Vietnam Journal of Mathematics 40:2&3(2012) 319-330
© VAST
An Iteration Method for Finding the Unique
Solution of a Variational Inequality Problem
Constrained by the Intersection of the Sets of
Fixed Points of a Countable Family of
Nonexpansive Mappings in Hilbert Spaces
Lai-Jiu Lin and Chih-Sheng Chuang
Department of Mathematics, National Changhua University of Education,
Changhua, 50058, Taiwan
Dedicated to Professor Phan Quoc Khanh on the occasion of his 65th-birthday
Received October 14, 2011
Revised January 18, 2012
Abstract. In this paper, we consider an iteration process which converges strongly
to a common fixed point of a countable family of nonexpansive mappings, and it is
the unique solution of a variational inequality constrained by this set of common fixed
points. Note that our results generalize the corresponding results by Ceng et al. (Comput. Math. Appl. 61 (2011), 2447-2455).
2000 Mathematics Subject Classification. 47H10, 47H05, 47H17.
Key words. Variational inequality, common fixed point, nonexpansive mapping.
1. Introduction
Throughout this paper, let N be the set of positive integers and let R be the set
of real numbers.
Let C be a nonempty closed convex subset of a real Hilbert space H. A
mapping T : C → H is nonexpansive if kT x − T yk ≤ kx − yk for all x, y ∈ C.
Let Fix(T ) := {x ∈ C : T x = x} denote the fixed point set of T . It is well-known
This research was supported by the National Science Council of Republic of China.
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L.-J. Lin, C.-S. Chuang
that Fix(T ) is a closed convex subset of C if T is a nonexpansive mapping with
Fix(T ) 6= ∅. An operator B ⊆ H ×H is said to be accretive if hx1 −x2 , y1 −y2 i ≥ 0
for each (x1 , y1 ) ∈ B and (x2 , y2 ) ∈ B. A mapping A : H → H is a strongly
positive operator with coefficient γ̄ > 0 if hAx, xi ≥ γ̄kxk2 for all x ∈ H. A
nonlinear mapping S whose domain D(S) ⊆ H and range R(S) ⊆ H is said to
be strongly monotone with coefficient η > 0 if hx − y, Sx − Syi ≥ ηkx − yk2 for
all x, y ∈ D(S).
Let f : C → C be a contraction mapping, and let T : C → C be a nonexpansive mapping, and let A : H → H be a strongly positive operator.
In 2004, Xu [13] proved that under certain appropriate conditions on {αn },
the sequence {xn } generated by
xn+1 := αn f (xn ) + (1 − αn )T xn ,
n ∈ N,
converges strongly to the unique solution x̄ ∈ Fix(T ) of the variational inequality
h(I − f )x̄, x − x̄i ≥ 0
for all x ∈ Fix(T ).
In 2006, Marino and Xu [7] considered the following general iterative process:
xn+1 := αn γf (xn ) + (I − αn A)T xn ,
n ∈ N.
(1)
They proved that if the sequence {αn } of parameters satisfies appropriate conditions, then the sequence {xn } generated by (1) converges strongly to a point
x̄ ∈ Fix(T ), and it is the unique solution of the following variational inequality
h(γf − A)x̄, x − x̄i ≤ 0 for all x ∈ C.
In 2010, Tian [9] introduced the following general iterative process:
xn+1 := αn γf (xn ) + (I − µαn F )T xn ,
n ∈ N.
(2)
The author proved that if {αn } satisfies appropriate conditions, then the sequence {xn } generated by (2) converges strongly to a point x̄ ∈ Fix(T ), and it
is the unique solution of the variational inequality
h(γf − µF )x̄, x − x̄i ≤ 0 for all x ∈ Fix(T ).
This scheme improves and extends the corresponding ones given by Marino and
Xu [7], and Yamada [14].
In 2011, Ceng et al. [5] consider the following general composite iterative
process:


 x1 ∈ C chosen arbitrary,
yn := (I − αn µF )T xn + αn γf (xn ),
(3)


xn+1 := (I − βn A)T xn + βn yn ,
where A is a strongly positive bounded linear operator on H with coefficient
An iteration method for finding the unique solution of a VIP
321
γ̄ ∈ (1, 2), and {αn } ⊆ [0, 1] and {βn } ⊆ (0, 1] satisfy appropriate conditions.
They proved that the sequence {xn } generated by (3) converges strongly to a
point x̄ ∈ Fix(T ), and it is the unique solution of the variational inequality
h(I − A)x̄, x − x̄i ≤ 0
for all x ∈ Fix(T ).
Recall that the metric projection from a real Hilbert space H onto a nonempty
closed convex subset C of H is the mapping PC : H → C which assigns to each
point x ∈ H the unique point PC x satisfying the property
kx − PC xk = inf kx − yk.
y∈C
Motivated by the above works, we consider the following iterative process. Let
C be a nonempty closed convex subset of a real Hilbert space H. For each n ∈ N,
let Tn : C → C be a nonexpansive mapping. Let f : H → H be a contraction
mapping with coefficient α ∈ (0, 1), F : H → H be Lipschitz with coefficient
k > 0 and a strongly monotone mapping with coefficient η > 0, A : H → H be a
strongly positive bounded linear operator with coefficient γ̄ ∈ (1, 2). Let PC be
the metric projection. Let {αn } ⊆ [0, 1] and {βn } ⊆ (0, 1]. Let 0 < µ < 2η/k 2 ,
2
0 < γ < µ(η − µk2 )/α = τ /α. Let {xn } be generated by


 x1 ∈ C chosen arbitrary,
yn := PC [(I − αn µF )Tn xn + αn γf (xn )],


xn+1 := PC [(I − βn A)Tn xn + βn yn ].
(4)
We prove that if {αn } and {βn } satisfy suitable conditions, then the sequence
{xn } generated by (4) converges strongly to a point x̄ ∈ ∩∞
n=1 Fix(Tn ), and it is
the unique solution of the variational inequality
h(A − I)x̄, x̄ − xi ≤ 0
for all x ∈ ∩∞
n=1 Fix(Tn ).
Note that our results are different from the corresponding ones in [4, 5, 7, 9, 10,
14] since these iteration processes are different.
Finally, we consider the problem of finding a zero of an accretive operator. Let
C be a nonempty closed convex subset of a real Hilbert space H. Let f : H → H
be a contraction mapping with coefficient α ∈ (0, 1), F : H → H be a Lipschitz
mapping with coefficient k > 0 and strongly monotone with coefficient η > 0,
A : H → H be a strongly positive bounded linear self-adjoint operator with
coefficient γ̄ ∈ (1, 2). Let B be an accretive operator such that B −1 (0) 6= ∅.
Let {αn } ⊆ [0, 1], {βn } ⊆ (0, 1], and let {λn } be a sequence in (0, ∞). Let
2
0 < µ < 2η/k 2 , 0 < γ < µ(η − µk2 )/α = τ /α. Let {xn } be generated by


 x1 ∈ C chosen arbitrary,
yn := PC [(I − αn µF )Jλn xn + αn γf (xn )],


xn+1 := PC [(I − βn A)Jλn xn + βn yn ],
(5)
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L.-J. Lin, C.-S. Chuang
where Jλ is the resolvent of B. (For details, see Section 4.) We prove that if {αn },
{βn } and {λn } satisfy suitable conditions, then the sequence {xn } generated by
(5) converges strongly to a point x̄ ∈ B −1 (0), and it is the unique solution of
the variational inequality
h(A − I)x̄, x̄ − zi ≤ 0
for all z ∈ B −1 (0).
Note that our result is different from Theorems 4.2 and 4.3 in [2] since these
iteration processes are different.
2. Preliminaries
Let H be a (real) Hilbert space with inner product h·, ·i and norm k · k. We
denote the strongly convergence and the weak convergence of {xn } to x ∈ H by
xn → x and xn ⇀ x, respectively. It is easy to see that for each x, y ∈ H and
λ ∈ [0, 1], we have
kλx + (1 − λ)yk2 = λkxk2 + (1 − λ)kyk2 − λ(1 − λ)kx − yk2 .
Furthermore, we have
2hx − y, u − vi = kx − vk2 + ky − uk2 − kx − uk2 − ky − vk2 .
Lemma 2.1. [11] Assume that {an }n∈N is a sequence of nonnegative real numbers such that an+1 ≤ (1 − γn )an + δn , n ∈ N, where {γn } ⊆ (0, 1) and {δn } is
∞
∞
P
P
a sequence in R such that (i)
γn = ∞; (ii) lim sup γδnn ≤ 0 or
|δn | < ∞.
n=1
Then lim an = 0.
n→∞
n=1
n→∞
Lemma 2.2. [3] Let C be a nonempty closed convex subset of a real Hilbert space
H. Let T be a nonexpansive mapping of C into itself, and let {xn } be a sequence
in C. If xn ⇀ w and lim kxn − T xn k = 0, then T w = w.
n→∞
Lemma 2.3. [8] Let C be a nonempty closed convex subset of a Hilbert space
H. Let PC be the metric projection from H onto C. Then we have
(i) y = PC x if and only if hx − y, y − zi ≥ 0 for all z ∈ C;
(ii) kx − PC xk2 + kPC x − zk2 ≤ kx − zk2 for all z ∈ C;
(iii) kPC x − PC yk2 ≤ hx − y, PC x − PC yi for all x, y ∈ H. Consequently, PC
is a nonexpansive and monotone mapping.
In 2007, Aoyama, Kimura, Takahashi, and Toyoda [2] gave the following definition and lemma.
Definition 2.4. [2] Let C be a nonempty subset of a real Hilbert space H. Let
{Tn } be a countable family of mappings from C into itself. We say that a family
An iteration method for finding the unique solution of a VIP
323
{Tn } satisfies AKTT-condition if
∞
X
sup kTn+1 x − Tn xk < ∞
n=1 x∈B
for each nonempty bounded subset B of C.
Lemma 2.5. [2] Let C be a nonempty closed subset of a real Hilbert space H,
and let {Tn } be a sequence of mappings from C into itself. Suppose that {Tn }
satisfies AKTT-condition. Then, for each x ∈ C, {Tn x} converges strongly to a
point in C. Furthermore, let T : C → C be defined by
T x := lim Tn x, x ∈ C.
n→∞
Then, for each bounded subset B of C,
lim sup{kT z − Tn zk : z ∈ B} = 0.
n→∞
In the sequel, we say that {Tn , T } satisfies AKTT-condition if T is defined as
above and {Tn } satisfies AKTT-condition.
3. Main results
Theorem 3.1. Let C be a nonempty closed convex subset of a real Hilbert space
H. For each n ∈ N, let Tn : C → C be a nonexpansive mapping. Let f : H → H
be a contraction mapping with coefficient α ∈ (0, 1), F : H → H be a Lipschitz
mapping with coefficient k > 0 and strongly monotone with coefficient η > 0,
A : H → H be a strongly positive bounded linear self-adjoint operator with
coefficient γ̄ ∈ (1, 2). Let {αn } ⊆ [0, 1] and {βn } ⊆ (0, 1]. Let 0 < µ < 2η/k 2 ,
2
0 < γ < µ(η − µk2 )/α = τ /α. Suppose that ∩∞
n=1 Fix(Tn ) 6= ∅. Let {xn } be
generated by


 x1 ∈ C chosen arbitrary,
yn := PC [(I − αn µF )Tn xn + αn γf (xn )],


xn+1 := PC [(I − βn A)Tn xn + βn yn ].
Assume that:
(i) lim αn = 0; lim βn = 0;
n→∞
(ii)
∞
P
n→∞
|αn+1 − αn | < ∞;
n=1
∞
P
βn = ∞;
n=1
∞
P
|βn+1 − βn | < ∞;
n=1
(iii) {Tn , T } satisfies AKTT-condition, and Fix(T ) = ∩∞
n=1 Fix(Tn ).
Then lim xn = x̄, where x̄ ∈ ∩∞
n=1 Fix(Tn ) and h(A−I)x̄, x̄−zi ≤ 0 for all z ∈
n→∞
∩∞
(2I − A)x̄ = x̄.
n=1 Fix(Tn ), that is, P∩∞
n=1 Fix(Tn )
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L.-J. Lin, C.-S. Chuang
Proof. Since kAk ≥ γ̄ > 1, without loss of generality, we may assume that
0 < βn ≤ kAk−1 for all n ∈ N. Since lim αn = 0, we may assume that 1 >
n→∞
αn (τ − γα) and 1 > 2αn τ for each n ∈ N. For each x, y ∈ X and n ∈ N, we have
k(I − αn µF )x − (I − αn µF )yk2
= k(x − y) − (αn µF x − αn µF y)k2
= kx − yk2 − 2hx − y, αn µF x − αn µF yi + kαn µF x − αn µF yk2
≤ kx − yk2 − 2αn µηkx − yk2 + α2n µ2 k 2 kx − yk2
= (1 − 2αn µη + α2n µ2 k 2 )kx − yk2
≤ (1 − αn µ(2η − µk 2 ))kx − yk2
= (1 − 2αn τ )kx − yk2
≤ (1 − αn τ )kx − yk2 .
(6)
Take any w ∈ ∩∞
n=1 Fix(Tn ) and let w be fixed. Hence
kyn − wk
≤ k(I − αn µF )Tn xn + αn γf (xn ) − wk
= kαn (γf (xn ) − µF (w)) + (I − αn µF )Tn xn − (I − αn µF )Tn wk
≤ kαn γf (xn ) − αn γf (w)k + kαn γf (w) − αn µF (w))k
+ k(I − αn µF )Tn xn − (I − αn µF )Tn wk
≤ αn γαkxn − wk + αn kγf (w) − µF (w))k + (1 − αn τ )kxn − wk
= (1 − αn (τ − γα))kxn − wk + αn kγf (w) − µF (w))k
≤ kxn − wk + αn kγf (w) − µF (w))k,
and
kxn+1 − wk
≤ k(I − βn A)Tn xn + βn yn − wk
= k(I − βn A)Tn xn − (I − βn A)Tn w + βn (yn − w) + βn (I − A)wk
≤ k(I − βn A)Tn xn − (I − βn A)Tn wk + βn kyn − wk + βn kI − Ak · kwk
≤ (1 − βn γ̄)kTn xn − Tn wk + βn kI − Ak · kwk
+ βn [kxn − wk + αn kγf (w) − µF (w))k]
≤ (1 − βn (γ̄ − 1))kxn − wk + βn kI − Ak · kwk + βn kγf (w) − µF (w))k
= (1 − βn (γ̄ − 1))kxn − wk + βn (γ̄ − 1)
kI − Ak · kwk + kγf (w) − µF (w))k
.
γ̄ − 1
By induction, we get
kI − Ak · kwk + kγf (w) − µF (w))k
, n ∈ N.
kxn − wk ≤ max kx1 − wk,
γ̄ − 1
An iteration method for finding the unique solution of a VIP
325
This implies that {xn } is a bounded sequence. So, {Tn xn }, {f (xn )}, {F Tn xn },
{ATn xn }, and {yn } are bounded sequences. Let
M:= sup {kxn k, kf (xn )k, kγf (xn )k, kTn xn k, kynk, kµF Tn xn k, kATnxn k : n ∈ N}.
Besides, we also have
lim kyn − Tn xn k lim kαn γf (xn ) − αn µF Tn xn k = 0,
n→∞
n→∞
and
lim kxn+1 − Tn xn k lim kβn ATn xn − βn yn k = 0.
n→∞
n→∞
(7)
By (7), lim ||xn+1 − yn || = 0. Furthermore, by (6), we have
n→∞
kyn − yn−1 k
≤ k(I − αn µF )Tn xn + αn γf (xn ) − (I − αn−1 µF )Tn−1 xn−1 − αn−1 γf (xn−1 )k
≤ kαn γf (xn ) − αn γf (xn−1 )k + kαn γf (xn−1 ) − αn−1 γf (xn−1 )k
+ k(I − αn µF )Tn xn − (I − αn µF )Tn−1 xn−1 k
+ k(I − αn µF )Tn−1 xn−1 − (I − αn−1 µF )Tn−1 xn−1 k
≤ αn γαkxn − xn−1 k + |αn − αn−1 | · kγf (xn−1 )k
+ (1 − αn τ )kTn xn − Tn−1 xn−1 k + |αn − αn−1 | · µkF Tn−1 xn−1 k
≤ αn γαkxn − xn−1 k + |αn − αn−1 | · kγf (xn−1 )k + (1 − αn τ )kTn xn − Tn xn−1 k
+ kTn xn−1 − Tn−1 xn−1 k + |αn − αn−1 | · µkF Tn−1 xn−1 k.
≤ (1 − αn (τ − γα)kxn − xn−1 k + 2M |αn − αn−1 | + kTn xn−1 − Tn−1 xn−1 k.
Next, we have
kxn+1 − xn k
≤ k(I − βn A)Tn xn + βn yn − (I − βn−1 A)Tn−1 xn−1 − βn−1 yn−1 k
≤ k(I − βn A)Tn xn − (I − βn A)Tn−1 xn−1 k + k(I − βn A)Tn−1 xn−1
− (I − βn−1 A)Tn−1 xn−1 k + βn kyn − yn−1 k + |βn − βn−1 | · kyn−1 k
≤ (1 − βn γ̄)kTn xn − Tn−1 xn−1 k + |βn − βn−1 | · kATn−1 xn−1 k
+ βn kyn − yn−1 k + |βn − βn−1 | · kyn−1 k
≤ (1 − βn γ̄)kxn − xn−1 k + kTn xn−1 − Tn−1 xn−1 k + 2M |βn − βn−1 |
+βn [(1− αn (τ − γα)kxn − xn−1 k+ 2M |αn − αn−1 |+ kTn xn−1 −Tn−1 xn−1 k]
≤ (1 − βn (γ̄ − 1))kxn − xn−1 k + kTn xn−1 − Tn−1 xn−1 k + 2M |βn − βn−1 |
+ 2M βn |αn − αn−1 | + βn kTn xn−1 − Tn−1 xn−1 k
≤ (1 − βn (γ̄ − 1))kxn − xn−1 k + 2kTn xn−1 − Tn−1 xn−1 k
+ 2M βn |αn − αn−1 | + 2M |βn − βn−1 |
≤ (1 − βn (γ̄ − 1))kxn − xn−1 k + 2 sup{kTn x − Tn−1 xk : x ∈ {xn }}
+ 2M |αn − αn−1 | + 2M |βn − βn−1 |.
(8)
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L.-J. Lin, C.-S. Chuang
By (i), (ii), (iii), (8), and Lemma 2.1, we know that lim kxn+1 − xn k = 0. This
n→∞
implies that
lim kTn xn − xn k = 0.
(9)
n→∞
Furthermore,
kxn − T xn k ≤ kxn − Tn xn k + kTn xn − T xn k
≤ kxn − Tn xn k + sup{kTn z − T zk : z ∈ {xn }}
∞
X
≤ kxn − Tn xn k +
sup{kTn z − Tn+1 zk : z ∈ {xn }}.
(10)
k=n
By (iii), (9), and (10), we get lim kxn − T xn k = 0.
n→∞
Clearly, T is a nonexpansive mapping. For this nonexpansive mapping T , by
Theorem 3.1 in [5], there exists x̄ ∈ Fix(T ) such that x̄ is the unique solution of
the problem:
x∗ ∈ Fix(T ) : h(A − I)x∗ , x∗ − zi ≤ 0 ∀z ∈ Fix(T ).
(11)
Since {xn } is a bounded sequence, there exists a subsequence {xnk } of {xn } such
that
lim suphxn − x̄, (I − A)x̄i = lim hxnk − x̄, (I − A)x̄i.
(12)
n→∞
k→∞
Without loss of generality, we may assume that xnk ⇀ z. By Lemma 2.2, z ∈
Fix(T ). Hence, by (11) and (12), we get
lim suphxn − x̄, (I − A)x̄i = hz − x̄, (I − A)x̄i ≤ 0.
(13)
n→∞
Let un := (I − αn µF )Tn xn + αn γf (xn ). By Lemma 2.3,
kyn − x̄k2
= kPC (un ) − x̄k2
= hPC (un ) − un , yn − x̄i + hun − x̄, yn − x̄i
≤ hun − x̄, yn − x̄i
≤ h(I − αn µF )Tn xn + αn γf (xn ) − x̄, yn − x̄i
= h(I − αn µF )Tn xn + αn γf (xn ) − x̄, yn − x̄i
= h(I − αn µF )Tn xn − (I − αn µF )Tn x̄ + αn (γf (xn ) − µF x̄), yn − x̄i
≤ k(I − αn µF )Tn xn − (I − αn µF )Tn x̄k · kyn − x̄k + αn hγf (xn ) − µF x̄, yn − x̄i
≤ (1 − αn τ )kxn − x̄k · kyn − x̄k + αn kγf (xn ) − µF x̄k · kyn − x̄k
≤ kxn − x̄k · kyn − x̄k + αn kγf (xn ) − µF x̄k · kyn − x̄k.
Without loss of generality, we may assume that kyn − x̄k 6= 0 for each n ∈ N. So,
kyn − x̄k ≤ kxn − x̄k + αn kγf (xn ) − µF x̄k.
(14)
An iteration method for finding the unique solution of a VIP
327
Let wn := (I − βn A)Tn xn + βn yn . By (14) and Lemma 2.3 again,
kxn+1 − x̄k2
≤ hwn − x̄, xn+1 − x̄i
= h(I − βn A)Tn xn + βn yn − x̄, xn+1 − x̄i
= h(I − βn A)Tn xn − (I − βn A)Tn x̄, xn+1 − x̄i + βn hyn − x̄, xn+1 − x̄i
+ βn h(I − A)x̄, xn+1 − x̄i
≤ k(I − βn A)(Tn xn − Tn x̄)k · kxn+1 − x̄k + βn hyn − x̄, xn+1 − x̄i
+ βn h(I − A)x̄, xn+1 − x̄i
≤ (1 − βn γ̄)kxn − x̄k · kxn+1 − x̄k + βn kyn − x̄k · kxn+1 − x̄k
+ βn h(I − A)x̄, xn+1 − x̄i
≤ (1 − βn γ̄)kxn − x̄k · kxn+1 − x̄k + βn kxn − x̄k · kxn+1 − x̄k
+ αn βn kγf (xn ) − µF x̄k · kxn+1 − x̄k + βn h(I − A)x̄, xn+1 − x̄i
≤ (1 − βn (γ̄ − 1))kxn − x̄k · kxn+1 − x̄k + αn βn kγf (xn ) − µF x̄k · kxn+1 − x̄k
+ βn h(I − A)x̄, xn+1 − x̄i
1 − βn (γ̄ − 1) kxn − x̄k2 + kxn+1 − x̄k2
≤
2
+ αn βn kγf (xn ) − µF x̄k · kxn+1 − x̄k + βn h(I − A)x̄, xn+1 − x̄i.
Therefore,
kxn+1 − x̄k2
1 − βn (γ̄ − 1)
βn
kxn − x̄k2 +
αn kγf (xn ) − µF x̄k · kxn+1 − x̄k
1 + βn (γ̄ − 1)
1 + βn (γ̄ − 1)
+ h(I − A)x̄, xn+1 − x̄i
2βn (γ̄ − 1)
kxn − x̄k2
= 1−
1 + βn (γ̄ − 1)
2βn (γ̄ − 1)
1
+
αn kγf (xn ) − µF x̄k · kyn − x̄k
·
1 + βn (γ̄ − 1) 2(γ̄ − 1)
+ 2h(I − A)x̄, xn+1 − x̄i .
≤
By the integral test for series, we know that
∞
X
2βn (γ̄ − 1)
= ∞.
1
+ βn (γ̄ − 1)
n=1
Hence, by (i), (13), and Lemma 2.1, we know that lim xn = x̄. Therefore, the
n→∞
proof is complete.
The following result is a special case of Theorem 3.1. Note that our results are
different from the corresponding ones given by Marino and Xu [7], Yamada [14]
328
L.-J. Lin, C.-S. Chuang
and Tian [9] since these iteration processes are different and A is not an identity
mapping.
Corollary 3.2. [5] Let C be a nonempty closed convex subset of a real Hilbert
space H. Let T : C → C be a nonexpansive mapping, f : H → H be a contraction
mapping with coefficient α ∈ (0, 1), F : H → H be a Lipschitz mapping with
coefficient k > 0 and strongly monotone with coefficient η > 0, A : H → H
be a strongly positive bounded linear operator with coefficient γ̄ ∈ (1, 2). Let
2
{αn } ⊆ [0, 1] and {βn } ⊆ (0, 1]. Let 0 < µ < 2η/k 2 , 0 < γ < µ(η− µk2 )/α = τ /α.
Suppose that Fix(T ) 6= ∅. Let {xn } be generated by


 x1 ∈ C chosen arbitrary,
yn := PC [(I − αn µF )T xn + αn γf (xn )],


xn+1 := PC [(I − βn A)T xn + βn yn ].
Assume that:
(i) lim αn = 0; lim βn = 0;
n→∞
(ii)
∞
P
n→∞
|αn+1 − αn | < ∞;
n=1
∞
P
βn = ∞;
n=1
∞
P
|βn+1 − βn | < ∞.
n=1
Then lim xn = x̄, where x̄ ∈ Fix(T ) and h(A−I)x̄, x̄−zi ≤ 0 for all z ∈ Fix(T ),
n→∞
that is, PFix(T ) (2I − A)x̄ = x̄.
4. Applications
Let H be a real Hilbert space, and let C be a nonempty closed convex subset of
H. An accretive operator B is said to satisfy the range condition if cl(D(B)) ⊆
R(I + λB) for all λ > 0, where D(B) is the domain of B, I is the identity
mapping on H, R(I + λB) is the range of I + λB, and cl(D(B)) is the closure
of D(B).
If B is an accretive operator which satisfies the range condition, then we can
define, for each λ > 0, a mapping Jλ : R(I + λB) → D(B) by Jλ := (I + λB)−1 ,
which is called the resolvent of B. We know that Fix(Jλ ) = B −1 (0) for all λ > 0,
and
kJλ x − Jλ yk2 ≤ kx − yk2 − k(I − Jλ )x − (I − Jλ )yk2
for all x, y ∈ R(I + λB). Hence, Jλ is a nonexpansive mapping. Furthermore, we
know that [6]: for each λ1 , λ2 > 0 and x ∈ R(I + λ1 B) ∩ R(I + λ2 B), we have
kJλ1 x − Jλ2 xk ≤
|λ1 − λ2 |
kx − Jλ1 xk.
λ1
Theorem 4.1. Let C be a nonempty closed convex subset of a real Hilbert space
H. Let f : H → H be a contraction mapping with coefficient α ∈ (0, 1), F :
An iteration method for finding the unique solution of a VIP
329
H → H be a Lipschitz mapping with coefficient k > 0 and strongly monotone
with coefficient η > 0, A : H → H be a strongly positive bounded linear selfadjoint operator with coefficient γ̄ ∈ (1, 2). Let B be an accretive operator such
that B −1 (0) 6= ∅ and cl(D(B)) ⊆ C ⊆ ∩λ>0 R(I + λB). Let {αn } ⊆ [0, 1],
{βn } ⊆ (0, 1], and let {λn } be a sequence in (0, ∞). Let 0 < µ < 2η/k 2 , 0 < γ <
2
µ(η − µk2 )/α = τ /α. Let {xn } be generated by
Assume that:


 x1 ∈ C chosen arbitrary,
yn := PC [(I − αn µF )Jλn xn + αn γf (xn )],


xn+1 := PC [(I − βn A)Jλn xn + βn yn ].
(i) lim αn = 0; lim βn = 0;
n→∞
(ii)
∞
P
n→∞
|αn+1 − αn | < ∞;
n=1
∞
P
βn = ∞;
n=1
∞
P
|βn+1 − βn | < ∞.
n=1
Then lim xn = x̄, where x̄ ∈ B −1 (0) and h(A − I)x̄, x̄ − zi ≤ 0 for all z ∈
n→∞
B −1 (0), that is, PB −1 (0) x̄ = x̄.
Proof. For each n ∈ N, let Tn : C → C be defined by Tn x := Jλn x for each x ∈ C.
By following the same argument in the proof of Theorem 4.3 [2], we know that
the condition (iii) of Theorem 3.1 holds. By Theorem 3.1, we get the conclusion
of Theorem 4.1.
Acknowledgments. The authors wish to express their gratitude to the referees
for their valuable suggestions during the preparation of this paper. This research
was supported by the National Science Council of Republic of China.
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