CONVERGENCE THEOREMS FOR TOTAL ASYMPTOTICALLY NONEXPANSIVE
MAPPINGS
Yan Hao
Abstract
In this paper, a demiclosed principle for total asymptotically non- expansive mappings is established. An implicit iterative method for the class of total asymptotically nonexpansive mappings is considered.
Weak and strong convergence theorems are established in a real Hilbert space. As applications of main results, an equilibrium problem is con- sidered based on implicit iterative process.
1. Introduction and Preliminaries
Throughout this paper, we assume thatH is a real Hilbert space, whose inner product and norm are denoted by h·,·i andk · k. We also assume that ψ: [0,∞)→[0,∞) is strictly increasing continuous function withψ(0) = 0. → and⇀are denoted by strong convergence and weak convergence, respectively.
Let C be a nonempty closed and convex subset of H and T : C → C a mapping. In this paper, we use F(T) to denote the fixed point set of the mapping T. Recall thatT is said to be nonexpansive if
kT x−T yk ≤ kx−yk, ∀x, y∈C. (1.1)
Key Words: nonexpansive mapping; asymptotically nonexpanvie mapping; total asymp- totically nonexpansive mapping; fixed point; equilibrium
Mathematics Subject Classification: 47H09, 47H10, 47J25.
Received: June, 2009 Accepted: January, 2010
163
T is said to be asymptotically nonexpansive if there exists a positive se- quencehn⊂[1,∞) with limn→∞hn= 1 such that
kTnx−Tnyk ≤hnkx−yk, ∀x, y∈C, n≥1. (1.2) The class of asymptotically nonexpansive mappings was introduced by Goebel and Kirk [14] as a generalization of the class of nonexpansive mappings. They proved that ifCis a nonempty closed convex and bounded subset of a real uni- formly convex Banach space andTis an asymptotically nonexpansive mapping onC, then T has a fixed point.
T is said to be asymptotically nonexpansive in the intermediate sense if it is continuous and the following inequality holds:
lim sup
n→∞
sup
x,y∈C
(kTnx−Tnyk − kx−yk)≤0. (1.3) Observe that if we define
σn= sup
x,y∈C
(kTnx−Tnyk − kx−yk) andνn= max{0, σn}, thenνn →0 asn→ ∞. It follows that (1.3) is reduced to
kTnx−Tnyk ≤ kx−yk+νn, ∀x, y∈C, n≥1. (1.4) The class of mappings which are asymptotically nonexpansive in the interme- diate sense was introduced by Bruck, Kuczumow and Reich [2]. It is known [16] that if C is a nonempty closed convex bounded subset of a uniformly convex Banach spaceE andT is asymptotically nonexpansive in the interme- diate sense, then T has a fixed point. It is worth mentioning that the class of mappings which are asymptotically nonexpansive in the intermediate sense contains properly the class of asymptotically nonexpansive mappings.
Recently, Alber, Chidume and Zegeye [1] introduced the concept of to- tal asymptotically nonexpansive mappings. Recall that T is said to be total asymptotically nonexpansive if
kTnx−Tnyk ≤ kx−yk+µnψ(kx−yk) +νn, ∀x, y∈C, (1.5) where {µn} and {νn} are nonnegative real sequences such that µn → 0 and νn →0 asn→ ∞.From the definition, we see that the class of total asymptot- ically nonexpansive mappings include the class of asymptotically nonexpansive mappings as a special case; see also [13] for more details.
Recently, weak convergence problems of implicit (or non-implicit) iterative processes to a common fixed point for a finite family of nonexpansive mappings
and asymptotically nonexpansive mappings have been considered by a number of authors (see, for example, [1],[8],[9],[12],[15],[19],[21],[23],[24],[27]-[32],[34]- [36],[38]).
In 2001, Xu and Ori [34] introduced the following implicit iteration process for a finite family of nonexpansive mappings{T1, T2, . . . , TN}, with{αn}a real sequence in (0,1), and an initial pointx0∈C:
x1=α1x0+ (1−α1)T1x1, x2=α2x1+ (1−α2)T2x2, ...
xN =αNxN−1+ (1−αN)TNxN, xN+1=αN+1xN + (1−αN+1)T1xN+1, ...
which can be re-written in the following compact form:
xn=αnxn−1+ (1−αn)Tnxn, ∀n≥1, (1.6)
whereTn =Tn(modN)(here the modN function takes values in{1,2, . . . , N}).
They obtained the following results in a real Hilbert space.
Theorem XO.LetH be a real Hilbert space,C be a nonempty closed convex subset of H, and T :C →C be a finite family of nonexpansive self-mappings on C such that F =∩Ni=1F(Ti)6=∅. Let{xn} be defined by (1.6). If {αn} is chosen so thatαn →0,as n→ ∞, then{xn} converges weakly to a common fixed point of the family of {Ti}Ni=1.
In 2006, Chang et al. [9] improved the results of Xu and Ori [34] from the class of nonexpansive mappings to the class of asymptotically nonexpansive mappings which is defined on a nonempty closed and convex subset C of H such that C+C ⊂ C. To be more precise, they considered the following implicit iterative process for a finite family of asymptotically nonexpansive mappings{T1, T2, . . . , TN}, with{αn}a real sequence in (0,1),{un}a bounded
sequence inC, and an initial pointx0∈C:
x1=α1x0+ (1−α1)T1x1+u1, x2=α2x1+ (1−α2)T2x2+u2, ...
xN =αNxN−1+ (1−αN)TNxN+uN,
xN+1=αN+1xN + (1−αN+1)T1xN+1+uN+1, ...
x2N =α2Nx2N−1+ (1−α2N)TN2x2N +u2N,
x2N+1=α2N+1x2N + (1−α2N+1)T13x2N+1+u2N+1, ...
Since for eachn≥1, it can be written asn= (k−1)N+i,where i=i(n)∈ {1,2, . . . , N}, k =k(n)≥1 is a positive integer and k(n)→ ∞as n→ ∞.
Hence the above table can be rewritten in the following compact form:
xn=αnxn−1+ (1−αn)Ti(n)k(n)xn+un, ∀n≥1. (1.7) They obtained weak convergence theorems of the implicit iterative scheme (1.7) for a finite family of asymptotically nonexpansive mappings{T1, T2, . . . , TN};
see [9] for more details.
The purpose of this paper is to establish weak and strong convergence theorems of the implicit iteration process (1.7) for a finite family of uniformly Lipschitz total asymptotically nonexpansive mappings in a real Hilbert space.
Next, we recall some well-known concepts.
Recall that a space X is said to satisfy Opial’s condition [18] if for each sequence{xn}in X, the condition that the sequencexn →xweakly implies that
lim inf
n→∞ kxn−xk<lim inf
n→∞ kxn−yk for ally∈E andy6=x.
Recall that a mappingT :C→Cis semicompact if any sequence{xn}in C satisfying limn→∞kxn−T xnk= 0 has a convergent subsequence.
Recall that the mappingT is said to be demiclosed at the origin if for each sequence {xn} in C, the condition xn → x0 weakly and T xn → 0 strongly impliesT x0= 0.
Next, we show that the process (1.7) is well-defined if the control sequence satisfies 0 < L−1L < αn < 1, where L = max1≤i≤N{Li}. Indeed, define a mapping
Wnx=αnxn−1+ (1−αn)Ti(n)k(n)x+un, ∀n≥1, ∀x∈C.
It follows that
kWnx−Wnyk ≤(1−αn)Lkx−yk, ∀x, y∈C.
Since (1−αn)L <1, it follows thatWn is a contractive mapping and hence has a unique fixed pointxn inC. This is, the process (1.7) is well-defined.
In order to prove our main results, we also need the following lemmas.
Lemma 1.1([30]). Let {an}, {bn} and {cn} be three nonnegative sequences satisfying the following condition:
an+1≤(1 +bn)an+cn, ∀n≥n0, wheren0is some nonnegative integer,P∞
n=0cn<∞andP∞
n=0bn<∞. Then limn→∞an exists.
Lemma 1.2([27]). Let H be a real Hilbert space and 0< p≤tn≤q <1 for all n≥0. Suppose that{xn} and{yn} are sequences of H such that
lim sup
n→∞
kxnk ≤r, lim sup
n→∞
kynk ≤r
and
n→∞lim ktnxn+ (1−tn)ynk=r hold for somer≥0.Thenlimn→∞kxn−ynk= 0.
Lemma 1.3. LetCbe a nonempty closed convex and bounded subset of a real Hilbert space H and T be a L-Lipschitz continuous and total asymptotically nonexpansive mapping with the functionψ and sequences{µn} and{νn} such that µn→0 andνn →0 asn→ ∞. ThenI−T is demiclosed at zero.
Proof. Let{xn} be a sequence inC such thatxn ⇀ x∗ and xn−T xn →0 as n→ ∞. Next, we show thatx∗ ∈C andx∗ =T x∗. SinceC is closed and convex, we see that x∗ ∈ C. It is sufficient to show that x∗ = T x∗. Choose α ∈ (0,1+L1 ) and define yα,m = (1−α)x∗+αTmx∗ for arbitrary but fixed m≥1.Note that
kxn−Tmxnk ≤ kxn−T xnk+kT xn−T2xnk+· · ·+kTm−1xn−Tmxnk
≤mLkxn−T xnk.
It follows that
n→∞lim kxn−Tmxnk= 0. (1.8) Note that
hx∗−yα,m, yα,m−Tmyα,mi
=hx∗−xn, yα,m−Tmyα,mi+hxn−yα,m, yα,m−Tmyα,mi=
=hx∗−xn, yα,m−Tmyα,mi+hxn−yα,m, Tmxn−Tmyα,mi−
− hxn−yα,m, xn−yα,mi+hxn−yα,m, xn−Tmxni
≤ hx∗−xn, yα,m−Tmyα,mi+
+kxn−yα,mk(kxn−yα,mk+µnψ(kxn−yα,mk) +νm)−
− kxn−yα,mk2+kxn−yα,mkkxn−Tmxnk ≤
≤ hx∗−xn, yα,m−Tmyα,mi+µmM ψ(M) +νmM+ +kxn−yα,mkkxn−Tmxnk,
(1.9)
whereM = supn≥0{kxn−yα,mk}. Sincexn ⇀ x∗ and (1.8), we arrive at hx∗−yα,m, yα,m−Tmyα,mi ≤µmM ψ(M) +νmM. (1.10) On the other hand, we have
hx∗−yα,m,(x∗−Tmx∗)−(yα,m−Tmyα,m)i ≤(1 +L)kx∗−yα,mk2
= (1 +L)α2kx∗−Tmx∗k2. (1.11) Note that
kx∗−Tmx∗k2=hx∗−Tmx∗, x∗−Tmx∗i=
= 1
αhx∗−yα,m, x∗−Tmx∗i=
= 1
αhx∗−yα,m,(x∗−Tmx∗)−(yα,m−Tmyα,m)i+
+ 1
αhx∗−yα,m, yα,m−Tmyα,mi.
(1.12)
Substituting (1.10) and (1.11) into (1.12), we arrive at
kx∗−Tmx∗k2≤(1 +L)αkx∗−Tmx∗k2+µmM ψ(M) +νmM
α .
This implies that
α[1−(1 +L)α]kx∗−Tmx∗k2≤µmM ψ(M) +νmM, ∀m≥1. (1.13)
Letting m → ∞ in (1.13), we see that Tmx∗ → x∗. Since T is uniformly L-Lipschitz, we obtain thatx∗=T x∗.This completes the proof.
2. Main results
Theorem 2.1. Let H be a real Hilbert space and C be a nonempty closed convex and bounded subset of H such that C+C ⊂ C. Let Ti : C → C be a uniformly Li-Lipschitz total asymptotically nonexpansive mapping with the function ψi and sequences {µ(i)n }, {νn(i)} for each i ∈ {1,2, . . . , N}. Assume that P∞
n=1µ(i)n <∞ andP∞
n=1νn(i)<∞ for each i∈ {1,2, . . . , N}. Let {un} be a bounded sequence in C such that P∞
n=1kunk<∞and {αn} a sequence in [L−1L , a], where L= max1≤i≤N{Li} >1 and a is some constant in(0,1).
Assume that F =∩Ni=1F(Ti)6=∅. Let {xn} be a sequence generated by (1.7).
Then the sequence {xn} converges weakly to some pointx∗∈ F.
Proof. Define the following sequences
µn= max{µ(1)n , µ(2)n , . . . , µ(Nn )} and
νn= max{νn(1), νn(2), . . . , νn(N)}.
It is easy to see that P∞
n=1µn < ∞ and P∞
n=1νn <∞. Fixing p ∈ F, we have
kxn−pk ≤αnkxn−1−pk+ (1−αn)kTi(n)k(n)xn−pk+kunk ≤
≤αnkxn−1−pk+ (1−αn) kxn−pk+µk(n)ψi(n)(kxn−pk) +νk(n)
+kunk ≤
≤ kxn−1−pk+µk(n)ψi(n)((diam C)) +νk(n)+kunkleq
≤ kxn−1−pk+µk(n)ψr((diam C)) +νk(n)+kunk,
(2.1) where
ψr((diam C)) = max{ψ1((diam C)), ψ2((diam C)), . . . , ψN((diam C))}.
In view of Lemma 1.1, we obtain that the limit of the sequence {kxn−pk}
exits. Next, we assume that
d= lim
n→∞kxn−pk, (2.2)
where d >0 is some constant. It follows that lim sup
n→∞
kxn−1−p−unk ≤lim sup
n→∞
kxn−1−pk+ lim sup
n→∞
kunk=d. (2.3)
Note that
kTi(n)k(n)xn−p+unk ≤ kTi(n)k(n)xn−pk+kunk ≤
≤ kxn−pk+µk(n)ψi(n)(kxn−pk) +νk(n)+kunk ≤
≤ kxn−pk+µk(n)ψr((diam C)) +νk(n)+kunk.
It follows that
lim sup
n→∞
kTi(n)k(n)xn−p+unk ≤d. (2.4) On the other hand, we have
d= lim
n→∞kxn−pk= lim
n→∞kαn(xn−1−p−un) + (1−αn)(Ti(n)k(n)xn−p+un)k.
(2.5) Combining (2.3), (2.4) with (2.4) and from Lemma 1.2, we obtain that
n→∞lim kxn−1−Ti(n)k(n)xnk= 0. (2.6) Note that
kxn−xn−1k ≤(1−αn)kxn−1−Ti(n)k(n)xnk+kunk.
From (2.6), we arrive at
n→∞lim kxn−xn−1k= 0. (2.7) On the other hand, we have
kTi(n)k(n)xn−xnk ≤ kTi(n)k(n)xn−xn−1k+kxn−1−xnk, which combined with (2.6) gives that
n→∞lim kTi(n)k(n)xn−xnk= 0. (2.8) From (2.7), we also have
n→∞lim kxn−xn+lk= 0, ∀l= 1,2, . . . , N. (2.9) For any positive n > N, it can be rewritten as n = (k(n)−1)N +i(n), i(n)∈ {1,2, . . . , N}. Note that
kxn−1−Tnxnk ≤ kxn−1−Ti(n)k(n)xnk+kTi(n)k(n)xn−Tnxnk ≤
≤ kxn−1−Ti(n)k(n)xnk+LkTi(n)k(n)−1xn−xnk ≤
≤ kxn−1−Ti(n)k(n)xnk+L{kTi(n)k(n)−1xn−Ti(n−Nk(n)−1)xn−Nk+
+kTi(n−Nk(n)−1)xn−N−x(n−N)−1k+kx(n−N)−1−xnk}.
(2.10)
On the other hand, we haven−N = ((k(n)−1)−1)N+i(n) = ((k(n)−1)− 1)N+i(n−N), i.e.,
k(n−N) =k(n)−1 and i(n−N) =i(n).
It follows that
kTi(n)k(n)−1xn−Ti(n−N)k(n)−1xn−Nk ≤Lkxn−xn−Nk (2.11) and
kTi(n−N)k(n)−1xn−N −x(n−N)−1k=kTi(n−Nk(n−N))xn−N −x(n−N)−1k. (2.12) Combining (2.6), (2.9) with (2.10), we see that
n→∞lim kxn−1−Tnxnk= 0. (2.13) On the other hand, we have that
kxn−Tnxnk ≤ kxn−xn−1k+kxn−1−Tnxnk.
From (2.7) and (2.13), we obtain that
n→∞lim kxn−Tnxnk= 0. (2.14) Consequently, for anyj= 1,2, . . . , N, we see that
kxn−Tn+jxnk ≤ kxn−xn+jk+kxn+j−Tn+jxn+jk+kTn+jxn+j−Tn+jxnk ≤
≤(1 +L)kxn−xn+jk+kxn+j−Tn+jxn+jk.
From (2.9) and (2.14), we arrive at
n→∞lim kxn−Tn+jxnk= 0.
Therefore, for∀i∈ {1,2, . . . , N}, there exists somee∈ {1,2, . . . , N}such that n+e=i(mod N). It follows that
n→∞lim kxn−Tixnk= lim
n→∞kxn−Tn+exnk= 0. (2.15) Since {xn} is bounded, we see that there exists a subsequence{xni} ⊂ {xn} such that xni ⇀ x∗. From Lemma 1.3, we can obtain that x∗∈ F. Next we prove that {xn} converges weakly tox∗. Suppose the contrary. Then we see that there exists some subsequence {xnj} ⊂ {xn} such that{xnj} converges
weakly to ¯x∈ C and ¯x 6=x∗. From Lemma 1.3, we also have ¯x∈ F. Put w= limn→∞kxn−x∗k.SinceH is an Opial’s space, we see that
w= lim
ni→∞kxni−x∗k< lim
ni→∞kxni−xk¯ =
= lim
nj→∞kxnj −xk¯ < lim
nj→∞kxnj−x∗k=
= lim
ni→∞kxni−x∗k=w.
This derives a contradiction. It follows that ¯x=x∗. This completes the proof.
Remark 2.2. Theorem 2.1 improves Theorem 1 of Chang et al. [9] from asymptotically nonexpansive mappings to total asymptotically nonexpansive mappings.
Corollary 2.3. Let H be a real Hilbert space and C be a nonempty closed convex and bounded subset ofH. Let Ti :C→C be a uniformly Li-Lipschitz total asymptotically nonexpansive mapping with the functionψi and sequences {µ(i)n }, {νn(i)} for each i ∈ {1,2, . . . , N}. Assume that P∞
n=1µ(i)n < ∞ and P∞
n=1νn(i)<∞for eachi∈ {1,2, . . . , N}. Let{αn} be a sequence in[L−1L , a], where L = max1≤i≤N{Li} > 1 and a is some constant in (0,1). Assume that F =∩Ni=1F(Ti)6=∅. Let {xn} be a sequence generated by the following manner:
x0∈C, xn =αnxn−1+ (1−αn)Ti(n)k(n)xn, ∀n≥1. (2.16) Then the sequence{xn}converges weakly to some point x∗∈ F.
Next, we prove a strong convergence theorem under the condition of semi- compactness.
Theorem 2.4. Let H be a real Hilbert space andC a nonempty closed con- vex and bounded subset of H such that C+C ⊂ C. Let Ti : C → C be a uniformly Li-Lipschitz total asymptotically nonexpansive mapping with the function ψi and sequences {µ(i)n }, {νn(i)} for each i ∈ {1,2, . . . , N}. Assume that P∞
n=1µ(i)n <∞andP∞
n=1νn(i)<∞ for eachi∈ {1,2, . . . , N}. Let {un} be a bounded sequence in C such that P∞
n=1kunk <∞ and {αn} a sequence in [L−1L , a], where L= max1≤i≤N{Li}>1 anda is some constant in (0,1).
Assume thatF=∩Ni=1F(Ti)6=∅and at least there exists a mapping Trwhich is semicompact. Let{xn}be a sequence generated by (1.7). Then the sequence {xn} converges weakly to some point x∗∈ F.
Proof. Without loss of generality, we may assume thatT1 is semicompact. It follows from (2.15) that
n→∞lim kxn−T1xnk= 0.
By the semicompactness of T1, we have there exists a subsequence {xni} of {xn} such thatxni →x∈C strongly. From (2.15), we have
nlimi→∞kxni−Tlxnik=kx−Tlxk= 0,
for all l= 1,2,· · · , N.This implies thatx∈ F.From Theorem 2.1, we know that limn→∞kxn−pkexists for eachp∈ F. This shows that limn→∞kxn−xk exists. From xni→x, we have
n→∞lim kxn−x∗k= 0.
This completes the proof of Theorem 2.4.
Corollary 2.5. Let H be a real Hilbert space and C be a nonempty closed convex bounded subset of H. Let Ti : C → C be a uniformly Li-Lipschitz total asymptotically nonexpansive mapping with the functionψiand sequences {µ(i)n }, {νn(i)} for each i ∈ {1,2, . . . , N}. Assume that P∞
n=1µ(i)n < ∞ and P∞
n=1νn(i)<∞for eachi∈ {1,2, . . . , N}. Let{αn}be a sequence in[L−1L , a], where L = max1≤i≤N{Li} > 1 and a is some constant in (0,1). Assume that F = ∩Ni=1F(Ti) 6= ∅ and at least there exists a mapping Tr which is semicompact. Let {xn} be a sequence generated by (2.16). Then the sequence {xn} converges weakly to some point x∗∈ F.
3. Applications
LetA:C→H be a mapping. Recall that the classical variational inequal- ity problem is to findx∈C such that
hAx, y−xi ≥0, ∀y∈C. (3.1) It is known that x ∈ C is a solution to the variational inequality problem (3.1) if and only if x is a fixed point of the mapping PC(I −ρA), where PC is the metric projection from H onto C, ρ > 0 is a constant and I is the identity mapping. This implies that the variational inequality problem (3.1) is equivalent to a fixed point problem. This alternative formula is very important form the numerical analysis point of view. Recently, many authors studied the variational inequality (3.1) by iterative methods.
LetF be a bifunction ofC×C intoR, whereRis the set of real numbers.
We consider the following equilibrium problem:
Findx∈C such thatF(x, y)≥0, ∀y∈C. (3.2)
In this paper, the set of suchx∈Cis denoted by EP(F), i.e., EP(F) ={x∈C:F(x, y)≥0, ∀y∈C}.
Numerous problems in physics, optimization and economics reduce to find a solution of (1.1); see ([3],[5],[10],[11]). Approximating solutions of the problem (3.2) based on iterative methods was studied by many authors, see, for exam- ple, ([4]-[7],[17],[20],[22],[25],[26],[33],[37]) and the reference therein. Putting F(x, y) = hAx, y−xi, ∀x, y ∈ C, we see that z ∈ F P(F) if and only if hAz, y−zi ≥0,∀y∈C. That is,z is a solution to the variational inequality (3.1). Numerous problems in physics, optimization, and economics reduce to find a solution of the problem (3.2). To study the problem (3.2), we may assume that the bifunctionF :C×C→Rsatisfies the following conditions:
(A1) F(x, x) = 0 for allx∈C;
(A2) F is monotone, i.e.,F(x, y) +F(y, x)≤0 for allx, y ∈C;
(A3) for eachx, y, z∈C, limt↓0F(tz+ (1−t)x, y)≤F(x, y);
(A4) for eachx∈C,y7→F(x, y) is convex and weakly lower semi-continuous.
Next, we consider the convergence of implicit iterative process (1.7) for the equilibrium problem (3.2). To prove the main results in this section, we need the following lemma which can be found in [3] and [4].
Lemma 3.1. Let C be a nonempty closed convex subset of H ad let F : C×C → R be a bifunction satisfying (A1)-(A4). Then, for any r > 0 and x∈ H, there existsz ∈C such that F(z, y) + 1rhy−z, z−xi ≥ 0, ∀y ∈ C.
Further, define a mapping
Trx={z∈C:F(z, y) +1
rhy−z, z−xi ≥0, ∀y∈C}
for allr >0 andx∈H.Then, the following hold:
(1) Tr is single-valued;
(2) Tr is firmly nonexpansive, i.e., for anyx, y∈H, kTrx−Tryk2≤ hTrx−Try, x−yi;
(3) F(Tr) =EP(F);
(4) EP(F) is closed and convex.
Theorem 3.2. LetCbe a nonempty closed and convex subset of a real Hilbert space H. Let F1 be a bifunction from C×C toR satisfying (A1)-(A4) such that EP(F1)is nonempty. Let {xn} be a sequence generated in the following manner:
x0∈H, chosen arbitrily
F1(yn, y) +r1nhy−yn, yn−xni ≥0, ∀y∈C, xn=αnxn−1+ (1−αn)yn+un, ∀n≥1.
(3.3)
where {rn} ⊂(0,∞), {αn} ⊂(0,1) and{un} is a bounded sequence. Assume that the following conditions are satisfied
(1) a≤αn≤b, where0< a < b <1;
(2) lim infn→∞rn >0;
(3) P∞
n=1kunk<∞.
Then the sequence{xn} generated in(3.3) converges weakly to some point in EP(F).
Proof. Putting yn = Trnxn for each n ≥ 1, we from Lemma 3.1 see that Trnis firmly nonexpansive. Whenever needed, we shall equivalently write the implicit iteration (3.3) as
x0∈H, xn =αnxn−1+ (1−αn)Trnxn+un, ∀n≥1, (3.4) Fixingp∈EP(F), we see that
kxn−pk ≤αnkxn−1−pk+ (1−αn)kTrnxn−pk+kunk
≤αnkxn−1−pk+ (1−αn)kxn−pk+kunk.
This in turn implies that
kxn−pk ≤ kxn−1−pk+kunk a .
From Lemma 1.1, we obtain that limn→∞kxn−pk exits. Next, we assume that limn→∞kxn−pk=r >0. Note that
lim sup
n→∞
kxn−1−p+unk ≤r (3.5) and
lim sup
n→∞
kTrnxn−p+unk ≤lim sup
n→∞
(kxn−pk+kunk)≤r. (3.6)
On the other hand, we have
n→∞lim kxn−pk= lim
n→∞kαn(xn−1−p+un)+(1−αn)(Trnxn−p+un)k=r. (3.7) Combining (3.5), (3.6) with (3.7), from Lemma 1.2, we obtain that
n→∞lim kxn−1−Trnxnk= 0. (3.8) From (3.1), we arrive at
xn−xn−1= (1−αn)(Trnxn−xn−1) +un. From the conditions (3) and (3.8), we see that
n→∞lim kxn−1−xnk= 0. (3.9) Note that
kxn−Trnxnk ≤ kxn−xn−1k+kxn−1−Trnxnk.
In view of (3.8) and (3.9), we obtain that
n→∞lim kxn−Trnxnk= 0. (3.9) Since the sequence{xn}is a bounded, we see that there exists a subsequence {xni}of{xn}such thatxni ⇀ q.
Next, we show that q∈EP(T). Sinceyn=Trnxn, we have F(yn, y) + 1
rn
hy−yn, yn−xni ≥0, ∀y∈C.
It follows from (A2) that
hy−yn,yn−xn
rn
i ≥F(y, yn) and hence
hy−yni,yni−xni
rni
i ≥F(y, yni).
Since ynir−xni
ni →0,yni ⇀ q and (A4), we haveF(y, q)≤0 for ally∈C.For twith 0< t≤1 and y∈C,letyt=ty+ (1−t)q. Sincey∈C andq∈C, we haveyt∈C and henceF(yt, q)≤0.So, from (A1) and (A4), we have
0 =F(yt, yt)≤tF(yt, y) + (1−t)F(yt, q)≤tF(yt, y).
That is,F(yt, y)≥0.It follows from (A3) that F(q, y)≥0 for ally ∈C and henceq∈EP(F). This proves that q∈EP(T).
Finally, we show that the sequence{xn} converges weakly toq. Suppose the contrary holds. It follows that there exists some subsequence {xnj} of {xn} such that xnj ⇀q¯andq6= ¯q. By the same method as given above, we can prove that ¯q∈EP(T). Put
n→∞lim kxn−qk=d1 and lim
n→∞kxn−qk¯ =d2,
where d1 andd2 are two nonnegative numbers. In view of Opial’s condition, we see that
d1= lim inf
i→∞ kxni−qk<lim inf
j→∞ kxni−¯qk= lim inf
j→∞ kxnj−¯qk<lim inf
j→∞ kxnj−qk=d1. This is a contradiction. Hence ¯q = q. This shows that the sequence {xn} converges weakly toq.The proof is completed.
Acknowledgments
The author are extremely grateful to the referee for useful suggestions that improved the contents of the paper.
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Zhejiang Ocean University
School of Mathematics, Physics and Information Science Zhoushan 316004, China
Email: [email protected]
