# On the Fixed Points of Correspondences in Modular Spaces

Authors: Fatemeh Lael; Kourosh Nourouzi
Journal: ISRN Geometry
 (2011)

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## Abstract

We give two fixed point results for contractive and nonexpansive
correspondences defined on modular spaces.


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## Body

## 1. Introduction and Preliminaries

A modular on a real linear spaceX is a real functional ρ on X which satisfies the conditions:(1)
ρ(x)=0 if and only if x=0,(2)
ρ(x)=ρ(-x),(3)
ρ(αx+βy)≤ρ(x)+ρ(y), for all x,y∈X and α,β≥0, α+β=1.Then(X,ρ) is called a modular space. Given a modular ρ, a corresponding vector space Xρ is given as (1.1)Xρ={x∈X:lim⁡λ→0ρ(λx)=0},
which is called modular linear space.It is easy to see that for every modularρ if α,β∈ℝ+ and α≤β, then ρ(αx)≤ρ(βx), for all x∈X.The theory of modular spaces was initiated by Nakano [1] in connection with the theory of ordered spaces. Musielak and Orlicz [2] redefined and generalized the notion of a modular space in order to obtain a generalization of the classical function spaces Lp. Even if a metric is not defined, many results in metric fixed point theory can be reformulated in modular spaces, we refer, for instance, to [3–5].In this work, we give some results on the existence of fixed points for contractive and nonexpansive correspondences on modular spaces.We first recall some basic concepts of modular spaces. We refer to [6, 7] for more details on modular spaces.Definition 1.1.
LetB be a subset of a modular space (X,ρ).  (1)
A sequence{xn} in X is said to be convergent to a point x∈X and denoted by xn→x, if for every ϵ>0, there is a positive integer N such that ρ(xn-x)<ϵ, for all n>N.(2)
Theclosure of B is denoted by B¯ and defined as the set of all x∈X such that there is a sequence {xn} of B which is convergent to x. We say that B is closed if B=B¯.(3)
A sequence{xn} in X is said to be Cauchy, if for every ϵ>0, there is a positive integer N such that ρ(xn-xm)<ϵ, for all m,n>N.(4)
B is said to be complete if each Cauchy sequence in B is convergent to a point of B.(5)
B is said to be compact if every sequence in B has a convergent subsequence in B.(6)
B is called sequentially bounded, if for each {xn}⊂B and each real sequence {ϵn} converging to zero we have ϵnxn→0, as n→∞.
By a correspondencef from a set X to a set Y we mean a relation that assigns to each x in X a nonempty subset f(x) of Y. For any subset C of X and correspondence f:C↠X, an element x∈C is said to be a fixed point if x∈f(x). Also, f(C)=⋃c∈Cf(c).Definition 1.2.
LetC be a subset of X and let k∈[0,1). We say that f:C↠C is a k-contraction if for each x,y∈C and p∈f(x) there is q∈f(y) which satisfies the condition:
(1.2)ρ(p-q)≤kρ(x-y).Definition 1.3 (see [8]).
For a modular space(X,ρ), the function wρ which is called growth function is defined on [0,∞) as follows:
(1.3)wρ(t)=inf⁡{w:ρ(tx)≤wρ(x):x∈X,0<ρ(x)}.


## 2. Main Results

In the sequel, it is assumed thatB is a closed subset of complete modular space (X,ρ) and f:B↠B is a correspondence with compact values.Lemma 2.1.
Let(X,ρ) be a modular space satisfying wρ(2)<∞. Then every convergent sequence in (X,ρ) is a Cauchy sequence in (X,ρ).Proof.
Let{xn} be a convergent sequence in (X,ρ). Then for ϵ>0, there exists k∈ℕ such that
(2.1)ρ(xn-x)<ϵ2wρ(2),
for all n>k. Thus
(2.2)ρ(xn+r-xn)≤wρ(2)ρ(xn+r-x)+wρ(2)ρ(xn-x)<ϵ2+ϵ2<ϵ,
for every n>k, r∈ℕ. Hence {xn} is a Cauchy sequence.Definition 2.2.
LetC be a subset of a modular space X and let ϵ>0 be given. A set Mϵ⊆X is called an ϵ-net for C if for every point z∈C there is a point x of Mϵ such that ρ(z-x)<ϵ. The set C is said to be totally bounded if for every ϵ there is a finite ϵ-net for C.By a relatively sequentially compact set in a modular space(X,ρ) we mean that its closure is sequentially compact.The following lemma is a counterpart of totally boundedness in metric spaces and has a same argument which is omitted here.Lemma 2.3.
Let A be a subset of a modular spaceX satisfying wρ(2)<∞. Then,  (1)
if A is totally bounded and X is complete, each sequence in A has a convergent subsequence inX;(2)
if A is relatively sequentially compact, A is totally bounded.Let(X,ρ) be a modular space and let A be a nonempty subset of X. The diameter of A is defined by
(2.3)D(A)=sup⁡a,b∈Aρ(a-b),
and the set A is bounded if and only if D(A)<∞.Definition 2.4.
Let(X,ρ) be a modular space and let A be a bounded subset of X. Then the Kuratowski constant α(A) of the set A is defined as the greatest upper bound for ϵ>0 that A can be covered with a finite number of sets of diameter less than ϵ.Theorem 2.5.
Everyk-contraction f:B↠B with wρ(2)2k<1/2 has a fixed point.Proof.
Choosep0∈B and p1∈f(p0). Since f is k-contraction, there exists p2∈f(p1) for which
(2.4)ρ(p1-p2)≤kρ(p0-p1).
By induction, there exists a sequence {pn} in B such that
(2.5)pn+1∈f(pn),ρ(pn+1-pn)≤kρ(pn-pn-1).
Put P={pn:n∈ℕ} and choose l∈(2kwρ(2)2,1). We claim that
(2.6)α(f(P))≤lα(P).
To see this, let r′>r>0 and lα(P)<r. By Definition 2.4, there exists a finite number of sets Ni(i∈{1,2,…,n}) with diameter less than r/l which covers {pn}. Let xi∈Ni. By part (2) of Lemma 2.3, the totally boundedness of f(xi) implies that for
(2.7)ϵ=r2wρ(2)2(1-2kwρ(2)2l),
there exists a finite ϵ-net Mϵ,i⊆B for f(xi). Let Mϵ,i={z1i,z2i,…,zn(i)i} and Z=⋃i=1nMϵ,i. We will show that
(2.8)f(P)⊆⋃z∈ZB(z),D(B(z))<r,
where
(2.9)B(z)={x∈B:ρ(x-z)<r2wρ(2)}.
Let y∈f(pn), for some n∈ℕ. Since P⊆⋃i=1nNi, then pn∈Ni for some i. If pn,xi∈Ni, y∈f(pn) there exists x∈f(xi) such that ρ(y-x)≤kρ(xi-pn). Also there exists zji(j∈{1,2,…,n(i)}) such that
(2.10)ρ(x-zji)<r2wρ(2)2(1-2kwρ(2)2l).
Therefore,
(2.11)ρ(y-zji)≤wρ(2)ρ(y-x)+wρ(2)ρ(x-zji)≤wρ(2)kρ(pn-xi)+wρ(2)ρ(x-zji)<wρ(2)krl+rwρ(2)2wρ(2)2(1-2kwρ(2)2l)<r2wρ(2).
It implies that f(P)⊆⋃z∈PB(z). Also for each z′,z′′∈B(z), we have
(2.12)ρ(z′-z′′)≤wρ(2)ρ(z′-z)+wρ(2)ρ(z′′-z)≤wρ(2)r2wρ(2)+wρ(2)r2wρ(2)<r,
therefore, D(B(z))<r′, that is, α(f(P))<r′. Hence α(f(P))≤lα(P). Since for each n, pn+1∈f(pn), we get α(P)≤α(f(P)). Consequently, for every n∈ℕ(2.13)α(P)≤α(f(P))≤lα(P)≤lα(f(P))≤⋯≤lnα(P).
It implies that α(P)=0 and hence P is totally bounded. By Lemma 2.3, {pn} has a convergent subsequence {pnl}. Let liml→∞pnl=p. We have
(2.14)ρ(pnl+1-pnl)≤knlρ(p1-p0),
also
(2.15)ρ(pnl+1-p)≤wρ(2)ρ(pnl+1-pnl)+wρ(2)ρ(pnl-p),
therefore, liml→∞ρ(pnl+1-p)=0. On the other hand, for each l, there exists an xnl∈f(p) such that
(2.16)ρ(pnl+1-xnl)≤kρ(pnl-p),
therefore, liml→∞ρ(pnl+1-xnl)=0. Since f(p) is a compact set, {xnl} has a convergent subsequence. Let limi→∞xnli=x. Given ϵ>0, there exists i0∈ℕ such that for i>i0,
(2.17)ρ(x-p3)≤ρ(x-xnli)+ρ(xnli-pnli+1)+ρ(pnli+1-p)<ϵ.
It shows that x=p which completes the proof.Definition 2.6.
A sequence{yn} in C is said to be approximate fixed point sequence of f:C↠C, if for every n∈ℕ there exists pn∈f(yn) such that ρ(yn-pn)→0, as n→∞.
We recall that a subsetC of a vector space is called star shaped, if there exists z∈C (the center of C) such that λx+(1-λ)z∈C, for every x∈C, and λ∈[0,1].Lemma 2.7.
Suppose thatρ satisfies wρ(β)wρ2(2)<1/2 for every β∈(0,1), B is star shaped with the center z, and f:B↠B is nonexpansive, that is, for each x,y∈B and p∈f(x) there exists q∈f(y) such that
(2.18)ρ(p-q)≤ρ(x-y).
Then, (1)
for everyα,β∈ℝ+, where α+β=1, there exist p0∈B such that p0∈αz+βf(p0);(2)
iff(B) is sequentially bounded, then f has an approximate fixed point sequence.Proof.
(1) Letα,β∈ℝ+, with α+β=1. We define g:B↠B by g(x)=αz+βf(x). If x,y∈B and p∈f(x), there exists q∈f(y) such that ρ(p-q)≤ρ(x-y). This implies that αz+βq∈g(y) and
(2.19)ρ(αz+βp-αz-βq)=ρ(β(p-q))≤wρ(β)ρ(x-y).
Since wρ(β)wρ2(2)<1/2, Theorem 2.5 implies that g has a fixed point.
(2) Let{kn:n∈ℕ}⊂(0,1), kn→1. By part (1), for each n∈ℕ, there exist yn∈B and pn∈f(yn) such that yn=(1-kn)z+knpn. Since f(B) is sequentially bounded, ρ(2(1-kn)z)→0 and ρ(2(1-kn)pn)→0, as n→∞. Therefore,
(2.20)ρ(yn-pn)≤ρ(2(1-kn)pn)+ρ(2(1-kn)z)⟶0,
as n→∞, that is, f has an approximate fixed point sequence.Theorem 2.8.
LetXρ be a complete modular linear space, and let B be a compact and star shaped subset of Xρ. If ρ satisfies wρ(β)wρ2(2)<1/2  for every β∈(0,1), and f:B↠B is nonexpansive, then f has a fixed point.Proof.
First, we show thatB is sequentially bounded. To see this, if λk's are real numbers converging to zero and {pk}⊆B, then every subsequence of {λkpk} has a convergent subsequence to zero. Choose subsequence {λknpkn} of {λkpk}. Since B is relatively compact, there exist p∈Xρ and a subsequence {kni} of {kn} such that ρ(pkni-p)→0 as i→∞. Taking i so large that 2λkni<1, we obtain
(2.21)ρ(λnkipnki)≤ρ(2λnki(pnki-p))+ρ(2λnkip)≤ρ(pnki-p)+ρ(2λnkip).
Therefore, limi→∞λnkipnki=0. It implies that limk→∞λkpk=0. Otherwise there exists ϵ>0 and a subsequence {λkjpkj} of {λkpk} such that ρ(λkjpkj)>ϵ, for all j. This contradicts the fact that {λkjpkj} has a convergent subsequence to zero.
By part (2) of Lemma2.7, f has an approximate fixed point sequence, that is, there exist {yn} and {pn} in B such that pn∈f(yn) and ρ(pn-yn)→0. The sequences {pn} and {yn} have convergent subsequences {pnk} and {ynk}, say limk→∞pnk=p, and limk→∞ynk=y. Since
(2.22)ρ(p-y3)≤ρ(p-pnk)+ρ(pnk-ynk)+ρ(ynk-y),
so y=p. The nonexpansivity of f implies that for each k there exists znk∈f(y) such that
(2.23)ρ(znk-pnk)≤ρ(ynk-y).
Since f(y) is compact, consider a convergent subsequence of {znk}, znki→z. Again,
(2.24)ρ(y-z3)≤ρ(y-pnki)+ρ(pnki-znki)+ρ(znki-z)≤ρ(y-pnki)+ρ(ynki-y)+ρ(znki-z).
As i→∞, we get p∈f(p).



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