https://doi.org/10.1051/ro/2020085 www.rairo-ro.org
SPLIT VARIATIONAL INCLUSIONS FOR BREGMAN MULTIVALUED MAXIMAL MONOTONE OPERATORS
Mujahid Abbas
1, Faik G¨ ursoy
2,∗, Yusuf Ibrahim
3and Abdul Rahim Khan
4Abstract.We introduce a new algorithm to approximate a solution of split variational inclusion prob- lems of multivalued maximal monotone operators in uniformly convex and uniformly smooth Banach spaces under the Bregman distance. A strong convergence theorem for the above problem is established and several important known results are deduced as corollaries to it. As application, we solve a split minimization problem and provide a numerical example to support better findings of our result.
Mathematics Subject Classification. 47J25.
Received January 27, 2020. Accepted July 29, 2020.
1. Introduction
Censor [8] imposed the well known split feasibility problem (SFP), which is formulated as finding a point x∗∈Csuch thatAx∗∈Q, whereCandQare nonempty closed and convex subsets ofRn andRm, respectively, whereAis anm×nmatrix. Byrne [3], defined CQ-algorithm as follows:
xn+1=PC(xn+γAT(PQ−I)Axn), n≥0,
where x0 ∈ Rn is an initial value, γ ∈ (0,kAk22) and PC and PQ denote the metric projections onto C and Q, respectively. The split feasibility problem has been considered by many authors and in many aspects [1–3,5,8,9,13,16,25,26,30]. In practice, SFP serves as a model in the intensity-modulation radiation ther- apy (IMRT) treatment planning [2,5]. Censor et al. [10] introduced a concept of Split Variational Inequality Problem (SVIP), which is a problem of finding a pointx∗∈H1solves
hf(x∗), x−x∗i ≥0 for allx∈C, and the pointy∗=Ax∗∈H2 such that
hg(y∗), y−y∗i ≥0 for ally∈Q,
Keywords.Split variational inclusion problem, maximal monotone operators, Bregman distance, strong convergence, uniformly convex and uniformly smooth Banach space.
1 Department of Mathematics, Government College University, Katchery Road, Lahore 54000, Pakistan.
2 Department of Mathematics, Adiyaman University, Adiyaman 02040, Turkey.
3 Department of Mathematics, Sa’adatu Rimi College of Education, Kumbotso Kano, P.M.B. 3218 Kano, Nigeria.
4 Department of Mathematics and Statistics, King Fahd University of Petroleum and Minerals, Dhahran 31261, Saudi Arabia.
∗Corresponding author:faikgursoy02@hotmail.com
Article published by EDP Sciences c EDP Sciences, ROADEF, SMAI 2021
where C and Q are closed and convex subsets of Hilbert spaces H1 and H2, respectively,A : H1 → H2 is a bounded linear operator and A∗ : H2 → H1 is adjoint of A, f : H1 → H1 and g : H2 → H2 are two given operators. Furthermore, they proposed the following algorithm. Let λ >0 and x1 ∈ H1 be arbitrary chosen.
Define the sequence{xn} by
xn+1=PCf,λ(xn+γA∗(PQg,λ−I)Axn)),∀n≥0, (1.1) where γ∈(0,kAk12), and denoted byPCf,λ and PQg,λ the expressionsPC(I−λf) and PQ(I−λg), respectively.
By some assumptions imposed on the operatorsf andg, they proved weak convergence result for the sequence {xn} to a solution point of split variational inequality problem.
Let E be a real normed space with dual E∗ and J(x) ={x∗ ∈E∗;hx, x∗i=kxkkx∗k,kx∗k =kxk} be the normalized duality. A mapB :E →E∗ is called monotone if for each x, y∈E, the following inequality holds:
hη−ν, x−yi ≥ 0∀η ∈ Bx, ν ∈ By. It is called maximal monotone if, in addition, the graph of B is not properly contained in the graph of any other monotone operator. Also, B is maximal monotone if and only if it is monotone and for all t > 0,R(J +tB) = E∗, where R(J +tB) is the range of (J +tB); see [4]. By using maximal monotone mappings, Moudafi [15] introduced the following Split Monotone Variational Inclusion (SMVI).
(findx∗∈H1: 0∈f(x∗) +B1(x∗), and
y∗=Ax∗∈H2: 0∈g(y∗) +B2(y∗), (1.2) where B1 : H1 → 2H1 andB2 : H2 →2H2 are multi-valued maximal monotone mappings on Hilbert spaces, H1 andH2, respectively, and A:H1 →H2 is a bounded linear operator, f :H1→H1 and g :H2 →H2 are two given single-valued operators. Whenf andgare zero functions in (1.2), we have the usual Split Variational Inclusion Problem (SVIP). The algorithm introduced by Sch¨opferet al. [20] involves computations in terms of Bregman distance in the setting of p-uniformly convex and uniformly smooth real Banach spaces. Their iterative algorithm given below, converges weakly under some suitable conditions.
xn+1= ΠCJ−1(J xn+γA∗J(PQ−I)Axn), n≥0, (1.3) where ΠC denotes the Bregman Projection andA∗the adjoint operator ofA. It is obvious that, strong conver- gence is more useful than the weak convergence in some applications. Recently, strong convergence theorems for SFP have been studied in the setting of p-uniformly convex and uniformly smooth real Banach spaces; see for example [11,17,22,23].
In this paper, inspired by the above cited works, we use a modified version of (1.1) and (1.3) to approximate a solution of the problem proposed here. Both the iterative methods and the underlying space used here are improvements of those employed in [6,7,10,11,13,17,20,22,23,28] and the references therein.
Definition 1.1. For eachp >1, letg:R+−→R+ given byg(t) =tp−1be a gauge function such thatg(0) = 0 and limt→∞g(t) =∞.We define the generalized duality mapJp:E−→2E∗ given by
Jg(t)=Jp(x) ={x∗∈E∗;hx, x∗i=kxkkx∗k,kx∗k=g(kxk) =kxkp−1}.
Definition 1.2. Let E be a smooth Banach space, the Bregman distance 4p of xto y, with respect to the convex continuous functionf :E→Rgiven byf(x) =1pkxkp, is defined as
4p(x, y) = 1
qkxkp− hJp(x), yi+1 pkykp, for allx, y∈E andp, q∈(1,∞) such that 1p+1q = 1.
Definition 1.3. Let E be a smooth Banach space and E∗ its dual, the bifunctional Vp with respect to the convex continuous functionf :E→Rgiven byf(x) =1pkxkp, is defined by
Vp(x∗, x) = 1
qkx∗kq− hx∗, xi+1 pkxkp, for allx∈E,x∗∈E∗ andp, q∈(1,∞) such that 1p+1q = 1.
Definition 1.4. A Banach space E is said to be uniformly convex, if for x, y ∈ E, 0 < δE() ≤ 1, where δE() = inf{1− k12(x+y)k;kxk=kyk= 1,kx−yk ≥,where 0≤≤2}.
Definition 1.5 ([19]). A Banach space E is said to be uniformly smooth, if for x, y ∈ E and r > 0, limr→0(ρEr(r)) = 0 whereρE(r) =12sup{kx+yk+kx−yk −2 :kxk= 1,kyk ≤r}. Moreover,
(1) ρE is continuous, convex and nondecreasing withρE(0) = 0 andρE(r)≤r.
(2) The functionr7→ ρEr(r) is nondecreasing and fulfills ρEr(r) >0 for allr >0.
Lemma 1.6([19]). Let {xn} be a sequence in a smooth Banach spaceE. Consider the following assertions;
(1) limn→∞kxn−xk= 0
(2) limn→∞kxnk=kxk andlimn→∞hJp(xn), xi=hJp(x), xi (3) limn→∞4p(xn, x) = 0.
The assertions (1)=⇒(2)=⇒(3) are valid. If E is also uniformly convex, then the assertions are equivalent.
Lemma 1.7. Let E be a reflexive and smooth Banach space and E∗ its dual. Let 4p andVp be the mappings defined as above andJEp the generalized duality map onE. Then4p(x, y) =Vp(JEpx, y)for all x, y∈E.
Proof. Forp, q∈(1,∞) let JEq∗:E∗→E andJEp :E →E∗ be duality mappings, whereJEq∗JEp =I. It follows from 1p+1q = 1 thatp(q−1) =q. So, we have that
4p(x, y) =1
qkxkp− hJEpx, yi+1 pkykp
=1
qkJEq∗JEpxkp− hJEpx, yi+1 pkykp
=1
qkJEpxkp(q−1)− hJEpx, yi+1 pkykp
=1
qkJEpxkq− hJEpx, yi+1 pkykp
=Vp(JEpx, y).
Lemma 1.8([19]). LetE be a reflexive, strictly convex and smooth Banach space andJpbe the duality mapping of E. Then
(i) for every closed and convex subset C ⊂E andx∈E, there exists a unique element ΠpC(x)∈C such that 4p(x,ΠpC(x)) = miny∈C4p(x, y); ΠpC(x) is called the Bregman projection of xonto C, with respect to the function f(x) =1pkxkp. Moreover, x0∈C is the Bregman projection of xontoC if
hJp(x0−x), y−x0i ≥0 or equivalently
4p(x0, y)≤ 4p(x, y)− 4p(x, x0) for everyy∈C.
(ii) the Bregman projection and the metric projection are related via PC(x)−x= ΠpC−x(0),∀x∈E. Especially, we have PC(0) = ΠpC(0)and thuskΠpC(0)k= miny∈Ckyk.
The uniform convexity of E implies thatE is reflexive andE∗ is uniformly smooth. Therefore, Theorem 2 in [27], forx, y∈E andx∗, y∗∈E∗ andkx+ykp replaced bykx∗−y∗kq gives the following technical result.
Lemma 1.9. For the uniformly smooth space E∗, with the duality mapJq,∀x∗, y∗∈E∗, we have kx∗−y∗kq ≤ kx∗kq−qhJq(x∗), y∗i+ ¯σq(x∗, y∗) where
¯
σq(x∗, y∗) =qGq
Z 1 0
(kx∗−ty∗k ∨ kx∗k)q
t ρE∗
tky∗k
2(kx∗−ty∗k ∨ kx∗k)
dt (1.4)
andGq = 8∨64cKq−1 with c, Kq>0.
Lemma 1.10 ([19]). Let E be a reflexive, strictly convex and smooth Banach space. We write 4∗q(x, y) = 1pkx∗kq − hJEq∗x∗, y∗i+ 1qky∗kq for x∗ = JEp(x), y∗ = JEp(y) for the Bregman distance on the dual spaceE∗ with respect to the functionfq∗(x∗) = 1qkx∗kq. Then we have4p(x, y) =4∗q(x∗, y∗).
Lemma 1.11. Let E be a reflexive, smooth and strictly convex Banach space. Then for all x, y, z ∈ E and x∗=JEpx,z∗=JEpz, the following hold:
(1) 4p(x, y)≥0 and4p(x, y) = 0 ifx=y;
(2) 4p(x, y) =4p(x, z) +4p(z, y) +hx∗−z∗, z−yi.
Proof. The property (1) is proved in [19]. For (2) we have that 4p(x, z) +4p(z, y) = 1
qkxkp− hx∗, zi+1
pkzkp+1
qkzkp− hz∗, yi+1 pkykp
= 1
qkxkp− hx∗, zi+kzkp− hz∗, yi+1
pkykp+hx∗, yi − hx∗, yi
= 1
qkxkp− hx∗, yi+1 pkykp
+hz∗, zi − hz∗, yi+hx∗, yi − hx∗, zi
⇔ 4p(x, y) =4p(x, z) +4p(z, y) +hx∗−z∗, z−yi.
IfE is smooth andf(x) = 1pkxkp, then the following result holds (cf.Prop. 5 in [18]).
Lemma 1.12. Let E be a smooth Banach space and f : E → R be a continuous convex function given by f(x) = 1pkxkp. Ifx0∈E and the sequence{4p(xn, x0)}∞n=1 is bounded, then the sequence{xn}is also bounded.
2. Main results
Let E1 and E2 be uniformly convex and uniformly smooth Banach spaces and E1∗ and E2∗ be their duals, respectively. LetU :E1→2E1∗andT :E2→2E∗2 be multi-valued maximal monotone operators. ForK ⊂ E1, closed and convex, δ > 0 andp, q ∈(1,∞), let A : E1 → E2 be a bounded and linear operator, A∗ denotes the adjoint of A and AK be closed and convex. Suppose that ΠpAK : E2 → AK is the Bregman projection onto a closed and convex subset AK. Let BUδ : E1 → E1 be the generalized resolvent operator defined by BδU = (JEp
1 +δU)−1JEp
1 and BTδ : E2 → E2 be another generalized resolvent operator defined by BδT = (JEp
2 +δT)−1JEp
2. Let us denote the solutions of variational inclusion problem with respect to U and T by SOLV IP(U) and SOLV IP(T), respectively. Let the set of solutions of split variational inclusion problem be
given by Ω ={x∗∈SOLV IP(U);Ax∗∈SOLV IP(T)} 6=∅. Letx1∈E1be chosen arbitrarily and the sequence {xn} ⊂E1be defined as follows;
un=BδU
n
JEq∗
1 JEp
1xn−λnA∗JEp
2(I−ΠpAKBδT
n)Axn
, Kn+1={v∈Kn:4p(un, v)≤ 4p(xn, v)},
xn+1= ΠpK
n+1(x1), n≥1,
(2.1)
whereδn∈(0,∞). It is remarked that we have replaced the gradient algorithm in (1.1) [the projection maps in (1.3), respectively] with the resolvent operators and used the generalized duality map in our algorithm.
We shall strictly employ the above terminology in the sequel.
Lemma 2.1. Suppose that σ¯q is the function in (1.4) for the characteristic inequality of the uniformly smooth space E1∗. For the sequence {xn} ⊂ E1 defined by (2.1), let 0 6= xn ∈ E1, 0 6= A and 06=JEp
2(I−ΠpAKBδT
n)Axn ∈ E2∗. Let λn >0 andµn >0 be defined, respectively, by λn= 1
kAk
1 kJEp
2(I−ΠpAKBTδ
n)Axnk and µn= 1
kxnkp−1· (2.2)
Then 1 qσ¯q JEp
1xn, λnA∗JEp
2(I−ΠpAKBTδ
n)Axn
≤
(2qGqkJEp
1xnkqρE1∗(µn) if µn ∈(0,1],
2qGqρE∗1(µn) if µn ∈(1,∞), (2.3) whereGq is the constant defined in Lemma1.9andρE∗1 is the modulus of smoothness ofE1∗.
Proof. By Lemma1.9, we have 1
qσ¯q JEp
1xn, λnA∗JEp
2(I−ΠpAKBδTn)Axn
=Gq
Z 1 0
JEp
1xn−λnA∗JEp
2(I−ΠpAKBTδ
n)Axn
∨ kJEp
1xnkq t
×ρE∗ tkλnA∗JEp
2(I−ΠpAKBTδ
n)Axnk JEp
1xn−λnA∗JEp
2(I−ΠpAKBδT
n)Axn
∨ kJEp1xnk
! dt, (2.4) for every t∈[0,1].
We note that JEp
1xn−λn,iA∗JEp
2(I−ΠpAKBδTn)Axn
≤ kxnkp−1+
λnA∗JEp
2(I−ΠpAKBδTn)Axn
. By (2.2), withxn6= 0
λn= µn
kAk
kxnkp−1 JEp
2(I−ΠpAKBδT
n)Axn
(2.5) and so we have that
JEp
1xn−λnA∗JEp
2(I−ΠpAKBTδn)Axn
≤(1 +µn)kxnkp−1 and
(kxnkp−1≤ JEp
1xn−λnA∗JEp
2 I−ΠpAKBδT
n
Axn
∨ kJEp1xnk ≤2kxnkp−1 if µn∈(0,1]
kxnkp−1≤ JEp
1xn−λnA∗JEp
2 I−ΠpAKBδTn Axn
∨ kJEp
1xnk ≤2 if µn∈(1,∞). (2.6)
By (2.6), (2.5) and Definition 1.5(2), we get ρE1∗
t
λnA∗JEp
2(I−ΠpAKBδT
n)Axn JEp
1xn−tλnA∗JEp
2(I−ΠpAKBδT
n)Axn
∨ kJEp
1xnk
!
≤ρE∗1
t
λnA∗JEp
2(I−ΠpAKBδT
n)Axn kxnkp−1
!
=ρE∗
1(tµn). (2.7)
Substituting (2.7) and (2.6) into (2.4), and using nondecreasingness of ρE1∗, we get (2.3) as required.
Lemma 2.2. For the sequence {xn} ⊂E1 defined by (2.1), let 06=xn, 06=JEp
2(I−ΠpAKBδT
n)Axn∈E2∗, and λn>0 andµn>0 be defined by (2.2) andλn andµn are chosen such that
ρE1∗(µn) =
ι 2qGqkAk×
D JEp
2(I−ΠpAKBδnT )Axn,(I−ΠpAKBTδn)Axn E
kJEp
1xnkqkJEp
2(I−ΠpAKBT
δn)Axnk , if µn ∈(0,1],
ι 2qGqkAk×
D JEp
2(I−ΠpAKBδnT )Axn,(I−ΠpAKBTδn)Axn E
kJEp
2(I−ΠpAKBTδn)Axnk , if µn ∈(1,∞),
(2.8)
whereι∈(0,1). Then, for all v∈Ω, we get 4p(un, v)≤ 4p(xn, v)−[1−ι]
JEp
2(I−ΠpAKBδT
n)Axn,(I−ΠpAKBδT
n)Axn kAkkJEp
2(I−ΠpAKBδT
n)Axnk · (2.9)
Proof. Forv=BUγv andAv=BTγAv, by Lemma1.7, we have that 4p(un, v) =4p
BUδ
n
JEq∗
1 JEp
1xn−λnA∗JEp
2 I−ΠpAKBδT
n
Axn , v
=4p
BUδn
JEq∗ 1 JEp
1xn−λnA∗JEp
2 I−ΠpAKBδTn Axn
, BδUnv
≤Vp JEp
1xn−λnA∗JEp
2 I−ΠpAKBδTn
Axn, v
=1 q
JEp
1xn−λnA∗JEp
2 I−ΠpAKBδTn Axn
q+1 pkvkp
− hJEp
1xn, vi+
λnA∗JEp
2 I−ΠpAKBTδn
Axn, v
, (2.10)
where, λnA∗JEp
2 I−ΠpAKBδTn
Axn, v
= λnJEp
2 I−ΠpAKBδTn
Axn, Av−Axn+Axn−ΠpAKBδTnAxn +
λnJEp
2 I−ΠpAKBTδn
Axn,ΠpAKBδTnAxn−Axn+Axn
=− λnJEp
2 ΠpAKBTδ
n−I
Axn,(Av−Axn)− ΠpAKBδT
n−I Axn
− λnJEp
2 I−ΠpAKBTδ
n
Axn, I−ΠpAKBδT
n
Axn +
λnJEp
2 I−ΠpAKBTδn
Axn, Axn
.
As AK is closed and convex so by Lemma1.8(i) and the variational inequality for the Bregman projection of zero ontoAK−Axn, as in Lemma 1.8(ii), we arrive at
λnJEp
2(ΠpAKBδTn−I)Axn,(Av−Axn)−(ΠpAKBTδn−I)Axn
≥0 and therefore, we obtain
λnA∗JEp
2(I−ΠpAKBTδ
n)Axn, v
≤ − λnJEp
2(I−ΠpAKBδT
n)Axn,(I−ΠpAKBTδ
n)Axn +
λnJEp
2(I−ΠpΓBδTn)Axn, Axn
. (2.11)
In addition, by Lemma1.9, we have that 1
q JEp
1xn−λnA∗JEp
2(I−ΠpAKBδTn)Axn
q ≤ 1 qkJEp
1xnkq−λn
Axn, JEp
2(I−ΠpAKBTδn)Axn +1
qσ¯q JEp
1xn, λnA∗JEp
2(I−ΠpAKBδTn)Axn
(2.12) By Lemma2.1and (2.12), we have that
1 q
JEp
1xn−λnA∗JEp
2(I−ΠpAKBδTn)Axn
q ≤ 1 qkJEp
1xnkq−λn
Axn, JEp
2(I−ΠpAKBTδn)Axn
+ 2qGqkJEp
1xnkqρE1∗(µn). (2.13) Substituting (2.13) and (2.11) into (2.10), we have that
4p(un, v)≤ 1 qkJEp
1xnkq+1
pkvkp− hJEp
1xn, vi+ 2qGqkJEp
1xnkqρE∗1(µn)
− λnJEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axn
= 4p(xn, v) + 2qGqkJEp
1xnkqρE∗1(µn)
− λnJEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axn
. (2.14)
Substituting (2.2) and (2.8) into (2.14), we have that 4p(un, v)≤ 4p(xn, v) +ι
(JEp
2(I−ΠpAKBδT
n)Axn,(I−ΠpAKBTδ
n)Axn kAkkJEp
2(I−ΠpAKBTδ
n)Aunk
− (JEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axn kAkkJEp
2(I−ΠpAKBδT
n)Axnk
= 4p(xn, v)−[1−ι]
JEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axn kAkkJEp
2(I−ΠpAKBδT
n)Axnk ·
Thus, (2.9) holds.
We now prove our main result.
Theorem 2.3. For δ > 0 and p, q ∈ (1,∞), let (I−ΠpAKBδT) be demiclosed at zero. Let x1 ∈E1 be chosen arbitrarily and the sequence{xn}be defined by (2.1), where
λn =
1 kAk
1 kJEp
2(I−ΠpAKBT
δn)Axnk, xn 6= 0
1 kAkp
D JEp
2(I−ΠpAKBδnT )Axn,(I−ΠpAKBδnT )Axn Ep−1
kJEp
2(I−ΠpAKBT
δn)Axnkp , xn = 0
and µn= 1
kxnkp−1 (2.15) are chosen such that equation (2.8) holds. If Ω = {x∗ ∈SOLV IP(U);Ax∗ ∈ SOLV IP(T)} 6= ∅, then {xn} converges strongly to x∗∈Ω, whereΠpAKBδT
n(Ax∗) =BδT
n(Ax∗).
Proof. We will divide the proof into two steps.
Step one. We show that{xn}is a bounded sequence.
Assume thatkJEp
2(I−ΠpAKBδT
n)Axnk= 0. Then fromv=BγUv, Lemma1.7and v∈Ω, we get 4p(un, v) =4p
BδUn(JEq∗ 1(JEp
1xn)), BUδnv
≤Vp(JEp
1xn, v) =4p(xn, v). (2.16)
Next assume thatkJEp2(I−ΠpAKBTδ
n)Axnk 6= 0 and xn 6= 0. Then forv∈Ω, by Lemma2.2, we get 4p(un, v)≤ 4p(xn, v)−[1−ι]
JEp
2(I−ΠpAKBδT
n)Axn,(I−ΠpAKBδT
n)Axn kAkkJEp
2(I−ΠpAKBδT
n)Axnk (2.17)
≤ 4p(xn, v). (2.18)
Forxn = 0, we have
4p(xn, v) =1
pkvkp (2.19)
and so by (2.19), we have that
4p(un, v) =1 q
λnA∗JEp
2(I−ΠpAKBTδn)Axn
q
+4p(xn, v) +λn
JEp
2(I−ΠpAKBδTn)Axn, Av
. (2.20)
Substituting (2.11) in (2.20), we have that 4p(un, v)≤1
q
λnA∗JEp
2(I−ΠpAKBδT
n)Axn
q
+4p(xn, v) +λn JEp
2(I−ΠpAKBδTn)Axn, Axn
−λn JEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axn
. (2.21)
By (2.15), we have that 1
q
λnA∗JEp
2(I−ΠpAKBTδn)Axn
q =1 q
1 kAkp
JEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axnp JEp
2(I−ΠpAKBδT
n)Axn
p · (2.22)
Substituting (2.22) into (2.21), we have that 4p(un, v)≤1
q 1 kAkp
JEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axn
p
JEp
2(I−ΠpAKBδT
n)Axn
p
+4p(xn, v) +λnhJEp
2(I−ΠpAKBδT
n)Axn, Axni
−λnhJEp
2(I−ΠpAKBδTn)Axn,(I−ΠpAKBTδn)Axni
≤
1−1 p
1 kAkp
hJEp
2(I−ΠpAKBδT
n)Axn,(I−ΠpAKBδT
n)Axnip kJEp
2(I−ΠpAKBδT
n)Axnkp +4p(xn, v) +λnkJEp
2(I−ΠpAKBδTn)AxnkkAxnk
− 1 kAkp
hJEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axnip kJEp
2(I−ΠpAKBδT
n)Axnkp
= 4p(xn, v)− 1 pkAkp
hJEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axnip kJEp
2(I−ΠpAKBδT
n)Axnkp · (2.23)
This implies that
4p(un, v)≤ 4p(xn, v). (2.24) By (2.1), (2.16), (2.18) and (2.24),v∈Kn so that Ω⊂Kn.
We know from (2.1),xn= ΠpK
nx1. Then, by Lemma 1.8, we have
4p(xn, x1) =4p(ΠpKx1, x1)≤ 4p(v, x1)− 4p(v, xn)⇒ 4p(xn, x1)≤ 4p(v, x1)∀v∈Ω⊂Kn. (2.25) By (2.25), the sequence{4p(xn, x1)} is bounded and therefore by Lemma1.12,{xn} is bounded. Hence,{un} is also bounded. Consequently, there exists a subsequence xnj such that xnj * x∗ as j → ∞ (* stands for weak convergence).
Step two. We show thatxn→x∗∈Ω.
Sincexn+1= ΠpK
n+1x1⊂Kn+1⊂Kn andJpis weakly sequentially continuous, we have by Lemma1.11 4p(un, xn) = 4p(un, xn+1) +4p(xn+1, xn) +
un−xn+1, JEp
1xn+1−JEp
1xn
≤ 4p(xn, xn+1) +4p(xn+1, xn) +
un−xn+1, JEp
1xn+1−JEp
1xn
= 4p(xn, xn) +
un−xn, JEp
1xn+1−JEp
1xn
−→0 asn→ ∞. (2.26)
It follows from (2.1) that (JEp
1xn−JEp
1un)−λnA∗JEp
2(I−ΠpAKBTδ
n)Axn
δn ∈U(un). (2.27)
By (2.17), we have that
4p(un, v)≤ 4p(xn, v)−[1−ι]
JEp
2(I−ΠpAKBTδ
n)Axn,(I−ΠpAKBδT
n)Axn kAkkJEp
2(I−ΠpAKBδT
n)Axnk , and
k(I−ΠpAKBδT
n)Axnk ≤
4p(xn, v)− 4p(un, v) kAk−1[1−ι]
−→0 asn→ ∞. (2.28)
By (2.23), we have that
4p(un, v)≤ 4p(xn, v)− 1 pkAkp
hJEp
2(I−ΠpAKBδT
n)Axn,(I−ΠpAKBδT
n)Axnip kJEp
2(I−ΠpAKBδT
n)Axnkp and therefore
k(I−ΠpAKBδTn)Axnk ≤
4p(xn, v)− 4p(un, v) (pkAk)−1
1p
−→0 asn→ ∞. (2.29)
By (2.26) to (2.29) and weak sequential continuity property ofJp, we have that 0∈ U(x∗). This means that x∗∈SOLV IP(U).But, since4p(·, x) is lower semi continuous and convex and thus weakly lower semi contin- uous on int(domf) then from the fact thatxnj * x∗ asj → ∞, wee see that
4p(x∗, x1)≤lim inf
j→∞ 4p(xnj, x1)≤ 4p(v, x1).
From the definition ofv, that isv=BδU(v), we can conclude thatx∗=vand the sequencexn* x∗. In addition, it is clear thatAxn * Ax∗. So by using (2.28), (2.29) and applying the demicloseness of (I−ΠpAKBδT
n) at zero, we have that 0∈T(Ax∗) as ΠpAKBδT(Ax∗) =BTδ(Ax∗). ThereforeAx∗∈SOLV IP(T).Hence,x∗∈Ω.