NONEXPANSIVE AND
m-ACCRETIVE OPERATORS
RUDONG CHEN AND ZHICHUAN ZHUReceived 10 June 2006; Accepted 22 July 2006
LetX be a real reflexive Banach space, letCbe a closed convex subset ofX, and letA be anm-accretive operator with a zero. Consider the iterative method that generates the sequence{xn}by the algorithmxn+1=αnf(xn) + (1−αn)Jrnxn, whereαnandγnare two
sequences satisfying certain conditions,Jrdenotes the resolvent (I+rA)−1forr >0, and
let f :C→Cbe a fixed contractive mapping. The strong convergence of the algorithm {xn}is proved assuming thatXhas a weakly continuous duality map.
Copyright © 2006 R. Chen and Z. Zhu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
LetX be a real Banach space, letCbe a nonempty closed convex subset ofX, and let T:C→Cbe a nonexpansive mapping if for allx,y∈C, such that
Tx−T y ≤ x−y. (1.1)
We useF(T) to denote the set of fixed points ofT, that is,F(T)= {x∈C:x=Tx}. And
denotes weak convergence,→denotes strong convergence. Recall that a self-mapping f :C→Cis a contraction onCif there exists a constantβ∈(0, 1) such that
f(x)−f(y)≤βx−y, x,y∈C. (1.2)
Browder [2] considered an iteration in a Hilbert space as follows. Fixu∈Cand define a contractionTt:C→Cby
Ttx=tu+ (1−t)Tx, x∈C, (1.3)
wheret∈(0, 1). Banach’s contraction mapping principle guarantees thatTthas a unique
fixed pointxtinC.
Xu [7] defined the following one viscosity iteration for nonexpansive mappings in uniformly smooth Banach space.
Theorem1.1 [2, Theorem 4.1, page 287]. LetX be a uniformly smooth Banach space, letCbe a closed convex subset ofX,T:C→Cis a nonexpansive mapping withF(T)=φ, and f ∈ΠC, whereΠC denotes the set of all contractions onC. Then{xt}defined by the
following:
xt=t fxt+ (1−t)Txt, x∈C, (1.4)
converges strongly to a point in F (T). DefineQ:ΠC→F(T)by
Q(f) :=lim
t→0xt, f ∈
C
, (1.5)
then Q(f) solves the variational inequality
(I−f)Q(f),JQ(f)−p≤0, f ∈
C
, p∈F(T). (1.6)
Xu [8] proved the strong convergence of{xt}defined by (1.3) in a reflexive Banach
space with a weakly continuous duality mapJϕwith gaugeϕ.
Recall that an operatorAwithD(A) and rangeR(A) inXis said to be accretive, if for eachxi∈D(A) andyi∈Axi, (i=1, 2) such that
y2−y1,J
x2−x1
≥0, (1.7)
whereJis the duality map fromXto the dual spaceX∗given by
J(x)=f ∈X∗:x,f= x2= f2 , x∈X. (1.8)
An accretive operatorAism-accretive ifR(I+λA)=Xfor allλ >0. Denote byJrthe resolvent ofAforr >0,
Jr=(I+rA)−1. (1.9)
It is known thatJris a nonexpansive mapping fromXtoC:=D(A) which will be assumed
convex.
Also in [8], Xu considered the following algorithm:
xn+1=αnu+
1−αn
Jrnxn, n≥0, (1.10)
whereu∈Cis arbitrarily fixed, {αn} is a sequence in (0, 1), and{rn}is a sequence of
positive numbers. Xu proved that ifXis a reflexive Banach space with weakly continuous duality mapping, then the sequence{xn}given by (1.10) converges strongly to a point in
Fprovided the sequences{αn}and{rn}satisfy certain conditions.
The main purpose of this paper is to consider the following two iterations both in a reflexive Banach spaceXwhich has a weakly continuous duality mapping:
xt=t f
xt
+ (1−t)Txt, t∈(0, 1), (1.11)
xn+1=αnf
xn
+1−αn
2. Preliminaries
In order to prove our main results, we need the following lemmas. The proof ofLemma 2.1can be found in [5,6].Lemma 2.2is an immediate consequence of the subdifferential inequality of the function (1/2) · 2.
Lemma2.1. Let{an}nbe a sequence of nonnegative real numbers such that
an+1≤
1−αn
an+αnβn, n≥0, (2.1)
where{αn}n⊂(0, 1), andβnsatisfy the conditions:
(i) limn→∞αn=0,
(ii)∞n=1αn= ∞,
(iii) lim supn→∞βn≤0.
Thenlimn→∞an=0.
Lemma2.2. LetXbe an arbitrary real Banach space. Then
x+y2≤ x2+ 2y,J(x+y), x,y∈X. (2.2)
Recall that a gauge is a continuous strictly increasing functionϕ: [0, +∞)→[0, +∞) such thatϕ(0)=0 andϕ(t)→ ∞. Associated to a gaugeϕis the duality mapJϕ:X→X∗
defined by
Jϕ(x)=f ∈X∗:x,f= xϕx,f =ϕx , x∈X. (2.3)
Following Browder [3], we say that a Banach spaceX has a weakly continuous duality map if there exists a gaugeϕfor which the duality mapJϕis single valued and
weak-to-weak∗sequentially continuous, that is, if{xn}is a sequence inXweakly convergent to a
pointx, then the sequence{Jϕ(xn)}converges weakly∗toJϕ(x). It is known thatphas a
weakly continuous duality map for all 1< p <∞. Set
Φ(t)=
t
0ϕ(τ)dτ, τ≥0. (2.4)
Then
Jϕ(x)=∂Φ
x, x∈X, (2.5)
where∂denotes the subdifferential in the sense of convex analysis.
We also need the next lemma, and the first part ofLemma 2.3is an immediate conse-quence of the subdifferential inequality and the proof of the second part can be found in [4].
Lemma2.3. Assume thatXhas a weakly continuous duality mapJϕwith gaugeϕ.
(i)For allx,y∈X, there holds the inequality
Φx+y≤Φx+y,Jϕ(x+y)
(ii)Assume a sequence{xn}inXis weakly convergent to a pointx. Then there holds the
identity
lim sup
n→∞ Φ
xn−y=lim sup
n→∞ Φ
xn−x+Φ
y−x, x,y∈X. (2.7)
Lemma 2.4is the resolvent identity which can be found in [1].
Lemma2.4. Forλ,μ >0, there holds the identity
Jλx=Jμ
μ
λx+
1−μ λ
Jλx
, x∈X. (2.8)
3. Main results
Theorem3.1. LetXbe a real reflexive Banach space and have a weakly continuous duality mappingJϕwithϕ. SupposeCis a closed convex subset ofX, andT:C→Cis a nonexpansive
mapping, let f :C→C be a fixed contractive mapping. Fort∈(0, 1),{xt} is defined by
(1.11). ThenT has a fixed point if and only if{xt}remains bounded ast→0+, and in this
case,{xt}converges strongly to a fixed point ofTast→0+.
Proof. Assume first thatF(T)=φ. Takeu∈F(T), it follows that
xt−u=t f
xt
+ (1−t)Txt−u
≤tfxt
−u+ (1−t)Txt−u
≤tβxt−u+tf(u)−u+ (1−t)xt−u
=1−(1−β)txt−u+tf(u)−u.
(3.1)
Hence
xt−u≤ 1
1−βf(u)−u. (3.2)
Therefore,{xt}is bounded, so are{Txt}and{f(xt)}.
Next assume that{xt}is bounded ast→0+. Assumetn→0+and{xtn}is bounded.
SinceXis reflexive, we may assume thatxtnpfor some p∈C. SinceJϕis weakly
con-tinuous, we have byLemma 2.3,
lim sup
n→∞ Φ
xt
n−x=lim sup n→∞ Φ
xt
n−p+Φx−p, ∀x∈X. (3.3)
Put
g(x)=lim sup
n→∞ Φ
xt
n−x, x∈X. (3.4)
It follows that
Since
xt
n−Txtn= tn
1−tn
f(xt
n)−xtn−→0, (3.6)
we obtain
g(T p)=lim sup
n→∞ Φ
xt
n−T p≤lim sup n→∞ Φ
Txt
n−T p ≤lim sup
n→∞ Φ
xt
n−p=g(p).
(3.7)
On the other hand, however,
g(T p)=g(p) +ΦT p−p. (3.8)
From (3.7) and (3.8), we get
ΦT p−p≤0. (3.9)
HenceT p=pandp∈F(T).
Now we prove that{xt}converges strongly to a fixed point ofTprovided it remains
bounded whent→0.
Let{tn}be a sequence in (0, 1) such thattn→0 andxtnpasn→ ∞. Then the
argu-ment above shows thatp∈F(T). We next show thatxtn→p. As a matter of fact, we have
byLemma 2.3,
Φxtn−p=Φtn
Txtn−p
+1−tn
f(xtn
−p
≤ΦtnTxtn−p+
1−tn
fxtn
−p,Jϕ
xtn−p
≤tnΦxtn−p+
1−tnfxtn
−f(p),Jϕxtn−p
+1−tnf(p)−p,Jϕxtn−p
.
(3.10)
This implies that
Φxtn−p≤
fxtn
−f(p),Jϕxtn−p
+f(p)−p,Jϕxtn−p
≤βxtn−pJϕ
xtn−p+
f(p)−p,Jϕxtn−p
=βΦxtn−p+
f(p)−p,Jϕxtn−p
,
(3.11)
that is,
Φxtn−p≤
1 1−β
f(p)−p,Jϕ
xtn−p
. (3.12)
Now noting thatxtnpimpliesJϕ(xtn−p)0, we get
Φxtn−p−→0. (3.13)
We have proved for any sequence{xtn} in {xt:t∈(0, 1)} that there exists a
subse-quence which is still denoted by {xtn} that converges to some fixed point p of T. To
prove that the entire net{xt}converges strongly to p, we assume there exists another
sequence{sn} ∈(0, 1) such thatxsn→q, thenq∈F(T). We have to showp=q. Indeed,
foru∈F(T), it is easy to see that
xt−Txt,Jϕ
xt−u
=Φxt−u+
u−Txt,Jϕ
xt−u
≥Φxt−u−u−Txt·Jϕ
xt−u
≥Φxt−u−Φxt−u=0.
(3.14)
On the other hand, since
xt−Txt= t
1−t
fxt
−xt
, (3.15)
we get fort∈(0, 1) andu∈F(T),
xt−f
xt
,Jϕ
xt−u
≤0. (3.16)
Since the sets{xt−u}and{xt}are bounded and a Banach spaceXhas a weakly
contin-uous duality mapJϕ, thenJϕis single valued and weak-to-weak∗sequentially continuous,
for anyu∈F(T), byxsn→q(sn→0), we have
xs
n−f
xsn
−q−f(q)−→0 sn−→0
,
xs
n−f
xsn
,Jϕ
xsn−u
−q−f(q),Jϕ(q−u)
=xsn−f
xsn
−q−f(q),Jϕ
xsn−u
+q−f(q),Jϕ
xsn−u
−Jϕ(q−u)
≤xsn−f
xsn
−q−f(q)Jϕxsn−u
+q−f(q),Jϕxsn−u
−Jϕ(q−u) assn−→0.
(3.17)
Therefore, we get
q−f(q),Jϕ(q−u)=lim sn→0
xsn−f
xsn
,Jϕ
xsn−u
≤0. (3.18)
Interchangepanduto obtain
q−f(q),Jϕ(q−p)
≤0. (3.19)
Interchangeqanduto obtain
p−f(p),Jϕ(p−q)≤0. (3.20)
This implies that
(p−q)−f(p)−f(q),Jϕ(p−q)
That is,
p−qϕp−q≤βp−qϕp−q. (3.22)
This is a contradiction, so we havep=q.
The proof is complete.
Remark 3.2. Theorem 3.1is proved in a weaker condition than [7, Theorem 4.1], and the method of proof is different from [7], we introduce a continuous strict increasing function.
Next two main results are about accretive operators, we consider the problem of find-ing a zero of anm-accretive operatorAin a reflexive Banach spaceX, 0∈Ax. Denote by F(A) the zero set ofA, that is,
F(A) :=x∈D(A) : 0∈Ax =A−1(0). (3.23)
Theorem3.3. Suppose thatXis reflexive and has a weakly continuous duality mapJϕwith
gaugeϕ. Suppose thatAis anm-accretive operator inXsuch thatC=D(A)is convex with F(A)=φ, and f :C→Cis a fixed contractive map. Assume
(i)αn→0and
∞
n=0αn= ∞,
(ii)γn→ ∞.
Then the sequence{xn}defined by (1.12) converges strongly to a point inF(A).
Proof. First we prove{xt}is bounded. Indeed, takeu∈F(A), then
xn+1−u≤αnf
xn
−u+1−αnJrnxn−u
≤αnβxn−u+αnf(u)−u+1−αnxn−u
≤1−(1−β)αnxn−u+αnf(u)−u.
(3.24)
By induction, we get
xn−u≤maxx0−u, 1
1−βf(u)−u
∀n≥0. (3.25)
This implies that{xn}is bounded, so are{f(xn)}and{Jrnxn}, and hence
xn+1−Jr
nxn=αnf
xn
−Jrnxn−→0. (3.26)
We next prove
lim sup
n→∞
f(p)−p,Jϕ
xn−p
≤0, p∈F(A). (3.27)
ByTheorem 3.1, putp=limt→0xt, we take a subsequence{xnk}of{xn}such that
lim sup
n→∞
f(p)−p,Jϕ
xn−p
=lim sup
n→∞
f(p)−p,Jϕ
xnk−p
SinceXis reflexive, we may further assume thatxnkx. Moreover, since
xn+1−Jr
nxn−→0, (3.29)
we obtain
Jrnk−1xrnk−1 x. (3.30)
Taking the limit ask→ ∞in the relation
Jrnk−1xrnk−1,Arnk−1xrnk−1
∈A, (3.31)
we get [x, 0]∈A, that is,x∈F(A). Hence by (3.28) and (3.18), we have
lim sup
n→∞
f(p)−p,Jϕxn−p=f(p)−p,Jϕ(x−p)≤0. (3.32)
That is, (3.27) holds.
Finally, we prove thatxn→p.
We applyLemma 2.3to get
Φxn+1−p=Φ1−αn
Jrnxn−p
+αn
fxn
−p
=Φ1−αnJrnxn−p
+αnfxn−f(p)+αnf(p)−p)
≤Φ1−αnJrnxn−p
+αnfxn−f(p)
+αnf(p)−p,Jϕxn+1−p ≤1−(1−β)αn
Φxn−p) +αn
f(p)−p,Jϕ
xn+1−p
.
(3.33)
ApplyingLemma 2.1, we get
Φxn−p−→0. (3.34)
That is,xn−p →0, that is,xn→p.
The proof is complete.
Theorem3.4. Suppose thatXis reflexive and has a weakly continuous duality mapJϕwith
gaugeϕ. Suppose thatAis anm-accretive operator inXsuch thatC=D(A)is convex with F(A)=φ, and f :C→Cis a fixed contractive map. Assume
(i)αn−→0,
∞
n=0 αn= ∞,
∞
n=1
αn+1−αn<∞,
(ii)γn≥ε ∀n,
∞
n=1
γn+1−γn<∞. (3.35)
Proof. We only include the differences. We have
xn+1=αnf
xn
+1−αn
Jγnxn, xn=αn−1f
xn−1
+1−αn−1
Jγn−1xn−1. (3.36)
Thus,
xn+1−xn=αnfxn+1−αnJγnxn−
αn−1fxn−1+1−αn−1Jγn−1xn−1
=αn
fxn
−fxn−1
+αn−αn−1
fxn−1
−Jγn−1xn−1
+1−αn
Jγnxn−Jγn−1xn−1
.
(3.37)
Ifγrn−1≤γn, byLemma 2.4, we get
Jγnxn=Jγn−1
γ
n−1 γn xn+
1−γn−1 γn
Jγnxn
, (3.38)
we have
Jγ
nxn−Jγn−1xn−1≤ γn−1
γn
xn−xn−1+1−γn−1 γn
Jγnxn−xn−1
≤xn−xn−1+
γ
n−γn−1 γn
Jγnxn−xn−1
≤xn−xn−1+1
εγn−1−γnJγnxn−xn−1.
(3.39)
It follows from the above results that
xn+1−xn≤αn−αn−1f
xn−1
−Jγn−1xn−1+
1−αnxn−xn−1
+1 ε
1−αnγn−1−γnJγnxn−xn−1+αnβxn−xn−1
≤Mαn−αn−1+γn−1−γn+
1−(1−β)αnxn−xn−1,
(3.40)
whereM >0 is some appropriate constant. Similarly, we can prove the last inequality if γn−1≥γn. By assumptions (i) and (ii) andLemma 2.1, we have
xn+1−xn−→0. (3.41)
This implies that
xn−Jγ
nxn≤xn+1−xn+xn+1−Jγnxn. (3.42)
Sincexn+1−Jγnxn =αnf(xn)−Jγnxn →0. It follows from (3.42) that
Aγ
nxn=
1 γn
xn−Jγ
nxn≤
1
εxn−Jγnxn−→0. (3.43)
Now if{xnk}is a subsequence of{xn}converging weakly to a pointx, then taking the
limit ask→ ∞in the relation
Jγnkxnk,Aγnkxnk
we get [x, 0]∈A, that is,x∈F(A). We therefore conclude that all weak limit points of {xn}are zeros ofA.
The rest of the proof follows fromTheorem 3.3.
The proof is complete.
Acknowledgment
This work is supported by the National Science Foundation of China, Grants 10471033 and 10271011.
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Rudong Chen: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China
E-mail address:[email protected]
Zhichuan Zhu: Department of Mathematics, Tianjin Polytechnic University, Tianjin 300160, China
