# A Generalization of Suzuki's Lemma

Authors: B. Panyanak; A. Cuntavepanit
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

Let{zn}, {wn}, and {vn} be bounded sequences in a metric space of hyperbolic type
(X,d), and let {αn} be a sequence in [0,1] with 0<liminfnαn≤limsupnαn<1. If zn+1=αnwn⊕(1−αn)vn for all n∈ℕ, limnd(zn,vn)=0, and limsupn(d(wn+1,wn)-d(zn+1,zn))≤0, then limnd(wn,zn)=0. This is a generalization of Lemma 2.2 in (T. Suzuki, 2005). As
a consequence, we obtain strong convergence theorems for the modified Halpern iterations
of nonexpansive mappings in CAT(0) spaces.


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## Body

## 1. Introduction

Suppose that(X,d) is a metric space which contains a family ℒ of metric segments (isometric images of real line segments) such that distinct points x,y∈X lie on exactly one member S[x,y] of ℒ. Let α∈[0,1], we use the notation αx⊕(1-α)y to denote the point of the segment S[x,y] with distance αd(x,y) from y, that is,(1.1)d(αx⊕(1-α)y,y)=αd(x,y).
We will say that (X,d,ℒ) is of hyperbolic type if for each p,x,y∈X and α∈[0,1],(1.2)d(αp⊕(1-α)x,αp⊕(1-α)y)≤(1-α)d(x,y).
It is proved in [1] that (1.2) implies(1.3)d(p,(1-α)x⊕αy)≤(1-α)d(p,x)+αd(p,y).
It is well-known that Banach spaces are of hyperbolic type. Notice also that CAT(0) spaces and hyperconvex metric spaces are of hyperbolic type (see [2, 3]).In 1983, Goebel and Kirk [4] proved that if {zn} and {wn} are sequences in a metric space of hyperbolic type (X,d) and {αn}⊂[0,1] which satisfy for all i,n∈ℕ, (i) zn+1=αnwn⊕(1-αn)zn, (ii) d(wn+1,wn)≤d(zn+1,zn), (iii) d(wi+n,xi)≤a<∞, (iv) αn≤b<1, and (v) ∑n=1∞αn=∞, then limnd(wn,zn)=0. It was proved by Suzuki [5] that one obtains the same conclusion if the conditions (i)–(v) are replaced by the conditions (S1)–(S4) as follows:(S1)
zn+1=αnwn⊕(1-αn)zn,(S2)
limsupn(d(wn+1,wn)-d(zn+1,zn))≤0,(S3)
{zn} and {wn} are bounded sequences,(S4)
0<liminfnαn≤limsupnαn<1.Both Goebel-Kirk’s and Suzuki’s results have been used to prove weak and strong convergence theorems for approximating fixed points of various types of mappings. The purpose of this paper is to generalize Suzuki's result by relaxing the condition (S1), namely, we can definezn+1 in terms of wn and vn such that limnd(zn,vn)=0. Precisely, we are going to prove the following lemma.Lemma 1.1.
Let{zn}, {wn}, and {vn} be bounded sequences in a metric space of hyperbolic type (X,d), and let {αn} be a sequence in [0,1] with 0<liminfnαn≤limsupnαn<1. Suppose that zn+1=αnwn⊕(1-αn)vn for all n∈ℕ,limnd(zn,vn)=0, and limsupn(d(wn+1,wn)-d(zn+1,zn))≤0, then limnd(wn,zn)=0.


## 2. Proof of Lemma1.1

We begin by proving a crucial lemma.Lemma 2.1.
Let{zn}, {wn}, and {vn} be sequences in a metric space of hyperbolic type (X,d), and let {αn} be a sequence in [0,1] with limsupnαn<1. Put
(2.1)r=lim⁡⁡sup⁡n→∞⁡d(wn,zn)orr=lim⁡⁡inf⁡n→∞⁡d(wn,zn).
Suppose that r<∞,zn+1=αnwn⊕(1-αn)vn for all n∈ℕ,limnd(zn,vn)=0, and
(2.2)lim⁡⁡sup⁡n→∞⁡(d(wn+1,wn)-d(zn+1,zn))≤0,
then
(2.3)lim⁡⁡inf⁡n→∞⁡|d(wn+k,zn)-(1+αn+αn+1+⋯+αn+k-1)r|=0
holds for all k∈ℕ.Proof.
(This proof is patterned after the proof of [5, Lemma  1.1]). For each n∈ℕ, let un=αnwn⊕(1-αn)zn, then by (1.2), we have
(2.4)d(un,zn+1)≤(1-αn)d(zn,vn)≤d(zn,vn).
This implies
(2.5)d(wn+1,zn+1)≤d(wn+1,wn)+d(wn,un)+d(un,zn+1)≤d(wn+1,wn)+d(wn,un)+d(zn,vn).
Since d(wn,un)+d(un,zn)=d(wn,zn), then
(2.6)d(wn+1,zn+1)-d(wn,zn)≤d(wn+1,wn)+d(wn,un)+d(zn,vn)-d(wn,un)-d(un,zn)=d(wn+1,wn)+d(zn,vn)-d(un,zn).
This fact and (2.4) yield
(2.7)d(wn+1,zn+1)-d(wn,zn)-d(zn,vn)≤d(wn+1,zn+1)-d(wn,zn)-d(un,zn+1)≤d(wn+1,wn)+d(zn,vn)-d(un,zn)-d(un,zn+1)≤d(wn+1,wn)+d(zn,vn)-d(zn+1,zn)=d(wn+1,wn)-d(zn+1,zn)+d(zn,vn).
Since limnd(zn,vn)=0, we have
(2.8)lim⁡⁡sup⁡n→∞⁡(d(wn+1,zn+1)-d(wn,zn))≤lim⁡⁡sup⁡n→∞⁡(d(wn+1,wn)-d(zn+1,zn)).
By using this fact, we have, for j∈ℕ,
(2.9)lim⁡⁡sup⁡n→∞⁡(d(wn+j,zn+j)-d(wn,zn))=lim⁡⁡sup⁡n→∞⁡∑i=0j-1(d(wn+i+1,zn+i+1)-d(wn+i,zn+i))≤∑i=0j-1lim⁡⁡sup⁡n→∞⁡(d(wn+i+1,zn+i+1)-d(wn+i,zn+i))≤∑i=0j-1lim⁡⁡sup⁡n→∞⁡(d(wn+i+1,wn+i)-d(zn+i+1,zn+i))≤0.
Put a=(1-limsupnαn)/2. We note that 0<a≤1/2. Fix k,l∈ℕ and ε>0, then there exists m′≥l such that a≤1-αn,d(zn,vn)≤ε/2,d(wn+1,wn)-d(zn+1,zn)≤ε/2, and d(wn+j,zn+j)-d(wn,zn)≤ε/4, for all n≥m′ and j=1,2,…,k. In the case of r=limsupnd(wn,zn), we choose m≥m′ satisfying
(2.10)d(wm+k,zm+k)≥r-ε4,
and d(wn,zn)≤r+(ε/2) for all n≥m. We note that
(2.11)d(wm+j,zm+j)≥d(wm+k,zm+k)-ε4≥r-ε2,
for j=0,1,…,k-1. In the case of r=liminfnd(wn,zn), we choose m≥m′ satisfying
(2.12)d(wm,zm)≤r+ε4,
and d(wn,zn)≥r-(ε/2) for all n≥m. We note that
(2.13)d(wm+j,zm+j)≤d(wm,zm)+ε4≤r+ε2,
for j=1,2,…,k. In both cases, such m satisfies that m≥l,a≤1-αn≤1,d(zn,vn)≤ε/2,d(wn+1,wn)-d(zn+1,zn)≤ε/2 for all n≥m, and
(2.14)r-ε2≤d(wm+j,zm+j)≤r+ε2,
for j=0,1,…,k. We next show that
(2.15)d(wm+k,zm+j)≥(1+αm+j+αm+j+1+⋯+αm+k-1)r-(k-j)(2k+2)ak-jε,
for j=0,1,…,k-1. Since
(2.16)r-ε2≤d(wm+k,zm+k)=d(wm+k,αm+k-1wm+k-1⊕(1-αm+k-1)vm+k-1)≤αm+k-1d(wm+k,wm+k-1)+(1-αm+k-1)d(wm+k,vm+k-1)≤αm+k-1d(zm+k,zm+k-1)+ε2+(1-αm+k-1)d(wm+k,zm+k-1)+(1-αm+k-1)d(zm+k-1,vm+k-1)≤αm+k-1(d(zm+k,um+k-1)+d(um+k-1,zm+k-1))+ε2+(1-αm+k-1)d(wm+k,zm+k-1)+(1-αm+k-1)d(zm+k-1,vm+k-1)≤αm+k-1(1-αm+k-1)d(zm+k-1,vm+k-1)+αm+k-12d(wm+k-1,zm+k-1)+ɛ2+(1-αm+k-1)d(wm+k,zm+k-1)+(1-αm+k-1)d(zm+k-1,vm+k-1)≤αm+k-12(r+ε2)+ε2+(1-αm+k-1)d(wm+k,zm+k-1)+(1-αm+k-12)d(zm+k-1,vm+k-1)≤αm+k-12r+ε+(1-αm+k-1)d(wm+k,zm+k-1)+(1-αm+k-12)ε2
and a≤1-αm+k-1, we have
(2.17)d(wm+k,zm+k-1)≥(1-αm+k-12)r-(3/2)ε-(1-αm+k-12)ε/21-αm+k-1≥(1+αm+k-1)r-2k+1aε-εa=(1+αm+k-1)r-2k+2aε.
Hence, (2.15) holds for j=k-1. We assume that (2.15) holds for some j∈{1,2,…,k-1}. Then, since
(2.18)(1+∑i=jk-1αm+i)r-(k-j)(2k+2)ak-jε≤d(wm+k,zm+j)=d(wm+k,αm+j-1wm+j-1⊕(1-αm+j-1)vm+j-1)≤αm+j-1d(wm+k,wm+j-1)+(1-αm+j-1)d(wm+k,vm+j-1)≤αm+j-1∑i=j-1k-1d(wm+i+1,wm+i)+(1-αm+j-1)d(wm+k,vm+j-1)≤αm+j-1∑i=j-1k-1(d(zm+i+1,zm+i)+ε2)+(1-αm+j-1)d(wm+k,vm+j-1)≤αm+j-1∑i=j-1k-1d(zm+i+1,zm+i)+kε2+(1-αm+j-1)d(wm+k,vm+j-1)≤αm+j-1∑i=j-1k-1(αm+id(wm+i,zm+i)+(1-αm+i)d(zm+i,vm+i))+kɛ2+(1-αm+j-1)d(wm+k,vm+j-1)≤αm+j-1∑i=j-1k-1αm+id(wm+i,zm+i)+αm+j-1∑i=j-1k-1(1-αm+i)d(zm+i,vm+i)+kε2+(1-αm+j-1)d(wm+k,zm+j-1)+(1-αm+j-1)d(zm+j-1,vm+j-1)≤αm+j-1∑i=j-1k-1αm+i(r+ε2)+(k+1)ε2+kε2+(1-αm+j-1)d(wm+k,zm+j-1)≤αm+j-1∑i=j-1k-1αm+ir+(3k+1)ε2+(1-αm+j-1)d(wm+k,zm+j-1),
we obtain
(2.19)d(wm+k,zm+j-1)≥1+∑i=jk-1αm+i-αm+j-1∑i=j-1k-1αm+i1-αm+j-1r-(k-j)(2k+2)/ak-j/+(3k+1)/21-αm+j-1ɛ≥(1+∑i=j-1k-1αm+i)r-(k-j+1)(2k+2)ak-j+1ε.
Hence, (2.15) holds for j≔j-1. Therefore, (2.15) holds for all j=0,1,…,k-1. Specially, we have
(2.20)d(wm+k,zm)≥(1+αm+αm+1+⋯+αm+k-1)r-k(2k+2)akε.
On the other hand, we have
(2.21)d(wm+k,zm)≤d(wm+k,zm+k)+∑i=0k-1d(zm+i+1,zm+i)≤d(wm+k,zm+k)+∑i=0k-1d(zm+i+1,um+i)+∑i=0k-1d(um+i,zm+i)≤d(wm+k,zm+k)+∑i=0k-1d(vm+i,zm+i)+∑i=0k-1αm+id(wm+i,zm+i)≤r+ε2+kε2+∑i=0k-1αm+i(r+ε2)=(1+∑i=0k-1αm+i)r+(2k+12)ε.
This fact and (2.20) imply
(2.22)|d(wm+k,zm)-(1+αm+αm+1+⋯+αm+k-1)r|≤k(2k+2)akε.
Since l∈ℕ and ε>0 are arbitrary, we obtain the desired result.By using Lemma2.1 together with the argument in the proof of [5, Lemma  2.2], simply replacing ∥·∥ by d(·,·), we can obtain Lemma 1.1 as desired.


## 3. Applications

In this section, we apply Lemma1.1 to prove two strong convergence theorems for the modified Halpern iterations of nonexpansive mappings in CAT(0) spaces. The results we obtain are analogs of the Banach space results of Song and Li [6].A metric spaceX is a CAT(0) space if it is geodesically connected, and if every geodesic triangle in X is at least as “thin” as its comparison triangle in the Euclidean plane. It is well known that any complete, simply connected Riemannian manifold having nonpositive sectional curvature is a CAT(0) space. Other examples include pre-Hilbert spaces (see [7]), ℝ–trees (see [8]), Euclidean buildings (see [9]), the complex Hilbert ball with a hyperbolic metric (see [10]), and many others. For a thorough discussion of these spaces and of the fundamental role they play in geometry, we refer the reader to Bridson and Haefliger [7].Fixed-point theory in CAT(0) spaces was first studied by Kirk (see [2, 11]). He showed that every nonexpansive (single-valued) mapping defined on a bounded closed convex subset of a complete CAT(0) space always has a fixed point. Since then, the fixed-point theory for single-valued and multivalued mappings in CAT(0) spaces has been rapidly developed, and many papers have appeared (see, e.g., [12–24] and the references therein). It is worth mentioning that fixed-point theorems in CAT(0) spaces (specially in ℝ–trees) can be applied to graph theory, biology, and computer science (see, e.g., [8, 25–28]).Let(X,d) be a metric space. A geodesic path joining x∈X to y∈X (or, more briefly, a geodesic from x to y) is a map c from a closed interval [0,l]⊂ℝ to X such that c(0)=x,c(l)=y, and d(c(t),c(t′))=|t-t′| for all t,t′∈[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 it is unique, this geodesic is denoted by [x,y]. The space (X,d) is said to be a geodesic 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,y∈X. A subset Y⊆X is said to be convex if Y includes every geodesic segment joining any two of its points.Ageodesic triangle∆(x1,x2,x3) in a geodesic space (X,d) consists of three points x1,x2,x3 in X (thevertices 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)≔∆(x¯1,x¯2,x¯3) in the Euclidean plane 𝔼2 such that d𝔼2(x¯i,x¯j)=d(xi,xj) for i,j∈{1,2,3}.A geodesic space is said to be a CAT(0) space if all geodesic triangles satisfy the following comparison axiom.CAT(0): 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¯∈∆¯,(3.1)d(x,y)≤dE2(x¯,y¯).
We now collect some elementary facts about CAT(0) spaces.Lemma 3.1.
Let(X,d) be a CAT(0) space. (i)
(see [7, Proposition  2.4]) Let C be a closed-convex subset of X, then, for every x∈X, there exists a unique point Px∈C such that d(x,Px)=inf{d(x,y):y∈C}. The mapping P:X→C is called the nearest point (or metric) projection from X onto C.(ii)
(see [15, Lemma  2.5]) For x,y,z∈X and t∈[0,1], one has(3.2)d((1-t)x⊕ty,z)2≤(1-t)d(x,z)2+td(y,z)2-t(1-t)d(x,y)2.Recall that a mappingT on a CAT(0) space (X,d) is called nonexpansive if(3.3)d(Tx,Ty)≤d(x,y)∀x,y∈X.
A point x∈X is called a fixed point of T if x=Tx. We will denote by F(T) the set of fixed points of T. The following result can be found in [13] (see also [2, Theorem  12]).Theorem 3.2.
LetC be a convex subset of a CAT(0) space, and let T:C→C be a nonexpansive mapping whose fixed-point set is nonempty, then F(T) is closed and convex.A continuous linear functionalμ on ℓ∞, the Banach space of bounded real sequences, is called a Banach limit if ∥μ∥=μ(1,1,…)=1 and μn(an)=μn(an+1) for all {an}∈ℓ∞.Lemma 3.3 (see [29], Proposition  2).
Let{a1,a2,⋯}∈ℓ∞ be such that μn(an)≤0 for all Banach limits μ and limsupn(an+1-an)≤0, then limsupnan≤0.Lemma 3.4 (see [21], Lemma  2.1).
LetC be a closed-convex subset of a complete CAT(0) space X, and let T:C→C be a nonexpansive mapping. Let u∈C be fixed. For each t∈(0,1), the mapping St:C→C defined by
(3.4)Stz=tu⊕(1-t)Tzforz∈C
has a unique fixed-point zt∈C, that is,
(3.5)zt=St(zt)=tu⊕(1-t)T(zt).Lemma 3.5 (see [21], Lemma  2.2).
LetC and T be as the preceding lemma, then F(T)≠∅ if and only if {zt} given by the formula (3.5) remains bounded as t→0. In this case, the following statements hold:  (1)
{zt} converges to the unique fixed-point z of T which is the nearest u,(2)
d2(u,z)≤μnd2(u,xn) for all Banach limits μ and all bounded sequences {xn} with limnd(xn,Txn)=0.Lemma 3.6 (see [30], Lemma 2.1).
Let{αn}n=1∞ be a sequence of nonnegative real numbers satisfying the condition
(3.6)αn+1≤(1-γn)αn+γnσn,n≥1,
where {γn} and {σn} are sequences of real numbers, such that(i)
{γn}⊂[0,1] and ∑n=1∞γn=∞,(ii)
eitherlimsupn→∞σn≤0 or ∑n=1∞|γnσn|<∞,then limn→∞αn=0.The following result is an analog of [6, Theorem  3.1].Theorem 3.7.
LetC be a nonempty closed-convex subset of a complete CAT(0) space X, and let T:C→C be a nonexpansive mapping such that F(T)≠∅. Given a point u∈C and sequences {αn} and {λn} in [0,1], the following conditions are satisfied:  (C1)
limn→∞αn=0,(C2)
∑n=1∞αn=∞,(C3)
0<liminfn→∞λn≤limsupn→∞λn<1.Define a sequence {xn} in C by x1=x∈C arbitrarily, and
(3.7)xn+1=λnxn⊕(1-λn)T(αnu⊕(1-αn)xn),∀n≥1,
then {xn} converges to a fixed-point Pu of T, where P is the nearest point projection from C onto F(T).Proof.
For eachn≥1, we let yn=T(αnu⊕(1-αn)xn). We divide the proof into 3 steps. (i) We show that {xn} and {yn} are bounded sequences. (ii) We show that limnd(xn,Txn)=0. (iii) We show that {xn} converges to a fixed-point z∈F(T) which is the nearest u.
(i) Letp∈F(T), then we have
(3.8)d(xn+1,p)=d(λnxn⊕(1-λn)yn),p)≤λnd(xn,p)+(1-λn)d(T(αnu⊕(1-αn)xn),p)≤λnd(xn,p)+(1-λn)αnd(u,p)+(1-λn)(1-αn)d(xn,p)≤(λn+(1-λn)(1-αn))d(xn,p)+(1-λn)αnd(u,p)=[1-(1-λn)αn]d(xn,p)+(1-λn)αnd(u,p)≤max⁡{d(xn,p),d(u,p)}.
Now, an induction yields
(3.9)d(xn,p)≤max⁡{d(x1,p),d(u,p)},n≥1.
Hence, {xn} is bounded and so is {yn}.
(ii) First, we show thatlimnd(xn,yn)=0. Consider
(3.10)d(yn+1,yn)=d(T(αn+1u⊕(1-αn+1)xn+1),T(αnu⊕(1-αn)xn))≤d(αn+1u⊕(1-αn+1)xn+1,αnu⊕(1-αn)xn)≤αn+1d(u,αnu⊕(1-αn)xn)+(1-αn+1)d(xn+1,αnu⊕(1-αn)xn)≤αn+1(1-αn)d(u,xn)+(1-αn+1)αnd(u,xn+1)+(1-αn+1)(1-αn)d(xn+1,xn).
This implies
(3.11)d(yn+1,yn)-d(xn+1,xn)≤αn+1(1-αn)d(u,xn)+(1-αn+1)αnd(u,xn+1).
By the condition (C1), we have
(3.12)lim⁡⁡sup⁡n→∞⁡(d(yn+1,yn)-d(xn+1,xn))≤0.
It follows from Lemma 1.1 that limn→∞d(xn,yn)=0. Now,
(3.13)d(xn,Txn)≤d(xn,yn)+d(yn,Txn)≤d(xn,yn)+d(T(αnu⊕(1-αn)xn),Txn)≤d(xn,yn)+d(αnu⊕(1-αn)xn,xn)≤d(xn,yn)+αnd(u,xn)⟶0asn⟶∞.
(iii) From Lemma3.4, let z=limt→0zt where zt is given by the formula (3.5). Then z is the point of F(T) which is the nearest u. By applying Lemma 3.1, we have
(3.14)d2(xn+1,z)=d2(λnxn⊕(1-λn)yn,z)≤λnd2(xn,z)+(1-λn)d2(yn,z)-λn(1-λn)d2(xn,yn)=λnd2(xn,z)+(1-λn)d2(αnu⊕(1-αn)xn,z)-λn(1-λn)d2(xn,yn)≤λnd2(xn,z)+(1-λn)[αnd2(u,z)+(1-αn)d2(xn,z)-αn(1-αn)d2(u,xn)]≤[λn+(1-λn)(1-αn)]d2(xn,z)+αn(1-λn)[d2(u,z)-(1-αn)d2(u,xn)]=(1-(1-λn)αn)d2(xn,z)+(1-λn)αn[d2(u,z)-(1-αn)d2(u,xn)].
By Lemma 3.5, we have μn(d2(u,z)-d2(u,xn))≤0 for all Banach limit μ. Moreover, since
(3.15)d(xn+1,xn)=d(λnxn⊕(1-λn)yn,xn)≤(1-λn)d(yn,xn)⟶0asn⟶∞,lim⁡⁡sup⁡n→∞⁡[(d2(u,z)-d2(u,xn+1))-(d2(u,z)-d2(u,xn))]=0.
It follows from condition (C1) and Lemma 3.3 that
(3.16)lim⁡⁡sup⁡n→∞⁡(d2(u,z)-(1-αn)d2(u,xn))=lim⁡⁡sup⁡n→∞⁡(d2(u,z)-d2(u,xn))≤0.
Hence, the conclusion follows from Lemma 3.6.Remark 3.8.
In the proof of Theorem3.7, one may observe that it is not necessary to use Lemma 1.1 because Suzuki's original lemma is sufficient. However, in [6], there is a strong convergence theorem for another type of modified Halpern iteration (see [6, Theorem  3.2]). We show that the proof is quite easy when we use Lemma 1.1.Theorem 3.9.
LetC be a nonempty closed-convex subset of a complete CAT(0) space X, and let T:C→C be a nonexpansive mapping such that F(T)≠∅. Given a point u∈C and sequences {αn} and {λn} in [0,1], the following conditions are satisfied:  (C1)
limn→∞αn=0,(C2)
∑n=1∞αn=∞,(C3)
0<liminfn→∞λn≤limsupn→∞λn<1.Define a sequence {xn} in C by x1=x∈C arbitrarily, and
(3.17)xn+1=λn(αnu⊕(1-αn)xn)⊕(1-λn)Txn,∀n≥1,
Then {xn} converges to a fixed-point Pu of T, where P is the nearest point projection from C onto F(T).Proof.
Using the same technique as in the proof of Theorem3.7, we easily obtain that both {xn} and {Txn} are bounded. Let yn=αnu⊕(1-αn)xn, then xn+1=λnyn⊕(1-λn)Txn. By the condition (C1), we have
(3.18)d(xn,yn)=d(xn,αnu⊕(1-αn)xn)≤αnd(xn,u)⟶0asn⟶∞.
It follows from the nonexpansiveness of T that
(3.19)lim⁡⁡sup⁡n→∞⁡(d(Txn+1,Txn)-d(xn+1,xn))≤0.
By Lemma 1.1, we have
(3.20)lim⁡n→∞⁡d(Txn,xn)=0.
From (3.18) and (3.20), we get that
(3.21)d(xn+1,xn)=d(λnyn⊕(1-λn)Txn,xn)≤λnd(yn,xn)+(1-λn)d(Txn,xn)⟶0asn⟶∞.
Let z=limt→0zt where zt is given by (3.5), then z is the point of ∈F(T) which is the nearest u. Consider
(3.22)d2(xn+1,z)=d2(λnyn⊕(1-λn)Txn,z)≤λnd2(yn,z)+(1-λn)d2(Txn,z)-λn(1-λn)d2(yn,Txn)≤λnd2(αnu⊕(1-αn)xn,z)+(1-λn)d2(Txn,z)≤λn(αnd2(u,z)+(1-αn)d2(xn,z)-αn(1-αn)d2(u,xn))+(1-λn)d2(xn,z)≤(λn(1-αn)+(1-λn))d2(xn,z)+λnαnd2(u,z)-λnαn(1-αn)d2(u,xn)=(1-λnαn)d2(xn,z)+λnαn(d2(u,z)-(1-αn)d2(u,xn)).
By Lemma 3.5, we have μn(d2(u,z)-d2(u,xn))≤0 for all Banach limit μ. Moreover, since d(xn+1,xn)→0, then
(3.23)lim⁡⁡sup⁡n→∞⁡[(d2(u,z)-d2(u,xn+1))-(d2(u,z)-d2(u,xn))]=0.
It follows from condition (C1) and Lemma 3.3 that
(3.24)lim⁡⁡sup⁡n→∞⁡(d2(u,z)-(1-αn)d2(u,xn))=lim⁡⁡sup⁡n→∞⁡(d2(u,z)-d2(u,xn))≤0.
Hence, the conclusion follows from Lemma 3.6.



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*Source: 824718-2011-06-16.xml*