# Convergence of a General Composite Iterative Method for a

Authors: Shuang Wang
Journal: ISRN Applied Mathematics
 (2012)

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## Abstract

We propose a general composite iterative method for computing common fixed
points of a countable family of nonexpansive mappings in the framework of Hilbert spaces. Our results
improve and complement the corresponding ones announced by many others.


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## Body

## 1. Introduction and Main Result

LetH be a real Hilbert space with inner product 〈·,·〉 and norm ∥·∥. A mapping T:H→H is said to be nonexpansive if ∥Tx-Ty∥≤∥x-y∥ for  all x,y∈H. The set of fixed points, Fix(T)={x∈H:Tx=x}, of a nonexpansive mapping is always a closed and convex subset of H.In addition to nonexpansive mappings, we are going to use contractions andk-Lipschitzian and η-strongly monotone operators. A self-mapping f:H→H is a contraction on H, if there exists a constant α∈(0,1) such that ∥f(x)-f(y)∥≤α∥x-y∥, for all x,y∈H. We use ΠH to denote the collection of mappings f verifying the above inequality. That is, ΠH={f:H→H∣f is  a  contraction  with  constant  α}. A mapping F:H→H is called k-Lipschitzian if there exists a positive constant k such that
(1.1)‖Fx-Fy‖≤k‖x-y‖,∀x,y∈H.F is said to be η-strongly monotone if there exists a positive constant η such that
(1.2)〈Fx-Fy,x-y〉≥η‖x-y‖2,∀x,y∈H.Recently, Jung [1] introduced the following composite iterative scheme for the solution of a specific minimization problem, which involves a closed convex subset C⊆H, a nonexpansive mapping T:C→C, and a contraction f:C→C,
(1.3)x1=x∈Carbitrarilychosen,yn=αnf(xn)+(1-αn)Txn,xn+1=(1-βn)yn+βnTyn,n≥1.
Therein the control sequences {αn} and {βn} satisfy certain conditions. He proved that the sequence {xn} defined by (1.3) converges strongly to x~=Q(f)∈Fix(T), which is the unique solution of the variational inequality 〈x~-f(x~),x~-z〉≤0,z∈Fix(T). The results of Jung [1] are even stronger than stated here. In fact, they hold for a finite family of nonexpansive mappings and in the setting of Banach spaces.Very recently, Tian [2] considered the following iterative method: for nonexpansive mapping T:H→H with Fix(T)≠∅,
(1.4)xn+1=αnγf(xn)+(I-μαnF)Txn,n≥1,
where F is a k-Lipschitzian and η-strongly monotone operator. He obtained that the sequence {xn} generated by (1.4) converges to a point q in Fix(T), which is the unique solution of the variational inequality 〈(γf-μF)q,p-q〉≤0,p∈Fix(T).In this connection, notice that Aoyama et al. [3] proposed a Halpern approximation method for finding a common fixed point of a countable family of nonexpansive mappings. As their main result, they established the following strong convergence theorem.Theorem 1.1 (see [3, Theorem 3.4]).
LetE be a uniformly convex Banach space whose norm is uniformly Gâteaux differentiable, and let C be a nonempty closed convex subset of E. Let {Tn} be a sequence of nonexpansive mappings Tn:C→C such that
(1.5)∑n=1∞‍sup⁡{‖Tn+1z-Tnz‖:z∈B}<∞
for each bounded subset B of C. Suppose in addition that
(1.6)Fix⁡(T)=⋂n=1∞‍Fix⁡(Tn)≠∅,
where T:C→C is the nonexpansive mapping defined by Tz=limn→∞Tnz. Let αn∈(0,1] be a sequence satisfying the conditions (C1) limn→∞αn=0, (C2) ∑n=1∞αn=∞, and (C3) ∑n=1∞|αn+1-αn|<∞ or limn→∞(αn+1/αn)=1. Given x∈C, define {xn}n=1∞ by
(1.7)x1=x,xn+1=αnx+(1-αn)Tnxn.
Then {xn} converges strongly to Qx, where Q is a sunny nonexpansive retraction of E onto Fix(T).Inspired by Jung [1], Tian [2] and Aoyama et al. [3], the goal of this paper is to combine these three results into a single method. In this connection, observe that the iteration methods (1.3) and (1.4) just can compute a fixed point of one nonexpansive mapping (or, perhaps, finitely many), while the iteration methods (1.3) and (1.7) do not contain the k-Lipschitzian and η-strongly monotone operator F. On the other hand, in contrast to the result of Jung [1] and Aoyama et al. [3], our result will be restricted to the setting of Hilbert spaces, which on the other hand is natural when dealing with a k-Lipschitzian and η-strongly monotone operator F.First of all, let us remark that condition (1.5) implies that sequence of mappings {Tn} are uniformly Cauchy on each bounded subset B⊂C. Hence the limiting map T is well defined and is in fact the uniform limit (on B) of the maps Tn. In other words,
(1.8)lim⁡n→∞sup⁡{‖Tz-Tnz‖:z∈B}=0.Notice that condition (1.5) is quite strong so that, in general, we cannot apply the result directly to an arbitrary countable family of nonexpansive mappings. Furthermore, we have to assume the nontrivial condition (1.6) that the fixed point set of T coincides with the set of common fixed points of the family {Tn}.Fortunately, as pointed out in [3, Section 4], given an arbitrary countable family of nonexpansive mappings Sn:C→C, which have at least one common fixed point, one can find nonnegative numbers βnk (1≤k≤n,n∈ℕ) such that the mappings
(1.9)Tnx=∑k=1nβnkSkx
are nonexpansive self-maps of C satisfying (1.5) and (1.6). More specifically, the Tn are constructed as certain convex combinations of the S1,…,Sn. (For details, see the reference.) Another possibility of a construction will be mentioned below. Furthermore, we establish the following strong convergence theorem.Theorem 1.2.
LetH be a real Hilbert space. Let F be a k-Lipschitzian and η-strongly monotone operator on H with 0<μ<2η/k2 and f∈ΠH with 0<γ<μ(η-μk2/2)/α=τ/α and τ<1. Assume that {Tn} is a sequence of nonexpansive mappings from H into itself such that the condition (1.5). Suppose that T:H→H is defined by Tz=limn→∞Tnz such that the condition (1.6). Let {αn},{βn} be sequences in (0,1) satisfying the following conditions:(B1)
limn→∞αn=0 and ∑n=1∞αn=∞;(B2)
b:=limsupn→∞βn<1 and ∑n=1∞|βn+1-βn|<∞;(B3)
|αn+1-αn|≤o(αn+1)+σn with σn≥0 and ∑n=1∞σn<∞. Then, for arbitrary x1=x∈H, the sequence {xn}n=1∞, defined by
(1.10)yn=αnγf(xn)+(I-αnμF)Tnxn,xn+1=(1-βn)yn+βnTnyn,n≥1,
converges strongly to some q∈Fix(T), which satisfies the variational inequality 〈(μF-γf)q,q-z〉≤0, for all z∈Fix(T).The iterative scheme (1.10) is a direct generalization of the three iteration methods considered before. Thus Theorem 1.2 complements or improves Theorem 3.2 of Jung [1], Theorem 3.2 of Tian [2], Theorem 3.4 of Aoyama et al. [3], and Theorem 3.1 of Jung [4].An important special case is obtained forF=I, the identity mapping. Then η=k=1 and we can choose γ=μ=1 (compare with iteration scheme (1.3) above).Besides the basic conditions (B1) and (B2) on the sequencesαn and βn, we have the “control condition” (B3). It can obviously be replaced by one of the following ones:(B3-1)
∑n=1∞|αn+1-αn|<∞;(B3-2)
αn∈(0,1] for all n∈ℕ and limn→∞(αn+1/αn)=1. Indeed, (B3-1) implies (B3) by choosing σn=|αn+1-αn|, and (B3-2) implies (B3) by choosing σn=0. In this sense, (B3) is a weaker condition than the previous condition (C3).As has already noticed in Theorem1.1, the assumptions (1.5) and (1.6) do not apply to arbitrary families of nonexpansive mappings. Besides the construction above, we mention another construction, which has appeared in the literature. See [5–8], the references therein, and also Remark 3.1 of Peng and Yao [9]. Proofs are given there.Let{Si:H→H} be a countable family of nonexpansive mappings, and let {ξi} be a sequence of real numbers such that 0≤ξi≤c<1, for all i≥1. For any n≥1, define a mapping Tn:H→H as follows:
(1.11)Un,n+1=I,Un,n=ξnSnUn,n+1+(1-ξn)I,Un,n-1=ξn-1Sn-1Un,n+(1-ξn-1)I,⋮Un,k=ξkSkUn,k+1+(1-ξk)I,⋮Un,2=ξ2S2Un,3+(1-ξ2)I,Un,1=ξ1S1Un,2+(1-ξ1)I,Tn=Un,1.Proposition 1.3.
LetH be a real Hilbert space, {Si:H→H} a sequence nonexpansive mappings with ⋂i=1∞‍Fix(Si)≠∅, and {ξi}  a real sequence such that 0≤ξi≤c<1, for all i≥1. Define {Tn} as above. Then(1)
Tn is a nonexpansive and Fix(Tn)=⋂i=1n‍Fix(Si) for each n≥1;(2)
the mappings{Tn} satisfy the condition (1.5);(3)
definingT:H→H by Tx=limn→∞Tnx, we have
(1.12)Fix⁡(T)=⋂n=1∞‍Fix⁡(Tn)=⋂n=1∞‍Fix⁡(Sn).


## 2. Proof of the Main Results

Ahead of the proof, we start with recalling some known auxiliary results.Lemma 2.1 (see [10]).
LetF be a k-Lipschitzian and η-strongly monotone operator on a Hilbert space H with k>0,η>0,0<μ<2η/k2, and 0<t<1. Then S=(I-tμF):H→H is a contraction with contractive coefficient 1-tτ and τ=(1/2)μ(2η-μk2).Lemma 2.2 (see [2]).
LetH be a Hilbert space. Assume that {xt} is defined by
(2.1)xt=tγf(xt)+(I-tμF)Txt,
where F is a k-Lipschitzian and η-strongly monotone operator on a Hilbert space H with k>0,η>0,0<μ<2η/k2 and 0<t<1,f∈ΠH, and T:H→H is a nonexpansive mapping. Then xt converges strongly as t→0 to a fixed point q of T, which solves the variational inequality 〈(μF-γf)q,q-p〉≤0,p∈F(T).Some more auxiliary results are given next.Lemma 2.3 (see [11]).
LetH be a Hilbert space, C a closed convex subset of H, and T:C→C a nonexpansive mapping with Fix(T)≠∅. If {xn} is a sequence in C which converges weakly to x and if {(I-T)xn} converges strongly to y, then (I-T)x=y.Lemma 2.4 (see [3, Lemma 2.3]).
Let{sn} be a sequence of nonnegative real numbers, {δn} a sequence of numbers in [0,1] such that ∑n=1∞δn=∞, {μn} a sequence of nonnegative real numbers with ∑n=1∞μn<∞, and {θn} a sequence of real numbers with limsupn→∞θn≤0. Suppose that
(2.2)sn+1≤(1-δn)sn+δnθn+μn
for all n∈ℕ. Then limn→∞sn=0.Now we are prepared to prove the main result.Proof of Theorem1.2.
We divide the proof into seven steps.
Step1.We claim that {xn} is bounded. Taking any point q∈⋂n=1∞‍Fix(Tn) and using Lemma 2.1, we obtain
(2.3)‖xn+1-q‖=‖(1-βn)(yn-q)+βn(Tnyn-q)‖≤‖yn-q‖=‖αn(γf(xn)-μFq)+(I-αnμF)Tnxn-(I-αnμF)q‖≤(1-αnτ)‖xn-q‖+αn‖γf(xn)-μFq‖≤(1-αnτ)‖xn-q‖+αn‖γf(xn)-γf(q)‖+αn‖γf(q)-μFq‖≤(1-αnτ)‖xn-q‖+αnγα‖xn-q‖+αn‖γf(q)-μFq‖≤(1-αn(τ-γα))‖xn-q‖+αn(τ-γα)‖γf(q)-μFq‖τ-γα≤max⁡{‖xn-q‖,‖γf(q)-μFq‖τ-γα}.
By induction, it follows
(2.4)‖xn-q‖≤max⁡{‖x1-q‖,‖γf(q)-μFq‖τ-γα},n≥1,
and hence {xn} is bounded. From this, we also obtain that {yn},{Tnyn},{FTnxn}, and {f(xn)} are all bounded. In what follows, let B stand for some bounded set of H, which contains all of {xn},{yn},{Tnyn},{FTnxn}, {f(xn)}.
Step2. We show that limn→∞∥xn+1-xn∥=0.
LetM=sup{∥Tnyn-yn∥,∥f(xn)∥,∥FTnxn∥:n∈ℕ}<∞. From the definition of {xn}, we obtain
(2.5)‖xn+2-xn+1‖=‖(1-βn+1)yn+1+βn+1Tn+1yn+1-(1-βn)yn-βnTnyn‖=‖(1-βn+1)yn+1+βn+1Tn+1yn+1-(1-βn+1)yn-(βn+1-βn)yn-βn+1Tnyn+(βn+1-βn)Tnyn‖=‖(1-βn+1)(yn+1-yn)+βn+1(Tn+1yn+1-Tnyn)+(βn+1-βn)(Tnyn-yn)‖≤(1-βn+1)‖yn+1-yn‖+βn+1‖Tn+1yn+1-Tnyn‖+|βn+1-βn|M≤(1-βn+1)‖yn+1-yn‖+βn+1‖Tn+1yn+1-Tnyn+1‖+βn+1‖yn+1-yn‖+|βn+1-βn|M≤‖yn+1-yn‖+‖Tn+1yn+1-Tnyn+1‖+|βn+1-βn|M
for all n∈ℕ. From (1.10), we have
(2.6)yn=αnγf(xn)+(I-αnμF)Tnxn,yn+1=αn+1γf(xn+1)+(I-αn+1μF)Tn+1xn+1.
With M1=(μ+γ)M, we thus have
(2.7)‖yn+1-yn‖=‖αn+1γf(xn+1)+(I-αn+1μF)Tn+1xn+1-αnγf(xn)-(I-αnμF)Tnxn‖=‖(I-αn+1μF)Tn+1xn+1-(I-αn+1μF)Tnxn-(αn+1-αn)μFTnxn+αn+1γ(f(xn+1)-f(xn))+(αn+1-αn)γf(xn)‖≤(1-αn+1τ)‖Tn+1xn+1-Tnxn‖+|αn+1-αn|μ‖FTnxn‖+αn+1γα‖xn+1-xn‖+γ|αn+1-αn|‖f(xn)‖≤(1-αn+1τ)‖Tn+1xn+1-Tnxn‖+αn+1γα‖xn+1-xn‖+|αn+1-αn|M1≤(1-αn+1τ)(‖Tn+1xn+1-Tn+1xn‖+‖Tn+1xn-Tnxn‖)+αn+1γα‖xn+1-xn‖+M1|αn+1-αn|≤[1-αn+1(τ-γα)]‖xn+1-xn‖+‖Tn+1xn-Tnxn‖+M1|αn+1-αn|.
Combining (2.5) and (2.7), we conclude that
(2.8)‖xn+2-xn+1‖≤[1-αn+1(τ-γα)]‖xn+1-xn‖+M2(|αn+1-αn|+|βn+1-βn|)+2sup⁡{‖Tn+1z-Tnz‖:z∈B},
where M2=max{M,M1}. Using the assumption (B3), we obtain
(2.9)‖xn+2-xn+1‖≤[1-αn+1(τ-γα)]‖xn+1-xn‖+M2o(αn+1)+μn
with μn=M2(σn+|βn+1-βn|)+2sup{∥Tn+1z-Tnz∥:z∈B}. The various assumptions imply that ∑n=1∞‍μn<∞. Now it follows from Lemma 2.4 that limn→∞∥xn+1-xn∥=0.
Step3. Next we show that
(2.10)lim⁡n→∞‖Tnxn-yn‖=lim⁡n→∞‖xn-yn‖=0.
In fact, from Step 1, {γf(xn)-μFTnxn} is bounded. Moreover, it follows from limn→∞αn=0 that
(2.11)‖Tnxn-yn‖=‖αnγf(xn)-αnμFTnxn‖=αn‖γf(xn)-μFTnxn‖⟶0(n⟶∞).
This is the first part. From condition (B2), we obtain, for n sufficiently large and b<b1<1,
(2.12)‖xn+1-yn‖=βn‖Tnyn-yn‖≤βn(‖Tnyn-Tnxn‖+‖Tnxn-yn‖)≤b1(‖yn-xn‖+‖Tnxn-yn‖)≤b1(‖yn-xn+1‖+‖xn+1-xn‖+‖Tnxn-yn‖),
which implies
(2.13)‖xn+1-yn‖≤b11-b1(‖xn+1-xn‖+‖Tnxn-yn‖).
Hence, using Step 2 and the above, we obtain limn→∞∥xn+1-yn∥=0 and thus limn→∞∥xn-yn∥=0.
Step4. We show that limn→∞∥yn-Tnyn∥=0.
Indeed, we have(2.14)‖yn-Tnyn‖≤‖yn-Tnxn‖+‖Tnxn-Tnyn‖≤‖yn-Tnxn‖+‖xn-yn‖,
and now the statement follows from Step 3.
Step5. We show that ∥Tyn-yn∥→0 as n→∞.
Observe that(2.15)‖Tyn-yn‖≤‖Tyn-Tnyn‖+‖Tnyn-yn‖≤sup⁡{‖Tz-Tnz‖:z∈B}+‖Tnyn-yn‖,
and now apply Step 4 and the fact that Tn converges uniformly on B to T (which is a consequence of (1.5)).
Step6. We claim that limsupn→∞〈γf(q)-μFq,yn-q〉≤0, where q=limt→0xt with xt defined by (2.1) and the existence of q being guaranteed by Lemma 2.2.
Assume the contrary. Then there exists a subsequence{ynk} of {yn} such that the limit
(2.16)c:=lim⁡k→∞〈γf(q)-μFq,ynk-q〉
exists and is greater than zero. Since {yn} is bounded, there exists a subsequence of {ynk}, which converges weakly to some z∈H. We can assume without loss of generality that {ynk} is this subsequence. From ∥Tyn-yn∥→0, we obtain that Tynk converges weakly to z. Now Lemma 2.3 implies that z∈Fix(T). Hence from weak convergence and Lemma 2.2,
(2.17)lim⁡k→∞〈γf(q)-μFq,ynk-q〉=〈(γf-μF)q,z-q〉≤0.
This is a contradiction.
Step7. We show that {xn} converges strongly to q. From (1.10), we have
(2.18)‖xn+1-q‖2=‖(1-βn)(yn-q)+βn(Tnyn-q)‖2≤‖yn-q‖2=‖αnγf(xn)+(I-αnμF)Tnxn-q‖2=‖αn(γf(xn)-μFq)+(I-αnμF)Tnxn-(I-αnμF)q‖2.
Using the general inequality ∥x+y∥2≤∥x∥2+2〈y,x+y〉, we can further estimate the above by
(2.19)≤‖(I-αnμF)Tnxn-(I-αnμF)q‖2+2αn〈γf(xn)-μFq,yn-q〉≤(1-αnτ)2‖xn-q‖2+2αn〈γf(xn)-γf(q),yn-q〉+2αn〈γf(q)-μFq,yn-q〉≤(1-αnτ)2‖xn-q‖2+2αnγα‖xn-q‖(‖yn-xn‖+‖xn-q‖)+2αn〈γf(q)-μFq,yn-q〉≤(1-2αn(τ-γα))‖xn-q‖2+(αnτ)2‖xn-q‖2+2αnγα‖xn-q‖‖yn-xn‖+2αn〈γf(q)-μFq,yn-q〉.
Introducing M3=sup{∥xn-q∥:n≥1}, δn=2αn(τ-γα)>0, and
(2.20)θn=αnτ2M322(τ-γα)+γαM3τ-γα‖yn-xn‖+1τ-γα〈γf(q)-μFq,yn-q〉,
we obtain
(2.21)‖xn+1-q‖2≤(1-δn)‖xn-q‖2+δnθn.
Because δn→0, ∑n=1∞‍δn=∞, and limsupn→∞θn≤0, it follows from Lemma 2.4 that the sequence {xn} converges strongly to q. This finishes the proof of the main part of the theorem.
From the definition ofq as q=limt→0xt (see Step 6) and from Lemma 2.2, we obtain 〈(μF-γf)q,q-z〉≤0, for all z∈Fix(T)=⋂n=1∞‍Fix(Tn). This completes the proof.



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