# Strong Convergence Theorem of Implicit Iteration Process for Generalized Asymptotically Nonexpansive Mappings in Hilbert Space

Authors: Lili He; Lei Deng; Jianjun Liu
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

---

## Abstract

LetC be a nonempty closed and convex subset of a Hilbert space H, let T and S:C→C be two commutative generalized asymptotically nonexpansive mappings. We introduce an implicit iteration process of S and T defined by xn=αnx0+(1−αn)(2/((n+1)(n+2)))∑k=0n∑i+j=kSiTjxn, and then prove that {xn} converges strongly to a common fixed point of S and T. The results generalize and unify the corresponding results.


---

## Body

## 1. Introduction

LetH be  a Hilbert space, C a nonempty closed and convex subset of H.
A mapping S is said to be generalized asymptotically
nonexpansive if(1.1)∥Snx−Sny∥≤kn∥x−y∥+rnwith kn≥0, rn≥0, kn→1, rn→0 (n→∞),
for each x,y∈C, n=0,1,2,…. If kn=1 and rn=0, (1.1) reduces to nonexpansive mapping; if rn=0, (1.1) reduces to asymptotically nonexpansive
mapping; if kn=1, (1.1) reduces to asymptotically nonexpansive-type
mapping. So, a generalized asymptotically nonexpansive mapping is much more
general than many other mappings.Browder [1] introduced the following implicit
iteration process of a nonexpansive self-mapping for arbitrary x0∈C:(1.2)xn=αnx0+(1−αn)Txn,and then proved that {xn} converges strongly to the fixed point of T which is nearest to x0.After that, Baillon [2] proved the first nonlinear
ergodic theorem: let T be a nonexpansive self-mapping defined on a
nonempty bounded closed and convex set. Then for arbitrary x∈C, {(1/(n+1))∑i=0nTix} converges weakly to a fixed point of T.
Those results have been extended by several authors (see, e.g., [3–7]).Recently, Shimizu and Takahashi [8] studied the
following implicit iteration process of asymptotically nonexpansive mappings
for arbitrary x0∈C:(1.3)xn=αnx0+(1−αn)1n+1∑i=0nTixn,where αn=(bn−1)/(bn−1+a), bn=(1/(n+1))∑i=0n(1+|1−ki|+e−i), {kn} is the asymptotical coefficient of T.
And then proved that {xn} converges strongly to the fixed point of T which is nearest to x0.They also studied the following explicit iteration
process of two commutative nonexpansive self-mappings in [9]:(1.4)xn+1=αnx0+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=kSiTjxn,where {αn}⊆[0,1], limn→∞αn=0, ∑n=0∞αn=∞. And then proved that {xn} converges strongly to the common fixed point
of S and T which is nearest to x0.In this paper, we prove the strong convergence theorem
of implicit iteration process for generalized asymptotically nonexpansive
mappings which is defined by(1.5)xn=αnx0+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=kSiTjxn,where {αn}⊆[0,1], limn→∞αn=0 and limsupn→∞(1−αn)(2/((n+1)(n+2)))∑k=0n∑i+j=kkitj<1.It is well known that a Hilbert spaceH satisfies Opial's condition [5], that is, if a
sequence {xn} converges weakly to an element y∈H and y≠z,
then(1.6)liminfn→∞∥xn−y∥<liminfn→∞∥xn−z∥.Throughout this paper,S and T are two commutative generalized asymptotically
nonexpansive mappings. We denote by F(S) and F(T) the set of fixed points of S and T,
respectively; and suppose that F(S)⋂F(T)≠∅. S and T satisfy the following
conditions:(1.7)∥Snx−Sny∥≤kn∥x−y∥+rn,∥Tnx−Tny∥≤tn∥x−y∥+bn,where kn≥0, rn≥0, tn≥0, bn≥0,
and kn→1, rn→0, tn→1, bn→0 (n→∞).


## 2. Auxiliary Lemmas

This section collects some lemmas which will be used
to prove the main results in the next section.Lemma 2.1 (see [9]).
LettingLn=((n+1)(n+2))/2,
there holds the identity in a Hilbert space H:(2.1)∥yn−v∥2=1Ln∑k=0n∑i+j=k∥xi,j−v∥2−1Ln∑k=0n∑i+j=k∥xi,j−yn∥2for {xi,j}i,j=0∞⊆H, yn=(1/Ln)∑k=0n∑i+j=kxi,j∈H,
and v∈H.Lemma 2.2.
LetC be a nonempty bounded closed convex subset of
a Hilbert space H, S and T two generalized asymptotically nonexpansive
mappings of C into itself such that ST=TS.
For any x∈C,
put Fn(x)=(2/((n+1)(n+2)))∑k=0n∑i+j=kSiTjx.
Then(2.2)liml→∞limn→∞supx∈C∥Fn(x)−SlFn(x)∥=0,liml→∞limn→∞supx∈C∥Fn(x)−TlFn(x)∥=0.Proof.
Putxi,j=SiTjx, v=SlFn(x) and Ln=((n+1)(n+2))/2.
It follows from Lemma 2.1 that(2.3)∥Fn(x)−SlFn(x)∥2=1Ln∑k=0n∑i+j=k∥SiTjx−SlFn(x)∥2−1Ln∑k=0n∑i+j=k∥SiTjx−Fn(x)∥2=1Ln∑k=0l−1∑i+j=k∥SiTjx−SlFn(x)∥2+1Ln∑k=ln∑i+j=k,i≤l−1∥SiTjx−SlFn(x)∥2+1Ln∑k=ln∑i+j=k,i≥l∥SiTjx−SlFn(x)∥2−1Ln∑k=0n∑i+j=k∥SiTjx−Fn(x)∥2≤1Ln∑k=0l−1∑i+j=k∥SiTjx−SlFn(x)∥2+1Ln∑k=ln∑i+j=k,i≤l−1∥SiTjx−SlFn(x)∥2+1Ln∑k=ln∑i+j=k,i≥l(kl∥Si−lTjx−Fn(x)∥+rl)2−1Ln∑k=0n∑i+j=k∥SiTjx−Fn(x)∥2=1Ln∑k=0l−1∑i+j=k∥SiTjx−SlFn(x)∥2+1Ln∑k=ln∑i+j=k,i≤l−1∥SiTjx−SlFn(x)∥2+1Ln∑k=0n−l∑i+j=k(kl2∥SiTjx−Fn(x)∥2+2klrl∥SiTjx−Fn(x)∥+rl2)−1Ln∑k=0n∑i+j=k∥SiTjx−Fn(x)∥2≤1Ln∑k=0l−1∑i+j=k∥SiTjx−SlFn(x)∥2+1Ln∑k=ln∑i+j=k,i≤l−1∥SiTjx−SlFn(x)∥2+1Ln∑k=0n−l∑i+j=k(kl2−1)∥SiTjx−Fn(x)∥2+2klrlLn∑k=0n−l∑i+j=k∥SiTjx−Fn(x)∥+(n+2−l)(n+1−l)(n+2)(n+1)rl2.Choose p∈F(S)⋂F(T),
then there exists a constant M>0 such that (2.4)∥SiTjx−p∥≤ki∥Tjx−p∥+ri≤kitj∥x−p∥+kibj+ri≤M2,∥Fn(x)−p∥≤1Ln∑k=0n∑i+j=k∥SiTjx−p∥≤M2,∥SlFn(x)−p∥≤kl∥Fn(x)−p∥+rl≤M2 for all nonnegative integer i, j, l, and n.
Hence, ∥SiTjx−SlFn(x)∥≤M, ∥SiTjx−Fn(x)∥≤M for all nonnegative integer i,j,l, and n.
So(2.5)supx∈C∥Fn(x)−SlFn(x)∥2≤(l+1)l(n+2)(n+1)M2+2(n+1−l)l(n+2)(n+1)M2+(kl2−1)(n+2−l)(n+1−l)(n+2)(n+1)M2+2klrl(n+2−l)(n+1−l)(n+2)(n+1)M+(n+2−l)(n+1−l)(n+2)(n+1)rl2→0(n→∞,l→∞).Similarly, we can prove
that(2.6)liml→∞limn→∞supx∈C∥Fn(x)−TlFn(x)∥=0.Remark 2.3.
Lemma2.2 extends [8, Lemma 3] and [9, Lemma 1].Lemma 2.4.
LetS and T be two continuous generalized asymptotically
nonexpansive mappings defined on a nonempty bounded closed convex subset C of a Hilbert space H with ST=TS.
Let Ln=((n+1)(n+2))/2.
If {xn} is a sequence in C such that {xn} converges weakly to some x∈C and {xn−(1/Ln)∑k=0n∑i+j=kSiTjxn} converges strongly to 0, then x∈F(S)⋂F(T).Proof.
We claim that{Slx} converges strongly to x as l→∞.
If not, there exist a positive number ε0 and a subsequence {lm} of {l} such that ∥Slmx−x∥≥ε0 for all m.
However, we have(2.7)∥xni−Slmx∥≤∥xni−1Lni∑k=0ni∑i+j=kSiTjxni∥+∥1Lni∑k=0ni∑i+j=kSiTjxni−Slm(1Lni∑k=0ni∑i+j=kSiTjxni)∥+∥Slm(1Lni∑k=0ni∑i+j=kSiTjxni)−Slmx∥≤∥xni−1Lni∑k=0ni∑i+j=kSiTjxni∥+∥1Lni∑k=0ni∑i+j=kSiTjxni−Slm(1Lni∑k=0ni∑i+j=kSiTjxni)∥+klm∥1Lni∑k=0ni∑i+j=kSiTjxni−x∥+rlm≤∥xni−1Lni∑k=0ni∑i+j=kSiTjxni∥+∥1Lni∑k=0ni∑i+j=kSiTjxni−Slm(1Lni∑k=0ni∑i+j=kSiTjxni)∥+klm∥1Lni∑k=0ni∑i+j=kSiTjxni−xni∥+klm∥xni−x∥+rlm.By Opial's condition, for any y∈C with y≠x,
we have(2.8)liminfn→∞∥xn−x∥<liminfn→∞∥xn−y∥.Let r=liminfn→∞∥xn−x∥ and choose a positive number ρ such that(2.9)ρ<r2+ε024−r.Then, there exists a subsequence {xni} of {xn} such that limi→∞∥xni−x∥=r and ∥xni−x∥<r+ρ/5 for all i.
By definition of {klm} and {rlm},
there exists a positive integer m0 such that(2.10)klm∥xni−x∥<r+ρ5,rlm<ρ5for all m>m0.
Since(2.11)limn→∞∥xn−1Ln∑k=0n∑i+j=kSiTjxn∥=0and {klm} is bounded , there exists a positive integer i0 such that(2.12)∥xni−1Lni∑k=0ni∑i+j=kSiTjxni∥<ρ5,klm∥1Lni∑k=0ni∑i+j=kSiTjxni−xni∥<ρ5for all m and i>i0.
As {xni}⊂C is bounded and by Lemma 2.2, there exist m1>m0 and i1>0 such that(2.13)∥1Lni∑k=0ni∑i+j=kSiTjxni−Slm1(1Lni∑k=0ni∑i+j=kSiTjxni)∥<ρ5for all i>i1.
By (2.7), (2.10), (2.12), and (2.13),
we have(2.14)∥xni−Slm1x∥<ρ5+ρ5+ρ5+r+ρ5+ρ5=r+ρfor all i>max{i0,i1}.
However,(2.15)∥xni−Slm1x+x2∥2=12∥xni−Slm1x∥2+12∥xni−x∥2−14∥Slm1x−x∥2<(r+ρ)22+(r+ρ/5)22−ε024<(r+ρ)2−ε024<r2for all i>max{i0,i1}.
This contradicts with (2.8).
So {Slx} converges strongly to x and then x∈F(S).
Similarly, we can get x∈F(T).
Hence, x is a common fixed point of S and T.


## 3. Main Results

In this section, we prove our main theorem.Theorem 3.1.
LetC be a nonempty closed convex subset of a
Hilbert space H and let S, T  be two continuous generalized asymptotically
nonexpansive mappings of C into itself such that(i)
∥Snx−Sny∥≤kn∥x−y∥+rn, kn≥0, rn≥0, kn→1, rn→0 (n→∞),(ii)
∥Tnx−Tny∥≤tn∥x−y∥+bn, tn≥0, bn≥0, tn→1, bn→0 (n→∞),(iii)
ST=TS and F(S)⋂F(T)≠∅. For arbitrary x0∈C,
the sequence {xn}n=0∞ is defined by(3.1)xn=αnx0+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=kSiTjxn,n≥0, with αn∈[0,1] and limn→∞αn=0.
If limsupn→∞(1−αn)(2/((n+1)(n+2)))∑k=0n∑i+j=kkitj<1,
then {xn}n=0∞ converges strongly to the common fixed point Px0 of S and T,
where P is the metric projection of H onto F(S)⋂F(T).Proof.
Firstly, we show that{xn}n=0∞ is bounded. By limsupn→∞(1−αn)(2/((n+1)(n+2)))∑k=0n∑i+j=kkitj<1,
there exist N>0 and 0<a<1 such that (1−αn)(2/((n+1)(n+2)))∑k=0n∑i+j=kkitj≤a for all n>N. Choose p∈F(S)⋂F(T),
then we have(3.2)∥xn−p∥=∥αn(x0−p)+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=k(SiTjxn−SiTjp)∥≤αn∥x0−p∥+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=k(kitj∥xn−p∥+kibj+ri)=αn∥x0−p∥+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=kkitj∥xn−p∥+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=kkibj+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=kri.Hence(3.3)(1−a)∥xn−p∥≤αn∥x0−p∥+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=kkibj+(1−αn)2(n+1)(n+2)∑k=0n∑i+j=krifor all n>N.
So {xn} is bounded.
Letting{xni} be any subsequence of {xn},
there exists a subsequence {xmt} of {xni} such that {xmt} converges weakly to some u∈C.
For {xmt} is bounded, so is {(2/((mt+1)(mt+2)))∑k=0mt∑i+j=kSiTjxmt}.
Thus(3.4)xmt−2(mt+1)(mt+2)∑k=0mt∑i+j=kSiTjxmt=αmt(x0−2(mt+1)(mt+2)∑k=0mt∑i+j=kSiTjxmt)→0(t→∞).It follows from Lemma 2.4 that u∈F(S)⋂F(T).
For(3.5)αmtPx0=Px0−(1−αmt)2(mt+1)(mt+2)∑k=0mt∑i+j=kSiTjPx0,we have(3.6)αmt〈x0−Px0,xmt−Px0〉=〈xmt−Px0−(1−αmt)2(mt+1)(mt+2)∑k=0mt∑i+j=k(SiTjxmt−SiTjPx0),xmt−Px0〉=∥xmt−Px0∥2−(1−αmt)〈2(mt+1)(mt+2)∑k=0mt∑i+j=k(SiTjxmt−SiTjPx0),xmt−Px0〉≥∥xmt−Px0∥2−(1−αmt)2(mt+1)(mt+2)×∑k=0mt∑i+j=k(kitj∥xmt−Px0∥+kibj+ri)∥xmt−Px0∥=(1−(1−αmt)2(mt+1)(mt+2)∑k=0mt∑i+j=kkitj)∥xmt−Px0∥2−(1−αmt)2(mt+1)(mt+2)∑k=0mt∑i+j=k(kibj+ri)∥xmt−Px0∥.By hypothesis, there exist M>0 and 0<a<1 such that (1−αmt)(2/((mt+1)(mt+2)))∑k=0mt∑i+j=kkitj≤a for all t>M.
So(3.7)(1−a)∥xmt−Px0∥2≤αmt〈x0−Px0,xmt−Px0〉+(1−αmt)2(mt+1)(mt+2)∑k=0mt∑i+j=k(kibj+ri)∥xmt−Px0∥for all t>M.
We can easily prove that limn→∞(1−αmt)(2/((mt+1)(mt+2)))∑k=0mt∑i+j=k(kibj+ri)∥xmt−Px0∥=0.
Since P is metric projection, 〈x0−Px0,x−Px0〉≤0 for all x∈F(S)⋂F(T).
Hence(3.8)〈x0−Px0,xmt−Px0〉=〈x0−Px0,xmt−u〉+〈x0−Px0,u−Px0〉≤〈x0−Px0,xmt−u〉.Since {xmt} converges weakly to u, limsupt→∞〈x0−Px0,xmt−Px0〉≤0. It follows from (3.7) that {xmt} converges strongly to Px0.
Hence, {xn} converges strongly to Px0.
This completes the proof.



---

*Source: 649510-2008-09-22.xml*