Convergence theorems based on hybrid methods for generalized equilibrium problems and fixed point problems

Nonlinear Analysis 71 (2009) 4203–4214

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Nonlinear Analysis journal homepage: www.elsevier.com/locate/na

Convergence theorems based on hybrid methods for generalized equilibrium problems and fixed point problems Yeol Je Cho a , Xiaolong Qin b,∗ , Jung Im Kang c a

Department of Mathematics Education and the RINS, Gyeongsang National University, Chinju 660-701, Republic of Korea

b

Department of Mathematics, Gyeongsang National University, Chinju 660-701, Republic of Korea

c

National Institute for Mathematical Sciences, 385-16, 3F Tower Koreana, Doryong-dong, Yuseong-gu, Daejeon 305-340, Republic of Korea

article

abstract

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Article history: Received 16 October 2008 Accepted 25 February 2009

The purpose of this work is to introduce a hybrid projection method for finding a common element of the set of a generalized equilibrium problem, the set of solutions to a variational inequality and the set of fixed points of a strict pseudo-contraction in a real Hilbert space. © 2009 Elsevier Ltd. All rights reserved.

Keywords: Nonexpansive mapping Inverse-strongly monotone mapping Equilibrium problem Variational inequality Strict pseudo-contraction

1. Introduction and preliminaries Let H be a real Hilbert space, whose inner product and norm are denoted by h·, ·i and k · k, respectively. Let C be a nonempty closed convex subset of H and B : C → H a nonlinear mapping. Let PC be the projection of H onto the closed convex subset C . Recall the following definitions: (1) B is said to be monotone if

hBx − By, x − yi ≥ 0,

∀ x, y ∈ C .

(2) B is said to be strongly monotone if there exists a constant α > 0 such that

hBx − By, x − yi ≥ αkx − yk2 ,

∀ x, y ∈ C .

For such a case, B is said to be α -strongly-monotone. (3) B is said to be inverse-strongly monotone if there exists a constant α > 0 such that

hBx − By, x − yi ≥ αkBx − Byk2 ,

∀ x, y ∈ C .

For such a case, B is said to be α -inverse-strongly monotone. (4) A set-valued mapping T : H → 2H is said to be monotone if, for all x, y ∈ H, f ∈ Tx and g ∈ Ty imply hx − y, f − g i ≥ 0. (5) A monotone mapping T : H → 2H is maximal if the graph G(T ) of T is not properly contained in the graph of any other monotone mapping.

∗

Corresponding author. E-mail addresses: [email protected] (Y.J. Cho), [email protected], [email protected] (X. Qin), [email protected] (J.I. Kang).

0362-546X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2009.02.106

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It is known that a monotone mapping T is maximal if and only if, for any (x, f ) ∈ H × H, hx − y, f − g i ≥ 0 for all

(y, g ) ∈ G(T ) implies f ∈ Tx.

Let B be a monotone mapping of C into H and NC v be the normal cone to C at v ∈ C , i.e., NC v = {w ∈ H : hv − u, wi ≥ 0, ∀u ∈ C }

and define a mapping T on C by Tv =



Bv + NC v, v ∈ C ∅, v 6∈ C .

(1 )

Then T is maximal monotone and 0 ∈ T v if and only if hBv, u − vi ≥ 0 for all u ∈ C (see [1] for more details). The classical variational inequality problem is to find u ∈ C such that

hBu, v − ui ≥ 0,

∀v ∈ C . (1.1) In this paper, we use VI (C , B) to denote the set of solutions of the problem (1.1). One can easily see that the variational inequality problem is equivalent to a fixed point problem. A point u ∈ C is a solution to the problem (1.1) if and only if u is a fixed point of the mapping PC (I − λB), where λ > 0 is a constant. For finding solutions to a variational inequality for an inverse-strongly monotone mapping, Iiduka, Takahashi and Toyoda [2] proved the following theorem. Theorem ITT. Let C be a nonempty closed convex subset of a real Hilbert space H and let A be an α -inverse strongly monotone operator of H into H with VI (C , A) 6= ∅. Let {xn } be a sequence defined as follows:



x1 = x ∈ C , xn+1 = PC (αn xn + (1 − αn )PC (xn − λn Axn )),

∀n ≥ 1,

where C is the metric projection from H onto C , {αn } is a sequence in [−1, 1] and {λn } is a sequence in [0, 2α]. If {αn } and {λn } are chosen so that {αn } ∈ [a, b] for some a, b with −1 < a < b < 1 and {λn } ∈ [c , d] for some c , d with 0 < c < d < 2(1 + a)α , then {xn } converges weakly to an element of VI (C , A). Now, we recall the following definitions. (1) A mapping T : C → C is said to be nonexpansive if

kTx − Tyk ≤ kx − yk,

∀x, y ∈ C .

(2) T is said to be strictly pseudo-contractive with the coefficient k ∈ [0, 1) if

kTx − Tyk2 ≤ kx − yk2 + kk(I − T )x − (I − T )yk2 ,

∀x, y ∈ C .

(3) T is said to be pseudo-contractive if

kTx − Tyk2 ≤ kx − yk2 + k(I − T )x − (I − T )yk2 ,

∀ x, y ∈ C .

Clearly, the class of strict pseudo-contractions falls into the one between classes of nonexpansive mappings and pseudocontractions. Recently, many authors considered the problem of finding a common element of the set of solutions to the variational inequality problem (1.1) and of the set of fixed points of nonexpansive mappings in Hilbert spaces (see, for example, [3–11] and the references therein). In 2005, Takahashi and Toyoda [10] obtained the following results. Theorem TT1. Let C be a closed convex subset of a real Hilbert space H. Let α > 0. Let A be an α inverse strongly-monotone mapping of C into H and S be a nonexpansive mapping of C into itself such that F (S ) ∩ VI (C , A) 6= ∅. Let {xn } be a sequence generated by



x0 ∈ C , xn+1 = αn xn + (1 − αn )SPC (xn − λn Axn ),

∀n ≥ 0,

where {λn } ∈ [a, b] for some a, b ∈ (0, 2α) and {αn } ⊂ [c , d] for some c , d ∈ (0, 1). Then {xn } converges weakly to a point z ∈ F (S ) ∩ VI (C , A), where z = limn→∞ PF (S )∩VI (C ,A) xn . Let f be a bifunction of C × C into R, where R is the set of real numbers. The equilibrium problem for the function f : C × C → R is to find y ∈ C such that f (y, u) ≥ 0,

∀u ∈ C .

(1.2)

The set of solution of (1.2) is denoted by EP (f ). For any mapping T : C → H, let f (y, u) = hTy, u − yi for all y, u ∈ C . Then z ∈ EP (f ) if and only if hTz , u − z i ≥ 0 for all u ∈ C , i.e., z is a solution of the variational inequality. Numerous problem in physics, optimization and economics reduce to find a solution of (1.2). Some methods have been proposed to solve the equilibrium problems (see, for instance, [12,13]). For solving the equilibrium problem for a bifunction f : C × C → R, we may assume that f satisfies the following conditions:

Y.J. Cho et al. / Nonlinear Analysis 71 (2009) 4203–4214

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(A1) f (x, x) = 0 for all x ∈ C ; (A2) f is monotone, i.e., f (x, y) + f (y, x) ≤ 0 for all x, y ∈ C ; (A3) for each x, y, z ∈ C , lim f (tz + (1 − t )x, y) ≤ f (x, y); t ↓0

(A4) for each x ∈ C , y 7→ f (x, y) is convex and lower semi-continuous. Recently, many authors studied methods of finding a solution to the problem (1.2) (see, for example, [14,12,15–18,13,19– 27] and the references therein). Takahashi and Takahashi [25] studied the problem (1.2) by using contractions and proved some strong convergence theorems in real Hilbert spaces. To be more precise, they proved the following result. Theorem TT2. Let C be a nonempty closed convex subset of H. Let f be a bifunction from C × C to R satisfying (A1)–(A4) and S be a nonexpansive mapping of C into H such that F (S ) ∩ EP (f ) 6= ∅. Let f1 be a contraction of H into itself and {xn }, {un } be sequences generated by

 x ∈ H,   1 1 f (yn , u) + hu − yn , yn − xn i ≥ 0, ∀u ∈ C ,  r n  xn+1 = αn f1 (xn ) + (1 − αn )Syn , ∀n ≥ 1, where {αn } ∈ [0, 1] and {rn } ⊂ (0, ∞) satisfy the following conditions: lim αn = 0,

n→∞

∞ X

αn = ∞,

n=1

lim inf rn > 0, n→∞

∞ X

|αn+1 − αn | < ∞,

n=1

lim |rn+1 − rn | = 0.

n→∞

Then {xn } and {yn } converge strongly to a point z ∈ F (S ) ∩ EP (f ), where z = PF (S )∩EP (f ) f (z ). Let A : C → H be an inverse-strongly monotone mapping and f be a bifunction of C × C into R. We consider the following equilibrium problem: Find z ∈ C such that f (z , y) + hAz , y − z i ≥ 0,

∀y ∈ C .

(1.3)

In this paper, the set of such z ∈ C is denoted by EP (f , A), i.e., EP (f , A) = {z ∈ C : f (z , y) + hAz , y − z i ≥ 0, ∀y ∈ C }. If the case of A ≡ 0 (: the zero mapping), then the problem (1.3) is reduced to the problem (1.2). In the case of f ≡ 0, the problem (1.3) reduces to the classical variational inequality problem (1.1). The problem (1.3) is very general in the sense that it includes, as special cases, some optimization problems, variational inequalities, minimax problems, the Nash equilibrium problem in noncooperative games and others (see, for instance, [14,28]). In this paper, we introduce a hybrid projection method for the variational inequality problem (1.1), the equilibrium problem (1.3) and a fixed point problem for a strict pseudo-contraction. Also, some strong convergence theorems and applications are established in the framework of real Hilbert spaces. In order to prove our main results, we need the following lemmas. Lemma 1.1 ([14,12]). Let C be a nonempty closed convex subset of H ad let f : C × C → R be a bifunction satisfying (A1)–(A4). Then, for any r > 0 and x ∈ H, there exists z ∈ C such that f (z , y) +

1 r

hy − z , z − xi ≥ 0,

∀y ∈ C .

Further, if Tr x = {z ∈ C : f (z , y) + 1r hy − z , z − xi ≥ 0, ∀y ∈ C }, then the following hold: (1) Tr is single-valued; (2) Tr is firmly nonexpansive, i.e., for any x, y ∈ H,

kTr x − Tr yk2 ≤ hTr x − Tr y, x − yi; (3) F (Tr ) = EP (f ); (4) EP (f ) is closed and convex. Lemma 1.2 ([30]). Let C be a nonempty closed convex subset of a real Hilbert space H and T : C → C be a k-strict pseudocontraction. Define a mapping S : C → C by Sx = α x + (1 − α)Tx for all x ∈ C and α ∈ [k, 1). Then S is a nonexpansive mapping such that F (S ) = F (T ).

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2. Main results Now, we are ready to give our main results in this paper. Theorem 2.1. Let C be a closed convex subset of a real Hilbert space H and let f : C ×C → R be a bifunction satisfying (A1)–(A4). Let A be an α -inverse-strongly monotone mapping of C into H and B be a β -inverse-strongly monotone mapping of C into H, respectively. Let S : C → C be a k-strict pseudo-contraction with a fixed point. Define a mapping Sk : C → C by Sk x = kx + (1 − k)Sx for all x ∈ C . Assume that F := F (S ) ∩ VI (C , B) ∩ EP (f , A) 6= ∅. Let {xn } be a sequence generated by the following algorithm:

 x1 ∈ C ,    C1 = C ,   1   f (un , y) + hAxn , y − un i + hy − un , un − xn i ≥ 0, rn

∀y ∈ C ,

zn = PC (un − λn Bun ),     yn = αn xn + (1 − αn )Sk zn ,     Cn+1 = {w ∈ Cn : kyn − wk ≤ kxn − wk}, xn+1 = PCn+1 x1 , ∀n ≥ 1,

(ϒ )

where αn ⊂ [0, 1], {λn } ⊂ (0, 2β) and {rn } ⊂ (0, 2α) satisfy the following conditions: 0 ≤ αn ≤ a < 1,

0 < b ≤ λn ≤ c < 2β,

0 < d ≤ rn ≤ e < 2α

for some a, b, c , d, e ∈ R. Then the sequence {xn } defined by the algorithm (ϒ ) converges strongly to a point x¯ = PF x1 , where PF is the metric projection of H onto F . Proof. First, we show that the mapping PC (I − λn B) is nonexpansive for each n ≥ 1. Indeed, for any x, y ∈ C , from the assumption {λn } ⊂ [0, 2β], we have

kPC (I − λn B)x − PC (I − λn B)yk2 ≤ k(I − λn B)x − (I − λn B)yk2 = k(x − y) − λn (Bx − By)k2 = kx − yk2 − 2λn hx − y, Bx − Byi + λ2n kBx − Byk2 ≤ kx − yk2 − 2λn βkBx − Byk2 + λ2n kBx − Byk2 = kx − yk2 + λn (λn − 2β)kBx − Byk2 ≤ kx − yk2 . From the assumption {rn } ⊂ [0, 2α], for any p ∈ F , we also have

kun − pk2 = kTrn (xn − rn Axn ) − Trn (p − rn Ap)k2 ≤ k(xn − rn Axn ) − (p − rn Ap)k2 = k(xn − p) − rn (Axn − Ap)k2 = kxn − pk2 − 2rn hxn − p, Axn − Api + rn2 kAxn − Apk2 ≤ kxn − pk2 − 2rn αkAxn − Apk2 + rn2 kAxn − Apk2 = kxn − pk2 + rn (rn − 2α)kAxn − Apk2 ≤ kxn − pk2 .

(2.1)

Now, we show that Cn is closed and convex for each n ≥ 1. From the assumption, we see that C1 = C is closed and convex. Suppose that Cm is closed and convex for some m ≥ 1. Next, we show that Cm+1 is closed and convex for the same m. Indeed, for any w ∈ Cm , we see that

kym − wk ≤ kxm − wk is equivalent to

kym k2 − kxm k2 − 2hw, ym − xm i ≥ 0. Thus Cm+1 is closed and convex. Next, we show that F ⊂ Cn for each n ≥ 1. From the assumption, we see that F ⊂ C = C1 . Suppose F ⊂ Cm for some m ≥ 1. From Lemma 1.2, we see that Sk is a nonexpansive mapping. For any w ∈ F = Cm , we have

kym − wk = kαm xm + (1 − αm )Sk zm − wk ≤ αm kxm − wk + (1 − αm )kzm − wk

Y.J. Cho et al. / Nonlinear Analysis 71 (2009) 4203–4214

= ≤ ≤ =

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αm kxm − wk + (1 − αm )kPC (um − λm Bum ) − wk2 αm kxm − wk + (1 − αm )kum − wk2 αm kxm − wk + (1 − αm )kxm − wk2 kxm − wk.

It follows that w ∈ Cm+1 . This implies that F ⊂ Cn for each n ≥ 1. Since xn = PCn x1 and xn+1 = PCn+1 x1 ∈ Cn+1 ⊂ Cn , we have 0 ≤ hx1 − xn , xn − xn+1 i

= hx1 − xn , xn − x1 + x1 − xn+1 i ≤ −kx1 − xn k2 + kx1 − xn kkx1 − xn+1 k. It follows that

kxn − x1 k ≤ kxn+1 − x1 k.

(2.2)

Noting xn = PCn x1 , for any w ∈ F ⊂ Cn , we have

kx1 − xn k ≤ kx1 − wk. In particular, we have

kx1 − xn k ≤ kx1 − PF x1 k.

(2.3)

Therefore, it follow that limn→∞ kxn − x1 k exists. This implies that {xn } is bounded. Notice

k x n − x n +1 k 2 = = = = ≤

kxn − x1 + x1 − xn+1 k2 kxn − x1 k2 + 2hxn − x1 , x1 − xn+1 i + kx1 − xn+1 k2 kxn − x1 k2 + 2hxn − x1 , x1 − xn + xn − xn+1 i + kx1 − xn+1 k2 kxn − x1 k2 − 2kxn − x1 k2 + 2hxn − x1 , xn − xn+1 i + kx1 − xn+1 k2 kx1 − xn+1 k2 − kxn − x1 k2 .

It follows that lim kxn − xn+1 k = 0.

n→∞

(2.4)

Since xn+1 = PCn+1 x1 ∈ Cn+1 , we see that

kyn − xn+1 k ≤ kxn − xn+1 k. It follows that

kyn − xn k ≤ kyn − xn+1 k + kxn − xn+1 k ≤ 2kxn − xn+1 k. From (2.4), we obtain that lim kxn − yn k = 0.

n→∞

(2.5)

On the other hand, we have

kxn − yn k = kxn − αn xn − (1 − αn )Sk zn k = (1 − αn )kxn − Sk zn k. From the assumption 0 ≤ αn ≤ a < 1 and (2.5), we have lim kxn − Sk zn k = 0.

n→∞

(2.6)

For any w ∈ F , we have

kyn − wk2 = kαn xn + (1 − αn )Sk zn − wk2 ≤ αn kxn − wk2 + (1 − αn )kSk zn − wk2 = αn kxn − wk2 + (1 − αn )kzn − wk2 . It follows from (2.1) that

kyn − wk2 ≤ ≤ ≤ ≤ ≤

αn kxn − wk2 + (1 − αn )kPC (I − λn B)un − wk2 αn kxn − wk2 + (1 − αn )k(I − λn B)un − (I − λn B)wk2 αn kxn − wk2 + (1 − αn )(kun − wk2 + λn (λn − 2β)kBun − Bwk2 ) αn kxn − wk2 + (1 − αn )(kxn − wk2 + λn (λn − 2β)kBun − Bwk2 ) kxn − wk2 + (1 − αn )λn (λn − 2β)kBun − Bwk2 .

(2.7)

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Y.J. Cho et al. / Nonlinear Analysis 71 (2009) 4203–4214

This implies that

(1 − a)b(2β − c )kBun − Bwk2 ≤ kxn − wk2 − kyn − wk2 ≤ (kxn − wk − kyn − wk)(kxn − wk + kyn − wk) ≤ kxn − yn k(kxn − wk + kyn − wk). From (2.5), we obtain that lim kBun − Bwk = 0.

n→∞

(2.8)

On the other hand, we have

kzn − wk2 = kPC (I − λn B)un − PC (I − λn B)wk2 ≤ h(I − λn B)un − (I − λn B)w, zn − wi = ≤ =

1 2 1 2 1 2

{k(I − λn B)un − (I − λn B)wk2 + kzn − wk2 − k(I − λn B)un − (I − λn B)w − (zn − w)k2 } {kun − wk2 + kzn − wk2 − kun − zn − λn (Bun − Bw)k2 } {kun − wk2 + kzn − wk2 − kun − zn k2 + 2λn hun − zn , Bun − Bwi − λ2n kBun − Bwk2 },

which implies that

kzn − wk2 ≤ kun − wk2 − kun − zn k2 + 2λn hun − zn , Bun − Bwi − λ2n kBun − Bwk2 ≤ kxn − wk2 − kun − zn k2 + 2λn kun − zn kkBun − Bwk.

(2.9)

Substituting (2.9) into (2.7), we arrive at

kyn − wk2 ≤ kxn − wk2 − (1 − αn )kun − zn k2 + 2(1 − αn )λn kun − zn kkBun − Bwk. This implies that

(1 − αn )kun − zn k2 ≤ kxn − wk2 − kyn − wk2 + 2(1 − αn )λn kun − zn kkBun − Bwk ≤ kxn − yn k(kxn − wk + kyn − wk) + 2(1 − αn )λn kun − zn kkBun − Bwk. From the assumption 0 ≤ αn ≤ a < 1, (2.5) and (2.8), we obtain that lim kun − zn k = 0.

n→∞

(2.10)

Also, from (2.7), we have

kyn − wk2 ≤ αn kxn − wk2 + (1 − αn )kun − wk2 .

(2.11)

It follows that

kyn − wk2 ≤ αn kxn − wk2 + (1 − αn )kTrn (xn − rn Axn ) − Trn (w − rn Aw)k2 ≤ ≤ ≤ ≤

αn kxn − wk2 + (1 − αn )k(xn − rn Axn ) − (w − rn Aw)k2 αn kxn − wk2 + (1 − αn )(kxn − wk2 + rn (rn − 2α)kAxn − Awk2 ) αn kxn − wk2 + (1 − αn )(kxn − wk2 + rn (rn − 2α)kAxn − Awk2 ) kxn − wk2 + (1 − αn )rn (rn − 2α)kAxn − Awk2 .

From the assumptions 0 ≤ αn ≤ a < 1 and 0 < d ≤ rn ≤ e < 2β , we see that

(1 − a)d(2a − e)kAxn − Awk2 ≤ kxn − wk2 − kyn − wk2 ≤ kxn − yn k(kxn − wk + kyn − wk). From (2.5), we arrive at lim kAxn − Awk = 0.

n→∞

On the other hand, we have

kun − wk2 = kTrn (I − rn A)xn − Trn (I − rn A)wk2 ≤ h(I − rn A)xn − (I − rn A)w, un − wi

(2.12)

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= ≤ =

1 2 1 2 1 2

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(k(I − rn A)xn − (I − rn A)wk2 + kun − wk2 − k(I − rn A)xn − (I − rn A)w − (un − w)k2 ) (kxn − wk2 + kun − wk2 − k(xn − un ) − rn (A3 xn − A3 w)k2 ) (kxn − wk2 + kun − wk2 − kxn − un k2 + 2rn hxn − un , Axn − Awi − rn2 kAxn − Awk2 )

and hence

kun − wk2 ≤ kxn − wk2 − kxn − un k2 + 2rn hxn − un , Axn − Awi − rn2 kAxn − Awk2 ≤ kxn − wk2 − kxn − un k2 + 2rn kxn − un kkAxn − Awk.

(2.13)

Substituting (2.13) into (2.11), we have

kyn − wk2 ≤ kxn − wk2 − (1 − αn )kxn − un k2 + 2(1 − αn )rn kxn − un kkAxn − Awk. It follows that

(1 − αn )kxn − un k2 ≤ kxn − wk2 − kyn − wk2 + 2rn kxn − un kkAxn − Awk ≤ kxn − yn k(kxn − wk + kyn − wk) + 2rn kxn − un kkAxn − Awk. From the assumptions 0 ≤ αn ≤ a < 1, 0 < d ≤ rn ≤ e < 2β , (2.5) and (2.12), we obtain that lim kxn − un k = 0.

(2.14)

n→∞

On the other hand, we have

kxn − Sk xn k ≤ kSk xn − Sk zn k + kSk zn − xn k ≤ kxn − zn k + kSk zn − xn k ≤ kxn − un k + kun − zn k + kSk zn − xn k. It follows from (2.6), (2.10) and (2.14) that lim kxn − Sk xn k = 0.

(2.15)

n→∞

Since {xn } is bounded, we assume that a subsequence {xni } of {xn } converges weakly to ξ . Next, we show that ξ ∈ F (S )∩ VI (C , B)∩ EF (f , A). First, we prove that ξ ∈ VI (C , B). In fact, let T be the maximal monotone mapping defined by (1):

 Tx =

Bx + NC x, x ∈ C , ∅, x 6∈ C .

For any given (x, y) ∈ G(T ), we have y − Bx ∈ NC x. Since zn ∈ C , by the definition of NC , we have

hx − zn , y − Bxi ≥ 0.

(2.16)

On the other hand, from zn = PC (I − λn B)un , it follows that

hx − zn , zn − (I − λn B)un i ≥ 0 and hence



x − zn ,

zn − un

λn



+ Bun ≥ 0.

From (2.16) and the β -inverse monotonicity of B, we see that

hx − zni , yi ≥ hx − zni , Bxi   zn − uni ≥ hx − zni , Bxi − x − zni , i + Buni λni   zn − uni = hx − zni , Bx − Bzni i + hx − zni , Bzni − Buni i − x − zni , i λni   zni − uni ≥ hx − zni , Bzni − Buni i − x − zni , . λni

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Notice that

kzn − xn k ≤ kzn − un k + kun − xn k. From (2.10) and (2.14), we have lim kzn − xn k = 0.

n→∞

It follows that zni * ξ . On the other hand, from the definition of β -inverse-strongly monotone, we have that B is β1 -Lipschitz continuous. It follows that

hx − ξ , yi ≥ 0. Notice that T is maximal monotone and hence 0 ∈ T ξ . This shows that ξ ∈ VI (C , B). Now, we show that ξ ∈ EF (f , A). Since un = Trn (xn − rn Axn ) for any y ∈ C , we have f (un , y) + hAxn , y − un i +

1 rn

hy − un , un − xn i ≥ 0.

From the condition (A2), we see that

hAxn , y − un i +

1 rn

hy − un , un − xn i ≥ f (y, un ).

Replacing n by ni , we arrive at



hAxni , y − uni i + y − uni ,

un i − x n i



rni

≥ f (y, uni ).

(2.17)

For any t with 0 < t ≤ 1 and y ∈ C , let ρt = ty + (1 − t )ξ . Since y ∈ C and ξ ∈ C , we have ρt ∈ C . It follows from (2.17) that

  un − x n i hρt − uni , Aρt i ≥ hρt − uni , Aρt i − hAxni , ρt − uni i − ρt − uni , i + f (ρt , uni ) rni



= hρt − uni , Aρt − Aun,i i + hρt − uni , Aun,i − Axni i − ρt − uni ,

uni − xni rni



+ f (ρt , uni ).

(2.18)

Since A is Lipschitz continuous, from (2.14), we have Aun,i − Axni → 0 as i → ∞. On the other hand, from the monotonicity of A, we get that

hρt − uni , Aρt − Aun,i i ≥ 0. It follows from (A4) and (2.18) that

hρt − ξ , Aρt i ≥ f (ρt , ξ ) (i → ∞). From (A1), (A4) and (2.19), we see that 0 = f (ρt , ρt ) ≤ tf (ρt , y) + (1 − t )f (ρt , ξ )

≤ tf (ρt , y) + (1 − t )hρt − ξ , Aρt i = tf (ρt , y) + (1 − t )t hy − ξ , Aρt i, which yields that f (ρt , y) + (1 − t )hy − ξ , Aρt i ≥ 0. Letting t → 0 in the above inequality, we arrive at f (ξ , y) + hy − ξ , Aξ i ≥ 0. This shows that ξ ∈ EP (f , A). Next, we show that ξ ∈ F (S ) = F (Sk ). Suppose that ξ 6= Sk ξ . From Opial’s condition and (2.15), we arrive at lim inf kxni − ξ k < lim inf kxni − Sk ξ k i→∞

i→∞

= lim inf kxni − Sk xni + Sk xni − Sk ξ k i→∞

≤ lim inf kxni − ξ k, i→∞

(2.19)

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which derives a contradiction. Therefore, we obtain ξ ∈ F (Sk ) = F (S ). This shows that ξ ∈ F . Put x¯ = PF x1 . Since x¯ = PF x1 ⊂ Cn+1 and xn+1 = PCn+1 x1 , we have

kx1 − xn+1 k ≤ kx1 − x¯ k. On the other hand, we have

kx1 − x¯ k ≤ kx1 − ξ k ≤ lim inf kx1 − xni k i→∞

≤ lim sup kx1 − xni k i→∞

≤ kx1 − x¯ k. We, therefore, obtain that

kx1 − ξ k = lim kx1 − xni k = kx1 − x¯ k. i→∞

This implies xni → ξ = x¯ . Since {xni } is an arbitrary subsequence of {xn }, we have xn → x¯

(n → ∞).

This completes the proof.



3. Applications In this section, we prove some strong convergence theorems by using Theorem 2.1. Theorem 3.1. Let C be a closed convex subset of a real Hilbert space H and f : C × C → R be a bifunction satisfying (A1)–(A4). Let A be an α -inverse-strongly monotone mapping of C into H and S : C → C be a k-strict pseudo-contraction with a fixed point. Define a mapping Sk : C → C by Sk x = kx + (1 − k)Sx for all x ∈ C . Assume that F := F (S ) ∩ EP (f , A) 6= ∅. Let {xn } be a sequence generated by the following algorithm:

 x1 ∈ C ,    C1 = C ,    1  f (un , y) + hAxn , y − un i + hy − un , un − xn i ≥ 0, rn   yn = αn xn + (1 − αn )Sk un ,     Cn+1 = {w ∈ Cn : kyn − wk ≤ kxn − wk}, xn+1 = PCn+1 x1 , ∀n ≥ 1,

∀y ∈ C ,

where αn ⊂ [0, 1] and {rn } ⊂ (0, 2α) satisfy the following conditions: 0 ≤ αn ≤ a < 1,

0 < d ≤ rn ≤ e < 2 α

for some a, d, e ∈ R. Then the sequence {xn } defined by the above algorithm converges strongly to a point x¯ = PF x1 , where PF is the metric projection of H onto F . Proof. In Theorem 2.1, putting B = 0 (: the zero mapping), we can conclude the desired conclusion immediately.



Theorem 3.2. Let C be a closed convex subset of a real Hilbert space H and f : C × C → R be a bifunction satisfying (A1)–(A4). Let S : C → C be a nonexpansive mapping with a fixed point. Assume that F := F (S ) ∩ EP (f ) 6= ∅. Let {xn } be a sequence generated by the following algorithm:

 x1 ∈ C ,    C1 = C ,    1  f (un , y) + hy − un , un − xn i ≥ 0, ∀y ∈ C , rn   yn = αn xn + (1 − αn )Sun ,     Cn+1 = {w ∈ Cn : kyn − wk ≤ kxn − wk}, xn+1 = PCn+1 x1 , ∀n ≥ 1, where αn ⊂ [0, 1] and {rn } ⊂ (0, ∞) satisfy the following conditions: 0 ≤ αn ≤ a < 1,

0 < d ≤ rn ≤ e < ∞

for some a, d, e ∈ R. Then the sequence {xn } defined by above algorithm converges strongly to a point x¯ = PF x1 , where PF is the metric projection of H onto F .

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Proof. In Theorem 2.1, put A = B = 0 (: the zero mapping). Then, for any α > 0, we see that

hx − y, Ax − Ayi ≥ αkAx − Ayk2 ,

∀x, y ∈ C .

Let {rn } be a sequence satisfying the restriction: d ≤ rn ≤ e, where d, e ∈ (0, ∞). Then we can obtain the desired conclusion easily from Theorem 2.1. This completes the proof.  Theorem 3.3. Let C be a closed convex subset of a real Hilbert space H. Let A be an α -inverse-strongly monotone mapping of C into H and B be a β -inverse-strongly monotone mapping of C into H. Let S : C → C be a k-strict pseudo-contraction with a fixed point. Define a mapping Sk : C → C by Sk x = kx + (1 − k)Sx for all x ∈ C . Assume that F := F (S ) ∩ VI (C , B) ∩ VI (C , A) 6= ∅. Let {xn } be a sequence generated by the following algorithm:

 x1 ∈ C ,    C1 = C ,   1   hAxn , y − un i + hy − un , un − xn i ≥ 0,

rn zn = PC (un − λn Bun ),     yn = αn xn + (1 − αn )Sk zn ,     Cn+1 = {w ∈ Cn : kyn − wk ≤ kxn − wk}, xn+1 = PCn+1 x1 , ∀n ≥ 1,

∀y ∈ C ,

where αn ⊂ [0, 1], {λn } ⊂ (0, 2β) and {rn } ⊂ (0, 2α) satisfy the following conditions: 0 ≤ αn ≤ a < 1,

0 < b ≤ λn ≤ c < 2β,

0 < d ≤ rn ≤ e < 2 α

for some a, b, c , d, e ∈ R. Then the sequence {xn } defined by the above algorithm converges strongly to a point x¯ = PF x1 , where PF is the metric projection of H onto F . Proof. In Theorem 2.1, put f (x, y) = 0 for all x, y ∈ C . From

hAxn , y − un i +

1 rn

hy − un , un − xn i,

we have

hy − un , xn − un − rn Axn i ≥ 0,

∀y ∈ C .

This implies that un = PC (xn − rn Axn ). The algorithm in Theorem 3.3 is reduced to the following:

 x1 ∈ C ,    C1 = C ,    un = PC (xn − rn Axn ), zn = PC (un − λn Bun ),   yn = αn xn + (1 − αn )Sk zn ,    Cn+1 = {w ∈ Cn : kyn − wk ≤ kxn − wk}, xn+1 = PCn+1 x1 , ∀n ≥ 1. From the proof of Theorem 3.1, we can obtain the desired conclusion easily. This completes the proof.



From the first section in this paper, we see that a mapping T : C → C is said to be strictly pseudo-contractive if there exists k ∈ [0, 1) such that

kTx − Tyk2 ≤ kx − yk2 + kk(I − T )x − (I − T )yk2 ,

∀x, y ∈ C .

In this case, we say that T : C → C is a k-strict pseudo-contraction (see, for example, [29]). Putting A = I − T , we obtain

hx − y, Ax − Ayi ≥

1−k 2

kAx − Ayk2 ,

x, y ∈ C .

Thus the following theorem is not hard to derive. Theorem 3.4. Let C be a closed convex subset of a real Hilbert space H and f : C × C → R be a bifunction satisfying (A1)–(A4). Let TA : C → C be an kα -strict pseudo-contraction and TB : C → C be a kβ -strict pseudo-contraction. Let S : C → C be a

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k-strict pseudo-contraction with a fixed point. Define a mapping Sk : C → C by Sk x = kx + (1 − k)Sx for all x ∈ C . Assume that F := F (S ) ∩ F (TB ) ∩ EP (f , (I − T )) 6= ∅. Let {xn } be a sequence generated by the following algorithm:

 x1 ∈ C ,    C 1 = C ,    f (un , y) + h(I − TA )xn , y − un i + 1 hy − un , un − xn i ≥ 0,  rn

∀y ∈ C ,

zn = (1 − λn )un + λn TB un ,     yn = αn xn + (1 − αn )Sk zn ,     Cn+1 = {w ∈ Cn : kyn − wk ≤ kxn − wk}, xn+1 = PCn+1 x1 , ∀n ≥ 1, where αn ⊂ [0, 1], {λn } ⊂ (0, 1 − kβ ) and {rn } ⊂ (0, 1 − kα ) satisfy the following conditions: 0 ≤ αn ≤ a < 1,

0 < b ≤ λn ≤ c < 1 − kβ ,

0 < d ≤ rn ≤ e < 1 − kα

for some a, b, c , d, e ∈ R. Then the sequence {xn } converges strongly to a point x¯ = PF x1 , where PF is the metric projection of H onto F . 1−kβ Proof. Put A = I − TA and B = I − TB , respectively. Then we see that A is 1−2kα -inverse-strongly monotone and B is 2 inverse-strongly monotone, respectively. We have F (TB ) = VI (C , B) and

PC (un − λn Bun ) = (1 − λn )un + λn TB un . Therefore, it is easy to obtain the desired conclusion from Theorem 3.1.



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