Fixed Point Theory and Applications Volume 2010, Article ID 590278,33pages doi:10.1155/2010/590278
Research Article
Strong and Weak Convergence Theorems for
Common Solutions of Generalized Equilibrium
Problems and Zeros of Maximal
Monotone Operators
L.-C. Zeng,
1, 2Q. H. Ansari,
3David S. Shyu,
4and J.-C. Yao
51Department of Mathematics, Shanghai Normal University, Shanghai 200234, China 2Scientific Computing Key Laboratory of Shanghai Universities, Shanghai 200234, China 3Department of Mathematics, Aligarh Muslim University, Aligarh 202 002, India 4Department of Finance, National Sun Yat-Sen University, Kaohsiung 80424, Taiwan
5Department of Applied Mathematics, National Sun Yat-Sen University, Kaohsiung 80424, Taiwan
Correspondence should be addressed to J.-C. Yao,[email protected]
Received 27 October 2009; Accepted 12 January 2010
Academic Editor: Tomonari Suzuki
Copyrightq2010 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.
The purpose of this paper is to introduce and study two modified hybrid proximal-point algorithms for finding a common element of the solution setEPof a generalized equilibrium problem and the setT−10∩T−10 for two maximal monotone operatorsTandTdefined on a Banach
spaceX. Strong and weak convergence theorems for these two modified hybrid proximal-point algorithms are established.
1. Introduction
LetXbe a real Banach space with its dualX∗. The mappingJ :X → 2X∗defined by
Jx:x∗∈X∗:x∗, xx2x∗2, ∀x∈X, 1.1
is called the normalized duality mapping. From the Hahn-Banach theorem, it follows that Jx/∅for eachx∈X.
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. 1.2
It is well known that ifX∗is strictly convex, thenJis single-valued. In the sequel, we shall still denote the single-valued normalized duality mapping byJ. LetCbe a nonempty closed convex subset ofX,f :C×C → Ra bifunction, andA:C → X∗a nonlinear mapping. Very recently, Zhang 1 considered and studied the generalized equilibrium problem of findingx∈Csuch that
fx, y Ax, y −x ≥0, ∀y∈C. 1.3
The set of solutions of 1.3is denoted by EP. Problem 1.3and related problems have been studied and investigated extensively in the literature; See, for example,2–12and references therein. WheneverA ≡ 0, problem 1.3 reduces to the equilibrium problem of findingx∈Csuch that
fx, y ≥0, ∀y∈C. 1.4
The set of solutions of1.4is denoted byEPf. Wheneverf ≡0, problem1.3reduces to the variational inequality problem of findingx∈Csuch that
Ax, y −x ≥0, ∀y∈C. 1.5
The set of solutions of1.5is denoted byV IC, A.
WheneverX Ha Hilbert space, problem 1.3was very recently introduced and considered by S. Takahashi and W. Takahashi13. Problem1.3is very general in the sense that it includes, as spacial cases, optimization problems, variational inequalities, minimax problems, Nash equilibrium problem in noncooperative games, and others; See, for example, 1,2,4,6–9,14–17which are references therein.
A mappingS:C → Xis called nonexpansive ifSx−Sy ≤ x−yfor allx, y ∈C. Denote byFSthe set of fixed points ofS, that is,FS {x∈ C: Sx x}. Very recently, W. Takahashi and K. Zembayashi18proposed an iterative algorithm for finding a common element of the solution set of the equilibrium problem1.4and the set of fixed points of a relatively nonexpansive mappingS in a Banach spaceX. They also studied the strong and weak convergence of the sequences generated by their algorithm. In particular, they proposed the following iterative algorithm:
x0∈C,
un∈Csuch thatfun, yr1 n
y−un, Jun−Jyn ≥0, ∀y∈C,
Hnz∈C:φz, un≤φz, xn,
Wn{z∈C:xn−z, Jx−Jxn ≥0},
xn1 ΠHn∩Wnx, n≥0,
1.6
whereφx, y x2−2x, Jyy2 for allx, y ∈ X,{α
n} ⊂ 0,1, and{rn} ⊂ a,∞for somea >0. They proved that the sequence{xn}generated by the above algorithm converges strongly to ΠFS∩EPfx0, where ΠFS∩EPf is the generalized projection of X onto FS∩
EPf. They have also studied the weak convergence of the sequence{xn}generated by the following algorithm:
u0 ∈X,
xn∈Csuch thatfxn, yr1 n
y−xn, Jxn−Jun ≥0, ∀y∈C,
un1J−1αnJxn 1−αnJSxn, n≥0,
1.7
toz∈FS∩EPf, wherezlimn→ ∞ΠFS∩EPfxn.
Let C be a nonempty closed convex subset of a uniformly smooth and uniformly convex Banach spaceX. LetA :C → X∗ be anα-inverse-strongly monotone mapping and f:C×C → Ra bifunction satisfying 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 supt↓0ftz 1−tx, y≤fx, y; A4for allx∈C,fx,·is convex and lower semicontinuous.
LetS1, S2:C → Cbe two relatively nonexpansive mappings such thatFS1∩FS2∩
EP /∅. Let{xn}be the sequence generated by
x0∈C, C0C;
znJ−1αnJxn 1−αnJS1xn,
ynJ−1βnJxn1−βnJS2zn,
un∈Csuch thatfun, yAun, y−un r1 n
y−un, Jun−Jyn ≥0, ∀y∈C,
Cn1v∈Cn:φv, un≤βnφv, xn 1−βnφv, zn≤φv, xn;
xn1 ΠCn1x0, ∀n≥0.
1.8
Zhang 1 proved the strong convergence of the sequence {xn} toΠFS1∩FS2∩EPx0 under
On the other hand, a classic method of solving 0 ∈ Tx in a Hilbert space H is the proximal point algorithm which generates, for any starting pointx0 ∈H, a sequence{xn}in Hby the iterative scheme
xn1Jrnxn, n0,1,2, . . . , 1.9
where{rn} is a sequence in0,∞,Jr IrT−1 for eachr > 0 is the resolvent operator forT, andI is the identity operator onH. This algorithm was first introduced by Martinet 19and further studied by Rockafellar20in the framework of a Hilbert space H. Later several authors studied1.9and its variants in the setting of a Hilbert spaceHor in a Banach spaceX; See, for example,15,21–25and references therein. Very recently, Li and Song24
introduced and studied the following iterative scheme:
x0∈X chosen arbitrarily,
ynJ−1βnJxn1−βnJJrnxn
,
xn1J−1αnJx0 1−αnJyn
, n0,1,2, . . . ,
1.10
whereJr JrT−1JandJis the duality mapping onX.
Algorithm1.10covers, as special cases, the algorithms introduced by Kohsaka and Takahashi23and Kamimura et al.22in a smooth and uniformly convex Banach spaceX. Let X be a uniformly smooth and uniformly convex Banach space, and let C be a nonempty closed convex subset ofX. LetT :X → 2X∗be a maximal monotone operator such that:
A5T−10∩EPf/∅.
In addition, for eachr >0, define a mappingTr :X → Cas follows:
Trx
z∈C:fz, y1 r
y−z, Jz−Jx ≥0, ∀y∈C
1.11
for allx∈X.
Very recently, utilizing the ideas of the above algorithms in15,16,18,21,22,24, we
17introduced two iterative methods for finding an element ofT−10∩EPfand established
the following strong and weak convergence theorems.
Theorem 1.1see17. Suppose that conditions (A1)–(A5) are satisfied and letx0 ∈X be chosen arbitrarily. Consider the sequence
where
Hnz∈C:φz, Trnyn
≤αnφz, x0 1−αnφz, xn, Wn{z∈C:xn−z, Jx0−Jxn ≥0},
ynJ−1αnJx
0 1−αn
βnJxn1−βnJJrnxn
,
1.13
Tr is defined by 1.11,{αn},{βn} ⊂ 0,1 satisfy limn→ ∞αn 0, lim infn→ ∞βn1−βn > 0, and {rn} ⊂ 0,∞ satisfies lim infn→ ∞rn > 0. Then, the sequence {xn} converges strongly to ΠT−10∩EPfx0, whereΠT−10∩EPfis the generalized projection ofXontoT−10∩EPf.
Theorem 1.2see17. Suppose that conditions (A1)–(A5) are satisfied and letx0 ∈X be chosen arbitrarily. Consider the sequence
xn1J−1αnJx0 1−αnβnJTrnxn
1−βnJJrnTrnxn
, n0,1,2, . . . , 1.14
where Tr is defined by 1.11, {αn},{βn} ⊂ 0,1 satisfy the conditions ∞n0αn < ∞ and lim infn→ ∞βn1 − βn > 0, and {rn} ⊂ 0,∞ satisfies lim infn→ ∞rn > 0. If J is weakly sequentially continuous, then {xn} converges weakly to an element z ∈ T−10 ∩ EPf, where zlimn→ ∞ΠT−10∩EPfxn.
The purpose of this paper is to introduce and study two new iterative methods for finding a common element of the solution setEP of generalized equilibrium problem1.3
and the setT−10 ∩T−10 for maximal monotone operators T and T in a uniformly smooth
and uniformly convex Banach space X. Firstly, motivated by Theorem 1.1 and a result of Zhang 1, we introduce a sequence{xn} that converges strongly toΠT−10∩T−10∩EPx0 under
some appropriate conditions.
Secondly, inspired byTheorem 1.2and a result of Zhang 1, we define a sequence
that converges weakly to an elementz∈T−10∩T−10∩EP, wherezlimn
→ ∞ΠT−10∩T−10∩EPxn Section 4.
Our results represent a generalization of known results in the literature, including those in 16–18, 24. Our Theorems 3.1 and 4.2 are the extension and improvements of Theorems1.1and1.2in the following way:
ithe problem of finding an element ofT−10∩T−10∩EP includes the one of finding an element ofT−10∩EPfas a special case;
iithe algorithms in this paper are very different from those in 17 because of considering the complexity involving the problem of finding an element ofT−10∩
T−10∩EP.
2. Preliminaries
Throughout the paper, we denote the strong convergence, weak convergence, and weak∗ convergence of a sequence{xn}to a pointx∈Xbyxn → x,xn xandxn x, respectively.∗
mapping and letf : C×C → Rbe a bifunction satisfying the conditions A1–A4. Let T,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, Jy y2, ∀x, y∈X. 2.1
It is clear that in a Hilbert spaceH,2.1reduces toφx, y x−y2, ∀x, y∈H.
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. 2.2
The existence and uniqueness of the operatorΠC follow from the properties of the functional φx, y and strict monotonicity of the mapping J; See, for example, 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. 2.3
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. 2.4
LetSbe a mapping fromCinto itself. A pointpinCis called an asymptotic fixed point ofS28ifCcontains a sequence{xn}which converges weakly topsuch thatSxn−xn → 0. The set of asymptotic fixed points ofSis denoted byFS. A mapping SfromSinto itself is called relatively nonexpansive18,29,30ifFS FSandφp, Sx≤φp, x, for allx∈C andp∈FS.
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, from2.3, we havexy, which implies thatx, Jyx2y2.
From the definition ofJ, we haveJxJyand therefore,xy. For further details, we refer to31.
Lemma 2.2see32. LetXbe a smooth and uniformly convex Banach space and let{xn}and{yn} be two sequences ofX. Ifφxn, yn → 0and either{xn}or{yn}is bounded, thenxn−yn → 0.
Lemma 2.3see26,32. LetCbe a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach spaceX,x∈Xandz∈C. Then
z ΠCx⇐⇒y−z, Jx−Jz ≤0, ∀y∈C. 2.5
Lemma 2.4see26,32. LetCbe a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach spaceX. Then
φx,ΠCyφΠCy, y≤φx, y, ∀x∈C, y∈X. 2.6
Lemma 2.5 see33. LetX be a reflexive, strictly convex and smooth Banach space and letT : X → 2X∗be a multivalued operator. Then
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 2.6see34. 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, 2.7
for allx, y∈Br andt∈0,1, whereBr {z∈X:z ≤r}.
Lemma 2.7see32. LetXbe a smooth and uniformly convex Banach space and letr >0. Then there exists a strictly increasing, continuous, and convex functiong :0,2r → Rsuch thatg0 0 and
gx−y≤φx, y, ∀x, y∈Br. 2.8
The following result is due to Blum and Oettli14.
Lemma 2.8see14. LetCbe a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach spaceX,f :C×C → Ra bifunction satisfying conditions (A1)–(A4), andr >0 andx∈X. Then, there existsz∈Csuch that
fz, y1ry−z, Jz−Jx ≥0, ∀y∈C. 2.9
Lemma 2.9 see18. LetCbe a nonempty closed convex subset of a uniformly smooth, strictly convex and reflexive Banach spaceX, andf :C×C → Ra bifunction satisfying conditions (A1)– (A4). Forr >0andx∈X, define a mappingTr :X → Cas follows:
Trx
z∈C:fz, y1 r
y−z, Jz−Jx ≥0, ∀y∈C
2.10
for allx∈X. Then
iTr is single-valued;
iiTr is a firmly nonexpansive-type mapping, that is, for allx, y∈X,
Trx−Try, JTrx−JTry ≤Trx−Try, Jx−Jy ; 2.11
iiiFTr FT r EPf; ivEPfis closed and convex.
UsingLemma 2.9, we have the following result.
Lemma 2.10see18. LetCbe a nonempty closed convex subset of a smooth, strictly convex and reflexive Banach spaceX,f :C×C → Ra bifunction satisfying conditions (A1)–(A4), andr >0. Then, forx∈Xandq∈FTr,
φq, TrxφTrx, x≤φq, x. 2.12
Utilizing Lemmas2.8,2.9, and2.10, Zhang1derived the following result.
Proposition 2.11see1. LetX be a smooth, strictly convex and reflexive Banach space and let Cbe a nonempty closed convex subset ofX. LetA : C → X∗ be anα-inverse-strongly monotone mapping,f:C×C → Ra bifunction satisfying conditions (A1)–(A4), andr >0. Then
Iforx∈X, there existsu∈Csuch that
fu, yAu, y−u 1ry−u, Ju−Jx ≥0, ∀y∈C; 2.13
IIifXis additionally uniformly smooth andKr:C → Cis defined as
Krx
u∈C:fu, yAu, y−u 1 r
y−u, Ju−Jx ≥0, ∀y∈C
, ∀x∈C, 2.14
then the mappingKrhas the following properties:
iKris single-valued,
iiKris a firmly nonexpansive-type mapping, that is,
iiiFKr FK r EP,
ivEPis a closed convex subset ofC,
vφp, Krx φKrx, x≤φp, x,for allp∈FKr.
Proof. Define a bifunctionF:C×C → Rby
Fx, yfx, yAx, y−x , ∀x, y∈C. 2.16
It is easy to verify thatFsatisfies the conditionsA1–A4. Therefore, the conclusionsIand IIfollow immediately from Lemmas2.8,2.9, and2.10.
LetT,T:X → 2X∗be two maximal monotone operators in a smooth Banach spaceX. We denote the resolvent operators ofTandTbyJr JrT−1JandJr JrT −1Jfor each r >0, respectively. ThenJr :X → DTandJr :X → DT are two single-valued mappings. Also,T−10FJrandT−10FJrfor eachr >0, whereFJrandFJrare the sets of fixed
points ofJr andJr, respectively. For eachr > 0, the Yosida approximations ofT andTare defined byAr J−JJr/randAr J−JJr/r, respectively. It is known that
Arx∈TJrx, Arx∈T
Jrx
, for eachr >0, x∈X. 2.17
Lemma 2.12see23. LetXbe 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 2.13see36. Let{an}and{bn}be two sequences of nonnegative real numbers such that 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.
Theorem 3.1. Suppose thatAssumption 2.1is satisfied. Letx0 ∈X be chosen arbitrarily. Consider the sequence
where
Hnz∈C:φz, Krnyn
≤αnαn−αnαnφz, x0 1−αn1−αnφz, xn, Wn{z∈C:xn−z, Jx0−Jxn ≥0},
xnJ−1αnJx0 1−αnβnJxn1−βnJJrnxn
,
ynJ−1
αnJx0 1−αn
βnJxn
1−βn
JJrnxn
,
3.2
Kr is defined by2.14,{αn},{βn},{αn},{βn} ⊂0,1satisfy
lim
n→ ∞αn0, nlim→ ∞αn0, lim infn→ ∞ βn
1−βn>0, lim infn
→ ∞ βn
1−βn
>0, 3.3
and {rn} ⊂ 0,∞ satisfies lim infn→ ∞rn > 0. Then, the sequence {xn} converges strongly to ΠT−10∩T−10∩EPx0, whereΠT−10∩T−10∩EPis the generalized projection ofXontoT−10∩T−10∩EP.
Proof. For the sake of simplicity, we define
un:Krnyn, zn:J−1
βnJxn1−βnJJrnxn
, zn:J−1
βnJxn
1−βn
JJrnxn
, 3.4
so that
xnJ−1αnJx0 1−αnJzn, ynJ−1αnJx0 1−αnJzn. 3.5
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 us show thatHnis convex. Forz1, z2∈Hnandt∈0,1, putztz1 1−tz2. It is sufficient
to show thatz∈Hn. We first writeγnαnαn−αnαnfor eachn≥0. Next, we prove that
φz, un≤γnφz, x0
1−γnφz, xn 3.6
is equivalent to
2γnz, Jx02
1−γnz, Jxn −2z, Jun ≤γnx02
1−γnxn2− un2. 3.7
Indeed, from2.1we deduce that there hold the following:
φz, x0 z2−2z, Jx0x02,
φz, xn z2−2z, Jxnxn2,
φz, un z2−2z, Junun2,
which combined with3.6yield that3.6is equivalent to3.7. Thus we have
2γnz, Jx02
1−γnz, Jxn −2z, Jun
2γntz1 1−tz2, Jx02
1−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.
3.9
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∈Csuch thatfun, yAun, y−un r1 n
y−un, Jun−Jyn ≥0, ∀y∈C. 3.10
Then fromLemma 2.12, we obtain
φw, zn φ
w, J−1β
nJxn1−βnJJrnxn
w2−2w, β
nJxn1−βnJJrnxn βnJxn 1−βnJJrnxn 2
≤ w2−2β
nw, Jxn −21−βnw, JJrnxnβnxn
21−β
nJrnxn 2
βnφw, xn 1−βnφw, Jrnxn
≤βnφw, xn 1−βnφw, xn φw, xn,
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.
Moreover, we have
φw,zn φ
w, J−1β nJxn
1−βn
JJrnxn
≤βnφw,xn
1−βn
φw,Jrnxn
≤βnφw,xn
1−βn
φw,xn φw,xn,
φw, ynφ
w, J−1α
nJx0 1−αnJzn
≤ w2−2α
nw, Jx0 −21−αnw, Jznαnx02 1−αnzn2
αnφw, x0 1−αnφw,zn
≤αnφw, x0 1−αnφw,xn
≤αnφw, x0 1−αnαnφw, x0 1−αnφw, xn
αn 1−αnαnφw, x0 1−αn1−αnφw, xn
≤αnαn−αnαnφw, x0 1−αn1−αnφw, xn,
3.12
and hence byProposition 2.11,
φw, un φw, Krnyn
≤φw, yn
≤αnαn−αnαnφw, x0 1−αn1−αnφw, xn.
3.13
Sow∈Hnfor alln≥0. Now, let us show that
T−10∩T−10∩EP ⊂W
n ∀n≥0. 3.14
We prove this by induction. Forn 0, we haveT−10∩T−10∩EP ⊂ C W
0. Assume that
T−10∩T−10∩EP ⊂Wn. Sincexn
1is the projection ofx0ontoHn∩Wn, byLemma 2.3we have
xn1−z, Jx0−Jxn1 ≥0, ∀z∈Hn∩Wn. 3.15
AsT−10∩T−10∩EP ⊂ Hn∩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 ⊂W
n1. Hence3.14holds 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 ΠWnx0andLemma 2.4that
φxn, x0 φΠWnx0, x0≤φ
p, x0
−φp, xn≤φp, x0
for eachp∈T−10∩T−10∩EP ⊂Wnfor eachn≥0. Therefore,{φxn, x
0}is bounded, which
implies that the limit of{φxn, x0}exists. Since
xn − x02≤φxn, x0≤xnx02, ∀n≥0, 3.17
so{xn}is bounded. FromLemma 2.4, we have
φxn1, xn φxn1,ΠWnx0≤φxn1, x0−φΠWnx0, x0
φxn1, x0−φxn, x0,
3.18
for eachn≥0. This implies that
lim
n→ ∞φxn1, xn 0. 3.19
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≤αnαn−αnαnφxn1, x0 1−αn1−αnφxn1, xn, ∀n≥0. 3.20
Therefore, fromαn → 0αn → 0 andφxn1, xn → 0, it follows that limn→ ∞φxn1, un 0. Since limn→ ∞φxn1, xn limn→ ∞φxn1, un 0 and X is uniformly convex and smooth, we have fromLemma 2.2that
lim
n→ ∞xn1−xnnlim→ ∞xn1−un0, 3.21
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 toLemma 2.5andProposition 2.11, we know thatΩis a nonempty closed convex subset ofXsuch thatΩ⊂C. Fixu∈Ωarbitrarily. As in the proof ofStep 2, we can show thatφu, zn≤φu, xn,
φu,xn≤αnφu, x0 1−αnφu, xn,
φu,zn≤φu,xn,
φu, yn≤αnαn−αnαnφu, x0 1−αn1−αnφu, xn,
φu, un≤αnαn−αnαnφu, x0 1−αn1−αnφu, xn.
Hence it follows from the boundedness of{xn}that{zn},{xn},{zn},{yn}, and{un}are also bounded. Letr sup{xn,xn,Jrnxn,Jrnxn : n ≥ 0}. SinceX is a uniformly smooth
Banach space, we know thatX∗is a uniformly convex Banach space. Therefore, byLemma 2.6
there exists a continuous, strictly increasing, and convex functiong, withg0 0, such that
αx∗ 1−αy∗2≤αx∗2 1−αy∗2−α1−αgx∗−y∗, 3.23
forx∗, y∗∈Br∗andα∈0,1. So, we have that
φu, zn φ
u, J−1β
nJxn1−βnJJrnxn
u2−2u, β
nJxn1−βnJJrnxn βnJxn 1−βnJJrnxn 2
≤ u2−2β
nu, Jxn −21−βnu, JJrnxn
βnxn21−βnJrnxn 2−β
n1−βngJxn−JJrnxn
βnφu, xn 1−βnφu, Jrnxn−βn
1−βngJxn−JJrnxn
≤βnφu, xn 1−βnφu, xn−βn1−βngJxn−JJrnxn
φu, xn−βn1−βngJxn−JJrnxn,
φu,zn φ
u, J−1β nJxn
1−βn
JJrnxn
u2−2u,β nJxn
1−βn
JJrnxn
βnJxn 1−βnJJrnxn 2
≤ u2−2β
nu, Jxn −2
1−βn
u, JJrnxn
βnxn2
1−βnJrnxn 2
−βn
1−βn
gJxn−JJrnxn
βnφu,xn
1−βn
φu,Jrnxn
−βn
1−βn
gJxn−JJrnxn
≤βnφu,xn
1−βn
φu,xn−βn
1−βn
gJxn−JJrnxn
φu,xn−βn
1−βn
gJxn−JJrnxn
,
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−αnβn
1−βngJxn−JJrnxn,
φu, un φu, Krnyn
≤φu, yn using Proposition 2.10
φu, 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−αn
φu,xn−βn
1−βn
gJxn−JJrnxn
αnφu, x0 1−αnφu,xn−1−αnβn
1−βn
gJxn−JJrnxn
≤αnφu, x0 φu,xn,
3.25
for alln≥0. Consequently, we have
1−αnβn1−βngJxn−JJrnxn
≤αnϕu, x0 1−αnϕu, xn−ϕu,xn
≤αnϕu, x0 ϕu, xn−ϕu,xn
αnϕu, x0 ϕu, xn−ϕu, un ϕu, un−ϕu,xn
αnϕu, x0 xn2− un2−2u, Jxn−Junϕu, un−ϕu,xn
≤αnϕu, x0 xn2− un22|u, Jxn−Jun|ϕu, un−ϕu,xn
≤αnϕu, x0 αnϕu, x0 |xn − un|xnun 2uJxn−Jun
≤αnαnϕu, x0 xn−unxnun 2uJxn−Jun,
Sincexn−un → 0 andJis uniformly norm-to-norm continuous on bounded subsets ofX, we obtainJxn−Jun → 0. From lim infn→ ∞βn1−βn>0 and limn→ ∞αnαn 0, we have
lim
n→ ∞gJxn−JJrnxn 0. 3.27
Therefore, from the properties ofg, we get
lim
n→ ∞Jxn−JJrnxnnlim→ ∞xn−Jrnxn0, 3.28a
recalling thatJ−1is uniformly norm-to-norm continuous on bounded subsets ofX∗. Next let
us show that
lim n→ ∞
Jxn−JJrnxnnlim→ ∞xn−Jrnxn0. 3.28b
Observe first that
φun, xn−φxn1, un
xn2− xn12−2un, Jxn2xn1, Jun
xn − xn1xnxn1 2xn1−un, Jxn2xn1, Jun−Jxn
≤ xn−xn1xnxn1 2xn1−unxn2xn1Jun−Jxn.
3.29
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.
3.30
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 βnJxn 1−βnJJrnxn 2
≤ un2−2βnun, Jxn −21−βnun, JJrnxnβnxn2
1−βnJrnxn2
βnφun, xn 1−βnφun, Jrnxn ≤φun, xn φun, Jrnxn,
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.
3.32
Sinceαn → 0, φun, xn → 0 andφun, Jrnxn → 0, it follows from the boundedness of {un}thatφun,xn → 0. Thus, in terms ofLemma 2.2, we have thatun−xn → 0 and so
xn−xn → 0. Furthermore, it follows from3.25that
φu, un≤αnφu, x0 1−αnφu,xn−1−αnβn
1−βn
gJxn−JJrnxn
≤αnφu, x0 φu,xn−1−αnβn
1−βn
gJxn−JJrnxn
, 3.33
and hence
1−αnβn
1−βn
gJxn−JJrnxn
≤αnφu, x0 φu,xn−φu, un
αnφu, x0 xn2− un22u, Jun−Jxn
αnφu, x0 xn − unxnun 2u, Jun−Jxn
≤αnφu, x0 xn−unxnun 2uJun−Jxn.
3.34
Since J is uniformly norm-to-norm continuous on bounded subsets ofX, it follows from
xn−un → 0 thatJun−Jxn → 0. Thus fromαn → 0, lim infn→ ∞βn1−βn > 0, and the boundedness of both{xn}and{un}, we deduce thatgJxn−JJrnxn → 0. Utilizing
the properties ofg, we have thatJxn−JJrnxn → 0. SinceJ−1is uniformly norm-to-norm
continuous on bounded subsets ofX∗, it follows thatxn−Jrnxn → 0.
Step 5. We claim thatωw{xn}⊂T−10∩T−10∩EP, where
ωw{xn}:x∈C:xnk xfor some subsequence{nk} ⊂ {n}withnk↑ ∞
. 3.35
Indeed, since{xn}is bounded andXis reflexive, we know thatωw{xn}/∅. Takex∈ ωw{xn}arbitrarily. Then there exists a subsequence{xnk}of{xn}such thatxnk x. Hence
and{Jrnkxnk}converge weakly to the same pointx. On the other hand, from 3.28a,3.28b
and lim infn→ ∞rn>0, we obtain that
lim
n→ ∞Arnxnnlim→ ∞
1
rnJxn−JJrnxn0,
lim n→ ∞
A
rnxnnlim → ∞
1 rn
Jxn−JJrnxn0.
3.36
Ifz∗∈Tzandz∗∈Tz, then it follows from 2.17and the monotonicity of the operatorsT, T that for allk≥1
z−Jrnkxnk, z∗−Arnkxnk
≥0, z−Jrnkxnk,z∗−Arnkxnk
≥0. 3.37
Lettingk → ∞, we have thatz−x, z ∗ ≥0 andz−x, z∗ ≥0. Then the maximality of the operatorsT, Timplies thatx∈T−10 andx∈T−10.
Next, let us show thatx∈EP. Since
φu, yn≤αnαn−αnαnφu, x0 1−αn1−αnφu, xn, 3.38
fromun KrnynandProposition 2.11it follows that
φun, ynφKrnyn, yn
≤φu, yn−φu, Krnyn
≤αnαn−αnαnφu, x0 1−αn1−αnφu, xn−φu, Krnyn
≤αnαn−αnαnφu, x0 φu, xn−φu, un.
3.39
Also, since
ϕu, xn−ϕu, unxn2− un22u, Jun−Jxn
≤xn2− un22|u, Jun−Jxn|
|xn − un|xnun 2uJun−Jxn
≤ xn−unxnun 2uJun−Jxn,
3.40
so we get
lim n→ ∞
φu, xn−φu, un0. 3.41
SinceXis uniformly convex and smooth, we conclude fromLemma 2.2that
lim
n→ ∞un−yn0. 3.42
Fromxnk x, xn−un → 0, and3.42, we haveynk xandunk x.
SinceJis uniformly norm-to-norm continuous on bounded subsets ofX, from3.42
we derive
lim
n→ ∞Jun−Jyn0. 3.43
From lim infn→ ∞rn>0, it follows that
lim n→ ∞
Jun−Jyn
rn 0. 3.44
By the definition ofun :Krnyn, we have
Fun, yr1 n
y−un, Jun−Jyn ≥0, ∀y∈C, 3.45
where
Fun, yfun, yAun, y−un . 3.46
Replacingnbynk, we have fromA2that
1 rnk
y−unk, Junk −Jynk ≥ −F
unk, y
≥Fy, unk
, ∀y∈C. 3.47
Sincey→fx, y Ax, y−xis convex and lower semicontinuous, it is also weakly lower semicontinuous. Lettingnk → ∞in the last inequality, from3.44andA4, we have
Fy,x≤0, ∀y∈C. 3.48
Fort, with 0< t≤1, andy∈C, letytty 1−tx. Since y∈Candx∈C, thenyt∈Cand henceFyt,x ≤0. So, fromA1we have
Dividing byt, we have
Fyt, y≥0, ∀y∈C. 3.50
Lettingt↓0, fromA3it follows that
Fx, y ≥0, ∀y∈C. 3.51
So,x∈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. 3.52
Since the norm is weakly lower semicontinuous, then
φx, x 0 x2−2x, Jx 0x02≤lim infk
→ ∞
xnk2−2xnk, Jx0x02
lim inf
k→ ∞ φxnk, x0≤lim supk→ ∞ φxnk, x0≤φw, x0.
3.53
From the definition ofΠT−10∩T−10∩EP, we havexw. Hence limk→ ∞φxnk, x0 φw, x0, and
0 lim k→ ∞
φxnk, x0−φw, x0
lim k→ ∞
xnk2− w2−2xnk−w, Jx0
lim k→ ∞
xnk
2− w2, 3.54
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 3.2. InTheorem 3.1, letA ≡ 0,T ≡ 0, andαn 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, there hold the following
Hnz∈C:φz, Krnyn
≤αnαn−αnαnφz, x0 1−αn1−αnφz, xn z∈C:φz, Trnyn
≤αnφz, x0 1−αnφz, xn
,
ynJ−1
αnJx0 1−αn
βnJxn
1−βn
JJrnxn
J−1β nJxn
1−βn
JJrnxn
J−1β nJxn
1−βn
Jxn
J−1Jx nxn,
3.56
and hence
ynxnJ−1αnJx0 1−αn
βnJxn1−βnJJrnxn
. 3.57
In this case,Theorem 3.1reduces to17, Theorem 3.1.
4. Weak Convergence Theorem
In this section, we present the following algorithm for finding a common element of the solution set of 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−αn
βnJKrnxn
1−βnJJrnKrnxn
,
xn1J−1
αnJx0 1−αn
βnJKrnxn
1−βn
JJrnKrnxn
, n0,1,2, . . . , 4.1
where{αn},{βn},{αn},{βn} ⊂0,1, {rn} ⊂0,∞, andKr, r >0 is defined by2.14. Before proving a weak convergence theorem, we need the following proposition.
Proposition 4.1. Suppose thatAssumption 2.1is fulfilled and let{xn}be a sequence defined by4.1, where{αn},{βn},{αn},{βn} ⊂0,1satisfy the following conditions:
∞
n0
αn<∞,
∞
n0
αn<∞, lim infn
→ ∞ βn
1−βn>0, lim infn
→ ∞ βn
1−βn
>0. 4.2
Proof. We setΩ:T−10∩T−10∩EPand
un:Krnxn, yn:J−1
βnJun1−βnJJrnun
,
un:Krnxn, yn:J−1
βnJun
1−βn
JJrnun
, 4.3
so that
xnJ−1αnJx0 1−αnJyn,
xn1J−1αnJx0 1−αnJyn
, n0,1,2, . . . . 4.4
Then, in terms ofLemma 2.5andProposition 2.11,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∈Cand
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 fromProposition 2.11, we obtain
φu, ynφ
u, J−1β
nJun1−βnJJrnun
u2−2u, β
nJun1−βnJJrnun βnJun 1−βnJJrnun 2
≤ u2−2β
nu, Jun −21−βnu, JJrnunβnun2
1−βnJrnun2
βnφu, un 1−βnφu, Jrnun ≤βnφu, un 1−βnφu, un φu, un φu, Krnxn≤φu, xn,
4.6a
φu,ynφ
u, J−1β nJun
1−βn
JJrnun
u2−2u,β nJun
1−βn
JJrnun
βnJun 1−βnJJrnun 2
≤ u2−2β
nu, Jun −2
1−βn
u, JJrnun
βnun2
1−βnJrnun 2
βnφu,un
1−βn
φu,Jrnun
≤βnφu,un
1−βn
φu,un
φu,un φu, Krnxn≤φu,xn,
and hence byProposition 2.11
φu,xn φ
u, J−1α
nJx0 1−αnJyn
u2−2u, α
nJx0 1−αnJynαnJx0 1−αnJyn2
≤ u2−2α
nu, Jx0 −21−αnu, Jynαnx02 1−αnyn2
αnφu, x0 1−αnφu, yn
≤αnφu, x0 φ
u, yn
≤φu, xn αnφu, x0,
4.6c
φu, xn1 φ
u, J−1α
nJx0 1−αnJyn
u2−2u,α
nJx0 1−αnJynαnJx0 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.
4.6d
Consequently, the last two inequalities yield that
φu, xn1≤φu,xn αnφu, x0
≤φu, xn αnφu, x0 αnφu, x0
φu, xn αnαnφu, x0,
4.6e
for all n ≥ 0. So, from ∞n0αn < ∞, ∞n0αn < ∞, and Lemma 2.13, 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,
4.7
for eachp ∈ Ω. This, together with the boundedness of {xn}, implies that{zn}is bounded and so isφzn, x0. Furthermore, fromzn∈Ωand4.6e, we have
SinceΠΩis the generalized projection, then, fromLemma 2.4we obtain
φzn1, xn1 φΠΩxn1, xn1≤φzn, xn1−φzn,ΠΩxn1
φzn, xn1−φzn, zn1≤φzn, xn1.
4.9
Hence, from4.8, it follows thatφzn1, xn1≤φzn, xn αnαnφzn, x0.
Note that∞n0αn < ∞,∞n0αn < ∞, and{φzn, x0}is bounded, so that∞n0αn
αnφzn, x0 <∞. Therefore,{φzn, xn}is a convergent sequence. On the other hand, from 4.6ewe derive, for allm≥0,
φu, xnm≤φu, xn m−1
j0
αnjαnjφu, x0. 4.10
In particular, we have
φzn, xnm≤φzn, xn m−1
j0
αnjαnjφzn, x0. 4.11
Consequently, fromznm ΠΩxnmandLemma 2.4, we have
φzn, znm φznm, xnm≤φzn, xnm≤φzn, xn m−1
j0
αnjαnjφzn, x0 4.12
and hence
φzn, znm≤φzn, xn−φznm, xnm m−1
j0
αnjαnjφzn, x0. 4.13
Letrsup{zn:n≥0}. FromLemma 2.7, there exists a continuous, strictly increasing, and convex functiongwithg0 0 such that
gx−y≤φx, y, ∀x, y∈Br. 4.14
So, we have
gzn−znm≤φzn, znm
≤φzn, xn−φznm, xnm m−1
j0
αnjαnjφzn, x0.
4.15
Since {φzn, xn} is a convergent sequence, {φzn, x0} is bounded and ∞
n0αn αn is convergent; from the property ofg, we have that{zn}is a Cauchy sequence. SinceΩis closed,
Now, we are in a position to prove the following theorem.
Theorem 4.2. Suppose thatAssumption 2.1is fulfilled and let{xn}be a sequence defined by4.1, where{αn},{βn},{αn},{βn} ⊂0,1satisfy the following conditions:
∞
n0
αn<∞,
∞
n0
αn<∞, lim infn
→ ∞ βn
1−βn>0, lim infn
→ ∞ βn
1−βn
>0, 4.16
and {rn} ⊂ 0,∞satisfieslim infn→ ∞rn > 0. If J is weakly sequentially continuous, then{xn} converges weakly toz∈T−10∩T−10∩EP, wherezlimn
→ ∞ΠT−10∩T−10∩EPxn.
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,Jrnun:n≥0
. 4.17
From Lemma 2.6 and as in the proof of Theorem 3.1, there exists a continuous, strictly increasing, and convex functiongwithg0 0 such that
αx∗ 1−αy∗2 ≤αx∗2 1−αy∗2−α1−αgx∗−y∗ 4.18
forx∗, y∗∈Br∗andα∈0,1. Observe that foru∈Ω:T−10∩T−10∩EP,
φu, yn
φu, J−1β
nJun1−βnJJrnun
u2−2u, β
nJun1−βnJJrnun βnJun 1−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,ynφ
u, J−1β nJun
1−βn
JJrnun
u2−2u,β nJun
1−βn
JJrnun
βnJun 1−βnJJrnun 2
≤ u2−2β
nu, Jun −2
1−βn
u, JJrnun
βnun2
1−βnJrnun 2
−βn
1−βn
gJun−JJrnun
≤βnφu,un
1−βn
φu,Jrnun
−βn
1−βn
gJun−JJrnun
≤βnφu,un
1−βn
φu,un−βn
1−βn
gJun−JJrnun
φu,un−βn
1−βn
gJun−JJrnun
.
4.19
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, Jyn αnx02 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,
φu, xn1 φ
u, J−1α
nJx0 1−αnJyn
u2−2u,α
nJx0 1−αnJyn αnJx0 1−αnJyn2
≤ u2−2α
nu, Jx0 −21−αnu, Jyn αnx02 1−αnyn2
αnφu, x0 1−αnφu,yn
≤αnφu, x0 φ
u,yn
≤αnφu, x0 φu,un−βn
1−βn
gJun−JJrnun
αnφu, x0 φu, Krnxn−βn
1−βn
gJun−JJrnun
≤αnφu, x0 φu,xn−βn
1−βn
gJun−JJrnun
.
Consequently, the last two inequalities yield that
φu, xn1≤αnφu, x0 φu,xn−βn
1−βn
gJun−JJrnun
≤αnφu, x0 αnφu, x0 φu, xn−βn1−βngJun−JJrnun
−βn
1−βn
gJun−JJrnun
φu, xn αnαnφu, x0−βn1−βngJun−JJrnun
−βn
1−βn
gJun−JJrnun
.
4.21
Thus, we have
βn1−βngJun−JJrnun βn
1−βn
gJun−JJrnun
≤φu, xn−φu, xn1 αnαnφu, x0.
4.22
By the proof ofProposition 4.1, it is known that{φu, xn}is convergent; since limn→ ∞αn0, limn→ ∞αn0, lim infn→ ∞βn1−βn>0, and lim infn→ ∞βn1−βn>0, then we have
lim
n→ ∞gJun−JJrnun nlim→ ∞gJun−JJrnun
0. 4.23
Taking into account the properties ofg, as in the proof ofTheorem 3.1, we have
lim
n→ ∞Jun−JJrnunnlim→ ∞un−Jrnun0,
lim n→ ∞
Jun−JJrnunnlim→ ∞un−Jrnun0,
4.24
sinceJ−1is uniformly norm-to-norm continuous on bounded subsets ofX∗.
Now let us show that
lim
n→ ∞φu, xn nlim→ ∞φu,xn nlim→ ∞φu, un nlim→ ∞φu,un. 4.25
Indeed, from4.6ewe get
φu, xn1−αnφu, x0≤φu,xn≤φu, xn αnφu, x0, 4.26
which, together with limn→ ∞αnlimn→ ∞αn0, yields that
lim
From4.6dit follows that
φu, xn1−αnφu, x0≤φ
u,yn≤φu,xn, 4.28
which, together with limn→ ∞φu,xn limn→ ∞φu, xn, yields that
lim n→ ∞φ
u,ynnlim→ ∞φu, xn. 4.29
From4.6cit follows that
φu,xn−αnφu, x0≤φ
u, yn≤φu, xn, 4.30
which, together with limn→ ∞φu,xn limn→ ∞φu, xn, yields that
lim n→ ∞φ
u, ynnlim
→ ∞φu, xn. 4.31
From4.6cit follows that
φu,yn≤φu,un≤φu,xn, 4.32
which together with
lim
n→ ∞φu,xn nlim→ ∞φ
u,ynnlim
→ ∞φu, xn, 4.33
yields that
lim
n→ ∞φu,un nlim→ ∞φu, xn. 4.34
From4.6ait follows that
φu, yn≤φu, un≤φu, xn 4.35
which, together with limn→ ∞φu, yn limn→ ∞φu, xn, yields that
lim
n→ ∞φu, un nlim→ ∞φu, xn. 4.36
On the other hand, let us show that
lim
Indeed, lets sup{xn,un,xn,un : n ≥ 0}. FromLemma 2.7, there exists a continuous, strictly increasing, and convex functiong1withg10 0 such that
g1x−y≤φ
x, y, ∀x, y∈Bs. 4.38
SinceunKrnxnandunKrnxn, we deduce fromProposition 2.11that foru∈Ω,
g1un−xn≤φun, xn≤φu, xn−φu, un,
g1un−xn≤φun,xn≤φu,xn−φu,un.
4.39
This implies that
lim
n→ ∞g1un−xn nlim→ ∞g1un−xn 0. 4.40
SinceJis uniformly norm-to-norm continuous on bounded subsets ofX, from the properties ofg1, we obtain
lim
n→ ∞un−xnnlim→ ∞Jun−Jxn0,
lim
n→ ∞un−xnnlim→ ∞Jun−Jxn0.
4.41
Note that
φxn, un−φun, xn xn2−2xn, Junun2−
xn2−2un, Jxnun2
−2xn, Jun2un, Jxn
2xn, Jxn−Jun2un−xn, Jxn
≤2xnJxn−Jun2un−xnxn,
φxn, Jrnun xn2−2xn, JJrnunJrnun2
xn2− xn2Jrnun2− xn22xn, Jxn−JJrnun Jrnun − xnJrnunxn 2xn, Jxn−JJrnun ≤ Jrnun−xnJrnunxn 2xnJxn−JJrnun
Jrnun−unun−xnJrnunxn
2xnJxn−JunJun−JJrnun ≤Jrnun−unun−xnJrnunxn
2xnJxn−JunJun−JJrnun.
4.42
Sinceφun, xn → 0, it follows from4.24and4.41thatφxn, un → 0 andφxn, Jrnun →
