Available online at www.tjnsa.com
J. Nonlinear Sci. Appl. 7 (2014), 325–344
Research Article
Visco-resolvent algorithms for monotone operators
and nonexpansive mappings
Peize Lia , Shin Min Kangb,∗, Li-Jun Zhuc
a
Department of Mathematics, Tianjin Polytechnic University, Tianjin 300387, China.
b
Department of Mathematics and the RINS, Gyeongsang National University, Jinju 660-701, Korea.
c
School of Mathematics and Information Science, Beifang University of Nationalities, Yinchuan 750021, China.
Communicated by Y. J. Cho
Special Issue In Honor of Professor Ravi P. Agarwal
Abstract
Two new type of visco-resolvent algorithms for finding a zero of the sum of two monotone operators and a
fixed point of a nonexpansive mapping in a Hilbert space are investigated. The algorithms consist of the
zeros and the fixed points of the considered problems in which one operator is replaced with its resolvent
and a viscosity term is added. Strong convergence of the algorithms are shown. As special cases, we can
approach to the minimum norm common element of the zero of the sum of two monotone operators and the
fixed point of a nonexpansive mapping without using the metric projection. Some applications are included.
c
2014
All rights reserved.
Keywords: Monotone operator, nonexpansive mapping, zero point, fixed point, resolvent.
2010 MSC: 49J40, 47J20, 47H09, 65J15.
1. Introduction
Let H be a real Hilbert space. Let A : H → H be a single-valued nonlinear mapping and B : H → 2H
be a set-valued mapping. Now we concern the following variational inclusion, which is to find a zero x ∈ H
of the sum of two monotone operators A and B such that
0 ∈ A(x) + B(x),
∗
(1.1)
Corresponding author
Email addresses: tjlipeize@163.com (Peize Li), smkang@gnu.ac.kr (Shin Min Kang), zhulijun1995@sohu.com (Li-Jun
Zhu)
Received 2014-09-10
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
326
where 0 is the zero vector in H. The set of solutions of problem (1.1) is denoted by (A + B)−1 (0). If
H = Rm , then problem (1.1) becomes the generalized equation introduced by Robinson [22]. If A = 0, then
problem (1.1) becomes the inclusion problem introduced by Rockafellar [24]. It is known that (1.1) provides a convenient framework for the unified study of optimal solutions in many optimization related areas
including mathematical programming, complementarity, variational inequalities, optimal control, mathematical economics, equilibria, game theory, etc. Also various types of variational inclusions problems have
been extended and generalized. Recently, Zhang et al. [32] introduced a new iterative scheme for finding
a common element of the set of solutions to the problem (1.1) and the set of fixed points of nonexpansive mappings in Hilbert spaces. Peng et al. [21] introduced another iterative scheme by the viscosity
approximate method for finding a common element of the set of solutions of a variational inclusion with
set-valued maximal monotone mapping and inverse strongly monotone mappings, the set of solutions of an
equilibrium problem and the set of fixed points of a nonexpansive mapping. Some related works, please see
[3, 4, 7, 9, 10, 12, 13, 15, 16, 19, 20] and the references therein.
Recently, Takahashi et al. [27] introduced the following iterative algorithm for finding a zero of the sum
of two monotone operators and a fixed point of a nonexpansive mapping
(1.2)
xn+1 = βn xn + (1 − βn )S αn x + (1 − αn )JλBn (xn − λn Axn )
for all n ≥ 0. Under some assumptions, they proved that the sequence {xn } converges strongly to a point
of F (S) ∩ (A + B)−1 0.
Remark 1.1. We note that in their result, the authors added an additional assumption: the domain of B is
included in C (The reader can refer to Lemma 4.3 in the last section for a possible example which satisfies
this assumption). This assumption is indeed not restrict in order to guarantee JλBn (xn − λn Axn ) ∈ C.
Remark 1.2. From the listed references, there exist a large number of problems which need to find the
minimum norm solution, see, e.g., [11, 17, 25, 30, 31]. A useful path to circumvent this problem is to use
projection. Bauschke and Browein [2] and Censor and Zenios [6] provide reviews of the field. The main
difficult is in computation. Hence, it is an interesting problem of finding the minimum norm element without
using the projection. We note that the algorithm (1.1) can not use to find the minimum norm element.
Motivated and inspired by the works in this field, we first suggest the following two algorithms without
using projection:
xt = JλB tγf (xt ) + (1 − t)Sxt − λASxt , t ∈ (0, 1)
and
xn+1 = βn xn + (1 − βn )JλBn αn γf (xn ) + (1 − αn )Sxn − λn ASxn ,
n ≥ 0.
(Notice that these two algorithms are indeed well-defined (see the next section).) We will show the suggested
algorithms converge strongly to to a point x̃ = PF (S)∩(A+B)−1 0 (γf (x̃)) which solves the following variational
inequality
hγf (x̃) − x̃, x̃ − zi ≥ 0, ∀z ∈ F (S) ∩ (A + B)−1 0.
As special cases, we can approach to the minimum norm element in F (S) ∩ (A + B)−1 0 without using the
metric projection. Some applications are also included.
2. Preliminaries
Let H be a real Hilbert space with inner product h·, ·i and norm k · k, respectively. Let C be a nonempty
closed convex subset of H. Recall that a mapping S : C → C is said to be nonexpansive if
kSx − Syk ≤ kx − yk
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
327
for all x, y ∈ C. We denote by F (S) the set of fixed points of S. A mapping A : C → H is said to be
α-inverse strongly-monotone iff
hAx − Ay, x − yi ≥ αkAx − Ayk2
for some α > 0 and for all x, y ∈ C. It is known that if A is α-inverse strongly-monotone, then kAx − Ayk ≤
1
α kx − yk for all x, y ∈ C.
Let B be a mapping of H into 2H . The effective domain of B is denoted by dom(B), that is, dom(B) =
{x ∈ H : Bx 6= ∅}. A multi-valued mapping B is said to be a monotone operator on H iff
hx − y, u − vi ≥ 0
for all x, y ∈ dom(B), u ∈ Bx, and v ∈ By. A monotone operator B on H is said to be maximal iff its graph
is not strictly contained in the graph of any other monotone operator on H. Let B be a maximal monotone
operator on H and let B −1 0 = {x ∈ H : 0 ∈ Bx}.
For a maximal monotone operator B on H and λ > 0, we may define a single-valued operator JλB =
(I + λB)−1 : H → dom(B), which is called the resolvent of B for λ. It is known that the resolvent JλB is
firmly nonexpansive, i.e.,
B
J x − J B y 2 ≤ J B x − J B y, x − y
λ
λ
λ
λ
for all x, y ∈ C and B −1 0 = F (JλB ) for all λ > 0.
The following resolvent identity is well-known: for λ > 0 and µ > 0, there holds the identity
µ B
B µ
B
Jλ x = Jµ
x+ 1−
J x , x ∈ H.
λ
λ λ
(2.1)
We use the following notation:
• xn * x stands for the weak convergence of (xn ) to x;
• xn → x stands for the strong convergence of (xn ) to x.
We need the following lemmas for the next section.
Lemma 2.1. ([28]) Let C be a nonempty closed convex subset of a real Hilbert space H. Let the mapping
A : C → H be α-inverse strongly monotone and λ > 0 be a constant. Then, we have
k(I − λA)x − (I − λA)yk2 ≤ kx − yk2 + λ(λ − 2α)kAx − Ayk2 , ∀x, y ∈ C.
In particular, if 0 ≤ λ ≤ 2α, then I − λA is nonexpansive.
Lemma 2.2. ([14]) Let C be a closed convex subset of a Hilbert space H. Let S : C → C be a nonexpansive
mapping. Then F (S) is a closed convex subset of C and the mapping I − S is demiclosed at 0, i.e. whenever
{xn } ⊂ C is such that xn * x and (I − S)xn → 0, then (I − S)x = 0.
Lemma 2.3. ([18]) Let C be a nonempty closed convex subset of a real Hilbert space H. Assume that the
mapping F : C → H is monotone and weakly continuous along segments, that is, F (x + ty) → F (x) weakly
as t → 0. Then the variational inequality
x∗ ∈ C,
hF x∗ , x − x∗ i ≥ 0,
∀x ∈ C.
is equivalent to the dual variational inequality
x∗ ∈ C,
hF x, x − x∗ i ≥ 0,
∀x ∈ C.
Lemma 2.4. ([26]) Let {xn } and {yn } be bounded sequences in a Banach space X and let {βn } be a sequence
in [0, 1] with 0 < lim inf n→∞ βn ≤ lim supn→∞ βn < 1. Suppose xn+1 = (1 − βn )yn + βn xn for all n ≥ 0 and
lim supn→∞ (kyn+1 − yn k − kxn+1 − xn k) ≤ 0. Then, limn→∞ kyn − xn k = 0.
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
328
Lemma 2.5. ([29]) Assume {an } is a sequence of nonnegative real numbers such that
an+1 ≤ (1 − γn )an + δn γn ,
where {γn } is a sequence in (0, 1) and {δn } is a sequence such that
P
(1) ∞
n=1 γn = ∞;
P
(2) lim supn→∞ δn ≤ 0 or ∞
n=1 |δn γn | < ∞.
Then limn→∞ an = 0.
3. Main results
In this section, we will prove our main results.
Theorem 3.1. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A be an
α-inverse strongly-monotone mapping of C into H. Let f : C → H be a ρ-contraction and γ be a constant
such that 0 < γ < ρ1 . Let B be a maximal monotone operator on H, such that the domain of B is included
in C. Let JλB = (I + λB)−1 be the resolvent of B for λ > 0 and let S be a nonexpansive mapping of C into
−1
itself, such that
F (S) ∩ (A + B) 0 6= ∅. Let λ be a constant satisfying a ≤ λ ≤ b where [a, b] ⊂ (0, 2α). For
λ
t ∈ 0, 1 − 2α , let {xt } ⊂ C be a net generated by
xt = JλB tγf (xt ) + (1 − t)Sxt − λASxt .
(3.1)
Then the net {xt } converges strongly, as t → 0+, to a point x̃ = PF (S)∩(A+B)−1 0 (γf (x̃)) which solves the
following variational inequality
∀z ∈ F (S) ∩ (A + B)−1 0.
λ
Proof. First, we show the net {xt } is well-defined. For any t ∈ 0, 1 − 2α
, we define a mapping W :=
λ
JλB tγf + (1 − t)S − λAS . Note that JλB , S and I − 1−t
A (see Lemma 2.1) are nonexpansive. For any
x, y ∈ C, we have
hγf (x̃) − x̃, x̃ − zi ≥ 0,
kW x − W yk
B
= tγf
(x)
+
(1
−
t)
I−
J
λ
λ
λ
A Sx − JλB tγf (y) + (1 − t) I −
A Sy 1−t
1−t
λ
λ
A Sx − I −
A Sy ≤ tγ(f (x) − f (y)) + (1 − t) I −
1−t
1−t
λ
λ
≤ tγkf (x) − f (y)k + (1 − t)(I −
A)Sx − I −
A Sy 1−t
1−t
≤ tγρkx − yk + (1 − t)kx − yk
= [1 − (1 − γρ)t]kx − yk,
which implies the mapping W is a contraction on C. We use xt to denote the unique fixed point of W in
C. Therefore, {xt } is well-defined.
Take any z ∈ F (S) ∩ (A + B)−1 0. It is obvious that z = Sz = JλB (z − λAz) for all λ > 0. So, we have
λ
B
B
z = Sz = Jλ (z − λAz) = Jλ tz + (1 − t) I −
A Sz
1−t
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
329
for all t ∈ (0, 1). Since JλB is nonexpansive for all λ > 0, we have
kxt − zk
B
λ
= Jλ tγf (xt ) + (1 − t) I −
A Sxt − z 1−t
B
λ
λ
B
= Jλ tγf (xt ) + (1 − t) Sxt −
− Jλ tz + (1 − t) Sz −
ASxt
ASz
1−t
1−t
λ
λ
≤ tγf (xt ) + (1 − t) Sxt −
− tz + (1 − t) Sz −
ASxt
ASz
1−t
1−t
λ
λ
+ t(γf (xt ) − z)
= (1 − t) Sxt − 1 − t ASxt − Sz − 1 − t ASz
λ
λ
≤ (1 − t) I −
A Sxt − I −
A Sz + tγkf (xt ) − f (z)k + tkγf (z) − zk
1−t
1−t
≤ (1 − t)kxt − zk + tγρkxt − zk + tkγf (z) − zk.
(3.2)
It follows that
kxt − zk ≤
1
kγf (z) − zk.
1 − γρ
Therefore, {xt } is bounded. We deduce immediately that {f (xt )}, {Axt } {Sxt } and {ASxt } are also
bounded.
By using the convexity of k · k and the α-inverse strong monotonicity of A, from (3.2), we derive
kxt − zk2
2
λ
λ
≤ (1 − t) Sxt −
ASxt − Sz −
ASz
+ t(γf (xt ) − z)
1−t
1−t
2
λ
λ
+ tkγf (xt ) − zk2
ASx
ASz
−
Sz
−
≤ (1 − t)
Sx
−
t
t
1−t
1−t
= (1 − t)
(Sxt − Sz) −
2
λ
2
(ASxt − ASz)
+ tkγf (xt ) − zk
1−t
2λ
λ2
2
2
= (1 − t) kSxt − Szk −
hASxt − ASz, Sxt − Szi +
kASxt − ASzk
1−t
(1 − t)2
+tkγf (xt ) − zk2
2αλ
λ2
2
2
2
≤ (1 − t) kSxt − Szk −
kASxt − ASzk +
kASxt − ASzk
1−t
(1 − t)2
+tkγf (xt ) − zk2
= (1 − t) kSxt − Szk2 +
≤ (1 − t)kxt − zk2 +
λ
(λ − 2(1 − t)α)kASxt − ASzk2
(1 − t)2
+ tkγf (xt ) − zk2
λ
(λ − 2(1 − t)α)kASxt − ASzk2 + tkγf (xt ) − zk2 .
1−t
(3.3)
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
330
So,
λ
(2(1 − t)α − λ)kASxt − Azk2 ≤ tkγf (xt ) − zk2 − tkxt − zk2 → 0.
1−t
By the assumption, we have 2(1 − t)α − λ > 0 for all t ∈ (0, 1 −
λ
2α ).
Then, we obtain
lim kASxt − Azk = 0.
(3.4)
t→0+
Next, we show kxt − Sxt k → 0. By using the firm nonexpansivity of JλB , we have
kxt − zk2
2
= JλB tγf (xt ) + (1 − t)Sxt − λASxt − z 2
= JλB tγf (xt ) + (1 − t)Sxt − λASxt − JλB z − λAz ≤
=
tγf (xt ) + (1 − t)Sxt − λASxt − (z − λAz), xt − z
1
ktγf (xt ) + (1 − t)Sxt − λASxt − (z − λAz)k2 + kxt − zk2
2
2 −tγf (xt ) + (1 − t)Sxt − λ(ASxt − λAz) − xt .
By the nonexpansivity of I −
λ
1−t A,
we have
ktγf (xt ) + (1 − t)Sxt − λASxt − (z − λAz)k2
= (1 − t) Sxt −
2
λ
λ
ASxt − Sz −
ASz
+ t(γf (xt ) − z)
1−t
1−t
2
λ
λ
2
ASxt − Sz −
ASz ≤ (1 − t) Sxt −
+ tkγf (xt ) − zk
1−t
1−t
≤ (1 − t)kxt − zk2 + tkγf (xt ) − zk2 .
Thus,
1
(1 − t)kxt − zk2 + tkγf (xt ) − zk2 + kxt − zk2
2
2 −tγf (xt ) + (1 − t)Sxt − λ(ASxt − Az) − xt .
kxt − zk2 ≤
That is,
kxt − zk2
≤ (1 − t)kxt − zk2 + tkγf (xt ) − zk2
2
−tγf (xt ) + (1 − t)Sxt − xt − λ(ASxt − Az)
= (1 − t)kxt − zk2 + tkγf (xt ) − zk2 − ktγf (xt ) + (1 − t)Sxt − xt k2
+2λ tγf (xt ) + (1 − t)Sxt − xt , ASxt − Az − λ2 kASxt − Azk2
≤ (1 − t)kxt − zk2 + tkγf (xt ) − zk2 − ktγf (xt ) + (1 − t)Sxt − xt k2
+2λktγf (xt ) + (1 − t)Sxt − xt kkASxt − Azk.
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
331
Hence,
ktγf (xt ) + (1 − t)Sxt − xt k2 ≤ tkγf (xt ) − zk2 + 2λktγf (xt ) + (1 − t)Sxt − xt kkASxt − Azk.
Since kASxt − Azk → 0, we deduce
lim ktγf (xt ) + (1 − t)Sxt − xt k = 0.
t→0+
Which implies that
lim kxt − Sxt k = 0.
(3.5)
t→0+
From (3.2), we have
kxt − zk2
2
λ
λ
≤ (1 − t) Sxt −
ASxt − z −
Az
+ t(γf (xt ) − z)
1−t
1−t
2
λ
λ
ASxt − z −
Az 1−t
1−t
λ
λ
ASxt − z −
Az
+ t2 kγf (xt ) − zk2
+ 2t(1 − t) γf (xt ) − z, Sxt −
1−t
1−t
λ
(ASxt − ASz) − z
≤ (1 − t)2 kxt − zk2 + 2t(1 − t) γf (xt ) − z, Sxt −
1−t
= (1 − t) Sxt −
2
+t2 kγf (xt ) − zk2
λ
= (1 − t) kxt − zk + 2t(1 − t)γ f (xt ) − f (z), Sxt −
(ASxt − ASz) − z
1−t
λ
+2t(1 − t) γf (z) − z, Sxt −
(ASxt − ASz) − z + t2 kγf (xt ) − zk2
1−t
λ
2
2
≤ (1 − t) kxt − zk + 2t(1 − t)γkf (xt ) − f (z)k kSxt − zk + (ASxt − ASz)
1−t
λ
(ASxt − ASz) − z + t2 kγf (xt ) − zk2
+2t(1 − t) γf (z) − z, Sxt −
1−t
2
2
≤ (1 − t)2 kxt − zk2 + 2t(1 − t)γρkxt − zk2 + 2tλγρkxt − zkkASxt − ASzk
λ
+2t(1 − t) γf (z) − z, Sxt −
(ASxt − ASz) − z + t2 kγf (xt ) − zk2
1−t
λ
2
≤ [1 − 2(1 − γρ)t]kxt − zk + 2t (1 − t) γf (z) − z, Sxt −
(ASxt − Az) − z
1−t
t
2
2
+ (kγf (xt ) − zk + kxt − zk ) + λγρkxt − zkkASxt − ASzk .
2
It follows that
kxt − zk2
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
332
γf (z) − z, Sxt −
λ
≤
(ASxt − Az) − z
1−t
t
λ
2
2
(ASxt − Az) − z + (kγf (xt ) − zk + kxt − zk ) + tkγf (z) − zk
Sxt −
2
1−t
+λγρkxt − zkkASxt − ASzk
1
1 − γρ
1
γf (z) − z, Sxt − z + (t + kASxt − Azk)M,
1 − γρ
≤
(3.6)
where M is some constant such that
1
1
λ
2
2
sup
(kγf (xt ) − zk + kxt − zk ), kγf (z) − zkSxt −
(ASxt − Az) − z ,
1 − γρ 2
1−t
λ
λγρkxt − zk, t ∈ 0, 1 −
≤ M.
2α
Next we show that {xt } is relatively norm-compact as t → 0+. Assume {tn } ⊂ (0, 1 −
tn → 0+ as n → ∞. Put xn := xtn . From (3.6), we have
1 kxn − zk2 ≤
γf (z) − z, Sxn − z + (tn + kASxn − Azk)M.
1 − γρ
λ
2α )
is such that
(3.7)
Since {xn } is bounded, without loss of generality, we may assume that xnj * x̃ ∈ C. Hence, xnj −
λ
1−tn (ASxnj − Az) * x̃ because of kASxn − Azk → 0 by (3.4). From (3.5), we have
j
lim kxn − Sxn k = 0.
(3.8)
n→∞
We can use Lemma 2.2 to (3.8) to deduce x̃ ∈ F (S). Further, we show that x̃ is also in (A + B)−1 0. Let
v ∈ Bu. Note that xn = JλB tn γf (xn ) + (1 − tn )Sxn − λASxn for all n. Then, we have
tn γf (xn ) + (1 − tn )Sxn − λASxn ∈ (I + λB)xn
tn γf (xn ) 1 − tn
xn
+
Sxn − ASxn −
∈ Bxn .
λ
λ
λ
Since B is monotone, we have, for (u, v) ∈ B,
tn γf (xn ) 1 − tn
xn
+
Sxn − ASxn −
− v, xn − u ≥ 0
λ
λ
λ
=⇒
=⇒
htn γf (xn ) + (1 − tn )Sxn − λASxn − xn − λv, xn − ui ≥ 0
=⇒
hASxn + v, xn − ui ≤
=⇒
hAS x̃ + v, xn − ui ≤
tn
1
hSxn − xn , xn − ui − hSxn − γf (xn ), xn − ui
λ
λ
1
tn
hSxn − xn , xn − ui − hSxn − γf (xn ), xn − ui
λ
λ
+hAS x̃ − ASxn , xn − ui
=⇒
hAS x̃ + v, xn − ui ≤
1
tn
kSxn − xn kkxn − uk + kSxn − γf (xn )kkxn − uk
λ
λ
+kAS x̃ − ASxn kkxn − uk.
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
333
It follows that
hAS x̃ + v, x̃ − ui
≤
tn
1
kSxnj − xnj kkxnj − uk + j kSxnj − γf (xnj )kkxnj − uk
λ
λ
+kAS x̃ − ASxnj kkxnj − uk + hAS x̃ + v, x̃ − xnj i.
(3.9)
Since
hxnj − x̃, ASxnj − AS x̃i ≥ αkASxnj − AS x̃k2 ,
ASxnj → ASz and xnj * x̃, we have ASxnj → AS x̃. We also observe that tn → 0 and kSxn − xn k → 0.
Then, from (3.9), we derive
hAS x̃ + v, x̃ − ui ≤ 0.
That is, h−Ax̃ − v, x̃ − ui ≥ 0. Since B is maximal monotone, we have −Ax̃ ∈ B x̃. This shows that
0 ∈ (A + B)x̃. Hence, we have x̃ ∈ F (S) ∩ (A + B)−1 0. Therefore we can substitute x̃ for z in (3.7) to get
1 kxn − x̃k2 ≤
γf (x̃) − x̃, Sxn − x̃ + (tn + kASxn − Ax̃k)M.
1 − γρ
Consequently, the weak convergence of {xn } to x̃ actually implies that xn → x̃. This has proved the relative
norm-compactness of the net {xt } as t → 0+.
Now we return to (3.7) and take the limit as n → ∞ to get
kx̃ − zk2 ≤
1
hγf (z) − z, x̃ − zi,
1 − γρ
∀z ∈ F (S) ∩ (A + B)−1 0.
In particular, x̃ solves the following variational inequality
x̃ ∈ F (S) ∩ (A + B)−1 0, hγf (z) − z, x̃ − zi ≥ 0,
for all z ∈ F (S) ∩ (A + B)−1 0, or the equivalent dual variational inequality (see Lemma 2.3)
x̃ ∈ F (S) ∩ (A + B)−1 0, hγf (x̃) − x̃, x̃ − zi ≥ 0,
for all z ∈ F (S) ∩ (A + B)−1 0. Hence,
x̃ = PF (S)∩(A+B)−1 0 (γf (x̃)).
Clearly this is sufficient to conclude that the entire net {xt } converges to x̃. This completes the proof.
Theorem 3.2. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A be an
α-inverse strongly-monotone mapping of C into H. Let f : C → H be a ρ-contraction and γ be a constant
such that 0 < γ < ρ1 . Let B be a maximal monotone operator on H, such that the domain of B is included
in C. Let JλB = (I + λB)−1 be the resolvent of B for λ > 0 and let S be a nonexpansive mapping of C into
itself, such that
F (S) ∩ (A + B)−1 0 6= ∅.
For given x0 ∈ C, let {xn } ⊂ C be a sequence generated by
xn+1 = βn xn + (1 − βn )JλBn αn γf (xn ) + (1 − αn )Sxn − λn ASxn
for all n ≥ 0, where {λn } ⊂ (0, 2α),{αn } ⊂ (0, 1) and {βn } ⊂ (0, 1) satisfy
P
(i) limn→∞ αn = 0 and n αn = ∞;
(ii) 0 < lim inf n→∞ βn ≤ lim supn→∞ βn < 1;
(iii) a(1 − αn ) ≤ λn ≤ b(1 − αn ) where [a, b] ⊂ (0, 2α) and limn→∞ (λn+1 − λn ) = 0.
(3.10)
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
334
Then {xn } generated by (3.10) converges strongly to a point x̃ = PF (S)∩(A+B)−1 0 (γf (x̃)) which solves the
following variational inequality
hγf (x̃) − x̃, x̃ − zi ≥ 0,
for all z ∈ F (S) ∩ (A + B)−1 0.
Proof. Pick up z ∈ F (S) ∩ (A + B)−1 0. It is obvious that
z = JλBn (z − λn Az) = JλBn αn z + (1 − αn )(z − λn Az/(1 − αn ))
λ
for all n ≥ 0. Since JλB , S and I − 1−α
A are nonexpansive for all λ > 0 and n, we have
n
B
J αn γf (xn ) + (1 − αn )Sxn − λn ASxn − z λn
B
= Jλn αn γf (xn ) + (1 − αn ) Sxn −
λn
ASxn
1 − αn
λn
B
−Jλn αn z + (1 − αn ) z −
Az
1 − αn
λn
≤ αn γf (xn ) + (1 − αn Sxn −
ASxn
1 − αn
− αn z + (1 − αn ) z −
= (1 − αn ) Sxn −
λn
Az
1 − αn
2
λn
λn
ASxn − z −
Az + αn (γf (xn ) − z)
1 − αn
1 − αn
(3.11)
≤ (1 − αn )kxn − zk + αn kγf (xn ) − γf (z)k + αn kγf (z) − zk
≤ [1 − (1 − γρ)αn ]kxn − zk + αn kγf (z) − zk.
Hence, we have
kxn+1 − zk ≤ βn kxn − zk + (1 − βn )[1 − (1 − γρ)αn ]kxn − zk + (1 − βn )αn kγf (z) − zk
= [1 − (1 − γρ)αn (1 − βn )]kxn − zk + (1 − βn )αn kγf (z) − zk.
By induction, we have
1
kγf (z) − zk .
kxn+1 − zk ≤ max kx0 − zk,
1 − γρ
Therefore, {xn } is bounded. Since A is α-inverse strongly monotone, it is α1 -Lipschitz continuous. We deduce
immediately that {f (xn )}, {Sxn } and {ASxn } are also bounded. Set un = αn γf (xn )+(1−αn )Sxn −λn ASxn
and yn = JλBn un for all n ≥ 0. Noticing that JλBn is nonexpansive, we can check easily that {un } and {yn }
are bounded.
By using the convexity of k · k and the α-inverse strong monotonicity of A, from (3.11), we derive
2
λ
λ
n
n
(1 − αn ) Sxn −
ASxn − z −
Az
+ αn (γf (xn ) − z)
1 − αn
1 − αn
λn
λn
≤ (1 − αn ) Sxn −
ASxn − z −
Az k2 + αn kγf (xn ) − zk2
1 − αn
1 − αn
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
= (1 − αn )
(Sxn − z) −
335
2
λn
2
(ASxn − Az)
+ αn kγf (xn ) − zk
1 − αn
λ2n
2λn
2
2
hASxn − Az, Sxn − zi +
kASxn − Azk
= (1 − αn ) kSxn − zk −
1 − αn
(1 − αn )2
+αn kγf (xn ) − zk2
2αλn
λ2n
2
≤ (1 − αn ) kxn − zk2 −
kASxn − Azk2 +
kASx
−
Azk
n
1 − αn
(1 − αn )2
+αn kγf (xn ) − zk2
= (1 − αn ) kxn − zk2 +
λn
(λn − 2(1 − αn )α)kASxn − Azk2
(1 − αn )2
+αn kγf (xn ) − zk2 .
(3.12)
By condition (iii), we get
λn − 2(1 − αn )α ≤ 0
for all n ≥ 0. Then, from (3.11) and (3.12), we obtain
B
λn
2
J un − z 2 ≤ (1 − αn ) kxn − zk2 +
(λ
−
2(1
−
α
)α)kASx
−
Azk
n
n
n
λn
(1 − αn )2
+αn kγf (xn ) − zk2 .
(3.13)
From (3.10), we have
2
kxn+1 − zk2 = βn (xn − z) + (1 − βn ) JλBn un − z 2
≤ βn kxn − zk2 + (1 − βn )J B un − z .
λn
(3.14)
We can rewrite (3.10) as xn+1 = βn xn + (1 − βn )yn for all n ≥ 0. Next, we estimate kxn+1 − xn k. In fact,
we have
kyn+1 − yn k
= JλBn+1 un+1 − JλBn un ≤ JλBn+1 un+1 − JλBn+1 un + JλBn+1 un − JλBn un ≤ αn+1 γf (xn+1 ) + (1 − αn+1 )Sxn+1 − λn+1 ASxn+1
− αn γf (xn ) + (1 − αn )Sxn − λn ASxn + JλBn+1 un − JλBn un = k(I − λn+1 A)Sxn+1 − (I − λn+1 A)Sxn + (λn − λn+1 )ASxn
+αn (Sxn − γf (xn )) − αn+1 (Sxn+1 − γf (xn+1 ))k + kJλBn+1 un − JλBn un k
≤ k(I − λn+1 A)Sxn+1 − (I − λn+1 A)Sxn k + |λn+1 − λn |kASxn k
+αn kSxn − γf (xn )k + αn+1 kSxn+1 − γf (xn+1 )k + kJλBn+1 un − JλBn un k.
Since I − λn+1 A is nonexpansive for λn+1 ∈ (0, 2α), we have
k(I − λn+1 A)Sxn+1 − (I − λn+1 A)Sxn k
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
336
≤ kSxn+1 − Sxn k ≤ kxn+1 − xn k.
By the resolvent identity (2.1), we have
λn
λn
B
B
B
un .
Jλn+1 un = Jλn
un + 1 −
J
λn+1
λn+1 λn+1
It follows that
kJλB un − JλBn un k
n+1
B
λn
λn
B
B
= Jλn
un + (1 −
)Jλn+1 un − Jλn un λn+1
λn+1
λn
λn
B
≤ un + (1 −
)Jλn+1 un − un λn+1
λn+1
|λn+1 − λn |
kun − JλBn+1 un k.
λn+1
≤
So,
kyn+1 − yn k ≤ kxn+1 − xn k + |λn+1 − λn |kASxn k + αn kSxn − γf (xn )k
+αn+1 kSxn+1 − γf (xn+1 )k +
|λn+1 − λn |
kun − JλBn+1 un k.
λn+1
Then,
kyn+1 − yn k − kxn+1 − xn k ≤ |λn+1 − λn |kASxn k + αn kSxn − γf (xn )k
+αn+1 kSxn+1 − γf (xn+1 )k +
|λn+1 − λn |
kun − JλBn+1 un k.
λn+1
Since αn → 0, λn+1 − λn → 0 and lim inf n→∞ λn > 0, we obtain
lim sup(kyn+1 − yn k − kxn+1 − xn k) ≤ 0.
n→∞
From Lemma 2.4, we get
lim kyn − xn k = 0.
(3.15)
n→∞
Consequently, we obtain
lim kxn+1 − xn k = lim (1 − βn )kyn − xn k = 0.
n→∞
n→∞
From (3.13) and (3.14), we have
kxn+1 − zk2
2
≤ βn kxn − zk2 + (1 − βn )JλBn un − z ≤ (1 − βn ) (1 − αn ) kxn − zk2 +
2
+αn kγf (xn ) − zk
λn
(λn − 2(1 − αn )α)kASxn − Azk2
(1 − αn )2
+ βn kxn − zk2
= [1 − (1 − βn )αn ]kxn − zk2 +
(1 − βn )λn
(λn − 2(1 − αn )α)kASxn − Azk2
1 − αn
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
337
+(1 − βn )αn kγf (xn ) − zk2
≤ kxn − zk2 +
(1 − βn )λn
(λn − 2(1 − αn )α)kASxn − Azk2
1 − αn
+(1 − βn )αn kγf (xn ) − zk2 .
Then, we obtain
(1 − βn )λn
(2(1 − αn )α − λn )kASxn − Azk2
1 − αn
≤ kxn − zk2 − kxn+1 − zk2 + (1 − βn )αn kγf (xn ) − zk2
≤ (kxn − zk − kxn+1 − zk)kxn+1 − xn k + (1 − βn )αn kγf (xn ) − zk2 .
Since
lim αn = 0,
n→∞
lim kxn+1 − xn k = 0
n→∞
and
lim inf
n→∞
(1 − βn )λn
(2(1 − αn )α − λn ) > 0,
1 − αn
we have
lim kASxn − Azk = 0.
(3.16)
n→∞
Next, we show kxn − Sxn k → 0. By using the firm nonexpansivity of JλBn , we have
B
J un − z 2 = J B αn γf (xn ) + (1 − αn )Sxn − λn ASxn − J B z − λn Az 2
λn
λn
λn
≤
=
αn γf (xn ) + (1 − αn )Sxn − λn ASxn − (z − λn Az), JλBn un − z
2
1
kαn γf (xn ) + (1 − αn )Sxn − λn ASxn − (z − λn Az)k2 + JλBn un − z 2
2 −αn γf (xn ) + (1 − αn )Sxn − λn (ASxn − Az) − JλBn un .
From condition (iii) and the α-inverse strongly monotonicity of A, we know that I − λn A/(1 − αn ) is
nonexpansive. Hence
kαn γf (xn ) + (1 − αn )Sxn − λn ASxn − (z − λn Az)k2
= (1 − αn ) Sxn −
≤ (1 − αn )
Sxn −
2
λn
λn
ASxn − z −
Az
+ αn (γf (xn ) − z)
1 − αn
1 − αn
2
λn
λn
2
ASxn − z −
Az + αn kγf (xn ) − zk
1 − αn
1 − αn
≤ (1 − αn )kxn − zk2 + αn kγf (xn ) − zk2 .
Thus,
B
J un − z 2
λn
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
≤
338
2
1
(1 − αn )kxn − zk2 + αn kγf (xn ) − zk2 + JλBn un − z 2
2 −αn γf (xn ) + (1 − αn )Sxn − JλBn un − λn (ASxn − Az) .
That is,
B
J un − z 2
λn
≤ (1 − αn )kxn − zk2 + αn kγf (xn ) − zk2
2
−αn γf (xn ) + (1 − αn )Sxn − JλBn un − λn (ASxn − Az)
2
= (1 − αn )kxn − zk2 + αn kγf (xn ) − zk2 − αn γf (xn ) + (1 − αn )Sxn − JλBn un +2λn αn γf (xn ) + (1 − αn )Sxn − JλBn un , ASxn − Az − λ2n kASxn − Azk2
2
≤ (1 − αn )kxn − zk2 + αn kγf (xn ) − zk2 − αn γf (xn ) + (1 − αn )Sxn − JλBn un +2λn αn γf (xn ) + (1 − αn )Sxn − JλBn un kASxn − Azk.
This together with (3.14) imply that
kxn+1 − zk2 ≤ βn kxn − zk2 + (1 − βn )(1 − αn )kxn − zk2 + (1 − βn )αn kγf (xn ) − zk2
2
−(1 − βn )αn γf (xn ) + (1 − αn )Sxn − JλBn un +2λn (1 − βn )αn γf (xn ) + (1 − αn )Sxn − JλBn un kASxn − Azk
= [1 − (1 − βn )αn ]kxn − zk2 + (1 − βn )αn kγf (xn ) − zk2
2
−(1 − βn )αn γf (xn ) + (1 − αn )Sxn − JλBn un +2λn (1 − βn )αn γf (xn ) + (1 − αn )Sxn − JλBn un kASxn − Azk.
Hence,
2
(1 − βn )αn γf (xn ) + (1 − αn )Sxn − JλBn un ≤ kxn − zk2 − kxn+1 − zk2 − (1 − βn )αn kxn − zk2 + (1 − βn )αn kγf (xn ) − zk2
+2λn (1 − βn )αn γf (xn ) + (1 − αn )Sxn − JλBn un kASxn − Azk
≤ (kxn − zk + kxn+1 − zk)kxn+1 − xn k + (1 − βn )αn kγf (xn ) − zk2
+2λn (1 − βn )αn γf (xn ) + (1 − αn )Sxn − JλBn un kASxn − Azk.
Since lim supn→∞ βn < 1, kxn+1 − xn k → 0, αn → 0 and kASxn − Azk → 0 (by (3.16)), we deduce
lim αn γf (xn ) + (1 − αn )Sxn − JλBn un = 0.
n→∞
This implies that
lim Sxn − JλBn un = 0.
n→∞
(3.17)
Combining (3.15) and (3.17), we get
lim kxn − Sxn k = 0.
n→∞
(3.18)
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
339
Put x̃ = limt→0+ xt = PF (S)∩(A+B)−1 0 (γf (x̃)) where xt is the net defined by (1). We will finally show that
xn → x̃.
λn
(ASxn − Ax̃) for all n. Taking z = x̃ in (3.16) to get kASxn − Ax̃k → 0. First,
Setting vn = xn − 1−α
n
we prove lim supn→∞ hγf (x̃) − x̃, Sxn − x̃i ≤ 0. We take a subsequence {Sxni } of {Sxn } such that
lim suphγf (x̃) − x̃, Sxn − x̃i = lim hγf (x̃) − x̃, Sxni − x̃i.
i→∞
n→∞
It is clear that {Sxni } is bounded due to the boundedness of {Sxn } and kASxn −Ax̃k → 0. Then, there exists
a subsequence {Sxnij } of {Sxni } which converges weakly to some point w ∈ C. Hence, {xnij } and {ynij }
also converge weakly to w because of kSxnij − xnij k → 0 and kxnij − ynij k → 0. By the demi-closedness
principle of the nonexpansive mapping (see Lemma 2.2) and (3.18), we deduce w ∈ F (S). Furthermore, by
the similar argument as that of Theorem 3.1, we can show that w is also in (A + B)−1 0. Hence, we have
w ∈ F (S) ∩ (A + B)−1 0. This implies that
lim suphγf (x̃) − x̃, Sxn − x̃i = lim hγf (x̃) − x̃, Sxnij − x̃i = hγf (x̃) − x̃, w − x̃i.
j→∞
n→∞
Note that x̃ = PF (S)∩(A+B)−1 0 (γf (x̃)). Then, hγf (x̃) − x̃, w − x̃i ≤ 0, w ∈ F (S) ∩ (A + B)−1 0. Therefore,
lim suphγf (x̃) − x̃, Sxn − x̃i ≤ 0.
n→∞
From (3.10), we have
kxn+1 − x̃k2
≤ βn kxn − x̃k2 + (1 − βn )kJλBn un − x̃k2
= βn kxn − x̃k2 + (1 − βn )kJλBn un − JλBn (x̃ − λn Ax̃)k2
≤ βn kxn − x̃k2 + (1 − βn )kun − (x̃ − λn Ax̃)k2
= βn kxn − x̃k2 + (1 − βn )kαn γf (xn ) + (1 − αn )Sxn − λn ASxn − (x̃ − λn Ax̃)k2
λn
λn
= (1 − βn )(1 − αn ) Sxn −
ASxn − x̃ −
Ax̃
1 − αn
1 − αn
2
2
+αn (γf (xn ) − x̃)
+ βn kxn − x̃k
2
λn
λn
ASxn − x̃ −
Ax̃ 1 − αn
1 − αn
λn
λn
ASxn − x̃ −
Ax̃
+2αn (1 − αn ) γf (xn ) − x̃, Sxn −
1 − αn
1 − αn
2
2
+αn kγf (xn ) − x̃k
2
= βn kxn − x̃k + (1 − βn ) (1 − αn ) Sxn −
2
≤ βn kxn − x̃k2 + (1 − βn ) (1 − αn )2 kxn − x̃k2 + 2αn λn hγf (xn ) − x̃, ASxn − Ax̃i
+2αn (1 − αn )γhf (xn ) − f (x̃), Sxn − x̃i + 2αn (1 − αn )hγf (x̃) − x̃, Sxn − x̃i
+αn2 kγf (xn ) − x̃k2
≤ βn kxn − x̃k2 + (1 − βn ) (1 − αn )2 kxn − x̃k2 + 2αn λn kγf (xn ) − x̃kkASxn − Ax̃k
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
+2αn (1 − αn )γρkxn − x̃k2 + 2αn (1 − αn )hγf (x̃) − x̃, Sxn − x̃i + αn2 kγf (xn ) − x̃k2
340
≤ [1 − 2(1 − βn )(1 − γρ)αn ]kxn − x̃k2 + 2αn (1 − βn )λn kγf (xn ) − x̃kkASxn − Ax̃k
+2αn (1 − βn )(1 − αn )hγf (x̃) − x̃, Sxn − x̃i + (1 − βn )αn2 (kγf (xn ) − x̃k2 + kxn − x̃k2 )
= [1 − 2(1 − βn )(1 − γρ)αn ]kxn − x̃k2
λn
+2(1 − βn )(1 − γρ)αn
kγf (xn ) − x̃kkASxn − Ax̃k
1 − γρ
1 − αn
αn
2
2
+
hγf (x̃) − x̃, Sxn − x̃i +
(kγf (xn ) − x̃k + kxn − x̃k ) .
1 − γρ
2(1 − γρ)
P
It is clear that n 2(1 − βn )(1 − γρ)αn = ∞ and
1 − αn
λn
kγf (xn ) − x̃kkASxn − Ax̃k +
hγf (x̃) − x̃, Sxn − x̃i
lim sup
1 − γρ
1 − γρ
n→∞
αn
2
2
+
(kγf (xn ) − x̃k + kxn − x̃k ) ≤ 0.
2(1 − γρ)
We can therefore apply Lemma 2.5 to conclude that xn → x̃. This completes the proof.
Corollary 3.3. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A be an
α-inverse strongly-monotone mapping of C into H. Let B be a maximal monotone operator on H, such that
the domain of B is included in C. Let JλB = (I + λB)−1 be the resolvent of B for λ > 0 and let S be a
nonexpansive mapping of C into itself, such that F (S) ∩ (A + B)−1 0 6= ∅. Let λ be a constant satisfying
λ
a ≤ λ ≤ b where [a, b] ⊂ (0, 2α). For t ∈ 0, 1 − 2α
, let {xt } ⊂ C be a net generated by
xt = JλB (1 − t)Sxt − λASxt .
Then the net {xt } converges strongly, as t → 0+, to a point x̃ = PF (S)∩(A+B)−1 0 (0) which is the minimum
norm element in F (S) ∩ (A + B)−1 0.
Corollary 3.4. Let C be a closed and convex subset of a real Hilbert space H. Let A be an α-inverse
strongly-monotone mapping of C into H and let B be a maximal monotone operator on H, such that the
domain of B is included in C. Let JλB = (I +λB)−1 be the resolvent of B for λ > 0 such that (A+B)−1 0 6= ∅.
λ
Let λ be a constant satisfying a ≤ λ ≤ b where [a, b] ⊂ (0, 2α). For t ∈ (0, 1 − 2α
), let {xt } ⊂ C be a net
generated by
xt = JλB (1 − t)xt − λAxt .
Then the net {xt } converges strongly, as t → 0+, to a point x̃ = P(A+B)−1 0 (0) which is the minimum norm
element in (A + B)−1 0.
Corollary 3.5. Let C be a nonempty closed and convex subset of a real Hilbert space H. Let A be an
α-inverse strongly-monotone mapping of C into H. Let B be a maximal monotone operator on H, such that
the domain of B is included in C. Let JλB = (I + λB)−1 be the resolvent of B for λ > 0 and let S be a
nonexpansive mapping of C into itself, such that F (S) ∩ (A + B)−1 0 6= ∅. For given x0 ∈ C, let {xn } ⊂ C
be a sequence generated by
xn+1 = βn xn + (1 − βn )JλBn (1 − αn )Sxn − λn ASxn
for all n ≥ 0, where {λn } ⊂ (0, 2α),{αn } ⊂ (0, 1) and {βn } ⊂ (0, 1) satisfy
P
(i) limn→∞ αn = 0 and n αn = ∞;
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
341
(ii) 0 < lim inf n→∞ βn ≤ lim supn→∞ βn < 1;
(iii) a(1 − αn ) ≤ λn ≤ b(1 − αn ) where [a, b] ⊂ (0, 2α) and limn→∞ (λn+1 − λn ) = 0.
Then {xn } converges strongly to a point x̃ = PF (S)∩(A+B)−1 0 (0) which is the minimum norm element in
F (S) ∩ (A + B)−1 0.
Corollary 3.6. Let C be a closed and convex subset of a real Hilbert space H. Let A be an α-inverse
strongly-monotone mapping of C into H and let B be a maximal monotone operator on H, such that the
domain of B is included in C. Let JλB = (I +λB)−1 be the resolvent of B for λ > 0 such that (A+B)−1 0 6= ∅.
For given x0 ∈ C, let {xn } ⊂ C be a sequence generated by
xn+1 = βn xn + (1 − βn )JλBn (1 − αn )xn − λn Axn
for all n ≥ 0, where {λn } ⊂ (0, 2α),{αn } ⊂ (0, 1) and {βn } ⊂ (0, 1) satisfy
P
(i) limn→∞ αn = 0 and n αn = ∞;
(ii) 0 < lim inf n→∞ βn ≤ lim supn→∞ βn < 1;
(iii) a(1 − αn ) ≤ λn ≤ b(1 − αn ) where [a, b] ⊂ (0, 2α) and limn→∞ (λn+1 − λn ) = 0.
Then {xn } converges strongly to a point x̃ = P(A+B)−1 0 (0) which is the minimum norm element in (A +
B)−1 0.
Remark 3.7. The present paper provides some methods which do not use projection for finding the minimum
norm solution problem.
4. Applications
Next, we consider the problem for finding the minimum norm solution of a mathematical model related
to equilibrium problems. Let C be a nonempty, closed and convex subset of a Hilbert space and let
G : C × C → R be a bifunction satisfying the following conditions:
(E1)
(E2)
(E3)
(E4)
G(x, x) = 0 for all x ∈ C;
G is monotone, i.e., G(x, y) + G(y, x) ≤ 0 for all x, y ∈ C;
for all x, y, z ∈ C, lim supt↓0 G(tz + (1 − t)x, y) ≤ G(x, y);
for all x ∈ C, G(x, ·) is convex and lower semicontinuous.
Then, the mathematical model related to equilibrium problems (with respect to C) is to find x̃ ∈ C such
that
G(x̃, y) ≥ 0
(4.1)
for all y ∈ C. The set of such solutions x̃ is denoted by EP (G). The following lemma appears implicitly in
Blum and Oettli [5]:
Lemma 4.1. Let C be a nonempty, closed and convex subset of H and let G be a bifunction of C × C into
R satisfying (E1)-(E4). Let r > 0 and x ∈ H. Then, there exists z ∈ C such that
1
G(z, y) + hy − z, z − xi ≥ 0,
r
∀y ∈ C.
The following lemma was given in Combettes and Hirstoaga [8]:
Lemma 4.2. Assume that G : C × C → R satisfies (E1)-(E4). For r > 0 and x ∈ H, define a mapping
Tr : H → C as follows:
n
o
1
Tr (x) = z ∈ C : G(z, y) + hy − z, z − xi ≥ 0, ∀y ∈ C
r
for all x ∈ H. Then, the following hold:
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
342
(1) Tr is single-valued;
(2) Tr is a firmly nonexpansive mapping, i.e., for all x, y ∈ H,
kTr x − Tr yk2 ≤ hTr x − Tr y, x − yi;
(3) F (Tr ) = EP (G);
(4) EP (G) is closed and convex.
We call such Tr the resolvent of G for r > 0. Using Lemmas 4.1 and 4.2, we have the following lemma.
See [1] for a more general result.
Lemma 4.3. Let H be a Hilbert space and let C be a nonempty, closed and convex subset of H. Let
G : C × C → R satisfy (E1)-(E4). Let AG be a multivalued mapping of H into itself defined by
(
{{z ∈ H : G(x, y) ≥ hy − x, zi, ∀y ∈ C}, x ∈ C,
AG x =
∅,
x∈
/ C.
Then, EP (G) = A−1
G (0) and AG is a maximal monotone operator with dom(AG ) ⊂ C . Further, for any
x ∈ H and r > 0, the resolvent Tr of G coincides with the resolvent of AG ; i.e.,
Tr x = (I + rAG )−1 x.
Form Lemma 4.3, Theorems 3.1 and 3.2, we have the following results.
Theorem 4.4. Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let G be a
bifunction from C × C → R satisfying (E1)-(E4) and let Tr be the resolvent of G for r > 0. Let S be a
nonexpansive mapping from C into itself such that F (S) ∩ EP (G) 6= ∅. For t ∈ (0, 1), let {xt } ⊂ C be a net
generated by
xt = Tr (1 − t)Sxt , t ∈ (0, 1).
Then the net {xt } converges strongly, as t → 0+, to a point x̃ = PF (S)∩EP (G) (0) which is the minimum
norm element in F (S) ∩ EP (G).
Corollary 4.5. Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let G be a
bifunction from C × C → R satisfying (E1)-(E4) and let Tr be the resolvent of G for r > 0. Suppose
EP (G) 6= ∅. For t ∈ (0, 1), let {xt } ⊂ C be a net generated by
xt = Tr (1 − t)xt , t ∈ (0, 1).
Then the net {xt } converges strongly, as t → 0+, to a point x̃ = PEP (G) (0) which is the minimum norm
element in EP (G).
Theorem 4.6. Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let G be a
bifunction from C × C → R satisfying (E1)-(E4) and let Tλ be the resolvent of G for λ > 0. Let S be a
nonexpansive mapping from C into itself such that F (S) ∩ EP (G) 6= ∅. For given x0 ∈ C, let {xn } ⊂ C be
a sequence generated by
xn+1 = βn xn + (1 − βn )Tλn (1 − αn )Sxn
for all n ≥ 0, where {λn } ⊂ (0, ∞),{αn } ⊂ (0, 1) and {βn } ⊂ (0, 1) satisfy
P
(i) limn→∞ αn = 0 and n αn = ∞;
(ii) 0 < lim inf n→∞ βn ≤ lim supn→∞ βn < 1;
(iii) a ≤ λn ≤ b where [a, b] ⊂ (0, ∞) and limn→∞ (λn+1 − λn ) = 0.
Then {xn } converges strongly to a point x̃ = PF (S)∩EP (G) (0) which is the minimum norm element in F (S) ∩
EP (G).
P. Li, S. M. Kang, L. J. Zhu, J. Nonlinear Sci. Appl. 7 (2014), 325–344
343
Corollary 4.7. Let C be a nonempty, closed and convex subset of a real Hilbert space H. Let G be a
bifunction from C × C → R satisfying (E1)-(E4) and let Tλ be the resolvent of G for λ > 0. Suppose
EP (G) 6= ∅. For given x0 ∈ C, let {xn } ⊂ C be a sequence generated by
xn+1 = βn xn + (1 − βn )Tλn (1 − αn )xn
for all n ≥ 0, where {λn } ⊂ (0, ∞),{αn } ⊂ (0, 1) and {βn } ⊂ (0, 1) satisfy
P
(i) limn→∞ αn = 0 and n αn = ∞;
(ii) 0 < lim inf n→∞ βn ≤ lim supn→∞ βn < 1;
(iii) a ≤ λn ≤ b where [a, b] ⊂ (0, ∞) and limn→∞ (λn+1 − λn ) = 0.
Then {xn } converges strongly to a point x̃ = PEP (G) (0) which is the minimum norm element in EP (G).
Acknowledgements
Li-Jun Zhu was supported in part by NSFC of China (61362033) and NZ13087.
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