Filomat 28:7 (2014), 1393–1415
DOI 10.2298/FIL1407393S
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
An Iterative Algorithm for Finding the Solution of a
General Equilibrium Problem System
H. R. Sahebia , A. Razanib
b Department
a Department of Mathematics, Ashtian Branch, Islamic Azad University, Ashtian, Iran.
of Mathematics, Faculty of Science, Imam Khomeini International University, P.O. Box 34149-16818, Qazvin, Iran.
Abstract. In this paper, we introduce a new iterative method for finding a common element of the set
of solution of a general equilibrium problem system (GEPS) and the set of fixed points of a nonexpansive
semigroup. Furthermore, we present some numerical examples (by using MATLAB software) to guarantee
the main result of this paper.
1. Preliminaries
Throughout this paper, we assume that H is a real Hilbert space with inner product and norm are
denoted by h., .i and k.k, respectively and let C be a nonempty closed convex subset of H. A mapping
T : C → C is called nonexpansive if kTx − Tyk ≤ kx − yk, for all x, y ∈ C. We use F(T) to denote the set of
fixed points of T, that is, F(T) = {Tx = x}. Recall that a self-mapping f : H → H is a contraction on H if there
exists a constant ρ ∈ (0, 1) and x, y ∈ H such that k f (x) − f (y)k ≤ ρkx − yk. We denote weak convergence and
strong convergence by notations * and →, respectively. Moreover, H satisfies the Opial’s condition [27], if
for any sequence {xn } with xn * x, the inequality
lim infkxn − xk < lim infkxn − yk,
n→∞
n→∞
holds for every y ∈ H with x , y.
A nonexpansive semigroup is a family {T(s) : s ∈ [0, ∞)} of self-mappings on C such that:
(i) T(0)x = x, for all x ∈ C;
(ii) T(s + t) = T(s)T(t), for all s, t ≥ 0;
(iii) kT(s)x − T(s)yk ≤ kx − yk, for all x, y ∈ C and s ≥ 0;
(iv) s 7→ T(s)x is continuous for all x ∈ C.
2010 Mathematics Subject Classification. Primary 47H09, 47H10; Secondary 47J20
Keywords. Nonexpansive semigroup, general equilibrium problems system, strongly positive linear bounded operator, α−inverse
strongly monotone mapping, fixed point, Hilbert space
Received: 24 June 2013; Revised: 19 June 2014; Accepted: 08 August 2014
Communicated by Dragan S. Djordjević
Email addresses: [email protected] (H. R. Sahebi), [email protected] (A. Razani)
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1394
T
We denoted by F(S) the common fixed points set of nonexpansive semigroup S, that is, F(S) = s≥0 F(T(s)).
It is well known that F(S) is closed and convex [1]. Recall that given a closed convex subset C of a real
Hilbert space H, the nearest point projection PC from H onto C assigns to each x ∈ H its nearest point
denoted by PC x in C from x to C; that is, PC x is the unique point in C with the property kx − PC xk ≤ kx − yk
for all y ∈ C. PC is call the metric projection of H into C.
Lemma 1.1. ( [2]) Let H be a real Hilbert space. Then the following hold:
(a) kx + yk2 ≤ kyk2 + 2hx, x + yi for all x, y ∈ H;
(b) kαx + (1 − α)yk2 = αkxk2 + (1 − α)kyk2 − α(1 − α)kx − yk2 ;
(c) kx − yk2 = kxk2 + kyk2 − 2hx, yi.
Let A be a strongly positive linear bounded operator on H: that is, there exists a constant γ̄ > 0 such that
hAx, xi ≥ γ̄kxk2 , for all x ∈ H.
Lemma 1.2. ([25]) Assume A is a strongly positive linear bounded operator on a Hilbert space H with coefficient
γ > 0 and f : H → H is a contractive mapping with coefficient ρ , 0 < ρ < kAk−1 . Then kI − ρAk ≤ I − ργ.
A mapping B : C → H is called α−inverse strongly monotone [26, 36] if there exists a positive real
number α > 0 such that for all x, y ∈ C
hBx − By, x − yi ≥ αkBx − Byk2 .
Shimizu et al. [30] studied the strongly convergent of the sequence {xn } which is defined by
Z tn
1
xn+1 = αn x + (1 − αn )
T(s)xn ds, x ∈ C,
tn 0
in a real Hilbert space, where {T(s) : s ∈ [0, ∞)} is a strongly continuous semigroup of nonexpansive
mappings on a closed convex subset C of a Hilbert space and lim tn = ∞.
n→∞
Later, Plubtieng et al. [29] introduced the following iterative method for nonexpansive semigroup
{T(s) : s ∈ [0, ∞)}.
Let f : C → C be a contraction and the sequence {xn } be defined by x1 ∈ C,
Z tn
1
xn+1 = αn f (xn ) + βn xn + (1 − αn − βn )
T(s)xn ds,
tn 0
where {αn }, {βn } are the sequences in (0, 1) and {sn } is a positive real divergent sequence. Under the conditions
∞
X
αn = ∞, αn + βn < 1, lim αn = 0 and lim βn = 0, they proved the strong convergence of the sequence.
n=1
n→∞
n→∞
Many authors used iterative method to find the solution of equilibrium problem, mixed equilibrium
problem for nonexpansive semigroups ([6–9, 14, 15, 17–22, 38]).
In 2007, Plubtieng et al. [28] introduced the following iterative scheme for finding a common element of
the set equilibrium problem (EP) and the set of fixed points of a nonexpansive mapping in a Hilbert space.
Let S : C → H be a nonexpansive mapping, defined sequences {xn } and {un } by
(
F(un , y) + r1n hy − un , un − xn i ≥ 0;
xn+1 = αn γ f (xn ) + (I − αn A)Sun , ∀y ∈ H.
They proved, under the certain appropriate conditions, the sequence {xn } converges strongly to the unique
solution of the variational inequality
h(A − γ f )z, x − zi ≥ 0, ∀x ∈ F(S) ∩ EP(F),
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which is the optimality condition for the minimization problem
min
x∈F(S)∩EP(F)
1
hAx, xi − h(x),
2
where h is a potential function for γ f .
In 2009, Kumam [18] proved that the sequence xn generated by


F(un , y) + r1n hy − un , un − xn i ≥ 0, ∀y ∈ C,



yn = PC (un − λn Aun ),



 xn+1 = αn u + βn xn + γn SPC (xn − λn Ayn ),
is strongly convergent to a common fixed point of a nonexpansive mapping, the set of solutions of an
equilibrium problem, and the solution set of the variational inequality problem for a monotone k-Lipschitzcontinues mapping in a real Hilbert space.
In 2010, Kumam et al. [24] introduced the following iterative scheme an iterative scheme by the shrinking
projection method for finding a common element of the set of solutions of generalized mixed equilibrium
problems, the set of fixed points of a finite family of quasi-nonexpansive mappings and the set of solutions
of variational inclusion problems in a real Hilbert space. Starting with an arbitrary x0 ∈ H, x1 = PC1 x0 ,
un ∈ C define sequences {xn }, {yn }, {vn } and {zn } by





















F(un , y) + ϕ(y) − ϕ(un ) + hAxn , y − un i +
yn = JM,δn (un − δn Bun ),
vn = JM,λn (yn − λn Byn ),
zn = αn xn + (1 − αn )Kn vn ,
Cn+1 = {z ∈ Cn : ||zn − z|| ≤ ||xn − z||},
xn+1 = PCn+1 x0 , n ∈ N.
1
rn hy
− un , un−xn i ≥ 0, ∀y ∈ C,
They proved that under certain appropriate conditions imposed on {αn } and {βn }, {λn }, {δn } and {rn }, the
sequence {xn } is strongly convergent to z = PTN F(Ti )∩GMEP(F,ϕ,A)∩I(B,M) x0 .
i=1
In the same year, Kumam and Jaiboom [23] used a new approximation iterative method to prove a strong
convergence theorem for finding a common element of the set of fixed points of strict pseudocontractions,
the set of common solutions of the system of generalized mixed equilibrium problems, and the set of a
common solutions of the variational inequalities for inverse-strongly monotone mappings in a real Hilbert
space. Furthermore, they obtained a strong convergence theorem for the sequences generated by these
processes under some parameter controlling conditions.
Katchang et al.[13] introduced modified Mann iterative algorithms by the new hybrid projection method
for finding a common element of the set of fixed points of a countable family of nonexpansive mappings,
the set of solutions of the generalized mixed equilibrium problems and the set of solutions of the general
system of the variational inequality for two inverse-strongly monotone mappings in a real Hilbert space.
They proved some convergence theorems for the sequences generated by these process which connected
with minimize problems.
In 2011, Sunthrayuth and Kumam [32] introduced a new iterative scheme for finding common element of the
set of solutions of the variational inclusion with set-valued maximal monotone mapping and Lipschitzian
relaxed cocoercive mapping and the set of fixed point of nonexpansive semigroups in a uniformly convex
and 2-uniformly smooth Banach space. They proved the strong convergence of the proposed iterative
method under some certain control conditions.
Furthermore, Sunthrayuth and Kumam [31] proved that the sequence xn generated by

x0 ∈ C chosen arbitriraly,


R tn



 zn = γn xn + (1 − γn ) t1n 0 T(s)xn ds,



yn = αn γ f (zn ) + (I − αn A)zn ,



xn+1 = βn xn + (1 − βn )yn , ∀n ≥ 0,
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is strongly convergent to a common fixed point x∗ , in which x∗ ∈ F(S), is the unique solution of the variational
inequality hγ f (x∗ ) − Ax∗ , Jϕ (x − x∗ )i ≤ 0, ∀x ∈ F(S).
Jitpeera and Kumam [11] considered a shrinking projection method for finding the common element of the
set of common fixed points for nonexpansive semigroups, the set of common fixed points for an infinite
family of a ξ-strict pseudocontraction, the set of solutions of a system of mixed equilibrium problems, and
the set of solutions of the variational inclusion problem. They proved strong convergence theorems of the
iterative sequence generated by the shrinking projection method under some suitable conditions in a real
Hilbert space.
Later, Kamraska et al. [12] introduced a new iterative by viscosity approximation methods in a Hilbert
space. To be more precisely, they proved the following result:
Theorem 1.3. Let S = (T(s))s≥0 be a nonexpansive semigroup on a real Hilbert space H. Let f : H → H be an
α-contraction, A : H → H a strongly positive linear bounded self-adjoint operator with coefficient γ. Let γ be a real
γ
number such that 0 < γ < α . Let G : H × H → R be a mapping satisfying hypotheses (A1) − (A4) and Ψ : H → H
an inverse-strongly monotone mapping with coefficients δ > 0 such that F(S) ∩ GEP(G, Ψ) , ∅. Let the sequences
{xn }, {un } and {yn } be generated by

x1 ∈ H chosen arbitrary,



 G(u , y) + hΨx , y − u i + 1 hy − u , u − x i;
n n
n
n
n
n

R rn


 y = β x + (1 − β ) 1 sn T(s)u ds, x
=
α
n
n n
n sn 0
n
n+1
n γ f (xn ) + (I − αn A)yn , ∀n ≥ 1.
Under the certain appropriate conditions, they proved that the sequences {xn }, {un } and {yn } is strongly
convergent to z, which is a unique solution in F(S) ∩ GEP(G, Ψ) of the variational inequality
h(γ f − A)z, p − zi ≤ 0, ∀p ∈ F(S) ∩ GEP(G, Ψ).
Recently, Wattanawitoon and Kumam [39] introduced the following new hybrid proximal-point algorithms
defined by x1 = x ∈ C:

Q

wn = C J−1 (Jxn − λn Axn ),



 zn = J−1 (βn J(xn ) + (1 − βn )J(Jr wn )),

n


 y = J−1 (α J(x ) + (1 − α )J(z )),

n
n
1
n
n


 u ∈ C such that,
n




 Θ(un , y) + ϕ(y) − ϕ(un ) + r1n hy − un , Jun − Jyn i ≥ 0, ∀y ∈ C,




Cn+1 = {z ∈ C : φ(z, un ) ≤ αn φ(z, x1 ) + (1 − αn )φ(z, xn )},



 xn+1 = Q
Cn+1 x
and

















un ∈ C such that,
Θ(un , y) + ϕ(y) − ϕ(un ) + r1n hy − un , Jun − Jyn i ≥ 0, ∀y ∈ C,
Q
zn = C J−1 (Jun − λn Aun ),
yn = J−1Q(βn J(xn ) + (1 − βn )J(Jrn zn )),
xn+1 = C J−1 (αn J(x1 ) + (1 − αn )J(yn )).
Under appropriate conditions,
they proved that the sequence {xn } generated by above Q
algorithms is strongly
Q
convergent to the point VI(C,A)∩T−1 (0)∩MEP(Θ,ϕ) x and converges weakly to the point lim VI(C,A)∩T−1 (0)∩MEP(Θ,ϕ) xn ,
n→∞
respectively. Very recently, Sunthrayuth and Kumam [33] introduced a general implicit iterative scheme
base on viscosity approximation method with a φ-strongly pseudocontractive mapping for finding a common element of the set of solutions for a system of mixed equilibrium problems, the set of common fixed
point for a nonexpansive semigroup, and the set of solutions of system of variational inclusions with setvalued maximal monotone mapping and Lipschitzian relaxed cocoercive mappings in Hilbert spaces. They
proved that the proposed iterative algorithm is strongly convergent to a common element of the above
three sets, which is a solution of the optimization problem related to a strongly positive bounded linear
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
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operator.
In this paper, by intuition from [5, 10, 12], a new iterative scheme is introduced. Indeed, we consider C1 and
C2 be closed convex subsets in H. Let F(x, y) be a bifunction satisfy conditions (A1) − (A4) with C replaced
by C1 and let {T(s) : s ∈ [0, ∞)} be a nonexpansive semigroup on C2 . By this scheme find a common element
of the set of solution of the generalized equilibrium problem system (GEPS) and the set of all common fixed
points of a nonexpansive semigroup in the framework of a real Hilbert space. Furthermore, by using these
results, we obtain mean eragodic theorems for a nonexpansive mapping in a real Hilbert space. Finally,
some numerical examples are also given.
2. Generalized Equilibrium Problems System
A typical problem is to minimize a quadratic function over the set of the fixed points of a nonexpansive
mappings on a real Hilbert space H:
min
1
hAx, xi − h(x),
2
where A is strongly positive linear bounded operator and h is a potential function for γ f , i. e., h0 (x) = γ f,
for all x ∈ H.
Let A : H → H be an inverse strongly monotone mapping and F : C × C → R be a bifunction. Then we
consider the following GEPS
Find x̃ ∈ C such that F(x̃, y) + hAx, y − xi ≥ 0, for all y ∈ C.
(1)
The set of such x ∈ C is denoted by GEPS(F, A), i.e.,
GEPS(F, A) = {x ∈ C : F(x, y) + hAx, y − xi ≥ 0, ∀y ∈ C}.
To study the generalized equilibrium problem (1), let F satisfies the following conditions:
(A1) F(x, x) = 0, for all x ∈ C;
(A2) F is monotone, i.e., F(x, y) + F(y, x) ≤ 0 for all x, y ∈ C;
(A3) for each x, y, z ∈ C, lim supt→0− F(tz + (1 − t)x, y) ≤ F(x, y);
(A4) for each x ∈ C y 7→ F(x, y) is convex and weakly lower semi-continuous.
In case that C1 = H, F(x, y) = 0, C2 = C, and T(s) = T is a nonexpansive mapping on C, (1) is the fixed point
problem of a nonexpansive mapping.
Lemma 2.1. ([3]) Let C be a nonempty closed convex subset of H and F : C × C → R be a bifunction satisfying
(A1) − (A4). Then, for any r > 0 and x ∈ H there exists z ∈ C such that
1
F(z, y) + hy − z, z − xi ≥ 0, ∀y ∈ C.
r
Further, define
1
Tr x = {z ∈ C : F(z, y) + hy − z, z − xi ≥ 0, ∀y ∈ C}
r
for all r > 0 and x ∈ H. Then
(a) Tr is single-valued;
(b) Tr is firmly nonexpansive, i.e., for any x, y ∈ H
kTr x − Tr yk2 ≤ hTr x − Tr y, x − yi;
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1398
(c) F(Tr ) = GEP(F);
(d) kTs x − Tr xk ≤
s−r
s kTs x
− xk;
(e) GEP(F) is closed and convex.
Remark 2.2. It is clear that for any x ∈ H and r > 0, by Lemma 2.1(a), there exists z ∈ H such that
1
F(z, y) + hy − z, z − xi ≥ 0, for all y ∈ H.
r
(2)
Replacing x with x − rΨx in (2), we obtain
1
F(z, y) + hΨx, y − zi + hy − z, z − xi ≥ 0, for all y ∈ H.
r
Lemma 2.3. ([35]) Let {xn } and {yn } be bounded sequences in a Banach space E and {βn } be a sequence in [0, 1] with
0 < lim infβn ≤ lim supβn < 1. Suppose
n→∞
n→∞
xn+1 = (1 − βn )yn + βn xn for all integers n ≥ 1 and lim sup(kyn+1 − yn k − kxn+1 − xn k) ≤ 0. Then lim kyn − xn k = 0.
n→∞
n→∞
Lemma 2.4. ([30]) Let C be a nonempty bounded closed convex subset of H and let S = {T(s) : s ∈ [0, ∞)} be a
nonexpansive semigroup on C. Then for any h ∈ [0, ∞),
Z
Z t
1 t
1
T(s)xds
−
T(h)(
T(s)xds)
= 0,
t
t
0
0
x∈C
lim sup
t→∞
for x ∈ C and t > 0.
Lemma 2.5. ([40]) Assume {an } is a sequence of nonnegative numbers such that
an+1 ≤ (1 − αn )an + δn ,
where {αn } is a sequence in (0, 1) and {δn } is a sequence in real number such that
(i) lim αn = 0,
n→∞
∞
X
αn = ∞;
n=1
∞
(ii) lim sup
n→∞
X
δn
≤ 0 or
|δn | < ∞;
αn
n=1
Then lim an = 0.
n→∞
Lemma 2.6. ([4]) If C is a closed convex subset of H, T is a nonexpansive mapping on C, {xn } is a sequence in C such
that xn * x ∈ C and xn − Txn → 0, then x − Tx = 0
3. Strong Convergence for an Iterative Algorithm
In this section, we introduce a new iterative for finding a common element of the set of solution for an
equilibrium problem involving a bifunction defined on a closed convex subset and the set of fixed points
for a nonexpansive semigroup.
Theorem 3.1. Let H be a real Hilbert space. Asuume that
• C1 , C2 are two nonempty convex closed subsets H,
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1399
• F1 , F2 , . . . , Fk be bifunctions from C1 × C1 to R satisfying (A1) − (A4),
• Ψ1 , Ψ2 , . . . , Ψk is µi −inverse strongly monotone mapping on H,
• f : H → H is a ρ−contraction,
• A is a strongly positive linear bounded operator on H with coefficient λ and
0 < γ < λρ ,
• F(S) = {T(s) : s ∈ [0, ∞)} is a nonexpansive semigroup on C2 such that
Tk
i=1 F(S) ∩ GEPS(Fi , Ψi ) , ∅,
• {xn } is a sequence generated in the following manner:

(i)

x1 ∈ H, and un ∈ C1 ,



(1) (1)
(1)
(1)


F1 (un , y) + hΨ1 xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C1 ,



(2)
(2)
(2) (2)


F2 (un , y) + hΨ2 xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C1 ,




 ..

.



(k)
(k)
(k) (k)


Fk (un , y) + hΨk xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C1 ,



P

(i)

 ωn = 1k ki=1 un ,

R tn


1
 x
n+1 = αn γ f (xn ) + βn xn + ((1 − βn )I − αn A) tn 0 T(s)PC2 ωn ds,
where {αn }, {βn } are the sequences in [0, 1] and {rn } ⊂ (0, ∞) is a real sequence which satisfy the following
conditions:
(C1) lim αn = 0,
n→∞
∞
X
αn = ∞;
n=1
(C2) 0 < lim infβn ≤ lim supβn < 1;
n→∞
n→∞
(C3) lim |rn+1 − rn | = 0 and 0 < b < rn < a < 2µi for i ∈ {1, 2, . . . , k};
n→∞
(C4) lim tn = ∞, and sup |tn+1 − tn | is bounded.
n→∞
Then
(i) the sequence {xn } is bounded;
(ii) lim kΨi xn − Ψi qk = 0, for q ∈
Tk
i=1
n→∞
F(S) ∩ GEPS(Fi , Ψi ) , i ∈ {1, 2, . . . , k};
Z tn
Z tn
1
1
T(s)PC2 ωn ds = 0 and lim ωn −
T(s)PC2 ωn ds = 0.
(iii) lim xn −
n→∞
n→∞
tn 0
tn 0
Proof. (i) By the same argument in [10, 16],
k(1 − βn )I − αn Ak ≤ 1 − βn − αn λ.
T
Let q ∈ ki=1 F(S) ∩ GEPS(Fi , Ψi ). Observe that I − rn Ψi for any i = 1, 2, . . . , k is a nonexpansive mapping.
Indeed, for any x, y ∈ H,
k(I − rn Ψi )x − (I − rn Ψi )yk2
=
k(x − y) − rn (Ψi x − Ψi y)k2
= kx − yk2 − 2rn hx − y, Ψi x − Ψi yi + r2n kΨi x − Ψi yk2
≤
kx − yk2 − rn (2µi − rn )kΨi x − Ψi yk2
≤
kx − yk2 .
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1400
So
(i)
kun − qk ≤ kxn − qk,
(3)
and hence
kωn − qk ≤ kxn − qk.
(4)
Thus
=
kxn+1 − qk
kαn γ f (xn ) + βn xn + ((1 − βn )I − αn A)
1
tn
tn
Z
T(s)PC2 ωn ds − qk
0
αn kγ f (xn ) − Aqk + βn kxn − qk
Z tn
1
T(s)PC2 ωn ds − qk
+k(1 − βn )I − αn A)kk
tn 0
αn {kγ f (xn ) − γ f (q)k + kγ f (q) − Aqk} + βn kxn − qk
Z tn
1
+(1 − βn − αn λ)
kT(s)PC2 ωn − PC2 qkds
tn 0
αn ργkxn − qk + αn kγ f (q) − Aqk + βn kxn − qk
≤
≤
≤
+(1 − βn − αn λ)kωn − qk
αn ργkxn − qk + αn kγ f (q) − Aqk + βn kxn − qk
≤
+(1 − βn − αn λ)kxn − qk
=
(1 − αn (λ − γρ))kxn − qk + αn kγ f (q) − Aqk
kγ f (q) − Aqk
}.
max{kxn − qk,
λ − γρ
≤
By induction
kxn − qk ≤ max{kx1 − qk,
kγ f (q) − Aqk
}.
λ − γρ
Therefore, the sequence {xn } is bounded and also { f (xn )}, {ωn } and { t1n
(ii) Note that
(i)
(i)
un
(i)
can be written as
kun+1 − un k
≤
(i)
un
R tn
0
T(s)PC2 ωn ds} are bounded.
= Tr(i) (xn − rn Ψi xn ). By Lemma 2.1, for any i = 1, 2, . . . , k,
n
kTr(i) (I − rn+1 Ψi )xn+1 − Tr(i) (I − rn Ψi )xn k
n+1
n+1
+kTr(i) (I − rn Ψi )xn − Tr(i) (I − rn Ψi )xn k
n
n+1
≤
k(I − rn+1 Ψi )xn+1 − (I − rn Ψi )xn k
+kTr(i) (I − rn Ψi )xn − Tr(i) (I − rn Ψi )xn k
n
n+1
≤
kxn+1 − xn k + |rn+1 − rn |kΨi xn k
rn+1 − rn
+
kTr(i) (I − rn Ψi )xn − Tr(i) (I − rn Ψi )xn k.
n
n+1
rn+1
Then
(i)
(i)
kun+1 − un k ≤ kxn+1 − xn k + 2Mi |rn+1 − rn |,
where
Mi = max{sup{
kTr(i) (I − rn Ψi )xn − Tr(i) (I − rn Ψi )xn k
n
n+1
rn+1
(5)
, sup{kΨi xn k}}}.
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
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Also
Z tn+1
Z tn
1
1
T(s)PC2 ωn+1 ds −
T(s)PC2 ωn ds
tn+1
t
n 0
0
Z tn+1
Z tn
1
1
1
[T(s)ω
−
T(s)ω
]ds
+
(
−
)
[T(s)ωn − T(s)q]ds
n+1
n
tn+1
tn+1 tn 0
0
Z tn+1
1
+
[T(s)ωn − T(s)q]ds
tn+1 tn
2|tn+1 − tn |
kωn+1 − ωn k +
kωn − qk.
tn+1
=
≤
Let M =
1
k
k
X
2Mi < ∞, since
i=1
k
kωn+1 − ωn k ≤
1 X (i)
(i)
kun+1 − un k ≤ kxn+1 − xn k + M|rn+1 − rn |,
k
i=1
hence
1 R tn+1
T(s)PC2 ωn+1 ds −
t
0
n+1
1
tn
R tn
0
≤ kxn+1 − xn k + M|rn+1 − rn | +
Suppose zn =
αn γ f (xn )+((1−βn )I−αn A)Λn
,
1−βn
kzn+1 − zn k
=
=
=
≤
≤
T(s)PC2 ωn ds
2|tn+1 −tn |
tn+1 kωn
(6)
− qk.
where Λn :=
1
tn
R tn
0
T(s)PC2 ωn ds. It follows from (5), (6)
αn+1 γ f (xn+1 ) + ((1 − βn+1 )I − αn+1 A)Λn+1
1 − βn+1
αn γ f (xn ) + ((1 − βn )I − αn A)Λn
−
k
1 − βn
αn+1 γ f (xn+1 ) (1 − βn+1 )Λn+1 αn+1 AΛn+1
k
+
−
1 − βn+1
1 − βn+1
1 − βn+1
αn γ f (xn ) (1 − βn )Λn αn AΛn
−
+
k
−
1 − βn
1 − βn
1 − βn
αn
αn+1
k
(γ f (xn+1 ) − AΛn+1 ) +
(AΛn − γ f (xn )) + (Λn+1 − Λn )k
1 − βn+1
1 − βn
αn+1
αn
kγ f (xn+1 ) − AΛn+1 k +
kAΛn − γ f (xn )k + kΛn+1 − Λn k
1 − βn+1
1 − βn
αn+1
αn
kγ f (xn+1 ) − AΛn+1 k +
kAΛn − γ f (xn )k + kxn+1 − xn k
1 − βn+1
1 − βn
2|tn+1 − tn |
+M|rn+1 − rn | +
kωn − qk.
tn+1
k
(C1), (C3) and (C4) implies that
lim sup(kzn+1 − zn k − kxn+1 − xn k) ≤ 0.
n→∞
By Lemma 2.3, lim kzn − xn k = 0.
n→∞
Consequently
lim kxn+1 − xn k = lim (1 − βn )kzn − xn k = 0.
n→∞
n→∞
(7)
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1402
Moreover, for any i ∈ {1, 2, . . . , k},
(i)
kun − qk2
≤
k(xn − q) − rn (Ψi xn − Ψi q)k2
= kxn − qk2 − 2rn hxn − q, Ψi xn − Ψi qi + r2n kΨi xn − Ψi qk2
≤
kxn − qk2 − rn (2µi − rn )kΨi xn − Ψi qk2 ,
and then
k
k
X
1 (i)
1 X (i)
k
(un − q)k2 ≤
kun − qk2
k
k
i=1
i=1
k
X
rn (2µi − rn )kΨi xn − Ψi q)k2 .
kxn − qk2 − 1k
kωn − qk2 =
≤
(8)
i=1
By (8), we have
kxn+1 − qk2
=
kαn (γ f (xn ) − Aq) + βn (xn − q) + ((1 − βn )I − αn A)(Λn − q)k2
≤
αn kγ f (xn ) − Aqk2 + βn kxn − qk2 + (1 − βn − αn λ)kΛn − qk2
≤
αn kγ f (xn ) − Aqk2 + βn kxn − qk2 + (1 − βn − αn λ)kωn − qk2
≤
αn kγ f (xn ) − Aqk2 + βn kxn − qk2 + (1 − βn − αn λ){kxn − qk2
k
1X
−
rn (2µi − rn )kΨi xn − Ψi q)k2 }
k
i=1
αn kγ f (xn ) − Aqk2 + kxn − qk2
k
1X
−(1 − βn − αn λ)
rn (2µi − rn )kΨi xn − Ψi qk2 ,
k
≤
i=1
and hence
(1 − βn − αn λ) 1k
k
X
b(2µi − a)kΨi xn − Ψi qk2
i=1
≤
≤
αn kγ f (xn ) − Aqk2 + kxn − qk2 − kxn+1 − qk2
αn kγ f (xn ) − Aqk2 + kxn+1 − xn k(kxn+1 − qk − kxn − qk).
Since αn → 0 and kxn+1 − xn k → 0, it follows that
lim kΨi xn − Ψi qk = 0, ∀i = 1, 2, . . . , k.
(9)
n→∞
(iii) By Lemma 2.1, for any i = 1, 2, . . . , k,
(i)
kun − qk2
≤
=
≤
=
(i)
h(I − rn Ψi )xn − (I − rn Ψi )q, un − qi
1
{k(I − rn Ψi )xn − (I − rn Ψi )qk2 + kun,i − qk2
2
(i)
−k(I − rn Ψi )xn − (I − rn Ψi )q − (un − q)k2 }
1
(i)
(i)
{kxn − qk2 + kun − qk2 − kxn − un − rn (Ψi xn − Ψi q)k2 }
2
1
(i)
(i)
{kxn − qk2 + kun − qk2 − (kxn − un k2
2
(i)
−2rn hxn − un , Ψi xn − Ψi qi + r2n kΨi xn − Ψi qk2 ).
This implies
(i)
(i)
(i)
kun − qk2 ≤ kxn − qk2 − kxn − un k2 + 2rn kxn − un kkΨi xn − Ψi qk,
(10)
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1403
and hence
k
X
1 (i)
k
(u − q)k2
k n
i=1
k
X
(i)
1
kun − qk2
k
kωn − qk2 =
≤
(11)
i=1
kxn − qk2 −
≤
1
k
k
X
k
(i)
kun − xn k2 +
i=1
1X
(i)
2rn kxn − un kkΨi xn − Ψi qk.
k
i=1
Observe that
kxn+1 − qk2
≤
αn kγ f (xn ) − Aqk2 + βn kxn − qk2 + (1 − βn − αn λ)kωn − qk2
≤
αn kγ f (xn ) − Aqk2 + βn kxn − qk2 + (1 − βn − αn λ){kxn − qk2
k
k
1 X (i)
1X
(i)
−
kun − xn k2 +
2rn kxn − un kkΨi xn − Ψi qk}.
k
k
i=1
i=1
It follows that
(1 − βn −
αn λ) 1k
k
X
(i)
kun − xn k2 ≤ αn kγ f (xn ) − Aqk2 + kxn − qk2 − kxn+1 − qk2
i=1
+(1 − βn −
αn λ) 1k
k
X
(i)
2rn kxn − un kkΨi xn − Ψi qk
i=1
≤ αn kγ f (xn ) − Aqk2 + kxn+1 − xn k(kxn − qk − kxn+1 − qk)
k
X
(i)
+(1 − βn − αn λ) 1k
2rn kxn − un kkΨi xn − Ψi qk.
i=1
Since αn → 0 and kxn+1 − xn k → 0, we have
(i)
lim kun − xn k = 0.
n→∞
(12)
It is easy to prove
lim kωn − xn k = 0.
n→∞
(13)
The definition of {xn } shows
kΛn − xn k
≤
kxn+1 − xn k + kxn+1 − Λn k
≤
kxn+1 − xn k + kαn γ f (xn ) + βn xn + ((1 − βn )I − αn A)Λn − Λn k
≤
kxn+1 − xn k + αn kγ f (xn ) − AΛn k + βn kxn − Λn k.
That is
kΛn − xn k ≤
αn
1
kxn+1 − xn k +
kγ f (xn ) − AΛn k.
1 − βn
1 − βn
The condition (C1) together (7) implies that
lim kΛn − xn k = 0.
n→∞
(14)
Moreover, kωn − Λn k ≤ kωn − xn k + kxn − Λn k, we get
lim kΛn − ωn k = 0.
n→∞
(15)
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1404
Then,
Z tn
1
lim xn −
T(s)PC2 ωn ds = 0,
n→∞
tn 0
Z tn
1
T(s)PC2 ωn ds = 0.
lim ωn −
n→∞
tn
0
Theorem 3.2. Suppose all assumptions of Theorem 3.1 are hold. Then the sequence {xn } is strongly convergent to a
T
point x̄, where x̄ ∈ ki=1 F(S) ∩ GEPS(Fi , Ψi ), which solves the variational inequality
h(A − γ f )x̄, x̄ − xi ≤ 0.
Equivalently, x̄ = PTk
i=1
F(S)∩GEPS(Fi ,Ψi ) (I
− A + γ f )(x̄).
Proof. For all x, y ∈ H, we have
kPTk
i=1
F(S)∩GEPS(Fi ,Ψi ) (I
− A + γ f )(x) − PTk
i=1
≤
k(I − A + γ f )(x) − (I − A + γ f )(y)k
≤
kI − Akkx − yk + γk f (x) − f (y)k
≤
(1 − λ)kx − yk + γρkx − yk
=
(1 − (λ − γρ))kx − yk.
F(S)∩GEPS(Fi ,Ψi ) (I
− A + γ f )(y)k
This implies that PTk F(S)∩GEPS(Fi ,Ψi ) (I − A + γ f ) is a contraction of H into itself. Since H is complete, then
i=1
there exists a unique element x̄ ∈ H such that
x̄ = PTk
i=1
F(S)∩GEPS(Fi ,Ψi ) (I
− A + γ f )(x̄).
Next, we prove
1
lim suph(A − γ f )x̄, x̄ −
t
n
n→∞
Let x̃ = PTk
i=1
F(S)∩GEPS(Fi ,Ψi ) x1 ,
tn
Z
T(s)PC2 ωn dsi ≤ 0.
0
set
Σ = { ȳ ∈ C2 : k ȳ − x̃k ≤ kx1 − x̃k +
kγ f (x̃) − Ax̃k
}.
λ − γρ
It is clear, Σ is nonempty closed bounded convex subset of C2 and S = {T(s) : s ∈ [0, ∞)} is a nonexpansive
semigroup on Σ.
Let Λn =
1
tn
R tn
0
T(s)PC2 ωn ds, since {Λn } ⊂ Σ is bounded, there is a subsequence {Λn j } of {Λn } such that
lim suph(A − γ f )x̄, x̄ − Λn i = lim h(A − γ f )x̄, x̄ − Λn j i.
n→∞
j→∞
(16)
As {ωn j } is also bounded, there exists a subsequence {ωn jl } of {ωn j } such that ωn jl * ξ. Without loss of
generality, let ωn j * ξ. From (iii) in Theorem 3.1, we have Λn j → ξ.
Since {ωn } ⊂ C1 and {Λn } ⊂ C2 and C1 , C2 are two closed convex subsets in H, we obtain that ξ ∈ C1 ∩ C2 .
Now, we prove the following items:
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1405
T
(i) ξ ∈ F(S) = s≥1 F(T(s)).
Since {Λn } ⊂ C2 , we have
kΛn − PC2 ωn k
=
kPC2 Λn − PC2 ωn k
≤
kΛn − ωn k.
By (iii) in Theorem 3.1, we have
lim kΛn − PC2 ωn k = 0.
(17)
n→∞
By using (iii) in Theorem 3.1 and (17), we obtain
lim kωn − PC2 ωn k = 0.
(18)
n→∞
This shows that the sequence PC2 ωn j * ξ as j → ∞.
For each h > 0, we have
kT(h)PC2 ωn − PC2 ωn k
≤
kT(h)PC2 ωn − T(h)Λn k + kT(h)Λn − Λn k + kΛn − PC2 ωn k
≤ 2kΛn − PC2 ωn k + kT(h)Λn − Λn k.
The lemma 2.4 implies that
lim kT(h)Λn − Λn k = 0,
(19)
n→∞
the equalities (17, 18) and (19)implies that
lim kT(h)PC2 ωn − ωn k = 0.
n→∞
Note that F(TPC ) = F(T) for any mapping T : C → C. The Lemma 2.6 implies that ξ ∈ F(T(h)PC2 ) =
F(T(h)) for all h > 0. This shows that ξ ∈ F(S).
(ii) ξ ∈
Tk
i=1
GEPS(Fi , Ψi ).
Since {ωn } is bounded and as respects (15), there exists a subsequence {ωn j } of {ωn } such that ωn j * ξ.
By intuition from [12],
(i)
(i)
Fi (un , y) + hΨi xn , y − un i +
1
(i) (i)
hy − un , un − xn i ≥ 0, for all y ∈ C1 .
rn
By (A2), we have
(i)
hΨi xn , y − un i +
1
(i) (i)
(i)
hy − un , un − xn i ≥ Fi (y, un ).
rn
Substitute n by n j , we get
(i)
hΨi xn j , y −
(i)
un j i
+ hy −
(i)
un j ,
un j − xn j
rn j
(i)
i ≥ Fi (y, un j ).
(20)
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1406
For 0 < l ≤ 1 and y ∈ C1 , set yl = ly + (1 − l)ξ. We have yl ∈ C1 and
(i)
hyl − un j , Ψi yl i
≥
(i)
(i)
hyl − un j , Ψi yl i − hΨi xn j , yl − un j i
(i)
−hyl −
=
(i)
un j ,
un j − xn j
rn j
(i)
i + Fi (yl , un j )
(i)
(i)
(i)
(i)
hyl − un j , Ψi yl − Ψi un j i + hyl − un j , Ψi un j − Ψi xn j i
(i)
−hyl −
(i)
un j ,
un j − xn j
rn j
(i)
i + Fi (yl , un j ).
The condition (A4), monotonicity of Ψi and (12) implies that
(i)
(i)
(i)
hyl − un j , Ψi yl − Ψi un j i ≥ 0 and kΨi un j − Ψi xn j k → 0 as j → ∞. Hence
hyl − ξ, Ψi yl i ≥ Fi (yl , ξ).
(21)
Now, (A1) and (A4) together (21) show
0 = Fi (yl , yl ) ≤
lFi (yl , y) + (1 − l)Fi (yl , ξ)
≤
lFi (yl , y) + (1 − l)hyl − ξ, Ψi yl i
=
lFi (yl , y) + (1 − l)lhy − ξ, Ψi yl i,
which yields Fi (yl , y) + (1 − l)hy − ξ, Ψi yl i ≥ 0.
By taking l → 0, we have
Fi (ξ, y) + hy − ξ, Ψi ξi ≥ 0.
This shows ξ ∈ GEPS(Fi , Ψi ), for all i = 1, 2, . . . , k. Then, ξ ∈
Tk
i=1
GEPS(Fi , Ψi ).
Now, in view of (16), we see
lim suph(A − γ f )x̄, x̄ − Λn i = h(A − γ f )x̄, x̄ − ξi ≤ 0.
n→∞
Finally, we prove {xn } is strongly convergent to x̄.
kxn+1 − x̄k2
=
kαn γ f (xn ) + βn xn + ((1 − βn )I − αn A)Λn − x̄k2
=
kαn (γ f (xn ) − Ax̄) + βn (xn − x̄) + ((1 − βn )I − αn A)(Λn − x̄)k2
=
kβn (xn − x̄) + ((1 − βn )I − αn A)(Λn − x̄)k2 + α2n kγ f (xn ) − Ax̄k2
+2αn βn hxn − x̄, γ f (xn ) − Ax̄i
+2αn h((1 − βn )I − αn A)(Λn − x̄), γ f (xn ) − Ax̄i
≤
{(1 − βn − αn λ)kΛn − x̄k + βn kxn − x̄k}2 + α2n kγ f (xn ) − Ax̄k2
+2αn βn γhxn − x̄, f (xn ) − f (x̄)i + 2αn βn hxn − x̄, γ f (x̄) − Ax̄i
+2(1 − βn )γαn hΛn − x̄, f (xn ) − f (x̄)i
+2(1 − βn )αn hΛn − x̄, γ f (x̄) − Ax̄i − 2α2n hA(Λn − x̄), γ f (xn ) − Ax̄i.
(22)
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1407
Consequently
kxn+1 − x̄k2
≤
{(1 − αn λ)2 + 2ραn βn γ + 2ρ(1 − βn )γαn }kxn − x̄k2
+α2n kγ f (xn ) − Ax̄k2 + 2αn βn hxn − x̄, γ f (x̄) − Ax̄i
+2αn (1 − βn )hΛn − x̄, γ f (x̄) − Ax̄i − 2α2n hA(Λn − x̄), γ f (xn ) − Ax̄i
≤
(1 − 2αn (λ − ργ))kxn − x̄k2 + λ2 α2n kxn − x̄k2 + α2n kγ f (xn ) − Ax̄k2
+2αn βn hxn − x̄, γ f (x̄) − Ax̄i + 2αn (1 − βn )hΛn − x̄, γ f (x̄) − Ax̄i
+2α2n kA(Λn − x̄)kkγ f (xn ) − Ax̄k
=
(1 − 2αn (λ − ργ))kxn − x̄k2 + αn {αn (λ2 kxn − x̄k2
+kγ f (xn ) − Ax̄k2 + 2kA(Λn − x̄)kkγ f (xn ) − Ax̄k
+2βn hxn − x̄, γ f (x̄) − Ax̄i + 2(1 − βn )hΛn − x̄, γ f (x̄) − Ax̄i}.
Since {xn }, { f (xn )} and {Λn } are bounded, one can take a constant Γ > 0 such that
Γ ≥ λ2 kxn − x̄k2 + kγ f (xn ) − Ax̄k2 + 2kA(Λn − x̄)kkγ f (xn ) − Ax̄k.
Let
Ξn = 2βn hxn − x̄, γ f (x̄) − Ax̄i + 2(1 − βn )hΛn − x̄, γ f (x̄) − Ax̄i + Γαn .
Hence
kxn+1 − x̄k2 ≤ (1 − 2αn (λ − ργ))kxn − x̄k2 + αn Ξn .
(23)
With respect to (22), lim supΞn ≤ 0 and so all conditions of Lemma 2.5 are satisfied for (23). Consequently,
n→∞
the sequence {xn } is strongly convergent to x̄.
As a result, by intuition from [12], the following mean ergodic theorem for a nonexpansive mapping in
Hilbert space is proved.
Corollary 3.3. Suppose all assumptions of Theorem 3.1 are holds. Let {Ti } be a family of nonexpansive mappings on
T
(i)
C1 for all i = 1, 2, . . . , k such that ki=1 F(Ti ) ∩ GEPS(Fi , Ψi ) , ∅. Let {xn } and {un } ⊂ C1 be sequences generated in
the following manner:































x1 ∈ H choosen arbitrary,
(1) (1)
(1)
(1)
F1 (un , y) + hΨ1 xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C1 ,
(2)
(2) (2)
(2)
F2 (un , y) + hΨ2 xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C1 ,
..
.
(k)
(k)
(k) (k)
Fk (un , y) + hΨk xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C1 ,
P
(i)
ωn = 1k ki=1 un ,
1 Pn
i
xn+1 = αn γ f (xn ) + βn xn + ((1 − βn )I − αn A) n+1
i=0 PC2 T ωn ,
where {αn }, {βn } are the sequences in [0, 1] and {rn } ⊂ (0, ∞) is a real sequence. Suppose the following conditions are
satisfied:
(C1) lim αn = 0,
n→∞
∞
X
αn = ∞;
n=1
(C2) 0 < lim infβn ≤ lim supβn < 1;
n→∞
n→∞
(C3) lim |rn+1 − rn | = 0 and 0 < b < rn < a < 2µi for i ∈ {1, 2, . . . , k}.
n→∞
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1408
Then the sequence {xn } is strongly convergent to a point x̄, where
x̄ = PTk
i=1
F(Ti )∩GEPS(Fi ,Ψi ) (I
− A + γ f )(x̄)
, is the unique solution of the variational inequality
h(A − γ f )x̄, x̄ − xi ≤ 0, ∀x ∈
k
\
F(Ti ) ∩ GEPS(Fi , Ψi ).
i=1
4. Application
If T(s) = T for all s > 0 and C1 = C2 = C, then we have the following corollary.
Corollary 4.1. Let H be a real Hilbert space, F1 , F2 , . . . , Fk be bifunctions from C × C to R satisfying (A1) − (A4),
Ψ1 , Ψ2 , . . . , Ψk be µi −inverse strongly monotone mapping on H, A be a strongly positive linear bounded operator on
H with coefficient λ and 0 < γ < λρ , f : H → H be a ρ−contraction. Suppose that T be a nonexpansive mapping on
T
C such that ki=1 F(T) ∩ GEPS(Fi , Ψi ) , ∅. Define the sequence {xn } as follows.































(i)
x1 ∈ H, and un ∈ C,
(1)
(1)
(1) (1)
F1 (un , y) + hΨ1 xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C,
(2)
(2) (2)
(2)
F2 (un , y) + hΨ2 xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C,
..
.
(k)
(k)
(k) (k)
Fk (un , y) + hΨk xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C,
P
(i)
ωn = 1k ki=1 un ,
xn+1 = αn γ f (xn ) + βn xn + ((1 − βn )I − αn A)TPC ωn ds,
where {αn }, {βn } ⊂ [0, 1] and {rn } ⊂ (0, ∞) are the sequences satisfying the conditions (C1) − (C3) in Theorem 3.2.
T
Then the sequence {xn } converges strongly to a point x̄, where x̄ ∈ ki=1 F(T) ∩ GEPS(Fi , Ψi ) solves the variational
inequality h(A − γ f )x̄, x̄ − xi ≤ 0.
We apply Theorem 3.2 for finding a common fixed point of a nonexpansive semigroup mappings and
strictly pseudo-contractive mapping and inverse strongly monotone mapping.
Recall that, a mapping T : C → C is called strictly pseudo-contractive if there exists k with 0 ≤ k ≤ 1 such
that
kTx − Tyk2 ≤ kx − yk2 + kk(I − T)x − (I − T)yk2 , for all x, y ∈ C.
If k = 0, then T is nonexpansive. Put J = I − T, where T : C → C is a strictly pseudo-contractive mapping. J
−1
is 1−k
2 -inverse strongly monotone and J (0) = F(T). Indeed, for all x, y ∈ C we have
k(I − J)x − (I − J)yk2 ≤ kx − yk2 + kkJx − Jyk2 .
Also
k(I − J)x − (I − J)yk2 ≤ kx − yk2 + kJx − Jyk2 − 2hx − y, Jx − Jyi.
So, we have
hx − y, Jx − Jyi ≥
1−k
kJx − Jyk2 .
2
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1409
Corollary 4.2. Let H be a real Hilbert space, F1 , F2 , . . . , Fk be bifunctions from C × C to R satisfying (A1) − (A4),
Ψ1 , Ψ2 , . . . , Ψk be µi −inverse strongly monotone mapping on H, A be a strongly positive linear bounded operator
on H with coefficient λ and 0 < γ < λρ , f : H → H be a ρ−contraction. Suppose that T : C → H be a k-strictly
T
pseudo-contractive mapping for some 0 ≤ k < 1 such that ki=1 F(T) ∩ GEPS(Fi , Ψi ) , ∅. Define the sequence {xn }
as follows.































(i)
x1 ∈ H, and un ∈ C,
(1) (1)
(1)
(1)
F1 (un , y) + hΨ1 xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C,
(2)
(2)
(2) (2)
F2 (un , y) + hΨ2 xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C,
..
.
(k)
(k)
(k) (k)
Fk (un , y) + hΨk xn , y − un i + r1n hy − un , un − xn i ≥ 0, for all y ∈ C,
P
(i)
ωn = 1k ki=1 un ,
xn+1 = αn γ f (xn ) + βn xn + ((1 − βn )I − αn A)PC Jωn ds,
where J : C → H is a mapping defined by Jx = kx + (1 − k)Tx and {αn }, {βn } ⊂ [0, 1] and {rn } ⊂ (0, ∞) are the
sequences satisfying the conditions (C1) − (C3) in Theorem 3.1. Then the sequence {xn } is strongly convergent to a
T
point x̄, where x̄ ∈ ki=1 F(T) ∩ GEPS(Fi , Ψi ) solves the variational inequality h(A − γ f )x̄, x̄ − xi ≤ 0.
Proof. Note that S : C → H is a nonexpansinve mapping and F(T) = F(S). By Lemma 2.3 in [41] and Lemma
2.2 in [37], we have PC S : C2 → C is a nonexpansive mapping and F(PC S) = F(S) = F(S). Therefore, the
result follows from Corollary 4.1.
5. Numerical Examples
In this section, we show numerical examples which grantee the main theorem. The programming has
been provided with Matlab according to the following algorithm.
Example 5.1. Suppose that H = R, C1 = [−1, 1], C2 = [0, 1] and
F1 (x, y) = −3x2 + xy + 2y2 , F2 (x, y) = −4x2 + xy + 3y2 , F3 (x, y) = −5x2 + xy + 4y2 .
x
x
x
. Suppose that A = 10
, f (x) = 10
with coefficient γ = 1
Also, we consider Ψ1 (x) = x, Ψ2 = 2x and Ψ3 (x) = 10
−s
and T(s) = e is a nonexpansive semigroup on C2 . It is easy to check that Ψ1 , Ψ2 , Ψ3 , A, f and T(s) satisfy
all conditions in Theorem 3.1. For each y ∈ C1 there exists z ∈ C1 such that
1
F1 (z, y) + hΨ1 x, y − zi + hy − z, z − xi ≥ 0
r
1
⇔ −3z2 + zy + 2y2 + x(y − z) + (y − z)(z − x) ≥ 0
r
⇔ 2ry2 + ((r + 1)z − (r − 1)x)y − 3rz2 − xzr − z2 + zx ≥ 0.
Set G(y) = 2ry2 + ((r + 1)z − (r − 1)x)y − 3rz2 − xzr − z2 + zx. Then G(y) is a quadratic function of y with
coefficients a = 2r, b = (r + 1)z − (r − 1)x and c = −3rz2 − xzr − z2 + zx. So
∆
= b2 − 4ac
=
[(r + 1)z − (r − 1)x]2 − 8r(−3rz2 − xzr − z2 + zx)
=
x2 (r − 1)2 + 2zx(r − 1)(5r + 1) + z2 (5r + 1)2
=
[(x(r − 1) + z(5r + 1))]2 .
Since G(y) ≥ 0 for all y ∈ C1 , if and only if ∆ = [(x(r − 1) + z(5r + 1))]2 ≤ 0. Therefore, z =
(1)
yields Tr(1) = un =
n
1 − rn
xn .
5rn + 1
1−r
x, which
5r + 1
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1410
By the same argument, for F2 and F3 , one can conclude
(2)
1 − 2rn
xn ,
7rn + 1
10 − rn
=
xn .
90rn + 10
Tr(2)
= un =
Tr(3)
= un
n
n
(3)
Then
(1)
ωn
=
=
By choosing rn =
sequence {xn }
xn+1 =
(2)
(3)
un + un + un
3
1 1 − rn
1 − 2rn
10 − rn
[
+
+
]xn .
3 5rn + 1 7rn + 1 90rn + 10
9
2n − 1
n+8
, tn = n, and αn =
, βn =
, we have the following algorithm for the
n
10n
10n − 9
!
200n2 − 10n − 81
800n2 − 890n + 81 1 − e−n
xn +
ωn .
n
100n2 − 90n
1000n2 − 900n
Choose x1 = 1000. By using MATLAB software, we obtain the following table and figure of the result.
n
1
2
3
4
5
6
7
8
9
10
xn
1000
1098.409511
305.4917461
73.2742215
16.71274356
3.722532274
0.8179356809
0.1781302867
0.03854703081
0.008300951942
n
11
12
13
14
15
16
17
18
19
20
xn
0.001780587335
0.0003806958425
0.00008116570943
0.00001726216226
0.000003663196562
0.0000007758152752
0.0000001640073634
0.00000003461280635
0.000000007293416021
0.000000001534585156
n
21
22
23
24
25
26
27
28
29
30
xn
0.0000000003224455074
0.00000000006766467135
0.00000000001418205647
0.000000000002969041259
0.0000000000006208956961
0.000000000000007698339387
0.0000000000001297089212
0.00000000000002707018951
0.000000000000005644205174
0.000000000000001175770771
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1411
Example 5.2. Theorem 3.2 can be illustrated by the following numerical example where the parameters are given as
follows:
H = [−10, 10]
,
Ψ1 (x) = x
,
1
2n
T(s) = e−s
αn =
C1 = [−1, 1], C2 = [0, 1], A = I, f (x) =
x
5
x
, Ψ4 (x) = 3x, Ψ5 (x) = 4x
10
n
n+1
, βn =
, rn =
2n + 1
n
, γ = 1, tn = n.
Ψ2 = 2x, Ψ3 (x) =
Moreover,
F1 (x, y) = −3x2 + xy + 2y2
,
F4 (x, y) = −6x2 + xy + 5y2
F2 (x, y) = −4x2 + xy + 3y2
,
F5 (x, y) = −8x2 + xy + 7y2
F3 (x, y) = −5x2 + xy + 4y2 .
(i)
By the same argument in Example 5.1, we compute un for i = 1, 2, 3, 4, 5 as follows:
(1)
=
(2)
=
(3)
=
(4)
=
(3)
=
Tr(1) = un
n
Tr(2) = un
n
Tr(3) = un
n
Tr(4) = un
n
Tr(5) = un
n
1 − rn
xn ,
5rn + 1
1 − 2rn
xn ,
7rn + 1
10 − rn
xn ,
90rn + 10
1 − 3rn
xn ,
11rn + 1
1 − 4rn
xn .
15rn + 1
Then
(1)
ωn
=
=
(2)
(5)
un + un + . . . + un
5
1 1 − rn
1 − 2rn
10 − rn
1 − 3rn
1 − 4rn
[
+
+
+
+
]xn .
5 5rn + 1 7rn + 1 90rn + 10 11rn + 1 15rn + 1
Choose x1 = 10. The detailed results of proposed iterative in Theorem 3.2 are presented in the following
table and figure.
n
1
2
3
4
5
6
7
8
9
10
xn
10
4.187523857
1.810011289
0.8111798915
0.37213169
0.173392393
0.08167872103
0.03878661445
0.0185321898
0.008897703796
n
11
12
13
14
15
16
17
18
19
20
xn
0.004288752976
0.002073897432
0.001005587742
0.0004887121222
0.0002379837774
0.0001160888527
0.00005671409128
0.00002774424594
0.00001358854452
0.000006662503243
n
21
22
23
24
25
26
27
28
29
30
xn
0.000003269796382
0.000001606140362
0.0000007895721035
0.0000003884328747
0.000000191218571
0.00000009419138076
0.00000004642361668
0.00000002289261628
0.0000000112944317
0.00000000557482625
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
Example 5.3. Let
H = [−10, 10]
1412
x
x
, f (x) =
10
10
x
Ψ3 (x) = x, Ψ4 (x) = 2x, Ψ5 (x) =
10
Ψ7 (x) = 4x
n
n+1
βn =
, rn =
3n + 1
n
γ = 1, tn = n.
, C1 = [0, 1], C2 = [−1, 1], A =
Ψ1 (x) = Ψ2 (x) = 0
,
Ψ6 (x) = 3x
1
αn =
n
T(s) = e−s
,
,
,
Moreover,
F1 (x, y) = (1 − x2 )(x − y) ,
F5 (x, y) = −5x2 + xy + 4y2 ;
F2 (x, y) = −x2 (x − y)2
, F6 (x, y) = −6x2 + xy + 5y2 ;
F3 (x, y) = −3x2 + xy + 2y2
, F7 (x, y) = −8x2 + xy + 7y2 ;
F4 (x, y) = −4x2 + xy + 3y2 .
(i)
By the same argument in Example 5.1, we compute un for i = 1, 2 as follows: For any y ∈ C1 and r > 0, we have
1
F(z, y) + hy − z, z − xi ≥ 0 ⇔ (y − z)(rz2 + z − r − x) ≥ 0.
r
√
−1+ 1+4r(r+x)
2
This implies that rz + z − r − x = 0. Therefore, z =
which yields Tr(1) =
2r
n
Also, we have
√
−1+
1+4rn (rn +xn )
.
2rn
1
F(z, y) + hy − z, z − xi ≥ 0 ⇔ −rz2 y2 + (2rz3 + z − x)y − rz4 − z2 + zx ≥ 0.
r
Set J(y) = −rz2 y2 + (2rz3 + z − x)y − rz4 − z2 + zx. Then J(y) is a quadratic function of y with coefficients
a = −rz2 , b = 2rz3 + z − x and c = −rz4 − z2 + zx. So
∆
= [2rz3 + z − x]2 + 4rz2 (−rz4 − z2 + zx)
= (z − x)2 .
Since J(y) ≥ 0 for all y ∈ H, if and only if ∆ = (z − x)2 = 0. Therefore, z = Tr(2) = x.
n
Then
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
Tr(1) =
n
−1 +
p
1 + 4rn (rn + xn )
,
2rn
(1)
un
=
(2)
=
xn ,
(3)
=
(4)
=
(5)
=
(6)
=
(7)
=
1 − rn
xn ,
5rn + 1
1 − 2rn
xn ,
7rn + 1
10 − rn
xn ,
90rn + 10
1 − 3rn
xn ,
11rn + 1
1 − 4rn
xn .
15rn + 1
Tr(2) = un
n
Tr(3) = un
n
Tr(4) = un
n
Tr(5) = un
n
Tr(6) = un
n
Tr(7) = un
n
1413
Then
(1)
ωn
=
(2)
(7)
un + un + . . . + un
.
7
We have
xn+1 =
!
20n2 + 7n − 1 1 − e−n
10n2 + 3n + 1
xn +
ωn .
n
30n2 − 10n
30n2 + 10n
Choose x1 = 10. The detailed results of proposed iterative in Theorem 3.2 are presented in the following
table and figure.
n
1
2
3
4
5
6
7
8
9
10
xn
10
7.414282455
5.25738677
3.188403646
1.720431319
0.8702902283
0.4279164132
0.2082805371
0.1011375703
0.04914122712
n
11
12
13
14
15
16
17
18
19
20
xn
0.0239156544
0.01166047726
0.005695329069
0.002786256225
0.001365047089
0.000669619601
0.0003288520367
0.0001616632095
0.00007954524563
0.00003917150569
n
21
22
23
24
25
26
27
28
29
30
xn
0.00001930391707
0.000009519419651
0.000004697209017
0.000002319056949
0.000001145527389
0.0000005661151488
0.0000002798942327
0.000000138439269
0.00000006849968096
0.00000003390554052
H. R. Sahebi, A. Razani / Filomat 28:7 (2014), 1393–1415
1414
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