• Aucun résultat trouvé

and Abdul Rahim Khan

N/A
N/A
Protected

Academic year: 2022

Partager "and Abdul Rahim Khan"

Copied!
15
0
0

Texte intégral

(1)

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

3

and Abdul Rahim Khan

4

Abstract.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−xi ≥0 for allx∈C, and the pointy=Ax∈H2 such that

hg(y), y−yi ≥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

(2)

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, xi=kxkkxk,kxk =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+γAJ(PQ−I)Axn), n≥0, (1.3) where ΠC denotes the Bregman Projection andAthe 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, xi=kxkkxk,kxk=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.

(3)

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

qkxkq− 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, whereJEqJEp =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

qkJEqJEpxkp− 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.

(4)

(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−ykq gives the following technical result.

Lemma 1.9. For the uniformly smooth space E, with the duality mapJq,∀x, y∈E, we have kx−ykq ≤ kxkq−qhJq(x), yi+ ¯σq(x, y) where

¯

σq(x, y) =qGq

Z 1 0

(kx−tyk ∨ kxk)q

t ρE

tkyk

2(kx−tyk ∨ kxk)

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 4q(x, y) = 1pkxkq − hJEqx, yi+ 1qkykq for x = JEp(x), y = JEp(y) for the Bregman distance on the dual spaceE with respect to the functionfq(x) = 1qkxkq. Then we have4p(x, y) =4q(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→2E1andT :E2→2E2 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

(5)

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−λnAJEp

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, λnAJEp

2(I−ΠpAKBTδ

n)Axn

(2qGqkJEp

1xnkqρE1n) if µn ∈(0,1],

2qGqρE1n) if µn ∈(1,∞), (2.3) whereGq is the constant defined in Lemma1.9andρE1 is the modulus of smoothness ofE1.

Proof. By Lemma1.9, we have 1

qσ¯q JEp

1xn, λnAJEp

2(I−ΠpAKBδTn)Axn

=Gq

Z 1 0

JEp

1xn−λnAJEp

2(I−ΠpAKBTδ

n)Axn

∨ kJEp

1xnkq t

×ρE tkλnAJEp

2(I−ΠpAKBTδ

n)Axnk JEp

1xn−λnAJEp

2(I−ΠpAKBδT

n)Axn

∨ kJEp1xnk

! dt, (2.4) for every t∈[0,1].

We note that JEp

1xn−λn,iAJEp

2(I−ΠpAKBδTn)Axn

≤ kxnkp−1+

λnAJEp

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−λnAJEp

2(I−ΠpAKBTδn)Axn

≤(1 +µn)kxnkp−1 and

(kxnkp−1≤ JEp

1xn−λnAJEp

2 I−ΠpAKBδT

n

Axn

∨ kJEp1xnk ≤2kxnkp−1 if µn∈(0,1]

kxnkp−1≤ JEp

1xn−λnAJEp

2 I−ΠpAKBδTn Axn

∨ kJEp

1xnk ≤2 if µn∈(1,∞). (2.6)

(6)

By (2.6), (2.5) and Definition 1.5(2), we get ρE1

t

λnAJEp

2(I−ΠpAKBδT

n)Axn JEp

1xn−tλnAJEp

2(I−ΠpAKBδT

n)Axn

∨ kJEp

1xnk

!

≤ρE1

t

λnAJEp

2(I−ΠpAKBδT

n)Axn kxnkp−1

!

E

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

ρE1n) =





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−λnAJEp

2 I−ΠpAKBδT

n

Axn , v

=4p

BUδn

JEq 1 JEp

1xn−λnAJEp

2 I−ΠpAKBδTn Axn

, BδUnv

≤Vp JEp

1xn−λnAJEp

2 I−ΠpAKBδTn

Axn, v

=1 q

JEp

1xn−λnAJEp

2 I−ΠpAKBδTn Axn

q+1 pkvkp

− hJEp

1xn, vi+

λnAJEp

2 I−ΠpAKBTδn

Axn, v

, (2.10)

where, λnAJEp

2 I−ΠpAKBδTn

Axn, v

= λnJEp

2 I−ΠpAKBδTn

Axn, Av−Axn+Axn−ΠpAKBδTnAxn +

λnJEp

2 I−ΠpAKBTδn

AxnpAKBδTnAxn−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

2pAKBδTn−I)Axn,(Av−Axn)−(ΠpAKBTδn−I)Axn

≥0 and therefore, we obtain

λnAJEp

2(I−ΠpAKBTδ

n)Axn, v

≤ − λnJEp

2(I−ΠpAKBδT

n)Axn,(I−ΠpAKBTδ

n)Axn +

λnJEp

2(I−ΠpΓBδTn)Axn, Axn

. (2.11)

(7)

In addition, by Lemma1.9, we have that 1

q JEp

1xn−λnAJEp

2(I−ΠpAKBδTn)Axn

q ≤ 1 qkJEp

1xnkq−λn

Axn, JEp

2(I−ΠpAKBTδn)Axn +1

qσ¯q JEp

1xn, λnAJEp

2(I−ΠpAKBδTn)Axn

(2.12) By Lemma2.1and (2.12), we have that

1 q

JEp

1xn−λnAJEp

2(I−ΠpAKBδTn)Axn

q ≤ 1 qkJEp

1xnkq−λn

Axn, JEp

2(I−ΠpAKBTδn)Axn

+ 2qGqkJEp

1xnkqρE1n). (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

1xnkqρE1n)

− λnJEp

2(I−ΠpAKBTδ

n)Axn,(I−ΠpAKBδT

n)Axn

= 4p(xn, v) + 2qGqkJEp

1xnkqρE1n)

− 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)

(8)

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

λnAJEp

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

λnAJEp

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

λnAJEp

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.

(9)

We know from (2.1),xn= ΠpK

nx1. Then, by Lemma 1.8, we have

4p(xn, x1) =4ppKx1, 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)−λnAJEp

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.

Références

Documents relatifs

3b shows the result of solving the inverse problem – the line-by-line resto- ration of the image by the quadrature method with Tikhonov's regularization according to (14) using

an estimate of the modulus of uniform continuity on balls of the duality map of a uniformly smooth Banach space in terms of the uniform convexity of..

It is clear that, due to semi-definiteness of the matrix P , there are certain static relations encoded in (1) which make it difficult to study the solutions of system class (1)

Nevertheless we could obtain further that the Hölder continuous densities of the equilibrium states with respect to the conformal mea- sures vary continuously with the dynamics in the

case when the (maximal monotone) subdifferential of a lower semicontinuous convex proper function f on E.. We start

Annales de l’Institut Henri Poincaré - Analyse non linéaire.. Let us state a theorem which links this definition of measurability of a multivalued function F to

If its Bowen-Margulis mea- sure is in the Lebesgue measure class, then, up to a constant change of time scale and finite covers, φ t is C ∞ flow equivalent either to the suspension of

The worst-case complexity bounds for the problem (1) with the exact and stochastic first order oracle are available for the case of strongly convex objective (see, e.g... [18, 1]