Hybrid ExtragradientArmijo Methods for Finite Families of Pseudomonotone Equilibrium problems and Nonexpansive Mappings

The purpose of this paper is to propose some hybrid extragradientArmijo algorithms for finding a common element of the set of solutions of a finite family of pseudomonotone equilibrium problems and the set of fixed points of a finite family of nonexpansive mappings in real Hilbert spaces. The proposed methods combine extragradient and Mann’s iterative methods as well as Armijo linesearch and hybrid techniques. The strong convergence of the proposed methods are established without assumption on the Lipschitztype condition of the bifunctions involved

Pseudomonotone Equilibrium problems and Nonexpansive Mappings

Le Q Thuy†, Pham K Anh‡, Le D Muu§

Abstract The purpose of this paper is to propose some hybrid extragradient-Armijo algorithms for finding a common element of the set of solutions of a finite family of pseudomonotone equilibrium problems and the set of fixed points of a finite family of nonexpansive mappings in real Hilbert spaces The proposed methods combine extragradient and Mann’s iterative methods as well as Armijo line-search and hybrid techniques The strong convergence of the proposed methods are established without assumption on the Lipschitz-type condition of the bifunctions involved.

Keywords: Equilibrium problem; Pseudomonotone bifuction; Nonexpansive mapping; Hybrid method; Armijo line search.

AMS Subject Classifications 47 H09, 47 H10, 47 J25, 65 K10, 65 Y05, 90 C25, 90 C33.

1 Introduction

Let C be a nonempty closed convex subset of a real Hilbert space H and f be a bifunction from C × C to R ∪ {+∞} satisfying condition f (x, x) = 0 for every x ∈ C Such a bifunction is called an equilibrium bifuction We consider the following problem:

Finding x∗ ∈ C such that f (x∗, y) ≥ 0, ∀y ∈ C (1.1) Problem (1.1) is refered to as the equilibrium problem or Ky Fan inequality [8] and its solution set is denoted by Sol(C, f )

The equilibrium problem (1.1) provides a unified framework for a wide class of problems, such as convex optimization, variational inequality, nonlinear complementarity, Nash equi-librium and fixed point problems The existence and solution methods for equiequi-librium problems have been extensively studied (see, e.g [3], [11], [16, 18], [20] and the references therein) A mapping T : C → C is said to be nonexpansive if kT (x) − T (y)k ≤ kx − yk for all x, y ∈ C The set of fixed points of T is denoted by F (T ) In recent year, the problem of finding a common solution of equilibrium problems, variational inequalities and fixed point problems for nonexpansive mappings in Hilbert spaces has attracted at-tention of several authors (see e.g [1], [5], [10], [12], [17], [19], [21], , and the references therein) The common approach in these papers is the use of the proximal point method for handling monotone equilibrium problems However, for pseudomonotone equilibrium problems, the auxiliary regularized subproblems may not strongly monotone, even not pseudomonotone, hence they cannot be solved by available methods requiring the mono-tonicity of these subproblems

In this article we propose three hybrid extragradient-Armijo algorithms for finding a common element of the set of solutions of a finite family of pseudomonotone equilibrium problems {fi}Ni=1 and the set of fixed points of a finite family of nonexpansive mappings

∗ This paper was complete during the authors’ stay at the Vietnam Institute for Advanced Study in Mathematics (VIASM).

† School of Applied Mathematics and Informatics Ha Noi University of Science and Technology, Hanoi, Vietnam (E-mail: thuy.lequang@hust.edu.vn).

‡ Department of Mathematics, Vietnam National University, Hanoi, Vietnam (E-mail: anhpk@vnu.edu.vn)

§ Institute of Mathematics, VAST, Hanoi, Vietnam (E-mail: ldmuu@math.ac.vn).

1

{Sj}Mj=1 in Hilbert spaces, without assuming the Lipschitz-type conditions on the bifunc-tions fi, i = 1, , N We combine the extragradient method and Armijo-type line search techniques for handling pseudomonotone equilibrium problems [1,7], and Mann’s iterative scheme for finding fixed points of nonexpansive mappings [13] This paper is organized as follows: In Section 2, we recall some definitions and preliminary results Section 3 deals with the convergence analysis of the proposed hybrid extragradient-Armijo methods

2 Preliminaries

We recall some definitions and results that will be used in the next section Let C be a nonempty closed convex of a real Hilbert space H with an inner product h., i and the induced norm k.k Let T : C → C be a nonexpansive mapping with the set of fixed points

F (T )

We begin by recalling the following properties of nonexpansive mappings

Lemma 2.1 [9] Assume that T : C → C is a nonexpansive mapping Then

(i) I − T is demiclosed, i.e., whenever {xn} is a sequence in C weakly converging to some x ∈ C and the sequence {(I − T )xn} strongly converges to some y , it follows that (I − T )x = y

(ii) if T has a fixed point, then F (T ) is a closed convex subset of H

It is well known that if C is a nonempty closed and convex subset of H, then for every

x ∈ H, there exists a unique element PCx, defined by

PCx = arg min {ky − xk : y ∈ C} The mapping PC : H → C is called the metric (orthogonal) projection of H onto C It enjoys the following remarkable properties:

(i) PC is firmly nonexpansive, or 1-inverse strongly monotone (1-ism), i.e.,

hPCx − PCy, x − yi ≥ kPCx − PCyk2 for all x, y ∈ C (2.1) (ii)

kx − PCxk2+ kPCx − yk2 ≤ kx − yk2 (2.2) (iii) z = PCx if only if

A bifunction f is called monotone on C if

f (x, y) + f (y, x) ≤ 0, ∀x, y ∈ C;

f is pseudomonotone on C if

f (x, y) ≥ 0 ⇒ f (y, x) ≤ 0, ∀x, y ∈ C

It is obvious that any monotone bifunction is a pseudomonotone one, but not vice versa Throughout this paper we consider bifunctions with the following assumptions:

A1 f is pseudomonotone on C;

A2 f is weakly continuous on C × C;

A3 f (x, ) is convex and subdifferentiable on C, for every fixed x ∈ C

The following results will be used in the next section

Lemma 2.2 [6] Let C be a convex subset of a real Hilbert space H and ϕ : C → R be a convex and subdifferentiable function on C Then, x∗ is a solution to the convex problem

min {ϕ(x) : x ∈ C}

if only if 0 ∈ ∂ϕ(x∗) + NC(x∗), where ∂ϕ(x∗) denotes the subdifferential of ϕ and NC(x∗)

is the normal cone of C at x∗

Lemma 2.3 [15] Let X be a uniformly convex Banach space, r be a positive number and

Br(0) ⊂ X be a closed ball centered at the origin with radius r Then, for any given subset {x1, x2, , xN} ⊂ Br(0) and for any positive numbers λ1, λ2, , λN with PN

i=1λi = 1, there exists a continuous, strictly increasing, and convex function g : [0, 2r) → [0, ∞) with g(0) = 0 such that, for any i, j ∈ {1, 2, , N} with i < j,

N X k=1

λkxk

2

≤

N X k=1

λkkxkk2− λiλjg(kxi− xjk)

Lemma 2.4 [2] Let C ⊂ H be a closed convex subset and f : C × C → R ∪ {+∞} be

an equilibrium bifuntion satisfying Assumptions A1 − A3 If the solution set Sol(C, f ) is nonempty, then it is weakly closed and convex

3 Main results

Throughout this section we assume that the common solution set is nonempty, i.e.,

F =

 N

∩ i=1Sol(C, fi)



\M

∩ j=1F (Sj)

 6= ∅,

and that each bifunction fi (i = 1, , N) satisfies Assumptions A1 − A3

Since F 6= ∅, all the sets F (Sj) j = 1, , M and Sol(C, fi) i = 1, , N are nonempty, hence according to Lemmas 2.1, 2.4, they are closed and convex and their intersection F

is a nonempty closed and convex subset of C Thus given any fixed element x0

∈ C there exists a unique element ˆx := PF(x0

)

Algorithm 3.1 Choose positive numbers β > 0, σ ∈ (0,β

2), γ ∈ (0, 1) and the sequence {αn} ⊂ (0, 1) satisfying the condition lim sup

n→∞

αn< 1 Let x0

∈ C and set n := 0

Step 1 Solve N strong convex programs

yni = argmin{fi(xn, y) +β

2kxn− yk

2 : y ∈ C} i = 1, , N

and set di(xn) = xn− yi

n Step 2 Let I(xn) = {i ∈ {1, 2, , N} : di(xn) 6= 0}

• For all i ∈ I(xn), find the smallest positive integer number mi

n such that

fi xn− γm i

ndi(xn), yi

n
≤ −σkdi(xn)k2

Define

Vi

n := {x ∈ H : hwi

n, x − ¯zi

ni ≤ 0}, where ¯zi

n = xn− γm i

ndi(xn), wi

n ∈ ∂2fi(¯zi

n, ¯zi

n) Compute

zi

n = PC∩V i

n(xn),

• For all i /∈ I(xn), set zi

n = xn

Step 3 Find in = argmax{kzi

n− xnk : i = 1, , N}, and set ¯zn := zi n

n Step 4 Compute

uj

n= αnxn+ (1 − αn)Sjz¯n, j = 1, , M

Step 5 Find jn= argmax{kuj

n− xnk : j = 1, , M}, and set ¯un:= uj n

n Step 6 Compute

xn+1= PC n ∩Q n(x0), where

Cn = {v ∈ C : k¯un− vk ≤ kxn− vk},

Qn = {v ∈ C : hx0− xn, v − xni ≤ 0}

Increase n by 1 and go back to Step 1

We now prove the strong convergence of Algorithm 3.1

Theorem 3.1 Let C be a nonempty closed convex subset of a real Hilbert space H Suppose that {fi}Ni=1 is a finite family of bifunctions satisfying conditions A1 − A3 and {Sj}Mj=1 is a finite family of nonexpansive mappings on C Moreover, suppose that the solution set F =

 N

∩ i=1Sol(C, fi)

 T

 M

∩ j=1F (Sj)



is nonempty Then, the sequence {xn} generated by Algorithm 3.1 converges strongly to ˆx = PFx0

Proof Consider two cases

• Case 1 For any natural number k there exists a number n > k such that I(xn) 6= ∅

• Case 2 There exists a number n0 such that I(xn) = ∅; ∀n ≥ n0

We begin with Case 1 by dividing the proof into several steps

Step 1 We prove that the linesearch is finite for every i ∈ I(xn), i.e., there exists the smallest nonnegative integer mi

n satisfying

fi(xn− γm i

ndi(xn), yi

n) ≤ −σkdi(xn)k2

∀n, i = 1, , N

Indeed, assuming by contradiction that for every nonnegative integer m, one has

fi xn− γm

di(xn), yni
+ σkdi

(xn)k2

> 0

Letting m → ∞, we obtain

fi xn, yi

n
+ σkdi(xn)k2

On the other hand, since yi

n is the unique solution of the strongly convex problem min{fi(xn, y) +β

2ky − xnk

2 : y ∈ C},

we have

fi(xn, y) +β

2ky − xnk

2

≥ fi(xn, yin) + β

2ky

i

n− xnk2

, ∀y ∈ C

With y = xn, the last inequality becomes

fi(xn, yni) + β

2kd

i (xn)k2

Combining (3.2) with (3.3) yields

σkdi(xn)k2

≥ β

2kd

i(xn)k2

Hence it must be either kdi(xn)k = 0 or σ ≥ β

2 The first case contradicts di(xn) 6= 0, while the second one contradicts the choice σ < β2

Step 2 We show that

F ⊆ C ∩ Vi

n, xn6∈ Vi

n ∀i ∈ I(xn)

Indeed, let x∗ ∈ F Since fi(x∗, x) ≥ 0 for all x ∈ C, by pseudomonotonicity of fi, we have

fi(¯zi

It follows from wi

n∈ ∂2fi(¯zi

n, ¯zi

n) that

f (¯zni, x∗) =f (¯zni, x∗) − f (¯zni, ¯zin)

≥hwi

n, x∗− ¯zi

Combining (3.4) and (3.5), we get

hwi

n, x∗− ¯zi

ni ≤ 0

On the other hand, from the definition of Vi

n, we have x∗ ∈ Vi

n Thus F ⊆ C ∩ Vi

n Now we show that xn∈ V/ i

n In fact, from ¯zi

n = xn− γm i

ndi(xn), it follows that

yni − ¯zni = 1 − γ

m i n

γm i

n (¯zni − xn)

Then using (3.1) and the assumption fi(x, x) = 0 for all x ∈ C, we have

0 > −σkdi(xn)k2

≥ fi(¯zni, yni)

= fi(¯zi

n, yi

n) − fi(¯zi

n, ¯zi

n)

≥ hwi

n, yni − ¯znii

= 1 − γ

m i n

γm i n h¯zni − xn, wnii

Hence

hxn− ¯zi

n, wi

ni > 0, which implies xn∈ V/ i

n Step 3 (Solodov-Svaiter) For all i ∈ I(xn), we claim that zi

n = PC∩V i

n(¯yi

n), where

¯i

n = PV i

n(xn) Indeed, let K := {x ∈ H : ht, x − x0

i ≤ 0} with ktk 6= 0 It is easy to check that

PK(y) = y −ht, y − x

0i ktk2 t, Hence,

¯i

n = PV i

n(xn)

= xn−hw

i

n, xn− ¯zi

ni

kwi

nk2 wni

= xn−γ

m i

nhwi

n, di(xn)i

kwi

nk2 wi

n Note that, for every y ∈ C ∩ Vi

n, there exists λi ∈ (0, 1) such that ˆ

x = λixn+ (1 − λi)y ∈ C ∩ ∂Vi

n,

∂Vi

n = {x ∈ H : hwi

n, x − ¯zi

ni = 0}

Since xn∈ C, ˆx ∈ ∂Vi

n and ¯yi

n= PV i

n(xn), we have

ky − ¯yink2

≥ (1 − λi)2

ky − ¯ynik2

= kˆx − λixn− (1 − λi)¯yi

nk2

= k(ˆx − ¯yni) − λi(xn− ¯yin)k2

= kˆx − ¯yi

nk2 + λ2

ikxn− ¯yi

nk2

− 2λihˆx − ¯yi

n, xn− ¯yi

ni

= kˆx − ¯yi

nk2 + λ2

ikxn− ¯yi

nk2

≥ kˆx − ¯yink2

At the same time

kˆx − xnk2

= kˆx − ¯yi

n+ ¯yi

n− xnk2

= kˆx − ¯ynik2

− 2hˆx − ¯yni, xn− ¯yini + k¯yni − xnk2

= kˆx − ¯yi

nk2 + k¯yi

n− xnk2

Using zi

n = PC∩V n(xn) and the Pythagorean theorem, we can write

kˆx − ¯yi

nk2

= kˆx − xnk2

− k¯yi

n− xnk2

≥ kzi

n− xnk2

− k¯yni − xnk2

= kzi

n− ¯yi

nk2

From (3.6) and (3.7), it follows that

kzi

n− ¯yi

nk ≤ ky − ¯yi

nk, ∀y ∈ C ∩ Vi

n Hence

zi

n= PC∩V n(¯yi

n)

Step 4 For all i = 1, 2, , N, we show that

kzi

n− x∗k2

≤ kxn− x∗k2

− kzi

n− xnk2

k¯zn− x∗k2

≤ kxn− x∗k2

− k¯zn− xnk2

It is clear that these inequalities hold true for all i ∈ I(xn)

If i /∈ I(xn), by Step 2 of Algorithm 3.1, one has zi

n= PC∩V i

n(xn), i.e.,

hxn− zi

n, z − znii ≤ 0, ∀z ∈ C ∩ Vi

n Substituting z = x∗ ∈ F ⊆ C ∩ Vi

n by Step 2, we have

hxn− zi

n, x∗− zi

ni ≤ 0 ⇔ hxn− zi

n, x∗− xn+ xn− zi

ni ≤ 0, which implies that

hzi

n− xn, xn− x∗i ≤ −kzi

n− xnk2

Hence

kzi

n− x∗k2

= k(zi

n− xn) + (xn− x∗)k2

= kzi

n− xnk2

+ kxn− x∗k2

+ 2hzi

n− xn, xn− x∗i

≤ kzi

n− xnk2

+ kxn− x∗k2

− 2kzi

n− xnk2

= kxn− x∗k2

− kzi

n− xnk2

Thus (3.9) follows from the definition of in

Step 5 It holds that F ⊂ Cn∩ Qn for every n ≥ 0 In fact, for each x∗ ∈ F , by convexity

of k.k2

, the nonexpansiveness of Sj, and (3.9), we can write

k¯un− x∗k2

= kαnxn+ (1 − αn)Sj nz¯n− x∗k2

≤ αnkxn− x∗k2

+ (1 − αn)kSj nz¯n− x∗k2

≤ αnkxn− x∗k2

+ (1 − αn)k¯zn− x∗k2

≤ αnkxn− x∗k2

+ (1 − αn)kxn− x∗k2

≤ kxn− x∗k2

which implies k¯un− x∗k ≤ kxn− x∗k or x∗ ∈ Cn Hence F ⊂ Cn for all n ≥ 0

Now we show that F ⊂ CnT Qn by induction Indeed, it is clear that F ⊂ C0 Besides,

F ⊂ C = Q0, hence F ⊂ C0T Q0 Assume that F ⊂ Cn−1T Qn−1 for some n ≥ 1 Then from xn = PC n −1 T

Q n −1x0 and (2.3), we get

hxn− z, x0− xni ≥ 0, ∀z ∈ Cn−1

\

Qn−1 Since F ⊂ Cn−1T Qn−1, we have hxn− z, x0− xni ≥ 0 for all z ∈ F , which together with definition of Qn imply that F ⊂ Qn Hence F ⊂ CnT Qn for all n ≥ 1 Since

F and Cn∩ Qn are nonempty closed convex subsets, PFx0 and xn+1 := PC n ∩Q n(x0) are well-defined

Step 6 There hold the relations lim

n→∞kxn− Sjxnk = 0, and lim

n→∞kxn+1− xnk = lim

n→∞kxn− ujnk = lim

n→∞kxn− znik = lim

n→∞kxn− yink = 0

Indeed, from the definition of Qn and (2.2), we see that xn = PQ nx0 Therefore, for every

u ∈ F ⊂ Qn, we get

This implies that the sequence {xn} is bounded From (3.10), the sequence {¯un}, and hence, the sequence {uj

n} are also bounded

Observing that xn+1 = PC n T

Q nx0 ∈ Qn and xn = PQ nx0, applying (2.2) with x = x0 and

y = xn+1, we have

kxn− x0k2 ≤ kxn+1− x0k2− kxn+1− xnk2 ≤ kxn+1− x0k2 (3.12) Thus, the sequence {kxn− x0k} is decreasing, hence it has a finite limit as n approaches infinity From (3.12) we obtain

kxn+1− xnk2 ≤ kxn+1− x0k2− kxn− x0k2, and thus

lim

Since xn+1 ∈ Cn, k¯un− xn+1k ≤ kxn+1− xnk Thus k¯un− xnk ≤ k¯un− xn+1k + kxn+1−

xnk ≤ 2kxn+1−xnk Combining the last inequality with (3.13) we find that k¯un−xnk → 0

as n → ∞ From the definition of jn, we conclude that

lim n→∞ uj

for all j = 1, , M Moreover, by Step 4, for any fixed x∗ ∈ F, we have

kuj

n− x∗k2

= kαnxn+ (1 − αn)Sjz¯n− x∗k2

≤ αnkxn− x∗k2

+ (1 − αn)kSjz¯n− x∗k2

≤ αnkxn− x∗k2

+ (1 − αn)k¯zn− x∗k2

≤ αnkxn− x∗k2

+ (1 − αn) kxn− x∗k2

− k¯zn− xnk2

= kxn− x∗k2

− (1 − αn)k¯zn− xnk2

Thus

(1 − αn)k¯zn− xnk2

≤ kxn− x∗k2

− kuj

n− x∗k2

= kxn− x∗k − kuj

n− x∗k

kxn− x∗k + kuj

n− x∗k

≤ kxn− uj

nk kxn− x∗k + kuj

Using the last inequality together with (3.14) and taking into account the boundedness

of the two sequences {uj

n}, {xn} as well as the condition lim supn→∞αn < 1, we obtain limn→∞k¯zn− xnk = 0 By the definition of in, we get

lim

for all i = 1, , N On the other hand, since uj

n= αnxn+ (1 − αn)Sjz¯n, we have

kuj

n− xnk = (1 − αn)kSjz¯n− xnk

= (1 − αn)k(Sjxn− xn) + (Sjz¯n− Sjxn)k

≥ (1 − αn) (kSjxn− xnk − kSjz¯n− Sjxnk)

≥ (1 − αn) (kSjxn− xnk − k¯zn− xnk) Therefore

kSjxn− xnk ≤ k¯zn− xnk + 1

1 − αn

kuj

n− xnk

The last inequality together with (3.14), (3.16) and the condition lim supn→∞αn < 1 implies that

lim

for all j = 1, , M

Since {xn} is bounded, there exists a subsequence of {xn} converging weakly to ¯x For the sake of simplicity, we denote the weakly convergent subsequence again by {xn} , i.e.,

xn⇀ ¯x

Step 7 We show that ¯x ∈ F =

N T i=1 Sol(C, fi)

 T M T j=1

F (Sj)

! Indeed, from (3.17) and the demiclosedness of I − Sj, we have ¯x ∈ F (Sj) Hence, ¯x ∈

TM

j=1F (Sj) Observing that

yi

n= argmin{fi(xn, y) +β

2kxn− yk

2 : y ∈ C}, from Lemma 2.2, we obtain

0 ∈ ∂2



fi(xn, y) +β

2kxn− yk

2

 (yi

n) + NC(yi

n)

Thus, there exists w ∈ ∂2fi(xn, yi

n) and ¯w ∈ NC(yi

n) such that

w + β(xn− yi

Since ¯w ∈ NC(yi

n), we have h ¯w, y − yi

ni ≤ 0 for all y ∈ C, which together with (3.18) implies that

i n

i

n− xn, y − yi

n

(3.19) for all y ∈ C Since w ∈ ∂2fi(xn, yi

n),

fi(xn, y) − fi(xn, yni) ≥ ni , ∀y ∈ C (3.20) From (3.19) and (3.20), it follows that

fi(xn, y) − fi(xn, yi

n) ≥ β i

n− xn, y − yi

Recalling that xn ⇀ ¯x and kxn − yi

nk → 0 as n → ∞, we get yi

n ⇀ ¯x Letting

n → ∞ in (3.21) and using assumptions A2, we conclude that fi(¯x, y) ≥ 0 for all y ∈ C (i = 1, , N) Thus, ¯x ∈

N T i=1 Sol(C, fi), hence ¯x ∈ F

Step 8 We show that the sequence {xn} converges strongly to ˆx := PFx0

Indeed, from ˆx ∈ F and (3.11), we obtain

kxn− x0k ≤ kˆx − x0k

The last inequality together with xn⇀ ¯x and the weak lower semicontinuity of the norm k.k implies that

k¯x − x0k ≤ lim inf

n→∞kxn− x0k ≤ lim supn→∞kxn− x0k ≤ kˆx − x0k

By the definition of ˆx we find ¯x = ˆx and limn→∞kxn−x0k = kˆx−x0k Thus limn→∞kxnk =

kˆxk, and together with the fact that xn⇀ ˆx , we conclude xn → ˆx as n → ∞

Finally, suppose that {xm} is another weakly convergent subsequence of {xn} By a sim-ilar argument as above, we can conclude that {xm} converges strongly to ˆx := PFx0 Therefore, the sequence {xn} generated by the Algorithm 3.1 converges strongly to PFx0 Now we consider Case 2, when I(xn) = ∅ for all n ≥ n0

If xn+1 = xn, then Algorithm 3.1 terminates at a finite iteration n < ∞, and xn is a common element of two sets N∩

i=1Sol(C, fi) and M∩

j=1F (Sj), i.e., xn∈ F Indeed, we have xn= xn+1= PC n ∩Q n(x0) ∈ Cn By the definition of Cn, ||¯un− xn|| ≤

||xn− xn|| = 0, hence ¯un= xn From the definition of jn, we obtain

uj

which together with uj

n = αnxn+ (1 − αn)Sjz¯n and 0 < αn < 1 imply that xn = Sjz¯n Let x∗ ∈ F By the nonexpansiveness of Sj, we get

||xn− x∗||2

= ||Sjz¯n− x∗||2

≤ ||¯zn− x∗||2

≤ ||xn− x∗||2

− k¯zn− xnk2

From the last inequality and the definition of d(xi n

n), we obtain xn = yi n

n = ¯zn Thus

xn = Sjz¯n = Sjxn or xn ∈ F (Sj) for all j = 1, , M Moreover, from the equality

xn= ¯zn and definition of in, we also get xn = zi

n for all i = 1, , N Combining this fact with definition of d(xi

n) we see that xn = yi

n for all i = 1, , N Thus,

xn= argmin{ρfi(xn, y) + 1

2||xn− y||

2 : y ∈ C}

By [14, Proposition 2.1], from the last relation we conclude that xn ∈ Sol(C, fi) for all

i = 1, , N, hence xn ∈ F

Otherwise xn+1 6= xn for all n, similarly as in the proof of Steps 6 and 8, the sequence {xn} converges strongly to PFx0

The proof of Theorem 3.1 is complete

Replacing Mann’s iteration in Step 4 of Algorithm 3.1 by Halpern’s one, we come to the following algorithm with modified sets Cn

Algorithm 3.2 Initialize: x0 ∈ C, β > 0; σ ∈ (0,β

2), γ ∈ (0, 1), n := 0 and the nonnegative sequences {αn,l} (l = 0, , M) satisfying the conditions: 0 ≤ αn,j ≤ 1, M

P

j=0

αn,j = 1, lim inf

n→∞ αn,0αn,l > 0 for all l = 1, , M

Step 1 Solve N strong convex programs

yi

n = argmin{fi(xn, y) +β

2kxn− yk

2 : y ∈ C}, i = 1, , N

and set di(xn) = xn− yi

n Step 2 Let I(xn) = {i ∈ {1, 2, , N} : di(xn) 6= 0}

• For all i ∈ I(xn), find the smallest positive integer number mi

n such that

fi xn− γm i

ndi(xn), yni
≤ −σkdi

(xn)k2 Compute

zi

n = PC∩Vi

n(xn), where

Vi

n = {x ∈ H : hwi

n, x − ¯zi

ni ≤ 0}

with ¯zi

n = xn− γm ndi(xn) and wi

n∈ ∂2fi(¯zi

n, ¯zi

n)

• For all i /∈ I(xn), set zi

n = xn Step 3 Find in = argmax{kzi

n− xnk : i = 1, , N}, and set ¯zn := zi n

n Step 4 Compute

ujn= αnx0+ (1 − αn)Sjz¯n, j = 1, , M

Step 5 Find jn= argmax{kuj

n− xnk : j = 1, , M}, and set ¯un:= uj n

n Step 6 Compute

xn+1= PC n ∩Q n(x0), where

Cn = {v ∈ C : k¯un− vk2

≤ αnkx0 − vk2

+ (1 − αn)kxn− vk2

},

Qn = {v ∈ C : hx0− xn, v − xni ≤ 0}

Increase n by 1 and go back to Step 1

Theorem 3.2 Let C be a nonempty closed convex subset of a real Hilbert space H Suppose that {fi}Ni=1 is a finite family of bifunctions satisfying assumptions A1 − A3, and {Sj}Mj=1 is a finite family of nonexpansive mappings on C Moreover, suppose that the solution set F is nonempty Then, the sequence {xn} generated by Algorithm 3.2 converges strongly to ˆx = PFx0