Convergence theorems of a hybrid algorithm for equilibrium problems

Nonlinear Analysis: Hybrid Systems 3 (2009) 386–394

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Convergence theorems of a hybrid algorithm for equilibrium problems Poom Kumam, Kriengsak Wattanawitoon ∗ Department of Mathematics, Faculty of Science, King Mongkut’s University of Technology Thonburi, Bang Mod, Bangkok 10140, Thailand

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Article history: Received 18 February 2009 Accepted 19 February 2009 Keywords: Hybrid method Hemi-relatively non-expansive mapping Equilibrium problem Monotone hybrid algorithm

abstract In this paper, we introduce a hybrid iterative scheme for finding a common element of the set of common fixed points of two hemi-relatively non-expansive mappings and the set of solutions of an equilibrium problem by the CQ hybrid method in Banach spaces. Our results improve and extend the corresponding results announced by Cheng and Tian [Y. Cheng, M. Tian, Strong convergence theorem by monotone hybrid algorithm for equilibrium problems, hemi-relatively nonexpansive mappings and maximal monotone operators, Fixed Point Theory Appl. 2008 (2008) 12 pages, doi:10.1155/2008/617248], Takahashi and Zembayashi [W. Takahashi, K. Zembayashi, Strong convergence theorem by a new hybrid method for equilibrium problems and relatively non-expansive mappings, Fixed Point Theory Appl. (2008) doi:10.1155/2008/528476] and some others. © 2009 Elsevier Ltd. All rights reserved.

1. Introduction Equilibrium problems which were introduced by Blum and Oettli [1] in 1994 have had a great impact and influence in the development of several branches of pure and applied sciences. It has been shown that the equilibrium problem theory provides a novel and unified treatment of a wide class of problems which arise in economics, finance, physics, image reconstruction, ecology, transportation, network, elasticity and optimization. Let E be a real Banach space, E ∗ the dual space of E. Let C be a non-empty closed convex subset of E and f a bifunction from C × C to R, where R denotes the set of real numbers. The equilibrium problem (for short, EP) is to find p∈C

such that f (p, y) ≥ 0, ∀y ∈ C .

(1.1)

The set of solutions of (1.1) is denoted by EP (f ). Given a mapping T : C → E , let f (x, y) = hTx, y − xi for all x, y ∈ C . Then p ∈ EP (f ) if and only if hTp, y − pi ≥ 0 for all y ∈ C ; i.e., p is a solution of the variational inequality, there are several other problems, for example, the complementarity problem, fixed point problem and optimization problem, which can also be written in the form of an EP. In other words, the EP is an unifying model for several problems arising in physics, engineering, science, optimization, economics, etc. In the last two decades, many papers have appeared in the literature on the existence of solutions of EP; see, for example [1–4] and references therein. Matsushita and Takahashi [5] introduced the following iteration: a sequence {xn } defined by ∗

xn+1 = ΠC J −1 (αn Jxn + (1 − αn )JTxn )

(1.2)

where the initial guess element x0 ∈ C is arbitrary, {αn } is a real sequence in [0, 1], T is a relatively nonexpansive mapping and ΠC denotes the generalized projection from E onto a closed convex subset C of E. They prove that the sequence {xn } converges weakly to a fixed point of T .

∗

Corresponding author. E-mail addresses: [email protected] (P. Kumam), [email protected], [email protected] (K. Wattanawitoon).

1751-570X/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.nahs.2009.02.006

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Many authors studied the problem of finding a common element of the set of fixed points of a non-expansive mapping and the set of solutions of an equilibrium problem in the framework of Hilbert spaces and Banach spaces, respectively; see, for instance, [6–13] and the references therein. Recently, Takahashi and Zembayashi [12], proposed the following modification of iteration (1.2) for a relatively nonexpansive mapping:

 x0 = x ∈ C , C0 = C ,    yn = J −1 (αn Jxn + (1 − αn )JSzn ),   1 un ∈ C such that f (un , y) + hy − un , Jun − Jyn i ≥ 0,  rn   Cn+1 = {z ∈ Cn : φ(z , un ) ≤ φ(z , xn )},   xn+1 = ΠCn+1 x0 ,

∀y ∈ C ,

(1.3)

where J is the duality mapping on E, and ΠC is the generalized projection from E onto a closed convex subset C of E and proved that the sequence {xn } converges strongly to ΠF (S )∩EP (f ) x0 . In this paper, motivated by Takahashi and Zembayashi [12] and Cheng and Tian [14], we modify iterations (1.3) to obtain strong convergence theorems for finding a common element of the set of solutions of an equilibrium problem and the set of common fixed points of two hemi relatively non-expansive mappings in Banach spaces. The results obtained in this paper improve and extend the recent ones announced by Takahashi and Zembayashi’s result [12] and many others. 2. Preliminaries ∗

Let E be a real Banach space with norm k · k and let J be the normalized duality mapping from E into 2E given by Jx = {x∗ ∈ E ∗ : hx, x∗ i = kxkkx∗ k, kxk = kx∗ k}

(2.1)

for all x ∈ E, where E ∗ denotes the dual space of E and h·, ·i the generalized duality pairing between E and E ∗ . It is well known that if E ∗ is uniformly convex, then J is uniformly continuous on bounded subsets of E. Let E be a smooth, strictly convex, and reflexive Banach space and let C be a non-empty closed convex subset of E. Throughout this paper, we denote by φ the function defined by

φ(x, y) = kxk2 − 2hx, Jyi + kyk2 for x, y ∈ E .

(2.2)

Following Alber [15], the generalized projection ΠC : E → C is a map that assigns to an arbitrary point x ∈ E the minimum point of the functional φ(x, y), that is, ΠC x = x¯ , where x¯ is the solution to the minimization problem

φ(¯x, x) = inf φ(y, x)

(2.3)

y∈C

existence and uniqueness of the operator ΠC follows from the properties of the functional φ(x, y) and strict monotonicity of the mapping J. It is obvious from the definition of function φ that

(kyk − kxk)2 ≤ φ(y, x) ≤ (kyk + kxk)2 ,

∀ x, y ∈ E .

(2.4)

If E is a Hilbert space, then φ(x, y) = kx − yk . If E is a reflexive, strictly convex and smooth Banach space, then for x, y ∈ E, φ(x, y) = 0 if and only if x = y. It is sufficient to show that if φ(x, y) = 0 then x = y. From (2.4), we have kxk = kyk. This implies that hx, Jyi = kxk2 = kJyk2 . From the definition of J, one has Jx = Jy. Therefore, we have x = y; for more details. Let C be a closed convex subset of E, and let T be a mapping from C into itself. We denote by F (T ) the set of fixed points of T . T is said to be hemi-relatively non-expansive φ(p, Tx) ≤ φ(p, x) for all x ∈ C and p ∈ F (T ). A point p in C is said to be an asymptotic fixed point of T [16] if C contains a sequence {xn } which converges weakly to p such that limn→∞ kxn − Txn k = 0. The set of asymptotic fixed points of T will be denoted by Fˆ (T ). A hemi-relatively non-expansive mapping T from C into itself is called relatively non-expansive [17–19] if Fˆ (T ) = F (T ). The class of hemi-relatively non-expansive mappings is more general than the class of relatively non-expansive mappings [17–19] which requires the strong restriction: F (T ) = F] (T ) (see also [23–25]). x +y A Banach space E is said to be strictly convex if k 2 k < 1 for all x, y ∈ E with kxk = kyk = 1 and x 6= y. It is said to be uniformly convex if limn→∞ kxn − yn k = 0 for any two sequences {xn } and {yn } in E such that kxn k = kyn k = 1 and x +y limn→∞ k n 2 n k = 1. Let U = {x ∈ E : kxk = 1} be the unit sphere of E. Then the Banach space E is said to be smooth provided 2

lim

t →0

kx + tyk − kxk t

exists for each x, y ∈ U. It is also said to be uniformly smooth if the limit is attained uniformly for x, y ∈ E. It is well known that if E is uniformly smooth, then J is uniformly norm-to-norm continuous on each bounded subset of E.

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We also need the following lemmas for the proof of our main results. Lemma 2.1 (Kamimura and Takahashi [20]). Let E be a uniformly convex and smooth Banach space and let {xn } and {yn } be two sequences of E. If φ(xn , yn ) → 0 and either {xn } or {yn } is bounded, then kxn − yn k → 0. Lemma 2.2 (Alber [15]). Let C be a nonempty closed convex subset of a smooth Banach space E and x ∈ E. Then, x0 = ΠC x if and only if

hx0 − y, Jx − Jx0 i ≥ 0 ∀y ∈ C . Lemma 2.3 (Alber [15]). Let E be a reflexive, strictly convex and smooth Banach space, let C be a nonempty closed convex subset of E and let x ∈ E. Then

φ(y, ΠC x) + φ(ΠC x, x) ≤ φ(y, x) ∀y ∈ C . Lemma 2.4 (Cho et al. [21]). Let X be a uniformly convex Banach space and Br (0) be a closed ball of X . Then there exists a continuous strictly increasing convex function g : [0, ∞) → [0, ∞) with g (0) = 0 such that

kλx + µy + γ z k2 ≤ λkxk2 + µkyk2 + γ kz k2 − λµg (kx − yk) for all x, y, z ∈ Br (0) and λ, µ, γ ∈ [0, 1] with λ + µ + γ = 1. Lemma 2.5 (Kamimure and Takahashi [20]). Let E be a smooth and uniformly convex Banach space and let r > 0. Then there exists a strictly increasing, continuous, and convex function g : [0, 2r ] → R such that g (0) = 0 and g (kx − yk) ≤ φ(x, y) for all x, y ∈ Br . For solving the equilibrium problem for a bifunction f : C × C → R, let us assume that f satisfies the following conditions: (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. Lemma 2.6 (Blum and Oettli [1]). Let C be a closed convex subset of a smooth, strictly convex, and reflexive Banach space E, let f be a bifunction from C × C to R satisfying (A1)–(A4), and let r > 0 and x ∈ E. Then, there exists z ∈ C such that f (z , y) +

1 r

hy − z , Jz − Jxi ≥ 0,

∀y ∈ C .

Lemma 2.7 (Takahashi and Zembayashi [11, Lemma 2.8.]). Let C be a closed convex subset of a uniformly smooth, strictly convex, and reflexive Banach space E, and let f be a bifunction from C × C to R satisfying (A1)–(A4). For r > 0 and x ∈ E, define a mapping Tr : E → C as follows:

 Tr x =

z ∈ C : f (z , y) +

1 r

hy − z , Jz − Jxi, ∀y ∈ C



for all x ∈ C . Then the following hold: (1) Tr is single-valued; (2) Tr is a firmly non-expansive-type mapping, i.e., for all x, y ∈ E,

hTr x − Tr , JTr x − JTr yi ≤ hTr x − Tr , Jx − Jyi; (3) F (Tr ) = EP (f ); (4) EP (f ) is closed and convex. Lemma 2.8 (Takahashi and Zembayashi [12]). Let C be a closed convex subset of a smooth, strictly convex, and reflexive Banach space E, let f be a bifunction from C × C to R satisfying (A1)–(A4), and let r > 0. Then, for x ∈ E and q ∈ F (Tr ),

φ(q, Tr x) + φ(Tr x, x) ≤ φ(q, x). 3. Main results In this section, we establish strong convergence theorem for equilibrium problems and fixed point problems of two hemirelatively non-expansive mappings which are more general than relatively non-expansive mappings in Banach spaces.

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Theorem 3.1. Let E a uniformly convex and uniformly smooth real Banach space, let C be a non-empty and closed convex subset of a E. Let f be a bifunction from C × C to R satisfying (A1)–(A4) and let T , S : C → C are closed hemi-relatively non-expansive mappings such that F := F (T ) ∩ F (S ) ∩ EP (f ) 6= ∅. Let {xn } be a sequence generated by the following manner:

x ∈ C chosen arbitrarity, 0    yn = J −1 (αn Jxn + (1 − αn )JSzn ),    zn = J −1 (βn Jxn + (1 − βn )JTxn ), 1

 un ∈ C such that f (un , y) + hy − un , Jun − Jyn i ≥ 0,   rn   Cn+1 = {z ∈ C : φ(z , un ) ≤ φ(z , xn )},  xn+1 = ΠCn+1 (x0 ),

∀y ∈ C ,

(3.1)

for every n ∈ N ∪ {0}, where J is the duality mapping on E. Assume that {αn } and {βn } are sequences in [0, 1] such that lim supn−→∞ αn < 1, limn−→∞ βn = 1, lim infn−→∞ (1 − αn )βn (1 − βn ) > 0 and {rn } ⊂ [a, ∞) for some a > 0. If S is uniformly continuous, then {xn } converges strongly to ΠF x0 , where ΠF is the generalized projection of E onto F := F (T ) ∩ F (S ) ∩ EP (f ). Proof. First, we show that Cn+1 is closed and convex for each n ≥ 0. From the definition of Cn+1 it is obvious that Cn is closed. Since

φ(z , un ) ≤ φ(z , xn ) ⇔ 2(hz , Jxn i − 2hz , Jun i) ≤ kxn k2 − kun k2 . It is easy to see that Cn+1 is convex. Then, for all n ≥ 0, Cn is closed and convex. This shows that F ⊂ Cn for all n ≥ 0. Let p ∈ F . Putting un = Trn yn for all n ≥ 0. On the other hand, from Lemma 2.7, one has Trn is hemi-relatively non-expansive mapping. Next, we prove F ⊂ Cn for all n ≥ 0. F ⊂ C1 = C is obvious. Suppose F ⊂ Ck for some k ∈ N. Then, for ∀p ∈ F ⊂ Ck , one has

φ(p, uk ) = ≤ = = ≤ = ≤

φ(p, Trk yk ) φ(p, yk ) φ(p, J −1 (αk Jxk + (1 − αk )JSzk )) kpk2 − 2hp, αk Jxk + (1 − αk )JSzk i + kαk Jxk + (1 − αk )JSzk k2 kpk2 − 2αk hp, Jxk i − 2(1 − αk )hp, JSzk i + αk kJxk k2 + (1 − αk )kJSzk k2 αk φ(p, xk ) + (1 − αk )φ(p, Szk ) αk φ(p, xk ) + (1 − αk )φ(p, zk ),

(3.2)

and then

φ(p, zk ) = = ≤ = = ≤

φ(p, J −1 (βk Jxk + (1 − βk )JTxk )) kpk2 − 2 hp, βk Jxk + (1 − βk )JTxk i + kβk Jxk + (1 − βk )Txk k2 kpk2 − 2βk hp, Jxk i − 2(1 − βk ) hw, JTxk i + βk kxk k2 + (1 − βk )kTxk k2 βk (kpk2 − 2 hp, Jxk i + kxk k2 ) + (1 − βk )(kpk2 − 2 hp, JTxk i + kTxk k2 ) βk φ(p, xk ) + (1 − βk )φ(p, Txk ) βk φ(p, xk ) + (1 − βk )φ(p, xk )

= φ(p, xk ).

(3.3)

Substituting (3.3) into (3.2), we have

φ(p, uk ) ≤ αk φ(p, xk ) + (1 − αk )φ(p, xk ) ≤ φ(p, xk ),

(3.4)

that is p ∈ Ck+1 . This implies that F ⊂ Cn for all n ≥ 0. From xn = ΠCn x0 , we have

hxn − z , Jx0 − Jxn i ≥ 0,

∀z ∈ Cn

(3.5)

hxn − p, Jx0 − Jxn i ≥ 0,

∀p ∈ F .

(3.6)

and

From Lemma 2.3, one has

φ(xn , x0 ) = φ(ΠCn x0 , x0 ) ≤ φ(p, x0 ) − φ(p, xn ) ≤ φ(p, x0 ), for each p ∈ F ⊂ Cn and n ≥ 1. Then, the sequence {φ(xn , x0 )} is bounded. Since xn = ΠCn x0 , we have

φ(xn , x0 ) ≤ φ(xn+1 , x0 ),

∀n ∈ N ∪ {0}.

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Therefore, {φ(xn , x0 )} is non-decreasing. It follows that the limit of {φ(xn , x0 )} exists. By the construction of Cn , one has that Cm ⊂ Cn and xm = ΠCm x0 ∈ Cn for any positive integer m ≥ n. It follows that

φ(xm , xn ) = φ(xm , ΠCn x0 ) ≤ φ(xm , x0 ) − φ(ΠCn x0 , x0 ) = φ(xm , x0 ) − φ(xn , x0 ).

(3.7)

Letting m, n −→ ∞ in (3.7), one has φ(xm , xn ) −→ 0. It follows from Lemma 2.1, that xm − xn −→ 0 as m, n −→ ∞. Hence, {xn } is a cauchy sequence. Since E is a Banach space and C is closed and convex, one can assume that xn −→ x˜ ∈ C as n −→ ∞. Since

φ(xn+1 , xn ) = φ(xn+1 , ΠCn x0 ) ≤ φ(xn+1 , x0 ) − φ(ΠCn x0 , x0 ) = φ(xn+1 , x0 ) − φ(xn , x0 ), for all n ∈ N ∪ {0}, we have limn−→∞ φ(xn+1 , xn ) = 0. From Lemma 2.2, we get limn−→∞ kxn+1 − xn k = 0. Since xn+1 = ΠCn+1 x0 ∈ Cn+1 , we have

φ(xn+1 , un ) ≤ φ(xn+1 , xn ),

∀n ∈ N ∪ {0}.

Therefore, we also have lim φ(xn+1 , un ) = 0.

n−→∞

Since limn−→∞ φ(xn+1 , xn ) = limn−→∞ φ(xn+1 , un ) = 0 and E is uniformly convex and smooth, we have from Lemma 2.1 that lim kxn+1 − xn k = lim kxn+1 − un k = 0.

n−→∞

n−→∞

So, we have lim kxn − un k = 0.

n−→∞

Since J is uniformly norm-to-norm continuous on bounded sets and limn−→∞ kxn − un k = 0, we have lim kJxn − Jun k = 0.

n−→∞

Since E is uniformly smooth Banach space, one knows that E ∗ is a uniformly convex Banach apace. Let r = supn∈N∪{0} {kxn k, kTxn k, kSzn k}. From Lemma 2.4, we have

φ(p, zn ) = = ≤ =

φ(p, J −1 (βn Jxn + (1 − βn )JTxn )) kpk2 − 2 hp, βn Jxn + (1 − βn )JTxn i + kβn Jxn + (1 − βn )JTxn k2 kpk2 − 2β hp, Jxn i − 2(1 − βn ) hp, JTxn i + βn kJxn k2 + (1 − βn )kJTxn k2 − βn (1 − βn )g (kJxn − JTxn k) βn φ(p, xn ) + (1 − βn )φ(p, Txn ) − βn (1 − βn )g (kJxn − JTxn k)

= φ(p, xn ) − βn (1 − βn )g (kJxn − JTxn k),

(3.8)

and

φ(p, un ) = φ(p, Trn yn ) ≤ φ(p, yn ) ≤ αn φ(p, xn ) + (1 − αn )φ(p, zn ).

(3.9)

Substituting (3.8) into (3.9), we have

φ(p, un ) ≤ αn φ(p, xn ) + (1 − αn )(φ(p, xn ) − βn (1 − βn )g (kJxn − JTxn k)) ≤ φ(p, xn ) − (1 − αn )βn (1 − βn )g (kJxn − JTxn k).

(3.10)

It follows that

(1 − αn )βn (1 − βn )g (kJxn − JTxn k) ≤ φ(p, xn ) − φ(p, un ). On the other hand, we have

φ(p, xn ) − φ(p, un ) = kxn k2 − kun k2 − 2 hp, Jxn − Jun i ≤ kxn − un k(kxn k + kun k) + 2kpkkJxn − Jun k. It follow from kxn − un k −→ 0 and kJxn − Jun k −→ 0 that

φ(p, xn ) − φ(p, un ) −→ 0,

as n −→ ∞.

(3.11)

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By assumption lim infn−→∞ (1 − αn )βn (1 − βn ) > 0 and by Lemma 2.5, we also g (kJxn − JTxn k) −→ 0,

as n −→ ∞.

From the property of g that

kJxn − JTxn k −→ 0, Since J

−1

as n −→ ∞.

is also uniformly norm-to-norm continuous on bounded sets, we see that

lim kxn − Txn k = 0.

n−→∞

Since T is a closed operator and xn −→ x˜ , the x˜ is a fixed point of T . Since limn−→∞ βn = 1 and {xn } is bounded, we obtain

φ(xn+1 , zn ) = = ≤ =

φ(xn+1 , J −1 (βn Jxn + (1 − βn )JTxn )) kxn+1 k2 − 2hxn+1 , βn Jxn + (1 − βn )JTn xn i + kβn Jxn + (1 − βn )JTxn k2 kxn+1 k2 − 2βn hxn+1 , Jxn i − 2(1 − βn )hxn+1 , JTxn i + βn kxn k2 + (1 − βn )kTxn k2 βn φ(xn+1 , xn ) + (1 − βn )φ(xn+1 , Txn )

≤ φ(xn+1 , xn ).

(3.12)

Since φ(xn+1 , xn ) −→ 0 as n −→ ∞, therefore, φ(xn+1 , zn ) → 0 as n → ∞. Since xn+1 = ΠCn+1 x0 ∈ Cn+1 , from (3.2) and (3.3), we have

φ(xn+1 , un ) ≤ φ(xn+1 , yn ) ≤ φ(xn+1 , xn ), for all n ≥ 0. Thus

φ(xn+1 , yn ) → 0,

as n → ∞.

By using Lemma 2.1, we also have lim kxn+1 − yn k = lim kxn+1 − xn k = lim kxn+1 − zn k = 0.

n→∞

n→∞

n→∞

(3.13)

Since J is uniformly norm-to-norm continuous on bounded sets, we have lim kJxn+1 − Jyn k = lim kJxn+1 − Jxn k = lim kJxn+1 − Jzn k = 0.

n→∞

n→∞

n→∞

(3.14)

For each n ∈ N ∪ {0}, we observe that

kJxn+1 − Jyn k = = = ≥

kJxn+1 − (αn Jxn + (1 − αn )JSzn )k kαn (Jxn+1 − Jxn ) + (1 − αn )(Jxn+1 − JSzn )k k(1 − αn )(Jxn+1 − JSzn ) − αn (Jxn − Jxn+1 )k (1 − αn )kJxn+1 − JSzn k − αn kJxn − Jxn+1 k.

It follows that

kJxn+1 − JSzn k ≤

1 1 − αn

(kJxn+1 − Jyn k + αn kJxn − Jxn+1 k).

By (3.14) and lim supn−→∞ αn < 1, we obtain lim kJxn+1 − JSzn k = 0.

n→∞

Since J −1 is uniformly norm-to-norm continuous on bounded sets, we have lim kxn+1 − Szn k = 0.

n→∞

(3.15)

Since

kzn − xn k ≤ kzn − xn+1 k + kxn+1 − xn k. By (3.13), we obtain lim kzn − xn k = 0.

n−→∞

By using the triangle inequality, we get

kxn − Sxn k ≤ kxn − xn+1 k + kxn+1 − Szn k + kSzn − Sxn k. Since S is uniformly continuous. It follow from (3.13), (3.15) and (3.16) that limn−→∞ kxn − Sxn k = 0.

(3.16)

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Again using J is uniformly norm-to-norm continuous on bounded sets, we obtain lim kJxn − JSxn k = 0.

n−→∞

Since S is closed operator and xn −→ x˜ , then x˜ is a fixed point of S. Hence x˜ ∈ F (T ) ∩ F (S ). Next, we show x˜ ∈ EF (f ) = F (Tr ). From un = Trn yn and Lemma 2.8, we obtain

φ(un , yn ) = ≤ ≤ =

φ(Trn yn , yn ) φ(˜x, yn ) − φ(˜x, Trn yn ) φ(˜x, xn ) − φ(˜x, Trn yn ) φ(˜x, xn ) − φ(˜x, un ).

It follows from (3.11) that

φ(un , yn ) → 0,

as n → ∞.

Noticing that Lemma 2.1, we get

kun − yn k → 0,

as n → ∞.

Since J is uniformly norm-to-norm continuous on bounded sets, we obtain lim kJun − Jyn k = 0.

n→∞

From the (A2), we note that

k y − un k

kJun − Jyn k rn

≥

1 rn

hy − un , Jun − Jyn i ≥ −f (un , y) ≥ f (y, un ),

∀y ∈ C .

By taking the limit as n → ∞ in above inequality and from (A4) and un −→ x˜ , we have f (y, x˜ ) ≤ 0, ∀y ∈ C . For 0 < t < 1 and y ∈ C , define yt = ty + (1 − t )˜x. Noticing that y, x˜ ∈ C , we obtains yt ∈ C , which yields that f (yt , x˜ ) ≤ 0. It follows from (A1) that 0 = f (yt , yt ) ≤ tf (yt , y) + (1 − t )f (yt , x˜ ) ≤ tf (yt , y). That is, f (yt , y) ≥ 0. Let t ↓ 0, from (A3), we obtain f (˜x, y) ≥ 0, for ∀y ∈ C . This implies that x˜ ∈ EP (f ). This shows that x˜ ∈ F . Finally, we prove x˜ = ΠF x0 . By taking limit in (3.5), one has

h˜x − p, Jx0 − Jpi ≥ 0,

∀p ∈ F .

At this point, in view of Lemma 2.2, one sees that x˜ = ΠF x0 . This completes the proof.



Corollary 3.2. Let E a uniformly convex and uniformly smooth real Banach space, let C be a non-empty and closed convex subset of a E. Let f be a bifunction from C × C to R satisfying (A1)–(A4) and let T : C → C is a closed hemi-relatively non-expansive mappings such that F := F (T ) ∩ EP (f ) 6= ∅. Let {xn } be a sequence generated by the following manner:

x ∈ C chosen arbitrarity, 0  −1   yn = J (αn Jxn + (1 − αn )JTzn ),  −  zn = J 1 (βn Jxn + (1 − βn )JTxn ), 1

 un ∈ C such that f (un , y) + hy − un , Jun − Jyn i ≥ 0,   rn   Cn+1 = {z ∈ C : φ(z , un ) ≤ φ(z , xn )},  xn+1 = ΠCn+1 (x0 ),

(3.17)

∀y ∈ C ,

for every n ∈ N ∪ {0}, where J is the duality mapping on E. Assume that {αn } and {βn } are sequences in [0, 1] such that lim supn−→∞ αn < 1, limn−→∞ βn = 1, lim infn−→∞ (1 − αn )βn (1 − βn ) > 0 and {rn } ⊂ [a, ∞) for some a > 0. Then {xn } converges strongly to ΠF x0 , where ΠF is the generalized projection of E onto F := F (T ) ∩ EP (f ). Proof. In Theorem 3.1 if S ≡ T , the identity mapping, then (3.1) reduced to (3.17).



Corollary 3.3. Let E a uniformly convex and uniformly smooth real Banach space, let C be a non-empty and closed convex subset of a E. Let f be a bifunction from C × C to R satisfying (A1)–(A4) and let T : C → C is a closed hemi-relatively non-expansive

P. Kumam, K. Wattanawitoon / Nonlinear Analysis: Hybrid Systems 3 (2009) 386–394

393

mappings such that F := F (T ) ∩ EP (f ) 6= ∅. Let {xn } be a sequence generated by the following manner:

 x0 ∈ C chosen arbitrarity,    yn = J −1 (αn Jxn + (1 − αn )JTxn ),   1 un ∈ C such that f (un , y) + hy − un , Jun − Jyn i ≥ 0,  rn     Cn+1 = {z ∈ C : φ(z , un ) ≤ φ(z , xn )}, xn+1 = ΠCn+1 (x0 ),

∀y ∈ C ,

(3.18)

for every n ∈ N ∪{0}, where J is the duality mapping on E. Assume that {αn } is a sequence in [0, 1] such that lim supn−→∞ αn < 1, limn−→∞ αn = 1 and {rn } ⊂ [a, ∞) for some a > 0. Then {xn } converges strongly to ΠF x0 , where ΠF is the generalized projection of E onto F := F (T ) ∩ EP (f ). Proof. In Theorem 3.1 if S ≡ I, the identity mapping, then (3.1) reduced to (3.18).



Since every hemi-relatively non-expansive mappings are just relatively non-expansive, we obtain the following results: Theorem 3.4. Let C be a non-empty and closed convex subset of a Hilbert space H. Let f be a bifunction from C × C to R satisfying (A1)–(A4) and let T , S : C → C be two closed relatively non-expansive mappings such that F := F (T ) ∩ F (S ) ∩ EP (f ) 6= ∅. Let {xn } be a sequence generated by the following manner:

x ∈ C chosen arbitrarity, 0   yn = J −1 (αn Jxn + (1 − αn )JSzn ),    zn = J −1 (βn Jxn + (1 − βn )JTxn ), 1

 un ∈ C such that f (un , y) + hy − un , Jun − Jyn i ≥ 0,   rn  C   n+1 = {z ∈ C : φ(z , un ) ≤ φ(z , xn )}, xn+1 = PCn+1 (x0 ),

∀y ∈ C ,

(3.19)

for every n ∈ N ∪ {0}, where P is a projection from H into its subset. Assume that {αn } and {βn } are sequences in [0, 1] such that lim supn−→∞ αn < 1, limn−→∞ βn = 1, lim infn−→∞ (1 − αn )βn (1 − βn ) > 0 and {rn } ⊂ [a, ∞) for some a > 0. If S is uniformly continuous for all n ∈ N. Then {xn } converges strongly to PF x0 . Corollary 3.5. Let C be a non-empty and closed convex subset of a Hilbert space H. Let f be a bifunction from C × C to R satisfying (A1)–(A4) and let T : C → C be two closed relatively non-expansive mappings such that F := F (T ) ∩ EP (f ) 6= ∅. Let {xn } be a sequence generated by the following manner:

x ∈ C chosen arbitrarity, 0    yn = J −1 (αn Jxn + (1 − αn )JTzn ),    zn = J −1 (βn Jxn + (1 − βn )JTxn ), 1

 un ∈ C such that f (un , y) + hy − un , Jun − Jyn i ≥ 0,   rn   Cn+1 = {z ∈ C : φ(z , un ) ≤ φ(z , xn )},  xn+1 = ΠCn+1 (x0 ),

∀y ∈ C ,

(3.20)

for every n ∈ N ∪ {0}, where P is a projection from H into its subset. Assume that {αn } and {βn } are sequences in [0, 1] such that lim supn−→∞ αn < 1, limn−→∞ βn = 1, lim infn−→∞ (1 − αn )βn (1 − βn ) > 0 and {rn } ⊂ [a, ∞) for some a > 0. If S is uniformly continuous for all n ∈ N. Then {xn } converges strongly to PF x0 . 4. Maximal monotone Remark 4.1 ([22]). Let E be a reflexive, strictly convex, and smooth Banach space and let A be a monotone operator from E to E ∗ . Then, A is maximal if and only if R(J + rA) = E ∗ for all r > 0. Let E be a reflexive, strictly convex, and smooth Banach space and let A be a maximal monotone operator from E to E ∗ . Using Remark 4.1 and strict convexity of E, we obtain that for every r > 0 and x ∈ E, there exists a unique xr ∈ D(A) such that Jx ∈ Jxr + rAxr . If Jr x = xr , then we can define a single-valued mapping Jr : E −→ D(A) by Jr = (J + rA)−1 J, and such a Jr is called the resolvent of A. We know that A−1 0 = F (Jr ) for all r > 0 and Jr is relatively non-expansive mapping (see [12] for more details). Using Theorem 3.1, we can consider the problem of strong convergence concerning maximal monotone operators in a Banach space.

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Theorem 4.2. Let E be a uniformly convex and uniformly smooth real Banach space, let C be a non-empty closed convex subset of E, let f be a bifunction from C × C to R satisfying (A1)–(A4), and let Jr be a resolvent of A and a closed mapping such that A−1 0 ∩ EP (f ) 6= ∅, where r > 0. Define a sequence {xn } in C by the following:

x ∈ C chosen arbitrarity, 0  −1   yn = J (αn Jxn + (1 − αn )JJr zn ),  −  zn = J 1 (βn Jxn + (1 − βn )JJr xn ), 1

 un ∈ C such that f (un , y) + hy − un , Jun − Jyn i ≥ 0,   rn   Cn+1 = {z ∈ C : φ(z , un ) ≤ φ(z , xn )},  xn+1 = ΠCn+1 (x0 ),

(4.1)

∀y ∈ C ,

for every n ∈ N ∪ {0}, where J is the duality mapping on E. Assume that {αn } and {βn } are sequences in [0, 1] such that lim supn−→∞ αn < 1, limn−→∞ βn = 1, lim infn−→∞ (1 − αn )βn (1 − βn ) > 0 and {rn } ⊂ [a, ∞) for some a > 0. Then {xn } converges strongly to ΠA−1 0∩EP (f ) x. Proof. Since Jr is a closed relatively non-expansive mapping and A−1 0 Theorem 4.2. 

=

F (Jr ), from Corollary 3.2, we obtain

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