Some fixed point theorems for ρ-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
(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 [<xref ref-type="bibr" rid="B7">1</xref>]).

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

Sis continuous; S(K) resides in a compact subset of X;

T is an expansive mapping;

z∈S(K) implies that T(K)+z⊃K, where T(K)+z={y+z∣y∈T(K)}.

Then there exists a point x*∈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 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.

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

A functional ρ:X→[0,∞] is called modular if

ρ(x)=0 if and only if x=0;

ρ(αx)=ρ(x) for α∈K with |α|=1, for all x∈X;

ρ(αx+βy)≤ρ(x)+ρ(y) if α,β≥0, α+β=1, for all x,y∈X.

If (iii) is replaced by

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.

A modular ρ defines a corresponding modular space, that is, the space Xρ given by
(2)Xρ={x∈X∣ρ(αx)⟶0as α⟶0}.

If ρ 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 that ρ 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.

Let Xρ be a modular space. Then we have the following.

A sequence (xn)n∈ℕ in Xρ is said to be

ρ-convergent to x if ρ(xn-x)→0 as n→∞;

ρ-Cauchy if ρ(xn-xm)→0 as n,m→∞.

Xρ is ρ-complete if every ρ-Cauchy sequence is ρ-convergent.

A subset B⊂Xρ is said to be ρ-closed if for any sequence (xn)n∈ℕ⊂B and xn→x then x∈B.

A subset B⊂Xρ is called ρ-bounded if δρ(B)=supρ(x-y)<∞, for all x,y∈B, where δρ(B) is called the ρ-diameter of B.

ρ has the Fatou property if
(4)ρ(x-y)≤liminfρ(xn-yn),

whenever xn→x and yn→y as n→∞.

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.

Let Xρ 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.

Let Xρ=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 [<xref ref-type="bibr" rid="B7">1</xref>, <xref ref-type="bibr" rid="B3">8</xref>]).

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 [<xref ref-type="bibr" rid="B6">6</xref>, <xref ref-type="bibr" rid="B2">9</xref>]).

Let Xρ 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.

Let Xρ 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 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; 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, let z 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 Theorem 9 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 all x,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.

Let Xρ 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

S:B→Xρ is a ρ-continuous mapping and S(B) resides in a ρ-compact subset of Xρ;

T:B→Xρ is a ρ-expansive mapping satisfying inequality (5) such that c>2l;

x∈S(B) implies that B⊂x+T(B), where T(B)+x={y+x∣y∈T(B)}.

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

Proof.

Let w∈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 Theorem 11.

Theorem 12.

Let Xρ 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

S:B→Xρ is a ρ-continuous mapping and S(B) resides in a ρ-compact subset of Xρ;

T:B→Xρ or T:Xρ→Xρ is a ρ-expansive mapping satisfying inequality (5) such that c>2l;

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 point z∈B such that Sz+Tz=z.

Proof.

By condition (III), for each w∈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.

Let Xρ be a modular space and B a subset of Xρ. A mapping T:B→Xρ is said to be

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;

ρ-closed if for every sequence {xn}⊂B such that ρ(xn-x)→0 and ρ(Txn-y)→0, then Tx=y.

Definition 14.

Let Xρ 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.

Let Xρ 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

S:B→Xρ is ρ-sequentially continuous;

T:B→Xρ is a ρ-expansive mapping satisfying inequality (5) such that c>2l;

x∈S(B) implies that B⊂x+T(B), where T(B)+x={y+x∣y∈T(B)};

T is ρ-closed in F and F is relatively ρ-compact.

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

Proof.

Let w∈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 that ηS=(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 that ηS 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 Theorem 15.

Theorem 16.

Let Xρ 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

S:B→Xρ is ρ-sequentially continuous;

T:B→Xρ is a ρ-expansive mapping satisfying inequality (5), such that c>2l;

S(B)⊂(I-T)(Xρ) and [x=Tx+Sy,y∈B] implies that x∈B (or S(B)⊂(I-T)(B)).

T is ρ-closed in F and F is relatively ρ-compact.

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

Proof.

By (III) for each w∈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 for <italic><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M435"><mml:mrow><mml:mi>ρ</mml:mi></mml:mrow></mml:math></inline-formula></italic>-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)=supt∈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.

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))

ϕ is onto. Also for t∈I, ϕ(t,·):Lφ→Lφ is ρ-continuous.

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)limt→∞β(t)∫0tγ(s)ds=0,ρ(c(ψ(t,s,x)))≤β(t)γ(s),

for all t,s∈I, s≤t and x∈Lφ.

There exists measurable function η:I×I×I→ℝ+ such that
(26)ρ(ψ(t,s,x)-ψ(r,s,x))≤η(t,r,s),

for all t,r,s∈I and x∈Lφ; also limt→r∫0bη(t,r,s)ds=0.

ρ(ψ(t,s,x)-ψ(t,s,y))≤ρ(x-y) for all t,s∈I and x,y∈Lφ.

Remark 17 (see [<xref ref-type="bibr" rid="B1">7</xref>]).

We consider Lφ, 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 that Lφ 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 that S(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 limt→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.

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 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.

Let Lφ 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=supt∈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

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 all x,y∈Lφ and ϕ is onto; also for t∈I, ϕ(t,·):Lφ→Lφ is ∥·∥ρ-continuous;

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 all t∈I and x∈Lφ. Also for t∈I, x→ψ(t,x) is nondecreasing on Lφ;

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 all t∈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;

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 constant k≥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).

For x∈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 that S 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.

For x,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 Theorems 18 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].

For x,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.

Conflict of Interests

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

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