# Two Fixed-Point Theorems for Mappings Satisfying a General Contractive Condition of Integral Type in the Modular Space

Authors: Maryam Beygmohammadi; Abdolrahman Razani
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

First we prove existence of a fixed point for mappings defined on a complete modular space satisfying a general contractive
inequality of integral type. Then we generalize fixed-point theorem for a quasicontraction mapping given by Khamsi (2008) and Ciric (1974).


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## Body

## 1. Introduction

In [1], Branciari established that a function f defined on a complete metric space satisfying a contraction condition of the form (1.1)∫0d(fx,fy)φ(t)dt≤c∫0d(x,y)φ(t)dt
has a unique attractive fixed point where φ:ℝ+→ℝ+ is a Lebesgue-integrable mapping and c∈[0,1).In [2], Rhoades extended this result to a quasicontraction function f. The purpose of this paper is to extend these theorems in modular space.First, we introduce the notion of modular space.Definition 1.1.
LetX be an arbitrary vector space over K(=ℝ or ℂ). A functional ρ:X→[0,+∞) is called modular if(1)
ρ(x)=0 if and only if x=0;(2)
ρ(αx)=ρ(x) for α∈K with |α|=1,for all x,y∈X;(3)
ρ(αx+βy)≤ρ(x)+ρ(y) if α,β≥0, α+β=1, for allx,y∈X.If (2.14) in Definition 1.1 is replaced by(1.2)ρ(αx+βy)≤αsρ(x)+βsρ(y)
for α,β≥0, αs+βs=1 with an s∈(0,1], then the modular ρ is called an s-convex modular; and if s=1,ρ is called a convex modular.Definition 1.2.
A modularρ defines a corresponding modular space, that is, the space Xρ is given by
(1.3)Xρ={x∈X∣ρ(λx)⟶0asλ⟶0}.Definition 1.3.
LetXρ be a modular space.  (1)
A sequence{xn}n in Xρ is said to be
(a)
ρ-convergent to x if ρ(xn-x)→0 as n→+∞,(b)
ρ-Cauchy if ρ(xn-xm)→0 as n,m→+∞.(2)
Xρ is ρ-complete if any ρ-Cauchy sequence is ρ-convergent.(3)
A subsetB⊂Xρ is said to be ρ-closed if for any sequence {xn}n⊂B with xn→x then x∈B. B¯ρ denotes the closure of B in the sense of ρ.(4)
A subsetB⊂Xρ is called ρ-bounded if
(1.4)δρ(B)=sup⁡x,y∈Bρ(x-y)<+∞,
where δρ(B) is called the ρ-diameter of B.(5)
We say thatρ has Fatou property if
(1.5)ρ(x-y)≤liminf⁡ρ(xn-yn)
whenever
(1.6)xn→ρx,yn→ρy.(6)
ρ is said to satisfy the Δ2-condition if:ρ(2xn)→0 as n→+∞ whenever ρ(xn)→0 as n→+∞.Remark 1.4.
Note that sinceρ does not satisfy a priori the triangle inequality, we cannot expect that if {xn} and {yn} are ρ-convergent, respectively, to x and y then {xn+yn} is ρ-convergent to x+y, neither that a ρ-convergent sequence is ρ-Cauchy.


## 2. Main Result

Theorem 2.1.
LetXρ be a complete modular space, where ρ satisfies the Δ2-condition. Assume that ψ:ℝ+→[0,∞) is an increasing and upper semicontinuous function satisfying
(2.1)ψ(t)<t,∀t>0.
Let φ:[0,+∞)→[0,+∞] be a nonnegative Lebesgue-integrable mapping which is summable on each compact subset of [0,+∞[ and such that for ɛ>0, ∫0ɛφ(t)dt>0 and let f:Xρ→Xρ be a mapping such that there are c,l∈ℝ+ where l<c,
(2.2)∫0ρ(c(fx-fy))φ(t)dt≤ψ(∫0ρ(l(x-y))φ(t)dt),
for each x,y∈Xρ. Then f has a unique fixed point in Xρ.Proof.
First, we show that forx∈Xρ, the sequence {ρ(c(fnx-fn-1x))} converges to 0. For n∈ℕ, we have
(2.3)∫0ρ(c(fnx-fn-1x))φ(t)dt≤ψ(∫0ρ(l(fn-1x-fn-2x))φ(t)dt)<∫0ρ(l(fn-1x-fn-2x))φ(t)dt<∫0ρ(c(fn-1x-fn-2x))φ(t)dt.
Consequently, {∫0ρ(c(fnx-fn-1x))φ(t)dt} is decreasing and bounded from below. Therefore {∫0ρ(c(fnx-fn-1x))φ(t)dt} converges to a nonnegative point a.
Now, ifa≠0,(2.4)a=lim⁡n→∞∫0ρ(c(fnx-fn-1x))φ(t)dt≤lim⁡n→∞ψ(∫0ρ(l(fn-1x-fn-2x))φ(t)dt)≤lim⁡n→∞ψ(∫0ρ(c(fn-1x-fn-2x))φ(t)dt),
then
(2.5)a≤ψ(a),
which is a contradiction, so a=0 and
(2.6)∫0ρ(c(fnx-fn+1x))φ(t)dt⟶0+asn⟶+∞.
This concludesρ(c(fnx-fn+1x))→0. Suppose that(2.7)lim⁡n→∞sup⁡ρ(c(fnx-fn+1x))=ɛ>0
then there exist a νɛ∈ℕ and a sequence (fnνx)ν≥νɛ such that
(2.8)ρ(c(fnνx-fnν+1x))⟶ɛ>0,ν⟶∞,ρ(c(fnνx-fnν+1x))≥ɛ2,∀ν≥νɛ,
then we get the following contradiction:
(2.9)0=lim⁡ν→∞∫0ρ(c(fnνx-fnν+1x))φ(t)dt≥∫0ɛ/2φ(t)dt>0.
Now, we prove for eachx∈Xρ the sequence {fnx}n∈ℕ is a ρ-Cauchy sequence.
Assume that there is anɛ>0 such that for each ν∈ℕ there exist mν,nν∈ℕ that mν>nν>ν,(2.10)ρ(l(fmνx-fnνx))≥ɛ.
Then we choose the sequence(mν)ν∈ℕ and (nν)ν∈ℕ such that for each ν∈ℕ, mν is minimal in the sense that(2.11)ρ(l(fmνx-fnνx))≥ɛ.
But
(2.12)ρ(l(fhx-fnνx))<ɛ,
for each h∈{nν+1,…,mν-1}.
Now, letα∈ℝ+ be such that l/c+1/α=1, then we have(2.13)∫0ɛφ(t)dt≤∫0ρ(l(fmνx-fnνx))φ(t)dt≤∫0ρ(c(fmνx-fnν+1x))φ(t)dt+∫0ρ(αl(fnν+1x-fnνx))φ(t)dt≤ψ(∫0ρ(l(fmν-1x-fnνx))φ(t)dt)+∫0ρ(αl(fnν+1x-fnνx))φ(t)dt≤∫0ρ(l(fmν-1x-fnνx))φ(t)dt+∫0ρ(αl(fnν+1x-fnνx))φ(t)dt≤∫0ɛφ(t)dt+∫0ρ(αl(fnν+1x-fnνx))φ(t)dt.
Thus, asν→∞, by Δ2-condition, ∫0ρ(αl(fnν+1x-fnνx))φ(t)dt→0. Therefore(2.14)∫0ρ(l(fmνx-fnνx))φ(t)dt⟶ɛ+,ν⟶∞.
Now,(2.15)∫0ρ(l(fmνx-fnνx))φ(t)dt≤∫0ρ(c(fmν+1x-fnν+1x))φ(t)dt+∫0ρ(2αl(fmνx-fmν+1x))φ(t)dt+∫0ρ(2αl(fnν+1x-fnνx))φ(t)dt≤ψ(∫0ρ(l(fmνx-fnνx))φ(t)dt)+∫0ρ(2αl(fmνx-fmν+1x))φ(t)dt+∫0ρ(2αl(fnν+1x-fnνx))φ(t)dt.
If ν→∞ we get
(2.16)∫0ɛφ(t)dt≤ψ(∫0ɛφ(t)dt),
which is a contradiction for ɛ>0. Therefore {lfnx} is a ρ-Cauchy sequence and by Δ2-condition {fnx} is ρ-Cauchy. By the fact that Xρ is ρ-complete, there is a z∈Xρ such that ρ(fnz-z)→0 as n→∞. Furthermore, z is the fixed point for f. In fact
(2.17)ρ(c2(z-fz))≤ρ(c(z-fnz))+ρ(c(fnz-fz))⟶0,n⟶∞
then ρ((c/2)(z-fz))=0 and fz=z.
Now, assume that we have more than one fixed point forf. Let z and u be two distinct fixed points, then(2.18)∫0ρ(c(z-u))φ(t)dt=∫0ρ(c(fz-fu))φ(t)dt≤ψ(∫0ρ(l(z-u))φ(t)dt)<∫0ρ(l(z-u))φ(t)dt≤∫0ρ(c(z-u))φ(t)dt,
which is a contradiction. So z=u and the proof is complete.Corollary 2.2 (see [1]).
LetXρ be a complete modular space, where ρ satisfies the Δ2-condition. Let f:Xρ→Xρ be a mapping such that there exists an λ∈(0,1) and c,l∈ℝ+ where l<c and for each x,y∈Xρ,
(2.19)∫0ρ(c(fx-fy))φ(t)dt≤λ(∫0ρ(l(x-y))φ(t)dt),
then f has a unique fixed point.Corollary 2.3 (see [3]).
LetXρ be a complete modular space, where ρ satisfies the Δ2-condition. Assume that ψ:ℝ+→[0,∞) is an increasing and upper semicontinuous function satisfying
(2.20)ψ(t)<t,∀t>0.
Let B be a ρ-closed subset of Xρ and T:B→B be a mapping such that there exist c,l∈ℝ+ with c>l,
(2.21)ρ(c(Tx-Ty))≤ψ(ρ(l(x-y)))
for all x,y∈B. Then T has a fixed point.In the next theorem we use the following notation:(2.22)m(x,y)=max⁡{ρ(x-y),ρ(x-Tx),ρ(y-Ty),ρ(1/2(x-Ty))+ρ(1/2(y-Tx))2}.Theorem 2.4.
Let(Xρ,ρ) be a ρ-complete modular space that ρ satisfies the Δ2-condition and let T:Xρ→Xρ be a mapping such that for each x,y∈Xρ,
(2.23)∫0ρ(Tx-Ty)ϕ(t)dt≤ψ(∫0m(x,y)ϕ(t)dt),
where ϕ:ℝ+→ℝ+ and ψ:ℝ+→[0,∞) are as in Theorem 2.1. Then T has a unique fixed point.Proof.
Letx∈Xρ, we will show that {Tnx} is a Cauchy sequence. First, we prove that {ρ(Tnx-Tn-1x)} converges to 0. From (2.23),
(2.24)∫0ρ(Tnx-Tn-1x)ϕ(t)dt≤ψ(∫0m(Tn-1x,Tn-2x)ϕ(t)dt).
By the definition ofm(x,y),(2.25)m(Tn-1x,Tn-2x)=max⁡{ρ(Tnx-Tn-1x),ρ(Tn-1-Tn-2x),ρ(1/2(Tnx-Tn-2x))2},ρ(1/2(Tnx-Tn-2x))2≤ρ(Tnx-Tn-1x)+ρ(Tn-1-Tn-2x)2≤max⁡{ρ(Tnx-Tn-1x),ρ(Tn-1-Tn-2x)}.
Hence,(2.26)m(Tn-1x,Tn-2x)=max⁡{ρ(Tnx-Tn-1x),ρ(Tn-1-Tn-2x)}
and therefore,
(2.27)∫0ρ(Tnx-Tn-1x)ϕ(t)dt≤ψ(∫0m(Tn-1x,Tn-2x)ϕ(t)dt)≤∫0m(Tn-1x,Tn-2x)ϕ(t)dt=∫0max⁡{ρ(Tnx-Tn-1x),ρ(Tn-1-Tn-2x)}ϕ(t)dt=max⁡{∫0ρ(Tnx-Tn-1x)ϕ(t)dt,∫0ρ(Tn-1-Tn-2x)ϕ(t)dt}=∫0ρ(Tn-1-Tn-2x)ϕ(t)dt.
This means that {ρ(Tnx-Tn-1x)} is decreasing and since it is bounded from below, it is a convergent sequence. Similarly to Theorem 2.1, it is easy to show that
(2.28){ρ(Tnx-Tn-1x)}⟶0.
Now, we show that{Tnx} is Cauchy. If not, then there exist an ɛ>0 and subsequences {m(p)} and {n(p)} such that m(p)<n(p)<m(p+1) with(2.29)ρ(Tm(p)x-Tn(p)x)≥ɛ,ρ(2(Tm(p)x-Tn(p)-1x))<ɛ.
From (2.22),(2.30)m(Tm(p)-1x,Tn(p)-1x)=max⁡{ρ(1/2(Tm(p)x-Tn(p)-1x))+ρ(1/2(Tn(p)x-Tm(p)-1x))2ρ(Tm(p)-1x-Tn(p)-1x),ρ(Tm(p)x-Tm(p)-1x),ρ(Tn(p)x-Tn(p)-1x),ρ(1/2(Tm(p)x-Tn(p)-1x))+ρ(1/2(Tn(p)x-Tm(p)-1x))2}.
By using (2.28), we get(2.31)lim⁡p∫0ρ(Tm(p)x-Tm(p)-1x)ϕ(t)dt=lim⁡p∫0ρ(Tn(p)x-Tn(p)-1x)ϕ(t)dt=0.
On the other hand,(2.32)ρ(Tm(p)-1x-Tn(p)-1x)≤ρ(2(Tm(p)-1x-Tm(p)x))+ρ(2(Tm(p)x-Tn(p)-1x))≤ρ(2(Tm(p)-1x-Tm(p)x))+ɛ,
thus by the Δ2-condition,
(2.33)lim⁡p∫0ρ(Tm(p)-1x-Tn(p)-1x)ϕ(t)dt≤∫0ɛϕ(t)dt.
For the last term inm(Tm(p)-1x,Tn(p)-1x) by the fact that ρ(cx) is an increasing function of c we have(2.34)v(m,n):=ρ(1/2(Tm(p)x-Tn(p)-1x))+ρ(1/2(Tn(p)x-Tm(p)-1x))2≤ρ(Tm(p)x-Tm(p)-1x)+ρ(2(Tn(p)x-Tn(p)-1x))2+ρ(2(Tm(p)x-Tn(p)-1x))+ρ(1/2(Tm(p)x-Tn(p)-1x))2≤ɛ+ρ(Tm(p)x-Tm(p)-1x)+ρ(2(Tn(p)x-Tn(p)-1x))2.
Hence, from (2.28) we get(2.35)lim⁡p∫0v(m,n)ϕ(t)dt≤∫0ɛϕ(t)dt.
Therefore from (2.31), (2.33), and (2.35) it can be concluded that(2.36)∫0ɛϕ(t)dt≤∫0ρ(Tm(p)x-Tn(p)x)ϕ(t)dt≤ψ(∫0m(Tm(p)-1x,Tn(p)-1x)ϕ(t)dt)<∫0m(Tm(p)-1x,Tn(p)-1x)ϕ(t)dt≤∫0ɛϕ(t)dt
which is a contradiction, when p is large enough. Therefore, {Tnx} is Cauchy and since Xρ is ρ-complete there is an z∈Xρ that Tnx→z. Now, we should prove that z is the fixed point for T. In fact,
(2.37)∫0ρ(1/2(Tz-z))ϕ(t)dt≤∫0ρ(Tz-Tnz)ϕ(t)dt+∫0ρ(Tnz-z)ϕ(t)dt≤ψ(∫0m(z,Tn-1z)ϕ(t)dt)+∫0ρ(Tnz-z)ϕ(t)dt⟶0asn⟶∞,
by the definition of m. It follows that Tz=z.
Letw∈Xρ be another fixed point of T. Then,(2.38)∫0ρ(w-z)ϕ(t)dt=∫0ρ(Tw-Tz)ϕ(t)dt≤ψ(∫0m(w,z)ϕ(t)dt)<∫0m(w,z)ϕ(t)dt=∫0ρ(w-z)ϕ(t)dt.
That is because
(2.39)m(w,z)=max⁡{ρ(z-w),ρ(z-z),ρ(w-w),ρ(1/2(z-w))+ρ(1/2(w-z))2}=ρ(w-z),
thus z=w.Corollary 2.5 (see [2]).
Let(X,d) be complete metric space, k∈[0,1), f:X→X a mapping such that, for x,y∈X,
(2.40)∫0d(f(x),f(y))ϕ(t)dt≤k∫0m(x,y)ϕ(t)dt,
where ϕ:ℝ+→ℝ+ is a Lebesgue-integrable mapping which is summable, nonnegative, and such that
(2.41)∫0ɛϕ(t)dt>0∀ɛ>0,
and where
(2.42)m(x,y)=max⁡{d(x,y),d(x,fx),d(y,fy),d(x,fy)+d(y,fy)2}.
Then f has a unique fixed point.Corollary 2.6 (see [4]).
Let(X,ρ) be a modular space such that ρ satisfies the Fatou property. Let C be a ρ-complete nonempty subset of Xρ and T:C→C be quasicontraction. Let x∈C such that δρ(x)<∞. Then {Tnx}ρ-converges to ω∈C. Here δρ(x)=sup{ρ(Tnx-Tmx);n,m∈ℕ}.



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