Filomat 28:10 (2014), 2175–2184
DOI 10.2298/FIL1410175C
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Approximation of Viscosity Zero Points of Accretive Operators in a
Banach Space
Sun Young Choa , Shin Min Kangb
a Department
b Department
of Mathematics, Gyeongsang National University, Jinju 660-701, Korea
of Mathematics and RINS, Gyeongsang National University, Jinju 660-701, Korea
Abstract. In this paper, zero points of m-accretive operators are investigated based on a viscosity iterative
algorithm with double computational errors. Strong convergence theorems for zero points of m-accretive
operators are established in a Banach space.
1. Introduction
Zero points of accretive operators have been investigated based on iterative algorithms recently. Interest
in accretive operators stems mainly from their firm connection with equations of evolution is an important
class of nonlinear operators. It is known that many physically significant problems can be modelled by
initial value problems of the form
x0 (t) + Ax(t) = 0, x(0) = x0 ,
(1.1)
where A is an accretive operator in an appropriate Banach space. Typical examples where such evolution
equations occur can be found in the heat, wave or Schrödinger equations. If x(t) is dependent of t, then
(1.1) is reduced to
Au = 0,
(1.2)
whose solutions correspond to the equilibrium points of the system (1.1). Consequently, considerable
research efforts have been devoted, especially within the past 40 years or so, to methods for finding
approximate solutions of the equation (1.2). An early fundamental result in the theory of accretive operators,
due to Browder [1], states that the initial value problem (1.1) is solvable if A is locally Lipschitz and accretive
on E.
The organization of this paper is as follows. In Section 2, we provide some necessary preliminaries. In
Section 3, we study zero points of m-accretive operators based on a viscosity iterative algorithm. Strong
convergence theorems for zero points of m-accretive operators are established in a Banach space. It is
proved that the zero point is also a solution to some variational inequality.
2010 Mathematics Subject Classification. Primary 47H06; Secondary 47H09, 47H10
Keywords. accretive operator; fixed point; nonexpansive mapping; iterative algorithm; zero point.
Received: 27 September 2013; Accepted: 25 November 2013
Communicated by Ljubomir Ćirić
Email addresses: [email protected];[email protected] (Sun Young Cho), [email protected] (Shin Min Kang )
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2. Preliminaries
In what follows, we always assume that E is a Banach space with the dual E∗ . Let UE = {x ∈ E : kxk = 1}.
kx+tyk−kxk
E is said to be smooth or said to be have a Gâteaux differentiable norm if the limit limt→0
exists for
t
each x, y ∈ UE . E is said to have a uniformly Gâteaux differentiable norm if for each y ∈ UE , the limit is attained
uniformly for all x ∈ UE . E is said to be uniformly smooth or said to have a uniformly Fréchet differentiable norm
if the limit is attained uniformly for x, y ∈ UE . Let C be a nonempty closed convex subset of a real Banach
space E and E∗ be the dual space of E. Let h·, ·i denote the pairing between E and E∗ . The normalized duality
∗
mapping J : E → 2E is defined by
J(x) = { f ∈ E∗ : hx, f i = kxk2 = k f k2 }
for all x ∈ E. In the sequel, we use j to denote the single-valued normalized duality mapping. It is known
that if the norm of E is uniformly Gâteaux differentiable, then the duality mapping J is single valued and
uniformly norm to weak∗ continuous on each bounded subset of E.
Recall that a closed convex subset C of a Banach space E is said to have normal structure if for each
bounded closed convex subset K of C which contains at least two points, there exists an element x of K
which is not a diametral point of K, i.e., sup{kx − yk : y ∈ K} < d(K), where d(K) is the diameter of K. It is
well known that a closed convex subset of uniformly convex Banach space has the normal structure and a
compact convex subset of a Banach space has the normal structure; see [2] for more details.
Let T : C → C be a mapping. In this paper, we use F(T) to denote the set of fixed points of T. Recall that
T is said to be contractive if there exits a constant α ∈ (0, 1) such that
kTx − Tyk ≤ αkx − yk,
∀x, y ∈ C.
For such a case, we also call T an α-contraction. T is said to be nonexpansive if
kTx − Tyk ≤ kx − yk,
∀x, y ∈ C.
T is said to be pseudocontractive if there exists some j(x − y) ∈ J(x − y) such that
hTx − Ty, j(x − y)i ≤ kx − yk2 ,
∀x, y ∈ C.
2
Let D be a nonempty subset
of C. Let Q : C → D. Q is said to be contraction if Q = Q; sunny if for each
x ∈ C and t ∈ (0, 1), we have Q tx + (1 − t)Qx = Qx; sunny nonexpansive retractction if Q sunny, nonexpansive,
and contraction. K is said to be a nonexpansive retract of C if there exists a nonexpansive retraction from C
onto D.
The following result, which was established in [3,4], describes a characterization of sunny nonexpansive
retractions on a smooth Banach space.
Let E be a smooth Banach space and C be a nonempty subset of E. Let Q : E → C be a retraction and j
be the normalized duality mapping on E. Then the following are equivalent:
(1) Q is sunny and nonexpansive;
(2) kQx − Qyk2 ≤ hx − y, j(Qx − Qy)i, ∀x, y ∈ E;
(3) hx − Qx, j(y − Qx)i ≤ 0, ∀x ∈ E, y ∈ C.
Let I denote the identity operator on E. An operator A ⊂ E × E with domain D(A) = {z ∈ E : Az , ∅} and
range R(A) = ∪{Az : z ∈ D(A)} is said to be accretive if for each xi ∈ D(A) and yi ∈ Axi , i = 1, 2, there exists
j(x1 − x2 ) ∈ J(x1 − x2 ) such that hy1 − y2 , j(x1 − x2 )i ≥ 0. An accretive operator A is said to be m-accretive if
R(I + rA) = E for all r > 0. In a real Hilbert space, an operator A is m-accretive if and only if A is maximal
monotone. In this paper, we use A−1 (0) to denote the set of zero points of A.
For an accretive operator A, we can define a nonexpansive single-valued mapping Jr : R(I + rA) → D(A)
by Jr = (I + rA)−1 for each r > 0, which is called the resolvent of A.
One of classical methods of studying the problem 0 ∈ Ax, where A ⊂ E × E is an accretive operator, is
the following:
x0 ∈ E, xn+1 = Jrn xn , ∀n ≥ 0,
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where Jrn = (I + rn A)−1 and {rn } is a sequence of positive real numbers.
The classical method was first proposed by Martinet [5] and generalized by Rockaffellar [6] and [7]. This
method and its dual version in the context of convex programming have been extensively studied; see, for
instance, [8-10] and the references therein. This method is known to yield as special cases decomposition
methods such as the method of partial inverses, the Douglas-Rachford splitting method, and the alternating
direction method of multipliers. Regularization methods recently have been investigated for treating zero
points of accretive operators; for [11-25] and the references therein. However, as pointed in [26], the ideal
form of the above method is often impractical since, in may cases, to solve the problem exactly is either
impossible or the same difficult as the original inclusion problem. Therefore, one of the most interesting
and important problems in the theory of accretive operators is to find an efficient iterative methods to
compute approximately zeros of A.
In this paper, zero points of m-accretive operators are investigated based on a viscosity iterative algorithm
with double computational errors. Strong convergence theorems for zero points of m-accretive operators
are established in a reflexive Banach space.
In order to prove our main results, we also need the following lemmas.
Lemma 2.1 [27] Let {an }, {bn }, and {cn } be three nonnegative real sequences satisfying
an+1 ≤ (1 − tn )an + bn + cn ,
∀n ≥ 0,
where {tPn } is a sequence in (0, 1). Assume that the following conditions are satisfied
(a) P∞
n=0 tn = ∞ and bn = o(tn );
(b) ∞
n=0 cn < ∞.
Then limn→∞ an = 0.
Lemma 2.2 [28] Let E be a Banach space, and A an m-accretive operator. For λ > 0, µ > 0, and x ∈ E, we have
Jλ x = Jµ
µ x+ 1−
Jλ x ,
λ
λ
µ
where Jλ = (I + λA)−1 and Jµ = (I + µA)−1 .
Lemma 2.3 [29] Let {xn } and {yn } be bounded sequences in a Banach space E, and {βn } be a sequence in (0, 1) with
0 < lim inf βn ≤ lim sup βn < 1.
n→∞
n→∞
Suppose that xn+1 = (1 − βn )yn + βn xn , ∀n ≥ 1 and
lim sup(kyn+1 − yn k − kxn+1 − xn k) ≤ 0.
n→∞
Then limn→∞ kyn − xn k = 0.
Lemma 2.4 [30] Let E a real reflexive Banach space with the uniformly Gâteaux differentiable norm and the normal
structure, and C be a nonempty closed convex subset of E. Let S : C → C be a nonexpansive mapping with a fixed
point, and T : C → C be a fixed contraction with the coefficient α ∈ (0, 1). Let {xt } be a sequence generated by the
following xt = tTxt + (1 − t)Sxt , where t ∈ (0, 1). Then {xt } converges strongly as t → 0 to a fixed point x∗ of T, which
is the unique solution in F(T) to the following variational inequality hTx∗ − x∗ , j(x∗ − p)i ≥ 0, ∀p ∈ F(S).
3. Main Results
Now, we are in a position to state our main result.
Theorem 3.1. Let E be a real reflexive Banach space with the uniformly Gâteaux differentiable norm and let A be an
m-accretive operators in E. Assume that C := D(A) is convex and has the normal structure. Let T : C → C be a fixed
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α-contraction. Let {αn }, {βn }, {γn }, {αn 0 }, {βn 0 } and {γn 0 } be real number sequences in (0, 1). Let QC be the sunny
nonexpansive retraction from E onto C and {xn } be a sequence generated in the following manner: x0 ∈ C and


0

 yn = αn 0 xn + βn Jrn (xn + en+1 ) + γn 0 QC ( fn ),


xn+1 = αn Txn + βn xn + γn yn , ∀n ≥ 0,
where {en } is a sequence in E, { fn } is a bounded sequence in E, {rn } is a positive real numbers sequence, and
Jrn = (I + rn A)−1 . Assume that A−1 (0) is not empty and the above control sequences satisfy the following restrictions:
(a) αn + βn + γn = αn 0 + β0n + γ0n = 1;P
(b) limn→∞ αn = limn→∞ α0n = 0 and ∞
n=1 αn = ∞;
(c) 0P< lim infn→∞ βn ≤Plim supn→∞ βn < 1;
∞
0
(d) ∞
n=0 γn < ∞;
n=1 ken k < ∞ and
0
0
(e) limn→∞ |βn − βn+1 | = 0;
(f) rn ≥ for each n ≥ 1 and limn→∞ |rn − rn+1 | = 0.
Then the sequence {xn } converges strongly to x̄, which is the unique solution to the following variational inequality
hTx̄ − x̄, j(p − x̄)i ≤ 0,
∀p ∈ A−1 (0).
Proof. First, we prove that {xn } is bounded. Fixing p ∈ A−1 (0), we see that
ky0 − pk = kα0 0 (x0 − p) + β0 0 (Jr0 (x0 + e1 ) − p) + γ0 0 (QC ( f0 ) − p)k
≤ α0 0 kx0 − pk + β0 0 kJr0 (x0 + e1 ) − pk + γ0 0 kQC ( f0 ) − pk
≤ (1 − γ00 )kx0 − pk + ke1 k + γ0 0 k f0 − pk.
It follows that
kx1 − pk ≤ α0 kTx0 − pk + β0 kx0 − pk + γ0 ky0 − pk
≤ α0 αkx0 − pk + α0 kTp − pk + β0 kx0 − pk + γ0 ky0 − pk
≤ α0 αkx0 − pk + α0 kTp − pk + β0 kx0 − pk + (1 − γ00 )γ0 kx0 − pk
+ ke1 k + γ0 γ00 k f0 − pk
≤ 1 − α0 (1 − α) kx0 − pk + α0 kTp − pk + ke1 k + γ00 k f0 − pk.
Put M1 = max{kx0 − pk,
kTp−pk
1−α }
< ∞. Next, we prove that
kxn − pk ≤ M1 +
n
X
kei k +
i=1
n−1
X
γ0i k fi k.
i=1
It is easy to see that (3.1) holds for n = 1. We assume that the result holds for some n. Notice that
kyn − pk = kαn 0 (xn − p) + βn 0 (Jrn (xn + en+1 ) − p) + γn 0 (QC ( fn ) − p)k
≤ αn 0 kxn − pk + βn 0 kJrn (xn + en+1 ) − pk + γn 0 kQC ( fn ) − pk
≤ (1 − γ0n )kxn − pk + ken+1 k + γn 0 k fn − pk.
It follows that
kxn+1 − pk ≤ αn kTxn − pk + βn kxn − pk + γn kyn − pk
≤ αn αkxn − pk + αn kTp − pk + βn kxn − pk + γn kyn − pk
≤ αn αkxn − pk + αn kTp − pk + βn kxn − pk + (1 − γ0n )γn kxn − pk
+ ken+1 k + γn γ0n k fn − pk
≤ 1 − αn (1 − α) kxn − pk + αn kTp − pk + ken+1 k + γ0n k fn − pk
≤ M1 +
n+1
X
i=1
kei k +
n
X
i=1
γ0i k fi k.
(3.1)
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This yields that the sequence {xn } is bounded. Next, we show that
lim kxn+1 − xn k = 0.
(3.2)
kJrn (xn + en+1 ) − Jrn+1 (xn+1 + en+2 )k
rn+1
rn+1 Jrn (xn + en+1 ) − Jrn+1 (xn+1 + en+2 )
(xn + en+1 ) + 1 −
= Jrn+1
rn
rn
r
rn+1 n+1
≤ (xn + en+1 ) − (xn+1 + en+2 ) + 1 −
Jrn (xn + en+1 ) − (xn+1 + en+2 ) r
rn
n
r
n+1
Jrn (xn + en+1 ) − (xn + en+1 ) = (xn + en+1 ) − (xn+1 + en+2 ) + 1 −
rn
M2
≤ kxn − xn+1 k + ken+1 k + ken+2 k +
|rn+1 − rn |,
(3.3)
n→∞
It follows from Lemma 2.2 that
where M2 is an appropriate constant such that
M2 ≥ sup{kJrn (xn + en+1 ) − (xn + en+1 )k}.
n≥1
Notice that
yn+1 − yn
= α0n+1 (xn+1 − xn ) + xn (α0n+1 − α0n )
+ β0n+1 Jrn+1 (xn+1 + en+2 ) − Jrn (xn + en+1 ) + Jrn (xn + en+1 )(β0n+1 − β0n )
+ γ0n+1 (QC ( fn+1 ) − QC ( fn )) + QC ( fn )(γ0n+1 − γ0n ),
which yields that
kyn+1 − yn k
≤ α0n+1 kxn+1 − xn k + kxn k|α0n+1 − α0n |
+ β0n+1 kJrn+1 (xn+1 + en+2 ) − Jrn (xn + en+1 )k + kJrn (xn + en+1 )k|β0n+1 − β0n |
(3.4)
+ γ0n+1 kQC ( fn+1 ) − QC ( fn )k + kQC ( fn )k|γ0n+1 − γ0n |.
Substituting (3.3) into (3.4), we find that
kyn+1 − yn k ≤ kxn+1 − xn k + kxn k|α0n+1 − α0n |
M2
|rn+1 − rn |
+ ken+1 k + ken+2 k +
+ kJrn (xn + en+1 )k|β0n+1 − β0n |
(3.5)
+ γ0n+1 kQC ( fn+1 ) − QC ( fn )k + kQC ( fn )k|γ0n+1 − γ0n |.
Put 1n =
xn+1 −βn xn
1−βn .
That is,
xn+1 = (1 − βn )1n + βn xn ,
n ≥ 0.
Now, we compute k1n+1 − 1n k. Note that
αn+1 Txn+1 + γn+1 yn+1 αn Txn + γn yn
−
1 − βn+1
1 − βn
γn+1
γn
αn+1
αn
=
Txn+1 +
yn+1 −
Txn −
yn
1 − βn+1
1 − βn+1
1 − βn
1 − βn
αn+1
αn
=
(Txn+1 − yn+1 ) + yn+1 −
(Txn − yn ) − yn .
1 − βn+1
1 − βn
1n+1 − 1n =
(3.6)
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This yields that
k1n+1 − 1n k ≤
αn+1
αn
kTxn+1 − yn+1 k + kyn+1 − yn k +
kTxn − yn k.
1 − βn+1
1 − βn
(3.7)
Substituting (3.5) into (3.7), we find that
k1n+1 − 1n k − kxn+1 − xn k
αn+1
≤
kTxn+1 − yn+1 k + kxn k|α0n+1 − α0n |
1 − βn+1
M2
+ ken+1 k + ken+2 k +
|rn+1 − rn | + kJrn (xn + en+1 )k|β0n+1 − β0n |
αn
+ γ0n+1 kQC ( fn+1 ) − QC ( fn )k + kQC ( fn )k|γ0n+1 − γ0n | +
kTxn − yn k.
1 − βn
From the restrictions (b)-(f), we see that
lim sup k1n+1 − 1n k − kxn − xn+1 k ≤ 0.
n→∞
It follows from Lemma 2.3 that
lim k1n − xn k = 0.
n→∞
In view of (3.6), we have
xn+1 − xn = (1 − βn )(1n − xn ).
Hence, we find that
lim kxn+1 − xn k = 0.
(3.8)
n→∞
Notice that
yn − xn =
xn+1 − xn + αn xn − f (xn )
γn
.
In view of the restrictions (b) and (c), we find from (3.8) that
lim kyn − xn k = 0.
n→∞
Notice that
(3.9)
kxn − Jrn (xn + en+1 )k
≤ kxn − xn+1 k + kxn+1 − yn k + kyn − Jrn (xn + en+1 )k
≤ kxn − xn+1 k + αn k f (xn ) − yn k + βn kxn − yn k + kyn − Jrn (xn + en+1 )k
≤ kxn − xn+1 k + αn k f (xn ) − yn k + βn kxn − yn k
+ α0n kxn − Jrn (xn + en+1 )k + γ0n kQC ( fn ) − Jrn (xn + en+1 )k.
In view of the restrictions (b) and (d), we find from (3.8) and (3.9) that
lim kxn − Jrn (xn + en+1 )k = 0.
n→∞
Notice that
kxn − Jrn xn k ≤ kxn − Jrn (xn + en+1 )k + kJrn (xn + en+1 ) − Jrn xn k
≤ kxn − Jrn (xn + en+1 )k + ken+1 k.
Since
P∞
n=1
ken k < ∞, we see from (3.10) that
lim kxn − Jrn xn k = 0.
n→∞
(3.10)
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Take a fixed number r such that > r > 0. In view of Lemma 2.2, we obtain that
r
r
kJrn xn − Jr xn k = kJr xn + (1 − )Jrn xn − Jr xn k
rn
rn
r
≤ k(1 − )(Jrn xn − xn )k
rn
≤ kJrn xn − xn k.
Note that
(3.11)
kxn − Jr xn k ≤ kxn − Jrn xn k + kJrn xn − Jr xn k
≤ 2kxn − Jrn xn k.
Using (3.11), we see that
lim kxn − Jr xn k = 0.
(3.12)
n→∞
Next, we claim that lim suphx̄ − Tx̄, j(xn − x̄)i ≤ 0, where x̄ = limt→0 zt , and zt solves the fixed point
n→∞
equation zt = tTzt + (1 − t)Jr zt , ∀t ∈ (0, 1), from which it follows that
kzt − xn k = k(1 − t)(Jr zt − xn ) + t(Tzt − xn )k.
For any t ∈ (0, 1), we see that
kzt − xn k2 = (1 − t)hJr zt − xn , j(zt − xn )i + thTzt − xn , j(zt − xn )i
= (1 − t) hJr zt − Jr xn , j(zt − xn )i + hJr xn − xn , j(zt − xn )i
+ thTzt − zt , j(zt − xn )i + thzt − xn , j(zt − xn )i
≤ (1 − t) kzt − xn k2 + kJr xn − xn kkzt − xn k
+ thTzt − zt , j(zt − xn )i + tkzt − xn k2
≤ kzt − xn k2 + kJr xn − xn kkzt − xn k + thTzt − zt , j(zt − xn )i.
It follows that
hzt − Tzt , j(zt − xn )i ≤
In view of (3.12), we see that
1
kJr xn − xn kkzt − xn k
t
∀t ∈ (0, 1).
lim suphzt − Tzt , j(zt − xn )i ≤ 0.
(3.13)
n→∞
Since zt → x̄, as t → 0 and the fact that j is strong to weak∗ uniformly continuous on bounded subsets of E,
we see that
|hTx̄ − x̄, j(xn − x̄)i − hzt − Tzt , j(zt − xn )i|
≤ |hTx̄ − x̄, j(xn − x̄)i − hTx̄ − x̄, j(xn − zt )i|
+ |hTx̄ − x̄, j(xn − zt )i − hzt − Tzt , j(zt − xn )i|
≤ |hTx̄ − x̄, j(xn − x̄) − j(xn − zt )i| + |hTx̄ − x̄ + zt − Tzt , J(xn − zt )i|
≤ kTx̄ − x̄kk j(xn − x̄) − j(xn − zt )k + kTx̄ − x̄ + zt − Tzt kkxn − zt k → 0,
as t → 0.
Hence, for any > 0, there exists λ > 0 such that ∀t ∈ (0, λ) the following inequality holds
hTx̄ − x̄, j(xn − x̄)i ≤ hzt − Tzt , j(zt − xn )i + .
This implies that
lim suphTx̄ − x̄, j(xn − x̄)i ≤ lim suphzt − Tzt , j(zt − xn )i + .
n→∞
n→∞
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Since is arbitrary and (3.13), we see that lim supn→∞ hTx̄ − x̄, j(xn − x̄)i ≤ 0. This implies that
lim suphTx̄ − x̄, j(xn+1 − x̄)i ≤ 0.
(3.14)
n→∞
Finally, we show that xn → x̄ as n → ∞. Note that
kyn − x̄k ≤ α0n kxn − x̄k + β0n kJrn (xn + en+1 ) − x̄k + γ0n kQC fn − x̄k
≤ (α0n + β0n )kxn − x̄k + ken+1 k + γ0n k fn − x̄k
≤ kxn − x̄k + ken+1 k +
(3.15)
γ0n M2 ,
where M2 = supn≥1 k fn − x̄k. Put δn = ken+1 k + γ0n M2 . In view of the restriction (d), we see that
Notice that
kxn+1 − x̄k2 ≤ αn hTxn − x̄, j(xn+1 − x̄)i + βn kxn − x̄kkxn+1 − x̄k
P∞
n=1
δn < ∞.
+ γn kyn − x̄kkxn+1 − x̄k
≤ αn hTxn − x̄, j(xn+1 − x̄)i +
+
βn
(kxn − x̄k2 + kxn+1 − x̄k2 )
2
γn
(kyn − x̄k2 + kxn+1 − x̄k2 ).
2
It follows from (3.15) that
kxn+1 − x̄k2 ≤ 2αn hTxn − x̄, j(xn+1 − x̄)i + βn kxn − x̄k2 + γn kyn − x̄k2
≤ 2αn hTxn − x̄, j(xn+1 − x̄)i + βn kxn − x̄k2 + γn kxn − x̄k2
+ γn δ2n + 2γn δn kxn − x̄k
(3.16)
≤ (1 − αn )kxn − x̄k2 + 2αn hTxn − x̄, j(xn+1 − x̄)i + δn M3 ,
where M3 is an appropriate constant such that M3 ≥ supn≥0 {δn + 2kxn − x̄k}. Let µn = max{hTxn − x̄, j(xn+1 −
x̄)i, 0}. Next, we show that limn→∞ µn = 0. Indeed, from (3.14), for any give > 0, there exists a positive
integer n1 such that
hTxn − x̄, j(xn+1 − x̄)i < , ∀n ≥ n1 .
This implies that 0 ≤ µn < , ∀n ≥ n1 . Since > 0 is arbitrary, we see that limn→∞ µn = 0. It follows from
(3.16) that
kxn+1 − x̄k2 ≤ (1 − αn )kxn − x̄k2 + 2αn µn + δn M3 .
Put an = kxn − x̄k2 , tn = αn , bn = 2αn µn and cn = δn M3 for every n ≥ 0. In view of Lemma 2.1, we find the
desired conclusion.
If the mapping T maps every point in C to a fixed element, then we have the following results.
Corollary 3.2. Let E be a real reflexive Banach space with the uniformly Gâteaux differentiable norm and A be an
m-accretive operators in E. Assume that C := D(A) is convex and has the normal structure. Let {αn }, {βn }, {γn },
{αn 0 }, {βn 0 } and {γn 0 } be real number sequences in (0, 1). Let QC be the sunny nonexpansive retraction from E onto C
and {xn } be a sequence generated in the following manner:


x0 ∈ C,




y = αn 0 xn + βn 0 Jrn (xn + en+1 ) + γn 0 QC ( fn ),


 n

xn+1 = αn u + βn xn + γn yn , ∀n ≥ 0,
where u is a fixed element in C, {en } is a sequence in E, { fn } is a bounded sequence in E, {rn } is a positive real numbers
sequence, and Jrn = (I + rn A)−1 . Assume that A−1 (0) is not empty and the above control sequences satisfy the following
restrictions:
(a) αn + βn + γn = αn 0 + β0n + γ0n = 1;
S.Y. Cho, S.M. Kang / Filomat 28:10 (2014), 2175–2184
2183
P
(b) limn→∞ αn = limn→∞ α0n = 0 and ∞
n=1 αn = ∞;
(c) 0P< lim infn→∞ βn ≤Plim supn→∞ βn < 1;
∞
0
(d) ∞
n=0 γn < ∞;
n=1 ken k < ∞ and
0
0
(e) limn→∞ |βn − βn+1 | = 0;
(f) rn ≥ for each n ≥ 1 and limn→∞ |rn − rn+1 | = 0.
Then the sequence {xn } converges strongly to x̄, which is the unique solution to the following variational inequality
hu − QA−1 (0) u, j(p − QA−1 (0) u)i ≤ 0,
∀p ∈ A−1 (0).
If γ0n = 0, then Corollary 3.2 is reduced to the following.
Corollary 3.3. Let E be a real reflexive Banach space with the uniformly Gâteaux differentiable norm and let A be
an m-accretive operators in E. Assume that C := D(A) is convex and has the normal structure. Let {αn }, {βn }, {γn },
{αn 0 }, and {βn 0 } be real number sequences in (0, 1). Let {xn } be a sequence generated in the following manner:


x ∈ C,


 0

yn = αn 0 xn + βn 0 Jrn (xn + en+1 ),




xn+1 = αn u + βn xn + γn yn , ∀n ≥ 0,
where u is a fixed element in C, {en } is a sequence in E, {rn } is a positive real numbers sequence, and Jrn = (I + rn A)−1 .
Assume that A−1 (0) is not empty and the above control sequences satisfy the following restrictions:
(a) αn + βn + γn = αn 0 + β0n = 1;
P
(b) limn→∞ αn = limn→∞ α0n = 0 and ∞
n=1 αn = ∞;
(c) 0P< lim infn→∞ βn ≤ lim supn→∞ βn < 1;
(d) ∞
n=1 ken k < ∞;
(e) limn→∞ |β0n − β0n+1 | = 0;
(f) rn ≥ for each n ≥ 1 and limn→∞ |rn − rn+1 | = 0.
Then the sequence {xn } converges strongly to x̄, which is the unique solution to the following variational inequality
hu − QA−1 (0) u, j(p − QA−1 (0) u)i ≤ 0,
∀p ∈ A−1 (0).
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