Volume 2009, Article ID 896252,34pages doi:10.1155/2009/896252
Research Article
Iterative Schemes for Generalized Equilibrium
Problem and Two Maximal Monotone Operators
L. C. Zeng,
1, 2Y. C. Lin,
3and J. C. Yao
41Department of Mathematics, Shanghai Normal University, Shanghai 200234, China 2Science Computing Key Laboratory of Shanghai Universities, Shanghai 200234, China
3Department of Occupational Safety and Health, China Medical University, Taichung 404, Taiwan 4Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 804, Taiwan
Correspondence should be addressed to J. C. Yao,[email protected]
Received 20 July 2009; Accepted 27 October 2009
Recommended by Yeol Je Cho
The purpose of this paper is to introduce and study two new hybrid proximal-point algorithms for finding a common element of the set of solutions to a generalized equilibrium problem and the sets of zeros of two maximal monotone operators in a uniformly smooth and uniformly convex Banach space. We established strong and weak convergence theorems for these two modified hybrid proximal-point algorithms, respectively.
Copyrightq2009 L. C. Zeng et al. 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
LetXbe a real Banach space andX∗its dual space. The normalized duality mappingJ:X → 2X∗is defined as
Jx:x∗∈X∗:x∗, xx2x∗2, ∀x∈X, 1.1
where·,·denotes the generalized duality pairing. Recall that ifXis a smooth Banach space then J is singlevalued. Throughout this paper, we will still denote byJ the single-valued normalized duality mapping. LetCbe a nonempty closed convex subset ofX,fa bifunction fromC×CtoR, andA:C → X∗a nonlinear mapping. The generalized equilibrium problem is to findx∈Csuch that
The set of solutions of1.2is denoted byEP. Problem1.2and similar problems have been extensively studied; see, for example,1–11. WheneverA0, problem1.2reduces to the equilibrium problem of findingx∈Csuch that
fx, y ≥0, ∀y∈C. 1.3
The set of solutions of1.3is denoted byEPf. Wheneverf 0, problem1.2reduces to the variational inequality problem of findingx∈Csuch that
Ax, y −x≥0, ∀y∈C. 1.4
WheneverX Ha Hilbert space, problem 1.2was very recently introduced and considered by Kamimura and Takahashi12. Problem1.2is very general in the sense that it includes, as spacial cases, optimization problems, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games, and others; see, for example,13,
14. A mappingS : C → X is called nonexpansive ifSx−Sy ≤ x−yfor allx, y ∈ C. Denote byFSthe set of fixed points ofS, that is,FS {x∈C:Sxx}. Iterative schemes for finding common elements of EP and fixed points set of nonexpansive mappings have been studied recently; see, for example,12,15–17and the references therein.
On the other hand, a classical method of solving 0 ∈ Tx in a Hilbert spaceHis the proximal point algorithm which generates, for any starting pointx0 ∈H, a sequence{xn}in
Hby the iterative scheme
xn1Jrnxn, n0,1,2, . . . , 1.5
where{rn}is a sequence in0,∞,Jr IrT−1for eachr >0 is the resolvent operator for
T, andI is the identity operator onH. This algorithm was first introduced by Martinet14
and generally studied by Rockafellar18in the framework of a Hilbert spaceH. Later many authors studied1.5and its variants in a Hilbert spaceHor in a Banach spaceX; see, for example,13,19–23and the references therein.
Let X be a uniformly smooth and uniformly convex Banach space and let C be a nonempty closed convex subset ofX. Letf be a bifunction fromC×CtoR satisfying the following conditionsA1–A4which were imposed in24:
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 supt↓0ftz 1−tx, y≤fx, y;
A4for allx∈C, fx,·is convex and lower semicontinuous. LetT :X → 2X∗be a maximal monotone operator such that
A5T−10∩EPf/∅.
12, Theorem 3.1 , Ceng et al. 16, Theorem 3.1 , and Zhang 17, Theorem 3.1 , we introduce a sequence{xn}that, under some appropriate conditions, is strongly convergent
toΠT−10∩T−10∩EPx0 in Section3. Second, inspired by Kamimura and Takahashi12, Theorem 3.1 , Ceng et al. 16, Theorem 4.1, and Zhang 17, Theorem 3.1 , we define a sequence weakly convergent to an elementz∈T−10∩T−10∩EP, wherezlimn
→ ∞ΠT−10∩T−10∩EPxnin
Section4. Our results represent a generalization of known results in the literature, including Takahashi and Zembayashi15, Kamimura and Takahashi12, Li and Song22, Ceng and Yao25, and Ceng et al.16. In particular, compared with Theorems 3.1 and 4.1 in16, our resultsi.e., Theorems3.2and4.2in this paperextend the problem of finding an element of
T−10∩EPfto the one of finding an element ofT−10∩T−10∩EP. Meantime, the algorithms in this paper are very different from those in 16 because of considering the complexity involving the problem of finding an element ofT−10∩T−10∩EP.
2. Preliminaries
In the sequel, we denote the strong convergence, weak convergence and weak∗convergence of a sequence{xn}to a pointx∈Xbyxn → x,xn xandxn x∗ , respectively.
A Banach spaceXis said to be strictly convex, ifxy/2<1 for allx, y∈U{z∈
X : z 1}withx /y.X is said to be uniformly convex if for each ∈ 0,2there exists
δ >0 such thatxy/2≤1−δfor allx, y∈Uwithx−y ≥. Recall that each uniformly convex Banach space has the Kadec-Klee property, that is,
xn x
xn −→ x
⇒xn −→x 2.1
The proof of the main results of Sections 3 and 4 will be based on the following assumption.
Assumption A. LetXbe a uniformly smooth and uniformly convex Banach space and letC be a nonempty closed convex subset ofX. Letf be a bifunction fromC×CtoRsatisfying the same conditionsA1–A4as in Section1. LetT,T:X → 2X∗be two maximal monotone operators such that
A5T−10∩T−10∩EP /∅.
Recall that if Cis a nonempty closed convex subset of a Hilbert space H, then the metric projectionPC :H → CofHontoCis nonexpansive. This fact actually characterizes
Hilbert spaces and hence, it is not available in more general Banach spaces. In this connection, Alber 26 recently introduced a generalized projection operatorΠC in a Banach space X
which is an analogue of the metric projection in Hilbert spaces. Consider the functional defined as in26by
φx, yx2−2x, Jyy2, ∀x, y∈X. 2.2
The generalized projectionΠC :X → Cis a mapping that assigns to an arbitrary point
x∈Xthe minimum point of the functionalφy, x; that is,ΠCxx, wherexis the solution
to the minimization problem
φx, x min
y∈C φ
y, x. 2.3
The existence and uniqueness of the operatorΠC follows from the properties of the
functionalφx, yand strict monotonicity of the mappingJsee, e.g.,27. In a Hilbert space,
ΠC PC. From26, in a smooth strictly convex and reflexive Banach spaceX, we have
y− x2≤φy, x≤yx2, ∀x, y∈X. 2.4
Moreover, by the property of subdifferential of convex functions, we easily get the following inequality:
φx, y≤φx, J−1JyJz−2y−x, Jz, ∀x, y, z∈X. 2.5
LetSbe a mapping fromCinto itself. A pointpinCis called an asymptotically fixed point ofSifCcontains a sequence{xn}which converges weakly topsuch thatSxn−xn → 0
28. The set of asymptotically fixed points ofSwill be denoted byFS. A mappingCfrom
Sinto itself is called relatively nonexpansive ifFS FSandφp, Sx ≤ φp, x, for all
x∈Candp∈FS 15.
Observe that, ifXis a reflexive strictly convex and smooth Banach space, then for any
x, y∈X, φx, y 0 if and only ifxy. To this end, it is sufficient to show that ifφx, y 0
thenxy. Actually, from2.4, we havexywhich implies thatx, Jyx2y2. From the definition ofJ, we haveJxJyand therefore,xy; see29for more details.
We need the following lemmas for the proof of our main results.
Lemma 2.1Kamimura and Takahashi 12. LetX be a smooth and uniformly convex Banach space and let{xn}and {yn}be two sequences ofX. Ifφxn, yn → 0 and either{xn}or {yn}is bounded, thenxn−yn → 0.
Lemma 2.2Alber 26, Kamimura and Takahashi12. LetC be a nonempty closed convex subset of a smooth strictly convex and reflexive Banach spaceX. Letx∈Xand letz∈C. Then
z ΠCx⇐⇒y−z, Jx−Jz≤0, ∀y∈C. 2.6
Lemma 2.3Alber 26, Kamimura and Takahashi12. LetC be a nonempty closed convex subset of a smooth strictly convex and reflexive Banach spaceX. Then
Lemma 2.4Rockafellar18. LetXbe a reflexive strictly convex and smooth Banach space and let T :X → 2X∗be a multivalued operator. Then there hold the following hold:
iT−10is closed and convex ifTis maximal monotone such thatT−10/∅;
iiT is maximal monotone if and only ifTis monotone withRJrT X∗for allr >0. Lemma 2.5Xu30. LetXbe a uniformly convex Banach space and letr >0. Then there exists a strictly increasing, continuous, and convex functiong :0,2r → Rsuch thatg0 0and
tx 1−ty2≤tx2 1−ty2−t1−tgx−y, 2.8
for allx, y∈Br andt∈0,1, whereBr {z∈X:z ≤r}.
Lemma 2.6Kamimura and Takahashi 12. LetX be a smooth and uniformly convex Banach space and letr > 0. Then there exists a strictly increasing, continuous, and convex functiong :
0,2r → Rsuch thatg0 0and
gx−y≤φx, y, ∀x, y∈Br. 2.9
The following result is due to Blum and Oettli24.
Lemma 2.7Blum and Oettli24. LetCbe a nonempty closed convex subset of a smooth strictly convex and reflexive Banach spaceX, letfbe a bifunction fromC×CtoRsatisfying (A1)–(A4), and letr >0andx∈X. Then, there existsz∈Csuch that
fz, y1ry−z, Jz−Jx≥0, ∀y∈C. 2.10
Motivated by Combettes and Hirstoaga 31 in a Hilbert space, Takahashi and Zembayashi15established the following lemma.
Lemma 2.8Takahashi and Zembayashi15. LetCbe a nonempty closed convex subset of a smooth strictly convex and reflexive Banach space X, and letf be a bifunction fromC×C toR satisfying (A1)–(A4). Forr >0andx∈X, define a mappingTr :X → Cas follows:
Trx
z∈C:fz, y1
r
y−z, Jz−Jx≥0, ∀y∈C
2.11
for allx∈X. Then, the following hold:
iTr is singlevalued;
iiTr is a firmly nonexpansive-type mapping, that is, for allx, y∈X,
iiiFTr FT r EPf;
ivEPfis closed and convex.
Using Lemma2.8, one has the following result.
Lemma 2.9Takahashi and Zembayashi15. LetCbe a nonempty closed convex subset of a smooth strictly convex and reflexive Banach spaceX, letfbe a bifunction fromC×CtoRsatisfying (A1)–(A4), and letr >0. Then, forx∈Xandq∈FTr,
φq, TrxφTrx, x≤φq, x. 2.13
Utilizing Lemmas 2.7, 2.8 and 2.9 as previously mentioned, Zhang17derived the following result.
Proposition 2.10Zhang21, Lemma. LetXbe a smooth strictly convex and reflexive Banach space and letCbe a nonempty closed convex subset ofX. LetA:C → X∗be anα-inverse-strongly monotone mapping, letf be a bifunction fromC×CtoRsatisfying (A1)–(A4), and letr > 0. Then the following hold:
forx∈X, there existsu∈Csuch that
fu, yAu, y−u1
r
y−u, Ju−Jx≥0, ∀y∈C, 2.14
ifXis additionally uniformly smooth andKr :X → Cis defined as
Krx
u∈C:fu, yAu, y−u1
r
y−u, Ju−Jx≥0, ∀y∈C
, ∀x∈X, 2.15
then the mappingKrhas the following properties:
iKris singlevalued,
iiKris a firmly nonexpansive-type mapping, that is,
Krx−Kry, JKrx−JKry≤Krx−Kry, Jx−Jy, ∀x, y∈X, 2.16
iiiFKr FK r EP,
ivEPis a closed convex subset ofC,
vφp, Krx φKrx, x≤φp, x,for all p∈FKr.
LetT,T : X → 2X∗ be two maximal monotone operators in a smooth Banach spaceX. We denote the resolvent operators ofT andT by Jr JrT−1Jand Jr JrTJ for eachr > 0, respectively. ThenJr : X → DT andJr : X → DT are two single-valued mappings. Also,
T−10 FJrandT−10 FJrfor each r > 0, whereFJrandFJrare the sets of fixed points
ofJr andJr, respectively. For eachr >0, the Yosida approximations ofT andT are defined byAr
J−JJr/randAr J−JJr/r, respectively. It is known that
Arx∈TJrx, Arx∈T
Jrx
Lemma 2.11 Kohsaka and Takahashi 13. Let X be a reflexive strictly convex and smooth Banach space and letT :X → 2X∗be a maximal monotone operator withT−10/∅. Then
φz, Jrx φJrx, x≤φz, x, ∀r >0, z∈T−10, x∈X. 2.18
Lemma 2.12Tan and Xu32. Let{an}and{bn}be two sequences of nonnegative real numbers satisfying the inequality:an1≤anbnfor alln≥0. If
∞
n0bn<∞, thenlimn→ ∞anexists.
3. Strong Convergence Theorem
In this section, we prove a strong convergence theorem for finding a common element of the set of solutions for a generalized equilibrium problem and the setT−10∩T−10 for two maximal monotone operatorsTandT.
Lemma 3.1. LetXbe a reflexive strictly convex and smooth Banach space and letT :X → 2X∗be a maximal monotone operator. Then for eachr ∈0,∞, the following holds:
Ju−Jv, Jru−Jrv ≥ JJru−JJrv, Jru−Jrv, ∀u, v∈X, 3.1
whereJr JrT−1J andJ is the duality mapping onX. In particular, whenever X Ha real Hilbert space,Jr is a nonexpansive mapping onH.
Proof. Since for eachu, v∈X
Jru JrT−1Ju, Jrv JrT−1Jv, 3.2
we have that
1
r ·Ju−JJru∈TJru,
1
r ·Jv−JJrv∈TJrv. 3.3
Thus, from the monotonicity ofTit follows that
1
r ·Ju−JJru−
1
r ·Jv−JJrv, Jru−Jrv
≥0, 3.4
and hence
Ju−Jv, Jru−Jrv ≥ JJru−JJrv, Jru−Jrv. 3.5
Theorem 3.2. Suppose that Assumption A is fulfilled and letx0∈Xbe chosen arbitrarily. Consider
the sequence
where
Hn
z∈C:φz, Krnyn
≤αnβn−αnβn−αnβnαnαnβn
φz, x0
1−αn1−αn
1−βn
αn1−αn
φz, xn
,
Wn{z∈C:xn−z, Jx0−Jxn ≥0},
xnJ−1αnJx0 1−αnβnJxn1−βnJJrnxn
,
ynJ−1
αnJxn 1−αnJJrnJ−1
βnJx0
1−βn
Jxn
,
3.7
Kr is defined by2.15,{αn},{βn},{αn},{βn} ⊂0,1satisfy
lim
n→ ∞αn0, nlim→ ∞βn 0, lim infn→ ∞ βn
1−βn>0, lim infn
→ ∞ αn1−αn>0, 3.8
and {rn} ⊂ 0,∞ satisfies lim infn→ ∞rn > 0. Then, the sequence {xn} converges strongly to
ΠT−10∩T−10∩EPx0 providedJrnvn−Jrnxn → 0for any sequence{vn} ⊂ X withvn−xn → 0, whereΠT−10∩T−10∩EP is the generalized projection ofXontoT−10∩T−10∩EP.
Remark 3.3. In Theorem 3.2, if X H a real Hilbert space, then {Jrn} is a sequence of
nonexpansive mappings onH. This implies that asn → ∞,
Jrnvn−Jrnxn≤ vn−xn −→0. 3.9
In this case, we can remove the requirement thatJrnvn−Jrnxn → 0 for any sequence{vn} ⊂
Xwithvn−xn → 0.
Proof of Theorem3.2. For the sake of simplicity, we define
un :Krnyn, zn:J−1
βnJxn1−βnJJrnxn
, zn:JrnJ−1
βnJx0
1−βn
Jxn
, 3.10
so that
xnJ−1αnJx0 1−αnJzn, ynJ−1 αnJxn 1−αnJzn. 3.11
We divide the proof into several steps.
Step 1. We claim thatHn∩Wnis closed and convex for eachn≥0.
Indeed, it is obvious thatHnis closed andWnis closed and convex for eachn≥0. Let
to show thatz∈Hn. We first writeγn αnβn−αnβn−αnβnαnαnβnfor eachn≥0. Next,
we prove that
φz, un≤γnφz, x0
1−γnφz, xn 3.12
is equivalent to
2γnz, Jx02
1−γnz, Jxn −2z, Jun ≤γnx02
1−γnxn2− un2. 3.13
Indeed, from2.4we deduce that the following equations hold:
φz, x0 z2−2z, Jx0x02,
φz, xn z2−2z, Jxnxn2,
φz, un z2−2z, Junun2,
3.14
which combined with3.12yield that3.12is equivalent to3.13. Thus we have
2γnz, Jx02
1−γnz, Jxn −2z, Jun
2γntz1 1−tz2, Jx0
21−γntz1 1−tz2, Jxn
−2tz1 1−tz2, Jun
2tγnz1, Jx021−tγnz2, Jx02
1−γntz1, Jxn
21−γn1−tz2, Jxn −2tz1, Jun −21−tz2, Jun
≤γnx02
1−γnxn2− un2.
3.15
This implies thatz∈Hn. Therefore,Hnis closed and convex. Step 2. We claim thatT−10∩T−10∩EP⊂H
n∩Wnfor eachn≥0 and that{xn}is well defined.
Indeed, takew∈T−10∩T−10∩EParbitrarily. Note thatunKr
nynis equivalent to
un∈C such that fun, yAun, y−unr1 n
Then from Lemma2.11we obtain
φw, zn φ
w, J−1β
nJxn1−βnJJrnxn
w2−2w, β
nJxn1−βnJJrnxn
βnJxn1−βnJJrnxn
2
≤ w2−2β
nw, Jxn −21−βnw, JJrnxnβnxn2
1−βnJrnxn2
βnφw, xn 1−βnφw, Jrnxn
≤βnφw, xn 1−βnφw, xn φw, xn,
3.17
φw,xn φ
w, J−1α
nJx0 1−αnJzn
w2−2w, α
nJx0 1−αnJznαnJx0 1−αnJzn2
≤ w2−2α
nw, Jx0 −21−αnw, Jznαnx02 1−αnzn2
αnφw, x0 1−αnφw, zn
≤αnφw, x0 1−αnφw, xn.
3.18
Moreover, we have
φw,zn φ
w,JrnJ−1
βnJx0
1−βn
Jxn
≤φw, J−1β nJx0
1−βn
xn
w2−2w,β nJx0
1−βn
Jxn
βnJx0
1−βn
Jxn
2
≤ w2−2β
nw, Jx0 −2
1−βn
w, Jxnβnx02
1−βn
xn2
βnφw, x0
1−βn
φw,xn
≤βnφw, x0
1−βnαnφw, x0 1−αnφw, xn
βn
1−βn
αn
φw, x0
1−βn
1−αnφw, xn,
φw, ynφ
w, J−1 α
nJxn 1−αnJzn
≤ w2−2α
αnφw,xn 1−αnφw,zn
≤αnαnφw, x0 1−αnφw, xn
1−αn
βn
1−βn
αn
φw, x0
1−βn
1−αnφw, xn
αnβn−αnβn−αnβnαnαnβn
φw, x0
1−αn1−αn
1−βn
αn1−αn
φw, xn
γnφw, x0
1−γnφw, xn, 3.19
whereγnαnβn−αnβn−αnβnαnαnβn. Sow∈Hnfor alln≥0. Now, let us show that
T−10∩T−10∩EP ⊂W
n ∀n≥0. 3.20
We prove this by induction. Forn 0, we haveT−10∩T−10∩EP ⊂ C W
0. Assume that
T−10∩T−10∩EP ⊂W
n. Sincexn1is the projection ofx0ontoHn∩Wn, by Lemma2.2we have
xn1−z, Jx0−Jxn1 ≥0, ∀z∈Hn∩Wn. 3.21
AsT−10∩T−10∩EP ⊂ H
n∩Wn by the induction assumption, the last inequality holds, in
particular, for allz∈T−10∩T−10∩EP. This, together with the definition ofWn
1implies that
T−10∩T−10∩EP ⊂Wn
1. Hence3.20holds for alln≥0. So,T−10∩T−10∩EP⊂Hn∩Wnfor
alln≥0. This implies that the sequence{xn}is well defined.
Step 3. We claim that{xn}is bounded and thatφxn1, xn → 0 asn → ∞.
Indeed, it follows from the definition of Wn that xn ΠWnx0. Since xn ΠWnx0
and xn1 ΠHn∩Wnx0 ∈ Wn, soφxn, x0 ≤ φxn1, x0 for alln ≥ 0; that is,{φxn, x0} is nondecreasing. It follows fromxn ΠWnx0and Lemma2.3that
φxn, x0 φΠWnx0, x0≤φ
p, x0
−φp, xn≤φp, x0
3.22
for eachp ∈T−10∩T−10∩EP ⊂ W
nfor eachn≥0. Therefore,{φxn, x0}is bounded which implies that the limit of{φxn, x0}exists. Since
so{xn}is bounded. From Lemma2.3, we have
φxn1, xn φxn1,ΠWnx0≤φxn1, x0−φΠWnx0, x0
φxn1, x0−φxn, x0,
3.24
for eachn≥0. This implies that
lim
n→ ∞φxn1, xn 0. 3.25
Step 4. We claim that limn→ ∞xn−un0, limn→ ∞xn−Jrnxn0, and limn→ ∞xn−Jrnxn
0.
Indeed, fromxn1 ΠHn∩Wnx0∈Hn, we have
φxn1, un≤
αnβn−αnβn−αnβnαnαnβn
φxn1, x0
1−αn1−αn
1−βn
αn1−αn
φxn1, xn
3.26
for all n ≥ 0. Therefore, from αn → 0, βn → 0 and φxn1, xn → 0, it follows that
limn→ ∞φxn1, un 0. Since limn→ ∞φxn1, xn limn→ ∞φxn1, un 0 andXis uniformly
convex and smooth, we have from Lemma2.1that
lim
n→ ∞xn1−xnnlim→ ∞xn1−un0, 3.27
and therefore, limn→ ∞xn −un 0. Since J is uniformly norm-to-norm continuous on
bounded subsets ofXandxn−un → 0, then limn→ ∞Jxn−Jun0.
Let us setΩ:T−10∩T−10∩EP. Then, according to Lemma2.4and Proposition2.10, we know thatΩis a nonempty closed convex subset ofXsuch thatΩ⊂C. Fixu∈Ωarbitrarily. As in the proof of Step2we can show thatφu, zn ≤ φu, xn, φu,xn ≤ αnφu, x0 1−
αnφu, xn, φu,zn≤βnφu, x0 1−βnφu,xn, φu, yn≤αnφu,xn 1−αnφu,zn,
andφu, un≤αnφu,xn 1−αnφu,zn. Hence it follows from the boundedness of{xn}
that{zn},{xn},{zn},{yn}, and{un}are also bounded. Letr sup{xn,xn,Jrnxn,zn:
n≥0}. SinceXis a uniformly smooth Banach space, we know thatX∗is a uniformly convex Banach space. Therefore, by Lemma 2.5 there exists a continuous, strictly increasing, and convex functiongwithg0 0 such that
αx∗ 1−αy∗2≤αx∗2 1−αy∗2−α
forx∗, y∗∈Br∗andα∈0,1. So, we have that
φu, zn φ
u, J−1β
nJxn1−βnJJrnxn
u2−2u, β
nJxn1−βnJJrnxn
βnJxn1−βnJJrnxn
2
≤ u2−2β
nu, Jxn −21−βnu, JJrnxn
βnxn21−βnJrnxn2−βn
1−βngJxn−JJrnxn
βnφu, xn 1−βnφu, Jrnxn−βn
1−βngJxn−JJrnxn
≤βnφu, xn 1−βnφu, xn−βn1−βngJxn−JJrnxn
φu, xn−βn1−βngJxn−JJrnxn,
3.29
φu,zn φ
u,JrnJ−1
βnJx0
1−βn
Jxn
≤φu, J−1β nJx0
1−βn
Jxn
≤ u2−2u,β nJx0
1−βn
Jxn
βnx02
1−βn
xn2
βnφu, x0
1−βn
φu,xn
≤βnφu, x0
1−βnφu, xn−βn1−βngJxn−JJrnxn
βnφu, x0
1−βn
φu, xn−
1−βn
βn1−βngJxn−JJrnxn,
3.30
and hence
φu,xn φ
u, J−1α
nJx0 1−αnJzn
u2−2u, α
nJx0 1−αnJznαnJx0 1−αnJzn2
≤ u2−2α
nu, Jx0 −21−αnu, Jznαnx02 1−αnzn2
αnφu, x0 1−αnφu, zn
≤αnφu, x0 1−αnφu, xn−βn1−βngJxn−JJrnxn
αnφu, x0 1−αnφu, xn−1−αnβn1−βngJxn−JJrnxn,
3.31
φu, un φu, Krnyn
≤φu, yn using Proposition 2.10
φu, J−1 α
nJxn 1−αnJzn
u2−2u,α
nJxn 1−αnJznαnJxn 1−αnJzn2
≤ u2−2α
nu, Jxn −21−αnu, Jznαnxn2 1−αnzn2
−αn1−αngJxn−Jzn
αnφu,xn 1−αnφu,zn−αn1−αngJxn−Jzn
≤αnαnφu, x0 1−αnφu, xn−1−αnβn1−βngJxn−JJrnxn
1−αn
βnφu, x0
1−βn
φu, xn
−1−βn
βn1−βngJxn−JJrnxn
−αn1−αngJxn−Jzn
≤αnβn
φu, x0
αn1−αn 1−αn
1−βn
φu, xn
−αn1−αn 1−αn
1−βn
βn1−βngJxn−JJrnxn
−αn1−αngJxn−Jzn
αnβn
φu, x0
1−αnαn−βn1−αn
φu, xn
−1−αnαn−βn1−αn
βn1−βngJxn−JJrnxn
−αn1−αngJxn−Jzn
≤αnβn
φu, x0 φu, xn
−1−αnαn−βn1−αn
βn1−βngJxn−JJrnxn
−αn1−αngJxn−Jzn, 3.32
for alln≥0. Consequently we have
1−αnαn−βn1−αn
βn1−βngJxn−JJrnxn αn1−αngJxn−Jzn
≤αnβn
φu, x0 φu, xn−φu, un
αnβn
φu, x0 xn2− un2−2u, Jxn−Jun
≤αnβn
φu, x0 xn2− un22|u, Jxn−Jun|
≤αnβn
φu, x0 |xn − un|xnun 2uJxn−Jun
≤αnβn
φu, x0 xn−unxnun 2uJxn−Jun.
Sincexn−un → 0 andJ is uniformly norm-to-norm continuous on bounded subsets ofX,
we obtainJxn−Jun → 0. From lim infn→ ∞βn1−βn > 0, lim infn→ ∞αn1−αn > 0, and
limn→ ∞αnlimn→ ∞βn0 we have
lim
n→ ∞gJxn−JJrnxn nlim→ ∞gJxn−Jzn 0. 3.34
Therefore, from the properties ofgwe get
lim
n→ ∞Jxn−JJrnxnnlim→ ∞xn−Jrnxn0,
lim
n→ ∞Jxn−Jznnlim→ ∞xn−zn0,
3.35
recalling thatJ−1is uniformly norm-to-norm continuous on bounded sunsets ofX∗. Next let us show that
lim
n→ ∞
Jxn−JJrnxnnlim→ ∞xn−Jrnxn0. 3.36
Observe first that
φun, xn−φxn1, un
xn2− xn12−2un, Jxn2xn1, Jun
xn − xn1xnxn1 2xn1−un, Jxn2xn1, Jun−Jxn
≤ xn−xn1xnxn1 2xn1−unxn2xn1Jun−Jxn.
3.37
Sinceφxn1, un → 0, xn1−xn → 0, xn1−un → 0, Jun−Jxn → 0, and{xn}is
bounded, so it follows thatφun, xn → 0. Also, observe that
φun, Jrnxn−φun, xn Jrnxn2− xn22un, Jxn−JJrnxn
Jrnxn − xnJrnxnxn 2un, Jxn−JJrnxn
≤ Jrnxn−xnJrnxnxn 2unJxn−JJrnxn.
Since φun, xn → 0, Jrnxn − xn → 0, Jxn − JJrnxn → 0 and the sequences
{xn},{un},{Jrnxn}are bounded, so it follows thatφun, Jrnxn → 0. Meantime, observe that
φun, zn φ
un, J−1βnJxn1−βnJJrnxn
un2−2un, βnJxn1−βnJJrnxn
βnJxn1−βnJJrnxn
2
≤ un2−2βnun, Jxn −21−βnun, JJrnxnβnxn2
1−βnJrnxn2
βnφun, xn 1−βnφun, Jrnxn
≤φun, xn φun, Jrnxn,
3.39
and hence
φun,xn φ
un, J−1αnJx0 1−αnJzn
un2−2un, αnJx0 1−αnJznαnJx0 1−αnJzn2
≤ un2−2αnun, Jx0 −21−αnun, Jznαnx02 1−αnzn2
αnφun, x0 1−αnφun, zn
≤αnφun, x0 φun, zn
≤αnφun, x0 φun, xn φun, Jrnxn.
3.40
Sinceαn → 0, φun, xn → 0 andφun, Jrnxn → 0, it follows from the boundedness of
{un}thatφun,xn → 0. Thus, in terms of Lemma2.1, we have thatun−xn → 0 and so
xn−xn → 0. Furthermore, sinceβnJx0 1−βnJxn−JxnβnJx0−Jxn → 0, from the
uniform norm-to-norm continuity ofJ−1on bounded subsets ofX∗, we obtain
J−1β
nJx0
1−βn
Jxn
−xn−→0. 3.41
Observe that
x
n−Jrnxn≤ xn−znzn−Jrnxn
xn−zn JrnJ−1
βnJx0
1−βn
Jxn
−Jrnxn.
3.42
Thus, from 3.35 it follows that xn − Jrnxn → 0. Since J is uniformly norm-to-norm
continuous on bounded subsets ofX, it follows thatJxn−JJrnxn → 0. Step 5. We claim thatωw{xn}⊂T−10∩T−10∩EP, where
ωw{xn}:x∈C:xnk x for some subsequence{nk} ⊂ {n}with nk↑ ∞
Indeed, since{xn}is bounded and X is reflexive, we know thatωw{xn}/∅. Take
x ∈ ωw{xn}arbitrarily. Then there exists a subsequence{xnk}of{xn}such thatxnk x.
Hence it follows fromxn−xn → 0, xn−Jrnxn → 0, andxn−Jrnxn → 0 that{xnk},{Jrnkxnk},
and{Jrnkxnk}converge weakly to the same pointx. On the other hand, from3.35,3.36,
and lim infn→ ∞rn>0 we obtain that
lim
n→ ∞Arnxnnlim→ ∞ 1
rnJxn−JJrnxn0, 3.44
lim
n→ ∞
Arnxnnlim→ ∞
1
rn
Jxn−JJrnxn0. 3.45
Ifz∗∈Tzandz∗∈Tz, then it follows from2.17and the monotonicity of the operatorsT,T that for allk≥1
z−Jrnkxnk, z∗−Arnkxnk
≥0, z−Jrnkxnk,z∗−Arnkxnk
≥0. 3.46
Lettingk → ∞, we have thatz−x, z ∗ ≥0 andz−x, z∗ ≥0. Then the maximality of the operatorsT,T implies thatx ∈ T−10 andx ∈ T−10. Next, let us show thatx∈ EP. Since we have by3.32
φu, yn≤
αnβn
φu, x0 φu, xn, 3.47
fromun Krnynand Proposition2.10it follows that
φun, ynφKrnyn, yn
≤φu, yn−φu, Krnyn
≤αnβn
φu, x0 φu, xn−φu, un.
3.48
Also, since
φu, xn−φu, unxn2− un22u, Jun−Jxn
≤ |xn − un|xnun 2uJun−Jxn
≤ xn−unxnun 2uJun−Jxn,
3.49
so we get
lim
n→ ∞
Thus from3.47,αn → 0, βn → 0, andφu, xn−φu, un → 0, we have limn→ ∞φun, yn
0. SinceXis uniformly convex and smooth, we conclude from Lemma2.1that
lim
n→ ∞un−yn0. 3.51
Fromxnk x, x n−un → 0, and3.51, we haveynk xandunk x. SinceJis uniformly
norm-to-norm continuous on bounded subsets ofX, from3.51we derive
lim
n→ ∞Jun−Jyn0. 3.52
From lim infn→ ∞rn>0, it follows that
lim
n→ ∞
Jun−Jyn
rn 0. 3.53
By the definition ofun :Krnyn, we have
Fun, yr1 n
y−un, Jun−Jyn≥0, ∀y∈C, 3.54
where
Fun, yfun, yAun, y−un. 3.55
Replacingnbynk, we have fromA2that
1
rnk
y−unk, Junk −Jynk
≥ −Funk, y
≥Fy, unk
, ∀y∈C. 3.56
Sincey→fx, y Ax, y−xis convex and lower semicontinuous, it is also weakly lower semicontinuous. Lettingnk → ∞in the last inequality, from3.53andA4we have
Fy,x≤0, ∀y∈C. 3.57
Fortwith 0< t≤1 andy∈C, letytty 1−tx. Sincey∈Candx∈C,yt∈Cand hence
Fyt,x ≤0. So, fromA1we have
0Fyt, yt≤tFyt, y 1−tFyt,x≤tFyt, y. 3.58
Dividing byt, we have
Lettingt↓0, fromA3it follows that
Fx, y ≥0, ∀y∈C. 3.60
Thusx∈EP. Therefore, we obtain thatωw{xn}⊂T−10∩T−10∩EPby the arbitrariness ofx. Step 6. We claim that{xn}converges strongly tow ΠT−10∩T−10∩EPx0.
Indeed, fromxn1 ΠHn∩Wnx0andw∈T−10∩T−10∩EP ⊂Hn∩Wn, It follows that
φxn1, x0≤φw, x0. 3.61
Since the norm is weakly lower semicontinuous, we have
φx, x 0 x2−2x, Jx 0x02≤lim inf k→ ∞
xnk
2−2x
nk, Jx0x0
2
lim inf
k→ ∞ φxnk, x0≤lim supk→ ∞ φxnk, x0≤φw, x0.
3.62
From the definition ofΠT−10∩T−10∩EP, we havexw. Hence limk→ ∞φxnk, x0 φw, x0and
0 lim
k→ ∞
φxnk, x0−φw, x0
lim
k→ ∞
xnk2− w2−2xnk−w, Jx0
lim
k→ ∞
xnk2− w2
,
3.63
which implies that limk→ ∞xnk w. SinceX has the Kadec-Klee property, thenxnk →
w ΠT−10∩T−10∩EPx0. Therefore,{xn}converges strongly toΠT−10∩T−10∩EPx0.
Remark 3.4. In Theorem3.2, putA≡0, T ≡0, andβn 0, ∀n≥0. Then, for allα, r ∈0,∞
andx, y∈C, we have that
Ax−Ay, x−y≥αAx−Ay2,
Krx
u∈C:fu, yAu, y−u 1
r
y−u, Ju−Jx≥0, ∀y∈C
u∈C:fu, y1
r
y−u, Ju−Jx≥0, ∀y∈C
Trx.
Moreover, the following hold:
Hn
z∈C:φz, Krnyn
≤αnβn−αnβn−αnβnαnαnβn
φz, x0
1−αn1−αn
1−βn
αn1−αn
φz, xn
,
z∈C:φz, Trnyn
≤αnφz, x0 1−αnφz, xn,
yn J−1
αnJxn 1−αnJJrnJ−1
βnJx0
1−βn
Jxn
J−1 α
nJxn 1−αn0Jx0 1−0Jxn
J−1 α
nJxn 1−αnJxn
J−1Jx nxn,
3.65
and hence
ynxnJ−1αnJx0 1−αnβnJxn1−βnJJrnxn. 3.66
In this case, the previous Theorem3.2reduces to20, Theorem 3.1.
4. Weak Convergence Theorem
In this section, we present the following algorithm for finding a common element of the set of solutions for a generalized equilibrium problem and the setT−10∩T−10 for two maximal monotone operatorsTandT.
Letx0 ∈Xbe chosen arbitrarily and consider the sequence{xn}generated by
xnJ−1αnJx0 1−αnβnJKrnxn
1−βnJJrnKrnxn
,
xn1 J−1
αnJKrnxn 1−αnJJrnJ−1
βnJx0
1−βn
JKrnxn
, n0,1,2, . . . , 4.1
where{αn},{βn},{αn},{βn} ⊂0,1,{rn} ⊂0,∞, andKr, r >0, is defined by2.15.
Before proving a weak convergence theorem, we need the following proposition.
Proposition 4.1. Suppose that Assumption A is fulfilled and let{xn}be a sequence defined by4.1, where{αn},{βn},{αn},{βn} ⊂0,1satisfy the following conditions:
∞
n0
αn<∞,
∞
n0
βn<∞, lim infn
→ ∞ βn
1−βn>0, lim infn
Then, {ΠT−10∩T−10∩EPxn}converges strongly to z ∈ T−10∩T−10∩EP, where ΠT−10∩T−10∩EP is the generalized projection ofXontoT−10∩T−10∩EP.
Proof. We setΩ:T−10∩T−10∩EPand
un:Krnxn, yn:J−1
βnJun1−βnJJrnun
,
un:Krnxn, yn:JrnJ−1
βnJx0
1−βn
Jun
,
4.3
so that
xnJ−1αnJx0 1−αnJyn,
xn1J−1
αnJun 1−αnJyn, n0,1,2, . . . .
4.4
Then, in terms of Lemma2.4and Proposition2.10,Ωis a nonempty closed convex subset of
Xsuch thatΩ⊂C. We first prove that{xn}is bounded. Fixu∈Ω. Note that by the first and
third of4.3,un,un∈C, and
Fun, yr1 n
y−un, Jun−Jxn≥0, ∀y∈C,
Fun, yr1 n
y−un, Jun−Jxn≥0, ∀y∈C.
4.5
Here, eachKrn is relatively nonexpansive. Then from Proposition2.10we obtain
φu, ynφ
u, J−1β
nJun1−βnJJrnun
u2−2u, β
nJun1−βnJJrnun
βnJun1−βnJJrnun
2
≤ u2−2β
nu, Jun −21−βnu, JJrnunβnun
21−β
nJrnun
2
βnφu, un 1−βnφu, Jrnun
≤βnφu, un 1−βnφu, un
φu,ynφ
u,JrnJ−1
βnJx0
1−βn
Jun
≤φu, J−1β nJx0
1−βn
Jun
u2−2u,β nJx0
1−βn
Jun
βnJx0
1−βn
Jun
2
≤ u2−2β
nu, Jx0 −2
1−βn
u, Junβnx02
1−βn
un2
βnφu, x0
1−βn
φu,un
≤βnφu, x0 φu,un
βnφu, x0 φu, Krnxn
≤βnφu, x0 φu,xn,
4.7
and hence by Proposition2.10, we have
φu,xn φ
u, J−1α
nJx0 1−αnJyn
u2−2u, α
nJx0 1−αnJyn
αnJx0 1−αnJyn2
≤ u2−2α
nu, Jx0 −21−αnu, Jynαnx02 1−αnyn2
αnφu, x0 1−αnφ
u, yn
≤αnφu, x0 φ
u, yn
≤φu, xn αnφu, x0,
4.8
φu, xn1 φ
u, J−1α
nJun 1−αnJyn
u2−2u,α
nJun 1−αnJyn αnJun 1−αnJyn2
≤ u2−2α
nu, Jun −21−αnu, Jynαnun2 1−αn yn2
αnφu,un 1−αnφu,yn
≤αnφu,xn 1−αn
βnφu, x0 φu,xn
≤φu,xn βnφu, x0.
Consequently, the last two inequalities yield that
φu, xn1≤φu,xn βnφu, x0
≤φu, xn αnφu, x0 βnφu, x0
φu, xn
αnβn
φu, x0
4.10
for all n ≥ 0. So, from ∞n0αn < ∞, ∞n0βn < ∞, and Lemma 2.12, we deduce that limn→ ∞φu, xnexists. This implies that{φu, xn}is bounded. Thus,{xn}is bounded and
so are{un},{un},{Jrnun}, and{Jrnun}.
Definezn ΠΩxnfor alln≥0. Let us show that{zn}is bounded. Indeed, observe that
zn − xn2≤φzn, xn φΠΩxn, xn≤φp, xn−φp,ΠΩxn
φp, xn−φp, zn≤φp, xn,
4.11
for eachp ∈ Ω. This, together with the boundedness of {xn}, implies that{zn}is bounded
and so isφzn, x0. Furthermore, fromzn∈Ωand4.10we have
φzn, xn1≤φzn, xn
αnβn
φzn, x0. 4.12
SinceΠΩis the generalized projection, then, from Lemma2.3we obtain
φzn1, xn1 φΠΩxn1, xn1≤φzn, xn1−φzn,ΠΩxn1
φzn, xn1−φzn, zn1≤φzn, xn1.
4.13
Hence, from4.12, it follows thatφzn1, xn1≤φzn, xn αnβnφzn, x0.
Note that∞n0αn < ∞,∞n0βn < ∞, and{φzn, x0}is bounded, so that∞n0αn
βnφzn, x0 < ∞. Therefore,{φzn, xn}is a convergent sequence. On the other hand, from
4.10we derive, for allm≥0,
φu, xnm≤φu, xn m−1
j0
αnjβnj
φu, x0. 4.14
In particular, we have
φzn, xnm≤φzn, xn m−1
j0
αnjβnj
Consequently, fromznm ΠΩxnmand Lemma2.3, we have
φzn, znm φznm, xnm≤φzn, xnm≤φzn, xn m−1
j0
αnjβnj
φzn, x0 4.16
and hence
φzn, znm≤φzn, xn−φznm, xnm m−1
j0
αnjβnj
φzn, x0. 4.17
Letrsup{zn:n≥0}. From Lemma2.6, there exists a continuous, strictly increasing, and
convex functiongwithg0 0 such that
gx−y≤φx, y, ∀x, y∈Br. 4.18
So, we have
gzn−znm≤φzn, znm
≤φzn, xn−φznm, xnm m−1
j0
αnjβnj
φzn, x0.
4.19
Since {φzn, xn} is a convergent sequence, {φzn, x0} is bounded and ∞n0αn βn is convergent, from the property ofgwe have that{zn}is a Cauchy sequence. SinceΩis closed,
{zn}converges strongly toz∈Ω. This completes the proof.
Now, we are in a position to prove the following theorem.
Theorem 4.2. Suppose that Assumption A is fulfilled and let {xn}be a sequence defined by4.1, where{αn},{βn},{αn},{βn} ⊂0,1satisfy the following conditions:
∞
n0
αn<∞,
∞
n0
βn<∞, lim infn
→ ∞ βn
1−βn>0, lim infn
→ ∞ αn1−αn>0, 4.20
and {rn} ⊂ 0,∞satisfieslim infn→ ∞rn > 0. If J is weakly sequentially continuous, then{xn} converges weakly toz∈T−10∩T−10∩EP, wherezlim
Proof. We consider the notations 4.3. As in the proof of Proposition 4.1, we have that
{xn},{un},{Jrnun},{xn},{un}, and{Jrnun}are bounded sequences. Let
rsupun,Jrnun,un, yn:n≥0
. 4.21
From Lemma 2.5 and as in the proof of Theorem 3.2, there exists a continuous, strictly increasing, and convex functiongwithg0 0 such that
αx∗ 1−αy∗2 ≤αx∗2 1−αy∗2−α
1−αgx∗−y∗ 4.22
forx∗, y∗∈Br∗andα∈0,1. Observe that foru∈Ω:T−10∩T−10∩EP,
φu, ynφ
u, J−1β
nJun1−βnJJrnun
u2−2u, β
nJun1−βnJJrnun
βnJun1−βnJJrnun
2
≤ u2−2β
nu, Jun −21−βnu, JJrnun
βnun21−βnJrnun2−βn
1−βngJun−JJrnun
≤βnφu, un 1−βnφu, Jrnun−βn
1−βngJun−JJrnun
≤βnφu, un 1−βnφu, un−βn1−βngJun−JJrnun
φu, un−βn1−βngJun−JJrnun,
φu,ynφ
u,JrnJ−1
βnJx0
1−βn
Jun
≤φu, J−1β nJx0
1−βn
Jun
u2−2u,β nJx0
1−βn
Jun
βnJx0
1−βn
Jun
2
≤ u2−2β
nu, Jx0 −2
1−βn
u, Junβnx02
1−βn
un2
βnφu, x0
1−βn
φu,un
βnφu, x0
1−βn
φu, Krnxn
≤βnφu, x0 φu,xn.
Hence,
φu,xn φ
u, J−1α
nJx0 1−αnJyn
u2−2u, α
nJx0 1−αnJyn
αnJx0 1−αnJyn2
≤ u2−2α
nu, Jx0 −21−αnu, Jynαnx02 1−αnyn2
αnφu, x0 1−αnφ
u, yn
≤αnφu, x0 φ
u, yn
≤αnφu, x0 φu, un−βn1−βngJun−JJrnun
αnφu, x0 φu, Krnxn−βn
1−βngJun−JJrnun
≤αnφu, x0 φu, xn−βn1−βngJun−JJrnun,
4.24
φu, xn1 φ
u, J−1α
nJun 1−αnJyn
u2−2u,α
nJun 1−αnJyn αnJun 1−αnJyn2
≤ u2−2α
nu, Jun −21−αnu, Jynαnun2 1−αn yn2
−αn1−αngJun−Jyn
αnφu,un 1−αnφu,yn−αn1−αngJun−Jyn
≤αnφu, Krnxn 1−αn
βnφu, x0 φu,xn
−αn1−αngJun−Jyn
≤φu,xn βnφu, x0−αn1−αngJun−Jyn.
4.25
Consequently, the last two inequalities yield that
φu, xn1≤φu,xn βnφu, x0−αn1−αngJun−Jyn
≤αnφu, x0 φu, xn−βn
1−βngJun−JJrnun
βnφu, x0−αn1−αngJun−Jyn
φu, xn
αnβn
φu, x0−βn1−βngJun−JJrnun
−αn1−αngJun−Jyn.
4.26
Thus, we have
βn1−βngJun−JJrnun αn1−αngJun−Jyn
≤φu, xn−φu,xn1
αnβn
φu, x0.
By the proof of Proposition4.1, it is known that{φu, xn}is convergent; since limn→ ∞αn
0,limn→ ∞βn0,lim infn→ ∞βn1−βn>0, and lim infn→ ∞αn1−αn>0, then we have
lim
n→ ∞gJun−JJrnun nlim→ ∞gJun−Jyn0. 4.28
Taking into account the properties ofg, as in the proof of Theorem3.2, we have
lim
n→ ∞Jun−JJrnunnlim→ ∞un−Jrnun0,
lim
n→ ∞Jun−Jynnlim→ ∞ un−yn0,
4.29
sinceJ−1is uniformly norm-to-norm continuous on bounded subsets ofX∗. Note thatβnJx 0
1−βnJun−JunβnJx0−Jun → 0. Hence, from the uniform norm-to-norm continuity
ofJ−1on bounded subsets ofX∗we obtainJ−1βnJx0 1−βnJun−un → 0. Also, observe
that
Jrnun−un≤ Jrnun−JrnJ−1
βnJx0
1−βn
Jun
JrnJ−1
βnJx0
1−βn
Jun
−un
≤un−J−1
βnJx0
1−βn
Jun yn−un.
4.30
Fromun−yn → 0 it follows thatJrnun−un → 0. SinceJ is uniformly norm-to-norm
continuous on bounded subsets ofX, we have
lim
n→ ∞Jun−JJrnunnlim→ ∞un−Jrnun0,
lim
n→ ∞
Jun−JJrnunnlim→ ∞un−Jrnun0.
4.31
Now let us show that
lim
n→ ∞φu, xn nlim→ ∞φu,xn nlim→ ∞φu, un nlim→ ∞φu,un. 4.32
Indeed, from4.10we get
φu, xn1−βnφu, x0≤φu,xn≤φu, xn αnφu, x0, 4.33
which, together with limn→ ∞αnlimn→ ∞βn0, yields that
lim
From4.9it follows that
φu, xn1≤αnφu,un 1−αnφu,yn
φu,ynαnφu,un−φu,yn≤φu,xn βnφu, x0.
4.35
Note that
φu,un−φu,ynun2− yn22u, Jyn−Jun
≤un − ynun yn2uJyn−Jun
≤ un−ynun yn2uJyn−Jun.
4.36
Sinceun−yn → 0 andJun−Jyn → 0, we obtain limn→ ∞φu,un−φu,yn 0, which,
together with limn→ ∞φu,xn limn→ ∞φu, xn, yields that
lim
n→ ∞φ
u,ynnlim
→ ∞φu, xn. 4.37
We have from4.8that
φu,xn−αnφu, x0≤φ
u, yn≤φu, xn, 4.38
which, together with limn→ ∞φu,xn limn→ ∞φu, xn, yields that
lim
n→ ∞φ
u, ynnlim→ ∞φu, xn. 4.39
Also from4.7it follows that
φu,yn−βnφu, x0≤φu,un≤φu,xn, 4.40
which, together with limn→ ∞φu,xn limn→ ∞φu,yn limn→ ∞φu, xn, yields that
lim
Similarly from4.6it follows that
φu, yn≤φu, un≤φu, xn 4.42
which, together with limn→ ∞φu, yn limn→ ∞φu, xn, yields that
lim
n→ ∞φu, un nlim→ ∞φu, xn. 4.43
On the other hand, let us show that
lim
n→ ∞xn−xn0. 4.44
Indeed, lets sup{xn,un,xn,un : n ≥ 0}. From Lemma2.6, there exists a
continuous, strictly increasing, and convex functiong1withg10 0 such that
g1x−y≤φ
x, y, ∀x, y∈Bs. 4.45
SinceunKrnxnandunKrnxn, we deduce from Proposition2.10that foru∈Ω,
g1un−xn≤φun, xn≤φu, xn−φu, un,
g1un−xn≤φ un,xn≤φu,xn−φu,un.
4.46
This implies that
lim
n→ ∞g1un−xn nlim→ ∞g1un−xn 0. 4.47
SinceJis uniformly norm-to-norm continuous on bounded subsets ofX, from the properties ofg1we obtain
lim
n→ ∞un−xnnlim→ ∞Jun−Jxn0, lim
n→ ∞un−xnnlim→ ∞Jun−Jxn0.
