrdr≤limsupn→∞∫0pxn,uwrdr−∫0ψpxn,fuwrdr≤limsupn→∞∫0pxn,uwrdr−liminfn→∞∫0ψpxn,fuwrdr≤0−liminfn→∞∫0ψpxn,fuwrdr≤0.It follows that(77)limn→∞∫0pfxn,fuwrdr=0.Thus, Lemma 2.2 in [12] and the above equation ensure that(78)limn→∞pfxn,fu=limn→∞pxn+1,fu=0.In light of (1) in [8] and (17), we arrive at(79)0≤pxn,fu≤pxn,xn+1+pxn+1,fu→0asn→∞,that is to say, (35) holds. By virtue of (30), (35) and Lemma 1 in [8], we have u=fu.Secondly, we assert thatpu,u=0. Assume that pu,u>0. Owing to (38) and w,ψ∈Φ2×Φ4, we deduce that(80)0<∫0pu,uwrdr=∫0pfu,fuwrdr≤∫0pu,uwrdr−∫0ψpu,fuwrdr=∫0pu,uwrdr−∫0ψpu,uwrdr<∫0pu,uwrdr.It is ridiculous. Hence,pu,u=0.Thirdly, we show the uniqueness of fixed points off in X. Assume that f possesses two fixed points u,v∈X. We get, similar to the proof of (80), that pu,u=pv,v=0. Assume that pu,v>0. Taking account of w,ψ∈Φ2×Φ4 and (38), we get that(81)0<∫0pu,vwrdr=∫0pfu,fvwrdr≤∫0pu,vwrdr−∫0ψpu,fvwrdr=∫0pu,vwrdr−∫0ψpu,vwrdr<∫0pu,vwrdr,which is ridiculous. Therefore, pu,v=0. It follows from pu,u=0 and Lemma 1 in [8] that u=v.Theorem 7.
Letp be a ω-distance in a complete metric space X,d and let f:X→X satisfy that(82)∫0pfx,fywrdr≤∫0px,ywrdr−∫0ψpfx,ywrdr,∀x,y∈X,here w,ψ∈Φ2×Φ4. Then, f possesses a unique fixed point u∈X satisfying pu,u=0,(83)limn→∞pfnx0,u=0andlimn→∞fnx0=uforeachx0∈X.Proof.
Firstly, we demonstrate thatf possesses fixed points in X. Let x0∈X and xn=fnx0 for each n∈N0. Now, we consider two cases below:Case 8.
xn0=xn0−1 for some n0∈N. Since xn0−1 is a fixed point of f, it follows that limn→∞fnx0=xn0−1. Assume that pxn0−1,xn0−1>0. Due to (82) and w,ψ∈Φ2×Φ4, we have(84)∫0pxn0−1,xn0−1wrdr=∫0pfxn0−1,fxn0−1wrdr≤∫0pxn0−1,xn0−1wrdr−∫0ψpfxn0−1,xn0−1wrdr=∫0pxn0−1,xn0−1wrdr−∫0ψpxn0−1,xn0−1wrdr<∫0pxn0−1,xn0−1wrdr,which is absurd. Hence, pxn0−1,xn0−1=0 and(85)limn→∞pfnx0,xn0−1=pxn0−1,xn0−1=0.Case 9.
xn≠xn−1 for all n∈N. Assume that (6) holds. In view of (6), (82), and w,ψ∈Φ2×Φ4, we obtain that(86)0≤∫0pxn0,xn0+1wrdr=∫0pfxn0−1,fxn0wrdr≤∫0pxn0−1,xn0wrdr−∫0ψpfxn0−1,xn0wrdr=∫0pxn0−1,xn0wrdr−∫0ψpxn0,xn0wrdr≤0,which means that(87)∫0pxn0,xn0+1wrdr=0.Combiningw∈Φ2 and the above equation, we get (3.3). We gain from (6), (9), and p1 that(88)0≤pxn0−1,xn0+1≤pxn0−1,xn0+pxn0,xn0+1=0,in other words, (11) sets up. In terms of (6), (11), and Lemma 1 in [8], we know immediately that xn0=xn0+1, which is absurd, and, hence, (12) is true. Assume that there is q∈N satisfying (61). We conclude from (12), (61), (82), and w,ψ∈Φ2×Φ4 that(89)0<∫0pxq−1,xqwrdr<∫0pxq,xq+1wrdr=∫0pfxq−1,fxqwrdr≤∫0pxq−1,xqwrdr−∫0ψpfxq−1,xqwrdr=∫0pxq−1,xqwrdr−∫0ψpxq,xqwrdr≤∫0pxq−1,xqwrdr,which is impossible. Hence, (63) is true. It follows from (63) that there is a real number c≥0 with (15). Assume that there is j∈N with(90)pxj,xj+2>pxj−1,xj+1.In terms of (82), (90), and w,ψ∈Φ2×Φ4, we infer that(91)0≤∫0pxj−1,xj+1wrdr<∫0pxj,xj+2wrdr=∫0pfxj−1,fxj+1wrdr≤∫0pxj−1,xj+1wrdr−∫0ψpfxj−1,xj+1wrdr=∫0pxj−1,xj+1wrdr−∫0ψpxj,xj+1wrdr≤∫0pxj−1,xj+1wrdr.It is absurd, and, hence,(92)0≤pxn,xn+2≤pxn−1,xn+1,∀n∈N.(92) means that pxn,xn+2n∈N0 is both nonnegative and nonincreasing. Consequently,(93)limn→∞pxn,xn+2=zforsomeconstantz≥0.Assume thatc>0. Owing to (15), (82), (93), w,ψ∈Φ2×Φ4, and Lemma 2.1 in [12], we get that(94)0≤∫0zwrdr=limsupn→∞∫0pxn,xn+2wrdr=limsupn→∞∫0pfxn−1,fxn+1wrdr≤limsupn→∞∫0pxn−1,xn+1wrdr−∫0ψpfxn−1,xn+1wrdr=limsupn→∞∫0pxn−1,xn+1wrdr−∫0ψpxn,xn+1wrdr≤limsupn→∞∫0pxn−1,xn+1wrdr−liminfn→∞∫0ψpxn,xn+1wrdr≤∫0zwrdr−∫0ψcwrdr<∫0zwrdr.It is contradictive. Thus, (17) is true. Suppose that (42) holds. We infer from (42), (82), and w,ψ∈Φ2×Φ4 that(95)0≤∫0pxn0+1,xn0wrdr=∫0pfxn0,fxn0−1wrdr≤∫0pxn0,xn0−1wrdr−∫0ψpfxn0,xn0−1wrdr=∫0pxn0,xn0−1wrdr−∫0ψpxn0+1,xn0−1wrdr≤0,that is,(96)∫0pxn0+1,xn0wrdr=0.Thus, (45) follows from the above equation and w∈Φ2. We deduct from (42), (45), and (1) in [8] that(97)0≤pxn0+1,xn0−1≤pxn0+1,xn0+pxn0,xn0−1=0,that is to say, (47) holds. Thus, xn0=xn0−1 is easily obtained from (45), (47), and Lemma 1 in [8], which is ridiculous. As a result, (48) holds. Suppose that there exists q∈N satisfying (49). By virtue of (48), (49), (82), and w,ψ∈Φ2×Φ4, we know that(98)0<∫0pxq,xq−1wrdr<∫0pxq+1,xqwrdr=∫0pfxq,fxq−1wrdr≤∫0pxq,xq−1wrdr−∫0ψpfxq,xq−1wrdr=∫0pxq,xq−1wrdr−∫0ψpxq+1,xq−1wrdr≤∫0pxq,xq−1wrdr,which is ridiculous. By means of (48), we have (51). It follows from (51) that the sequence pxn+1,xnn∈N0 is both positive and decreasing, which yields (52) for some a constant v≥0. Suppose that v>0. Put limsupn→∞pxn,xn=w. Obviously, there is a subsequence xnkk∈N of xnn∈N0 with (64). Using (82), and w,ψ∈Φ2×Φ4, we deduce that(99)0≤∫0pxn,xn+1wrdr=∫0pfxn−1,fxnwrdr≤∫0pxn−1,xnwrdr−∫0ψpfxn−1,xnwrdr=∫0pxn−1,xnwrdr−∫0ψpxn,xnwrdr≤∫0pxn−1,xnwrdr,∀n∈N.Lettingn→∞ in (99) and using (17), w∈Φ2 and Lemma 2.2 in [12], we find that(100)limn→∞∫0pxn−1,xnwrdr−∫0ψpxn,xnwrdr=0.It follows that(101)limn→∞∫0ψpxn,xnwrdr=limn→∞∫0pxn−1,xnwrdr−limn→∞∫0pxn−1,xnwrdr−∫0ψpxn,xnwrdr=0.We attain from Lemma 2.2 in [12], (17), (64), (82), (99), (101), and w,ψ∈Φ2×Φ4 that(102)0=limsupk→∞∫0pxnk,xnk+1wrdr=limsupk→∞∫0pfxnk−1,fxnkwrdr≤limsupk→∞∫0pxnk−1,xnkwrdr−∫0ψpfxnk−1,xnkwrdr=limsupk→∞∫0pxnk−1,xnkwrdr−∫0ψpxnk,xnkwrdr≤limsupk→∞∫0pxnk−1,xnkwrdr−liminfk→∞∫0ψpxnk,xnkwrdr≤0−∫0ψwwrdr.It follows that∫0ψwwrdr≤0. By virtue of w,ψ∈Φ2×Φ4, we have ∫0ψwwrdr=0 and w=0. It means that (69) holds. In view of (52), (69), (82), w,ψ∈Φ2×Φ4, and Lemma 2.2 in [12], we infer that(103)0=limsupn→∞∫0pxn+1,xn+1wrdr=limsupn→∞∫0pfxn,fxnwrdr≤limsupn→∞∫0pxn,xnwrdr−∫0ψpfxn,xnwrdr=limsupn→∞∫0pxn,xnwrdr−∫0ψpxn+1,xnwrdr≤limsupn→∞∫0pxn,xnwrdr−liminfn→∞∫0ψpxn+1,xnwrdr≤0−∫0ψvwrdr,which yields that ∫0ψvwrdr≤0. Using w,ψ∈Φ2×Φ4, we obtain that ∫0ψvwrdr=0 and v=0. Thus, (3.9) holds.Now, we prove that (19) holds. Suppose that there is an ε>0 such that for arbitrary k∈N, (20), (21), and (22) hold for some mk,nk∈N. As k→∞ in (3.13) and by virtue of (17), (18), and (21), we acquire that(104)limk→∞pxnk,xmk=limk→∞pxnk−1,xmk−1=limk→∞pxnk,xmk−1=ε.In terms of Lemma 2.1 in [12], (82), (104), and w,ψ∈Φ2×Φ4, we are aware of the fact that(105)∫0εwrdr=limsupk→∞∫0pxnk,xmkwrdr=limsupk→∞∫0pfxnk−1,fxmk−1wrdr≤limsupk→∞∫0pxnk−1,xmk−1wrdr−∫0ψpfxnk−1,xmk−1wrdr=limsupk→∞∫0pxnk−1,xmk−1wrdr−∫0ψpxnk,xmk−1wrdr≤limsupk→∞∫0pxnk−1,xmk−1wrdr−liminfk→∞∫0ψpxnk,xmk−1wrdr≤∫0εwrdr−∫0ψεwrdr<∫0εwrdr,which is absurd. Thus, (19) is true.We infer, similar to the proof of Theorem1, that (27) holds. It follows from (19) that for each ε>0 there is M∈N with(106)pxn,xm<ε,∀m>n>M,which together with (2) in [8] and (27) gets that(107)0≤pxn,u≤liminfm→∞pxn,xm≤ε,∀n>M,that is, (30) holds. On account of Lemma 2.1 in [12], (30), (82), and w,ψ∈Φ2×Φ4, we deduct that(108)0≤limsupn→∞∫0pfxn,fuwrdr≤limsupn→∞∫0pxn,uwrdr−∫0ψpfxn,uwrdr≤limsupn→∞∫0pxn,uwrdr−liminfn→∞∫0ψpxn+1,uwrdr≤0−∫0ψ0wrdr=0,in other words,(109)limn→∞∫0pfxn,fuwrdr=0.Lemma 2.2 in [12] and the above equation give that(110)limn→∞pfxn,fu=limn→∞pxn+1,fu=0.In light of (1) in [8] and (17), we attain that(111)0≤pxn,fu≤pxn,xn+1+pxn+1,fu→0asn→∞,that is to say, (35) holds. Using (30), (35), and Lemma 1 in [8], we have u=fu.Secondly, we prove thatpu,u=0. Assume that pu,u>0. Because of (82) and w,ψ∈Φ2×Φ4, we deduce that(112)0<∫0pu,uwrdr=∫0pfu,fuwrdr≤∫0pu,uwrdr−∫0ψpfu,uwrdr=∫0pu,uwrdr−∫0ψpu,uwrdr<∫0pu,uwrdr,which is impossible. Hence, pu,u=0.Thirdly, we assert the uniqueness of fixed points off in X. Assume that f possesses two fixed points u,v∈X. We deduce, similar to the proof of (112), that pu,u=pv,v=0. Assume that pu,v>0. On account of (82) and w,ψ∈Φ2×Φ4, we get that(113)0<∫0pu,vwrdr=∫0pfu,fvwrdr≤∫0pu,vwrdr−∫0ψpfu,vwrdr=∫0pu,vwrdr−∫0ψpu,vwrdr<∫0pu,vwrdr,which is ridiculous. Therefore, pu,v=0. Using pu,u=0 and Lemma 1 in [8], we infer immediately that u=v.We have, similar to the proof of Theorem1, the result below and omit its proof.Theorem 10.
Letp be a ω-distance in a complete metric space X,d and let f:X→X satisfy that(114)∫0pfx,fywrdr≤∫0px,ywrdr−∫0ψpfx,fywrdr,∀x,y∈X.Here, w,ψ∈Φ1×Φ4. Then, f possesses a unique fixed point u∈X such that pu,u=0,(115)limn→∞pfnx0,u=0andlimn→∞fnx0=uforeachx0∈X.


## 4. Four Examples

Now, we give four examples to explain the fixed point results obtained in Section3.Remark 11.
Lettingpx,y=dx,y,∀x,y∈X, we deduce that Theorem 1 reduces to Theorem 2.1 in [14], which generalizes Theorem 1 in [15]. On the other hand, the example below proves that Theorem 1 extends indeed these results in [14, 15] and differs from Theorem 2.1 in [1].Example 12.
LetX=0,6,dx,y=∣x−y∣ and px,y=y/2,∀x,y∈X. Let f:X→X, w and ψ:R+→R+ be defined by, respectively,(116)fx=2x−x2,∀x∈65,2,0,∀x∈0,65∪2,6and(117)wr=2,ψr=13r2,∀r∈R+.It follows thatp is a ω-distance in X and w,ψ∈Φ1×Φ4. Put x,y∈X. In order to check (2), we consider two cases below:Case 13.
x,y∈X×6/5,2. It follows that(118)∫0pfx,fywrdr=∫0y−12y22dr=2y−y2≤y−16y2=∫0y22dr−∫0112y22dr=∫0px,ywrdr−∫0ψpx.ywrdr.Case 14.
x,y∈X×0,6/5∪2,6. Note that(119)∫0pfx,fywrdr=0≤y−16y2=∫0px,ywrdr−∫0ψpx.ywrdr.That is, (2) is true. Hence, the conditions of Theorem 1 are fulfilled. Thus, Theorem 1 ensures that f possesses a unique fixed point in X. Now, we need to prove that Theorem 2.1 in [1], Theorem 2.1 in [14], and Theorem 1 in [15] are useless in checking the existence of fixed points for the mapping f in X.If there isψ∈Φ3 satisfying the conditions of Theorem 1 in [15], we know that(120)2125=df75,f1≤d75,1−ψd75,1<25,which is absurd.If there arec∈0,1 and w∈Φ1 satisfying the conditions of Theorem 2.1 in [1], we attain that(121)0<∫034wrdr=∫0df34,f32wrdr≤c∫0d34,32wrdr<∫034wrdr,which is ridiculous.If there isw,ψ∈Φ1×Φ4 satisfying the conditions of Theorem 2.1 in [14], we conclude that(122)0<∫02425wrdr=limsupy→65+∫0∣0−2y−y2∣wrdr=limsupy→65+∫0df65,fywrdr≤limsupy→65+∫0d65,ywrdr−∫0ψd65,ywrdr≤limsupy→65+∫0d65,ywrdr−liminfy→65+∫0ψd65,ywrdr≤0−∫0ψ0wrdr=0,which is impossible.Remark 15.
Examples16, 21, and 26 explain that Theorems 4, 7, and 10 are different from Theorem 2.1 in [14].Example 16.
LetX=R+,dx,y=∣x−y∣ and px,y=x+y,∀x,y∈X. Let f:X→X, w and ψ:R+→R+ be defined by, respectively,(123)fx=x3,∀x∈0,1,x4,∀x∈1,+∞,and(124)wr=2r,ψr=r2,∀r∈R+.Evidently,p is a ω-distance in X and w,ψ∈Φ2×Φ4. Let x,y∈X. For the sake of verifying (38), we take into account the following four possible cases:Case 17.
x,y∈0,1×0,1. Note that(125)∫0pfx,fywrdr=∫0x+y32tdt=19x+y2≤x+y2−14x+y32=∫0x+y2tdt−∫0ψx+y32tdt=∫0px,ywrdr−∫0ψpx,fywrdr.Case 18.
x,y∈0,1×1,+∞. Obviously,(126)∫0pfx,fywrdr=∫0x3+y42tdt=x3+y42≤x+y2−14x+y42=∫0x+y2tdt−∫0ψx+y42tdt=∫0px,ywrdr−∫0ψpx,fywrdr.Case 19.
x,y∈1,+∞×0,1. Notice that(127)∫0pfx,fywrdr=∫0x4+y32tdt=x4+y32≤x+y2−14x+y32=∫0x+y2tdt−∫0ψx+y32tdt=∫0px,ywrdr−∫0ψpx,fywrdr.Case 20.
x,y∈1,+∞×1,+∞. It follows that(128)∫0pfx,fywrdr=∫0x+y42tdt=116x+y2≤x+y2−14x+y42=∫0x+y2tdt−∫0ψx+y42tdt=∫0px,ywrdr−∫0ψpx,fywrdr.That is to say, (38) is true. Therefore, the conditions of Theorem 4 are fulfilled. Consequently, Theorem 4 means that f possesses a unique fixed point in X. However, we cannot use Theorem 2.1 in [14] to show the existence of fixed points for the mapping f in X. Or else, there is w,ψ∈Φ1×Φ4 with(129)0<∫0112wrdr=limsupy→1+∫0∣13−y4∣wrdr=limsupy→1+∫0df1,fywrdr≤limsupy→1+∫0d1,ywrdr−∫0ψd1,ywrdr≤limsupy→1+∫0d1,ywrdr−liminfy→1+∫0ψd1,ywrdr≤0−∫0ψ0wrdr=0,which is absurd.Example 21.
LetX=R+,dx,y=∣x−y∣ and px,y=x+y,∀x,y∈X. Let f:X→X, w and ψ:R+→R+ be defined by, respectively,(130)fx=0,∀x∈0,3,34x,∀x∈3,+∞,and(131)wr=4r,ψr=32r,∀r∈R+.Obviously,p is a ω-distance in X and w,ψ∈Φ2×Φ4. Let x,y∈X. To demonstrate (82), we consider four cases below:Case 22.
x,y∈0,3×0,3. Evidently,(132)∫0pfx,fywrdr=0≤2x+y2−32y2=∫0px,ywrdr−∫0ψpfx,ywrdr.Case 23.
x,y∈0,3×3,+∞. Clearly,(133)∫0pfx,fywrdr=∫034y4rdr=38y2≤2x+y2−32y2=∫0x+y4rdr−∫0ψy4rdr=∫0px,ywrdr−∫0ψpfx,ywrdr.Case 24.
x,y∈3,+∞×0,3. It is obvious that(134)∫0pfx,fywrdr=∫034x4rdr=38x2≤2x+y2−3234x+y2=∫0x+y4rdr−∫0ψ34x+y4rdr=∫0px,ywrdr−∫0ψpfx,ywrdr.Case 25.
x,y∈3,+∞×3,+∞. Clearly(135)∫0pfx,fywrdr=∫034x+y4rdr=38x+y2≤2x+y2−3234x+y2=∫0x+y4rdr−∫0ψ34x+y4tdt=∫0px,ywrdr−∫0ψpfx,ywrdr.In other words, (82) is true, and consequently, the conditions of Theorem 7 are fulfilled. Thus, Theorem 7 yields that f possesses a unique fixed point in X. Next, we testify that Theorem 2.1 in [14] is unapplicable in ensuring the existence of fixed points for the mapping f in X.If there isw,ψ∈Φ1×Φ4 satisfying the conditions of Theorem 2.1 in [14], we have(136)0<∫034wrdr=limsupy→3+∫0∣0−34y∣wrdr=limsupy→3+∫0df3,fywrdr≤limsupy→3+∫0d3,ywrdr−∫0ψd3,ywrdr≤limsupy→3+∫0d3,ywrdr−liminfy→3+∫0ψd3,ywrdr≤0−∫0ψ0wrdr=0,which is ridiculous.Example 26.
LetX=R+,dx,y=∣x−y∣ and px,y=y,∀x,y∈X. Let f:X→X, w and ψ:R+→R+ be defined by, respectively,(137)fx=0,∀x∈0,25,x5,∀x∈25,+∞,and(138)wr=45r3,ψr=5r,∀r∈R+.It is easy to see thatp is a ω-distance in X and w,ψ∈Φ1×Φ4. Let x,y∈X. To prove (114), we have to consider two cases below:Case 27.
x,y∈X×0,25. Apparently,(139)∫0pfx,fywrdr=0≤15y4=∫0px,ywrdr−∫0ψpfx,fywrdr.Case 28.
x,y∈X×25,+∞. It is easy to demonstrate that(140)∫0pfx,fywrdr=∫0y545r3dr=155y4≤15y4−5y2=∫0y45r3dr−∫0ψy545r3dr=∫0px,ywrdr−∫0ψpfx,fywrdr.Hence, (114) is true, and the conditions of Theorem 10 are satisfied. Thus, Theorem 10 guarantees that f possesses a unique fixed point in X. Then, we certify that Theorem 2.1 in [14] is unfulfilled in showing the existence of fixed points for the mapping f in X. Otherwise, there is w,ψ∈Φ1×Φ4 satisfying the conditions Theorem 2.1 in [14]. It means that(141)0<∫05wrdr=limsupy→25+∫0∣0−y5∣wrdr=limsupy→25+∫0df25,fywrdr≤limsupy→25+∫0d25,ywrdr−∫0ψd25,ywrdr≤limsupy→25+∫0d25,ywrdr−liminfy→25+∫0ψd25,ywrdr≤0−∫0ψ0wrdr=0,which is ridiculous.


## 5. Applications

In this section, we utilize Theorems1 and 7 to investigate the solvability of the nonlinear Fredholm and Volterra integral equations below, respectively,(142)xt=ht+∫abKt,r,xrdr,∀t∈a,b,(143)xt=ht+∫atKt,r,xrdr,∀t∈a,b,where a and b are constants in R with a<b, h:a,b→R and K:a,b2×R→R are given functions.We assume thatCa,b,R denotes the Banach space of all continuous functions x:a,b→R with the norm ∥x∥=supt∈a,b∣xt∣. Let X=Ca,b,R and(144)dx,y=supr∈a,b∣xr−yr∣,∀x,y∈X.Obviously,X,d is a complete metric space. Define two mappings T and S as follows:(145)Txt=ht+∫abKt,r,xrdr,∀t,x∈a,b×X,(146)Sxt=ht+∫atKt,r,xrdr,∀t,x∈a,b×X.Theorem 29.
Leth:a,b→R and K:a,b2×R→R satisfy that
(a1)h and K are continuous;
(a2) there isψ∈Φ4 with(147)∣htb−a+Kt,r,yr∣≤supr∈a,b∣yr∣−ψsupr∈a,b∣yr∣b−a,∀t,r,y∈a,b2×X.Then, Eq. (142) possesses a unique solution in X.Proof.
Define two functionsw:R+→R+ and p:X×X→R+ by(148)wr=12,∀r∈R+,px,y=supt∈a,b∣yt∣,∀x,y∈X.
Obviously,p is a ω-distance and w∈Φ1. It follows from a1 and (145) that for arbitrary x∈X, Tx is continuous in a,b, which means that T maps X into itself. Taking account of (145) and a2, we get that(149)∫0∣Tyt∣wrdr=12∣Tyt∣=12∣ht+∫abKt,r,yrdr∣≤12∫ab∣htb−a+Kt,r,yr∣dr≤12∫absupr∈a,b∣yr∣−ψsupr∈a,b∣yr∣b−adr=12∫abpx,y−ψpx,y∣b−ads=∫0px,ywrdr−∫0ψpx,ywrdr,∀t,x,y∈a,b×X2.
It follows that(150)∫0pTx,Tywrdr=∫0supt∈a,b∣Tyt∣wrdr≤∫0px,ywrdr−∫0ψpx,ywrdr=∫0px,ywrdr−∫0ψpTx,ywrdr,∀x,y∈X.
That is, (2) and (82) hold. It follows from each of Theorems 1 and 7 that T possesses a unique fixed point x∈X, that is, Eq. (142) has a unique solution x∈X.
We get, similar to the proof of Theorem29, the following result and omit its proof.Theorem 30.
Leth:a,b→R and K:a,b2×R→R satisfy a1 and a2. Then, Eq. (143) possesses a unique solution in X.


## 6. Conclusion

By usingω-distance, we prove several fixed point results for a few contractive mappings of integral type, some of which are used to investigate the existence and uniqueness of solutions for certain nonlinear Fredholm and Volterra integral equations, respectively. Four examples are provided to testify that our results extend or differ from some known results in the literature.



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