# Expansive Mappings and Their Applications in Modular SpaceAcademic research paper on "Mathematics"

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## Academic research paper on topic "Expansive Mappings and Their Applications in Modular Space"

﻿Hindawi Publishing Corporation Abstract and Applied Analysis Volume 2014, Article ID 580508, 8 pages http://dx.doi.org/10.1155/2014/580508

Research Article

Expansive Mappings and Their Applications in Modular Space

A. Azizi, R. Moradi, and A. Razani

Department of Mathematics, Faculty of Science, Imam Khomeini International University, Qazvin 34149-16818, Iran Correspondence should be addressed to A. Razani; razani@ipm.ir

Received 19 September 2013; Revised 30 January 2014; Accepted 31 January 2014; Published 14 April 2014 Academic Editor: Mohamed Amine Khamsi

Copyright © 2014 A. Azizi et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Some fixed point theorems for p-expansive mappings in modular spaces are presented. As an application, two nonlinear integral equations are considered and the existence of their solutions is proved.

1. Introduction

Let (X, d) be a metric space and B a subset of X. A mapping T : B ^ X is said to be expansive with a constant k > 1 such that

d (Tx, Ty) > kd (x, y) Vx, y e B. (1)

Xiang and Yuan [1] state a Krasnosel'skii-type fixed point theorem as follows.

Theoreml (see [1]). Let (X, ||-||) be a Banach space and K c X a nonempty, closed, and convex subset. Suppose that T and S map K into X such that

(I) S is continuous; S(K) resides in a compact subset of X; (II) T is an expansive mapping;

(III) z e S(K) implies that T(K) + zd K, where T(K) + z = {y + z| yeT(K)}.

Then there exists a point x* e K with Sx* + Tx* = x*.

For other related results, see also [2, 3]. In this paper, we study some fixed point theorems for S+T, where T is ^-expansive and S(B) resides in a compact subset of Xp, where B is a closed, convex, and nonempty subset of Xp and T,S : B ^ Xp. Our results improve the classical version of Krasnosel'skii fixed point theorems in modular spaces.

Finally, as an application, we study the existence of a solution of some nonlinear integral equations in modular function spaces.

In order to do this, first, we recall the definition of modular space (see [4-6]).

Definition 2. Let X be an arbitrary vector space over K = (R or C). Then we have the following.

(a) A functional p : X ^ [0, ot] is called modular if

(i) p(x) = 0 if and only if x = 0;

(ii) p(ax) = p(x) for a e K with la\ = 1, for all x e X;

(iii) p(ax + py) < p(x) + p(y) if a,p>0, a + p = 1, for all x,y e X.

If (iii) is replaced by (iii)' p(ax + fiy)< ap(x) + fip(y) for a, p > 0, a + p = 1, for all x,y e X, then the modular p is called a convex modular.

(b) A modular p defines a corresponding modular space, that is, the space Xp given by

Xp = [x eXl p (ax) —> 0 as a 0}. (2)

(c) If p is convex modular, the modular Xp can be equipped with a norm called the Luxemburg norm defined by

||%||p = inf {a > 0; < 1} . (3)

Remark 3. Note that p is an increasing function. Suppose that 0 < a < b; then property (iii), with y = 0, shows that p(ax) = p((a/b)(bx)) < p(bx).

Definition 4. Let Xp be a modular space. Then we have the following.

(a) A sequence in Xp is said to be

(i) ^-convergent to x if p(xn - x) ^ 0 as n ^ to;

(ii) ^-Cauchyif p(xn - xm) ^ 0 as n,m ^ to.

(b) Xp is ^-complete if every ^-Cauchy sequence is p-convergent.

(c) A subset B c Xp is said to be ^-closed if for any sequence (xn)ngN c B and xn ^ x then x e B.

(d) A subset B c Xp is called ^-bounded if Sp(B) = sup p(x - y) < to, for all x,y e B, where op(B) is called the ^-diameter of B.

(e) p has the Fatou property if

p(x-y)< lim inf p (xn - yn), (4)

whenever xn ^ x and yn ^ y as n ^ to.

(f) p is said to satisfy the A 2-condition if p(2xn) ^ 0 whenever p(xn) ^ 0 as n ^ to.

2. Expansive Mapping in Modular Space

In 2005, Hajji and Hanebaly [7] presented a modular version of Krasnosel'skii fixed point theorem, for a ^-contraction and a ^-completely continuous mapping.

Using the same argument as in [1], we state the modular version of Krasnosel'skii fixed point theorem for S + T, where T is a ^-expansive mapping and the image of B under S ; that is, S(B) resides in a compact subset of Xp, where B is a subset

Due to this, we recall the following definitions and theorems.

Definition 5. Let Xp be a modular space and B a nonempty subset of Xp. The mapping T : B ^ Xp is called ^-expansive mapping, if there exist constants c,k,l e R+ such that c > I, k > 1 and

p(l(Tx-Ty))>kp(c(x-y)), (5)

for all x,y e B.

Example 6. Let Xp = B = R+ and consider T : B ^ B with Tx = xn + 4x + 5 for x e B and n e N. Then for all x,y e B, we have

\Tx - Ty\ = \xn - yn + 4(x- y)\

= - y) (xn-1 + yx"-2 +----+ y"-1) + 4 (x - y)\

> 4\x - y\.

Therefore T is an expansive mapping with constant k = 4.

Theorem 7 (Schauder's fixed point theorem, page 825; see [1, 8]). Let (X, || • ||) be a Banach space and K c X is a nonempty, closed, and convex subset. Suppose that the mapping S : K ^ K is continuous and S(K) residesinacompactsubsetof X.Then S has at least one fixed point in K.

We need the following theorem from [6, 9].

Theorem 8 (see [6,9]). Let Xp be a p-complete modular space. Assume that p is a convex modular satisfying the A 2-condition and B is a nonempty, p-closed, and convex subset of Xp. T : B ^ B is a mapping such that there exist c,k,l e R+ such that c > I, 0 < k < 1 and for all x,y e B one has

p(c(Tx-Ty))<kp(l(x-y)). (7)

Then there exists a unique fixed point z e B such that Tz = z.

Theorem 9. Let Xp be a p-complete modular space. Assume that p is a convex modular satisfying the A 2-condition and B is a nonempty p-closed, and convex subset of Xp. T : B ^ Xp is a p-expansive mapping satisfying inequality (5) and B c T(B). Then there exists a unique fixed point z e B such that Tz = z.

Proof. We show that operator T is a bijection from B to T(B). Let x1 and x2 be in B such that Tx1 = Tx2; by inequality (5), we have x1 = x2;alsosinceB c T(B) it follows that the inverse of T:B ^ T(B) exists. For all x,y e T(B),

p(c(fx-fy))<±p(l(x-y)), (8)

where f = T-1. We consider f = T-1\B : B ^ B, where T-1\B denotes the restriction of the mapping T-1 to the set B. Since B c T(B), then f is a ^-contraction. Also since B is a p-closed subset of Xp, then, by Theorem 8, there exists a z e B such that fz = z. Also z is a fixed point of T.

For uniqueness, let z and w be two arbitrary fixed points of T; then

p(c(z-w))>p(l(z-w)) = p (I (Tz - Tw))

> kp (c (z - w)); hence (k - 1)p(c(z - w)) < 0 and z = w. □

We need the following lemma for the main result.

Lemma 10. Suppose that all conditions of Theorem 9 are fulfilled. Then the inverse of f := I - T : B ^ (I - T)(B) exists and

p(c(f-1x-f-1y))<i--ip(l' (x-y)), (10)

for all x,ye f(B), where l' = al and a is conjugate of c/l; that is, (l/c) + (1/a) = 1 and c > 21.

Proof. For all x,y e B,

p(l(Tx-Ty)) = p(l((x-fx)-(y-fy)))

<p(c(x-y)) + p(al(fx-fy)); (11)

kp (c (x -y))-p (c (x -y))<p (al (fx - fy)),

(k-l)p(c(x-y))<p(l' (fx-fy)]. (12)

Now, we show that f is an injective operator. Let x,y e B and fx = fy; then by inequality (12), (k - 1)p(c(x - y)) < 0 and x = y. Therefore f is an injective operator from B into f(B), and the inverse of f : B — f(B) exists. Also for all x,ye f(B),wehave f-1x, f-1y e B.Thenforallx, y e f(B), by inequality (12) we get

p(c(f-1x-f-1y]]<-1-p(l' (x-y)]. (13) k-l n

Theorem 11. Let Xp be a p-complete modular space. Assume that p is a convex modular satisfying the A2-condition and B is a nonempty, p-closed, and convex subset of Xp. Suppose that

(I) S : B — Xp is a p-continuous mapping and S(B) resides in a p-compact subset of Xp;

(II) T : B — Xp is a p-expansive mapping satisfying inequality (5) such that c > 21;

(III) x e S(B) implies that B cx + T(B), where T(B) + x = {y + x I y e T(B)}.

There exists a point z e B such that Sz + Tz = z.

Proof. Let w e S(B) and Tw = T + w. Consider the mapping Tw : B — Xp; then by Theorem 9, the equation Tx + w = x has a unique solution x = q(w). Now, we show that q is a ^-contraction. For w1,w2 e S(B), T(^(w1)) + w1 = ^(w1) and T(q(w2)) + w2 = q(w2). Applying the same technique in Lemma 10,

(k-1)p(c(tl(wi)-t1 (W2))) < p (l' (wi - w2)] , (14)

where l' = al. Then

p(c(n(Wi)-n(w2))) < (wi -w2)]- (15)

Therefore, mapping q : S(B) — B is a ^-contraction and hence is a ^-continuous mapping. By condition (I), qS : B — B is also ^-continuous mapping and, by A 2-condition, qS is || • ||p-continuous mapping. Also qS(B) resides in a || • ||p-compact subset of Xp. Then using Theorem 7, there exists a z e B such that z = q(S(z)) which implies that Tz + Sz = z. □

The following theorem is another version of Theorem 11.

Theorem 12. Let Xp be a p-complete modular space. Assume that p is a convex modular satisfying the A2-condition and B is a nonempty, p-closed, and convex subset of Xp. Suppose that

(I) S : B ^ Xp is a p-continuous mapping and S(B) resides in a p-compact subset of Xp;

(II) T : B ^ Xp or T : Xp ^ Xp is a p-expansive mapping satisfying inequality (5) such that c > 21;

(III) S(B) c (I - T)(Xp) and [x = Tx + Sy,y e B implies that x e B] or S(B) c (I - T)(B).

Then there exists a point z e B such that Sz + Tz = z.

Proof. By condition (III), for each w e B, there exists x e Xp such that x-Tx = Sw. If S(B) c (I - T)(B), then x e B; if S(B) c (I - T)(Xp), then by Lemma 10 and condition (III), x = (I - T)-1Sw e B. Now (I - T)-1 is a ^-continuous and so (I - T)-1S is a ^-continuous mapping of B into B. Since S(B) resides in a ^-compact subset of Xp, so (I-T)-1S(B) resides in a ^-compact subset of the closed set B. By using Theorem 7, there exists a fixed point z e B such that z = (I- T)-1Sz. □

Using the same argument as in [2], we can state a new version of Theorem 11, where S is ^-sequentially continuous.

Definition 13. Let Xp be a modular space and B a subset of Xp. A mapping T : B ^ Xp is said to be

(1) ^-sequentially continuous on the set B if for every sequence {xn} c B and x e B such that p(xn-x) ^ 0, then p(Txn - Tx) ^ 0;

(2) ^-closed if for every sequence {xn} c B such that p(xn - x) ^ 0 and p(Txn - y) ^ 0, then Tx = y.

Definition 14. Let Xp be a modular space and B, C two subsets of Xp. Suppose that T : B ^ Xp and S : C ^ Xp are two mappings. Define

F = [x e B : x = Tx + Sy for some y e C}. (16)

Theorem 15. Let Xp be a p-complete modular space. Assume that p is a convex modular satisfying the A 2-condition and B is a nonempty p-closed, and convex subset of Xp. Suppose that

(I) S : B —^ Xp is p-sequentially continuous;

(II) T : B — Xp is a p-expansive mapping satisfying inequality (5) such that c > 21;

(III) x e S(B) implies that B cx + T(B), where T(B) + x = {y + xI ye T(B)};

(IV) T is p-closed in F and F is relatively p-compact. Then there exists a point z e B such that Sz + Tz = z.

Proof. Let w e B, and TSw = T + Sw. One considers the mapping TSw : B — Xp; by Theorem 9, the equation

Tx + Sw = x (17)

has a unique solution x = q(Sw) e B.

Now, we show that qS = (I - T)- exists. For any w1,w2 e B and by the same technique of Lemma 10, we have

p (c (n (SW1) - n (Sw2))) < ^p (I' (W1 - w2)] , (18)

where l' = al. This implies that qS = (I - T)-1 exists and for all weB, qSw = (I - T)-1Sw and t^S(B) c F.

We show that qS is ^-sequentially continuous in B. Let {xn} be a sequence in B and x e B such that p(xn - x) — 0. Since qS(xn) e F and F is relatively ^-compact, then there exists z e B such that p(qSxn - z) — 0. On the other hand, by condition (I), p(Sxn - Sx) — 0. Thusby (17), we get

T (nSxn) + Sxn = t]Sxn; (19)

T (nSxn) -(z - Sx)

( (nSxn - Sxn) -(z - Sx) = P{-2-

<p{nSxn -z)+p(Sxn -Sx);

therefore when n ^ to, condition (IV) implies that Tz = z - Sx; that is, z = qSx and

p (qSxn - qSx) —> 0;

then qS is ^-sequentially continuous in F. By A 2-condition, qS is || • ^^-sequentially continuous. Let H = co"'"pF, where

co"'"p denotes the closure of the convex hull in the sense of || • ||p. Then H c B and is a compact set. Therefore qS is || • ^^-sequentially continuous from H into H. Then using Theorem 7, qS has a fixed point z e H such that qSz = z. From (17), we have

T (qSz) + Sz = qSz; (22)

that is, Tz + Sz = z. □

The following theorem is another version of Theorem 15.

Theorem 16. Let Xp be a p-complete modular space. Assume that p is a convex modular satisfying the A 2-condition and B is a nonempty, p-closed, and convex subset of Xp. Suppose that

Xp is p-sequentially continuous; ^ Xp is a p-expansive mapping satisfying

(I) S:B ^

(II) T : B inequality (5), such that c > 2l;

(III) S(B) c (I- T)(Xp) and [x = Tx + Sy,y e B] implies that x eB (or S(B) c (I - T)(B)).

(IV) T is p-closed in F and F is relatively p-compact. Then there exists a point z e B such that Sz + Tz = z.

Proof. By (III) for each w e B, there exists x e Xp such that x - Tx = Sw and % = (I - T)-1Sw e B. By the same technique of Theorem 15, (I - T)-1S : B ^ B is p-sequentially continuous and there exists a z e B such that z=(I-T)-1Sz. □

3. Integral Equation for p-Expansive Mapping in Modular Function Spaces

In this section, we study the following integral equation:

x(t) = $(t, x (t)) + I y(t,s,x (s)) ds, xeC (I, L9), J0 where L9 is the Musielak-Orlicz space and I = [0, b] c R. C(I, L9) denote the space of all ^-continuous functions from I to L9 with the modular a(x) = supteIp(x(t)). Also C(I, L9) is a real vector space. If p is a convex modular, then a is a convex modular. Also, if p satisfies the Fatou property and A 2-condition, then a satisfies the Fatou property and A 2-condition (see [9]). To study the integral equation (23), we consider the following hypotheses. (1)$ : I x L9 ^ L9 is a ^-expansive mapping; that is, there exist constants c,k,l e R+ such that c > 21, k > 2 and for all x,yeL9

p (I ($(t, x)-$ (t, y))) > kp (c (x - y)) (24)

and <p is onto. Also for t e I, <p(t, ■) : L9 ^ L9 is p-continuous.

(2) f is a function from I x I x L9 into L9 such that f(t,s,■) : x ^ f(t,s,x) is ^-continuous on L9 for almost all t,s e I and f(t, -,x) : s ^ f(t, s, x) is measurable function on I for each x e L9 and for almost all tel. Also, there are nondecreasing continuous functions ß,y:I ^ R+ such that

lim ß(t) [ y(

t^rn Jo

s) ds = 0,

p(c(f(t,s,x)))<p(t)y(s),

for all t,s e I, s <t and x e L9.

(3) There exists measurable function q : I x I x I ^ R+ such that

p (f (t, s,x)-f (r, s, x)) < q (t, r, s),

for all t,r,s e I and x e L9; also limt^r JQ q(t,r,s)ds = 0.

(4) p(f(t, s, x) - f(t, s, y)) < p(x - y) for all t,s e I and x,y e L9.

Remark 17 (see [7]). We consider L9, the Musielak-Orlicz space. Since p is convex and satisfies the A 2-condition, then

Fn -*||p

p(Xn -X) 0,

as n ^ to on L9. This implies that the topologies generated by || • ||p and p are equivalent.

Theorem 18. Suppose that the conditions (1)-(4) are satisfied. Further assume that L9 satisfies the A 2-condition. Also w(t) = p(t) y(s)ds and «(0) = 0; also sup{p(c($(t, v))), tel, v e L9} < w(t). Then integral equation (23) has at least one solution x e C(I,L9). Proof. Suppose that Tx(t) =$(t,x(t)),

(t (28) Sx(t) = I y(t,s,x(s))ds. J0

Conditions (1) and (2) imply that T and S are well defined on C(I,L9). Define the set B = {x e C(I, L9); p(c(x(t))) <

œ(t) for all t e I}. Then B is a nonempty, ^-bounded, ^-closed, and convex subset of 0(1,1/). Equation (23) is equivalent to the fixed point problem x = Tx + Sx. By Theorem 12, we find the fixed point for T + S in B. Due to this, we prove that S satisfies the condition (I) of Theorem 12. For x e B,we show that Sx e B. Indeed,

p(c(Sx(t))) = ^(c(j y(t,s,x(s))ds))

< j p(c(f(t,s,x(s))))ds

< j P(t)y(s)ds

= u(t);

then Sx e B. Since S(B) c B and B is ^-bounded, S(B) is a-boundedand by À 2-condition || • ||CT-bounded.

We show that S(B) is ^-equicontinuous. For all t,r e I and x e Lr such that t < r,

Sx(t)-Sx(r)= \ y (t,s,x(s))ds - \ y(r,s,x(s))ds;

then by condition (3),

p (Sx (t) - Sx (r)) < \ q(t,r,s)ds; (31)

since limt^r j0 q(t,r,s)ds = 0, then S(B) is p-equicontinuous. By using the Arzela-Ascoli theorem, we obtain that S is a a-compact mapping. Next, we show that S is a-continuous. Suppose that e > 0 is given; we find a S > 0 such that a(x - y) < S, for some x,y e B. Note that

Sx(t)-Sy(t)= \ y(t,s,x(s))ds- \ y(t,s,y(s))ds;

p (Sx (t) - Sy (t)) < \ p (x (s) - y (s)) ds < \ a(x-y)ds; Jo Jo

' (Sx - Sy) < \ a(x - y)

- y)ds < e;

therefore S is a-continuous.

Since 0 is ^-continuous, it shows that T transforms C(I, Lr) into itself. In view of supremum p and condition (1), it is easy to see that T is a-expansive with constant k>2. For x,yeB,

p(l(Tx(t)-Ty(t)))

<P(c (x(t)-y(t)))

+ p(al((I-T)x(t)-(I-T)y(t)));

p(al((I-T)x(t)-(I-T)y(t)))

>(k-1)p(c(x(t)-y(t))),

where a is conjugate of c/l. Let r = al; since k > 2, then

p(r(I-T)x (t)) >(k-1)p(c(x (t))) >p(c(x (t))). (37)

Now, assume that x = Tx + Sy for some y e B. Since c > 21, then r < c, and

p (c (x (t))) <p(r(I-T)x (t)) = p(r (Sy (t)))

<p(c(Sy(t)))<cv(t), which shows that x e B. Now, define a map Tz as follows:

Tz : C (I, Lv) C (I, Lv) , (39)

for each z e C(I, Lf); by

Tzx (t) = Tx (t) +z(t), (40)

for all x,y e C(I,Lf),

p (I (Tzx (t) - Tzy (t))) = p(l (Tx (t) - Ty (t))) >kP(c (x(t)-y(t)));

therefore

a(l(Tzx-Tzy))>ka(c(x-y));

then Tz is a-expansive with constant k > 2 and Tz is onto. By Theorem 9, there exists w e C(I, Lr) such that Tzw = w; that is, (I - T)w = z. Hence S(B) c (I - T)(Lv) and condition (III) of Theorem 12 holds. Therefore by Theorem 12, S + T has a fixed point z e B with Tz + Sz = z; that is, z is a solution to (23). □

Now, we consider another integral equation.

Let Lv be the Musielak-Orlicz space and I = [0,b] c R. Suppose that p is convex and satisfies the A 2-condition. Since topologies generated by || • ||p and p are equivalent, then we consider Banach space (Lr,H • ||p) and C(I,LV) denote the space of all || • ||p-continuous functions from I to Lr with the modular HxHa = supt£j||%(i)||p; also C(I,LV) is a real vector space. Consider the nonlinear integral equation

x(t) = $(t,x(t)) + X(t,x(t)) \ w(t,s)y(s,x(s))ds, (43) xeC(l,Lf), (1)$ : I x Lr ^ Lr is a || • ||p-expansive mapping; that is, there exists constant I > 2 such that

\\$(t,x)-$(t,:

WP >l\\x-y\\p,

for all x,yeLr and 0 is onto; also for t e I, <p(t, ■) : For x e B,we show that Sx e B. Consider

V ^ V is || • ||p-continuous;

(2) f is function from I x V into V such that f(t, •) : V ^ If is a || • ||p-continuous and t ^ f(t, x) is measurable for every x e If. Also, there exist functions ß e L1 (I) and a nondecreasing continuous function y : [0, >x) ^ (0, rn) such that

(t,x)\\p <ß(t)y(UxUp),

for all t e I and x e V. Also for t e I, x ^ f(t, x) is nondecreasing on V;

(3) X is function from I x V into V such that X(t, •) : V ^ V is || • ||p-continuous and there exists a a> 0 such that

\\X(t,x)-X(t,y)\\p < a\\x-y\\

for all t e I and x e If; also for x e If, t ^ X(t, x) is nondecreasing on I and for t e I, x ^ X(t, x) is nondecreasing on V;

(4) to is function from Ixlinto R+. For each t e I, w(t, s) is measurable on [0, t]. Also to(t) = esssup \to(t, s)| is bounded on [0, b] and r = sup |to(i)|. The map to(■, s) : t ^ to(t, s) is continuous from I to Lm(I). Also for s e I, t ^ to(t, s) is nondecreasing on I.

Theorem 19. Suppose that the conditions (1)-(4) are satisfied and there exists a constant k > 0 such that for all t el,

f ßd) J0

(ak + h) rb J0 y (k)

i W)ds'

where h := sup|||A(i, x)\\p, t e I,x e If} and also sup|||0(i, x)\\p, t el, x e If} < k. Then integral equation (43) has at least one solution x e C(I, If).

Proof. Define

B={x eC(l,Lr);l\x(t)llp <kVt el}; (48)

then B is a nonempty, || ■ ||p-bounded, || ■ ||p-closed, and convex subset of C(I, If). Consider

Tx(t) = $(t,x(t)), Sx (t) = X(t,x(t)) \ to(t,s)y (s, x (s)) ds. It is easy that by the hypothesis T and S are well defined on C(I, If). HSx(t)Hp X(t,x(t)) \ to(t,s)f(s,x(s))ds (X (t, x(t))-X (t, 0) + X (t, 0)) \ to (t, s) f (s, x (s)) ds <(a\\x(t)Up + h)r i ß(s)y(Ux(s)Up)ds (ak + h)r[ ß (s) y (k) ds < (ak +1 < (ak + h)r \ (ak + h) rby (k) Let x e B and assume that t > t e I such that \t - t\ <8, for a given positive constant S. We have ||S*(f)-S*(T)|L X(t,x(t)) \ to(t,s)f(s,x(s))ds -X(t,x(t))\ to(r,s)y(s,x(s))i X(t,x(t)) \ to(t,s)f(s,x(s))ds ± X(t,x (t)) \ to(r,s)f(s,x(s))ds ±X(t,x(t))\ to(r,s)f(s,x(s))ds -X(t,x(t))\ to(r,s)y(s,x(s))i X(t, x(i))(| to(t,s)y(s,x(s))ds \ to(r,s)f(s,x(s))ds (X(r,x(r))-X(r,x(r))) < \ to(r,s)f(s,x(s))ds A(t,%(t))J" to(r,s)f(s,x(s))ds X (t, x (t )) (J œ(t,s)y(s,x(s))ds - \ w (t, s) f (s, x (s)) ds Jo X (t, x (t)) (J (œ (t, s)-œ (t, s)) f (s, x (s)) ds) (X (t,x(t))-x(t,0) + x(t,0)) x (J (œ(t,s)-œ (t, s)) f (s, x (s)) ds) <(ak + h)\w(t,0)-w(T,0)\L \ ß(s)y(k)ds < -r\w(t,0)-w(rML^, (X(t,x(t)) - X(t,x(t))) \ œ(r,s)y(s,x(s))ds Jo I (t, x(t))-X (t, x (t))) r\ ß(s)y (k) ds Jo -±-h(\\X(t,x(t))-X(t,x(r))\\p +\\X(T,x(T))-X(t,x(T))\\p] (a\\x(t)-x(T)\\p + h), ak + h X(t, x(t))J w(T,s)y(s,x(s))ds (x(t,x(t))-x(t,0)+x(t,0)) x J w(T,s)f(s,x (s)) ds <(ak + h)r\ ß(s)y(k)ds <-b\t-*\> then S(B) is || • ||p-equicontinuous. By using the Arzela-Ascoli Theorem, we obtain that S is a || • ||p-compact mapping. We show that S is || • ||p-continuous. Suppose that e > 0 is given. We find a S > 0 such that Hx - yHa < S. We have ||S*(f)-Sy(f)||p X(t,x(t)) \ œ(t,s)f(s,x(s))ds -X(t,y(t)) \ œ(t,s)y(s,y(s))ds (X(t,x(t)) - X(t,y(t))) \ w(t,s)y(s,x(s))ds + x(t,y(t)) \ (y(s,x(s))-y(s,y(s)))ds < y(f)^P + (ak + h)r\Q < - + (ak + h) rb^X - Since 0 is || • ||p-continuous, it shows that T transforms C(I, Lr) into itself. In view of supremum || • ||p and condition (1), it is easy to see that T is || • ||CT-expansive with constant I > 2. For x,yeB, \\Tx(t)-Ty(t)\\ <\\x(t)-y(t)\l + \\(I-T)x(t)-(I-T)y( \\(I - T)x(t) -(I- T)y(t)\\p >(1-1) ||x (t) - y (i)\\p; (55) since I > 2, then U(I-T)x(t)Up >(l-l)Ux(t)Up >||*(f)||p. (56) Now, assume that x = Tx + Sy for some y e B. Then Ux(t)Up <U(I-T)x(t)Up = \\Sy(f)\\p <k, (57) which shows that x e B. Now for each z e C(I, Lr) we define a map Tz as follows: Tzx (t) = Tx (t) +z(t); for all x,y e C(I,Lf ), \\Tzx(t) - Tzy(t)\\p = \\Tx(t) - Ty(t)\\p > l\\x (t) - y (f)||p; therefore \\TZX-Tzy\l >l\\x-y\l; then Tz is || • ||CT-expansive with constant I > 2 and Tz is onto. By Theorem 9, there exists w e C(I, Lr) such that Tzw = w; that is, (I - T)w = z. Hence S(B) c (I - T)(Lf). Therefore by Theorem 12, S + T has a fixed point z e B with Tz + Sz = z; that is, z is a solution of (43). □ Finally, some examples are presented to guarantee Theorems 18 and 19. Example 20. Consider the following integral equation: xV^+f arctan ( 1+i2 Jo \(l + t)3 (l + ^W))' where Lr = R+, I = [0,1]. For x,ye R+ and t e I,we have \t(t,x)-4(t,y)\= —2-Y+? (63) Therefore by Theorem 18, the integral equation (62) has at least one solution. Example 21. Consider the following integral equation: x (t) = <9X + l arcsin x (t) [ ——x (s) ds, (64) l +t2 8 Jo t + s where$(t,x) = (9x/(l+t2)),X(t,x) = (l/8) arcsinx, w(t,s) = t/(t + s), and y(t, x) = x. Also Lv = R+, I = [0, l]. Therefore by Theorem 19, the integral equation (64) has at least one solution.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

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