# Fixed Point Problems for Nonexpansive Mappings in Bounded Sets of Banach Spaces

Authors: Xianbing Wu
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

It is well known that nonexpansive mappings do not always have fixed points for bounded sets in Banach space. The purpose of this paper is to establish fixed point theorems of nonexpansive mappings for bounded sets in Banach spaces. We study the existence of fixed points for nonexpansive mappings in bounded sets, and we present the iterative process to approximate fixed points. Some examples are given to support our results.


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## Body

## 1. Introduction and Preliminaries

Throughout the paper, we assume thatX is a real Banach space and C⊂X is a subset. T:C⟶X is a mapping, if a point x∗∈C such that Tx∗=x∗ we call that x∗ is a fixed point of T. The set of fixed points of T is denoted by FT. Let N and N+ denote the set of natural numbers and the set of positive integers; let Q and R denote the set of rational numbers and the set of real numbers.We recall that nonexpansive mapping if for everyx,y∈C such that Tx−Ty≤x−y. A Banach space X is said to have the fixed point property if for every closed convex bounded subset C⊂X and for every nonexpansive T:C⟶C, there is a fixed point. Since Browder [1] obtained fixed point theorems for nonexpansive mapping, the fixed point theory of nonexpansive mappings has made great progress. A large number of results are obtained by authors (e.g., see [2–7]). Goebel et al. presented generalized nonexpansive mappings, if for every x,y∈C such that Tx−Ty≤ax−y+bx−Tx+Ty−y+cTx−y+y−Tx, where a+2b+2c≤1, a,b,c≥0. In recent years, generalizations of nonexpansive mappings have received attention, and their fixed point theory has been studied by many authors (see [8–14]).Amini-Harandi et al. [15] presented α,β-nonexpansive mappings and obtained fixed point theorems for α,β-nonexpansive mappings in Banach space.Definition 1.
LettingX be a Banach space, we say T:X⟶X is an α,β-nonexpansive mapping, if for every x,y∈X,α,β∈R we have
(1)Tx−Ty2≤αTx−y2+αTy−x2+βTx−x2+βTy−y2+1=2α−2βx−y2.Theorem 2.
LetX be a Banach space, C⊂X is a bounded subset, T:C⟶C is an α,β-nonexpansive mapping with α=0,β>0, then T has a fixed point.Remark 3.
Obviously,α,β-nonexpansive mappings are more generalized nonexpansive mappings, and many fixed point theorems are extended by Theorem 2, but the conditions of Theorem 2 are inadequate. See Example 1.Example 1.
Letting the setC=0,1, take Tx=1/2x.
We haveT:0,1⟶0,1 which is a mapping; moreover,
(2)Tx−Ty2=14x−y2≤12x−y2+12y−x2−x−y2.
So,T is a (1,0)-nonexpansive mapping in bounded set (0,1]. Hence, the conditions of Theorem 2 are satisfied, but T has no fixed point. In fact, it is easy that bounded subset C of conditions of Theorem 2 should be a closed bounded subset C.Remark 4.
IfC is a closed bounded set, T:C⟶C is an α,β-nonexpansive mapping, where α=0,β>0. Then, T has no fixed point. See Example 2.Example 2.
Let the setC=−1,1,
(3)Tx=1,x=0,−x,x≠0.Ifx=0 and for every y≠0, we have
(4)T0−Ty2=y+12≤2+22y2−30−y2=2T0−02+2Ty−y2−30−y2,if for all x,y∈C/0, we get
(5)Tx−Ty2=x−y2≤22x2+22y2−3x−y2=2Tx−x2+2Ty−y2−3x−y2.So,T is a (0,2)-nonexpansive mapping in [−1,1], which has no fixed point.Remark 5.
For anα,β-nonexpansive mapping in mapping, as α=0 and β=0, the α,β-nonexpansive mapping is the class of nonexpansive mapping. It is important about fixed point theory of the class of nonexpansive mapping, but which is not contained by Theorem 2.The following examples show that for some nonclosed bounded sets, nonexpansive mappings have fixed points in Banach space, but for some closed bounded sets, nonexpansive mappings have no fixed point.Example 3.
Let the setC=−1,1, and take Tx=−x.
It is easy thatT:C⟶C is a nonexpansive mapping and C is a closed bounded set. But T has not fixed point in C.Example 4.
Let the setC=0,1
(6)Tx=14,x=1,78,x≠1.
It is obvious thatT is not a nonexpansive mapping in closed bounded set C; however, it is a nonexpansive mapping in nonclosed bounded set 0,1, and there exists a fixed point 7/8∈0,1.Example 5.
Dirichlet function:(7)Dx=1,x∈0,1∩Q,0,x∈0,1Q.
Obviously,D:0,1⟶0,1 is not a nonexpansive mapping, but D is a nonexpansive mapping in bounded set 0,1∩Q, and 1∈0,1∩Q is a fixed point of D.Usually, authors study fixed point problems of nonexpansive mappings for closed bounded sets. The above examples show that for some bounded sets, nonexpansive mappings have fixed points in Banach space, and for others of bounded sets, nonexpansive mappings have no fixed point. So it is significant to study fixed point problems of nonexpansive mappings for bounded sets. The goal of this paper is to obtain fixed point theorems of nonexpansive mappings for bounded sets.


## 2. Main Results

Letl∞ denote the Banach space of bounded real sequence with the supremum norm. Let ψ be a bounded linear functional on l∞; ψ is called a Banach limit, if it satisfies ψ=ψ1=1 and ψtn+1=ψtn, tn∈l∞. Moreover, suppose ψ is a Banach limit, then the following conclusions are held:
(i)
If for alln∈N, sn≤tn means ψsn≤ψtn, sn,tn∈l∞
(ii)
For eachp∈N+,tn∈l∞, have ψtn=ψtn+p
(iii)
liminfn→∞tn≤ψtn≤limsupn→∞tn, tn∈l∞Lemma 6.
[16] Suppose that an,bn are two sequences of nonnegative numbers, and ∑bn<∞, if there exists some number N0∈N, for all n≥N0 such that
(8)an+1≤an+bn,then limn→∞an exists.Theorem 7.
LettingX be a Banach space, C⊂X is a bounded subset and T:C⟶C is a nonexpansive mapping. If the following two conditions are satisfied:
(i)
There exists a sequencesn⊂0,1 with limn→∞sn=1, ∑sn+p−sn/1−sn<∞, for all n,p∈N,x∈C, such that snx∈C.
(ii)
If the sequencexn⊂C and limk→∞xnk=x∗, we have x∗∈C, where k⟶∞ implies nk⟶∞.
Then, (1) there is at least a fixed point ofT in C; (2) take xn=sn−1Txn−1, then the iterative sequence xn approach to a fixed point of T.Proof.
Letx0∈C, take xn=sn−1Txn−1, where the sequence sn⊂0,1 satisfies the condition (i) of Theorem 7, that is, limn→∞sn=1 and ∑sn+p−sn/1−sn<∞, for each n,p∈N.
Firstly, we show that there exists a subsequenceχnk of χn which is convergent.
SinceT:C⟶C is a nonexpansive mapping, for each n,p∈N+, we have
(9)xn+p−xn=sn+p−1Txn+p−1−sn−1Txn−1=sn+p−1Txn+p−1−sn−1Txn+p−1+sn−1Txn+p−1−sn−1Txn−1≤sn+p−1−sn−1Txn+p−1+sn−1Txn+p−1−Txn−1≤sn+p−1−sn−1Txn+p−1+sn−1xn+p−1−xn−1.
According to the Banach limit, we have(10)ψxn+p−xn≤ψsn+p−1−sn−1Txn+p−1+sn−1xn+p−1−xn−1=sn+p−1−sn−1ψTxn+p−1+sn−1ψxn+p−1−xn−1=sn+p−1−sn−1ψTxn+p−1+sn−1ψxn+p−xn.
Also, forC is a bounded set, which means that there exists a constant M>0 such that for every x∈C, we have x≤M. Thus, by the inequality (10), we have
(11)ψxn+p−xn≤sn+p−1−sn−11−sn−1ψTxn+p−1≤sn+p−1−sn−11−sn−1ψM.
From the condition (ii) of Theorem7, we have
(12)limn→∞sn+p−1−sn−11−sn−1=0.
Letn→∞ in the inequality (11), by (12), we get
(13)limn→∞ψxn+p−xn=0,that is
(14)limn→∞ψxm−xn=0,m,n∈N.
So(15)0≤liminfn→∞xm−xn≤ψxm−xn=0,which implies that there exists a monotonically increasing sequence nk (that is, n1≤n2≤n3⋯) such that
(16)limk→∞xmk−xnk=0,m,n∈N.
It means thatxnk is a Cauchy sequence; thus, there exists some x∗ such that
(17)limk→∞xnk=x∗.
From the condition (ii) of Theorem7, we have x∗∈C.
Next, we show thatxn is convergent to x∗ and x∗∈FT.
From (9), we have
(18)xn+p−xn≤xn+p−1−xn−1+sn+p−1−sn−1Txn+p−1.
Also, by the condition (i) of Theorem7, we may obtain ∑sn+p−sn<∞. Thus, applying Lemma 6 in (18), limn→∞xn+p−xn exists; moreover, from (16), limk→∞xmk−xnk=0, m,n∈N. So it implies
(19)limn→∞xm−xn=0,m,n∈N.
Hence,xn is a Cauchy sequence, also from (17), that is, limk→∞xnk=x∗. Therefore,
(20)limk→∞xn=x∗.
SinceT is nonexpansive, then
(21)x∗−Tx∗=x∗−xn+xn−Tx∗≤x∗−xn+xn−Tx∗=x∗−xn+sn−1Txn−1−sn−1Tx∗−1−sn−1Tx∗≤x∗−xn+sn−1Txn−1−Tx∗+1−sn−1Tx∗≤x∗−xn+sn−1xn−1−x∗+1−sn−1Tx∗.
According to (20) and the condition (i) of Theorem 7, letting n→∞ in the inequality (21), we have x∗−Tx∗=0; that is, Tx∗=x∗; therefore, x∗ is a fixed point of T.Corollary 8.
LetX be a Banach space, C⊂X is a bounded closed subset, and T:C→C is a nonexpansive mapping. If there exists the sequence sn⊂0,1 with limn→∞sn=1, ∑sn+p−sn/1−sn<∞, for all n,p∈N,x∈C, such that snx∈C.
Then, (1) there is at least a fixed point ofT in C; (2) taking xn=sn−1Txn−1, then the iterative sequence xn approaches to a fixed point of T.Let us use the following example to support the above results.Example 6.
Let the setC=±1/nn∈N∪0
(22)Tx=0,x=0,−x,x≠0.If everyx∈C,y=0, we have
(23)Tx−T0=x=x−0.Moreover, if every x,y∈±1/nn∈N, we get
(24)Tx−Ty=x−y.Obviously,C is a bounded closed set and T:C⟶C is a nonexpansive mapping. And for all x∈C, that is, for all n, then x=±1/n, or x=0, so there exists the sequence sn=n/n+1, which satisfies the following three conditions:
(a)
limn→∞sn=1
(b)
∑sn+p−sn/1−sn<∞,n,p∈N
(c)
snx∈CThus, all conditions of Corollary8 are satisfied; therefore, T has at least a fixed point. Next, we use the iterative approximation methods to obtain a fixed point of T. Now, we take x0=1∈C and xn+1=snTxn, then we easily obtain
(25)x1=−12,x2=13,⋯,xn=−1n−11n.Lettingn⟶∞, we have limn→∞xn=0, in which 0 is a fixed point of T.The following results are extensions of fixed point theorems of nonexpansive mappings for convex bounded sets.Theorem 9.
LetX be a Banach space, C⊂X is a bounded subset, and T:C⟶C is a nonexpansive mapping. If the following two conditions are satisfied:
(i)
There exists somex0∈C and the sequence sn⊂0,1 with limn→∞sn=1 and ∑sn+p−sn/1−sn<∞, for all n,p∈N,x∈C, such that 1−snx0+snx∈C
(ii)
If the subsequencexnk⊂xnk⊂C and limk→∞xnk=x∗, we havex∗∈C, where k⟶∞ implies nk⟶∞
Then, (1) there is at least a fixed point ofT in C; (2) taking iterative sequence xn=1−sn−1x0+sn−1Txn−1, the sequence xn approaches to a fixed point of T.Proof.
From the condition (i) of Theorem9, there exists some x0∈C and the sequence sn∈0,1, for each x∈C, such that 1−snx0+snx∈C, where the sequence sn⊂0,1 satisfies limn→∞sn=1 and ∑sn+p−sn/1−sn<∞, for all n,p∈N. Now, we take xn=1−sn−1x0+sn−1Txn−1, which implies that xn∈C; moreover, since T:C⟶C is a nonexpansive mapping, thus we have
(26)xn+p−xn=1−sn+p−1x0+sn+p−1Txn+p−1−1−sn−1x0−sn−1Txn−1=sn+p−1−sn−1x0+sn+p−1Txn+p−1−sn−1Txn+p−1+sn−1Txn+p−1−sn−1Txn−1≤sn+p−1−sn−1x0+sn+p−1−sn−1Txn+p−1+sn−1Txn+p−1−Txn−1≤sn+p−1−sn−1x0+Txn+p−1+sn−1xn+p−1−xn−1.
Based on the Banach limit, we have(27)ψxn+p−xn≤ψsn+p−1−sn−1x0+Txn+p−1+sn−1xn+p−1−xn−1=sn+p−1−sn−1ψx0+Txn+p−1+sn−1ψxn+p−1−xn−1=sn+p−1−sn−1ψx0+Txn+p−1+sn−1ψxn+p−xn.
SinceC is a bounded set, then there exists a constant M>0 for each x∈C such that x≤M. So by (27), we may get
(28)ψxn+p−xn≤sn+p−1−sn−11−sn−1ψx0+Txn+p−1≤sn+p−1−sn−11−sn−1ψ2M.
From the condition (ii) of Theorem9, we have
(29)limn→∞sn+p−1−sn−11−sn−1=0,
Hence, asn⟶∞ in (28), we get
(30)limn→∞ψxn+p−xn=0,that is
(31)limn→∞ψxm−xn=0,m,n∈N.
This implies that(32)0≤liminfn→∞xm−xn≤ψxm−xn=0.
Hence, there exists a monotonically increasing sequencenk (that is, n1≤n2≤n3⋯) such that
(33)limn→∞xmk−xnk=0,m,n∈N.
It means thatxnk is a Cauchy sequence, so there exists some x∗ such that
(34)limk→∞xnk=x∗.
From the condition (ii) of Theorem9, we have x∗∈C.
Next, we show thatxn is convergent to x∗ and x∗ is a fixed point of T.
From (26), we have
(35)xn+p−xn≤xn+p−1−xn−1+sn+p−1−sn−1Txn+p−1+x0.
Also, by the condition (i) of Theorem9, we obtain ∑sn+p−sn<∞. Thus, applying Lemma 6 in (35), we have limn→∞xn+p−xn; moreover, from (33), limk→∞xmk−xnk=0,m,n∈N. It implies
(36)limn→∞xm−xn=0,m,n∈N.
Soxn is a Cauchy sequence, also from (34), that is, limk→∞xnk=x∗. Therefore,
(37)limk→∞xn=x∗.
By the triangle inequality, we have(38)x∗−Tx∗=x∗−xn+xn−Tx∗≤x∗−xn+xn−Tx∗≤x∗−xn+sn−1Txn−1+1−sn−1x0−sn−1Tx∗−1−sn−1Tx∗≤x∗−xn+sn−1Txn−1−Tx∗+1−sn−1Tx∗−x0≤x∗−xn+sn−1xn−1−x∗+1−sn−1Tx∗−x0.
Applying (37) and the condition (i) of Theorem 9 and letting n⟶∞ in (38), we have x∗−Tx∗=0. That is, Tx∗=x∗; therefore, x∗ is a fixed point of T.Corollary 10.
LetX be a Banach space, C⊂X is a bounded closed subset, and T:C⟶C is a nonexpansive mapping. If there exists some x0∈C and sequence sn⊂0,1 with limn→∞sn=1, and ∑sn+p−sn/1−sn<∞, for all n,p∈N,x∈C, such that
(39)1−snx0+snx∈C.
Then, (1) there is at least a fixed point ofT in C; (2) taking iterative sequence xn=1−sn−1x0+sn−1Txn−1, the sequence xn approaches to a fixed point of T.Definition 11.
LettingX be a Banach space and C⊂X be a subset, if there exists a x0∈C such that for every y∈C,λ∈0,1, we have
(40)1−λx0+λy∈C,then, C is a called a star-shaped set.Obviously, a star-shaped set satisfies the conditions of which setC of Corollary 10, so we haveCorollary 12.
LetX be a Banach space, C⊂X is a bounded closed star-shaped set, and T:C⟶C is a nonexpansive mapping. Then, T has at least a fixed point in C.Corollary 13.
LetX be a Banach space, C⊂X is a bounded closed convex set, and T:C⟶C is a nonexpansive mapping. Then, T has at least a fixed pont in C.Let us give the example to support our results.Example 7.
Let the setC=0,1∩Q, take Tx=−x+1,x∈C.It means thatC be a bounded set and T:C⟶C is a nonexpansive mapping. Now, we take x0=1 and a sequence sn=n/n+1, we have sn∈0,1, and for all n,p∈N+ such that
(41)limn→∞sn=1,∑sn+p−sn1−sn<∞.Moreover, for allx∈C, we have 1−snx0+snx∈C.As an overview, all conditions of Theorem9 are satisfied. So T has a fixed point.Next, we take(42)xn+1=1−snx0+snTxn,then we have
(43)x1=12,x2=23,x3=12,x4=35,x5=12,x6=47,⋯So we have(44)xn=12,n=2k+1,k∈N,n+22n+2,n=2k,k∈N.It impliesxn∈C, and limn→∞xn=1/2∈C. It is clear that 1/2 is a fixed point of T.Therefore, a fixed point ofT be iteratively approximated by the sequence xn.From Remark 2.1 of the reference [15], a subset C of Banach space X,T:C⟶X is an (α,β)-nonexpansive mapping; if α+β<0, then T has a fixed point. So by Theorem 7 and Theorem 9, we easily obtain the following two results.Theorem 14.
LetX be a Banach space, C⊂X is a bounded subset, and T:C⟶C is an (α,β)-nonexpansive mapping with α≤0,β≤0. If the following two conditions are satisfied:
(i)
There exists a sequencesn⊂0,1 with limn→∞sn=1 and ∑sn+p−sn/1−sn<∞, for all n,p∈N,x∈C such that snx∈C
(ii)
If the sequencexn⊂C and limk→∞xnk=x∗, we have x∗∈C, where k⟶∞ implies nk⟶∞
Then, (1) there is at least a fixed point ofT in C; (2) taking xn=sn−1Txn−1, the iterative sequence xn approaches to a fixed point of T.Theorem 15.
LetX be a Banach space, C⊂X is a bounded subset, and T:C⟶C is an (α,β)-nonexpansive mapping with α≤0,β≤0. If the following two conditions are satisfied:
(i)
There exists somex0∈C and the sequence sn⊂0,1 with limn→∞sn=1 and ∑sn+p−sn/1−sn<∞, for all n,p∈N,x∈C, such that 1−snx0+snx∈C
(ii)
If the subsequencexnk⊂xnk⊂C and limk→∞xnk=x∗, we have x∗∈C, where k⟶∞ implies nk⟶∞
Then, (1) there is at least a fixed point ofT in C; (2) taking iterative sequence xn=1−sn−1x0+sn−1Txn−1, the sequence xn approaches to a fixed point of T.



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