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Iterative common solutions for monotone inclusion problems, fixed point problems and equilibrium problems

Abstract

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let α>0, and let A be an α-inverse strongly-monotone mapping of C into H. Let T be a generalized hybrid mapping of C into H. Let B and W be maximal monotone operators on H such that the domains of B and W are included in C. Let 0<k<1, and let g be a k-contraction of H into itself. Let V be a γ ¯ -strongly monotone and L-Lipschitzian continuous operator with γ ¯ >0 and L>0. Take μ,γR as follows:

0<μ< 2 γ ¯ L 2 ,0<γ< γ ¯ L 2 μ 2 k .

Suppose that F(T) ( A + B ) 1 0 W 1 0, where F(T) and ( A + B ) 1 0, W 1 0 are the set of fixed points of T and the sets of zero points of A+B and W, respectively. In this paper, we prove a strong convergence theorem for finding a point z 0 of F(T) ( A + B ) 1 0 W 1 0, where z 0 is a unique fixed point of P F ( T ) ( A + B ) 1 0 W 1 0 (IV+γg). This point z 0 F(T) ( A + B ) 1 0 W 1 0 is also a unique solution of the variational inequality

( V γ g ) z 0 , q z 0 0,qF(T) ( A + B ) 1 0 W 1 0.

Using this result, we obtain new and well-known strong convergence theorems in a Hilbert space. In particular, we solve a problem posed by Kurokawa and Takahashi (Nonlinear Anal. 73:1562-1568, 2010).

MSC:47H05, 47H10, 58E35.

1 Introduction

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let and be the sets of positive integers and real numbers, respectively. A mapping T:CH is called generalized hybrid [1] if there exist α,βR such that

α T x T y 2 +(1α) x T y 2 β T x y 2 +(1β) x y 2

for all x,yC. We call such a mapping an (α, β)-generalized hybrid mapping. Kocourek, Takahashi and Yao [1] proved a fixed point theorem for such mappings in a Hilbert space. Furthermore, they proved a nonlinear mean convergence theorem of Baillon’s type [2] in a Hilbert space. Notice that the mapping above covers several well-known mappings. For example, an (α, β)-generalized hybrid mapping T is nonexpansive for α=1 and β=0, i.e.,

TxTyxy,x,yC.

It is also nonspreading [3, 4] for α=2 and β=1, i.e.,

2 T x T y 2 T x y 2 + T y x 2 ,x,yC.

Furthermore, it is hybrid [5] for α= 3 2 and β= 1 2 , i.e.,

3 T x T y 2 x y 2 + T x y 2 + T y x 2 ,x,yC.

We can also show that if x=Tx, then for any yC,

α x T y 2 +(1α) x T y 2 β x y 2 +(1β) x y 2 ,

and hence xTyxy. This means that an (α, β)-generalized hybrid mapping with a fixed point is quasi-nonexpansive. The following strong convergence theorem of Halpern’s type [6] was proved by Wittmann [7]; see also [8].

Theorem 1 Let C be a nonempty closed convex subset of H, and let T be a nonexpansive mapping of C into itself with F(T). For any x 1 =xC, define a sequence { x n } in C by

x n + 1 = α n x+(1 α n )T x n ,nN,

where { α n }(0,1) satisfies

lim n α n =0, n = 1 α n =and n = 1 | α n α n + 1 |<.

Then { x n } converges strongly to a fixed point of T.

Kurokawa and Takahashi [9] also proved the following strong convergence theorem for nonspreading mappings in a Hilbert space; see also Hojo and Takahashi [10] for generalized hybrid mappings.

Theorem 2 Let C be a nonempty closed convex subset of a real Hilbert space H. Let T be a nonspreading mapping of C into itself. Let uC and define two sequences { x n } and { z n } in C as follows: x 1 =xC and

{ x n + 1 = α n u + ( 1 α n ) z n , z n = 1 n k = 0 n 1 T k x n

for all n=1,2,, where { α n }(0,1), lim n α n =0 and n = 1 α n =. If F(T) is nonempty, then { x n } and { z n } converge strongly to Pu, where P is the metric projection of H onto F(T).

Remark We do not know whether Theorem 1 for nonspreading mappings holds or not; see [9] and [10].

In this paper, we provide a strong convergence theorem for finding a point z 0 of F(T) ( A + B ) 1 0 W 1 0 such that it is a unique fixed point of

P F ( T ) ( A + B ) 1 0 W 1 0 (IV+γg)

and a unique solution of the variational inequality

( V γ g ) z 0 , q z 0 0,qF(T) ( A + B ) 1 0 W 1 0,

where T, A, B, W, g and V denote a generalized hybrid mapping of C into H, an α-inverse strongly-monotone mapping of C into H with α>0, maximal monotone operators on H such that the domains of B and W are included in C, a k-contraction of H into itself with 0<k<1 and a γ ¯ -strongly monotone and L-Lipschitzian continuous operator with γ ¯ >0 and L>0, respectively. Using this result, we obtain new and well-known strong convergence theorems in a Hilbert space. In particular, we solve a problem posed by Kurokawa and Takahashi [9].

2 Preliminaries

Let H be a real Hilbert space with inner product , and norm , respectively. When { x n } is a sequence in H, we denote the strong convergence of { x n } to xH by x n x and the weak convergence by x n x. We have from [11] that for any x,yH and λR,

x + y 2 x 2 +2y,x+y
(2.1)

and

λ x + ( 1 λ ) y 2 =λ x 2 +(1λ) y 2 λ(1λ) x y 2 .
(2.2)

Furthermore, we have that for x,y,u,vH,

2xy,uv= x v 2 + y u 2 x u 2 y v 2 .
(2.3)

All Hilbert spaces satisfy Opial’s condition, that is,

(2.4)

if x n u and uv; see [12]. Let C be a nonempty closed convex subset of a Hilbert space H, and let T:CH be a mapping. We denote by F(T) the set of fixed points for T. A mapping T:CH is called quasi-nonexpansive if F(T) and Txyxy for all xC and yF(T). If T:CH is quasi-nonexpansive, then F(T) is closed and convex; see [13]. For a nonempty closed convex subset C of H, the nearest point projection of H onto C is denoted by P C , that is, for all xH and yC. Such P C is called the metric projection of H onto C. We know that the metric projection P C is firmly nonexpansive; for all x,yH. Furthermore, x P C x,y P C x0 holds for all xH and yC; see [14]. The following result is in [15].

Lemma 3 Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let T:CH be a generalized hybrid mapping. Suppose that there exists { x n }C such that x n z and x n T x n 0. Then zF(T).

Let B be a mapping of H into 2 H . The effective domain of B is denoted by D(B), that is, D(B)={xH:Bx}. A multi-valued mapping B is said to be a monotone operator on H if xy,uv0 for all x,yD(B), uBx, and vBy. A monotone operator B on H is said to be maximal if its graph is not properly contained in the graph of any other monotone operator on H. For a maximal monotone operator B on H and r>0, we may define a single-valued operator J r = ( I + r B ) 1 :HD(B), which is called the resolvent of B for r. We denote by A r = 1 r (I J r ) the Yosida approximation of B for r>0. We know from [8] that

A r xB J r x,xH,r>0.
(2.5)

Let B be a maximal monotone operator on H, and let B 1 0={xH:0Bx}. It is known that B 1 0=F( J r ) for all r>0 and the resolvent J r is firmly nonexpansive, i.e.,

J r x J r y 2 xy, J r x J r y,x,yH.
(2.6)

We also know the following lemma from [16].

Lemma 4 Let H be a real Hilbert space, and let B be a maximal monotone operator on H. For r>0 and xH, define the resolvent J r x. Then the following holds:

s t s J s x J t x, J s xx J s x J t x 2

for all s,t>0 and xH.

From Lemma 4, we have that

for all λ,μ>0 and xH; see also [14, 17]. To prove our main result, we need the following lemmas.

Lemma 5 ([18]; see also [19])

Let { s n } be a sequence of nonnegative real numbers, let { α n } be a sequence of [0,1] with n = 1 α n =, let { β n } be a sequence of nonnegative real numbers with n = 1 β n <, and let { γ n } be a sequence of real numbers with lim sup n γ n 0. Suppose that

s n + 1 (1 α n ) s n + α n γ n + β n

for all n=1,2,. Then lim n s n =0.

Lemma 6 ([20])

Let { Γ n } be a sequence of real numbers that does not decrease at infinity in the sense that there exists a subsequence { Γ n i } of { Γ n } which satisfies Γ n i < Γ n i + 1 for all iN. Define the sequence { τ ( n ) } n n 0 of integers as follows:

τ(n)=max{kn: Γ k < Γ k + 1 },

where n 0 N such that {k n 0 : Γ k < Γ k + 1 }. Then the following hold:

  1. (i)

    τ( n 0 )τ( n 0 +1) and τ(n);

  2. (ii)

    Γ τ ( n ) Γ τ ( n ) + 1 and Γ n Γ τ ( n ) + 1 , nN.

3 Strong convergence theorems

Let H be a real Hilbert space. A mapping g:HH is a contraction if there exists k(0,1) such that g(x)g(y)kxy for all x,yH. We call such a mapping g a k-contraction. A nonlinear operator V:HH is called strongly monotone if there exists γ ¯ >0 such that xy,VxVy γ ¯ x y 2 for all x,yH. Such V is also called γ ¯ -strongly monotone. A nonlinear operator V:HH is called Lipschitzian continuous if there exists L>0 such that VxVyLxy for all x,yH. Such V is also called L-Lipschitzian continuous. We know the following three lemmas in a Hilbert space; see Lin and Takahashi [21].

Lemma 7 ([21])

Let H be a Hilbert space, and let V be a γ ¯ -strongly monotone and L-Lipschitzian continuous operator on H with γ ¯ >0 and L>0. Let t>0 satisfy 2 γ ¯ >t L 2 and 1>2t γ ¯ . Then 0<1t(2 γ ¯ t L 2 )<1 and ItV:HH is a contraction, where I is the identity operator on H.

Lemma 8 ([21])

Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let P C be the metric projection of H onto C, and let V be a γ ¯ -strongly monotone and L-Lipschitzian continuous operator on H with γ ¯ >0 and L>0. Let t>0 satisfy 2 γ ¯ >t L 2 and 1>2t γ ¯ , and let zC. Then the following are equivalent:

  1. (1)

    z= P C (ItV)z;

  2. (2)

    Vz,yz0, yC;

  3. (3)

    z= P C (IV)z.

Such zC always exists and is unique.

Lemma 9 ([21])

Let H be a Hilbert space, and let g:HH be a k-contraction with 0<k<1. Let V be a γ ¯ -strongly monotone and L-Lipschitzian continuous operator on H with γ ¯ >0 and L>0. Let a real number γ satisfy 0<γ< γ ¯ k . Then Vγg:HH is a ( γ ¯ γk)-strongly monotone and (L+γk)-Lipschitzian continuous mapping. Furthermore, let C be a nonempty closed convex subset of H. Then P C (IV+γg) has a unique fixed point z 0 in C. This point z 0 C is also a unique solution of the variational inequality

( V γ g ) z 0 , q z 0 0,qC.

Now, we prove the following strong convergence theorem of Halpern’s type [6] for finding a common solution of a monotone inclusion problem for the sum of two monotone mappings, of a fixed point problem for generalized hybrid mappings and of an equilibrium problem for bifunctions in a Hilbert space.

Theorem 10 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let α>0, and let A be an α-inverse strongly-monotone mapping of C into H. Let B and W be maximal monotone operators on H such that the domains of B and W are included in C. Let J λ = ( I + λ B ) 1 and T r = ( I + r W ) 1 be resolvents of B and W for λ>0 and r>0, respectively. Let S be a generalized hybrid mapping of C into H. Let 0<k<1, and let g be a k-contraction of H into itself. Let V be a γ ¯ -strongly monotone and L-Lipschitzian continuous operator with γ ¯ >0 and L>0. Take μ,γR as follows:

0<μ< 2 γ ¯ L 2 ,0<γ< γ ¯ L 2 μ 2 k .

Suppose F(S) ( A + B ) 1 0 W 1 0. Let x 1 =xH, and let { x n }H be a sequence generated by

x n + 1 = β n x n +(1 β n ) { α n γ g ( x n ) + ( I α n V ) S J λ n ( I λ n A ) T r n x n }

for all nN, where { α n }(0,1), { β n }(0,1), { λ n }(0,) and { r n }(0,) satisfy

lim n α n = 0 , n = 1 α n = , 0 < lim inf n β n lim sup n β n < 1 , lim inf n r n > 0 and 0 < a λ n b < 2 α .

Then { x n } converges strongly to z 0 F(S) ( A + B ) 1 0 W 1 0, where z 0 is a unique fixed point in F(S) ( A + B ) 1 0 W 1 0 of P F ( S ) ( A + B ) 1 0 W 1 0 (IV+γg).

Proof Let zF(S) ( A + B ) 1 0 W 1 0. We have that z=Sz, z= J λ n (I λ n A)z and z= T r n z. Putting w n = J λ n (I λ n A) T r n x n and u n = T r n x n , we obtain that

(3.1)

Put τ= γ ¯ L 2 μ 2 . Using lim n α n =0, we have that for any x,yH,

(3.2)

Since 1 α n τ>0, we obtain that for any x,yH,

( I α n V ) x ( I α n V ) y (1 α n τ)xy.
(3.3)

Putting y n = α n γg( x n )+(I α n V)S J λ n (I λ n A) T r n x n , from z= α n Vz+z α n Vz, (3.1) and (3.3) we have that

Using this, we get

Putting K=max{ x 1 z, γ g ( z ) V z τ γ k }, we have that x n zK for all nN. Then { x n } is bounded. Furthermore, { u n }, { w n } and { y n } are bounded. Using Lemma 9, we can take a unique z 0 F(S) ( A + B ) 1 0 W 1 0 such that

z 0 = P F ( S ) ( A + B ) 1 0 W 1 0 (IV+γg) z 0 .

From the definition of { x n }, we have that

x n + 1 x n = β n x n +(1 β n ) { α n γ g ( x n ) + ( I α n V ) S w n } x n

and hence

x n + 1 x n ( 1 β n ) α n γ g ( x n ) = β n x n + ( 1 β n ) ( I α n V ) S w n x n = ( 1 β n ) { ( I α n V ) S w n x n } = ( 1 β n ) ( S w n x n α n V S w n ) .

Thus, we have that

(3.4)

From (2.3) and (3.1), we have that

2 x n S w n , x n z 0 = x n z 0 2 + S w n x n 2 S w n z 0 2 x n z 0 2 + S w n x n 2 x n z 0 2 = S w n x n 2 .
(3.5)

From (3.4) and (3.5), we also have that

2 x n x n + 1 , x n z 0 = 2 ( 1 β n ) α n γ g ( x n ) , x n z 0 2 ( 1 β n ) x n S w n , x n z 0 2 ( 1 β n ) α n V S w n , x n z 0 2 ( 1 β n ) α n γ g ( x n ) , x n z 0 ( 1 β n ) S w n x n 2 2 ( 1 β n ) α n V S w n , x n z 0 .
(3.6)

Furthermore, using (2.3) and (3.6), we have that

x n + 1 z 0 2 x n x n + 1 2 x n z 0 2 2 ( 1 β n ) α n γ g ( x n ) , x n z 0 ( 1 β n ) S w n x n 2 2 ( 1 β n ) α n V S w n , x n z 0 .

Setting Γ n = x n z 0 2 , we have that

(3.7)

Noting that

x n + 1 x n = ( 1 β n ) α n ( γ g ( x n ) V S w n ) + ( 1 β n ) ( S w n x n ) ( 1 β n ) ( S w n x n + α n γ g ( x n ) V S w n ) ,
(3.8)

we have that

x n + 1 x n 2 ( 1 β n ) 2 ( S w n x n + α n γ g ( x n ) V S w n ) 2 = ( 1 β n ) 2 S w n x n 2 + ( 1 β n ) 2 2 α n S w n x n γ g ( x n ) V S w n + ( 1 β n ) 2 α n 2 γ g ( x n ) V S w n 2 .
(3.9)

Thus, we have from (3.7) and (3.9) that

Γ n + 1 Γ n x n x n + 1 2 + 2 ( 1 β n ) α n γ g ( x n ) , x n z 0 ( 1 β n ) S w n x n 2 2 ( 1 β n ) α n V S w n , x n z 0 ( 1 β n ) 2 S w n x n 2 + ( 1 β n ) 2 2 α n S w n x n γ g ( x n ) V S w n + ( 1 β n ) 2 α n 2 γ g ( x n ) V S w n 2 + 2 ( 1 β n ) α n γ g ( x n ) , x n z 0 ( 1 β n ) S w n x n 2 2 ( 1 β n ) α n V S w n , x n z 0

and hence

(3.10)

We divide the proof into two cases.

Case 1: Suppose that Γ n + 1 Γ n for all nN. In this case, lim n Γ n exists and then lim n ( Γ n + 1 Γ n )=0. Using 0< lim inf n β n lim sup n β n <1 and lim n α n =0, we have from (3.10) that

lim n S w n x n =0.
(3.11)

Using (3.8), we also have that

lim n x n + 1 x n =0.
(3.12)

Since x n + 1 x n =(1 β n )( y n x n ), we have from (3.12) that

lim n y n x n =0.
(3.13)

We also have from (2.6) that

2 u n z 0 2 = 2 T r n x n T r n z 0 2 2 x n z 0 , u n z 0 = x n z 0 2 + u n z 0 2 u n x n 2

and hence

u n z 0 2 x n z 0 2 u n x n 2 .
(3.14)

From (3.1) we have that

S w n z 0 2 u n z 0 2 x n z 0 2 u n x n 2

and hence

u n x n 2 x n z 0 2 S w n z 0 2 M S w n x n 2 ,

where M=sup{ x n z 0 +S w n z 0 :nN}. Thus, from (3.11) we have that

lim n u n x n =0.
(3.15)

We show lim n S w n w n =0. Since 2 is a convex function, we have that

(3.16)

From z 0 = α n V z 0 + z 0 α n V z 0 and (2.1), we also have that

(3.17)

Using (3.16) and (3.17), we have that

(3.18)

Thus, we have that

(3.19)

Then we have that

lim n A u n A z 0 =0.
(3.20)

Since J λ n is firmly nonexpansive, we have that

Thus, we get

(3.21)

Using (3.17), we obtain

from which it follows that

Then we have

(3.22)

From (3.22) and (3.15), we have that

(3.23)

Since S w n w n S w n x n + x n w n , we have that

lim n S w n w n =0.
(3.24)

Take λ 0 R with 0<a λ 0 b<2α arbitrarily. Put s n =(I λ n A) u n . Using u n = T r n x n and w n = J λ n (I λ n A) u n , we have from Lemma 4 that

J λ 0 ( I λ 0 A ) u n w n = J λ 0 ( I λ 0 A ) u n J λ n ( I λ n A ) u n = J λ 0 ( I λ 0 A ) u n J λ 0 ( I λ n A ) u n + J λ 0 ( I λ n A ) u n J λ n ( I λ n A ) u n ( I λ 0 A ) u n ( I λ n A ) u n + J λ 0 s n J λ n s n | λ 0 λ n | A u n + | λ 0 λ n | λ 0 J λ 0 s n s n .
(3.25)

We also have from (3.25) that

u n J λ 0 ( I λ 0 A ) u n u n w n + w n J λ 0 ( I λ 0 A ) u n .
(3.26)

We will use (3.25) and (3.26) later.

Let us show that lim sup n (Vγg) z 0 , x n z 0 0. Put

A= lim sup n ( V γ g ) z 0 , x n z 0 .

Without loss of generality, we may assume that there exists a subsequence { x n i } of { x n } such that A= lim i (Vγg) z 0 , x n i z 0 and { x n i } converges weakly to some point wH. From x n w n 0 and x n u n 0, we also have that { w n i } and { u n i } converge weakly to wC. On the other hand, from { λ n i }[a,b] there exists a subsequence { λ n i j } of { λ n i } such that λ n i j λ 0 for some λ 0 [a,b]. Without loss of generality, we assume that w n i w, u n i w and λ n i λ 0 . From (3.24) we know lim n S w n w n =0. Thus, we have from Lemma 3 that w=Sw. Since W is a monotone operator and x n i u n i r n i W u n i , we have that for any (u,v)W,

u u n i , v x n i u n i r n i 0.

Since lim inf n r n >0, u n i w and x n i u n i 0, we have

uw,v0.

Since W is a maximal monotone operator, we have 0Ww and hence w W 1 0. Since λ n i λ 0 , we have from (3.25) that

J λ 0 ( I λ 0 A ) u n i w n i 0.

Furthermore, we have from (3.26) that

u n i J λ 0 ( I λ 0 A ) u n i 0.

Since J λ 0 (I λ 0 A) is nonexpansive, we have that w= J λ 0 (I λ 0 A)w. This means that 0Aw+Bw. Thus, we have

wF(T) ( A + B ) 1 0 W 1 0.

Then we have

A= lim i ( V γ g ) z 0 , x n i z 0 = ( V γ g ) z 0 , w z 0 0.
(3.27)

Since y n z 0 = α n (γg( x n )V z 0 )+(I α n V)S w n (I α n V) z 0 , we have

Thus, we have

Consequently, we have that

By (3.27) and Lemma 5, we obtain that x n z 0 , where

z 0 = P F ( S ) ( A + B ) 1 0 W 1 0 (IV+γg) z 0 .

Case 2: Suppose that there exists a subsequence { Γ n i }{ Γ n } such that Γ n i < Γ n i + 1 for all iN. In this case, we define τ:NN by

τ(n)=max{kn: Γ k < Γ k + 1 }.

Then we have from Lemma 6 that Γ τ ( n ) < Γ τ ( n ) + 1 . Thus, we have from (3.10) that for all nN,

(3.28)

Using lim n α n =0 and 0< lim inf n β n lim sup n β n <1, we have from (3.28) and Lemma 6 that

lim n S w τ ( n ) x τ ( n ) =0.
(3.29)

As in the proof of Case 1, we also have that

lim n x τ ( n ) + 1 x τ ( n ) =0
(3.30)

and

lim n y τ ( n ) x τ ( n ) =0.
(3.31)

Furthermore, we have that lim n u τ ( n ) x τ ( n ) =0, lim n A u τ ( n ) A z 0 =0, and . From these we have that lim n S w τ ( n ) w τ ( n ) =0. As in the proof of Case 1, we can show that

lim sup n ( V γ g ) z 0 , x τ ( n ) z 0 0.

We also have that

and hence

From Γ τ ( n ) < Γ τ ( n ) + 1 , we have that

Since (1 β τ ( n ) ) α τ ( n ) >0, we have that

Thus, we have that

and hence x τ ( n ) z 0 0 as n. Since x τ ( n ) x τ ( n ) + 1 0, we have x τ ( n ) + 1 z 0 0 as n. Using Lemma 6 again, we obtain that

x n z 0 x τ ( n ) + 1 z 0 0

as n. This completes the proof. □

4 Applications

In this section, using Theorem 10, we can obtain well-known and new strong convergence theorems in a Hilbert space. Let H be a Hilbert space, and let f be a proper lower semicontinuous convex function of H into (,]. Then the subdifferential ∂f of f is defined as follows:

f(x)= { z H : f ( x ) + z , y x f ( y ) , y H }

for all xH. From Rockafellar [22], we know that ∂f is a maximal monotone operator. Let C be a nonempty closed convex subset of H, and let i C be the indicator function of C, i.e.,

i C (x)={ 0 , x C , , x C .

Then, i C is a proper lower semicontinuous convex function on H. So, we can define the resolvent J λ of i C for λ>0, i.e.,

J λ x= ( I + λ i C ) 1 x

for all xH. We know that J λ x= P C x for all xH and λ>0; see [11].

Theorem 11 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let S be a generalized hybrid mapping of C into C. Suppose F(S). Let u, x 1 C, and let { x n }C be a sequence generated by

x n + 1 = β n x n +(1 β n ) { α n u + ( 1 α n ) S x n }

for all nN, where { β n }(0,1) and { α n }(0,1) satisfy

lim n α n =0, n = 1 α n =

and

0< lim inf n β n lim sup n β n <1.

Then the sequence { x n } converges strongly to z 0 F(S), where z 0 = P F ( S ) u.

Proof Put A=0, B=W= i C and λ n = r n =1 for all nN in Theorem 10. Then we have J λ n = T r n = P C for all nN. Furthermore, put g(x)=u and V(x)=x for all xH. Then we can take γ ¯ =L=1. Thus, we can take μ=1. On the other hand, since g(x)g(y)=0 1 3 xy for all x,yH, we can take k= 1 3 . So, we can take γ=1. Then for u, x 1 C, we get that

x n + 1 = β n x n +(1 β n ) { α n u + ( I α n ) S x n }

for all nN. So, we have { x n }C. We also have

z 0 = P F ( S ) C (IV+γg) z 0 = P F ( S ) ( z 0 z 0 +1u)= P F ( S ) u.

Thus, we obtain the desired result by Theorem 10. □

Theorem 11 solves the problem posed by Kurokawa and Takahashi [9]. The following result is a strong convergence theorem of Halpern’s type [6] for finding a common solution of a monotone inclusion problem for the sum of two monotone mappings, of a fixed point problem for nonexpansive mappings and of an equilibrium problem for bifunctions in a Hilbert space.

Theorem 12 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let α>0, and let A be an α-inverse strongly-monotone mapping of C into H. Let B and W be maximal monotone operators on H such that the domains of B and W are included in C. Let J λ = ( I + λ B ) 1 and T r = ( I + r W ) 1 be resolvents of B and W for λ>0 and r>0, respectively. Let S be a nonexpansive mapping of C into H. Let 0<k<1, and let g be a k-contraction of H into itself. Let V be a γ ¯ -strongly monotone and L-Lipschitzian continuous operator with γ ¯ >0 and L>0. Take μ,γR as follows:

0<μ< 2 γ ¯ L 2 ,0<γ< γ ¯ L 2 μ 2 k .

Suppose F(S) ( A + B ) 1 0 W 1 0. Let x 1 =xH, and let { x n }H be a sequence generated by

x n + 1 = β n x n +(1 β n ) { α n γ g ( x n ) + ( I α n V ) S J λ n ( I λ n A ) T r n x n }

for all nN, where { α n }(0,1), { β n }(0,1), { λ n }(0,) and { r n }(0,) satisfy

lim n α n = 0 , n = 1 α n = , 0 < lim inf n β n lim sup n β n < 1 , lim inf n r n > 0 and 0 < a λ n b < 2 α .

Then the sequence { x n } converges strongly to z 0 F(S) ( A + B ) 1 0 W 1 0, where z 0 = P F ( S ) ( A + B ) 1 0 W 1 0 (IV+γg) z 0 .

Proof We know that a nonexpansive mapping T of C into H is a (1,0)-generalized hybrid mapping. So, we obtain the desired result by Theorem 10. □

Let f:C×CR be a bifunction. The equilibrium problem (with respect to C) is to find x ˆ C such that

f( x ˆ ,y)0,yC.
(4.1)

The set of such solutions x ˆ is denoted by EP(f), i.e.,

EP(f)= { x ˆ C : f ( x ˆ , y ) 0 , y C } .

For solving the equilibrium problem, let us assume that the bifunction f:C×CR satisfies the following conditions:

(A1) f(x,x)=0 for all xC;

(A2) f is monotone, i.e., f(x,y)+f(y,x)0 for all x,yC;

(A3) for all x,y,zC,

lim sup t 0 f ( t z + ( 1 t ) x , y ) f(x,y);

(A4) for all xC, f(x,) is convex and lower semicontinuous.

The following lemmas were given in Combettes and Hirstoaga [23] and Takahashi, Takahashi and Toyoda [16]; see also [24, 25].

Lemma 13 ([23])

Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Assume that f:C×CR satisfies (A1)-(A4). For r>0 and xH, define a mapping T r :HC as follows:

T r x= { z C : f ( z , y ) + 1 r y z , z x 0 , y C }

for all xH. Then the following hold:

  1. (1)

    T r is single-valued;

  2. (2)

    T r is a firmly nonexpansive mapping, i.e., for all x,yH,

    T r x T r y 2 T r x T r y,xy;
  3. (3)

    F( T r )=EP(f);

  4. (4)

    EP(f) is closed and convex.

We call such T r the resolvent of f for r>0.

Lemma 14 ([16])

Let H be a Hilbert space, and let C be a nonempty closed convex subset of H. Let f:C×CR satisfy (A1)-(A4). Let A f be a set-valued mapping of H into itself defined by

A f x={ { z H : f ( x , y ) y x , z , y C } , x C , , x C .

Then EP(f)= A f 1 0 and A f is a maximal monotone operator with D( A f )C. Furthermore, for any xH and r>0, the resolvent T r of f coincides with the resolvent of A f , i.e.,

T r x= ( I + r A f ) 1 x.

Using Lemmas 13, 14 and Theorem 10, we also obtain the following result for generalized hybrid mappings of C into H with equilibrium problem in a Hilbert space; see also [2628].

Theorem 15 Let H be a real Hilbert space, and let C be a nonempty closed convex subset of H. Let S be a generalised hybrid mapping of C into H. Let f be a bifunction of C×C into satisfying (A1)-(A4). Let 0<k<1, and let g be a k-contraction of H into itself. Let V be a γ ¯ -strongly monotone and L-Lipschitzian continuous operator of H into itself with γ ¯ >0 and L>0. Take μ,γR as follows:

0<μ< 2 γ ¯ L 2 ,0<γ< γ ¯ L 2 μ 2 k .

Suppose that F(S)EP(f). Let x 1 =xH, and let { x n }H be a sequence generated by

f ( u n , y ) + 1 r n y u n , u n x n 0 , y C , x n + 1 = β n x n + ( 1 β n ) { α n γ g ( x n ) + ( I α n V ) S u n }

for all nN, where { β n }(0,1), { α n }(0,1) and { r n }(0,) satisfy

lim n α n = 0 , n = 1 α n = , lim inf n r n > 0 , and 0 < lim inf n β n lim sup n β n < 1 .

Then the sequence { x n } converges strongly to z 0 F(S)EP(f), where z 0 = P F ( S ) E P ( f ) (IV+γg) z 0 .

Proof Put A=0 and B= i C in Theorem 10. Furthermore, for the bifunction f:C×CR, define A f as in Lemma 14. Put W= A f in Theorem 10, and let T r n be the resolvent of A f for r n >0. Then we obtain that the domain of A f is included in C and T r n x n = u n for all nN. Thus, we obtain the desired result by Theorem 10. □

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Acknowledgements

The first author was partially supported by Grant-in-Aid for Scientific Research No. 23540188 from Japan Society for the Promotion of Science. The second and the third authors were partially supported by the grant Taiwan NSC 99-2115-M-110-007-MY3 and the grant Taiwan NSC 99-2115-M-037-002-MY3, respectively.

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Correspondence to Jen-Chih Yao.

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Takahashi, W., Wong, N. & Yao, J. Iterative common solutions for monotone inclusion problems, fixed point problems and equilibrium problems. Fixed Point Theory Appl 2012, 181 (2012). https://doi.org/10.1186/1687-1812-2012-181

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Keywords

  • maximal monotone operator
  • resolvent
  • inverse-strongly monotone mapping
  • generalized hybrid mapping
  • fixed point
  • iteration procedure
  • equilibrium problem