# Convergence to Common Fixed Point for Generalized Asymptotically Nonexpansive Semigroup in Banach Spaces

Authors: Yali Li; Jianjun Liu; Lei Deng
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

LetK be a nonempty closed convex subset of a reflexive and strictly convex Banach
space E with a uniformly Gâteaux differentiable norm, ℱ={T(h):h≥0} a generalized asymptotically
nonexpansive self-mapping semigroup of K, and f:K→K a fixed contractive mapping
with contractive coefficient β∈(0,1). We prove that the following implicit and modified implicit
viscosity iterative schemes {xn} defined by xn=αnf(xn)+(1−αn)T(tn)xn and xn=αnyn+(1−αn)T(tn)xn,yn=βnf(xn−1)+(1−βn)xn−1 strongly converge to p∈F as n→∞ and p is the unique solution to the following variational
inequality: 〈f(p)−p,j(y−p)〉≤0 for all y∈F.


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## Body

## 1. Introduction

LetC be  a closed convex   subset of a Hilbert space H and T a nonexpansive mapping from C into itself. We denote by F(T) the set of fixed points of T.
Let F(T) be nonempty and u an element of C.
For each t with 0<t<1,
let  xt∈C be the unique fixed point of the contraction x↦tu+(1−t)Tx. Browder [1] showed that {xt} defined by xt=tu+(1−t)Txt converges strongly to the element of F(T) which is nearest to u in F(T) as t→0.In 2004, for a contractionf:C→C and a nonexpansive mapping T:C→C,
Xu [2] proposed the  following viscosity approximation method in Banach space:(1.1)xt=tf(xt)+(1−t)Txt,t∈(0,1),t→0,and Song and Xu [3] studied the
convergence of the following implicit viscosity iterative scheme:(1.2)xn=αnf(xn)+(1−αn)T(tn)xn,where {αn}⊂(0,1).On the other hand, for a fixed Lipschitz strongly
pseudocontractive mappingf and a continuous pseudocontractive mapping T,
Song and Chen [4] proposed the following motivated implicit viscosity iterative
scheme:(1.3)xn=αnyn+(1−αn)Txn,yn=βnf(xn−1)+(1−βn)xn−1.In this paper, we will still study the implicit
viscosity iterative scheme (1.2) and propose the following iterative scheme:(1.4)xn=αnyn+(1−αn)T(tn)xn,yn=βnf(xn−1)+(1−βn)xn−1,where {T(h):h≥0} is a generalized asymptotically nonexpansive
self-mappings semigroup and f a fixed contractive mapping with contractive
coefficient β∈(0,1).


## 2. Preliminaries

Throughout this
paper, we assume thatE is a Banach space and K a nonempty closed convex subset of E.
Let E* be a dual space of E, J:E→2E* the normalized duality mapping defined by(2.1)J(x)={f∈E*,〈x,f〉=∥x∥⋅∥f∥,∥x∥=∥f∥},where 〈⋅,⋅〉 denotes the generalized duality pairing.Definition 2.1 (see {[5]}).
A mappingT:E→E is said to be
total asymptotically nonexpansive if there exist nonnegative real sequences {kn(1)} and {kn(2)},n>0,
with kn(1) and kn(2)→0 as n→∞,
and strictly increasing and continuous functions ϕ:R+→R+ with ϕ(0)=0 such that(2.2)∥Tnx−Tny∥≤∥x−y∥+kn(1)ϕ(∥x−y∥)+kn(2)∀x,y∈K.Remark 2.2.
Ifϕ(λ)=λ,
the total asymptotically nonexpansive mapping coincides with generalized
asymptotically nonexpansive mapping. In addition, for all n∈ℕ,
if kn(2)=0,
then generalized asymptotically nonexpansive mapping coincide with
asymptotically nonexpansive mapping; if kn(1)=0,kn(2)=max{0,pn},
where pn:=supx,y∈K(∥Tnx−Tny∥−∥x−y∥),
then generalized asymptotically nonexpansive mapping coincide with
asymptotically nonexpansive mapping in the intermediate sense; if kn(1)=0 and kn(2)=0,
then we obtain from (2.2) the class of nonexpansive mapping.Remark 2.3.
In [5], for
the total asymptotically nonexpansive mapping, the authors assume that there
exist M,M*>0 such that ϕ(λ)≤M*λ for all λ≥M,
so for M0=max{ϕ(M),M*}, ϕ(λ)≤M0(1+λ) for all λ≥0,
then the total asymptotically nonexpansive mapping studied by [5] coincides
with generalized asymptotically nonexpansive mapping.A (one-parameter) generalized asymptotically
nonexpansive semigroup is a familyℱ={T(h):h≥0} of self-mapping of K such that(i)
T(0)x=x for x∈K;(ii)
T(s+t)x=T(s)T(t)x for t,s≥0 and x∈K;(iii)
limt→0T(t)x=x for x∈K;(iv)
for eachh≥0,T(h) is generalized asymptotically nonexpansive,
that is,(2.3)∥T(h)x−T(h)y∥≤(1+kh(1))∥x−y∥+kh(2)∀x,y∈K.We will denote byF the common fixed point set of ℱ,
that is,(2.4)F:=Fix(ℱ)={x∈K:T(h)x=x,h≥0}=⋂h≥0Fix(T(h)).Definition 2.4.
A Banach spaceE is said to be strictly convex if ∥x+y∥/2<1 for ∥x∥=∥y∥=1 and x≠y.Definition 2.5.
LetU={x∈E:∥x∥=1},
the norm of E is said to be uniformly Ga^teaux differentiable, if for each y∈U, limt→0((∥x+ty∥−∥x∥)/t) exists uniformly for x∈U.Definition 2.6.
Letμ be a continuous   liner functional on l∞ and let (a0,a1,…)∈l∞.
One writes μn(an) instead of μ((a0,a1,…)).  One  calls μ a Banach limit when μ satisfies ∥μ∥=μn(1)=1 and μn(an+1)=μn(an) for each (a0,a1,…)∈l∞.For a Banach limitμ,
one knows that lim¯n→∞an≤μn(an)≤lim¯n→∞an for every a=(a0,a1,…)∈l∞.
So if a=(a0,a1,…)∈l∞,b=(b0,b1,…)∈l∞ and an−bn→0 as n→∞,
one has μn(an)=μn(bn).Definition 2.7.
LetK be a nonempty closed convex subset of a Banach
space E, ℱ={T(h):h≥0} a continuous operator semigroup on K.
Then ℱ is said to be uniformly asymptotically regular
(in short, u.a.r.) on K if for all h≥0 and any bounded subset C of K, limt→∞supx∈C∥T(h)(T(t)x)−T(t)x∥=0.Lemma 2.8 (see {[6]}).
LetE be a Banach space with a uniformly Gâteaux differentiable norm, then the normalized
duality mapping J:E→2E* defined by (2.1) is single-valued and
uniformly continuous from the norm topology of E to the weak* topology of E* on each bounded subset of E.The single-valued normalized duality mapping is
denoted byj.Lemma 2.9.
LetE be a reflexive and strictly convex Banach
space with a uniformly Gâteaux differentiable norm, K a nonempty closed convex subset of E.
Suppose that {xn} is a bounded sequence in K, {T(h):h≥0} a continuous generalized asymptotically
nonexpansive semigroup from K into itself such that limn→∞∥xn−T(h)xn∥=0 for all h≥0. Define the set(2.5)K*={x∈K:μn∥xn−x∥2=miny∈Kμn∥xn−y∥2}.If F≠∅,
then K*∩F≠∅.Proof.
Setg(y)=μn∥xn−y∥2,
then g(y) is a convex and continuous function, and g(y)→∞ as ∥y∥→∞.
Using [7, Theorem 1.3.11], there exists x∈K such that g(x)=infy∈Kg(y) by the reflexivity of E,
that is, K* is nonempty. Clearly, K* is closed convex by the convexity and
continuity of g(y).
Sincelimn→∞∥xn−T(h)xn∥=0,limh→∞kh(i)=0(i=1,2), and g(y) is continuous for all z∈K*, we have(2.6)g(limh→∞T(h)z)=limh→∞g(T(h)z)=limh→∞μn∥xn−T(h)z∥2≤limh→∞μn∥T(h)xn−T(h)z∥2≤limh→∞μn((1+kh(1))∥xn−z∥+kh(2))2=μn∥xn−z∥2.Hence limh→∞T(h)z∈K*.
Letp∈F.
Since K* is closed convex set, there exists a unique v∈K* such that(2.7)∥p−v∥=minx∈K*∥p−x∥.
Sincep=limh→∞T(h)p and limh→∞T(h)v∈K*,(2.8)∥p−limh→∞T(h)v∥=∥limh→∞T(h)p−limh→∞T(h)v∥=limh→∞∥T(h)p−T(h)v∥≤limh→∞(1+kh(1))∥p−v∥+kh(2)=∥p−v∥.Therefore, limh→∞T(h)v=v. Since T(s+t)x=T(s)T(t)x for all x∈K,
then we have(2.9)v=limt→∞T(t)v=limt→∞T(s+t)v=limt→∞T(s)T(t)v=T(s)limt→∞T(t)v=T(s)vfor all s≥0.
Therefore v∈F and the proof is complete.Lemma 2.10 (see {[8]}).
LetK be a nonempty convex subset of a Banach space E with a uniformly Gâteaux differentiable norm, and {xn} a bounded sequence of E.
If z0∈K, then(2.10)μn∥xn−z0∥2=miny∈Kμn∥xn−y∥2if and only if(2.11)μn〈y−z0,J(xn−z0)〉≤0∀y∈K.Lemma 2.11 (see {[9]}).
Let{an} be a sequence of nonnegative real numbers
satisfying the following conditions:(2.12)an+1≤(1−λn)an+bn+cn∀n≥n0,where n0 is some nonnegative integer, λn∈[0,1] with ∑n=1∞λn=∞,limsupn→∞(bn/λn)≤0, and ∑n=1∞cn<∞.
Then an→0 as n→∞.


## 3. Implicit Iteration Scheme

Theorem 3.1.
LetE be a real reflexive and strictly convex
Banach space with a uniformly Gâteaux differentiable norm, K a nonempty closed convex subset of E, ℱ={T(h):h≥0} a u.a.r generalized asymptotically
nonexpansive semigroup from K into itself with sequences {kh(1)}{kh(2)},h≥0, such that F≠∅,
and f:K→K a fixed contractive mapping with contractive
coefficient β∈(0,1).
If {xn} is given by (1.2), where limn→∞tn=∞,αn∈(0,1),limn→∞αn=0 and limn→∞(ktn(i)/αn)=0(i=1,2), then {xn} converges strongly to some common fixed point p of F such that p is the unique solution in F to variational inequality:(3.1)〈f(p)−p,j(y−p)〉≤0∀y∈F.Proof.
For any fixedy∈F,(3.2)∥xn−y∥2=〈αn(f(xn)−y)+(1−αn)(T(tn)xn−y),j(xn−y)〉=αn〈f(xn)−f(y),j(xn−y)〉+αn〈f(y)−y,j(xn−y)〉+(1−αn)〈T(tn)xn−T(tn)y,j(xn−y)〉≤αnβ∥xn−y∥2+αn〈f(y)−y,j(xn−y)〉+(1−αn)∥xn−y∥[(1+ktn(1))∥xn−y∥+ktn(2)]=(1−αn(1−β)+(1−αn)ktn(1))∥xn−y∥2+αn〈f(y)−y,j(xn−y)〉+(1−αn)ktn(2)∥xn−y∥. Let dn(i)=(ktn(i)/αn)(i=1,2). Since limn→∞(ktn(i)/αn)=0 for all ε∈(0,1−β),
there exists N∈ℕ such that ktn(i)/αn<ε<1−β<(1−β)/(1−αn) for all n≥N.
Furthermore,(3.3)∥xn−y∥2≤〈f(y)−y,j(xn−y)〉1−β−(1−αn)dn(1)+(1−αn)dn(2)∥xn−y∥1−β−(1−αn)dn(1)for all n≥N.
That is, ∥xn−y∥≤(∥f(y)−y∥+(1−αn)dn(2))/(1−β−(1−αn)dn(1)) for all n≥N.
Thus {xn} is bounded, so are {T(tn)xn} and {f(xn)}.
This imply that(3.4)limn→∞∥xn−T(tn)xn∥=limn→∞αn∥T(tn)xn−f(xn)∥=0.Since {T(h)} is u.a.r and limn→∞tn=∞, then for all h≥0,(3.5)limn→∞∥T(h)T(tn)xn−T(tn)xn∥≤limn→∞supx∈C∥T(h)T(tn)x−T(tn)x∥=0,where C is any bounded subset of K containing {xn}.
Since {T(h)} is continuous, hence(3.6)∥xn−T(h)xn∥≤∥xn−T(tn)xn∥+∥T(tn)xn−T(h)(T(tn)xn)∥+∥T(h)(T(tn)xn)−T(h)xn∥→0.That is, for all h≥0,limn→∞∥xn−T(h)xn∥=0. We claim that the set {xn} is sequentially compact. Indeed, define the
set(3.7)K*={x∈K:μn∥xn−x∥2=miny∈Kμn∥xn−y∥2}.By Lemma 2.9, we can found p∈K*∩F.
Using Lemma 2.10, we get that(3.8)μn〈y−p,j(xn−p)〉≤0∀y∈K.It follows from (3.3) that
(3.9)μn∥xn−p∥2≤μn〈f(p)−p,j(xn−p)〉1−β−(1−αn)dn(1)+μn(1−αn)dn(2)∥xn−p∥1−β−(1−αn)dn(1)→0.Then we have μn∥xn−p∥=0.
Hence, there exists a subsequence{xnk} of {xn} which strongly converges to p∈F as k→∞.
Next we show thatp is a solution in F to the variational inequality (3.1). In fact,
for any fixed y∈F,
there exists a constant Q>0 such that ∥xn−y∥≤Q, then(3.10)∥xn−y∥2=〈αn(f(xn)−y)+(1−αn)(T(tn)xn−y),j(xn−y)〉=αn〈f(xn)−f(p)+p−xn,j(xn−y)〉+αn〈f(p)−p,j(xn−y)〉+αn〈xn−y,j(xn−y)〉+(1−αn)〈T(tn)xn−T(tn)y,j(xn−y)〉≤αn(β+1)∥xn−p∥Q+αn〈f(p)−p,j(xn−y)〉+∥xn−y∥2+(1−αn)ktn(1)Q2+(1−αn)ktn(2)Q.Therefore,(3.11)〈f(p)−p,j(y−xn)〉≤(β+1)∥xn−p∥Q+(1−αn)dn(1)Q2+(1−αn)dn(2)Q.Taking limit as nk→∞ in two sides of (3.11), by Lemma 2.8 and {xnk}→p as k→∞, we obtain(3.12)〈f(p)−p,j(y−p)〉≤0∀y∈F.That is, p∈F is a solution of variational inequality (3.1).
Suppose thatp,q∈F satisfy (3.1),
we have (3.13)〈f(p)−p,j(q−p)〉≤0,(3.14)〈f(q)−q,j(p−q)〉≤0. Combining (3.13) and (3.14), it
follows that(3.15)(1−β)∥p−q∥2≤〈(p−q)−f(p)+f(q),j(p−q)〉≤0.Hence p=q,
that is, p∈F is the unique solution of variational
inequality (3.1), so each cluster point of sequence {xn} is equal to p.
Therefore, {xn} converges to p and the proof is complete.Remark 3.2.
LetE,K,F,f,{αn}, and {tn} be as in Theorem 3.1, ℱ={T(h):h≥0} a u.a.r nonexpansive semigroup from K into itself, then our result coincides with
Theorem 3.2 in [3].


## 4. Modified Implicit Iteration Scheme

Theorem 4.1.
LetE be a real reflexive and strictly convex
Banach space with a uniformly Gâteaux differentiable norm, K a nonempty closed convex subset of E, ℱ={T(h):h≥0} a u.a.r generalized asymptotically
nonexpansive semigroup from K into itself with sequences {kh(1)}{kh(2)}, h≥0, such that F≠∅,
and f:K→K a fixed contractive mapping with contractive
coefficient β∈(0,1).
If {xn} is given by (1.4), where limn→∞tn=∞, αn,βn∈(0,1], limn→∞αn=0, ∑n=1∞βn=∞,
and ∑n=1∞(ktn(1)/αn)<∞,∑n=1∞(ktn(2)/αn)<∞, then {xn} converges strongly to some common fixed point p of F such that p is the unique solution in F to variational inequality (3.1).Proof.
For any fixedy∈F,(4.1)∥xn−y∥=∥αn(yn−y)+(1−αn)(T(tn)xn−y)∥≤(1−αn)∥T(tn)xn−y∥+αn∥yn−y∥≤(1−αn)[(1+ktn(1))∥xn−y∥+ktn(2)]+αn∥yn−y∥.Let dn(i)=(ktn(i)/αn)(i=1,2). Hence,(4.2)∥xn−y∥≤(1−αn)dn(2)1−(1−αn)dn(1)+∥yn−y∥1−(1−αn)dn(1)≤dn(2)1−dn(1)+βn∥f(xn−1)−y∥+(1−βn)∥xn−1−y∥1−dn(1)≤dn(2)1−dn(1)+βn∥f(xn−1)−f(y)∥1−dn(1)+βn∥f(y)−y∥1−dn(1)+(1−βn)∥xn−1−y∥1−dn(1)≤(1−βn(1−β))(∥xn−1−y∥+dn(2))1−dn(1)+βn(∥f(y)−y∥+dn(2))1−dn(1)≤11−dn(1)max{∥xn−1−y∥+dn(2),∥f(y)−y∥+dn(2)1−β}.By induction, we  get
that(4.3)∥xn−y∥≤11−dn(1)max{∥xn−2−y∥+dn−1(2)1−dn−1(1)+dn(2),∥f(y)−y∥+dn−1(2)(1−dn−1(1))(1−β),∥f(y)−y∥+dn(2)1−β}⋯≤1(1−dn(1))⋯(1−d1(1))×max{∥x1−y∥+∑i=1ndi(2),∥f(y)−y∥+d1(2)1−β,…,∥f(y)−y∥+dn(2)1−β}.Since ∑n=1∞dn(i)<∞(i=1,2), we know from Abel–Dini theorem that there
exists r>0 such that limn→∞(1−dn(1))⋯(1−d1(1))=r.
Thus {xn} is bounded, so are {T(tn)xn}, {f(xn)} and {yn}.
This imply that(4.4)limn→∞∥xn−T(tn)xn∥=limn→∞αn∥yn−T(tn)xn∥=0.Since {T(h):h≥0} is u.a.r. and limn→∞tn=∞ for all h≥0,(4.5)limn→∞∥T(h)T(tn)xn−T(tn)xn∥≤limn→∞supx∈C∥T(h)T(tn)x−T(tn)x∥=0,where C is any bounded subset of K containing {xn}.
Since T(h) is continuous, hence(4.6)∥xn−T(h)xn∥≤∥xn−T(tn)xn∥+∥T(tn)xn−T(h)(T(tn)xn)∥+∥T(h)(T(tn)xn)−T(h)xn∥→0.That is, for all h≥0,(4.7)limn→∞∥xn−T(h)xn∥=0.From Theorem 3.1, there exists
the unique solution p∈F to the variational inequality (3.1). Since p=T(h)p for all h≥0, we have(4.8)∥xn+1−p∥2=αn+1〈yn+1−p,j(xn+1−p)〉+(1−αn+1)〈T(tn+1)xn+1−p,j(xn+1−p)〉=αn+1〈βn+1f(xn)+(1−βn+1)xn−p,j(xn+1−p)〉+(1−αn+1)〈T(tn+1)xn+1−p,j(xn+1−p)〉≤αn+1βn+1〈f(xn)−f(p),j(xn+1−p)〉+αn+1βn+1〈f(p)−p,j(xn+1−p)〉+αn+1(1−βn+1)〈xn−p,j(xn+1−p)〉+(1−αn+1)∥xn+1−p∥[(1+ktn+1(1))∥xn+1−p∥+ktn+1(2)]≤αn+1βn+1β2∥xn−p∥2+∥xn+1−p∥22+αn+1(1−βn+1)∥xn−p∥2+∥xn+1−p∥22+(1−αn+1)∥xn+1−p∥2+(1−αn+1)ktn+1(1)∥xn+1−p∥2+(1−αn+1)ktn+1(2)∥xn+1−p∥+αn+1βn+1〈f(p)−p,j(xn+1−p)〉=αn+12∥xn+1−p∥2+(1−αn+1)∥xn+1−p∥2+αn+12(1+β2βn+1−βn+1)∥xn−p∥2+(1−αn+1)ktn+1(1)∥xn+1−p∥2+(1−αn+1)ktn+1(2)∥xn+1−p∥+αn+1βn+1〈f(p)−p,j(xn+1−p)〉.Therefore(4.9)∥xn+1−p∥2≤(1−(1−β2)βn+1)∥xn−p∥2+2βn+1〈f(p)−p,j(xn+1−p)〉+2(1−αn+1)(dn+1(1)∥xn+1−p∥+dn+1(2))∥xn+1−p∥.That is,(4.10)∥xn+1−p∥2≤(1−λn)∥xn−p∥2+bn+cn,where λn=(1−β2)βn+1,bn=2βn+1〈f(p)−p,j(xn+1−p)〉 and cn=2(1−αn+1)(dn+1(1)∥xn+1−p∥+dn+1(2))∥xn+1−p∥. Since ∑n=1∞λn=∞,  ∑n=1∞dn(i)<∞(i=1,2),∥xn+1−p∥ is bounded, we have ∑n=1∞cn<∞. So we only need to show that limsupn→∞(bn/λn)≤0, that is,(4.11)lim¯n→∞〈f(p)−p,j(xn+1−p)〉≤0.Let zm=amf(zm)+(1−am)T(tm)zm,
where tm and αm satisfy the condition of Theorem 3.1. Then it
follows from Theorem 3.1 that p=limm→∞zm.
Since(4.12)∥xn+1−zm∥2=(1−αm)〈T(tm)zm−xn+1,j(zm−xn+1)〉+αm〈f(zm)−xn+1,j(zm−xn+1)〉=(1−αm)(〈T(tm)zm−T(tm)xn+1,j(zm−xn+1)〉+〈T(tm)xn+1−xn+1,j(zm−xn+1)〉)+αm〈f(zm)−zm−(f(p)−p),j(zm−xn+1)〉+αm〈f(p)−p,j(zm−xn+1)〉+αm〈zm−xn+1,j(zm−xn+1)〉≤∥xn+1−zm∥2+(1−αm)(ktm(1)Q+ktm(2))Q+(1−αm)∥T(tm)xn+1−xn+1∥Q+αm〈f(p)−p,j(zm−xn+1)〉+αm(1+β)∥zm−p∥Q.Furthermore,(4.13)〈f(p)−p,j(xn+1−zm)〉≤1−αmαm(ktm(1)Q+ktm(2))Q+∥T(tm)xn+1−xn+1∥αmQ+(1+β)∥zm−p∥Q,where Q is a constant such that Q≥∥zm−xn+1∥.
Hence, taking upper limit as n→∞ firstly, and then as m→∞ in (4.13), we have(4.14)lim¯m→∞lim¯n→∞〈f(p)−p,j(xn+1−zm)〉≤0.On the other hand, since p=limm→∞zm and by Lemma 2.8, we have(4.15)〈f(p)−p,j(xn+1−zm)〉→〈f(p)−p,j(xn+1−p)〉uniformly.Thus given ε>0, there exists N≥1 such that if m>N for all n we have(4.16)〈f(p)−p,j(xn+1−p)〉<〈f(p)−p,j(xn+1−zm)〉+ε.Hence, taking upper limit as n→∞ firstly, and then as m→∞ in two sides of (4.16), we get that(4.17)lim¯n→∞〈f(p)−p,j(xn+1−p)〉≤lim¯m→∞lim¯n→∞〈f(p)−p,j(xn+1−zm)〉+ε≤ε.For the arbitrariness of ε,
(4.11) holds. By Lemma 2.11, xn→p and the proof is complete.



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*Source: 687815-2008-12-03.xml*