Volume 2009, Article ID 719360,18pages doi:10.1155/2009/719360
Research Article
A Hybrid Iterative Scheme for Equilibrium
Problems, Variational Inequality Problems, and
Fixed Point Problems in Banach Spaces
Prasit Cholamjiak
School of Science and Technology, Naresuan University at Phayao, Phayao 56000, Thailand
Correspondence should be addressed to Prasit Cholamjiak,[email protected]
Received 5 February 2009; Accepted 10 April 2009
Recommended by Simeon Reich
The purpose of this paper is to introduce a new hybrid projection algorithm for finding a common element of the set of solutions of the equilibrium problem and the set of the variational inequality for an inverse-strongly monotone operator and the set of fixed points of relatively quasi-nonexpansive mappings in a Banach space. Then we show a strong convergence theorem. Using this result, we obtain some applications in a Banach space.
Copyrightq2009 Prasit Cholamjiak. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
LetEbe a real Banach space and letE∗be the dual ofE. LetCbe a closed convex subset ofE. LetA :C → E∗be an operator. The classical variational inequality problem forAis to find
x∈Csuch that
Ax, y −x ≥0, ∀y∈C. 1.1
The set of solutions of1.1is denoted by V IA, C. Such a problem is connected with the convex minimization problem, the complementarity, the problem of finding a pointx ∈ E satisfying 0Ax, and so on. First, we recall that
1an operatorAis calledmonotoneif
2an operatorAis calledα-inverse-strongly monotoneif there exists a constant α > 0 with
Ax−Ay, x−y ≥αAx−Ay2, ∀x, y∈C. 1.3
Assume that
C1Aisα-inverse-strongly monotone,
C2V IA, C/∅,
C3 Ay ≤ Ay−Au for ally∈Candu∈V IA, C.
Iiduka and Takahashi1introduced the following algorithm for finding a solution of the variational inequality for an operator A that satisfies conditions C1–C3 in a 2-uniformly convex and 2-uniformly smooth Banach spaceE. For an initial pointx1 x ∈ C, define a sequence{xn}by
xn1 ΠCJ−1Jxn−λnAxn, ∀n≥1, 1.4
where J is the duality mapping on E, and ΠC is the generalized projection from E onto
C. Assume that λn ∈ a, b for some a, b with 0 < a < b < c2α/2 where 1/c is thep
-uniformly convexity constant ofE. They proved that ifJis weakly sequentially continuous, then the sequence {xn} converges weakly to some element z in V IA, C where z
limn→ ∞ΠV IA,Cxn.
The problem of finding a common element of the set of the variational inequalities for monotone mappings in the framework of Hilbert spaces and Banach spaces has been intensively studied by many authors; see, for instance,2–4and the references cited therein. Letf :C×C → Rbe a bifunction. The equilibrium problem forfis to findx∈Csuch that
fx, y ≥0, ∀y∈C. 1.5
The set of solutions of1.5is denoted byEPf.
For solving the equilibrium problem, let us assume that a bifunctionf satisfies the following conditions:
A1fx, x 0 for allx∈C;
A2fis monotone, that is,fx, y fy, x≤0 for allx, y∈C;
A3for allx, y, z∈C,
lim sup
t↓0
ftz 1−tx, y≤fx, y; 1.6
Recently, Takahashi and Zembayashi 5, introduced the following iterative scheme which is called the shrinking projection method:
x0x∈C, C0C,
ynJ−1αnJxn 1−αnJTxn,
un∈C such that fun, yr1
ny−un, Jun−Jyn ≥0, ∀y∈C,
Cn1
z∈Cn:φz, un≤φz, xn,
xn1 ΠCn1x0, ∀n≥0,
1.7
whereJis the duality mapping onEandΠCis the generalized projection fromEontoC. They
proved that the sequence {xn}converges strongly to q ΠFT∩EPfx0 under appropriate conditions.
Very recently, Qin et al.6extend the iteration process1.7from a single relatively nonexpansive mapping to two relatively quasi-nonexpansive mappings:
x0∈E, chosen arbitrarily,
C1C, x1 ΠC1x0,
ynJ−1αnJxnβnJTxnγnJSxn,
un∈C such that fun, yr1
ny−un, Jun−Jyn ≥0, ∀y∈C,
Cn1
z∈Cn:φz, un≤φz, xn,
xn1 ΠCn1x0.
1.8
Under suitable conditions over {αn},{βn}, and {γn}, they obtain that the sequence {xn}
generated by1.8converges strongly toq ΠFT∩FS∩EPfx0.
The problem of finding a common element of the set of fixed points and the set of solutions of an equilibrium problem in the framework of Hilbert spaces and Banach spaces has been studied by many authors; see5,7–16.
2. Preliminaries
LetEbe a real Banach space and letU{x∈E: x 1}be the unit sphere ofE. A Banach spaceEis said to bestrictly convexif for anyx, y∈U,
x /y impliesxy 2
<1. 2.1
It is also said to beuniformly convexif for eachε∈0,2, there existsδ >0 such that for any
x, y∈U,
x−y≥ε impliesxy 2
<1−δ. 2.2
It is known that a uniformly convex Banach space is reflexive and strictly convex; and we define a functionδ:0,2 → 0,1called themodulus of convexityofEas follows:
δε inf
1−xy 2
: x, y∈E, x y1, x−y≥ε . 2.3
ThenE is uniformly convex if and only ifδε > 0 for all ε ∈ 0,2. Letp be a fixed real number withp≥2. A Banach spaceEis said to bep-uniformly convexif there exists a constant
c >0 such thatδε≥cεpfor allε∈0,2; see17–19for more details. A Banach spaceEis said to besmoothif the limit
lim
t→0
xty− x
t 2.4
exists for all x, y ∈ U. It is also said to be uniformly smooth if the limit 2.4 is attained uniformly forx, y ∈ U. One should note that no Banach space isp-uniformly convex for 1 < p < 2; see19. It is well known that a Hilbert space is 2-uniformly convex, uniformly smooth. For eachp >1, thegeneralized duality mappingJp:E → 2E∗is defined by
Jpx
x∗∈E∗:x, x∗ x p, x∗ x p−1 2.5
for allx∈E. In particular,J J2is called thenormalized duality mapping. IfEis a Hilbert space, thenJ I, whereI is the identity mapping. It is also known that ifEis uniformly smooth, thenJis uniformly norm-to-norm continuous on each bounded subset ofE. See20,21for more details.
Lemma 2.1See18,22. Letpbe a given real number withp ≥ 2 andEap-uniformly convex Banach space. Then, for allx, y∈E,jx∈Jpxandjy ∈Jpy,
x−y, jx−jy ≥ c p
2p−2px−y
p,
2.6
LetEbe a smooth Banach space. The functionφ:E×E → Ris defined by
φx, y x 2−2x, Jyy2
2.7
for allx, y∈E. In a Hilbert spaceH, we haveφx, y x−y 2for allx, y∈H.
Recall that a mappingT :C → Cis called nonexpansive if Tx−Ty ≤ x−y for all
x, y∈Cand relatively nonexpansive ifTsatisfies the following conditions:
1FT/∅, whereFTis the set of fixed points ofT;
2φp, Tx≤φp, xfor allp∈FTandx∈C;
3FT FT, whereFTis the set of all asymptotic fixed points ofT;
see10,23,24for more details.
Tis said to be relatively quasi-nonexpansive ifT satisfies the conditions1and2. It is easy to see that the class of relatively quasi-nonexpansive mappings is more general than the class of relatively nonexpansive mappings9,25,26.
We give some examples which are closed relatively quasi-nonexpansive; see6.
Example 2.2. LetEbe a uniformly smooth and strictly convex Banach space andA⊂E×E∗
be a maximal monotone mapping such that its zero setA−10/∅. Then,Jr JrA−1Jis a closed relatively quasi-nonexpansive mapping fromEontoDAandFJr A−10.
Example 2.3. Let ΠC be the generalized projection from a smooth, strictly convex, and
reflexive Banach spaceEonto a nonempty closed convex subsetCofE. Then,ΠCis a closed
relatively quasi-nonexpansive mapping withFΠC C.
Lemma 2.4 Kamimura and Takahashi27. LetEbe a uniformly convex and smooth Banach space and let{xn},{yn}be two sequences ofE. Ifφxn, yn → 0and either{xn}or{yn}is bounded,
then xn−yn → 0asn → ∞.
LetCbe a nonempty closed convex subset ofE. IfEis reflexive, strictly convex and smooth, then there existsx0 ∈ Csuch thatφx0, x minφy, xforx ∈ Eandy ∈C. The generalized projectionΠC :E → Cdefined byΠCx x0. The existence and uniqueness of the operatorΠCfollows from the properties of the functionalφand strict monotonicity of the
duality mappingJ; for instance, see20,27–30. In a Hilbert space,ΠCis coincident with the
metric projection.
Lemma 2.5Alber28. LetCbe a nonempty closed convex subset of a smooth Banach spaceEand
x∈E. Thenx0 ΠCxif and only ifx0−y, Jx−Jx0 ≥0for ally∈C.
Lemma 2.6Alber28. LetCbe a nonempty closed convex subset of a reflexive, strictly convex and smooth Banach spaceEand letx∈E. Then
φy,ΠCxφΠCx, x≤φy, x, ∀y∈C. 2.8
Lemma 2.8Cho et al.31. LetEbe a uniformly convex Banach space and letBr0be a closed ball ofE. Then there exists a continuous strictly increasing convex functiong : 0,∞ → 0,∞
withg0 0such that
αxβyγz2≤α x 2βy2γ z 2−αβgx−y, 2.9
for allx, y, z∈Br0, andα, β, γ ∈0,1withαβγ1.
Lemma 2.9Blum and Oettli7. LetCbe a closed convex subset of a smooth, strictly convex, and reflexive Banach spaceE, letfbe a bifunction fromC×CtoRsatisfying (A1)–(A4), and letr >0
andx∈E. Then, there existsz∈Csuch that
fz, y1ry−z, Jz−Jx ≥0, ∀y∈C. 2.10
Lemma 2.10Qin et al.6. LetCbe a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach spaceE, and letfbe a bifunction fromC×CtoRsatisfying (A1)–(A4). For all
r >0andx∈E, define a mappingTr :E → Cas follows:
Trx
z∈C:fz, y1
ry−z, Jz−Jx ≥0, ∀y∈C . 2.11
Then, the following hold:
1Tr is single-valued;
2Tr is a firmly nonexpansive-type mapping [32], that is, for allx, y∈E,
Trx−Try, JTrx−JTry ≤ Trx−Try, Jx−Jy; 2.12
3FTr EPf;
4EPfis closed and convex.
Lemma 2.11Takahashi and Zembayashi 14. Let C be a closed convex subset of a smooth, strictly, and reflexive Banach spaceE, letfbe a bifucntion fromC×CtoRsatisfying (A1)–(A4), let
r >0. Then, for allx∈Eandq∈FTr,
φq, TrxφTrx, x≤φq, x. 2.13
We make use of the following mappingV studied in Alber28:
Vx, x∗ x 2−2x, x∗ x∗ 2 2.14
Lemma 2.12Alber28. LetEbe a reflexive, strictly convex, smooth Banach space and letV be as in2.14. Then
Vx, x∗ 2J−1x∗−x, y∗ ≤Vx, x∗y∗ 2.15
for allx∈Eandx∗, y∗∈E∗.
An operatorAofCintoE∗is said to behemicontinuousif for allx, y∈C, the mapping
Fof0,1intoE∗defined byFt Atx 1−tyis continuous with respect to the weak∗ topology ofE∗. We define byNCvthenormal coneforCat a pointv∈C, that is,
NCv x∗∈E∗:v−y, x∗ ≥0, ∀y∈C. 2.16
Theorem 2.13Rockafellar33. LetCbe a nonempty, closed convex subset of a Banach spaceE andAa monotone, hemicontinuous operator ofCintoE∗. LetTe ⊂E×E∗be an operator defined as follows:
Tev
⎧ ⎨ ⎩
AvNCv, v∈C;
∅, otherwise. 2.17
ThenTeis maximal monotone andTe−10V IA, C.
3. Strong Convergence Theorems
Theorem 3.1. LetEbe a2-uniformly convex, uniformly smooth Banach space, letCbe a nonempty closed convex subset ofE. Letf be a bifunction fromC×CtoRsatisfying (A1)–(A4), letAbe an operator ofCintoE∗ satisfying (C1)–(C3), and letT, Sbe two closed relatively quasi-nonexpansive mappings fromCinto itself such thatF:FT∩FS∩EPf∩V IA, C/∅. For an initial point
x0∈Ewithx1 ΠC1x0andC1C, define a sequence{xn}as follows:
zn ΠCJ−1Jxn−λnAxn,
yn J−1αnJxnβnJTxnγnJSzn,
un∈C such thatfun, yr1
ny−un, Jun−Jyn ≥0, ∀y∈C,
Cn1
z∈Cn:φz, un≤φz, xn,
xn1 ΠCn1x0, ∀n≥1,
3.1
where J is the duality mapping on E. Assume that {αn},{βn}, and {γn} are sequences in 0,1 satisfying the restrictions:
B1αnβnγn1;
B3{rn} ⊂s,∞for somes >0;
B4{λn} ⊂ a, bfor some a, b with 0 < a < b < c2α/2, where 1/c is the 2-uniformly convexity constant ofE.
Then,{xn}and{un}converge strongly toq ΠFx0.
Proof. We divide the proof into eight steps.
Step 1. Show thatΠFx0andΠCn1x0are well defined.
It is obvious thatV IA, Cis a closed convex subset ofC. ByLemma 2.7, we know that
FT∩FSis closed and convex. FromLemma 2.104, we also haveEPfis closed and convex. HenceF:FT∩FS∩EPf∩V IA, Cis a nonempty, closed, and convex subset ofC; consequently,ΠFx0is well defined.
Clearly,C1Cis closed and convex. Suppose thatCkis closed and convex fork∈N.
For allz∈Ck, we knowφz, yk≤φz, xkis equivalent to
2z, Jxk−Jyk ≤ xk 2−yk2. 3.2
So,Ck1is closed and convex. By induction,Cnis closed and convex for alln≥1. This shows thatΠCn1x0is well-defined.
Step 2. Show thatF⊂Cnfor alln∈N.
PutvnJ−1Jxn−λnAxn. First, we observe thatunTrnynfor alln≥1 andF ⊂C1
C. SupposeF ⊂Ckfork∈N. Then, for allu∈F, we know fromLemma 2.6andLemma 2.12
that
φu, zk φu,ΠCvk
≤φu, vk
φu, J−1Jx
k−λkAxk
Vu, Jxk−λkAxk
≤Vu,Jxk−λkAxk λkAxk−2
J−1Jx
k−λkAxk−u, λkAxk
Vu, Jxk−2λkvk−u, Axk
φu, xk−2λkxk−u, Axk2vk−xk,−λkAxk.
3.3
Sinceu∈V IA, Cand fromC1, we have
−2λkxk−u, Axk−2λkxk−u, Axk−Au −2λkxk−u, Au
≤ −2αλk Axk−Au 2.
FromLemma 2.1andC3, we obtain
2vk−xk,−λkAxk2J−1Jxk−λkAxk−J−1Jxk,−λkAxk
≤2J−1Jxk−λkAxk−J−1Jxk λkAxk
≤ c42JJ−1Jx
k−λkAxk−JJ−1Jxk λkAxk
c42 Jxk−λkAxk−Jxk λkAxk
c42λ2
k Axk 2
≤ c42λ2
k Axk−Au 2.
3.5
Replacing3.4and3.5into3.3, we get
φu, zk≤φu, xk 2λk
2
c2λk−α
Axk−Au 2≤φu, xk. 3.6
By the convexity of · 2, for eachu∈F⊂Ck, we obtain
φu, uk φu, Trkyk
≤φu, yk
φu, J−1α
kJxkβkJTxkγkJSzk
u 2−2α
ku, Jxk −2βku, JTxk −2γku, JSzk
αkJxkβkJTxkγkJSzk2
≤ u 2−2α
ku, Jxk −2βku, JTxk −2γku, JSzk
αk Jxk 2βk JTxk 2γk JSzk 2
αkφu, xk βkφu, Txk γkφu, Szk
≤αkφu, xk βkφu, xk γkφu, zk
≤φu, xk.
3.7
This shows thatu∈Ck1; consequently,F⊂Ck1. HenceF ⊂Cnfor alln≥1.
Step 3. Show that limn→ ∞φxn, x0exists.
Fromxn ΠCnx0andxn1 ΠCn1x0 ∈Cn1⊂Cn, we have
FromLemma 2.6, we have
φxn, x0 φΠCnx0, x0≤φu, x0−φu, xn≤φu, x0. 3.9
Combining3.8and3.9, we obtain that limn→ ∞φxn, x0exists.
Step 4. Show that{xn}is a Cauchy sequence inC.
Sincexm ΠCmx0∈Cm⊂Cnform > n, byLemma 2.6, we also have
φxm, xn φxm,ΠCnx0
≤φxm, x0−φΠCnx0, x0
φxm, x0−φxn, x0.
3.10
Takingm, n → ∞, we obtain thatφxm, xn → 0. FromLemma 2.4, we have xm−xn → 0.
Hence{xn}is a Cauchy sequence. By the completeness ofEand the closedness ofC, one can
assume thatxn → q∈Casn → ∞. Further, we obtain
lim
n→ ∞φxn1, xn 0. 3.11
Sincexn1 ΠCn1x0∈Cn1, we have
φxn1, un≤φxn1, xn−→0, 3.12
asn → ∞. ApplyingLemma 2.4to3.11and3.12, we get
lim
n→ ∞ un−xn 0. 3.13
This implies that un → q asn → ∞. Since J is uniformly norm-to-norm continuous on
bounded subsets ofE, we also obtain
lim
n→ ∞ Jun−Jxn 0. 3.14
Letr supn≥1{ xn , Txn , Szn }. From 3.6and Lemma 2.8, we know that there
exists a continuous strictly increasing convex functiong :0,∞ → 0,∞withg0 0 such that
φu, un φu, Trnyn
≤φu, yn
φu, J−1α
nJxnβnJTxnγnJSzn
u 2−2α
nu, Jxn −2βnu, JTxn −2γnu, JSzn
αnJxnβnJTxnγnJSzn2
≤ u 2−2α
nu, Jxn −2βnu, JTxn −2γnu, JSzn
αn Jxn 2βn JTxn 2γn JSzn 2
−αnβng Jxn−JTxn
αnφu, xn βnφu, Txn γnφu, Szn
−αnβng Jxn−JTxn
≤φu, xn 2γnλn
2
c2λn−α
Axn−Au 2
−αnβng Jxn−JTxn .
3.15
This implies that
αnβng Jxn−JTxn ≤φu, xn−φu, un
xn 2− un 2−2u, Jxn−Jun
≤ xn−un xn un 2 u Jxn−Jun .
3.16
It follows from3.13,3.14, andB2that
lim
n→ ∞g Jxn−JTxn 0. 3.17
By the property ofg, we also obtain that
lim
n→ ∞ Jxn−JTxn 0. 3.18
SinceJis uniformly norm-to-norm continuous on bounded sets, so isJ−1. Then
lim
n→ ∞ xn−Txn nlim→ ∞
J−1Jx
In the same manner, we can show that
lim
n→ ∞ xn−Szn 0. 3.20
Again, by3.15, we have
2a
α− 2 c2b
Axn−Au 2 ≤ γ1 n
φu, xn−φu, un, 3.21
which yields that
lim
n→ ∞ Axn−Au 0. 3.22
FromLemma 2.6,Lemma 2.12, and3.5, we have
φxn, zn φxn,ΠCvn
≤φxn, vn
φxn, J−1Jxn−λnAxn
Vxn, Jxn−λnAxn
≤Vxn,Jxn−λnAxn λnAxn
−2J−1Jxn−λnAxn−xn, λnAxn
φxn, xn 2vn−xn,−λnAxn
2vn−xn,−λnAxn ≤ c42b2 Axn−Au 2.
3.23
It follows fromLemma 2.4and3.22that
lim
n→ ∞ xn−zn 0. 3.24
Hencezn → qasn → ∞and
lim
n→ ∞ Jxn−Jzn 0. 3.25
Combining3.20and3.24, we also obtain
lim
n→ ∞ Szn−zn 0. 3.26
Step 6. Show thatxn → q∈EPf.
From3.15, we see
φu, yn≤φu, xn. 3.27
From3.16, we observe
lim
n→ ∞φu, xn−φu, un 0. 3.28
Note thatunTrnyn. From3.27andLemma 2.11, we have
φun, ynφTrnyn, yn
≤φu, yn−φu, Trnyn
≤φu, xn−φu, Trnyn
φu, xn−φu, un.
3.29
From3.28, we get limn→ ∞φun, yn 0. ByLemma 2.4, we obtain
un−yn−→0 3.30
asn → ∞. Sincern≥s, we have
Jun−Jyn
rn −→0 3.31
asn → ∞. FromunTrnynwe have
fun, yr1
ny−un, Jun−Jyn ≥0, ∀y∈C. 3.32
ByA2, we have
y−un
Jun−Jyn
rn ≥
1
rny−un, Jun−Jyn
≥ −fun, y
≥fy, un, ∀y∈C.
3.33
FromA4andun → q, we getfy, q ≤ 0 for ally ∈ C. For 0 < t < 1 andy ∈ C. Define
yt ty 1−tq, thenyt ∈ C, which implies thatfyt, q ≤ 0. FromA1, we obtain that
0 fyt, yt≤ tfyt, y 1−tfyt, q ≤ tfyt, y. Thus,fyt, y ≥ 0. FromA3, we have
Step 7. Show thatxn → q∈V IA, C.
DefineTe ⊂ E×E∗ be as in 2.17. ByTheorem 2.13, Te is maximal monotone and
T−1
e 0V IA, C. Letv, w∈GTe. Sincew ∈TevAvNCv, we getw−Av∈NCv.
Fromzn∈C, we have
v−zn, w−Av ≥0. 3.34
On the other hand, sincezn ΠCJ−1Jxn−λnAxn. Then, byLemma 2.5, we havev−zn, Jzn−
Jxn−λnAxn ≥0 and thus
v−zn,Jxnλ−Jzn
n −Axn
≤0. 3.35
It follows from3.34and3.35that
v−zn, w ≥ v−zn, Av
≥ v−zn, Av
v−zn,Jxnλ−Jzn
n −Axn
v−zn, Av−Axn
v−zn,Jxnλ−Jzn n
v−zn, Av−Aznv−zn, Azn−Axn
v−zn,Jxnλ−Jzn n
≥ − v−zn zn−αxn − v−zn Jxn−aJzn
≥ −M
z
n−xn
α
Jxn−Jzn
a
,
3.36
whereM supn≥1{ v−zn }. By taking the limit asn → ∞and from3.24and3.25, we
obtainv−q, w ≥0. By the maximality ofTe, we haveq∈Te−10 and henceq∈V IA, C. Step 8. Show thatq ΠFx0.
Fromxn ΠCnx0, we have
Jx0−Jxn, xn−z ≥0, ∀z∈Cn. 3.37
SinceF ⊂Cn, we also have
Jx0−Jxn, xn−u ≥0, ∀u∈F. 3.38
By taking limit in3.38, we obtain that
ByLemma 2.5, we can conclude thatq ΠFx0. Furthermore, it is easy to see thatun → qas
n → ∞. This completes the proof.
As a direct consequence ofTheorem 3.1, we obtain the following results.
Corollary 3.2. LetEbe a2-uniformly convex and uniformly smooth Banach space, and let Cbe a nonempty closed convex subset ofE. Letfbe a bifunction fromC×CtoRsatisfying (A1)–(A4) and letTbe a closed relatively quasi-nonexpansive mapping fromCinto itself such thatFT∩EPf/∅. Assume that{αn} ⊂ 0,1satisfieslim infn→ ∞αn1−αn > 0and{rn} ⊂ s,∞for somes > 0. Then the sequence{xn}generated by1.7converges strongly toq ΠFT∩EPfx0.
Proof. PuttingST andA≡0 inTheorem 3.1, we obtain the result.
Remark 3.3. IfA ≡ 0 inTheorem 3.1, thenTheorem 3.1reduces to Theorem 3.1 of Qin et al.
6.
Remark 3.4. Corollary 3.2 improves Theorem 3.1 of Takahashi and Zembayashi 5 from
the class of relatively nonexpansive mappings to the class of relatively quasi-nonexpansive mappings, that is, we relax the strong restriction: FT FT. Further, the algorithm in Corollary 3.2is also simpler to compute than the one given in14.
4. Applications
Next, we consider the problem of finding a zero point of an inverse-strongly monotone operator ofEintoE∗. Assume thatAsatisfies the conditions:
D1Aisα-inverse-strongly monotone,
D2A−10{u∈E: Au0}/∅.
Theorem 4.1. LetEbe a2-uniformly convex, uniformly smooth Banach space. Letf be a bifunction fromE×EtoRsatisfying (A1)–(A4), letAbe an operator ofEintoE∗ satisfying (D1) and (D2), and letT, Sbe two closed relatively quasi-nonexpansive mappings fromEinto itself such thatF :
FT∩FS∩EPf∩A−10/∅. For an initial pointx
0∈Ewithx1 ΠC1x0andC1E, define a
sequence{xn}as follows:
znJ−1Jxn−λnAxn,
yn J−1αnJxnβnJTxnγnJSzn,
un∈E such thatfun, yr1
ny−un, Jun−Jyn ≥0, ∀y∈E,
Cn1
z∈Cn:φz, un≤φz, xn,
xn1 ΠCn1x0, ∀n≥1,
4.1
where J is the duality mapping on E. Assume that {αn},{βn}, and {γn} are sequences in 0,1 satisfying the conditions (B1)–(B4) ofTheorem 3.1.
Proof. PuttingC EinTheorem 3.1, we haveΠE I. We also haveV IA, E A−10 and
then the conditionC3ofTheorem 3.1holds for ally ∈Eandu∈ A−10. So, we obtain the result.
LetKbe a nonempty, closed convex cone inE,Aan operator ofKintoE∗. We define itspolarinE∗to be the set
K∗y∗∈E∗:x, y∗ ≥0, ∀x∈K. 4.2
Then the elementu∈Kis called a solution of thecomplementarity problemif
Au∈K∗, u, Au0. 4.3
The set of solutions of the complementarity problem is denoted byCK, A. Assume thatAis an operator satisfying the conditions:
E1Aisα-inverse-strongly monotone,
E2CK, A/∅,
E3 Ay ≤ Ay−Au for ally∈Kandu∈CK, A.
Theorem 4.2. LetEbe a2-uniformly convex, uniformly smooth Banach space, andK a nonempty, closed convex cone inE. Letf be a bifunction fromK×KtoR satisfying (A1)–(A4), letAbe an operator ofKintoE∗ satisfying (E1)–(E3), and letT, Sbe two closed relatively quasi-nonexpansive mappings fromKinto itself such thatF:FT∩FS∩EPf∩CK, A/∅. For an initial point
x0∈Ewithx1 ΠC1x0andC1K, define a sequence{xn}as follows:
zn ΠKJ−1Jxn−λnAxn,
yn J−1αnJxnβnJTxnγnJSzn,
un∈K such thatfun, yr1
ny−un, Jun−Jyn ≥0, ∀y∈K,
Cn1
z∈Cn:φz, un≤φz, xn,
xn1 ΠCn1x0, ∀n≥1,
4.4
where J is the duality mapping on E. Assume that {αn},{βn} and {γn} are sequences in 0,1 satisfying the conditions (B1)–(B4) ofTheorem 3.1.
Then,{xn}and{un}converge strongly toq ΠFx0.
Proof. From20, Lemma 7.1.1, we haveV IK, A CK, A. Hence, we obtain the result.
Acknowledgments
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