# Expansive Mappings and Their Applications in Modular Space

Authors: A. Azizi; R. Moradi; A. Razani
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

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


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## Body

## 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
(1)d(Tx,Ty)≥kd(x,y)∀x,y∈B.
Xiang and Yuan [1] state a Krasnosel’skii-type fixed point theorem as follows.Theorem 1 (see [1]).
Let(X,∥·∥) be a Banach space and K⊂X a nonempty, closed, and convex subset. Suppose that T and S map K into X such that (I)
Sis continuous; S(K) resides in a compact subset of X;(II)
T is an expansive mapping;(III)
z∈S(K) implies that T(K)+z⊃K, where T(K)+z={y+z∣y∈T(K)}.
Then there exists a pointx*∈K with Sx*+Tx*=x*.For other related results, see also [2, 3].In this paper, we study some fixed point theorems forS+T, where T is ρ-expansive and S(B) resides in a compact subset of Xρ, where B is a closed, convex, and nonempty subset of Xρ and T,S:B→Xρ. 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.
LetX be an arbitrary vector space over K=(ℝ or ℂ). Then we have the following.(a)
A functionalρ:X→[0,∞] is called modular if
(i)
ρ(x)=0 if and only if x=0;(ii)
ρ(αx)=ρ(x) for α∈K with |α|=1, for all x∈X;(iii)
ρ(αx+βy)≤ρ(x)+ρ(y) if α,β≥0, α+β=1, for all x,y∈X.If (iii) is replaced by(iii)′
ρ(αx+βy)≤αρ(x)+βρ(y) for α, β≥0, α+β=1, for all x,y∈X, then the modular ρ is called a convex modular.(b)
A modularρ defines a corresponding modular space, that is, the space Xρ given by
(2)Xρ={x∈X∣ρ(αx)⟶0asα⟶0}.(c)
Ifρ is convex modular, the modular Xρ can be equipped with a norm called the Luxemburg norm defined by
(3)∥x∥ρ=inf⁡{α>0;ρ(xα)≤1}.Remark 3.
Note thatρ is an increasing function. Suppose that 0<a<b; then property (iii), with y=0, shows that ρ(ax)=ρ((a/b)(bx))≤ρ(bx).Definition 4.
LetXρ be a modular space. Then we have the following.(a)
A sequence(xn)n∈ℕ in Xρ is said to be
(i)
ρ-convergent to x if ρ(xn-x)→0 as n→∞;(ii)
ρ-Cauchy if ρ(xn-xm)→0 as n,m→∞.(b)
Xρ is ρ-complete if every ρ-Cauchy sequence is ρ-convergent.(c)
A subsetB⊂Xρ is said to be ρ-closed if for any sequence (xn)n∈ℕ⊂B and xn→x then x∈B.(d)
A subsetB⊂Xρ is called ρ-bounded if δρ(B)=sup⁡ρ(x-y)<∞, for all x,y∈B, where δρ(B) is called the ρ-diameter of B.(e)
ρ has the Fatou property if
(4)ρ(x-y)≤liminf⁡ρ(xn-yn),wheneverxn→x and yn→y as n→∞.(f)
ρ is said to satisfy the Δ2-condition if ρ(2xn)→0 whenever ρ(xn)→0 as n→∞.


## 2. Expansive Mapping in Modular Space

In 2005, Hajji and Hanebaly [7] presented a modular version of Krasnosel’skii fixed point theorem, for a ρ-contraction and a ρ-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 ρ-expansive mapping and the image of B under S;that is, S(B) resides in a compact subset of Xρ, where B is a subset of Xρ.Due to this, we recall the following definitions and theorems.Definition 5.
LetXρ be a modular space and B a nonempty subset of Xρ. The mapping T:B→Xρ is called ρ-expansive mapping, if there exist constants c,k,l∈ℝ+ such that c>l, k>1 and
(5)ρ(l(Tx-Ty))≥kρ(c(x-y)),
for all x,y∈B.Example 6.
LetXρ=B=ℝ+ and consider T:B→B with Tx=xn+4x+5 for x∈B and n∈ℕ. Then for all x,y∈B, we have
(6)|Tx-Ty|=|xn-yn+4(x-y)|=|(x-y)(xn-1+yxn-2+⋯+yn-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⊂X is a nonempty, closed, and convex subset. Suppose that the mapping S:K→K is continuous and S(K) resides in a compact subset of X. Then S has at least one fixed point in K.We need the following theorem from [6, 9].Theorem 8 (see [6, 9]).
LetXρ be a ρ-complete modular space. Assume that ρ is a convex modular satisfying the Δ2-condition and B is a nonempty, ρ-closed, and convex subset of Xρ. T:B→B is a mapping such that there exist c,k,l∈ℝ+ such that c>l, 0<k<1 and for all x,y∈B one has
(7)ρ(c(Tx-Ty))≤kρ(l(x-y)).
Then there exists a unique fixed point z∈B such that Tz=z.Theorem 9.
LetXρ be a ρ-complete modular space. Assume that ρ is a convex modular satisfying the Δ2-condition and B is a nonempty, ρ-closed, and convex subset of Xρ. T:B→Xρ is a ρ-expansive mapping satisfying inequality (5) and B⊂T(B). Then there exists a unique fixed point z∈B such that Tz=z.Proof.
We show that operatorT 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; also since B⊂T(B) it follows that the inverse of T:B→T(B) exists. For all x,y∈T(B),
(8)ρ(c(fx-fy))≤1kρ(l(x-y)),
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⊂T(B), then f is a ρ-contraction. Also since B is a ρ-closed subset of Xρ, then, by Theorem 8, there exists a z∈B such that fz=z. Also z is a fixed point of T.
For uniqueness, letz and w be two arbitrary fixed points of T; then
(9)ρ(c(z-w))≥ρ(l(z-w))=ρ(l(Tz-Tw))≥kρ(c(z-w));
hence (k-1)ρ(c(z-w))≤0 and z=w.We need the following lemma for the main result.Lemma 10.
Suppose that all conditions of Theorem9 are fulfilled. Then the inverse of f∶=I-T:B→(I-T)(B) exists and
(10)ρ(c(f-1x-f-1y))≤1k-1ρ(l′(x-y)),
for all x,y∈f(B), where l′=αl and α is conjugate of c/l; that is, (l/c)+(1/α)=1 and c>2l.Proof.
For allx,y∈B,
(11)ρ(l(Tx-Ty))=ρ(l((x-fx)-(y-fy)))≤ρ(c(x-y))+ρ(αl(fx-fy));kρ(c(x-y))-ρ(c(x-y))≤ρ(αl(fx-fy)),
then
(12)(k-1)ρ(c(x-y))≤ρ(l′(fx-fy)).
Now, we show that f is an injective operator. Let x,y∈B and fx=fy; then by inequality (12), (k-1)ρ(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,y∈f(B), we have f-1x,f-1y∈B. Then for all x,y∈f(B), by inequality (12) we get
(13)ρ(c(f-1x-f-1y))≤1k-1ρ(l′(x-y)).Theorem 11.
LetXρ be a ρ-complete modular space. Assume that ρ is a convex modular satisfying the Δ2-condition and B is a nonempty, ρ-closed, and convex subset of Xρ. Suppose that(I)
S:B→Xρ is a ρ-continuous mapping and S(B) resides in a ρ-compact subset of Xρ;(II)
T:B→Xρ is a ρ-expansive mapping satisfying inequality (5) such that c>2l;(III)
x∈S(B) implies that B⊂x+T(B), where T(B)+x={y+x∣y∈T(B)}.
There exists a pointz∈B such that Sz+Tz=z.Proof.
Letw∈S(B) and Tw=T+w. Consider the mapping Tw:B→Xρ; then by Theorem 9, the equation Tx+w=x has a unique solution x=η(w). Now, we show that η is a ρ-contraction. For w1,w2∈S(B), T(η(w1))+w1=η(w1) and T(η(w2))+w2=η(w2). Applying the same technique in Lemma 10,
(14)(k-1)ρ(c(η(w1)-η(w2)))≤ρ(l′(w1-w2)),
where l′=αl. Then
(15)ρ(c(η(w1)-η(w2)))≤1k-1ρ(l′(w1-w2)).
Therefore, mapping η:S(B)→B is a ρ-contraction and hence is a ρ-continuous mapping. By condition (I), ηS:B→B is also ρ-continuous mapping and, by Δ2-condition, ηS is ∥·∥ρ-continuous mapping. Also ηS(B) resides in a ∥·∥ρ-compact subset of Xρ. Then using Theorem 7, there exists a z∈B such that z=η(S(z)) which implies that Tz+Sz=z.The following theorem is another version of Theorem11.Theorem 12.
LetXρ be a ρ-complete modular space. Assume that ρ is a convex modular satisfying the Δ2-condition and B is a nonempty, ρ-closed, and convex subset of Xρ. Suppose that(I)
S:B→Xρ is a ρ-continuous mapping and S(B) resides in a ρ-compact subset of Xρ;(II)
T:B→Xρ or T:Xρ→Xρ is a ρ-expansive mapping satisfying inequality (5) such that c>2l;(III)
S(B)⊂(I-T)(Xρ) and [x=Tx+Sy,y∈B implies that x∈B] or S(B)⊂(I-T)(B).
Then there exists a pointz∈B such that Sz+Tz=z.Proof.
By condition (III), for eachw∈B, there exists x∈Xρ such that x-Tx=Sw. If S(B)⊂(I-T)(B), then x∈B; if S(B)⊂(I-T)(Xρ), then by Lemma 10 and condition (III), x=(I-T)-1Sw∈B. Now (I-T)-1 is a ρ-continuous and so (I-T)-1S is a ρ-continuous mapping of B into B. Since S(B) resides in a ρ-compact subset of Xρ, so (I-T)-1S(B) resides in a ρ-compact subset of the closed set B. By using Theorem 7, there exists a fixed point z∈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 ρ-sequentially continuous.Definition 13.
LetXρ be a modular space and B a subset of Xρ. A mapping T:B→Xρ is said to be(1)
ρ-sequentially continuous on the set B if for every sequence {xn}⊂B and x∈B such that ρ(xn-x)→0, then ρ(Txn-Tx)→0;(2)
ρ-closed if for every sequence {xn}⊂B such that ρ(xn-x)→0 and ρ(Txn-y)→0, then Tx=y.Definition 14.
LetXρ be a modular space and B, C two subsets of Xρ. Suppose that T:B→Xρ and S:C→Xρ are two mappings. Define
(16)F={x∈B:x=Tx+Syforsomey∈C}.Theorem 15.
LetXρ be a ρ-complete modular space. Assume that ρ is a convex modular satisfying the Δ2-condition and B is a nonempty, ρ-closed, and convex subset of Xρ. Suppose that (I)
S:B→Xρ is ρ-sequentially continuous;(II)
T:B→Xρ is a ρ-expansive mapping satisfying inequality (5) such that c>2l;(III)
x∈S(B) implies that B⊂x+T(B), where T(B)+x={y+x∣y∈T(B)};(IV)
T is ρ-closed in F and F is relatively ρ-compact.
Then there exists a pointz∈B such that Sz+Tz=z.Proof.
Letw∈B, and TSw=T+Sw. One considers the mapping TSw:B→Xρ; by Theorem 9, the equation
(17)Tx+Sw=x
has a unique solution x=η(Sw)∈B.
Now, we show thatηS=(I-T)-1 exists. For any w1,w2∈B and by the same technique of Lemma 10, we have
(18)ρ(c(η(Sw1)-η(Sw2)))≤1k-1ρ(l′(w1-w2)),
where l′=αl. This implies that ηS=(I-T)-1 exists and for all w∈B, ηSw=(I-T)-1Sw and ηS(B)⊂F.
We show thatηS is ρ-sequentially continuous in B. Let {xn} be a sequence in B and x∈B such that ρ(xn-x)→0. Since ηS(xn)∈F and F is relatively ρ-compact, then there exists z∈B such that ρ(ηSxn-z)→0. On the other hand, by condition (I), ρ(Sxn-Sx)→0. Thus by (17), we get
(19)T(ηSxn)+Sxn=ηSxn;
then
(20)ρ(T(ηSxn)-(z-Sx)2)=ρ((ηSxn-Sxn)-(z-Sx)2)≤ρ(ηSxn-z)+ρ(Sxn-Sx);
therefore when n→∞, condition (IV) implies that Tz=z-Sx; that is, z=ηSx and
(21)ρ(ηSxn-ηSx)⟶0;
then ηS is ρ-sequentially continuous in F. By Δ2-condition, ηS is ∥·∥ρ-sequentially continuous. Let H=co⁡¯∥·∥ρF, where co⁡¯∥·∥ρ denotes the closure of the convex hull in the sense of ∥·∥ρ. Then H⊂B and is a compact set. Therefore ηS is ∥·∥ρ-sequentially continuous from H into H. Then using Theorem 7, ηS has a fixed point z∈H such that ηSz=z. From (17), we have
(22)T(ηSz)+Sz=ηSz;
that is, Tz+Sz=z.The following theorem is another version of Theorem15.Theorem 16.
LetXρ be a ρ-complete modular space. Assume that ρ is a convex modular satisfying the Δ2-condition and B is a nonempty, ρ-closed, and convex subset of Xρ. Suppose that (I)
S:B→Xρ is ρ-sequentially continuous;(II)
T:B→Xρ is a ρ-expansive mapping satisfying inequality (5), such that c>2l;(III)
S(B)⊂(I-T)(Xρ) and [x=Tx+Sy,y∈B] implies that x∈B (or S(B)⊂(I-T)(B)).(IV)
T is ρ-closed in F and F is relatively ρ-compact.
Then there exists a pointz∈B such that Sz+Tz=z.Proof.
By (III) for eachw∈B, there exists x∈Xρ such that x-Tx=Sw and x=(I-T)-1Sw∈B. By the same technique of Theorem 15, (I-T)-1S:B→B is ρ-sequentially continuous and there exists a z∈B such that z=(I-T)-1Sz.


## 3. Integral Equation forρ-Expansive Mapping in Modular Function Spaces

In this section, we study the following integral equation:(23)x(t)=ϕ(t,x(t))+∫0t‍ψ(t,s,x(s))ds,x∈C(I,Lφ),
where Lφ is the Musielak-Orlicz space and I=[0,b]⊂ℝ. C(I,Lφ) denote the space of all ρ-continuous functions from I to Lφ with the modular σ(x)=sup⁡t∈Iρ(x(t)). Also C(I,Lφ) is a real vector space. If ρ is a convex modular, then σ is a convex modular. Also, if ρ satisfies the Fatou property and Δ2-condition, then σ satisfies the Fatou property and Δ2-condition (see [9]).To study the integral equation (23), we consider the following hypotheses.(1)
ϕ:I×Lφ→Lφ is a ρ-expansive mapping; that is, there exist constants c,k,l∈ℝ+ such that c>2l, k≥2 and for all x,y∈Lφ(24)ρ(l(ϕ(t,x)-ϕ(t,y)))≥kρ(c(x-y))and
ϕ is onto. Also for t∈I,   ϕ(t,·):Lφ→Lφ is ρ-continuous.(2)
ψ is a function from I×I×Lφ into Lφ such that ψ(t,s,·):x→ψ(t,s,x) is ρ-continuous on Lφ for almost all t,s∈I and ψ(t,·,x):s→ψ(t,s,x) is measurable function on I for each x∈Lφ and for almost all t∈I. Also, there are nondecreasing continuous functions β,γ:I→ℝ+ such that
(25)lim⁡t→∞β(t)∫0t‍γ(s)ds=0,ρ(c(ψ(t,s,x)))≤β(t)γ(s),for allt,s∈I, s≤t and x∈Lφ.(3)
There exists measurable functionη:I×I×I→ℝ+ such that
(26)ρ(ψ(t,s,x)-ψ(r,s,x))≤η(t,r,s),for allt,r,s∈I and x∈Lφ; also lim⁡t→r∫0b‍η(t,r,s)ds=0.(4)
ρ(ψ(t,s,x)-ψ(t,s,y))≤ρ(x-y) for all t,s∈I and x,y∈Lφ.Remark 17 (see [7]).
We considerLφ, the Musielak-Orlicz space. Since ρ is convex and satisfies the Δ2-condition, then
(27)∥xn-x∥ρ⟶0⟺ρ(xn-x)⟶0,
as n→∞ on Lφ. This implies that the topologies generated by ∥·∥ρ and ρ are equivalent.Theorem 18.
Suppose that the conditions (1)–(4) are satisfied. Further assume thatLφ satisfies the Δ2-condition. Also ω(t)=β(t)∫0t‍γ(s)ds and ω(0)=0; also sup⁡{ρ(c(ϕ(t,v))),t∈I,v∈Lφ}≤ω(t). Then integral equation (23) has at least one solution x∈C(I,Lφ).Proof.
Suppose that(28)Tx(t)=ϕ(t,x(t)),Sx(t)=∫0t‍ψ(t,s,x(s))ds.
Conditions (1) and (2) imply that T and S are well defined on C(I,Lφ). Define the set B={x∈C(I,Lφ);ρ(c(x(t)))≤ω(t)forallt∈I}. Then B is a nonempty, ρ-bounded, ρ-closed, and convex subset of C(I,Lφ). 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∈B, we show that Sx∈B. Indeed,
(29)ρ(c(Sx(t)))=ρ(c(∫0t‍ψ(t,s,x(s))ds))≤∫0t‍ρ(c(ψ(t,s,x(s))))ds≤∫0t‍β(t)γ(s)ds=ω(t);
then Sx∈B. Since S(B)⊂B and B is ρ-bounded, S(B) is σ-bounded and by Δ2-condition ∥·∥σ-bounded.
We show thatS(B) is ρ-equicontinuous. For all t,r∈I and x∈Lφ such that t<r,
(30)Sx(t)-Sx(r)=∫0t‍ψ(t,s,x(s))ds-∫0r‍ψ(r,s,x(s))ds;
then by condition (3),
(31)ρ(Sx(t)-Sx(r))≤∫0b‍η(t,r,s)ds;
since lim⁡t→r∫0b‍η(t,r,s)ds=0, then S(B) is ρ-equicontinuous. By using the Arzela-Ascoli theorem, we obtain that S is a σ-compact mapping. Next, we show that S is σ-continuous. Suppose that ɛ>0 is given; we find a δ>0 such that σ(x-y)<δ, for some x,y∈B. Note that
(32)Sx(t)-Sy(t)=∫0t‍ψ(t,s,x(s))ds-∫0t‍ψ(t,s,y(s))ds;
also
(33)ρ(Sx(t)-Sy(t))≤∫0t‍ρ(x(s)-y(s))ds≤∫0t‍σ(x-y)ds;
then
(34)σ(Sx-Sy)≤∫0b‍σ(x-y)ds≤ɛ;
therefore S is σ-continuous.
Sinceϕ is ρ-continuous, it shows that T transforms C(I,Lφ) into itself. In view of supremum ρ and condition (1), it is easy to see that T is σ-expansive with constant k≥2. For x,y∈B,
(35)ρ(l(Tx(t)-Ty(t)))≤ρ(c(x(t)-y(t)))+ρ(αl((I-T)x(t)-(I-T)y(t)));
then
(36)ρ(αl((I-T)x(t)-(I-T)y(t)))≥(k-1)ρ(c(x(t)-y(t))),
where α is conjugate of c/l. Let r=αl; since k≥2, then
(37)ρ(r(I-T)x(t))≥(k-1)ρ(c(x(t)))≥ρ(c(x(t))).
Now, assume that x=Tx+Sy for some y∈B. Since c>2l, then r<c, and
(38)ρ(c(x(t)))≤ρ(r(I-T)x(t))=ρ(r(Sy(t)))≤ρ(c(Sy(t)))≤ω(t),
which shows that x∈B. Now, define a map Tz as follows:
(39)Tz:C(I,Lφ)⟶C(I,Lφ),
for each z∈C(I,Lφ); by
(40)Tzx(t)=Tx(t)+z(t),
for all x,y∈C(I,Lφ),
(41)ρ(l(Tzx(t)-Tzy(t)))=ρ(l(Tx(t)-Ty(t)))≥kρ(c(x(t)-y(t)));
therefore
(42)σ(l(Tzx-Tzy))≥kσ(c(x-y));
then Tz is σ-expansive with constant k≥2 and Tz is onto. By Theorem 9, there exists w∈C(I,Lφ) such that Tzw=w; that is, (I-T)w=z. Hence S(B)⊂(I-T)(Lφ) and condition (III) of Theorem 12 holds. Therefore by Theorem 12, S+T has a fixed point z∈B with Tz+Sz=z; that is, z is a solution to (23).Now, we consider another integral equation.LetLφ be the Musielak-Orlicz space and I=[0,b]⊂ℝ. Suppose that ρ is convex and satisfies the Δ2-condition. Since topologies generated by ∥·∥ρ and ρ are equivalent, then we consider Banach space (Lφ,∥·∥ρ) and C(I,Lφ) denote the space of all ∥·∥ρ-continuous functions from I to Lφ with the modular ∥x∥σ=sup⁡t∈I∥x(t)∥ρ; also C(I,Lφ) is a real vector space. Consider the nonlinear integral equation
(43)x(t)=ϕ(t,x(t))+λ(t,x(t))∫0t‍ω(t,s)ψ(s,x(s))ds,x∈C(I,Lφ),
where (1)
ϕ:I×Lφ→Lφ is a ∥·∥ρ-expansive mapping; that is, there exists constant l≥2 such that
(44)∥ϕ(t,x)-ϕ(t,y)∥ρ≥l∥x-y∥ρ,for allx,y∈Lφ and ϕ is onto; also for t∈I, ϕ(t,·):Lφ→Lφ is ∥·∥ρ-continuous;(2)
ψ is function from I×Lφ into Lφ such that ψ(t,·):Lφ→Lφ is a ∥·∥ρ-continuous and t→ψ(t,x) is measurable for every x∈Lφ. Also, there exist functions β∈L1(I) and a nondecreasing continuous function γ:[0,∞)→(0,∞) such that
(45)∥ψ(t,x)∥ρ≤β(t)γ(∥x∥ρ),for allt∈I and x∈Lφ. Also for t∈I, x→ψ(t,x) is nondecreasing on Lφ;(3)
λ is function from I×Lφ into Lφ such that λ(t,·):Lφ→Lφ is ∥·∥ρ-continuous and there exists a a≥0 such that
(46)∥λ(t,x)-λ(t,y)∥ρ≤a∥x-y∥ρ,for allt∈I and x∈Lφ; also for x∈Lφ, t→λ(t,x) is nondecreasing on I and for t∈I, x→λ(t,x) is nondecreasing on Lφ;(4)
ω is function from I×I into ℝ+. For each t∈I, ω(t,s) is measurable on [0,t]. Also ω(t)¯=esssup⁡|ω(t,s)| is bounded on [0,b] and r=sup⁡|ω(t)¯|. The map ω(·,s):t→ω(t,s) is continuous from I to L∞(I). Also for s∈I, t→ω(t,s) is nondecreasing on I.Theorem 19.
Suppose that the conditions (1)–(4) are satisfied and there exists a constantk≥0 such that for all t∈I,
(47)∫0t‍β(s)ds<k(ak+h)rb∫0t‍1γ(k)ds,
where h∶=sup⁡{∥λ(t,x)∥ρ,t∈I,x∈Lφ} and also sup⁡{∥ϕ(t,x)∥ρ,t∈I,x∈Lφ}≤k. Then integral equation (43) has at least one solution x∈C(I,Lφ).Proof.
Define(48)B={x∈C(I,Lφ);∥x(t)∥ρ≤k∀t∈I};
then B is a nonempty, ∥·∥ρ-bounded, ∥·∥ρ-closed, and convex subset of C(I,Lφ). Consider
(49)Tx(t)=ϕ(t,x(t)),Sx(t)=λ(t,x(t))∫0t‍ω(t,s)ψ(s,x(s))ds.
It is easy that by the hypothesis T and S are well defined on C(I,Lφ).
Forx∈B, we show that Sx∈B. Consider
(50)∥Sx(t)∥ρ=∥λ(t,x(t))∫0t‍ω(t,s)ψ(s,x(s))ds∥ρ=∥(λ(t,x(t))-λ(t,0)+λ(t,0))∫0t‍ω(t,s)ψ(s,x(s))ds∥ρ≤(a∥x(t)∥ρ+h)r∫0t‍β(s)γ(∥x(s)∥ρ)ds≤(ak+h)r∫0t‍β(s)γ(k)ds≤(ak+h)r∫0b‍kγ(k)(ak+h)rbγ(k)ds≤k.
Let x∈B and assume that t>τ∈I such that |t-τ|<δ, for a given positive constant δ. We have
(51)∥Sx(t)-Sx(τ)∥ρ=∥λ(t,x(t))∫0t‍ω(t,s)ψ(s,x(s))ds-λ(τ,x(τ))∫0τ‍ω(τ,s)ψ(s,x(s))ds∥ρ=∥λ(t,x(t))∫0t‍ω(t,s)ψ(s,x(s))ds±λ(t,x(t))∫0t‍ω(τ,s)ψ(s,x(s))ds±λ(τ,x(τ))∫0t‍ω(τ,s)ψ(s,x(s))ds-λ(τ,x(τ))∫0τ‍ω(τ,s)ψ(s,x(s))ds∥ρ≤∥λ(t,x(t))(∫0t‍ω(t,s)ψ(s,x(s))ds-∫0t‍ω(τ,s)ψ(s,x(s))ds)∥ρ+∥∫0t(λ(τ,x(τ))-λ(τ,x(τ)))×∫0t‍ω(τ,s)ψ(s,x(s))ds∥ρ+∥λ(τ,x(τ))∫τt‍ω(τ,s)ψ(s,x(s))ds∥ρ;
since
(52)∥λ(t,x(t))(∫0t‍ω(t,s)ψ(s,x(s))ds-∫0t‍ω(τ,s)ψ(s,x(s))ds)∥ρ=∥λ(t,x(t))(∫0t‍(ω(t,s)-ω(τ,s))ψ(s,x(s))ds)∥ρ≤∥∑22(λ(τ,x(τ))-λ(τ,0)+λ(τ,0))×(∫0t(ω(t,s)-ω(τ,s))ψ(s,x(s))ds)∑22∥ρ≤(ak+h)|ω(t,0)-ω(τ,0)|L∞∫0t‍β(s)γ(k)ds≤kr|ω(t,0)-ω(τ,0)|L∞,∥(λ(t,x(t))-λ(τ,x(τ)))∫0t‍ω(τ,s)ψ(s,x(s))ds∥ρ≤∥(λ(t,x(t))-λ(τ,x(τ)))r∫0t‍β(s)γ(k)ds∥ρ≤kak+h(∥λ(t,x(t))-λ(t,x(τ))∥ρ+∥λ(τ,x(τ))-λ(t,x(τ))∥ρ)≤kak+h(a∥x(t)-x(τ)∥ρ+h),∥λ(τ,x(τ))∫τt‍ω(τ,s)ψ(s,x(s))ds∥ρ=∥∑22(λ(τ,x(τ))-λ(τ,0)+λ(τ,0))×∫τt‍ω(τ,s)ψ(s,x(s))ds∑22∥ρ≤(ak+h)r∫τt‍β(s)γ(k)ds≤kb|t-τ|,
then S(B) is ∥·∥ρ-equicontinuous. By using the Arzela-Ascoli Theorem, we obtain that S is a ∥·∥ρ-compact mapping.
We show thatS is ∥·∥ρ-continuous. Suppose that ɛ>0 is given. We find a δ>0 such that ∥x-y∥σ<δ. We have
(53)∥Sx(t)-Sy(t)∥ρ=∥λ(t,x(t))∫0t‍ω(t,s)ψ(s,x(s))ds-λ(t,y(t))∫0t‍ω(t,s)ψ(s,y(s))ds∥ρ≤∥(λ(t,x(t))-λ(t,y(t)))∫0t‍ω(t,s)ψ(s,x(s))ds∥ρ+∥λ(t,y(t))∫0t‍(ψ(s,x(s))-ψ(s,y(s)))ds∥ρ≤kaak+h∥x(t)-y(t)∥ρ+(ak+h)r∫0t‍∥x(s)-y(s)∥ρds≤kaak+h∥x-y∥σ+(ak+h)rb∥x-y∥σ≤ɛ.
Since ϕ is ∥·∥ρ-continuous, it shows that T transforms C(I,Lφ) into itself. In view of supremum ∥·∥ρ and condition (1), it is easy to see that T is ∥·∥σ-expansive with constant l≥2.
Forx,y∈B,
(54)∥Tx(t)-Ty(t)∥ρ≤∥x(t)-y(t)∥ρ+∥(I-T)x(t)-(I-T)y(t)∥ρ;
then
(55)∥(I-T)x(t)-(I-T)y(t)∥ρ≥(l-1)∥x(t)-y(t)∥ρ;
since l≥2, then
(56)∥(I-T)x(t)∥ρ≥(l-1)∥x(t)∥ρ≥∥x(t)∥ρ.
Now, assume that x=Tx+Sy for some y∈B. Then
(57)∥x(t)∥ρ≤∥(I-T)x(t)∥ρ=∥Sy(t)∥ρ≤k,
which shows that x∈B. Now for each z∈C(I,Lφ) we define a map Tz as follows:
(58)Tz:C(I,Lφ)⟶C(I,Lφ);
by
(59)Tzx(t)=Tx(t)+z(t);
for all x,y∈C(I,Lφ),
(60)∥Tzx(t)-Tzy(t)∥ρ=∥Tx(t)-Ty(t)∥ρ≥l∥x(t)-y(t)∥ρ;
therefore
(61)∥Tzx-Tzy∥σ≥l∥x-y∥σ;
then Tz is ∥·∥σ-expansive with constant l≥2 and Tz is onto. By Theorem 9, there exists w∈C(I,Lφ) such that Tzw=w; that is, (I-T)w=z. Hence S(B)⊂(I-T)(Lφ). Therefore by Theorem 12, S+T has a fixed point z∈B with Tz+Sz=z; that is, z is a solution of (43).Finally, some examples are presented to guarantee Theorems18 and 19.Example 20.
Consider the following integral equation:(62)x(t)=9x(t)1+t2+∫0t‍arctan(5t(1+s)x(s)(1+t)3(1+x(s)))ds,
where Lφ=ℝ+, I=[0,1].
Forx,y∈ℝ+ and t∈I, we have
(63)|ϕ(t,x)-ϕ(t,y)|=|9x1+t2-9y1+t2|≥92|x-y|.
Therefore by Theorem 18, the integral equation (62) has at least one solution.Example 21.
Consider the following integral equation:(64)x(t)=9x(t)1+t2+18arcsinx(t)∫0t‍tt+sx(s)ds,
where ϕ(t,x)=(9x/(1+t2)), λ(t,x)=(1/8)arcsinx, ω(t,s)=t/(t+s), and ψ(t,x)=x. Also Lφ=ℝ+, I=[0,1]. Therefore by Theorem 19, the integral equation (64) has at least one solution.



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*Source: 580508-2014-04-14.xml*