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ISRN Applied Mathematics
Volume 2012 (2012), Article ID 576018, 10 pages
http://dx.doi.org/10.5402/2012/576018
Research Article

Convergence of a General Composite Iterative Method for a Countable Family of Nonexpansive Mappings

School of Mathematical Sciences, Yancheng Teachers University, Yancheng, 224051 Jiangsu, China

Received 17 March 2012; Accepted 10 May 2012

Academic Editors: A. Bellouquid, E. Kita, and J. Nedoma

Copyright © 2012 Shuang Wang. 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.

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.

1. Introduction and Main Result

Let 𝐻 be a real Hilbert space with inner product , and norm . A mapping 𝑇𝐻𝐻 is said to be nonexpansive if 𝑇𝑥𝑇𝑦𝑥𝑦 for all 𝑥,𝑦𝐻. The set of fixed points, Fix(𝑇)={𝑥𝐻𝑇𝑥=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 and 𝑘-Lipschitzian and 𝜂-strongly monotone operators. A self-mapping 𝑓𝐻𝐻 is a contraction on 𝐻, if there exists a constant 𝛼(0,1) such that 𝑓(𝑥)𝑓(𝑦)𝛼𝑥𝑦, for all 𝑥,𝑦𝐻. We use Π𝐻 to denote the collection of mappings 𝑓 verifying the above inequality. That is, Π𝐻={𝑓𝐻𝐻𝑓 is a contraction with constant 𝛼}. A mapping 𝐹𝐻𝐻 is called 𝑘-Lipschitzian if there exists a positive constant 𝑘 such that 𝐹𝑥𝐹𝑦𝑘𝑥𝑦,𝑥,𝑦𝐻.(1.1)𝐹 is said to be 𝜂-strongly monotone if there exists a positive constant 𝜂 such that 𝐹𝑥𝐹𝑦,𝑥𝑦𝜂𝑥𝑦2,𝑥,𝑦𝐻.(1.2)

Recently, Jung [1] introduced the following composite iterative scheme for the solution of a specific minimization problem, which involves a closed convex subset 𝐶𝐻, a nonexpansive mapping 𝑇𝐶𝐶, and a contraction 𝑓𝐶𝐶, 𝑥1𝑦=𝑥𝐶arbitrarilychosen,𝑛=𝛼𝑛𝑓𝑥𝑛+1𝛼𝑛𝑇𝑥𝑛,𝑥𝑛+1=1𝛽𝑛𝑦𝑛+𝛽𝑛𝑇𝑦𝑛,𝑛1.(1.3) 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 ̃𝑥𝑓(̃𝑥),̃𝑥𝑧0,𝑧Fix(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 𝑇𝐻𝐻 with Fix(𝑇), 𝑥𝑛+1=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝜇𝛼𝑛𝐹𝑇𝑥𝑛,𝑛1,(1.4) 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 (𝛾𝑓𝜇𝐹)𝑞,𝑝𝑞0,𝑝Fix(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]). Let 𝐸 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 𝑇𝑛𝐶𝐶 such that 𝑛=1𝑇sup𝑛+1𝑧𝑇𝑛𝑧𝑧𝐵<(1.5) for each bounded subset 𝐵 of 𝐶. Suppose in addition that Fix(𝑇)=𝑛=1𝑇Fix𝑛,(1.6) where 𝑇𝐶𝐶 is the nonexpansive mapping defined by 𝑇𝑧=lim𝑛𝑇𝑛𝑧. Let 𝛼𝑛(0,1] be a sequence satisfying the conditions (C1) lim𝑛𝛼𝑛=0, (C2) 𝑛=1𝛼𝑛=, and (C3) 𝑛=1|𝛼𝑛+1𝛼𝑛|< or lim𝑛(𝛼𝑛+1/𝛼𝑛)=1. Given 𝑥𝐶, define {𝑥𝑛}𝑛=1 by 𝑥1=𝑥,𝑥𝑛+1=𝛼𝑛𝑥+1𝛼𝑛𝑇𝑛𝑥𝑛.(1.7) 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 𝐵𝐶. Hence the limiting map 𝑇 is well defined and is in fact the uniform limit (on 𝐵) of the maps 𝑇𝑛. In other words, lim𝑛sup𝑇𝑧𝑇𝑛𝑧𝑧𝐵=0.(1.8)

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 𝑆𝑛𝐶𝐶, which have at least one common fixed point, one can find nonnegative numbers 𝛽𝑘𝑛 (1𝑘𝑛,𝑛) such that the mappings 𝑇𝑛𝑥=𝑛𝑘=1𝛽𝑘𝑛𝑆𝑘𝑥(1.9) 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. Let 𝐻 be a real Hilbert space. Let 𝐹 be a 𝑘-Lipschitzian and 𝜂-strongly monotone operator on 𝐻 with 0<𝜇<2𝜂/𝑘2 and 𝑓Π𝐻 with 0<𝛾<𝜇(𝜂𝜇𝑘2/2)/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 𝑇𝐻𝐻 is defined by 𝑇𝑧=lim𝑛𝑇𝑛𝑧 such that the condition (1.6). Let {𝛼𝑛},{𝛽𝑛} be sequences in (0,1) satisfying the following conditions:(B1)lim𝑛𝛼𝑛=0 and 𝑛=1𝛼𝑛=; (B2)𝑏=limsup𝑛𝛽𝑛<1 and 𝑛=1|𝛽𝑛+1𝛽𝑛|<; (B3)|𝛼𝑛+1𝛼𝑛|𝑜(𝛼𝑛+1)+𝜎𝑛 with 𝜎𝑛0 and 𝑛=1𝜎𝑛<. Then, for arbitrary 𝑥1=𝑥𝐻, the sequence {𝑥𝑛}𝑛=1, defined by 𝑦𝑛=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝑇𝜇𝐹𝑛𝑥𝑛,𝑥𝑛+1=1𝛽𝑛𝑦𝑛+𝛽𝑛𝑇𝑛𝑦𝑛,𝑛1,(1.10) converges strongly to some 𝑞Fix(𝑇), which satisfies the variational inequality (𝜇𝐹𝛾𝑓)𝑞,𝑞𝑧0, for all 𝑧Fix(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 for 𝐹=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 sequences 𝛼𝑛 and 𝛽𝑛, we have the “control condition” (B3). It can obviously be replaced by one of the following ones:(B3-1)𝑛=1|𝛼𝑛+1𝛼𝑛|<;(B3-2)𝛼𝑛(0,1] for all 𝑛 and lim𝑛(𝛼𝑛+1/𝛼𝑛)=1. Indeed, (B3-1) implies (B3) by choosing 𝜎𝑛=|𝛼𝑛+1𝛼𝑛|, 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 Theorem 1.1, 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 [58], the references therein, and also Remark 3.1 of Peng and Yao [9]. Proofs are given there.

Let {𝑆𝑖𝐻𝐻} be a countable family of nonexpansive mappings, and let {𝜉𝑖} be a sequence of real numbers such that 0𝜉𝑖𝑐<1, for all 𝑖1. For any 𝑛1, define a mapping 𝑇𝑛𝐻𝐻 as follows: 𝑈𝑛,𝑛+1𝑈=𝐼,𝑛,𝑛=𝜉𝑛𝑆𝑛𝑈𝑛,𝑛+1+1𝜉𝑛𝑈𝐼,𝑛,𝑛1=𝜉𝑛1𝑆𝑛1𝑈𝑛,𝑛+1𝜉𝑛1𝑈𝐼,𝑛,𝑘=𝜉𝑘𝑆𝑘𝑈𝑛,𝑘+1+1𝜉𝑘𝑈𝐼,𝑛,2=𝜉2𝑆2𝑈𝑛,3+1𝜉2𝑈𝐼,𝑛,1=𝜉1𝑆1𝑈𝑛,2+1𝜉1𝑇𝐼,𝑛=𝑈𝑛,1.(1.11)

Proposition 1.3. Let 𝐻 be a real Hilbert space, {𝑆𝑖𝐻𝐻} a sequence nonexpansive mappings with 𝑖=1Fix(𝑆𝑖), and {𝜉𝑖} a real sequence such that 0𝜉𝑖𝑐<1, for all 𝑖1. Define {𝑇𝑛} as above. Then(1)𝑇𝑛 is a nonexpansive and Fix(𝑇𝑛)=𝑛𝑖=1Fix(𝑆𝑖) for each 𝑛1;(2) the mappings {𝑇𝑛} satisfy the condition (1.5);(3) defining 𝑇𝐻𝐻 by 𝑇𝑥=lim𝑛𝑇𝑛𝑥, we have Fix(𝑇)=𝑛=1𝑇Fix𝑛=𝑛=1𝑆Fix𝑛.(1.12)

2. Proof of the Main Results

Ahead of the proof, we start with recalling some known auxiliary results.

Lemma 2.1 (see [10]). Let 𝐹 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=(𝐼𝑡𝜇𝐹)𝐻𝐻 is a contraction with contractive coefficient 1𝑡𝜏 and 𝜏=(1/2)𝜇(2𝜂𝜇𝑘2).

Lemma 2.2 (see [2]). Let 𝐻 be a Hilbert space. Assume that {𝑥𝑡} is defined by 𝑥𝑡𝑥=𝑡𝛾𝑓𝑡+(𝐼𝑡𝜇𝐹)𝑇𝑥𝑡,(2.1) 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,𝑓Π𝐻, and 𝑇𝐻𝐻 is a nonexpansive mapping. Then 𝑥𝑡 converges strongly as 𝑡0 to a fixed point 𝑞 of 𝑇, which solves the variational inequality (𝜇𝐹𝛾𝑓)𝑞,𝑞𝑝0,𝑝𝐹(𝑇).

Some more auxiliary results are given next.

Lemma 2.3 (see [11]). Let 𝐻 be a Hilbert space, 𝐶 a closed convex subset of 𝐻, and 𝑇𝐶𝐶 a nonexpansive mapping with Fix(𝑇). If {𝑥𝑛} is a sequence in 𝐶 which converges weakly to 𝑥 and if {(𝐼𝑇)𝑥𝑛} converges strongly to 𝑦, then (𝐼𝑇)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=1𝛿𝑛=, {𝜇𝑛} a sequence of nonnegative real numbers with 𝑛=1𝜇𝑛<, and {𝜃𝑛} a sequence of real numbers with limsup𝑛𝜃𝑛0. Suppose that 𝑠𝑛+11𝛿𝑛𝑠𝑛+𝛿𝑛𝜃𝑛+𝜇𝑛(2.2) for all 𝑛. Then lim𝑛𝑠𝑛=0.

Now we are prepared to prove the main result.

Proof of Theorem 1.2. We divide the proof into seven steps.
Step 1. We claim that {𝑥𝑛} is bounded. Taking any point 𝑞𝑛=1Fix(𝑇𝑛) and using Lemma 2.1, we obtain 𝑥𝑛+1=𝑞1𝛽𝑛𝑦𝑛𝑞+𝛽𝑛𝑇𝑛𝑦𝑛𝑦𝑞𝑛=𝛼𝑞𝑛𝑥𝛾𝑓𝑛+𝜇𝐹𝑞𝐼𝛼𝑛𝑇𝜇𝐹𝑛𝑥𝑛𝐼𝛼𝑛𝑞𝜇𝐹1𝛼𝑛𝜏𝑥𝑛𝑞+𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐹𝑞1𝛼𝑛𝜏𝑥𝑛𝑞+𝛼𝑛𝑥𝛾𝑓𝑛𝛾𝑓(𝑞)+𝛼𝑛𝛾𝑓(𝑞)𝜇𝐹𝑞1𝛼𝑛𝜏𝑥𝑛𝑞+𝛼𝑛𝑥𝛾𝛼𝑛𝑞+𝛼𝑛𝛾𝑓(𝑞)𝜇𝐹𝑞1𝛼𝑛𝑥(𝜏𝛾𝛼)𝑛𝑞+𝛼𝑛(𝜏𝛾𝛼)𝛾𝑓(𝑞)𝜇𝐹𝑞𝑥𝜏𝛾𝛼max𝑛,(𝑞𝛾𝑓𝑞)𝜇𝐹𝑞.𝜏𝛾𝛼(2.3) By induction, it follows 𝑥𝑛𝑥𝑞max1,(𝑞𝛾𝑓𝑞)𝜇𝐹𝑞𝜏𝛾𝛼,𝑛1,(2.4) 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)}.
Step 2. We show that lim𝑛𝑥𝑛+1𝑥𝑛=0.
Let 𝑀=sup{𝑇𝑛𝑦𝑛𝑦𝑛,𝑓(𝑥𝑛),𝐹𝑇𝑛𝑥𝑛𝑛}<. From the definition of {𝑥𝑛}, we obtain 𝑥𝑛+2𝑥𝑛+1=1𝛽𝑛+1𝑦𝑛+1+𝛽𝑛+1𝑇𝑛+1𝑦𝑛+11𝛽𝑛𝑦𝑛𝛽𝑛𝑇𝑛𝑦𝑛=1𝛽𝑛+1𝑦𝑛+1+𝛽𝑛+1𝑇𝑛+1𝑦𝑛+11𝛽𝑛+1𝑦𝑛𝛽𝑛+1𝛽𝑛𝑦𝑛𝛽𝑛+1𝑇𝑛𝑦𝑛+𝛽𝑛+1𝛽𝑛𝑇𝑛𝑦𝑛=1𝛽𝑛+1𝑦𝑛+1𝑦𝑛+𝛽𝑛+1𝑇𝑛+1𝑦𝑛+1𝑇𝑛𝑦𝑛+𝛽𝑛+1𝛽𝑛𝑇𝑛𝑦𝑛𝑦𝑛1𝛽𝑛+1𝑦𝑛+1𝑦𝑛+𝛽𝑛+1𝑇𝑛+1𝑦𝑛+1𝑇𝑛𝑦𝑛+||𝛽𝑛+1𝛽𝑛||𝑀1𝛽𝑛+1𝑦𝑛+1𝑦𝑛+𝛽𝑛+1𝑇𝑛+1𝑦𝑛+1𝑇𝑛𝑦𝑛+1+𝛽𝑛+1𝑦𝑛+1𝑦𝑛+||𝛽𝑛+1𝛽𝑛||𝑀𝑦𝑛+1𝑦𝑛+𝑇𝑛+1𝑦𝑛+1𝑇𝑛𝑦𝑛+1+||𝛽𝑛+1𝛽𝑛||𝑀(2.5) for all 𝑛. From (1.10), we have 𝑦𝑛=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝑇𝜇𝐹𝑛𝑥𝑛,𝑦𝑛+1=𝛼𝑛+1𝑥𝛾𝑓𝑛+1+𝐼𝛼𝑛+1𝑇𝜇𝐹𝑛+1𝑥𝑛+1.(2.6) With 𝑀1=(𝜇+𝛾)𝑀, we thus have 𝑦𝑛+1𝑦𝑛=𝛼𝑛+1𝑥𝛾𝑓𝑛+1+𝐼𝛼𝑛+1𝑇𝜇𝐹𝑛+1𝑥𝑛+1𝛼𝑛𝑥𝛾𝑓𝑛𝐼𝛼𝑛𝑇𝜇𝐹𝑛𝑥𝑛=𝐼𝛼𝑛+1𝑇𝜇𝐹𝑛+1𝑥𝑛+1𝐼𝛼𝑛+1𝑇𝜇𝐹𝑛𝑥𝑛𝛼𝑛+1𝛼𝑛𝜇𝐹𝑇𝑛𝑥𝑛+𝛼𝑛+1𝛾𝑓𝑥𝑛+1𝑥𝑓𝑛+𝛼𝑛+1𝛼𝑛𝑥𝛾𝑓𝑛1𝛼𝑛+1𝜏𝑇𝑛+1𝑥𝑛+1𝑇𝑛𝑥𝑛+||𝛼𝑛+1𝛼𝑛||𝜇𝐹𝑇𝑛𝑥𝑛+𝛼𝑛+1𝑥𝛾𝛼𝑛+1𝑥𝑛||𝛼+𝛾𝑛+1𝛼𝑛||𝑓𝑥𝑛1𝛼𝑛+1𝜏𝑇𝑛+1𝑥𝑛+1𝑇𝑛𝑥𝑛+𝛼𝑛+1𝑥𝛾𝛼𝑛+1𝑥𝑛+||𝛼𝑛+1𝛼𝑛||𝑀11𝛼𝑛+1𝜏𝑇𝑛+1𝑥𝑛+1𝑇𝑛+1𝑥𝑛+𝑇𝑛+1𝑥𝑛𝑇𝑛𝑥𝑛+𝛼𝑛+1𝑥𝛾𝛼𝑛+1𝑥𝑛+𝑀1||𝛼𝑛+1𝛼𝑛||1𝛼𝑛+1𝑥(𝜏𝛾𝛼)𝑛+1𝑥𝑛+𝑇𝑛+1𝑥𝑛𝑇𝑛𝑥𝑛+𝑀1||𝛼𝑛+1𝛼𝑛||.(2.7)
Combining (2.5) and (2.7), we conclude that 𝑥𝑛+2𝑥𝑛+11𝛼𝑛+1𝑥(𝜏𝛾𝛼)𝑛+1𝑥𝑛+𝑀2||𝛼𝑛+1𝛼𝑛||+||𝛽𝑛+1𝛽𝑛||𝑇+2sup𝑛+1𝑧𝑇𝑛𝑧,𝑧𝐵(2.8) where 𝑀2=max{𝑀,𝑀1}. Using the assumption (B3), we obtain 𝑥𝑛+2𝑥𝑛+11𝛼𝑛+1𝑥(𝜏𝛾𝛼)𝑛+1𝑥𝑛+𝑀2𝑜𝛼𝑛+1+𝜇𝑛(2.9) with 𝜇𝑛=𝑀2(𝜎𝑛+|𝛽𝑛+1𝛽𝑛|)+2sup{𝑇𝑛+1𝑧𝑇𝑛𝑧𝑧𝐵}. The various assumptions imply that 𝑛=1𝜇𝑛<. Now it follows from Lemma 2.4 that lim𝑛𝑥𝑛+1𝑥𝑛=0.
Step 3. Next we show that lim𝑛𝑇𝑛𝑥𝑛𝑦𝑛=lim𝑛𝑥𝑛𝑦𝑛=0.(2.10) In fact, from Step 1, {𝛾𝑓(𝑥𝑛)𝜇𝐹𝑇𝑛𝑥𝑛} is bounded. Moreover, it follows from lim𝑛𝛼𝑛=0 that 𝑇𝑛𝑥𝑛𝑦𝑛=𝛼𝑛𝑥𝛾𝑓𝑛𝛼𝑛𝜇𝐹𝑇𝑛𝑥𝑛=𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐹𝑇𝑛𝑥𝑛0(𝑛).(2.11) This is the first part. From condition (B2), we obtain, for 𝑛 sufficiently large and 𝑏<𝑏1<1, 𝑥𝑛+1𝑦𝑛=𝛽𝑛𝑇𝑛𝑦𝑛𝑦𝑛𝛽𝑛𝑇𝑛𝑦𝑛𝑇𝑛𝑥𝑛+𝑇𝑛𝑥𝑛𝑦𝑛𝑏1𝑦𝑛𝑥𝑛+𝑇𝑛𝑥𝑛𝑦𝑛𝑏1𝑦𝑛𝑥𝑛+1+𝑥𝑛+1𝑥𝑛+𝑇𝑛𝑥𝑛𝑦𝑛,(2.12) which implies 𝑥𝑛+1𝑦𝑛𝑏11𝑏1𝑥𝑛+1𝑥𝑛+𝑇𝑛𝑥𝑛𝑦𝑛.(2.13) Hence, using Step 2 and the above, we obtain lim𝑛𝑥𝑛+1𝑦𝑛=0 and thus lim𝑛𝑥𝑛𝑦𝑛=0.
Step 4. We show that lim𝑛𝑦𝑛𝑇𝑛𝑦𝑛=0.
Indeed, we have 𝑦𝑛𝑇𝑛𝑦𝑛𝑦𝑛𝑇𝑛𝑥𝑛+𝑇𝑛𝑥𝑛𝑇𝑛𝑦𝑛𝑦𝑛𝑇𝑛𝑥𝑛+𝑥𝑛𝑦𝑛,(2.14) and now the statement follows from Step 3.
Step 5. We show that 𝑇𝑦𝑛𝑦𝑛0 as 𝑛.
Observe that 𝑇𝑦𝑛𝑦𝑛𝑇𝑦𝑛𝑇𝑛𝑦𝑛+𝑇𝑛𝑦𝑛𝑦𝑛sup𝑇𝑧𝑇𝑛𝑧+𝑇𝑧𝐵𝑛𝑦𝑛𝑦𝑛,(2.15) and now apply Step 4 and the fact that 𝑇𝑛 converges uniformly on 𝐵 to 𝑇 (which is a consequence of (1.5)).
Step 6. We claim that limsup𝑛𝛾𝑓(𝑞)𝜇𝐹𝑞,𝑦𝑛𝑞0, where 𝑞=lim𝑡0𝑥𝑡 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 𝑐=lim𝑘𝛾𝑓(𝑞)𝜇𝐹𝑞,𝑦𝑛𝑘𝑞(2.16) exists and is greater than zero. Since {𝑦𝑛} is bounded, there exists a subsequence of {𝑦𝑛𝑘}, which converges weakly to some 𝑧𝐻. We can assume without loss of generality that {𝑦𝑛𝑘} is this subsequence. From 𝑇𝑦𝑛𝑦𝑛0, we obtain that 𝑇𝑦𝑛𝑘 converges weakly to 𝑧. Now Lemma 2.3 implies that 𝑧Fix(𝑇). Hence from weak convergence and Lemma 2.2, lim𝑘𝛾𝑓(𝑞)𝜇𝐹𝑞,𝑦𝑛𝑘𝑞=(𝛾𝑓𝜇𝐹)𝑞,𝑧𝑞0.(2.17) This is a contradiction.
Step 7. We show that {𝑥𝑛} converges strongly to 𝑞. From (1.10), we have 𝑥𝑛+1𝑞2=1𝛽𝑛𝑦𝑛𝑞+𝛽𝑛𝑇𝑛𝑦𝑛𝑞2𝑦𝑛𝑞2=𝛼𝑛𝑥𝛾𝑓𝑛+𝐼𝛼𝑛𝑇𝜇𝐹𝑛𝑥𝑛𝑞2=𝛼𝑛𝑥𝛾𝑓𝑛+𝜇𝐹𝑞𝐼𝛼𝑛𝑇𝜇𝐹𝑛𝑥𝑛𝐼𝛼𝑛𝑞𝜇𝐹2.(2.18) Using the general inequality 𝑥+𝑦2𝑥2+2𝑦,𝑥+𝑦, we can further estimate the above by 𝐼𝛼𝑛𝑇𝜇𝐹𝑛𝑥𝑛𝐼𝛼𝑛𝑞𝜇𝐹2+2𝛼𝑛𝑥𝛾𝑓𝑛𝜇𝐹𝑞,𝑦𝑛𝑞1𝛼𝑛𝜏2𝑥𝑛𝑞2+2𝛼𝑛𝑥𝛾𝑓𝑛𝛾𝑓(𝑞),𝑦𝑛𝑞+2𝛼𝑛𝛾𝑓(𝑞)𝜇𝐹𝑞,𝑦𝑛𝑞1𝛼𝑛𝜏2𝑥𝑛𝑞2+2𝛼𝑛𝑥𝛾𝛼𝑛𝑦𝑞𝑛𝑥𝑛+𝑥𝑛𝑞+2𝛼𝑛𝛾𝑓(𝑞)𝜇𝐹𝑞,𝑦𝑛𝑞12𝛼𝑛𝑥(𝜏𝛾𝛼)𝑛𝑞2+𝛼𝑛𝜏2𝑥𝑛𝑞2+2𝛼𝑛𝑥𝛾𝛼𝑛𝑦𝑞𝑛𝑥𝑛+2𝛼𝑛𝛾𝑓(𝑞)𝜇𝐹𝑞,𝑦𝑛𝑞.(2.19) Introducing 𝑀3=sup{𝑥𝑛𝑞𝑛1}, 𝛿𝑛=2𝛼𝑛(𝜏𝛾𝛼)>0, and 𝜃𝑛=𝛼𝑛𝜏2𝑀23+2(𝜏𝛾𝛼)𝛾𝛼𝑀3𝑦𝜏𝛾𝛼𝑛𝑥𝑛+1𝜏𝛾𝛼𝛾𝑓(𝑞)𝜇𝐹𝑞,𝑦𝑛𝑞,(2.20) we obtain 𝑥𝑛+1𝑞21𝛿𝑛𝑥𝑛𝑞2+𝛿𝑛𝜃𝑛.(2.21) Because 𝛿𝑛0, 𝑛=1𝛿𝑛=, and limsup𝑛𝜃𝑛0, 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 of 𝑞 as 𝑞=lim𝑡0𝑥𝑡 (see Step 6) and from Lemma 2.2, we obtain (𝜇𝐹𝛾𝑓)𝑞,𝑞𝑧0, for all 𝑧Fix(𝑇)=𝑛=1Fix(𝑇𝑛). This completes the proof.

Acknowledgment

This author is supported by the Natural Science Foundation of Yancheng Teachers University under Grant (11YCKL009).

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