Extragradient subgradient methods for solving bilevel equilibrium problems

(1)

R E V I E W Open Access

Extragradient subgradient methods for solving bilevel equilibrium problems

Tadchai Yuying¹, Bui Van Dinh², Do Sang Kim³and Somyot Plubtieng^1,4*

*Correspondence:

1Department of Mathematics, Faculty of Science, Naresuan University, Phitsanulok, Thailand

4Center of Excellence in Nonlinear Analysis and Optimization, Faculty of Science, Naresuan University, Phitsanulok, Thailand Full list of author information is available at the end of the article

Abstract

In this paper, we propose two algorithms for finding the solution of a bilevel equilibrium problem in a real Hilbert space. Under some sufficient assumptions on the bifunctions involving pseudomonotone and Lipschitz-type conditions, we obtain the strong convergence of the iterative sequence generated by the first algorithm.

Furthermore, the strong convergence of the sequence generated by the second algorithm is obtained without a Lipschitz-type condition.

MSC: 47J25; 65K10; 65K15; 90C25

Keywords: Bilevel equilibrium problems; Extragradient Subgradient-Halpern methods; Armijo line search; Strong convergence

1 Introduction

LetCbe a nonempty closed convex subset of a real Hilbert spaceH, and letf andgbe bifunctions fromH×HtoRsuch thatf(x,x) = 0 andg(x,x) = 0 for allx∈H. The equilibrium problem associated withgandCis denoted byEP(C,g) : Findx^∗∈Csuch that

g x^∗,y

≤0 for everyy∈C. (1)

The solution set of problem (1) is denoted byΩ. The equilibrium problem is very important because many problems arise in applied areas such as the ﬁxed point problem, the (generalized) Nash equilibrium problem in game theory, the saddle point problem, the variational inequality problem, the optimization problem and others.

The basic method for solving the monotone equilibrium problem is the proximal method (see [14,16,19]). In 2008, Tran et al. [25] proposed the extragradient algorithm for solving the equilibrium problem by using the strongly convex minimization problem to solve at each iteration. Furthermore, Hieu [9] introduced subgradient extragradient methods for pseudomonotone equilibrium problem and the other methods (see the details in [1,8,10,15,17,22,28]).

In this paper, we consider the equilibrium problem whose constraints are the solution sets of equilibrium problems (bilevel equilibrium problems) : Findx^∗∈Ωsuch that

f x^∗,y

≥0 for everyy∈Ω. (2)

©The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

(2)

The solution set of problem (2) is denoted byΩ^∗. The bilevel equilibrium problems were introduced by Chadli et al. [4] in 2000. This kind of problems is very important and inter- esting because it is a generalization class of problems such as optimization problems over equilibrium constraints, variational inequality over equilibrium constraints, hierarchical minimization problems, and complementarity problems. Furthermore, the particular case of the bilevel equilibrium can be applied to a real word model such as the variational inequality over the ﬁxed point set of a ﬁrmly nonexpansive mapping applied to the power control problem of CDMA networks which were introduced by Iiduka [11]. For more on the relation of bilevel equilibrium with particular cases, see [7,12,21].

Methods for solving bilevel equilibrium problems have been studied extensively by many authors. In 2010, Moudaﬁ [20] introduced a simple proximal method and proved the weak convergence to a solution of problem (2). In 2014, Quy [23] introduced the algorithm by combining the proximal method with the Halpern method for solving bilevel monotone equilibrium and ﬁxed point problem. For more details and most recent works on the methods for solving bilevel equilibrium problems, we refer the reader to [2,5,24]. The authors considered the method for monotone and pseudoparamonotone equilibrium problem. If a bifunction is more generally monotone, we cannot use the above methods for solving bilevel equilibrium problem, for example, the pseudomonotone property.

Inspired by the above work, in this paper, we propose a method for ﬁnding the solution for bilevel equilibrium problems wheref is strongly monotone andgis pseudomonotone and Lipschitz-type continuous. Firstly, we obtain the convergent sequence by combining an extragradient subgradient method with the Halpern method. Second, we obtain the convergent sequence without Lipschitz-type continuity on bifunctiongby combining an Armijo line search with the extragradient subgradient method.

2 Preliminaries

LetCbe a nonempty closed convex subset of a real Hilbert spaceH. Denote thatxnx andxn→xare the weak convergence and the strong convergence of a sequence{xn}tox, respectively. For everyx∈H, there exists a unique elementPCxdeﬁned by

PCx=inf

x–y:y∈C .

It is also known thatPCis a nonexpansive mapping fromHontoC, i.e.,PC(x) –PC(y) ≤ x–y ∀x,y∈H. For everyx∈Handy∈C, we have

x–y²≥ x–PCx²+y–PCx² and x–PCx,PCx–y ≥0.

A bifunctionψ:H×H→Ris called:

(i) β-strongly monotone onCif

ψ(x,y) +ψ(y,x)≤–βx–y² ∀x,y∈C;

(ii) monotone onCif

ψ(x,y) +ψ(y,x)≤0 ∀x,y∈C;

(3)

(iii) pseudomonotone onCif

ψ(x,y)≥0⇒ψ(y,x)≤0, ∀x,y∈C.

It is easy to check that the monotone bifunction implies the pseudomonotone bifunction.

On the other hand, if the bifunction is pseudomonotone, then we cannot guarantee that the bifunction is monotone, for example, letφ(x,y) =^2y–x_1–x for allx,y∈R. It follows thatφis pseudomonotone onR⁺\ {0}butφis not monotone onR⁺\ {0}. Letψ(x,·) be convex for everyx∈H. For each fixedx∈H, the subdifferential ofψ(x, .) atx, denoted by∂₂ψ(x,x), is defined by

∂₂ψ(x,x) =

w∈H:ψ(x,y) –ψ(x,x)≥ w,y–x ∀y∈H

=

w∈H:ψ(x,y)≥ w,y–x ∀y∈H

, (3)

studied in [13]. In this paper, we consider the bifunctionsf andg under the following conditions.

Condition A

(A1) f(x,·)is convex, weakly lower semicontinuous and subdiﬀerentiable onHfor every ﬁxedx∈H.

(A2) f(·,y)is weakly upper semicontinuous onHfor every ﬁxedy∈H.

(A3) f isβ-strongly monotone onH.

(A4) For eachx,y∈H, there existsL> 0such that

w–v ≤Lx–y, ∀w∈∂₂f(x,x),v∈∂₂f(y,y).

(A5) The functionx→∂₂f(x,x)is bounded on the bounded subsets ofH.

Condition B

(B1) g(x,·)is convex, weakly lower semicontinuous, and subdiﬀerentiable onHfor every ﬁxedx∈H.

(B2) g(·,y)is weakly upper semicontinuous onHfor every ﬁxedy∈H.

(B3) gis pseudomonotone onCwith respect toΩ, i.e., g

x,x^∗

≤0, ∀x∈C,x^∗∈Ω.

(B4) gis Lipschitz-type continuous, i.e., there are two positive constantsL1,L2such thatg(x,y) +g(y,z)≥g(x,z) –L1x–y²–L2y–z²,∀x,y,z∈H.

(B5) gis jointly weakly continuous onH×Hin the sense that, ifx,y∈Hand {xn},{yn} ∈Hconverge weakly toxandy, respectively, theng(xn,yn)→g(x,y)as n→+∞.

Example2.1 Let f,g:R×R→Rbe deﬁned byf(x,y) = 5y²– 7x²+ 2xyandg(x,y) = 2y²– 7x²+ 5xy. It follows thatf andgsatisfy ConditionAand ConditionB, respectively.

Lemma 2.2([3, Propositions 3.1, 3.2]) If the bifunction g satisﬁes Assumptions(B1), (B2), and(B3),then the solution setΩis closed and convex.

(4)

Remark2.3 Let the bifunctionf satisfy Assumption A and the bifunctiongsatisfy B. IfΩ=

∅, then the bilevel equilibrium problem (2) has a unique solution, see the details in [23].

Lemma 2.4([6]) Let C be a nonempty closed convex subset of a real Hilbert space H and φ:C→Rbe a convex,lower semicontinuous,and subdiﬀerentiable function on C.Then x^∗ is a solution to the convex optimization problemmin{φ(x) :x∈C}if and only if0∈

∂φ(x^∗) +NC(x^∗),where∂φ(·)denotes the subdiﬀerential ofφand NC(x^∗)is the normal cone of C at x^∗.

Lemma 2.5([27]) Let{an}be a sequence of nonnegative real numbers,{α_n}be a sequence in(0, 1),and{ξ_n}be a sequence inRsatisfying the condition

an+1≤(1 –α_n)an+α_nξ_n ∀n≥0, where_∞

n=0α_n=∞andlim sup_n→∞ξ_n≤0.Thenlim_n→∞a_n= 0.

Lemma 2.6([18]) Let{an}be a sequence of real numbers that does not decrease at inﬁnity, in the sense that there exists a subsequence{anj}of{an}such that

anj<anj+1 for all j≥0.

Also consider the sequence of integers{τ(n)}n≥n0 deﬁned,for all n≥n₀,by τ(n) =max{k≤n|ak<ak+1}.

Then{τ(n)}n≥n0is a nondecreasing sequence verifying

n→∞lim τ(n) =∞,

and,for all n≥n0,the following two estimates hold:

aτ(n)≤a_τ(n)+1 and a_n≤a_τ(n)+1.

Lemma 2.7 Suppose that f isβ-strongly monotone on H and satisﬁes(A4).Let0 <α< 1, 0≤η≤1 –α,and0 <μ<^2β_L2.For each x,y∈H,w∈∂₂f(x,x),and v∈∂₂f(y,y),we have

(1 –η)x–αμw–

(1 –η)y–αμv ≤(1 –η–ατ)x–y, whereτ = 1 –

1 –μ(2β–μL²)∈(0, 1].

Proof Letx,y∈H,w∈∂₂f(x,x), andv∈∂₂f(y,y). Thus (1 –η)x–αμw–

(1 –η)y–αμv

≤(1 –η)(x–y) –αμ(w–v)

=(1 –η–α)(x–y) +α

(x–y) –μ(w–v)

≤(1 –η–α)x–y+α(x–y) –μ(w–v). (4)

(5)

Sincef:H×H→Risβ-strongly monotone,w∈∂₂f(x,x), andv∈∂₂f(y,y), we have

–βx–y²≥f(x,y) +f(y,x)≥– x–y,w–v. (5)

From (5) and (A4), we have

(x–y) –μ(w–v)²=x–y²– 2μ x–y,w–v+μ²w–v²

≤ x–y²– 2μβx–y²+μ²L²x–y²

=

1 – 2μβ+μ²L²

x–y². This implies that

(x–y) –μ(w–v)≤

1 – 2μβ+μ²L²x–y. (6)

By using (4) and (6), we can conclude that (1 –η)x–αμw–

(1 –η)y–αμv

≤(1 –η–α)x–y+ α

1 – 2μβ+μ²L² x–y

= (1 –η–ατ)x–y, whereτ= 1 –

1 –μ(2β–μL²)∈(0, 1].

3 The extragradient subgradient Halpern methods

In this section, we propose the algorithm for ﬁnding the solution of a bilevel equilibrium problem under the strong monotonicity off and the pseudomonotonicity and Lipschitz- type continuous conditions ong.

Algorithm 1

Initialization: Choosex0∈H, 0 <μ<^2β_L2, the sequences{α_n} ⊂(0, 1),{η_n}, and{λ_n}are such that

⎧⎪

⎪⎨

⎪⎪

⎩

lim_n→∞α_n= 0, _∞

n=0α_n=∞,

0≤η_n≤1 –α_n ∀n≥0, lim_n→∞η_n=η< 1, 0 <λ≤λ_n≤λ<min(_2L¹

1,_2L¹

2).

Setn= 0 and go to Step 1.

Step 1. Compute y_n=arg min

y∈C

λ_ng(x_n,y) +1

2y–x_n²

, z_n=arg min

y∈C

λ_ng(y_n,y) +1

2y–x_n²

. Step 2. Computewn∈∂₂f(zn,zn) and

x_n+1=η_nx_n+ (1 –η_n)z_n–α_nμw_n. Setn=n+ 1 and go back to Step 1.

(6)

Theorem 3.1 Let bifunctions f and g satisfy ConditionAand ConditionB,respectively.

Assume thatΩ=∅.Then the sequence{xn}generated by Algorithm1converges strongly to the unique solution of the bilevel equilibrium problem(2).

Proof Under assumptions of two bifunctionsf andg, we get the unique solution of the bilevel equilibrium problem (2), denoted byx^∗.

Step 1: Show that

z_n–x^∗²≤x_n–x^∗²– (1 – 2λ_nL₁)x_n–y_n²– (1 – 2λ_nL₂)y_n–z_n². (7)

The deﬁnition ofynand Lemma2.4imply that 0∈∂₂

λ_ng(x_n,y) +1

2y–x_n²

(y_n) +N_C(y_n).

There arew∈∂₂g(xn,yn) andw¯ ∈NC(yn) such that

λ_nw+yn–xn+w¯= 0. (8)

Sincew¯ ∈NC(yn), we have

¯w,y–yn ≤0 for ally∈C. (9)

By using (8) and (9), we obtainλ_n w,y–yn ≥ xn–yn,y–ynfor ally∈C. Sincezn∈C, we have

λ_n w,zn–yn ≥ xn–yn,zn–yn. (10) It follows fromw∈∂₂g(xn,yn) that

g(x_n,y) –g(x_n,y_n)≥ w,y–y_n for ally∈H. (11) By using (10) and (11), we get

λn

g(xn,zn) –g(xn,yn)

≥ xn–yn,zn–yn. (12) Similarly, the deﬁnition ofz_nimplies that

0∈∂₂

λ_ng(yn,y) +1

2y–xn²

(zn) +NC(zn).

There areu∈∂₂g(y_n,z_n) andu¯∈N_C(z_n) such that

λ_nu+z_n–x_n+u¯= 0. (13)

Sinceu¯∈NC(zn), we have

¯u,y–zn ≤0 for ally∈C. (14)

(7)

By using (13) and (14), we obtainλ_n u,y–zn ≥ xn–zn,y–znfor ally∈C. Sincex^∗∈C, we have

λ_n

u,x^∗–zn

≥

xn–zn,x^∗–zn

. (15)

It follows fromu∈∂₂g(yn,zn) that

g(yn,y) –g(yn,zn)≥ u,y–zn for ally∈H. (16) By using (15) and (16), we get

λ_n g

yn,x^∗

–g(yn,zn)

≥

xn–zn,x^∗–zn

. (17)

Sincex^∗∈Ω, we haveg(x^∗,yn)≥0. If follows from the pseudomonotonicity ofgonCwith respect toΩthatg(yn,x^∗)≤0. This implies that

x_n–z_n,z_n–x^∗

≥λ_ng(y_n,z_n). (18) Sincegis Lipschitz-type continuous, there exist two positive constantsL₁,L₂such that

g(y_n,z_n)≥g(x_n,z_n) –g(x_n,y_n) –L₁x_n–y_n²–L₂y_n–z_n². (19) By using (18) and (19), we get

≥λ_n

g(x_n,z_n) –g(x_n,y_n)

–λ_nL₁x_n–y_n²–λ_nL₂y_n–z_n². From (12) and the above inequality, we obtain

2

≥2x_n–y_n,z_n–y_n–λ_nL₁x_n–y_n²–λ_nL₂y_n–z_n². (20) We know that

2

xn–zn,zn–x^∗

=xn–x^∗²–zn–xn²–zn–x^∗², 2 xn–yn,zn–yn=xn–yn²+zn–yn²–xn–zn². From (20), we can conclude that

zn–x^∗²≤xn–x^∗²– (1 – 2λnL1)xn–yn²– (1 – 2λnL2)yn–zn². Step 2: The sequences{xn},{wn},{yn}, and{zn}are bounded.

Since 0 <λ_n<a, wherea=min(_2L¹

1,_2L¹

2), we have (1 – 2λnL1) > 0 and (1 – 2λnL2) > 0.

It follows from (7) and the above inequalities that

zn–x^∗≤xn–x^∗ for alln∈N. (21)

(8)

By Lemma2.7and (21), we obtain

x_n+1–x^∗=η_nx_n+ (1 –η_n)z_n–α_nμw_n–x^∗+η_nx^∗–η_nx^∗+α_nμv–α_nμv

=(1 –η_n)zn–α_nμwn– (1 –η_n)x^∗+α_nμv+η xn–x^∗

–α_nμv

≤(1 –η_n)zn–α_nμwn–

(1 –η_n)x^∗–α_nμv +η_nxn–x^∗+α_nμv

≤(1 –η_n–α_nτ)zn–x^∗+η_nxn–x^∗+α_nμv

≤(1 –η_n–α_nτ)xn–x^∗+η_nxn–x^∗+α_nμv

= (1 –α_nτ)xn–x^∗+α_nμv

= (1 –α_nτ)xn–x^∗+α_nτ μv

τ

, (22)

wherewn∈∂₂f(zn,zn) andv∈∂₂f(x^∗,x^∗). This implies that xn+1–x^∗≤maxxn–x^∗,μv

τ

. By induction, we obtain

xn–x^∗≤maxx0–x^∗,μv τ

.

Thus the sequence{xn}is bounded. By using (21), we have{zn}, and using Condition (A5), we can conclude that{wn}is also bounded.

Step 3: Show that the sequence{xn}converges strongly tox^∗.

Sincex∈Ω^∗, we havef(x^∗,y)≥0 for ally∈Ω. Thusx^∗ is a minimum of the convex functionf(x^∗,·) over Ω. By Lemma2.4, we obtain 0∈∂₂f(x^∗,x^∗) +NΩ(x^∗). Then there existsv∈∂₂f(x^∗,x^∗) such that

v,z–x^∗

≥0 for allz∈Ω. (23)

Note that

x–y²≤ x²– 2 y,x–y for allx,y∈H. (24)

From Lemma (2.7) and (24), we obtain

x_n+1–x^∗² =η_nx_n+ (1 –η_n)z_n–α_nμw_n–x^∗²

=(1 –η_n)z_n–α_nμw_n–

(1 –η_n)x^∗–α_nμv +η_n

x_n–x^∗

–α_nμv²

xn–x^∗²– 2αnμ

v,xn+1–x^∗

xn–x^∗²– 2αnμ

v,xn+1–x^∗

(9)

≤

(1 –η_n–α_nτ)zn–x^∗+η_nxn–x^∗ ²– 2αnμ

v,xn+1–x^∗

≤(1 –η_n–α_nτ)zn–x^∗²+η_nxn–x^∗²– 2αnμ

v,xn+1–x^∗

≤(1 –η_n–α_nτ)xn–x^∗²+η_nxn–x^∗²– 2αnμ

v,xn+1–x^∗

= (1 –α_nτ)xn–x^∗²+ 2αnμ

v,x^∗–xn+1

.

It follows that

x_n+1–x^∗²≤(1 –α_nτ)x_n–x^∗²+ 2α_nμ

v,x^∗–x_n+1

. (25)

Let us consider two cases.

Case 1: There existsn0such that{xn–x^∗}is decreasing forn≥n0. Therefore the limit of sequence{x_n–x^∗}exists. By using (21) and (25), we obtain

0≤xn–x^∗²–zn–x^∗²

≤– α_nτ

1 –η_nz_n–x^∗²– 2αnμ 1 –η_n

v,x_n+1–x^∗

+ 1

1 –η_nxn–x^∗²–xn+1–x^∗²

. (26)

Sincelim_n→∞η_n=η< 1,lim_n→∞α_n= 0 and the limit of{xn–x^∗}exists, we have

n→∞limx_n–x^∗²–z_n–x^∗²

= 0. (27)

From 0 <λn<aand inequality (7), we get

(1 – 2a)xn–yn²≤(1 – 2λnL1)xn–yn²≤xn–x^∗²–zn–x^∗².

By using (27), we obtainlim_n→∞xn–yn= 0. Next, we show that lim sup

n→∞

v,x^∗–xn+1

≤0. (28)

Take a subsequence{xn_k}of{xn}such that lim sup

n→∞

v,x^∗–xn+1

=lim sup

k→∞

v,x^∗–xnk

.

Since{xn_k}is bounded, we may assume that{xn_k}converges weakly to somex¯∈H. There- fore

lim sup

n→∞

v,x^∗–xn+1

=lim sup

k→∞

v,x^∗–xn_k

= v,x^∗–x¯

. (29)

Sincelim_n→∞x_n–y_n= 0 andx_n_k x, we have¯ y_n_kx. Since¯ Cis closed and convex, it is also weakly closed and thusx¯∈C. Next, we show thatx¯∈Ω. From the deﬁnition of {y_n}and Lemma2.4, we obtain

0∈∂₂

λ_ng(x_n,y) +1

2x_n–y_n²

(y_n) +N_C(y_n).

(10)

There existw¯∈NC(yn) andw∈∂₂g(xn,yn) such that

λ_nw+y_n–x_n+w¯= 0. (30)

Sincew¯ ∈NC(yn), we have ¯w,y–yn ≤0 for ally∈C. From (30), we obtain

λn w,y–yn ≥ xn–yn,y–yn for ally∈C. (31) Sincew∈∂₂g(xn,yn), we have

g(xn,y) –g(xn,yn)≥ w,y–yn for ally∈H. (32) Combining (31) and (32), we get

λ_n

g(xn,y) –g(xn,yn)

≥ xn–yn,y–yn for ally∈C. (33) Takingn=n_kandk→ ∞in (33), the assumption ofλ_nand (B5), we obtaing(x,¯ y)≥0 for ally∈C. This implies thatx¯∈Ω. By inequality (23), we obtain v,x¯–x^∗ ≥0. It follows from (29) that

lim sup

n→∞

v,x^∗–xn+1

≤0. (34)

We can write inequality (25) in the following form:

xn+1–x^∗²≤(1 –αnτ)xn–x^∗²–αnτ ξn,

whereξ_n=^2μ_τ v,x^∗–xn+1. It follows from (34) thatlim sup_n→∞ξ_n≤0. By Lemma2.5, we can conclude thatlim_n→∞xn–x^∗²= 0. Hencexn→x^∗asn→ ∞.

Case 2: There exists a subsequence{x_n_j}of{x_n}such thatx_n_j–x^∗ ≤ x_n_j₊₁–x^∗for all j∈N. By Lemma2.6, there exists a nondecreasing sequence {τ(n)} of Nsuch that lim_n→∞τ(n) =∞, and for each suﬃciently largen∈N, we have

xτ(n)–x^∗≤xτ(n)+1–x^∗ and xn–x^∗≤xτ(n)+1–x^∗. (35)

Combining (22) and (35), we have xτ(n)–x^∗≤xτ(n)+1–x^∗

≤(1 –ητ(n)–α_τ(n)τ)zτ(n)–x^∗+ητ(n)xτ(n)–x^∗+α_τ(n)μv. (36) From (21) and (36), we get

0≤x_τ(n)–x^∗–zτ(n)–x^∗≤– ατ(n)τ 1 –ητ(n)

z_τ(n)–x^∗+ ατ(n)μ

1 –ητ(n)v. (37) Sincelim_n→∞α_n= 0,lim_n→∞η_n=η< 1,{z_n}is bounded, and (37), we havelim_n→∞(xτ(n)– x^∗–zτ(n)–x^∗) = 0. It follows from the boundedness of{x_n}and{z_n}that

n→∞limxτ(n)–x^∗²–zτ(n)–x^∗²

= 0. (38)

(11)

By using the assumption of{λ_n}, we get the following two inequalities:

1 – 2λτ(n)L1> 1 – 2aL1> 0 and 1 – 2λτ(n)L2> 1 – 2aL2> 0.

From (7), we obtain

zτ(n)–x^∗²≤xτ(n)–x^∗²– (1 – 2λτ(n)L1)xτ(n)–yτ(n)² – (1 – 2λτ(n)L2)yτ(n)–zτ(n)²

≤xτ(n)–x^∗²– (1 – 2aL₁)xτ(n)–yτ(n)² – (1 – 2aL2)yτ(n)–zτ(n)².

This implies that

0 < (1 – 2aL₁)x_τ(n)–yτ(n)²+ (1 – 2aL₂)yτ(n)–z_τ(n)²

≤x_τ(n)–x^∗²–zτ(n)–x^∗².

It follows from (38) and the above inequality that

n→∞lim xτ(n)–y_τ(n)= 0 and lim

n→∞yτ(n)–z_τ(n)= 0. (39)

Note thatxτ(n)–zτ(n) ≤ xτ(n)–yτ(n)+yτ(n)–zτ(n). From (39), we have

n→∞lim xτ(n)–zτ(n)= 0. (40)

By using the deﬁnition ofxn+1and Lemma2.7, we obtain

xτ(n)+1–xτ(n)=η_τ(n)xτ(n)+ (1 –ητ(n))zτ(n)–ατ(n)μtτ(n)–xτ(n)

=(1 –ητ(n))zτ(n)–α_τ(n)μtτ(n)

–

(1 –ητ(n))xτ(n)–ατ(n)wτ(n) –ατ(n)wτ(n)

≤(1 –ητ(n))zτ(n)–α_τ(n)tτ(n)

–

(1 –ητ(n))xτ(n)–ατ(n)wτ(n) +α_τ(n)wτ(n)

≤(1 –ητ(n)–ατ(n)τ)zτ(n)–xτ(n)+ατ(n)wτ(n)

≤ zτ(n)–xτ(n)+ατ(n)wτ(n),

wheretτ(n)∈∂₂f(zτ(n),zτ(n)) andwτ(n)∈∂₂f(xτ(n),xτ(n)). Sincelim_n→∞α_n= 0, the boundedness of{wτ(n)}and (40), we havelim_n→∞xτ(n)+1–xτ(n)= 0. As proved in the ﬁrst case, we can conclude that

lim sup

n→∞

v,x^∗–xτ(n)+1

=lim sup

k→∞

v,x^∗–xτ(n)

≤0. (41)

(12)

Combining (25) and (35), we obtain

xτ(n)+1–x^∗²≤(1 –α_τ(n)τ)xτ(n)–x^∗²+ 2ατ(n)μ

v,x^∗–xτ(n)+1

≤(1 –ατ(n)τ)xτ(n)+1–x^∗²+ 2ατ(n)μ

v,x^∗–xτ(n)+1

.

By using (35) again, we have

xn–x^∗²≤xτ(n)+1–x^∗²≤2μ τ

v,x^∗–xτ(n)+1

. (42)

From (41), we can conclude thatlim sup_n→∞x_n–x^∗²≤0. Hencex_n→x^∗ asn→ ∞.

This completes the proof.

4 The extragradient subgradient methods with line searches

In this section, we introduce the algorithm for ﬁnding the solution of a bilevel equilibrium problem without the Lipschitz condition for the bifunctiong.

Algorithm 2

Initialization: Choosex0∈C, 0 <μ<^2β

L²,ρ∈(0, 2),γ ∈(0, 1), the sequences{λ_n},{ξ_n}, and{α_n} ⊂(0, 1) such that

⎧⎨

⎩

lim_n→∞α_n= 0, _∞

n=0α_n=∞, _∞

n=0α_n²<∞, λ_n∈[λ,λ]⊂(0,∞), ξ_n∈[ξ,ξ]⊂(0, 2).

Setn= 0, and go to Step 2.

Step 1. Compute

y_n=arg min

y∈C

λ_ng(x_n,y) +1

2y–x_n²

.

Ifyn=xn, then setun=xnand go to Step 4. Otherwise, go to Step 2.

Step 2. (Armijo line search rule) Findmas the smallest positive integer number satisfying

⎧⎨

⎩

zn,m= (1 –γ^m)xn+γ^myn,

g(zn,m,xn) –g(zn,m,yn)≥_2λ^ρ_nxn–yn². Setz_n=z_n,mandγ_n=γ^m.

Step 3. Choosetn∈∂₂g(zn,xn) and computeun=PC(xn–ξ_nσ_ntn) whereσ_n=^g(z_tⁿ^,xⁿ⁾

n²

Step 4. Computewn∈∂₂f(un,un) and xn+1=PC(un–α_nμwn).

Setn=n+ 1, and go back to Step 1.

Lemma 4.1([26]) Suppose that yn=xnfor some n∈N.Then the line search corresponding to x_nand y_n(Step2)is well deﬁned,g(z_n,x_n) > 0,and0 /∈∂₂g(z_n,x_n).

(13)

Lemma 4.2([26]) Let g:Θ×Θ→ ∞be a bifunction satisfying conditions(B1)on C and (B5)onΘ,whereΘis an open convex set containing C.Letx,¯ y¯∈Θand{x_n},{y_n}be two sequences inΘconverging weakly tox,¯ y¯∈Θ,respectively.Then,for anyε> 0,there exist η> 0and nε∈Nsuch that

∂₂g(xn,yn)⊂∂₂g(¯x,y) +¯ ε ηB

for all n≥nε,where B denotes the closed unit ball in H.

Lemma 4.3([8]) Let the bifunction g satisfy assumption(B1)on C×C, (B5)onΘ×Θ.

Suppose that{xn}is a bounded sequence in C,ρ> 0and{yn}is a sequence such that y_n=arg min

y∈C

g(x_n,y) +ρ

2y–x_n²

.

Then{y_n}is also bounded.

Theorem 4.4 Let the bifunction f satisfy ConditionAand g satisfy Conditions(B1)–(B3) and(B5).Assume thatΩ=∅.Then the sequences{xn}generated by Algorithm2converge strongly to the unique solution of the bilevel equilibrium problem(2).

Proof Letx^∗be the unique solution of the bilevel equilibrium problem (2). Then we have x^∗∈Ωand there existsv∈∂₂f(x^∗,x^∗) such that

v,z–x^∗

≥0 for allz∈Ω. (43)

Step 1: Show that

u_n–x^∗²≤x_n–x^∗²–ξ_n(2 –ξ₂) σ_nt_n2

. (44)

By the deﬁnition ofun, we have

un–x^∗²≤PC(xn–ξ_nσ_ntn) –PC

x^∗²

≤xn–ξ_nσ_ntn–x^∗²

=x_n–x^∗²– 2ξ_nσ_n

t_n,x_n–x^∗ +

ξ_nσ_nt_n2

. (45)

Sincetn∈∂₂g(zn,xn) andg(zn,·) is convex onC, we haveg(zn,x^∗) –g(zn,xn)≥ tn,x^∗–xn. It follows that

tn,xn–x^∗

≥g(zn,xn) –g zn,x^∗

. (46)

Sincegis pseudomonotone onCwith respect toΩ, we haveg(zn,x^∗)≤0. It follows from (46) and the deﬁnition ofσnthat

tn,xn–x^∗

≥g(zn,xn) =σ_ntn². (47)

(14)

Combining (45) with (47), we obtain un–x^∗²≤xn–x^∗²– 2ξnσ_n

σ_ntn² +

ξ_nσ_ntn2

=xn–x^∗²– 2ξn

σ_ntn2

+ξ_n² σ_ntn2

=x_n–x^∗²–ξ_n(2 –ξ_n)

σ_nt_n2

.

Step 2: The sequences{x_n},{y_n},{u_n}, and{w_n}are bounded.

Sinceξ_n∈[ξ,ξ]⊂(0, 2) and (44), we have

u_n–x^∗≤x_n–x^∗. (48) By the deﬁnition ofx_n+1, we get

xn+1–x^∗=PC(un–α_nμwn) –PC

x^∗

≤u_n–α_nμw_n–x^∗

≤un–x^∗

–α_nμ(wn–v) –α_nμv

=un–x^∗

–αnμ(wn–v)+αnμv. (49)

From Lemma2.7, (48), and (49), we can conclude that xn+1–x^∗= (1 –α_nτ)un–x^∗+α_nμv

≤(1 –α_nτ)xn–x^∗+α_nτ μv

τ

. (50)

This implies that

x_n+1–x^∗≤maxx_n–x^∗,μv τ

. By induction, we obtain

xn–x^∗≤maxx0–x^∗,μv τ

.

Thus the sequence{x_n}is bounded. Hence we can conclude from (48) and Lemma4.3 that{yn}and{un}are bounded, respectively. From condition (A4), we have{wn}is also bounded.

Step 3: We show that if there is a subsequence{xnk}of{xn}converging weakly tox¯and lim_k→∞(σn_ktn_k) = 0, then we havex¯∈Ω.

Firstly, we will show that{t_n_k}is bounded. Since{z_n}is bounded, there is a subsequence {znk}of{zn}converging weakly toz¯By using Lemma4.2, for anyε> 0, there existη> 0 andk0such that

∂₂g(zn_k,xn_k)⊂∂₂g(¯z,x) +¯ ε ηB