# Some Fixed Point Results for Generalized Weak Contraction Mappings in Modular Spaces

Authors: Chirasak Mongkolkeha; Poom Kumam
Journal: International Journal of Analysis
 (2013)

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## Abstract

We prove the existence theorem of fixed points for a generalized
weak contractive mapping in modular spaces.


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## Body

## 1. Introduction

In 1997, Alber and Guerre-Delabriere [1] introduced the concept of weak contraction in Hilbert spaces. Later, Rhoades [2] proved that the result which Alber et al. is also valid in complete metric spaces, the result of Rhoades in the following: A mapping T:X→X where (X,d) is a metric space, is said to be weakly contractive if
(1)d(T(x),T(y))≤d(x,y)-ϕ(d(x,y)),
where ϕ:[0,∞)→[0,∞) is continuous and nondecreasing function such that ϕ(t)=0 if and only if t=0. In 2008, Dutta and Choudhury [3] introduced a new generalization of contraction in metric spaces and proved the following theorem.Theorem 1.
Let(X,d) be a complete metric space, and let T:X→X be a self-mapping satisfying the following inequality:
(2)ψ(d(Tx,Ty))≤ψ(d(x,y))-ϕ(d(x,y)),
where ψ,ϕ:[0,∞)→[0,∞) are both continuous and monotone nondecreasing function with ψ(t)=ϕ(t)=0 if and only if t=0. Then T has a unique fixed point.We note that, if one takesψ(t)=t, then (2) reduces to (1).Recall that the theory of modular on linear spaces and the corresponding theory of modular linear spaces were founded by Nakano [4, 5] and redefined by Musielak and Orlicz [6]. Furthermore, the most complete development of these theories is due to Mazur, Luxemburg, and Turpin [7–9]. In the present time, the theory of modular and modular spaces is extensively applied, in particular, in the study of various Orlicz spaces which in their turn have broad applications [10–14]. In many cases, particularly in applications to integral operators, approximation and fixed point theory, modular-type conditions are much more natural as modular-type assumptions can be more easily verified than their metric or norm counterparts. Even though a metric is not defined, many problems in metric fixed point theory can be reformulated in modular spaces. For instance, fixed point theorems are proved in [15, 16] for nonexpansive mappings. The existences for contraction mapping in modular spaces has been studied in [3, 17–24] and the references therein.From the above mentioned, we will study the existence of fixed point theorems for mappings satisfying generalized weak contraction mappings in modular spaces.


## 2. Preliminaries

First, we start with a brief recollection of basic concepts and facts in modular spaces.Definition 2.
LetX be a vector space over ℝ (or ℂ). A functional ρ:X→[0,∞] is called a modular if for arbitrary x and y, elements of X, it satisfies the following conditions:  (1)
ρ(x)=0 if and only if x=0;(2)
ρ(αx)=ρ(x) for all scalar α with |α|=1;(3)
ρ(αx+βy)≤ρ(x)+ρ(y), whenever α,β≥0 and α+β=1.  If one replaces (3) by(4)
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 s-convex modular, and if s=1, ρ is called convex modular.Ifρ is modular in X, then the set
(3)Xρ={x∈X:ρ(λx)⟶0asλ⟶0},
is called a modular space. Xρ is a vector subspace of X.Definition 3.
A modularρ is said to satisfy the Δ2-condition if ρ(2xn)→0, whenever ρ(xn)→0 as n→∞.Definition 4.
LetXρ be a modular space.  (1)
The sequence{xn}n∈ℕ in Xρ is said to be ρ-convergent to x∈Xρif ρ(xn-x)→0, as n→∞.(2)
The sequence{xn}n∈ℕ in Xρ is said to be ρ-Cauchy if ρ(xn-xm)→0, as n,m→∞.(3)
A subsetC of Xρ is said to be ρ-closed if the ρ-limit of a ρ-convergent sequence of C always belong to C.(4)
A subsetC of Xρ is said to be ρ-complete if any ρ-Cauchy sequence in C is ρ-convergent sequence and its ρ-limit is in C.(5)
A subsetC of Xρ is said to be ρ-bounded if any δρ(C)=sup{ρ(x-y);x,y∈C}<∞.Definition 5.
LetX be a nonempty set and T:X→X. A point x∈X is a fixed point  of T if and only if Tx=x.Definition 6.
LetC be a subset of a real numbers ℝ. A mapping T:C→ℝ is called monotone increasing (or monotone nondecreasing) x≤y if and only if T(x)≤T(y), for all x and y are elements in C. A mapping T:C→ℝ is called monotone decreasing (or monotone nonincreasing), x≥y if and only if T(x)≥T(y) for all x and y are elements in C.Definition 7.
A sequence{an} of a real number is said to be monotone increasing (or monotone nondecreasing), if it satisfies a1≤a2≤a3≤…. It is also said to be monotone decreasing (or monotone nonincreasing) if it satisfies a1≥a2≥a3≥….


## 3. A Generalized Weak Contraction in Modular Spaces

In this section, we prove fixed point theorems for mappings satisfying generalized weak contractions in modular spaces.Proposition 8.
Letρ be a modular space on X. If a,b∈ℝ+ with b≥a, then ρ(ax)≤ρ(bx).Proof.
In casea=b, we are done. Suppose b>a, and then one has a/b<1 and
(4)ρ(ax)=ρ(abbx)=ρ(abbx+(1-ab)(0))≤ρ(bx)+ρ(0)=ρ(bx).Proposition 9.
LetXρ be a modular space in which ρ satisfies the Δ2-condition and let {xn}n∈ℕ be a sequence in Xρ. If ρ(c(xn-xn-1))→0 as n→∞, then ρ(αl(xn-xn-1))→0 as n→∞, where c,l,α∈ℝ+ with c>l and l/c+1/α=1.Proof.
Sinceρ(c(xn-xn-1))→0  as n→∞, by the Δ2-condition, we get
(5)ρ(2mc(xn-xn-1))⟶0asn⟶∞,
for m∈ℕ. Using (5), Proposition 8, and the sandwich theorem, we conclude that
(6)ρ(2Nc(xn-xn-1))⟶0asn⟶∞,
for N∈ℕ. From the fact that l/c+1/α=1, we get αl=(α-1)c≥c, then there exist Nα∈ℕ such that
(7)2(Nα-1)c≤(α-1)c≤2(Nα)c.
By Proposition 8, we get
(8)ρ(2(Nα-1)c(xn-xn-1))≤ρ((α-1)c(xn-xn-1))≤ρ(2(Nα)c(xn-xn-1)).
From (6) and (8), we obtain
(9)limn→∞ρ(αl(xn-xn-1))=limn→∞ρ((α-1)c(xn-xn-1))=0.Theorem 10.
LetXρ be a ρ-complete modular space, where ρ satisfies the Δ2-condition. Let c,l∈ℝ+, c>l, and T:Xρ→Xρ be a mapping satisfying the inequality
(10)ψ(ρ(c(Tx-Ty)))≤ψ(ρ(l(x-y)))-ϕ(ρ(l(x-y))),
for all x,y∈Xρ, where ψ,ϕ:[0,∞)→[0,∞) are both continuous and monotone nondecreasing functions with ψ(t)=ϕ(t)=0 if and only if t=0. Then, T has a unique fixed point.Proof.
Letx0∈Xρ, and we construct the sequence {xn}n∈ℕ by xn=Txn-1, n=1,2,3,…. First, we prove that the sequence {ρ(c(Txn-Txn-1))} converges to 0. Indeed
(11)ψ(ρ(c(xn-xn+1)))≤ψ(ρ(l(xn-1-xn)))-ϕ(ρ(l(xn-1-xn)))≤ψ(ρ(l(xn-1-xn))).
By monotone nondecreasing of ψ and Proposition 8, we have
(12)ρ(c(xn-xn+1))≤ρ(l(xn-1-xn))≤ρ(c(xn-1-xn)).
This means that the sequence{ρ(c(xn-xn-1))} is monotone decreasing and bounded below. Hence there exists r≥0 such that
(13)limn→∞ρ(c(xn-xn-1))=r.
If r>0, taking n→∞ in the inequality (11), we get
(14)ψ(r)≤ψ(r)-ϕ(r)<ψ(r),
which is a contradiction, thus r=0. So we have
(15)ρ(c(xn-xn-1))→0asn→∞.
Next, we prove that the sequence {cxn}n∈ℕ is a ρ-Cauchy. Suppose that {cxn}n∈ℕ is not ρ-Cauchy, then there exist ε>0 and subsequence {xmk}, {xnk} with mk>nk≥k such that
(16)ρ(c(xmk-xnk))≥ε,ρ(c(xmk-1-xnk))<ε.
Now, let α∈R+ such that l/c+1/α=1, then we get
(17)ψ(ρ(c(xmk-xnk)))≤ψ(ρ(l(xmk-1-xnk-1)))-ϕ(ρ(l(xmk-1-xnk-1)))≤ψ(ρ(l(xmk-1-xnk-1)))
which implies that
(18)ρ(c(xmk-xnk))≤ρ(l(xmk-1-xnk-1)).
We have
(19)ρ(l(xmk-1-xnk-1))=ρ(l(xmk-1-xnk+xnk-xnk-1))=ρ(lcc(xmk-1-xnk)+1ααl(xnk-xnk-1))≤ρ(c(xmk-1-xnk))+ρ(αl(xnk-xnk-1))<ε+ρ(αl(xnk-xnk-1)).
By (16), (18), and (19), we get
(20)ε≤ρ(c(xmk-xnk))≤ρ(l(xmk-1-xnk-1))<ε+ρ(αl(xnk-xnk-1)).
Using (15) and Proposition 9, we have
(21)limk→∞ρ(αl(xnk-xnk-1))=0.
From (20) and (21), we obtain
(22)limk→∞ρ(c(xmk-xnk))=limk→∞ρ(l(xmk-1-xnk-1))=ε.
Lettingk→∞ in (17), by property of ψ and (22), we get
(23)ψ(ε)≤ψ(ε)-ϕ(ε)<ψ(ε)
which is a contradiction. Therefore, {cxn}n∈ℕ is ρ-Cauchy. Since Xρ is ρ-complete there exists a point u∈Xρ such that ρ(c(xn-u))→0 as n→∞. Consequently, ρ(l(xn-u))→0 as n→∞. Next, we prove that u is a unique fixed point of T. Putting x=xn-1 and y=u in (10), we obtain
(24)ψ(ρ(c(xn-Tu)))≤ψ(ρ(l(xn-1-u)))-ϕ(ρ(l(xn-1-u))).
Taking n→∞ in the inequality (24), we have
(25)ψ(ρ(c(u-Tu)))≤ψ(0)-ϕ(0)=0,
which implies that ρ(c(Tu-u))=0 and Tu=u. Suppose that there exists v∈Xρ such that Tv=v and v≠u, and then we have
(26)ψ(ρ(c(u-v)))=ψ(ρ(c(Tu-Tv)))≤ψ(ρ(l(u-v)))-ϕ(ρ(l(u-v)))<ψ(ρ(l(u-v)))≤ψ(ρ(c(u-v))),
which is a contradiction. Hence u=v and the proof is complete.Corollary 11.
LetXρ be a ρ-complete modular space, where ρ satisfies the Δ2-condition. Let c,l∈ℝ+, c>l, and T:Xρ→Xρ be a mapping satisfying the inequality
(27)ρ(c(Tx-Ty))≤ρ(l(x-y))-ϕ(ρ(l(x-y))),
for all x,y∈Xρ, where ϕ:[0,∞)→[0,∞) is continuous and monotone nondecreasing function with ϕ(t)=0 if and only if t=0. Then, T has a unique fixed point.Proof.
Takeψ(t)=t, and then we obtain the Corollary 11.Theorem 12.
LetXρ be a ρ-complete modular space, where ρ satisfies the Δ2-condittion and let T:Xρ→Xρ be a mapping satisfying the inequality
(28)ψ(ρ((Tx-Ty)))≤ψ(m(x,y))-ϕ(m(x,y))
for all x,y∈Xρ,where
(29)m(x,y)=max{12ρ(x-y),ρ(x-Tx),ρ(y-Ty),(ρ(12(x-Ty))+ρ(12(y-Tx)))(2)-1}
and ψ,ϕ:[0,∞)→[0,∞) are both continuous and monotone nondecreasing functions with ψ(t)=ϕ(t)=0 if and only if t=0. Then, T has a unique fixed point.Proof.
First, we prove that the sequence{ρ(c(Tnx-Tn-1x))} converges to 0. Since,
(30)ψ(ρ(Tnx-Tn-1x))≤ψ(m(Tn-1x,Tn-2x))-ϕ(m(Tn-1x,Tn-2x)).≤ψ(m(Tn-1x,Tn-2x)).
By monotone nondecreasing ofψ, we have
(31)ρ(Tnx-Tn-1x)≤m(Tn-1x,Tn-2x).
From the definition of m(x,y), we get
(32)m(Tn-1x,Tn-2x)=max{ρ(Tn-1x-Tn-2x)2ρ(Tn-1x-Tn-2x),ρ(Tnx-Tn-1x),ρ((1/2)(Tnx-Tn-2x))2}=max{ρ(Tn-1x-Tn-2x)2ρ(Tn-1x-Tn-2x),ρ(Tnx-Tn-1x),ρ(Tnx-Tn-1x)+ρ(Tn-1x-Tn-2x)2}=max{ρ(Tn-1x-Tn-2x),ρ(Tnx-Tn-1x)}.
If ρ(Tnx-Tn-1x)>ρ(Tn-1x-Tn-2x)≥0, then m(Tn-1x,Tn-2x)=ρ(Tnx-Tn-1x). Furthermore it is implied that
(33)ψ(ρ(Tnx-Tn-1x))≤ψ(m(Tn-1x,Tn-2x))-ϕ(m(Tn-1x,Tn-2x))≤ψ(ρ(Tnx-Tn-1x))-ϕ(ρ(Tnx-Tn-1x))<ψ(ρ(Tnx-Tn-1x))
which is a contradiction, and, hence,
(34)ρ(Tnx-Tn-1x)≤m(Tn-1x-Tn-2x)=ρ(Tn-1x-Tn-2x).
So, we have that the sequence {ρ(Tnx-Tn-1x)} is monotone decreasing and bounded below. Hence there exists r≥0 such that
(35)limn→∞ρ(Tnx-Tn-1x)=r.
If r>0, taking n→∞ in the inequality (30), we get
(36)ψ(r)≤ψ(r)-ϕ(r)<ψ(r)
which is a contradiction, and thus r=0. So, we have
(37)limn→∞ρ(Tnx-Tn-1x)=0.
Next, we prove that the sequence {Tn(x)}n∈ℕ is ρ-Cauchy. Suppose {Tn(x)}n∈ℕ is not ρ-Cauchy, and there exist ε>0 and sequence of integers {mk}, {nk} with {mk}>{nk}≥k such that
(38)ρ((Tmkx-Tnkx))≥ε,ρ(2(Tmk-1x-Tnkx))<ε.
Since,
(39)ψ(ρ((Tmkx-Tnkx)))≤ψ(m(Tmk-1x,Tnk-1x))-ϕ(m(Tmk-1x,Tnk-1x))≤ψ(m(Tmk-1x,Tnk-1x))
which implies that
(40)ρ((Tmkx-Tnkx))≤m(Tmk-1x,Tnk-1x).
On the other hand,
(41)m(Tmk-1x,Tnk-1x)=max{(12)ρ(Tmk-1x-Tnk-1x),ρ(Tmkx-Tmk-1x),ρ(Tnkx-Tnk-1x),(ρ(12(Tmkx-Tnk-1x))+ρ(12(Tmk-1x-Tnkx)))(2)-1},ρ((Tmk-1x-Tnk-1x))=ρ((Tmk-1x-Tnkx+Tnkx-Tnk-1x))≤ρ(2(Tmk-1x-Tnkx))+ρ(2(Tnkx-Tnk-1x))<ε+ρ(2(Tnkx-Tnk-1x)).
For the last term in m(Tmk-1x,Tnk-1x), by Proposition 8, we have
(42)(ρ(12(Tmkx-Tnk-1x))+ρ(12(Tmk-1x-Tnkx)))(2)-1=(ρ(12(Tmk-1x-Tnkx))+ρ(12(Tmkx-Tmk-1x+Tmk-1x-Tnkx)+12(Tnkx-Tnk-1x)))(2)-1≤(ρ(12(Tmk-1x-Tnkx))+ρ(Tmkx-Tmk-1x+Tmk-1x-Tnkx)+ρ(Tnkx-Tnk-1x)(12(Tmk-1x-Tnkx)))(2)-1≤(ρ(12(Tmk-1x-Tnkx))+ρ(2(Tmkx-Tmk-1x))+ρ(2(Tmk-1x-Tnkx))+ρ(Tnkx-Tnk-1x)(12(Tmk-1x-Tnkx)))(2)-1<ε+ρ(Tnkx-Tnk-1x)+ρ(2(Tmkx-Tmk-1x)2.
It follow from (41) and (42) that
(43)m(Tmk-1x,Tnk-1x)=max{(12(Tmk-1x-Tnkx))ρ(Tmk-1x-Tnk-1x),ρ(Tmkx-Tmk-1x),ρ(Tnkx-Tnk-1x),(ρ(12(Tmkx-Tnk-1x))+ρ(12(Tmk-1x-Tnkx)))(2)-1}<max{Tnk-12ε+ρ(2(Tnkx-Tnk-1x)),ρ(Tmkx-Tmk-1x),ρ(Tnkx-Tnk-1x),ε+ρ(Tnkx-Tnk-1x)+ρ(2(Tmkx-Tmk-1x)2}.
By (37), (38), (40), (43), and the Δ2-condition of ρ, we have
(44)limk→∞ρ((Tmkx-Tnkx))=limk→∞m(Tmk-1x,Tnk-1x)=ε.
Taking k→∞ in (39), by (44) and the continuity of ψ, we get
(45)ψ(ε)≤ψ(ε)-ϕ(ε)<ψ(ε)
which is a contradiction. Hence, {Tn(x)}n∈ℕ is ρ-Cauchy. Since Xρ is ρ-complete, there exists a point u∈Xρ such that ρ(Tnx-u)→0 as n→∞. Next, we prove that u is a unique fixed point of T. Suppose that Tu≠u, then ρ(u-Tu)>0.
Since,(46)ψ(ρ(Tnx-Tu))≤ψ(m(Tn-1x,u))-ϕ(m(Tn-1x,u)),(47)m(Tn-1x,u)=max{ρ((1/2))2ρ(Tn-1x-u),ρ(Tn-1x-Tnx),ρ(u-Tu),(ρ(12(Tn-1x-Tu))+ρ(12(u-Tnx)))(2)-1ρ((1/2))2}⟶max{0,0,ρ(u-Tu),ρ((1/2)(u-Tu))2}=ρ(u-Tu)asn⟶∞.
Taking n→∞ in (46), by using (47), we get
(48)ψ(ρ(u-Tu))≤ψ(ρ(u-Tu))-ϕ(ρ(u-Tu))<ψ(ρ(u-Tu))
which is a contradiction. Hence, ρ(u-Tu)=0 and Tu=u. If there exists point v∈Xρ such that Tv=v and u≠v, then using an argument similar to the above we get
(49)ψ(ρ(u-v))=ψ(ρ(Tu-Tv))≤ψ(m(u,v))-ϕ(m(u,v))≤ψ(ρ(u-v))-ϕ(ρ(u-v))<ψ(ρ(u-v))
which is a contradiction. Hence, u=v and the proof is complete.Corollary 13.
LetXρ be a ρ-complete modular space, where ρ satisfies the Δ2-condition, and let T:Xρ→Xρ be a mapping satisfying the inequality
(50)ρ((Tx-Ty))≤m(x,y)-ϕ(m(x,y))
for all x,y∈Xρ, where m(x,y)=max{ρ(x-y), ρ(x-Tx), ρ(y-Ty), (ρ((1/2)(x-Ty))+ρ((1/2)(y-Tx)))/2} and ϕ:[0,∞)→[0,∞) is continuous and monotone nondecreasing function with ϕ(t)=0 if and only if t=0. Then, T has a unique fixed point.Proof.
Takingψ(t)=t, we obtain the Corollary 13.



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