Abstract and Applied Analysis

Abstract and Applied Analysis / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 929037 | 18 pages | https://doi.org/10.1155/2011/929037

Strong Convergence Theorems for a Countable Family of Nonexpansive Mappings in Convex Metric Spaces

Academic Editor: Stefan Siegmund
Received03 Feb 2011
Accepted20 Jun 2011
Published18 Sep 2011

Abstract

We introduce a new modified Halpern iteration for a countable infinite family of nonexpansive mappings {𝑇𝑛} in convex metric spaces. We prove that the sequence {𝑥𝑛} generated by the proposed iteration is an approximating fixed point sequence of a nonexpansive mapping when {𝑇𝑛} satisfies the AKTT-condition, and strong convergence theorems of the proposed iteration to a common fixed point of a countable infinite family of nonexpansive mappings in CAT(0) spaces are established under AKTT-condition and the SZ-condition. We also generalize the concept of W-mapping for a countable infinite family of nonexpansive mappings from a Banach space setting to a convex metric space and give some properties concerning the common fixed point set of this family in convex metric spaces. Moreover, by using the concept of W-mappings, we give an example of a sequence of nonexpansive mappings defined on a convex metric space which satisfies the AKTT-condition. Our results generalize and refine many known results in the current literature.

1. Introduction

Let 𝐶 be a nonempty closed convex subset of a metric space (𝑋,𝑑), and let 𝑇 be a mapping of 𝐶 into itself. A mapping 𝑇 is called nonexpansive if 𝑑(𝑇𝑥,𝑇𝑦)𝑑(𝑥,𝑦) for all 𝑥,𝑦𝐶. The set of all fixed points of 𝑇 is denoted by 𝐹(𝑇), that is, 𝐹(𝑇)={𝑥𝐶𝑥=𝑇𝑥}.

In 1967, Halpern [1] introduced the following iterative scheme in Hilbert spaces which was referred to as Halpern iteration for approximating a fixed point of 𝑇:𝑥𝑛+1=𝛼𝑛𝑢+1𝛼𝑛𝑇𝑥𝑛𝑛,(1.1) where 𝑥1,𝑢𝐶 are arbitrarily chosen, and {𝛼𝑛} is a sequence in [0,1]. Wittmann [2] studied the iterative scheme (1.1) in a Hilbert space and obtained the strong convergence of the iteration. Reich [3] and Shioji and Takahashi [4] extended Wittmann's result to a real Banach space.

The modified version of Halpern iteration was investigated widely by many mathematicians. For instance, Kim and Xu [5] studied the sequence {𝑥𝑛} generated as follows:𝑦𝑛=𝛼𝑛𝑥𝑛+1𝛼𝑛𝑇𝑥𝑛,𝑥𝑛+1=𝛽𝑛𝑢+1𝛽𝑛𝑦𝑛𝑛,(1.2) where 𝑥1,𝑢𝐶 are arbitrarily chosen and {𝛼𝑛}, {𝛽𝑛} are two sequences in [0,1]. They proved the strong convergence of iterative scheme (1.2) in the framework of a uniformly smooth Banach space. In 2007, Aoyama et al. [6] introduced a Halpern iteration for finding a common fixed point of a countable infinite family of nonexpansive mappings in a Banach space as follows:𝑥𝑛+1=𝛼𝑛𝑢+1𝛼𝑛𝑇𝑛𝑥𝑛𝑛,(1.3) where 𝑥1,𝑢𝐶 are arbitrarily chosen, {𝛼𝑛} is a sequence in [0,1], and {𝑇𝑛} is a sequence of nonexpansive mappings with some conditions. They proved that the sequence {𝑥𝑛} generated by (1.3) converges strongly to a common fixed point of {𝑇𝑛}. In 2010, Saejung [7] extended the results of Halpern [1], Wittmann [2], Reich [3], Shioji and Takahashi [4], and Aoyama et al. [6] to the case of a CAT(0) space which is an example of a convex metric space. Recently, Cuntavepanit and Panyanak [8] extended the result of Kim and Xu [5] to a CAT(0) space.

Takahashi [9] introduced the concept of convex metric spaces by using the convex structure as follows. Let (𝑋,𝑑) be a metric space. A mapping 𝑊𝑋×𝑋×[0,1]𝑋 is said to be a convex structure on 𝑋 if for each 𝑥,𝑦𝑋 and 𝜆[0,1], 𝑑(𝑧,𝑊(𝑥,𝑦,𝜆))𝜆𝑑(𝑧,𝑥)+(1𝜆)𝑑(𝑧,𝑦),(1.4) for all 𝑧𝑋. A metric space (𝑋,𝑑) together with a convex structure 𝑊 is called a convex metric space which will be denoted by (𝑋,𝑑,𝑊). A nonempty subset 𝐶 of 𝑋 is said to be convex if 𝑊(𝑥,𝑦,𝜆)𝐶 for all 𝑥,𝑦𝐶 and 𝜆[0,1]. Clearly, a normed space and each of its convex subsets are convex metric spaces, but the converse does not hold.

Motivated by the above results, we introduce a new iterative scheme for finding a common fixed point of a countable infinite family of nonexpansive mappings {𝑇𝑛} of 𝐶 into itself in a convex metric space as follows:𝑦𝑛=𝑊𝑢,𝑇𝑛𝑥𝑛,𝛼𝑛,𝑥𝑛+1𝑦=𝑊𝑛,𝑇𝑛𝑦𝑛,𝛽𝑛𝑛,(1.5) where 𝑥1,𝑢𝐶 are arbitrarily chosen, and {𝛼𝑛}, {𝛽𝑛} are two sequences in [0,1]. The main propose of this paper is to prove the convergence theorem of the sequence {𝑥n} generated by (1.5) to a common fixed point of a countable infinite family of nonexpansive mappings in convex metric spaces and CAT(0) spaces under certain suitable conditions.

2. Preliminaries

We recall some definitions and useful lemmas used in the main results.

Lemma 2.1 (see [9, 10]). Let (𝑋,𝑑,𝑊) be a convex metric space. For each 𝑥,𝑦𝑋 and 𝜆,𝜆1,𝜆2[0,1], we have the following. (i)𝑊(𝑥,𝑥,𝜆)=𝑥,𝑊(𝑥,𝑦,0)=𝑦 and 𝑊(𝑥,𝑦,1)=𝑥.(ii)𝑑(𝑥,𝑊(𝑥,𝑦,𝜆))=(1𝜆)𝑑(𝑥,𝑦) and 𝑑(𝑦,𝑊(𝑥,𝑦,𝜆))=𝜆𝑑(𝑥,𝑦). (iii)𝑑(𝑥,𝑦)=𝑑(𝑥,𝑊(𝑥,𝑦,𝜆))+𝑑(𝑊(𝑥,𝑦,𝜆),𝑦). (iv)|𝜆1𝜆2|𝑑(𝑥,𝑦)𝑑(𝑊(𝑥,𝑦,𝜆1),𝑊(𝑥,𝑦,𝜆2)).

We say that a convex metric space (𝑋,𝑑,𝑊) has the property: (C)if 𝑊(𝑥,𝑦,𝜆)=𝑊(𝑦,𝑥,1𝜆) for all 𝑥,𝑦𝑋 and 𝜆[0,1], (I)if 𝑑(𝑊(𝑥,𝑦,𝜆1),𝑊(𝑥,𝑦,𝜆2))|𝜆1𝜆2|𝑑(𝑥,𝑦) for all 𝑥,𝑦𝑋 and 𝜆1,𝜆2[0,1], (H)if 𝑑(𝑊(𝑥,𝑦,𝜆),𝑊(𝑥,𝑧,𝜆))(1𝜆)𝑑(𝑦,𝑧) for all 𝑥,𝑦,𝑧𝑋 and 𝜆[0,1], (S)if 𝑑(𝑊(𝑥,𝑦,𝜆),𝑊(𝑧,𝑤,𝜆))𝜆𝑑(𝑥,𝑧)+(1𝜆)𝑑(𝑦,𝑤) for all 𝑥,𝑦,𝑧,𝑤𝑋 and 𝜆[0,1].

From the above properties, it is obvious that the property (C) and (H) imply continuity of a convex structure 𝑊𝑋×𝑋×[0,1]𝑋. Clearly, the property (S) implies the property (H). In [10], Aoyama et al. showed that a convex metric space with the property (C) and (H) has the property (S).

In 1996, Shimizu and Takahashi [11] introduced the concept of uniform convexity in convex metric spaces and studied some properties of these spaces. A convex metric space (𝑋,𝑑,𝑊) is said to be uniformly convex if for any 𝜀>0, there exists 𝛿=𝛿(𝜀)>0 such that for all 𝑟>0 and 𝑥,𝑦,𝑧𝑋 with 𝑑(𝑧,𝑥)𝑟, 𝑑(𝑧,𝑦)𝑟 and 𝑑(𝑥,𝑦)𝑟𝜀 imply that 𝑑(𝑧,𝑊(𝑥,𝑦,1/2))(1𝛿)𝑟. Obviously, uniformly convex Banach spaces are uniformly convex metric spaces. In fact, the property (I) holds in uniformly convex metric spaces, see [12].

Lemma 2.2. Property (C) holds in uniformly convex metric spaces.

Proof. Suppose that (𝑋,𝑑,𝑊) is a uniformly convex metric space. Let 𝑥,𝑦𝑋 and 𝜆[0,1]. It is obvious that the conclusion holds if 𝜆=0 or 𝜆=1. So, suppose 𝜆(0,1). By Lemma 2.1(ii), we have 𝑑𝑑(𝑥,𝑊(𝑥,𝑦,𝜆))=(1𝜆)𝑑(𝑥,𝑦),𝑑(𝑦,𝑊(𝑥,𝑦,𝜆))=𝜆𝑑(𝑥,𝑦),(𝑥,𝑊(𝑦,𝑥,1𝜆))=(1𝜆)𝑑(𝑥,𝑦),𝑑(𝑦,𝑊(𝑦,𝑥,1𝜆))=𝜆𝑑(𝑥,𝑦).(2.1)
We will show that 𝑊(𝑥,𝑦,𝜆)=𝑊(𝑦,𝑥,1𝜆). To show this, suppose not. Put 𝑧1=𝑊(𝑥,𝑦,𝜆) and 𝑧2=𝑊(𝑦,𝑥,1𝜆). Let 𝑟1=(1𝜆)𝑑(𝑥,𝑦)>0, 𝑟2=𝜆𝑑(𝑥,𝑦)>0, 𝜀1=𝑑(𝑧1,𝑧2)/𝑟1, and 𝜀2=𝑑(𝑧1,𝑧2)/𝑟2. It is easy to see that 𝜀1,𝜀2>0. Since (𝑋,𝑑,𝑊) is uniformly convex, we have 𝑑𝑧𝑥,𝑊1,𝑧2,12𝑟1𝜀1𝛿1𝑧,𝑑𝑦,𝑊1,𝑧2,12𝑟2𝜀1𝛿2.(2.2) By 𝜆(0,1), we get 𝑥𝑦. Since 𝛿(𝜀1)>0 and 𝛿(𝜀2)>0, then 𝑧𝑑(𝑥,𝑦)𝑑𝑥,𝑊1,𝑧2,12𝑧+𝑑𝑦,𝑊1,𝑧2,12𝑟1𝜀1𝛿1+𝑟2𝜀1𝛿2<𝑟1+𝑟2=𝑑(𝑥,𝑦).(2.3) This is a contradiction. Hence, 𝑊(𝑥,𝑦,𝜆)=𝑊(𝑦,𝑥,1𝜆).

By Lemma 2.2, it is clear that a uniformly convex metric space (𝑋,𝑑,𝑊) with the property (H) has the property (S), and the convex structure 𝑊 is also continuous.

Next, we recall the special space of convex metric spaces, namely, CAT(0) spaces. Let (𝑋,𝑑) be a metric space. A geodesic path joining 𝑥𝑋 to 𝑦𝑋 (or, more briefly, a geodesic from 𝑥 to 𝑦) is a map 𝑐 from a closed interval [0,𝑙] to 𝑋 such that 𝑐(0)=𝑥,𝑐(𝑙)=𝑦 and 𝑑(𝑐(𝑡1),𝑐(𝑡2))=|𝑡1𝑡2| for all 𝑡1,𝑡2[0,𝑙]. In particular, 𝑐 is an isometry and 𝑑(𝑥,𝑦)=𝑙. The image 𝛼 of 𝑐 is called a geodesic (or metric) segment joining 𝑥 and 𝑦. When unique, this geodesic is denoted [𝑥,𝑦]. The space (𝑋,𝑑) is said to be a geodesic metric space if every two points of 𝑋 are joined by a geodesic, and 𝑋 is said to be uniquely geodesic if there is exactly one geodesic joining 𝑥 and 𝑦 for each 𝑥,𝑦𝑋. A subset 𝑌 of 𝑋 is said to be convex if 𝑌 includes every geodesic segment joining any two of its points.

A geodesic triangle (𝑥1,𝑥2,𝑥3) in a geodesic metric space (𝑋,𝑑) consists of three points 𝑥1,𝑥2,𝑥3 in 𝑋 (the vertices of ) and a geodesic segment between each pair of vertices (the edges of ). A comparison triangle for geodesic triangle (𝑥1,𝑥2,𝑥3) in (𝑋,𝑑) is a triangle (𝑥1,𝑥2,𝑥3)=(𝑥1,𝑥2,𝑥3) in the Euclidean plane 𝔼2 such that 𝑑𝔼2(𝑥𝑖,𝑥𝑗)=𝑑(𝑥𝑖,𝑥𝑗) for 𝑖,𝑗{1,2,3}.

A geodesic metric space is said to be a CAT(0) space if all geodesic triangles satisfy the following comparison axiom. Let be a geodesic triangle in 𝑋, and let be a comparison triangle for . Then is said to satisfy the CAT(0) inequality if for all 𝑥,𝑦 and all comparison points 𝑥,𝑦, 𝑑(𝑥,𝑦)𝑑𝔼2(𝑥,𝑦).

If 𝑧,𝑥,𝑦 are points in a CAT(0) space and if 𝑚 is the midpoint of the segment [𝑥,𝑦], then the CAT(0) inequality implies 𝑑(𝑧,𝑚)212𝑑(𝑧,𝑥)2+12𝑑(𝑧,𝑦)214𝑑(𝑥,𝑦)2.(CN) This is the (CN) inequality of Bruhat and Tits [13], which is equivalent to 𝑑(𝑧,𝜆𝑥(1𝜆)𝑦)2𝜆𝑑(𝑧,𝑥)2+(1𝜆)𝑑(𝑧,𝑦)2𝜆(1𝜆)𝑑(𝑥,𝑦)2,(CN) for any 𝜆[0,1], where 𝜆𝑥(1𝜆)𝑦 denotes the unique point in [𝑥,𝑦]. The (CN*) inequality has appeared in [14]. By using the (CN) inequality, it is easy to see that the CAT(0) spaces are uniformly convex. In fact [15], a geodesic metric space is a CAT(0) space if and only if it satisfies the (CN) inequality. Moreover, if 𝑋 is CAT(0) space and 𝑥,𝑦𝑋, then for any 𝜆[0,1], there exists a unique point 𝜆𝑥(1𝜆)𝑦[𝑥,𝑦] such that𝑑(𝑧,𝜆𝑥(1𝜆)𝑦)𝜆𝑑(𝑧,𝑥)+(1𝜆)𝑑(𝑧,𝑦),(2.4) for any 𝑧𝑋. It follows that CAT(0) spaces have convex structure 𝑊(𝑥,𝑦,𝜆)=𝜆𝑥(1𝜆)𝑦. It is clear that the properties (C), (I), and (S) are satisfied for CAT(0) spaces, see [15, 16]. This is also true for Banach spaces.

Let 𝜇 be a continuous linear functional on 𝑙, the Banach space of bounded real sequences, and let (𝑎1,𝑎2,)𝑙. We write 𝜇𝑛(𝑎𝑛) instead of 𝜇((𝑎1,𝑎2,)). We call 𝜇 a Banach limit if 𝜇 satisfies 𝜇=𝜇(1,1,)=1 and 𝜇𝑛(𝑎𝑛)=𝜇𝑛(𝑎𝑛+1) for each (𝑎1,𝑎2,)𝑙. For a Banach limit 𝜇, we know that liminf𝑛𝑎𝑛𝜇𝑛(𝑎𝑛)limsup𝑛𝑎𝑛 for all (𝑎1,𝑎2,)𝑙. So if (𝑎1,𝑎2,)𝑙 with lim𝑛𝑎𝑛=𝑐, then 𝜇𝑛(𝑎𝑛)=𝑐, see also [17].

Lemma 2.3 ([4], Proposition 2). Let (𝑎1,𝑎2,)𝑙 be such that 𝜇𝑛(𝑎𝑛)0 for all Banach limit 𝜇. If limsup𝑛(𝑎𝑛+1𝑎𝑛)0, then limsup𝑛𝑎𝑛0.

Lemma 2.4 ([6], Lemma 2.3). Let {𝑠𝑛} be a sequence of nonnegative real numbers, let {𝛼𝑛} be a sequence of real numbers in [0,1] with 𝑛=1𝛼𝑛=, let {𝛿𝑛} be a sequence of nonnegative real numbers with 𝑛=1𝛿𝑛<, and let {𝛾𝑛}be a sequence of real numbers with limsup𝑛𝛾𝑛0. Suppose that 𝑠𝑛+11𝛼𝑛𝑠𝑛+𝛼𝑛𝛾𝑛+𝛿𝑛𝑛.(2.5) Then lim𝑛𝑠𝑛=0.

Lemma 2.5 ([18], Lemma 1). Let (𝑋,𝑑,𝑊) be a uniformly convex metric space with a continuous convex structure 𝑊𝑋×𝑋×[0,1]𝑋. Then for arbitrary positive number 𝜀 and 𝑟, there exists 𝜂=𝜂(𝜀)>0 such that 𝑑(𝑧,𝑊(𝑥,𝑦,𝜆))𝑟(12min{𝜆,1𝜆}𝜂),(2.6) for all 𝑥,𝑦,𝑧𝑋, 𝑑(𝑧,𝑥)𝑟, 𝑑(𝑧,𝑦)𝑟, 𝑑(𝑥,𝑦)𝑟𝜀, and 𝜆[0,1].

Remark 2.6. The above lemma also holds for a uniformly convex metric space with the property (H).

3. Main Results

The following condition was introduced by Aoyama et al. [6]. Let 𝐶 be a subset of a complete convex metric space (𝑋,𝑑,𝑊), and let {𝑇𝑛} be a countable infinite family of mappings from 𝐶 into itself. We say that {𝑇𝑛} satisfies AKTT-condition if𝑛=1𝑑𝑇sup𝑛+1𝑧,𝑇𝑛𝑧𝑧𝐵<,(3.1) for each bounded subset 𝐵 of 𝐶. If 𝐶 is a closed subset and {𝑇𝑛} satisfies AKTT-condition, then we can define a mapping 𝑇𝐶𝐶 such that 𝑇𝑥=lim𝑛𝑇𝑛𝑥 for all 𝑥𝐶. In this case, we also say that ({𝑇𝑛},𝑇) satisfies AKTT-condition. By using the same argument as in [6, Lemma 3.2], we have the following lemma.

Lemma 3.1. If ({𝑇𝑛},𝑇) satisfies AKTT-condition, then lim𝑛sup{𝑑(𝑇𝑧,𝑇𝑛𝑧)𝑧𝐵}=0 for all bounded subsets 𝐵 of 𝐶.

Theorem 3.2. Let 𝐶 be a nonempty closed convex subset of a complete convex metric space (𝑋,𝑑,𝑊) with the properties (I) and (S). Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself such that 𝑛=1𝐹(𝑇𝑛). Suppose that {𝑥𝑛} is a sequence of 𝐶 generated by (1.5), and let {𝛼𝑛} and {𝛽𝑛} be sequences in [0,1] which satisfy the conditions: (C1)0<𝛼𝑛<1, lim𝑛𝛼𝑛=0, 𝑛=1𝛼𝑛= and 𝑛=1|𝛼𝑛+1𝛼𝑛|<,(C2)𝛽𝑛(𝑏,1] for some 𝑏(0,1) and 𝑛=1|𝛽𝑛+1𝛽𝑛|<. Suppose that ({𝑇𝑛},𝑇) satisfies AKTT-condition. Then lim𝑛𝑑(𝑥𝑛+1,𝑥𝑛)=0 and lim𝑛𝑑(𝑇𝑥𝑛,𝑥𝑛)=0.

Proof. Let 𝑝𝑛=1𝐹(𝑇𝑛). By the definition of {𝑥𝑛} and {𝑦𝑛}, we have 𝑑𝑥𝑛+1𝑊𝑦,𝑝=𝑑𝑛,𝑇𝑛𝑦𝑛,𝛽𝑛,𝑝𝛽𝑛𝑑𝑦𝑛+,𝑝1𝛽𝑛𝑑𝑇𝑛𝑦𝑛𝑦,𝑝𝑑𝑛𝑊,𝑝=𝑑𝑢,𝑇𝑛𝑥𝑛,𝛼𝑛,𝑝𝛼𝑛𝑑(𝑢,𝑝)+1𝛼𝑛𝑑𝑇𝑛𝑥𝑛,𝑝𝛼𝑛𝑑(𝑢,𝑝)+1𝛼𝑛𝑑𝑥𝑛𝑥,𝑝max𝑑(𝑢,𝑝),𝑑𝑛.,𝑝(3.2) By induction on 𝑛, we obtain that 𝑑(𝑥𝑛,𝑝)max{𝑑(𝑢,𝑝),𝑑(𝑥1,𝑝)} for all 𝑛 and all 𝑝𝑛=1𝐹(𝑇𝑛). Hence, the sequence {𝑥𝑛} is bounded and so {𝑦𝑛}, {𝑇𝑛𝑥𝑛}, {𝑇𝑛𝑦𝑛} are bounded.
It follows by condition (𝐶1) that 𝑑𝑦𝑛,𝑇𝑛𝑥𝑛𝑊=𝑑𝑢,𝑇𝑛𝑥𝑛,𝛼𝑛,𝑇𝑛𝑥𝑛=𝛼𝑛𝑑𝑢,𝑇𝑛𝑥𝑛0.(3.3) By the definition of {𝑥𝑛} and {𝑦𝑛}, we have 𝑑𝑦𝑛,𝑦𝑛1𝑊=𝑑𝑢,𝑇𝑛𝑥𝑛,𝛼𝑛,𝑊𝑢,𝑇𝑛1𝑥𝑛1,𝛼𝑛1𝑊𝑑𝑢,𝑇𝑛𝑥𝑛,𝛼𝑛,𝑊𝑢,𝑇𝑛𝑥𝑛1,𝛼𝑛𝑊+𝑑𝑢,𝑇𝑛𝑥𝑛1,𝛼𝑛,𝑊𝑢,𝑇𝑛1𝑥𝑛1,𝛼𝑛𝑊+𝑑𝑢,𝑇𝑛1𝑥𝑛1,𝛼𝑛,𝑊𝑢,𝑇𝑛1𝑥𝑛1,𝛼𝑛11𝛼𝑛𝑑𝑇𝑛𝑥𝑛,𝑇𝑛𝑥𝑛1+1𝛼𝑛𝑑𝑇𝑛𝑥𝑛1,𝑇𝑛1𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑑𝑢,𝑇𝑛1𝑥𝑛11𝛼𝑛𝑑𝑥𝑛,𝑥𝑛1+1𝛼𝑛𝑑𝑇𝑛𝑥𝑛1,𝑇𝑛1𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑑𝑢,𝑇𝑛1𝑥𝑛11𝛼𝑛𝑑𝑥𝑛,𝑥𝑛1𝑇+𝑑𝑛𝑥𝑛1,𝑇𝑛1𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑑𝑢,𝑇𝑛1𝑥𝑛1,𝑑𝑥𝑛+1,𝑥𝑛𝑊𝑦=𝑑𝑛,𝑇𝑛𝑦𝑛,𝛽𝑛𝑦,𝑊𝑛1,𝑇𝑛1𝑦𝑛1,𝛽𝑛1𝑊𝑦𝑑𝑛,𝑇𝑛𝑦𝑛,𝛽𝑛𝑦,𝑊𝑛1,𝑇𝑛1𝑦𝑛1,𝛽𝑛𝑊𝑦+𝑑𝑛1,𝑇𝑛1𝑦𝑛1,𝛽𝑛𝑦,𝑊𝑛1,𝑇𝑛1𝑦𝑛1,𝛽𝑛1𝛽𝑛𝑑𝑦𝑛,𝑦𝑛1+1𝛽𝑛𝑑𝑇𝑛𝑦𝑛,𝑇𝑛1𝑦𝑛1+||𝛽𝑛𝛽𝑛1||𝑑𝑦𝑛1,𝑇𝑛1𝑦𝑛1𝛽𝑛𝑑𝑦𝑛,𝑦𝑛1+1𝛽𝑛𝑑𝑇𝑛𝑦𝑛,𝑇𝑛𝑦𝑛1𝑇+𝑑𝑛𝑦𝑛1,𝑇𝑛1𝑦𝑛1+||𝛽𝑛𝛽𝑛1||𝑑𝑦𝑛1,𝑇𝑛1𝑦𝑛1𝛽𝑛𝑑𝑦𝑛,𝑦𝑛1+1𝛽𝑛𝑑𝑦𝑛,𝑦𝑛1𝑇+𝑑𝑛𝑦𝑛1,𝑇𝑛1𝑦𝑛1+||𝛽𝑛𝛽𝑛1||𝑑𝑦𝑛1,𝑇𝑛1𝑦𝑛1𝑦𝑑𝑛,𝑦𝑛1𝑇+𝑑𝑛𝑦𝑛1,𝑇𝑛1𝑦𝑛1+||𝛽𝑛𝛽𝑛1||𝑑𝑦𝑛1,𝑇𝑛1𝑦𝑛11𝛼𝑛𝑑𝑥𝑛,𝑥𝑛1𝑇+𝑑𝑛𝑥𝑛1,𝑇𝑛1𝑥𝑛1+||𝛼𝑛𝛼𝑛1||𝑑𝑢,𝑇𝑛1𝑥𝑛1𝑇+𝑑𝑛𝑦𝑛1,𝑇𝑛1𝑦𝑛1+||𝛽𝑛𝛽𝑛1||𝑑𝑦𝑛1,𝑇𝑛1𝑦𝑛11𝛼𝑛𝑑𝑥𝑛,𝑥𝑛1+||𝛼𝑛𝛼𝑛1||+||𝛽𝑛𝛽𝑛1||𝑀𝑇+𝑑𝑛𝑥𝑛1,𝑇𝑛1𝑥𝑛1𝑇+𝑑𝑛𝑦𝑛1,𝑇𝑛1𝑦𝑛1,(3.4) where 𝑀=max{sup𝑛𝑑(𝑢,𝑇𝑛1𝑥𝑛1),sup𝑛𝑑(𝑦𝑛1,𝑇𝑛1𝑦𝑛1)}.
Putting 𝛿𝑛=(|𝛼𝑛𝛼𝑛1|+|𝛽𝑛𝛽𝑛1|)𝑀+𝑑(𝑇𝑛𝑥𝑛1,𝑇𝑛1𝑥𝑛1)+𝑑(𝑇𝑛𝑦𝑛1,𝑇𝑛1𝑦𝑛1), we have 𝑛=2𝛿𝑛𝑀𝑛=2||𝛼𝑛𝛼𝑛1||+||𝛽𝑛𝛽𝑛1||+𝑛=2𝑑𝑇sup𝑛𝑧,𝑇𝑛1𝑧𝑥𝑧𝑘+𝑛=2𝑑𝑇sup𝑛𝑧,𝑇𝑛1𝑧𝑦𝑧𝑘.(3.5) Hence, it follows from conditions (𝐶1), (𝐶2), AKTT-condition, and Lemma 2.4 that lim𝑛𝑑𝑥𝑛+1,𝑥𝑛=0.(3.6) Now, observe that 𝑑𝑥𝑛+1,𝑦𝑛𝑊𝑦=𝑑𝑛,𝑇𝑛𝑦𝑛,𝛽𝑛,𝑦𝑛=1𝛽𝑛𝑑𝑦𝑛,𝑇𝑛𝑦𝑛𝑑𝑦(1𝑏)𝑛,𝑇𝑛𝑥𝑛𝑇+𝑑𝑛𝑥𝑛,𝑇𝑛𝑥𝑛+1𝑇+𝑑𝑛𝑥𝑛+1,𝑇𝑛𝑦𝑛𝑑𝑦(1𝑏)𝑛,𝑇𝑛𝑥𝑛𝑥+𝑑𝑛,𝑥𝑛+1𝑥+𝑑𝑛+1,𝑦𝑛.(3.7) We obtain 𝑑𝑥𝑛+1,𝑦𝑛1𝑏𝑏𝑑𝑦𝑛,𝑇𝑛𝑥𝑛𝑥+𝑑𝑛,𝑥𝑛+1.(3.8) This implies by (3.3) and (3.6) that lim𝑛𝑑(𝑥𝑛+1,𝑦𝑛)=0. Therefore, we have 𝑑𝑥𝑛,𝑦𝑛𝑥𝑑𝑛,𝑥𝑛+1𝑥+𝑑𝑛+1,𝑦𝑛0.(3.9) Since 𝑑𝑇𝑛𝑥𝑛,𝑥𝑛𝑇𝑑𝑛𝑥𝑛,𝑦𝑛𝑦+𝑑𝑛,𝑥𝑛,(3.10) it follows by (3.3) and (3.9) that lim𝑛𝑑𝑇𝑛𝑥𝑛,𝑥𝑛=0.(3.11) By (3.11) and Lemma 3.1, we get 𝑑𝑇𝑥𝑛,𝑥𝑛𝑑𝑇𝑥𝑛,𝑇𝑛𝑥𝑛𝑇+𝑑𝑛𝑥𝑛,𝑥𝑛𝑑sup𝑇𝑧,𝑇𝑛𝑧𝑥𝑧𝑘𝑇+𝑑𝑛𝑥𝑛,𝑥𝑛0.(3.12)

Next, we consider a convergence theorem in CAT(0) spaces. The following two lemmas obtained by Saejung [7] are useful for our main results.

Lemma 3.3. Let 𝐶 be a closed convex subset of a complete CAT(0) space 𝑋, and let 𝑇𝐶𝐶 be a nonexpansive mapping. Let 𝑢𝐶 be fixed. For each 𝑡(0,1), the mapping 𝑆𝑡𝐶𝐶 defined by 𝑆𝑡𝑥=𝑡𝑢(1𝑡)𝑇𝑥 for 𝑥𝐶 has a unique fixed point 𝑥𝑡𝐶, that is, 𝑥𝑡=𝑆𝑡𝑥𝑡=𝑡𝑢(1𝑡)𝑇𝑥𝑡.

Lemma 3.4. Let 𝐶, 𝑇 be as the preceding lemma. Then 𝐹(𝑇) if and only if {𝑥𝑡} remains bounded as 𝑡0. In this case, the following statements hold: (i){𝑥𝑡} converges to the unique fixed point 𝑧 of 𝑇 which is nearest to 𝑢; (ii)𝑑(𝑢,𝑧)2𝜇𝑛𝑑(𝑢,𝑥𝑛)2 for all Banach limit 𝜇 and all bounded sequences {𝑥𝑛} with lim𝑛𝑑(𝑥𝑛,𝑇𝑥𝑛)=0.

Previously, we know that CAT(0) spaces have convex structure 𝑊(𝑥,𝑦,𝜆)=𝜆𝑥(1𝜆)𝑦 and also have the properties (C), (I), and (S). Thus, we have the following result.

Theorem 3.5. Let 𝐶 be a nonempty closed convex subset of a complete CAT(0) space 𝑋. Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself such that 𝑛=1𝐹(𝑇𝑛). Suppose that 𝑢,𝑥1𝐶 are arbitrarily chosen and {𝑥𝑛} is a sequence of 𝐶 generated by 𝑦𝑛=𝛼𝑛𝑢1𝛼𝑛𝑇𝑛𝑥𝑛,𝑥𝑛+1=𝛽𝑛𝑦𝑛1𝛽𝑛𝑇𝑛𝑦𝑛𝑛,(3.13) where {𝛼𝑛} and {𝛽𝑛} are sequences in [0,1] which satisfy the conditions (𝐶1) and (𝐶2) as in Theorem 3.2. Suppose that ({𝑇𝑛},𝑇) satisfies AKTT-condition. Then lim𝑛𝑑(𝑥𝑛+1,𝑥𝑛)=0 and lim𝑛𝑑(𝑇𝑥𝑛,𝑥𝑛)=0.

Theorem 3.6. Let 𝐶 be a nonempty closed convex subset of a complete CAT(0) space 𝑋. Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself such that 𝑛=1𝐹(𝑇𝑛). Suppose that {𝑥𝑛} is a sequence of 𝐶 generated by (3.13), and let {𝛼𝑛} and {𝛽𝑛} be sequences in [0,1] which satisfy the conditions (𝐶1) and (𝐶2) as in Theorem 3.2. Suppose that ({𝑇𝑛},𝑇) satisfies AKTT-condition and 𝐹(𝑇)=𝑛=1𝐹(𝑇𝑛). Then {𝑥𝑛} converges strongly to a common fixed point of {𝑇𝑛} which is nearest to 𝑢.

Proof. By Theorem 3.5, we have lim𝑛𝑑(𝑇𝑥𝑛,𝑥𝑛)=0. For each 𝑡(0,1), let 𝑧𝑡 be a unique point of 𝐶 such that 𝑧𝑡=𝑡𝑢(1𝑡)𝑇𝑧𝑡. It follows from Lemma 3.4 that {𝑧𝑡} converges to a point 𝑧𝐹(𝑇) which is nearest to 𝑢, and 𝑑(𝑢,𝑧)2𝜇𝑛𝑑𝑢,𝑥𝑛2forallBanachlimits𝜇,(3.14) that is, 𝜇𝑛(𝑑(𝑢,𝑧)2𝑑(𝑢,𝑥𝑛)2)0. Moreover, by Theorem 3.5, we get lim𝑛𝑑(𝑥𝑛+1,𝑥𝑛)=0. It follows that limsup𝑛𝑑(𝑢,𝑧)2𝑑𝑢,𝑥𝑛+12𝑑(𝑢,𝑧)2𝑑𝑢,𝑥𝑛2=0.(3.15) By lim𝑛𝑑(𝑇𝑛𝑥𝑛,𝑥𝑛)=0 and Lemma 2.3, we obtain limsup𝑛𝑑(𝑢,𝑧)21𝛼𝑛𝑑𝑢,𝑇𝑛𝑥𝑛2=limsup𝑛𝑑(𝑢,𝑧)2𝑑𝑢,𝑥𝑛20.(3.16) Finally, we show that lim𝑛𝑑(𝑥𝑛,𝑧)=0. By the definition of {𝑥𝑛} and {𝑦𝑛}, we have 𝑑𝑥𝑛+1,𝑧2𝛽=𝑑𝑛𝑦𝑛1𝛽𝑛𝑇𝑛𝑦𝑛,𝑧2𝛽𝑛𝑑𝑦𝑛+,𝑧1𝛽𝑛𝑑𝑇𝑛𝑦𝑛,𝑧2𝑦𝑑𝑛,𝑧2𝛼=𝑑𝑛𝑢1𝛼𝑛𝑇𝑛𝑥𝑛,𝑧2𝛼𝑛𝑑(𝑢,𝑧)2+1𝛼𝑛𝑑𝑇𝑛𝑥𝑛,𝑧2𝛼𝑛1𝛼𝑛𝑑𝑢,𝑇𝑛𝑥𝑛2𝛼𝑛𝑑(𝑢,𝑧)2+1𝛼𝑛𝑑𝑥𝑛,𝑧2𝛼𝑛1𝛼𝑛𝑑𝑢,𝑇𝑛𝑥𝑛2=1𝛼𝑛𝑑𝑥𝑛,𝑧2+𝛼𝑛𝑑(𝑢,𝑧)21𝛼𝑛𝑑𝑢,𝑇𝑛𝑥𝑛2.(3.17) This implies by 𝑛=1𝛼𝑛=, inequality (3.16), and Lemma 2.4 that lim𝑛𝑑(𝑥𝑛,𝑧)2=0. Hence, {𝑥𝑛} converges to 𝑧𝐹(𝑇)=𝑛=1𝐹(𝑇𝑛) which is nearest to 𝑢.

Corollary 3.7 (see [7], Theorem 8). Let 𝐶 be a nonempty closed convex subset of a complete CAT(0) space 𝑋. Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself such that 𝑛=1𝐹(𝑇𝑛). Suppose that 𝑢,𝑥1𝐶 are arbitrarily chosen and {𝑥𝑛} is a sequence of 𝐶 generated by 𝑥𝑛+1=𝛼𝑛𝑢1𝛼𝑛𝑇𝑛𝑥𝑛𝑛,(3.18) where {𝛼𝑛} is a sequence in [0,1] which satisfies the condition (𝐶1) as in Theorem 3.2. Suppose that ({𝑇𝑛},𝑇) satisfies AKTT-condition and 𝐹(𝑇)=𝑛=1𝐹(𝑇𝑛). Then {𝑥𝑛} converges strongly to a common fixed point of {𝑇𝑛} which is nearest to 𝑢.

Proof. By putting 𝛽𝑛=1 for all 𝑛 in Theorem 3.6, we obtain the desired result.

In 2009, Song and Zheng [19] introduced a condition in Banach spaces for a countable infinite family of nonexpansive mappings which is different from AKTT-condition and also give some examples of a family of mappings that satisfies this condition. Now, we state this condition in CAT(0) spaces, and it is referred as SZ-condition as follows. Let 𝐶 be a nonempty closed convex subset of a complete CAT(0) space 𝑋. Suppose that {𝑇𝑛} is a family of nonexpansive mappings from 𝐶 into itself with 𝑛=1𝐹(𝑇𝑛). We say that {𝑇𝑛} satisfies SZ-condition if, for any bounded subset 𝐾 of 𝐶, there exists a nonexpansive mapping 𝑇 of 𝐶 into itself such thatlim𝑛𝑑𝑇𝑇sup𝑛𝑥,𝑇𝑛𝑥𝑥𝐾=0,𝐹(𝑇)=𝑛=1𝐹𝑇𝑛.(3.19)

Theorem 3.8. Let 𝐶 be a nonempty closed convex subset of a complete CAT(0) space 𝑋. Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself such that 𝑛=1𝐹(𝑇𝑛) and satisfies SZ-condition. Suppose that {𝑥𝑛} is a sequence of 𝐶 defined by (3.13) with lim𝑛𝑑(𝑥𝑛+1,𝑥𝑛)=0. Let {𝛼𝑛} and {𝛽𝑛} be sequences in [0,1] which satisfy the following conditions: (C3)0<𝛼𝑛<1, lim𝑛𝛼𝑛=0, and 𝑛=1𝛼𝑛=, (C4)lim𝑛𝛽𝑛=1. Then {𝑥𝑛} converges strongly to a common fixed point of {𝑇𝑛} which is nearest to 𝑢.

Proof. As in the proof of Theorem 3.2, we have that {𝑥𝑛} and {𝑇𝑛𝑥𝑛} are bounded. Since {𝑇𝑛} satisfies SZ-condition, there exists a nonexpansive mapping 𝑇 of 𝐶 into itself such that lim𝑛sup{𝑑(𝑇(𝑇𝑛𝑥),𝑇𝑛𝑥)𝑥{𝑥𝑘}}=0 and 𝐹(𝑇)=𝑛=1𝐹(𝑇𝑛). By the definition of {𝑥𝑛} and {𝑦𝑛}, we have 𝑑𝑥𝑛+1,𝑇𝑛𝑥𝑛𝛽=𝑑𝑛𝑦𝑛1𝛽𝑛𝑇𝑛𝑦𝑛,𝑇𝑛𝑥𝑛𝛽𝑛𝑑𝑦𝑛,𝑇𝑛𝑥𝑛+1𝛽𝑛𝑑𝑇𝑛𝑦𝑛,𝑇𝑛𝑥𝑛𝛽𝑛𝑑𝑦𝑛,𝑇𝑛𝑥𝑛+1𝛽𝑛𝑑𝑦𝑛,𝑥𝑛=𝛽𝑛𝑑𝛼𝑛𝑢1𝛼𝑛𝑇𝑛𝑥𝑛,𝑇𝑛𝑥𝑛+1𝛽𝑛𝑑𝛼𝑛𝑢1𝛼𝑛𝑇𝑛𝑥𝑛,𝑥𝑛𝛽𝑛𝛼𝑛𝑑𝑢,𝑇𝑛𝑥𝑛+1𝛽𝑛𝛼𝑛𝑑𝑢,𝑥𝑛+1𝛼𝑛𝑑𝑇𝑛𝑥𝑛,𝑥𝑛.(3.20) It follows from condition (𝐶3) and (𝐶4) that lim𝑛𝑑𝑥𝑛+1,𝑇𝑛𝑥𝑛=0.(3.21) Since 𝑑𝑥𝑛+1,𝑇𝑥𝑛+1𝑥𝑑𝑛+1,𝑇𝑛𝑥𝑛𝑇+𝑑𝑛𝑥𝑛𝑇,𝑇𝑛𝑥𝑛𝑇𝑇+𝑑𝑛𝑥𝑛,𝑇𝑥𝑛+1𝑥2𝑑𝑛+1,𝑇𝑛𝑥𝑛𝑑𝑇𝑇+sup𝑛𝑥,𝑇𝑛𝑥𝑥𝑥𝑘,(3.22) this implies by (3.21) and SZ-condition, we have lim𝑛𝑑𝑥𝑛,𝑇𝑥𝑛=0.(3.23) From lim𝑛𝑑(𝑥𝑛+1,𝑥𝑛)=0 and 𝑑𝑥𝑛,𝑇𝑛𝑥𝑛𝑥𝑑𝑛,𝑥𝑛+1𝑥+𝑑𝑛+1,𝑇𝑛𝑥𝑛,(3.24) it follows that lim𝑛𝑑𝑥𝑛,𝑇𝑛𝑥𝑛=0.(3.25) By using the same arguments and techniques as those of Theorem 3.6, we can show that {𝑥𝑛} converges to a common fixed point of {𝑇𝑛} which is nearest to 𝑢.

Corollary 3.9. Let 𝐶 be a nonempty closed convex subset of a complete CAT(0) space 𝑋. Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself such that 𝑛=1𝐹(𝑇𝑛) and satisfies SZ-condition. Suppose that {𝑥𝑛} is a sequence of 𝐶 defined by (3.18) with lim𝑛𝑑(𝑥𝑛+1,𝑥𝑛)=0. Let {𝛼𝑛} be a sequence in [0,1] which satisfies the condition (𝐶3) as in Theorem 3.8. Then {𝑥𝑛} converges strongly to a common fixed point of {𝑇𝑛} which is nearest to 𝑢.

Proof. By putting 𝛽𝑛=1 for all 𝑛 in Theorem 3.8, we obtain the desired result.

4. W-Mapping in Convex Metric Spaces

In Theorems 3.2, 3.5, and 3.6 and Corollary 3.7, to obtain a convergence result, we have to assume that ({𝑇𝑛},𝑇) satisfies AKTT-condition. In general, one cannot apply these results for a sequence of nonexpansive mappings. However, we give an example of a sequence {𝑇𝑛} of nonexpansive mappings satisfying AKTT-condition.

Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself, where 𝐶 is a convex subset of a convex metric space (𝑋,𝑑,𝑊). We now define mappings 𝑈𝑛;1,𝑈𝑛;2,,𝑈𝑛;𝑛 and 𝑆𝑛 as follows. For {𝜆𝑛} a sequence in [0,1] and 𝑥𝑋, 𝑈𝑛;𝑛𝑇𝑥=𝑊𝑛𝑥,𝑥,𝜆𝑛,𝑈𝑛;𝑛1𝑇𝑥=𝑊𝑛1𝑈𝑛;𝑛𝑥,𝑥,𝜆𝑛1,𝑈𝑛;𝑛2𝑇𝑥=𝑊𝑛2𝑈𝑛;𝑛1𝑥,𝑥,𝜆𝑛2,𝑈𝑛;𝑘𝑇𝑥=𝑊𝑘𝑈𝑛;𝑘+1𝑥,𝑥,𝜆𝑘,𝑈𝑛;𝑘1𝑇𝑥=𝑊𝑘1𝑈𝑛;𝑘𝑥,𝑥,𝜆𝑘1,𝑈𝑛;2𝑇𝑥=𝑊2𝑈𝑛;3𝑥,𝑥,𝜆2,𝑆𝑛x=𝑈𝑛;1𝑇𝑥=𝑊1𝑈𝑛;2𝑥,𝑥,𝜆1.(4.1) Such a mapping 𝑆𝑛 is called the 𝑊-mapping generated by 𝑇1,𝑇2,,𝑇𝑛 and 𝜆1,𝜆2,,𝜆𝑛.

In 2007, Shimizu [18] generalized 𝑊-mapping which was introduced by Takahashi [20] from Banach spaces to convex metric spaces. Then, the following result is obtained by using the same proof as in of [18, Lemma 2].

Lemma 4.1. Let 𝐶 be a nonempty closed convex subset of a uniformly convex metric space (𝑋,𝑑,𝑊) with a continuous convex structure 𝑊𝑋×𝑋×[0,1]𝑋. Let 𝑇1,𝑇2,,𝑇𝑁 be nonexpansive mappings of 𝐶 into itself such that 𝑁𝑛=1𝐹(𝑇𝑛) and let 𝜆1,𝜆2,,𝜆𝑁 be real numbers such that 0<𝜆𝑛<1 for every 𝑛=1,2,,𝑁. Let 𝑆𝑁 be the 𝑊-mapping of 𝐶 into itself generated by 𝑇1,𝑇2,,𝑇𝑁 and 𝜆1,𝜆2,,𝜆𝑁. Then 𝐹(𝑆𝑁)=𝑁𝑛=1𝐹(𝑇𝑛).

Next, we consider the 𝑊-mapping given by a countable infinite family of nonexpansive mappings in a uniformly convex metric space.

Lemma 4.2. Let 𝐶 be a nonempty closed convex subset of a complete uniformly convex metric space (𝑋,𝑑,𝑊) with the property (H). Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself such that 𝑛=1𝐹(𝑇𝑛), and let 𝜆1,𝜆2, be real numbers such that 0<𝜆𝑛𝑏<1 for every 𝑛. Then for every 𝑥𝐶, and 𝑘, lim𝑛𝑈𝑛;𝑘𝑥 exists.

Proof. Let 𝑥𝐶 and 𝑝𝑛=1𝐹(𝑇𝑛). Fix 𝑘. Then for any 𝑛 with 𝑛>𝑘, we have 𝑑𝑈𝑛+1;𝑘𝑥,𝑈𝑛;𝑘𝑥𝑊𝑇=𝑑𝑘𝑈𝑛+1;𝑘+1𝑥,𝑥,𝜆𝑘𝑇,𝑊𝑘𝑈𝑛;𝑘+1𝑥,𝑥,𝜆𝑘𝜆𝑘𝑑𝑇𝑘𝑈𝑛+1;𝑘+1𝑥,𝑇𝑘𝑈𝑛;𝑘+1𝑥𝜆𝑘𝑑𝑈𝑛+1;𝑘+1𝑥,𝑈𝑛;𝑘+1𝑥=𝜆𝑘𝑑𝑊𝑇𝑘+1𝑈𝑛+1;𝑘+2𝑥,𝑥,𝜆𝑘+1𝑇,𝑊𝑘+1𝑈𝑛;𝑘+2𝑥,𝑥,𝜆𝑘+1𝜆𝑘𝜆𝑘+1𝑑𝑈𝑛+1;𝑘+2𝑥,𝑈𝑛;𝑘+2𝑥𝜆𝑘𝜆𝑘+1𝜆𝑛1𝑑𝑈𝑛+1;𝑛𝑥,𝑈𝑛;𝑛𝑥=𝜆𝑘𝜆𝑘+1𝜆𝑛1𝑑𝑊𝑇𝑛𝑈𝑛+1;𝑛+1𝑥,𝑥,𝜆𝑛𝑇,𝑊𝑛𝑥,𝑥,𝜆𝑛𝜆𝑘𝜆𝑘+1𝜆𝑛𝑑𝑇𝑛𝑈𝑛+1;𝑛+1𝑥,𝑇𝑛𝑥𝜆𝑘𝜆𝑘+1𝜆𝑛𝑑𝑈𝑛+1;𝑛+1𝑥,𝑥=𝜆𝑘𝜆𝑘+1𝜆𝑛𝑑𝑊𝑇𝑛+1𝑥,𝑥,𝜆𝑛+1,𝑥=𝜆𝑘𝜆𝑘+1𝜆𝑛+1𝑑𝑇𝑛+1𝑥,𝑥𝜆𝑘𝜆𝑘+1𝜆𝑛+1𝑑𝑇𝑛+1𝑥,𝑝+𝑑(𝑝,𝑥)2𝑑(𝑝,𝑥)𝑏𝑛𝑘+2.(4.2) Thus for 𝑚>𝑛, 𝑑𝑈𝑚;𝑘𝑥,𝑈𝑛;𝑘𝑥𝑈𝑑𝑚;𝑘𝑥,𝑈𝑚1;𝑘𝑥𝑈+𝑑𝑚1;𝑘𝑥,𝑈𝑚2;𝑘𝑥𝑈++𝑑𝑛+1;𝑘𝑥,𝑈𝑛;𝑘𝑥2𝑑(𝑝,𝑥)𝑏(𝑚1)𝑘+2+2𝑑(𝑝,𝑥)𝑏(𝑚2)𝑘+2++2𝑑(𝑝,𝑥)𝑏𝑛𝑘+2=2𝑑(𝑝,𝑥)𝑚1𝑗=𝑛𝑏𝑗𝑘+2.(4.3) It follows that {𝑈𝑛;𝑘𝑥} is a Cauchy sequence. Hence, lim𝑛𝑈𝑛;𝑘𝑥 exists.

Using the above lemma, one can define mappings 𝑈;𝑘 and 𝑆 of 𝐶 into itself as𝑈;𝑘𝑥=lim𝑛𝑈𝑛;𝑘𝑥,𝑆𝑥=lim𝑛𝑆𝑛𝑥=lim𝑛𝑈𝑛;1𝑥,(4.4) for every 𝑥𝐶. Such a mapping 𝑆 is called the 𝑊-mapping generated by 𝑇1,𝑇2, and 𝜆1,𝜆2,.

Lemma 4.3. Let 𝐶 be a nonempty closed convex subset of a complete uniformly convex metric space (𝑋,𝑑,𝑊) with the property (H). Let {𝑇𝑛} be a family of nonexpansive mappings of 𝐶 into itself such that 𝑛=1𝐹(𝑇𝑛), and let 𝜆1,𝜆2, be real numbers such that 0<𝜆𝑛𝑏<1 for every 𝑛. Let 𝑆 be the 𝑊-mapping generated by 𝑇1,𝑇2, and 𝜆1,𝜆2,. Then, 𝑆 is a nonexpansive mapping and 𝐹(𝑆)=𝑛=1𝐹(𝑇𝑛).

Proof. First, we show that 𝑆 is a nonexpansive mapping. For 𝑥,𝑦𝐶, we have 𝑑𝑆𝑛𝑥,𝑆𝑛𝑦𝑊𝑇=𝑑1𝑈𝑛;2𝑥,𝑥,𝜆1𝑇,𝑊1𝑈𝑛;2𝑦,𝑦,𝜆1𝜆1𝑑𝑇1𝑈𝑛;2𝑥,𝑇1𝑈𝑛;2𝑦+1𝜆1𝑑(𝑥,𝑦)𝜆1𝑑𝑈𝑛;2𝑥,𝑈𝑛;2𝑦+1𝜆1𝑑(𝑥,𝑦)𝜆1𝜆2𝜆𝑛1𝑑𝑈𝑛;𝑛𝑥,𝑈𝑛;𝑛𝑦+1𝜆1𝜆2𝜆𝑛1𝑑(𝑥,𝑦)=𝜆1𝜆2𝜆𝑛1𝑑𝑊𝑇𝑛𝑥,𝑥,𝜆𝑛𝑇,𝑊𝑛𝑦,𝑦,𝜆𝑛+1𝜆1𝜆2𝜆𝑛1𝑑(𝑥,𝑦)𝜆1𝜆2𝜆𝑛1𝜆𝑛𝑑𝑇𝑛𝑥,𝑇𝑛𝑦+𝜆1𝜆2𝜆𝑛11𝜆𝑛+𝑑(𝑥,𝑦)1𝜆1𝜆2𝜆𝑛1𝑑(𝑥,𝑦)𝑑(𝑥,𝑦).(4.5) This implies that 𝑆𝑛 is a nonexpansive mapping, and we have 𝑑(𝑆𝑥,𝑆𝑦)=lim𝑛𝑑(𝑆𝑛𝑥,𝑆𝑛𝑦)𝑑(𝑥,𝑦). Thus, 𝑆 is also a nonexpansive mapping.
Finally, we show that 𝐹(𝑆)=𝑛=1𝐹(𝑇𝑛). Let 𝑝𝑛=1𝐹(𝑇𝑛). Then, it is obvious that 𝑈𝑛;𝑘𝑝=𝑝 for all 𝑛,𝑘 with 𝑛>𝑘. So we have 𝑈;𝑘𝑝=𝑝 for all 𝑘. Therefore, we have 𝑆𝑝=𝑈;1𝑝=𝑝, and hence, 𝑛=1𝐹(𝑇𝑛)𝐹(𝑆). We now show that 𝐹(𝑆)𝑛=1𝐹(𝑇𝑛). Let 𝑥𝐹(𝑆) and let 𝑝𝑛=1𝐹(𝑇𝑛). Then we have 𝑑𝑆𝑛𝑝,𝑆𝑛𝑥𝑈=𝑑𝑛;1𝑝,𝑈𝑛;1𝑥𝑇=𝑑𝑝,𝑊1𝑈𝑛;2𝑥,𝑥,𝜆1𝜆1𝑑𝑝,𝑇1𝑈𝑛;2𝑥+1𝜆1𝑑(𝑝,𝑥)𝜆1𝑑𝑝,𝑈𝑛;2𝑥+1𝜆1𝑑(𝑝,𝑥)𝜆1𝜆2𝜆𝑘1𝑑𝑝,𝑈𝑛;𝑘𝑥+1𝜆1𝜆2𝜆𝑘1𝑑(𝑝,𝑥)=𝜆1𝜆2𝜆𝑘1𝑑𝑇𝑝,𝑊𝑘𝑈𝑛;𝑘+1𝑥,𝑥,𝜆𝑘+1𝜆1𝜆2𝜆𝑘1𝑑(𝑝,𝑥)𝜆1𝜆2𝜆𝑘1𝜆𝑘𝑑𝑝,𝑇𝑘𝑈𝑛;𝑘+1𝑥+𝜆1𝜆2𝜆𝑘11𝜆𝑘𝑑+(𝑝,𝑥)1𝜆1𝜆2𝜆𝑘1𝑑(𝑝,𝑥)=𝜆1𝜆2𝜆𝑘𝑑𝑝,𝑇𝑘𝑈𝑛;𝑘+1𝑥+1𝜆1𝜆2𝜆𝑘𝑑(𝑝,𝑥)𝜆1𝜆2𝜆𝑘𝑑𝑝,𝑈𝑛;𝑘+1𝑥+1𝜆1𝜆2𝜆𝑘𝑑(𝑝,𝑥)𝜆1𝜆2𝜆𝑛1𝑑𝑝,𝑈𝑛;𝑛𝑥+1𝜆1𝜆2𝜆𝑛1𝑑(𝑝,𝑥)=𝜆1𝜆2𝜆𝑛1𝑑𝑇𝑝,𝑊𝑛𝑥,𝑥,𝜆𝑛+1𝜆1𝜆2𝜆𝑛1𝑑(𝑝,𝑥)𝜆1𝜆2𝜆𝑛1𝜆𝑛𝑑𝑝,𝑇𝑛𝑥+𝜆1𝜆2𝜆𝑛11𝜆𝑛+𝑑(𝑝,𝑥)1𝜆1𝜆2𝜆𝑛1𝑑(𝑝,𝑥)=𝜆1𝜆2𝜆𝑛𝑑𝑝,𝑇𝑛𝑥+1𝜆1𝜆2𝜆𝑛𝑑(𝑝,𝑥)𝑑(𝑝,𝑥).(4.6) Taking 𝑛, we obtain 𝑑(𝑆𝑝,𝑆𝑥)𝜆1𝜆2𝜆𝑘1𝑑𝑇𝑝,𝑊𝑘𝑈;𝑘+1𝑥,𝑥,𝜆𝑘+1𝜆1𝜆2𝜆𝑘1𝑑(𝑝,𝑥)𝜆1𝜆2𝜆𝑘1𝜆𝑘𝑑𝑝,𝑇𝑘𝑈;𝑘+1𝑥+𝜆1𝜆2𝜆𝑘11𝜆𝑘+𝑑(𝑝,𝑥)1𝜆1𝜆2𝜆𝑘1𝑑(𝑝,𝑥)=𝜆1𝜆2𝜆𝑘𝑑𝑝,𝑇𝑘𝑈;𝑘+1𝑥+1𝜆1𝜆2𝜆𝑘𝑑(𝑝,𝑥)𝑑(𝑝,𝑥).(4.7) Since 𝑝𝑛=1𝐹(𝑇𝑛)𝐹(𝑆), we have 𝑑(𝑆𝑝,𝑆𝑥)=𝑑(𝑝,𝑥). Then, for 𝜆𝑛(0,1), 𝑛, we have 𝑑𝑝,𝑇𝑘𝑈;𝑘+1𝑥𝑇=𝑑(𝑝,𝑥),𝑑𝑝,𝑊𝑘𝑈;𝑘+1𝑥,𝑥,𝜆𝑘=𝑑(𝑝,𝑥),(4.8) for every 𝑘. Suppose that 𝑇𝑘𝑈;𝑘+1𝑥𝑥. Then 𝑑(𝑇𝑘𝑈;𝑘+1𝑥,𝑥)>0. It follows by Lemma 2.5, we have 𝑑𝑇𝑝,𝑊𝑘𝑈;𝑘+1𝑥,𝑥,𝜆𝑘<𝑑(𝑝,𝑥).(4.9) This is a contradiction. Hence, 𝑇𝑘𝑈;𝑘+1𝑥=𝑥. Since 𝑈𝑛;𝑘+1𝑥=𝑊(𝑇𝑘+1𝑈𝑛;𝑘+2𝑥,𝑥,𝜆𝑘+1), we have 𝑈;𝑘+1𝑥=lim𝑛𝑈𝑛;𝑘+1𝑇𝑥=𝑊𝑘+1𝑈;𝑘+2𝑥,𝑥,𝜆𝑘+1=𝑥.(4.10) So, we have 𝑥=𝑇𝑘𝑈;𝑘+1𝑥=𝑇𝑘𝑥 for every 𝑘. This implies that 𝑥𝑛=1𝐹(𝑇𝑛). Therefore, we have 𝐹(𝑆)𝑛=1𝐹(𝑇𝑛).

Lemma 4.4. Suppose that 𝑋,𝐶,{𝑇𝑛},{𝜆𝑛} are as in Lemma 4.3. Let 𝑆𝑛 and 𝑆 be the 𝑊-mappings generated by 𝑇1,𝑇2,,𝑇𝑛 and 𝜆1,𝜆2,,𝜆𝑛, and 𝑇1,𝑇2, and 𝜆1,𝜆2,, respectively. Then ({𝑆𝑛},𝑆) satisfies AKTT-condition, and 𝐹(𝑆)=𝑛=1𝐹(𝑆𝑛).

Proof. Let 𝐵 be a bounded subset of 𝐶 and 𝑥𝐵. For 𝑝𝑛=1𝐹(𝑇𝑛), we have 𝑑𝑆𝑛+1𝑥,𝑆𝑛𝑥𝑈=𝑑𝑛+1;1𝑥,𝑈𝑛;1𝑥𝑊𝑇=𝑑1𝑈𝑛+1;2𝑥,𝑥,𝜆1𝑇,𝑊1𝑈𝑛;2𝑥,𝑥,𝜆1𝜆1𝑑𝑇1𝑈𝑛+1;2𝑥,𝑇1𝑈𝑛;2𝑥𝜆1𝑑𝑈𝑛+1;2𝑥,𝑈𝑛;2𝑥𝜆1𝜆2𝜆𝑛1𝑑𝑈𝑛+1;𝑛𝑥,𝑈𝑛;𝑛𝑥=𝜆1𝜆2𝜆𝑛1𝑑