Two-Steps Hybrid Iterative Schemes With Errors For Generalized Equilibrium Problems And

Two-Steps Hybrid Iterative Schemes With Errors For Generalized Equilibrium Problems And

Common Fixed Point Problems

Seung-Hyun Kim

, Mee-Kwang Kang

, Byung-Soo Lee

Received 5 April 2015

Abstract

In this paper, we consider a two-steps hybrid iterative scheme with errors for generalized equilibrium problems and common …xed point problems, and prove the weak limits of sequences fxng and fvng obtained under the given scheme for N …nite asymptotically ki-strictly pseudo-contractive mappingsfTig^Ni=1and a …rmly nonexpansive mappingSr are the same and hence the point is a common

…xed point offTig^Ni=1andSr.

1 Introduction

The applications of equilibrium problems and …xed point theory to many branches have been well-known for a long time in nonlinear analysis including optimization theory, economics, etc. (see [2, 3, 5, 7, 9, 12]).

Recently there have been many researches on approximating convergence of …xed points under iteration schemes with errors concerning equilibrium problems and vari- ational inequalities, etc. (see [2, 3, 4, 5, 9]).

On the other hand, Qin et al. [11] and Kumam et al. [6] considered equilibrium problems with …xed point problems under one-step hybrid iterative schemes and two- step hybrid iterative schemes, respectively, in Hilbert spaces.

Inspired by those results, we consider the following two-step hybrid iterative scheme with errors for generalized equilibrium problems and common …xed point problems, and obtain a result that the weak limits of sequences fx_ng andfv_ng obtained under the given scheme for N …nite asymptoticallyki-strictly pseudo-contractive mappings fTig^Ni=1 and a …rmly nonexpansive mappingSrare the same and that the same point is a common …xed point of fTig^Ni=1 andSr.

ALGORITHM 1.1. Let C be a closed convex subset of a Hilbert space H; T_i; : C ! C (i= 1;2; ; N) be mappings and :C C ! Rbe a bifunction. For any

Mathematics Sub ject Classi…cations: 47H09, 47H10.

yDepartment of Mathematics, Kyungsung University, Busan 608-736, Korea

zCorresponding author. Department of Mathematics, Dongeui University, Busan 614-714, Korea

xDepartment of Mathematics, Kyungsung University, Busan 608-736, Korea

276

x₀2C;letfx_ngandfv_ngbe sequences generated by ( (vn 1; y) +h vn 1; y vn 1i+_r¹

n 1hy vn 1; vn 1 xn 1i 0;

x_n=a_{n 1}v_{n 1}+b_{n 1}T_i(n)^h(n)v_{n 1}+c_{n 1}u_{n 1}, (1) for all y 2C andn2N, wherefang, fbng andfcng are sequences in[0;1)such that a_n +b_n +c_n = 1, a_n k+", b_n " for some " 2 (0;1), P¹

n=1

c_n < 1, fu_ng is a bounded sequence in C, frng is a sequence in (0;1) such that lim

n!1infrn > 0 and i(n) n(modN), h(n) =dNⁿewith a ceiling function d e.

REMARK 1.1. (a) Putting 0in (1), we obtain an algorithm ( (v_{n 1}; y) +_r¹

n 1hy v_{n 1}; v_{n 1} x_{n 1}i 0 for ally2C;

xn=an 1vn 1+bn 1T_i(n)^h(n)vn 1+cn 1un 1 for eachn2N: (2) (b) Puttingc_n= 0for alln2Nin (2), we obtain the algorithm considered in [6]

( (vn 1; y) +_r¹

n 1hy vn 1; vn 1 xn 1i 0 for ally 2C;

xn =an 1vn 1+ (1 an 1)T_i(n)^h(n)vn 1 or eachn2N: (3) (c) Putting 0 andv_n=x_n (n2N)in (3), we obtain the algorithm considered in [11]

xn=an 1xn 1+ (1 an 1)T_i(n)^h(n)xn 1for eachn2N. (4)

2 Preliminaries

First of all, we recall some de…nitions and results needed in the main results.

DEFINITION 2.1. Let :C C!Rbe a function and :C!Cbe a nonlinear mapping. (a) is said to be monotone if (x; y) + (y; x) 0 for allx; y2C. (b) is said to be monotone ifh x y; x yi 0 for allx; y2C.

DEFINITION 2.2. A mapping T : C ! C is asymptotically k-strictly pseudo- contractive if there existk2[0;1)and a sequencefkng [1;1)with lim

n!1kn= 1such that

kTⁿx Tⁿyk² k²_nkx yk²+kk(I Tⁿ)x (I Tⁿ)yk² for allx; y2C andn2N:

LEMMA 2.1 ([7, 10]). LetH be a real Hilbert space. Then we have the following identities:

(i) kx yk²=kxk² kyk² 2hx y; yifor allx; y2H.

(ii) For allx; y2H anda; b; c2[0;1]with a+b+c= 1,

kax+by+czk²=akxk²+bkyk²+ckzk² abkx yk² bcky zk² cakz xk²: (iii) Iffxngis a sequence inH weakly converging toz, then

nlim!1supkxn yk²= lim

n!1supkxn zk²+kz yk² for ally2H:

LEMMA 2.2 ([10]). Letfa_ng, fc_ng andf ng be nonnegative real sequences satis- fying the conditiona_n+1 (1 + _n)a_n+c_n for eachn2N. If

X1 n=1

n <1 and X1 n=1

c_n<1; then lim

n!1a_n exists.

3 Main Results

We assume that the mapping :C C!Rsatis…es the following conditions:

(i) (x; x) = 0for allx2C;

(ii) is monotone;

(iii) is upper hemi-continuous in the …rst variable;

(iv) is convex and lower semi-continuous in the second variable.

We have the following theorems.

THEOREM 3.1. Let C be a closed convex subset of a Hilbert space H. Let :C!Cbe a monotone nonlinear mapping. Forr >0andx2H, de…ne a mapping Sr:H!2^C by

S_rx= z2C: (z; y) +h z; y zi+1

rhy z; z xi 0 for ally2C : Then the following statements (i)–(iii) hold.

(i) S_rxis a singleton for eachx2H. (ii) S_r is …rmly nonexpansive, i.e.,

kSrx Sryk² hSrx Sry; x yi forx; y2H:

(iii) The set F(S_r)of all …xed points of S_r is a closed and convex subset of C as a solution set of the following equilibrium problem considered in [9]: …ndingx2C such that (x; y) +h x; y xi 0 for ally2C.

PROOF. (i) We put (x; y) = (x; y)+h x; y xifor allx; y2C:By the conditions of and [1, Theorem 1], we see thatSrx6=; for anyx2C. Next, we show that Srx is a singleton forx2C. Suppose thatz1; z22Srx. Then

8<

:

(z1; y) +h z1; y z1i+¹_rhy z1; z1 xi 0 fory2C, (z2; y) +h z2; y z2i+¹_rhy z2; z2 xi 0 fory2C.

(5)

Putting y=z₂ in the …rst inequality andy=z₁ in the second inequality (5), respec- tively and adding them, we have

(z₁; z₂) + (z₂; z₁) +h z₁ z₂; z₂ z₁i 1

rkz₁ z₂k²:

Since (z1; z2) + (z2; z1) 0 and h z1 z2; z2 z1i 0, we have z1=z2. So we prove statement (i).

(ii) Letz=Srxandz⁰=Srx⁰. Then (

(z; z⁰) +h z; z⁰ zi+¹_rhz⁰ z; z xi 0;

(z⁰; z) +h z⁰; z z⁰i+¹_rhz z⁰; z⁰ x⁰i 0:

Adding two inequalities and applying the monotonicity of and , we have hS_rx S_rx⁰; x x⁰i=hz z⁰; x x⁰i kz z⁰k²=kS_rx S_rx⁰k²: Hence,Sr is a …rmly nonexpansive mapping. So we prove statement (ii).

(iii) Ifx2F(Sr), then

(x; y) +h x; y xi= (x; y) +h x; y xi+1

rhy x; x xi 0

for ally2C. Soxis a solution of the equilibrium problem in [9]. Next, letfx_ngbe a convergent sequence in F(Sr) with a limitx2H. Since F(Sr) C and C is closed, we have x2C. Also,Sr is continuous. Then we have

x= lim

n!1xn= lim

n!1Srxn =Srx:

It means that x2F(S_r), that is,F(S_r)is closed.

To show that F(S_r) is convex, we let z = x+ (1 )y for x; y 2 F(S_r) and 2[0;1]. By Lemma 2.1(ii) and the nonexpansiveness ofS_r, we have

kz Srzk² = k (x Srz) + (1 )(y Srz)k²

= kx S_rzk²+ (1 )ky S_rzk² (1 )kx yk² kx zk²+ (1 )ky zk² (1 )kx yk²

= kx ( x+ (1 )y)k²+ (1 )ky ( x+ (1 )y)k² (1 )kx yk²

= (1 )²kx yk²+ (1 ) ²kx yk² (1 )kx yk²= 0:

Hence, S_rz =z and z 2 F(S_r). Therefore F(S_r) is convex. So we prove statement (iii).

The proof of Theorem 3.1 is complete.

REMARK 3.1. By putting 0 in Theorem 3.1, we obtain [4, Lemma 2.12].

Next, we consider our main result.

THEOREM 3.2. Assume that the mappingsT_i : C !C fori = 1; ; N satisfy the following conditions:

(i) Cis a closed convex subset of a Hilbert space H;

(ii) T_i is asymptoticallyk_i-strictly pseudo-contractive fork_i2[0;1),i= 1;2; ; N and for eachi2 f1;2; ; Ng,fkn;igis a sequence in[1;1)such that P¹

n=1

(k_n;i² 1)<1;

(iii) k= maxfk_i: 1 i Ngandk⁰_n= maxfk_n;i: 1 i Ng for eachn2N. Let :C!C be a monotone nonlinear mapping with

F :=

\N i=1

F(Ti)

!\

F(Sr)6=;:

For any x0 2 C, letfxng and fvng be sequences generated by Algorithm 1.1. Then fxngandfvngconverge weakly to the unique same element ofF.

PROOF. Let p2 F. By Algorithm 1.1 and Theorem 3.1(i), we see that vn 1 = Srn 1xn 1and

kvn 1 pk=kSrn 1xn 1 Srn 1pk kxn 1 pk

for eachn2N. By Algorithm 1.1 and Lemma 2.1(ii), we have

kx_n pk² = a_{n 1}(v_{n 1} p) +b_{n 1}(T_i(n)^h(n)v_{n 1} p) +c_{n 1}(u_{n 1} p) ² a_{n 1}kv_{n 1} pk²+b_{n 1} T_i(n)^h(n)v_{n 1} T_i(n)^h(n)p ²+c_{n 1}ku_{n 1} pk²

a_{n 1}b_{n 1} T_i(n)^h(n)v_{n 1} v_{n 1} ²

an 1kvn 1 pk²+bn 1 (k_h(n)⁰ )²kvn 1 pk²+k (I T_i(n)^h(n))vn 1

(I T_i(n)^h(n))p ² +c_{n 1}ku_{n 1} pk² a_{n 1}b_{n 1} T_i(n)^h(n)v_{n 1} v_{n 1} ² (k⁰_h(n))²kvn 1 pk² bn 1(an 1 k) T_i(n)^h(n)vn 1 vn 1

2

+c_{n 1}ku_{n 1} pk² (6) h

1 + ((k_h(n)⁰ )² 1) i

kxn 1 pk²+cn 1kun 1 pk²: (7) Since P¹

n=1

(k_n;i² 1)<1, and by Lemma 2.2, we see that lim

n!1kx_n pk exists. On the other hand, since an k+"andbn "forn2Nand some "2(0;1), we have

(k⁰_h(n))²kxn 1 pk² kxn pk²+cn 1kun 1 pk² bn 1(an 1 k)kT_i(n)^h(n)vn 1 vn 1k²

"²kT_i(n)^h(n)v_{n 1} v_{n 1}k²: Since lim

n!1k⁰_h(n) = 1 and lim

n!1cn = 0, taking the limits as n ! 1 in the above inequality, we have

nlim!1 T_i(n)^h(n)v_{n 1} v_{n 1} ²= 0: (8) Observe that

kx_n v_{n 1}k = a_{n 1}v_{n 1}+b_{n 1}T_i(n)^h(n)v_{n 1}+c_{n 1}u_{n 1} v_{n 1}

= (1 a_{n 1}) v_{n 1} T_i(n)^h(n)v_{n 1} +c_{n 1} u_{n 1} T_i(n)^h(n)v_{n 1}

(1 a_{n 1}) v_{n 1} T_i(n)^h(n)v_{n 1} +c_{n 1} u_{n 1} T_i(n)^h(n)v_{n 1} :

By (8), we see that

nlim!1kx_n v_{n 1}k= 0: (9)

By the …rm nonexpansiveness of Srn 1 and Lemma 2.1(i), we have

kvn 1 pk²=kSrn 1xn 1 Srn 1pk² hSrn 1xn 1 Srn 1p; xn 1 pi

=hv_{n 1} p; x_{n 1} pi= h (x_{n 1} v_{n 1}) (x_{n 1} p); x_{n 1} pi

= 1

2 kxn 1 vn 1k² kxn 1 pk² kvn 1 pk² ; and hence

kv_{n 1} pk² kx_{n 1} pk² kx_{n 1} v_{n 1}k²: Applying this inequality to (6), we have

kx_n pk² k_h(n)^{0 2}(kx_{n 1} pk² kx_{n 1} v_{n 1}k²) +c_{n 1}ku_{n 1} pk²: Since lim

n!1kxn pkexists and lim

n!1k_h(n)⁰ = 1, we see that

nlim!1kxn 1 vn 1k= 0: (10)

Applying (9) and (10) to the triangle inequality, we have

kvn vn 1k kvn xnk+kxn vn 1k !0 asn! 1; which implies that

nlim!1kvn vn+jk= 0 forj2 f1; ; Ng: (11) Similarly, applying (10) and (11) to the triangle inequality, we obtain

kx_n x_{n 1}k kx_n v_nk+kv_n v_{n 1}k+kv_{n 1} x_{n 1}k !0 asn! 1; which implies that lim

n!1kxn xn+jk= 0 forj 2 f1; ; Ng. On the other hand, kvn 1 Tnvn 1k vn 1 T_i(n)^h(n)vn 1 + T_i(n)T_i(n)^{h(n) 1}vn 1 T_i(n)vn 1

vn 1 T_i(n)^h(n)vn 1 +L T_i(n)^{h(n) 1}vn 1 T_{i(n N)}^{h(n) 1}vn N

+ T_{i(n N)}^{h(n) 1}vn N vn N 1 +kvn N 1 vn 1k ; (12) where

L= sup

(k+p

1 + (k²_n 1)(1 k)

1 k :n2N

) :

Since, for eachn > N,n= (h(n) 1)N+i(n),i(n N) =i(n)andh(n N) =h(n) 1, T_i(n)^{h(n) 1}vn 1 T_{i(n N)}^{h(n) 1}vn N = T_i(n)^{h(n) 1}vn 1 T_i(n)^{h(n) 1}vn N

Lkvn 1 vn Nk (13)

and

T_{i(n N}^{h(n) 1}₎vn N vn N 1

T_{i(n N}^{h(n N)}₎vn N T_{i(n N)}^{h(n N)}vn N 1 + T_{i(n N)}^{h(n N)}vn N 1 vn N 1

L kvn N vn N 1k+ T_{i(n N)}^{h(n N)}vn N 1 vn N 1 : (14) So by (12)–(14), we see that

kv_{n 1} T_nv_{n 1}k

v_{n 1} T_i(n)^h(n)v_{n 1} +L n

T_i(n)^{h(n) 1}v_{n 1} T_{i(n N)}^{h(n) 1}v_{n N}

+ T_{i(n N}^{h(n) 1}₎vn N vn N 1 +kvn N 1 vn 1ko

vn 1 T_i(n)^h(n)vn 1 +L fLkvn 1 vn Nk+L kvn N vn N 1k + T_{i(n N}^{h(n N}₎⁾vn N 1 vn N 1 +kvn N 1 vn 1k

o :

By (8) and (11), we have that lim

n!1kv_{n 1} T_nv_{n 1}k= 0. Since

kvn Tnvnk kvn vn 1k+kvn 1 Tnvn 1k+kTnvn 1 Tnvnk (1 +L) kv_n v_{n 1}k+kv_{n 1} T_nv_{n 1}k !0asn! 1; for anyj = 1; ; N, we have

kv_n T_n+jv_nk kv_n v_n+jk+kv_n+j T_n+jv_n+jk+kT_n+jv_n+j T_n+jv_nk (1 +L) kvn vn+jk+kvn+j Tn+jvn+jk !0asn! 1; which gives that lim

n!1kv_n T_lv_nk = 0 for l 2 f1; ; Ng. Moreover, for each l 2 f1; ; Ng, we have

kxn Tlxnk kxn vnk+kvn Tlvnk+kTlvn Tlxnk

(1 +L) kx_n v_nk+kv_n T_lv_nk !0 asn! 1: Put

W(xn) =fx2H :xni * xfor some subsequencefxnigoffxngg: ThenW(xn)6=;by the fact thatfxngis bounded inH. Next, we claim thatW(xn) F. Letw2W(xn)be an arbitrary element. Then there exists a subsequencefxnigof fxngconverging weakly to w. Since lim

n!1kxn vnk= 0, we can obtain thatvni* w asi! 1. By the fact that lim

n!1kvn Tlvnk= 0,Tlvni !wforl2 f1; ; Ng. Now, we show that wis a …xed point ofS_r. Sincev_n =T_rnv_n for each n2N, we have

(vn; y) +h vn; y vni+ 1

rnhy vn; vn xni 0 for ally2Candn2N: By the monotonicity of , we have

hy vni;v_ni x_ni rni

i (y; vni) +h vni; vni yifori2N. Since ^vni_r ^xni

ni !0 andvni* was i! 1, and by the condition (iv) of , we have (y; w) +h w; w yi 0 fory2C:

By the conditions (i) and (iv) of , we see that 0 = (y_t; y_t) t (y_t; y) + (1 t) (y_t; w)

t (yt; y) + (1 t)h w; yt wi=t (yt; y) + (1 t)th w; y wi (yt; y) + (1 t)h w; y wi;

where t2(0;1],y2C, andyt=ty+ (1 t)w. By the condition (iii) of , 0 (w; y) +h w; y wi for ally2C;

which shows that w2F(Sr). Moreover,w2 TN l=1

F(Tl). In fact, ifw62F(Tl)for some l2 f1; ; Ng, then from the Opial’s condition and the fact that lim

n!1kxn Tlxnk= 0,

ilim!1infkxni wk< lim

i!1infkxni Tlwk lim

i!1inffkxni Tlxnik+kTlxni Tlwkg

ilim!1infL kx_ni wk;

which derives a contradiction. Consequently, we have w2F =

\N l=1

F(Tl)

!\ F(Sr):

Finally, we show thatfxng andfvngconverge weakly to the unique same element ofF. Indeed, it is su¢ cient to show that W(x_n)is a singleton. We take anyw₁; w₂2 W(x_n) and let fx_nig and fx_njg be subsequences of fx_ng such that x_ni * w₁ and x_nj * w₂. Since lim

n!1kx_n pk exists for each p 2 F and w₁; w₂ 2 F, by Lemma 2.1(iii), we obtain

lim sup

n!1 kxn w1k²= lim sup

j!1 kxnj w1k²= lim sup

j!1 kxnj w2k²+kw2 w1k²

= lim sup

i!1 kxni w2k²+kw2 w1k²

= lim sup

i!1 kx_ni w₁k²+ 2kw₂ w₁k²

= lim sup

n!1 kx_n w₁k²+ 2kw₂ w₁k²:

Hence w1=w2, which shows thatW(xn)is a singleton. The proof of Theorem 3.2 is complete.

We have the following theorems in [6, 11] as corollaries of Theorem 3.2.

THEOREM 3.3 ([6]). Assume that the conditions (i)–(iii) in Theorem 3.2 hold and that satis…es

F :=

\N i=1

F(Ti)

!\

S( )6=;:

For any x₀ 2 C, let fx_ng and fv_ng be sequences generated by (3), where n = (h 1)N +i(n 1), i = i(n) 2 f1;2; ; Ng, h = h(n) 1 is a positive integer and h(n)! 1 as n ! 1. Let fa_ng and fr_ng be sequences satisfying fa_ng [ ; ] for some ; 2(k;1), fr_ng (0;1)and lim

n!1infr_n >0. Then fx_ng andfv_ng converge weakly to an element ofF.

THEOREM 3.4 ([11]). Assume that the conditions (i)–(iii) in Theorem 3.2 hold and

F :=

\N i=1

F(T_i)

!

6

=;:

For any x₀2C, letfx_ngbe a sequence generated by (4), where fa_ng is a sequence in (0;1)such thatk+" an 1 "for some"2(0;1),n= (h 1)N+i(n 1);where i=i(n)2 f1;2; ; Ng, h=h(n) 1 is a positive integer andh(n)! 1 asn! 1. Thenfxngconverges weakly to an element ofF.

REMARK 3.2. Our result is a weak convergence under Algorithm 1.1 for a …nite family of asymptoticallyk_i-strictly pseudo-contractive mappings in Hilbert spaces. The convergences, mappings and spaces need to be more weakened, for examples, strongly convergences, asymptotically nonexpansive mappings andCAT(0)-spaces, respectively.

Till now, many kinds of strong convergence results are well-known, but the weak con- vergence results are few. So, we suggest the following open problem.

Open problem. Dofx_ngandfv_ng weakly converge for a …nite family of asymptot- ically nonexpansive mappings with Algorithm 1.1 under suitable conditions?

Acknowledgment. This work was supported by Dong-eui University Foundation Grant (2013).

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