CUBO A Mathematical Journal
Vol.15, No¯ 03, (09–18). October 2013

Approximating a solution of an equilibrium problem by
Viscosity iteration involving a nonexpansive semigroup

Binayak S. Choudhury & Subhajit Kundu

Bengal Engineering and Science University, Shibpur

Department of Mathematics,

P.O.: B. Garden, Shibpur, Howrah - 711103, West Bengal, India.

binayak12@yahoo.co.in, subhajit.math@gmail.com

ABSTRACT

In this paper we have defined a new iteration in order to solve an equilibrium problem

in Hilbert spaces. The iteration we have introduced is a viscosity type iteration and

involves a semigroup of nonexpansive operators. We have established that depending on

some control conditions, our iteration strongly converges to a solution of the equilibrium

problem.

RESUMEN

En este art́ıculo hemos definido una iteración nueva para resolver un problema de equi-

librio en espacios de Hilbert. La iteración que introducimos es de tipo viscoso e involucra

un semigrupo de operadores no expansivos. Hemos establecido que dependiendo de las

condiciones de control, nuestra iteración converge fuertemente a una solución de un

problema de equilibrio.

Keywords and Phrases: Equilibrium problem, Nonexpansive semigroup, Viscosity iteration,

Fixed point, Weak convergence, Hilbert space.

2010 AMS Mathematics Subject Classification: 46C05, 47H10, 91B50.



10 Binayak S. Choudhury & Subhajit Kundu CUBO
15, 3 (2013)

1 Introduction and Mathematical Preliminaries

The equilibrium problem we consider in this paper is formulated in the framework of real Hilbert

spaces. This problem is a generalization of several problems in physics, optimization and eco-

nomics. References [3, 9] give a good account of this feature. There are several iterative methods

for obtaining solutions of this equilibrium problem in Hilbert spaces and also in the more general

settings of Banach spaces [19, 10, 22]. A particular category of these iterations is viscosity itera-

tion which was first developed by Moudafi [15] to obtain fixed points of nonexpansive mappings.

Viscosity iterations have been used for solving equilibrium problems in works noted in [20, 16].

Semigroup of nonexpansive operators have been considered in the context of constructing fixed

point iteration in Banach and Hilbert spaces [8, 21, 1, 6, 5, 13, 7, 11, 12, 17]. The purpose of this

paper is to use nonexpansive semigroups in a viscosity iteration scheme in order to construct a two

step iteration for approximating a solution of an equilibrium problem in real Hilbert spaces. Pre-

cisely, we have shown that under suitable choices of the control conditions, our iteration strongly

converges to solution of the equilibrium problem.

Let H be a Hilbert space and C be a nonempty closed convex subset of H.

A mapping T : C → C is said to be a nonexpansive mapping if for all x, y ∈ C

‖Tx − Ty‖ ≤ ‖x − y‖. (1.1)

A mapping f : C→ C is said to be a θ contraction if for each x, y ∈ C,

‖fx − fy‖ ≤ θ‖x − y‖ when 0 < θ < 1. (1.2)

For any x ∈ H, the metric projection PC from H into C is defined as

PCx = {z ∈ C : ‖z − x‖ = inf
y∈C

‖y − x‖}. (1.3)

Obviously, ‖x − PCx‖ ≤ ‖x − y‖. It is well known that PC is a firmly nonexpansive mapping from

H onto C, that is,

‖PCx − PCy‖
2 ≤ 〈PCx − PCy, x − y〉 for all x, y ∈ H. PC is also nonexpansive mapping from H

onto C.

The set of fixed point of an operator T from H to H is denoted by Fix(T), that is, Fix(T)={x ∈ H :

Tx = x}.

A family S = (T(s))
s≥ 0 is a nonexpansive semigroup on H if it satisfies the following conditions:

(A1) T(0)x = x for all x ∈ H,

(A2) T(s + t) = T(s)T(t) for all s, t ≥ 0,

(A3) ‖T(s)x − T(s)y‖ ≤ ‖x − y‖ for all x, y ∈ H and s ≥ 0,

(A4) for all x ∈ H, s → T(s)x is continuous.

A sequence {xn} of elements of a Banach space X is said to converge weakly to an element x ∈ X

if f(xk) → f(x) as k → ∞ for all f ∈ X
′ where f is a continuous linear functional from X to R or C

where R is the set of real numbers and C is the set of complex number, and X′ is the dual of X.

A sequence {xn} is said to have a weak limit point l if there exists a subsequence {xnk} of {xn}

which converges weakly to l.



CUBO
15, 3 (2013)

Approximating a solution of an equilibrium problem . . . 11

We denote wω(xn) as the set of all weak limit point of {xn} and ws(xn) as the set of all strong

limit point of {xn} .

We denote the set of fixed point of T(s) by Fix(T(s)). The set of all common fixed points of S is

denoted by Fix(S). So Fix(S)= ∩s≥0Fix(T(s)) .

Baillon proved the following nonlinear ergodic theorem:

Theorem 1.1. [1] If T is a nonexpansive mapping from C into itself such that Fix(T) 6= φ and

x ∈ C, then 1
n

n−1∑

k=0

Tkx converges weakly to a fixed point of T.

Later Baillon and Brezis proved the following theorem for semigroup of nonexpansive operator:

Theorem 1.2. [2] If S = (T(s))
s≥ 0 is a nonexpansive semigroup on C, then {

1
t

∫t
0
T(s)xtds}t>0,

t ∈ (0, 1), s ∈ R+, where R+ is the set of positive real numbers, converges weakly to a common

fixed point of S.

Let F : C×C→ R be a bifunction where R is the set of real numbers. The equilibrium problem is

to find some x ∈ C such that

F(x, y) ≥ 0, for all y ∈ C . (1.4)

The set of solutions of (1.4) is denoted by EP(F), that is, EP(F) = {x ∈ C : F(x, y) ≥ 0 for all y ∈ C}.

In the equilibrium problem for the bifunction F from C×C→R , we assume that F satisfies following

conditions:

(C1) F(x, x) = 0 for all x ∈ C,

(C2) F is monotone, that is, F(x, y) + F(y, x) ≤ 0,

(C3) for each x, y, z ∈ C,

lim
t→0+

F(tz + (1 − t)x, y) ≤ F(x, y),

(C4) for each x ∈ C, y → F(x, y) is convex and lower semicontinuous.

Lemma 1.1. [9] Let C be a nonempty closed convex subset of H and let F be a bifunction from

C×C into R satisfying conditions (C1)- (C4). Then for any r > 0 and x ∈ H there exists z ∈ C

such that

F(z, y) + 1
r
〈y − z, z − x〉 ≥ 0 for all y ∈ C.

Further, if Trx = {z ∈ C : F(z, y) +
1
r
〈y − z, z − x〉 ≥ 0 for all y ∈ C} then the following hold:

(1) Tr is single valued,

(2) Tr is firmly nonexpansive, that is, for any x, y ∈ H

‖Trx − Try‖
2 ≤ 〈Trx − Try, x − y〉,

(3) Fix(Tr) = EP(F),

(4) EP(F) is closed and convex.

Lemma 1.2. [4] Let C be a nonempty closed convex subset of a real Hilbert space H. Given

z ∈ H and x ∈ C, the inequality 〈x − z, y − x〉 ≥ 0, for all y ∈ C holds if and only if x = PCz, where

PC denotes the metric projection from H onto C.

Lemma 1.3. [14] Let {an},{bn} and {cn} be three nonnegative real sequences satisfying

an+1 ≤ (1 − λn)an + bn + cn, n ≥ n0,

where n0 is some nonnegative integer, λn ∈ [0,1] ,

∞∑

n=1

λn=∞ , bn=o(λn) , and

∞∑

n=1

cn < ∞.



12 Binayak S. Choudhury & Subhajit Kundu CUBO
15, 3 (2013)

Then an → 0 as n → ∞.

Lemma 1.4. [18] Let C be a nonempty bounded closed convex subset of a Hilbert space H and

let (T(s))
s≥ 0 be a nonexpansive semigroup on C. Then for every h ≥ 0,

lim
t→∞

sup
x∈C

‖1
t

∫t

0

T(s)xds − T(h)
1

t

∫t

0

T(s)xds‖= 0.

Lemma 1.5. [4] Let X be a uniformly convex Banach space, C be a nonempty closed convex

subset of X and T : C → X be a nonexpansive mapping. Then, the mapping (I − T) is demiclosed

on C, that is, if {xn} is weakly convergent to x and {(I − T)xn} is strongly convergent to y, then

(I − T)x = y.

Lemma 1.6. [8] Let us suppose (C1)-(C4) hold. Let x, y ∈ H, r1, r2 > 0. Then

‖Tr2y − Tr1x‖ ≤ ‖y − x‖ + |
r2−r1

r2
|‖Tr2y − y‖.

Lemma 1.7. [23] Let {sn} be a sequence of nonnegative real numbers satisfying

sn+1 ≤ (1 − γn)sn + σn + δn, for all n ≥ 0, where {γn} is a sequence in (0,1) and {σn}, {δn} are

sequences of real numbers such that

(a) lim
n→∞

γn = 0 and

∞∑

n=0

γn = ∞,

(b) lim sup
n→∞

σn

γn
≤ 0,

(c) δn ≥ 0 and

∞∑

n=1

δn < ∞.

Then {sn} converges to zero.

The following lemma is a well known result of functional analysis.

Lemma 1.8. Let X be a reflexive Banach space. Then every bounded sequence in X has a weakly

convergent subsequence.

2 Main Result

Theorem 2.1. Let S = (T(s))
s≥ 0 be a nonexpansive semigroup on a real Hilbert space H. Let

f : H → H be a θ-contraction, with 0 < θ < 1. Let F : H × H → R be a mapping satisfying

hypothesis (C1)-(C4). Assume that Fix(S) ∩ EP(F) 6= φ. Let x0 ∈ H, {zn} ⊂ H and {xn} ⊂ H be

the sequences generated by






xn+1 = βnxn + (1 − βn)
1
sn

∫sn

0

T(s)ynds,

yn = αnf(xn) + (1 − αn)zn,

F(zn, y) +
1
rn

〈y − zn, zn − xn〉 ≥ 0, for all y ∈ H

where {αn}, {βn}, {sn} and {rn} satisfy the following conditions.



CUBO
15, 3 (2013)

Approximating a solution of an equilibrium problem . . . 13

(i) αn ∈ [0, 1], lim
n→∞

αn= 0,
∞∑

n=1

αn = ∞ and
∞∑

n=1

|αn − αn−1| < ∞,

(ii) lim
n→∞

sn = ∞ and lim
n→∞

| sn − sn−1|

sn

1

αn
= 0,

(iii) lim inf
n→∞

rn > 0,

∞∑

n=0

|rn+1 − rn| < ∞,

(iv) 0 < βn ≤ d < 1, lim
n→∞

βn = 0,

∞∑

n=0

| βn+1 − βn| < ∞.

Then {xn} converges strongly to a point p ∈ Fix(S) ∩ EP(F).

Proof. Here PFix(S)∩ EP(F) f is a mapping of H into Fix(S) ∩ EP(F) ⊂ H such that

‖PFix(S)∩ EP(F)f(x) − PFix(S)∩ EP(F)f(y)‖

≤ ‖f(x) − f(y)‖ ≤ θ‖x − y‖.

Therefore PFix(S)∩ EP(F)f is a contractive mapping and hence, by Banach’s contraction principle,

there exists a unique element p ∈ Fix(S) ∩ EP(F) such that p = PFix(S)∩ EP(F)f(p). Now, for this

p ∈ Fix(S) ∩ EP(F), n ≥ 0, we have,

‖ xn+1 − p ‖

= ‖ βnxn + (1 − βn)
1
sn

∫sn

0

T(s)ynds − p ‖

= ‖ βnxn + (1 − βn)
1
sn

∫sn

0

T(s)ynds −
1

sn

∫sn

0

T(s)p ds ‖

= ‖ βn(xn − p) + (1 − βn)
1
sn

∫sn

0

(T(s)yn − T(s)p)ds ‖

≤ βn‖ xn − p ‖ + (1 − βn)‖ yn − p ‖.

Now, ‖ yn − p ‖ = ‖ αnf(xn) + (1 − αn)zn − p ‖

≤ αn‖f(xn) − f(p)‖ + αn‖f(p) − p‖ + (1 − αn)‖ zn − p ‖.

Since by Lemma 1.1 we have zn = Trnxn, p = Trnp it follows that for all n ≥ 0, ‖ zn − p ‖ =

‖Trnxn − Trnp‖ ≤ ‖ xn − p ‖.

Therefore, for all n ≥ 0, ‖ yn − p ‖ ≤ αnθ‖ xn − p ‖ + αn‖ f(p) − p ‖ + (1 − αn)‖ xn − p ‖

= {1 − αn(1 − θ)}‖ xn − p ‖ + αn(1 − θ)
‖ f(p)−p ‖

1−θ
.

Therefore, for all n ≥ 0, ‖ yn − p ‖ ≤ max { ‖ xn − p ‖,
‖ f(p)−p ‖

1−θ
}.

Therefore, for all n ≥ 0, {yn} is bounded. So { zn } and {f(xn)} are also bounded.

Hence for all n ≥ 0, ‖ xn+1 − p ‖ ≤ max { ‖ xn − p ‖,
‖ f(p)−p ‖

1−θ
}.

Proceeding in the same way we get for all n ≥ 0, ‖ xn+1 − p ‖ ≤ max { ‖ x0 − p ‖,
‖ f(p)−p ‖

1−θ
}.

Therefore, { xn } is bounded.

Again, for all n ≥ 0, we have, xn+1 = βnxn + (1 − βn)un where un=
1
sn

∫sn

0

T(s)ynds.

Again, for all n ≥ 0, ‖un − p‖ = ‖
1
sn

∫sn

0

T(s)ynds −
1

sn

∫sn

0

T(s)p ds‖

≤ ‖yn − p‖ ≤max{ ‖ xn − p ‖,
‖ f(p)−p ‖

1−θ
}.

So {un} is also bounded.



14 Binayak S. Choudhury & Subhajit Kundu CUBO
15, 3 (2013)

Now, for all n ≥ 0, xn+1 − xn =βnxn + (1 − βn)un − βn−1xn−1 − (1 − βn−1)un−1.

Therefore, for all n ≥ 0, ‖xn+1 − xn‖

=‖(1 − βn)(un − un−1) − (βn − βn−1)un−1 + βn(xn − xn−1) + (βn − βn−1)xn−1‖

≤ (1 − βn)‖un − un−1‖ + |βn − βn−1|{ ‖ un−1 ‖ + ‖ xn−1 ‖ } + βn‖ xn − xn−1 ‖. (2.1)

Now, for all n ≥ 0, ‖ un − un−1 ‖

=‖ 1
sn

∫sn

0

T(s)ynds −
1

sn−1

∫sn−1

0

T(s)yn−1ds ‖

=‖ 1
sn

∫sn

0

(T(s)yn − T(s)yn−1)ds + (
1

sn
−

1

sn−1
)

∫sn−1

0

T(s)yn−1ds +
1

sn

∫sn

sn−1

T(s)yn−1ds ‖.

If p ∈ Fix(S) where S is the nonexpansive semigroup, then for all n ≥ 0, we have

‖ un − un−1 ‖

= ‖ 1
sn

∫sn

0

(T(s)yn − T(s)yn−1)ds + (
1

sn
−

1

sn−1
)

∫sn−1

0

(T(s)yn−1 − T(s)p)ds

+
1

sn

∫sn

sn−1

T(s)yn−1 − T(s)pds ‖

≤ ‖ yn − yn−1 ‖ + (
2|sn−sn−1|

sn
)‖ yn−1 − p ‖. (2.2)

Now, for all n ≥ 0, ‖ yn − yn−1 ‖

= ‖ αnf(xn) + (1 − αn)zn − αn−1f(xn−1) − (1 − αn−1)zn−1 ‖

=‖αn(f(xn) − f(xn−1)) + (αn − αn−1)(f(xn−1) − zn−1) + (1 − αn)(zn − zn−1) ‖.

Therefore, for all n ≥ 0,

‖ yn − yn−1 ‖ ≤ αn‖ f(xn) − f(xn−1)‖ + |αn − αn−1|‖ f(xn−1) − zn−1 ‖

+ (1 − αn)‖zn − zn−1 ‖. (2.3)

Again, for all n ≥ 0, ‖zn − zn−1‖

≤ ‖xn − xn−1‖ +
|rn−rn−1|

rn
‖zn − xn‖ (using Lemma 1.6)

We have lim inf
n→∞

rn > 0. Therefore there exists b > 0 such that rn > b for large n ∈ N where N is

the set of positive integers. Then, for all n ≥ 0,

‖zn − zn−1‖ ≤ ‖xn − xn−1‖ +
|rn−rn−1|

b
‖zn − xn‖. (2.4)

Using (2.3), (2.4) in (2.2), we get, for all n ≥ 0,

‖un − un−1‖

≤ αnθ‖xn − xn−1‖ + |αn − αn−1|‖f(xn−1) − zn−1‖ + (1 − αn)‖xn − xn−1‖

+ (1 − αn)
|rn−rn−1|

b
‖zn − xn‖ +

2|sn−sn−1|

sn
‖yn−1 − p ‖. (2.5)

Using (2.5) in (2.1) we get, for all n ≥ 0, ‖xn+1 − xn‖

≤ (1−βn)αnθ‖xn −xn−1‖+(1−βn)|αn −αn−1|‖f(xn−1)−zn−1‖+(1−βn)(1−αn)‖xn −xn−1‖

+(1 − βn)(1 − αn)
|rn−rn−1|

b
‖zn − xn‖ + (1 − βn)

2|sn−sn−1|

sn
‖yn−1 − p ‖ + |βn − βn−1|{ ‖ un−1 ‖ +

‖ xn−1 ‖ }+βn‖ xn − xn−1 ‖

={(1−βn)αnθ+(1−βn)(1−αn)+βn}‖xn −xn−1‖+(1−βn)|αn −αn−1|‖f(xn−1)−zn−1‖+(1−

βn)(1−αn)
|rn−rn−1|

b
‖zn −xn‖+(1−βn)

2|sn−sn−1|

sn
‖yn−1 −p ‖+ |βn −βn−1|{ ‖ un−1 ‖+‖ xn−1 ‖ }

= {1 − αn(1 − θ)(1 − βn)}‖ xn − xn−1 ‖ + (1 − βn)|αn − αn−1|‖f(xn−1) − zn−1‖ + (1 − βn)(1 −

αn)
|rn−rn−1|

b
‖zn − xn‖ + (1 − βn)

2|sn−sn−1|
sn

‖yn−1 − p ‖ + |βn − βn−1|{ ‖ un−1 ‖ + ‖ xn−1 ‖ }

≤ {1 − αn(1 − θ)(1 − d)}‖ xn − xn−1 ‖ + |αn − αn−1|‖f(xn−1) − zn−1‖ +
|rn−rn−1|

b
‖yn − xn‖ +



CUBO
15, 3 (2013)

Approximating a solution of an equilibrium problem . . . 15

2|sn−sn−1|

sn
‖yn−1 − p ‖ + |βn − βn−1|{ ‖ un−1 ‖ + ‖ xn−1 ‖ }

Let M= max{ sup
n∈N

‖f(xn−1) − zn−1‖, sup
n∈N

‖zn − xn‖, sup
n∈N

‖yn−1 − p‖, sup
n∈N

(‖(un−1‖ + ‖xn−1‖)}.

Therefore, for all n ≥ 0, ‖xn+1 − xn‖

≤ {1 − αn(1 − θ)(1 − d)}‖ xn − xn−1 ‖ + M[|αn − αn−1| +
|rn−rn−1|

b
+

2|sn−sn−1|

sn
+ |βn − βn−1|].

Let γn = αn(1 − θ)(1 − d), σn = 2M
|sn−sn−1|

sn
, δn = M[|αn − αn−1| +

|rn−rn−1|

b
+ |βn − βn−1|].

Using the Lemma 1.7 we get

lim
n→∞

‖xn+1 − xn‖ = 0.

Now,

‖ yn − zn ‖ = ‖ αnf(xn) + (1 − αn)zn − zn ‖ = αn‖ f(xn) − zn‖ → 0 as n → ∞. (2.6)

Again, for all n ≥ 0, ‖ xn − un ‖

=‖ βn−1xn−1 + (1 − βn−1)un−1 − un ‖

≤ ‖ un − un−1‖ + βn−1‖ xn−1 − un−1‖.

Since βn → 0 and ‖ un − un−1‖ → 0 as n → ∞, we have,

‖ xn − un ‖ → 0 as n → ∞. (2.7)

Also, for all n ≥ 0, ‖ zn − p ‖
2 = ‖Trnxn − Trnp ‖

2

≤ 〈Trnxn − Trnp, xn − p〉 ( by Lemma 1.1)

=〈zn − p, xn − p〉

=1
2
[‖ zn − p ‖

2 + ‖ xn − p ‖
2 − ‖ xn − zn ‖

2].

Therefore, for all n ≥ 0,

‖ zn − p ‖
2 ≤ ‖ xn − p ‖

2 − ‖ xn − zn ‖
2. (2.8)

Now, for all n ≥ 0,

‖ xn+1 − p ‖
2 =‖ βnxn + (1 − βn)un − p ‖

2 ≤ βn‖ xn − p ‖
2 + (1 − βn)‖ un − p ‖

2 (2.9)

Also for all n ≥ 0,

‖ yn − p ‖
2 =‖ αnf(xn) + (1 − αn)zn − p ‖

2 ≤ αn‖f(xn) − p‖
2 + (1 − αn)‖ zn − p ‖

2 (2.10)

Using (2.8) and (2.10) in (2.9), for all n ≥ 0, we get

‖ xn+1 − p ‖
2

≤ βn‖ xn − p ‖
2 + (1 − βn)[αn‖f(xn) − p‖

2 + (1 − αn)‖ zn − p ‖
2] (since ‖ un − p ‖ ≤ ‖ yn − p ‖)

≤ βn‖ xn − p ‖
2 + (1 − βn)αn‖f(xn) − p‖

2 + (1 − βn)(1 − αn)‖ zn − p ‖
2

≤ ‖ xn − p ‖
2 + αn‖ f(xn) − p ‖

2 − (1 − βn)‖ xn − zn ‖
2 [by (2.8)]

Therefore, (1 − βn)‖ xn − zn ‖
2 ≤ ‖ xn − p ‖

2 − ‖ xn+1 − p ‖
2 + αn‖ f(xn) − p ‖

2

≤ {‖ xn − p ‖ + ‖ xn+1 − p ‖}‖ xn − xn+1 ‖ + αn‖ f(xn) − p ‖
2

Therefore,

‖ xn − zn ‖ → 0 as n → ∞. (2.11)

Again ‖ zn − un ‖ ≤ ‖ zn − xn ‖ + ‖ xn − un ‖

By (2.7) and (2.11) we have

‖ zn − un ‖ → 0 as n → ∞. (2.12)

By (2.6), (2.7), (2.11) and (2.12) we can say that one of the sequences {xn}, {un}, {zn}, {yn} converge

if and only if the other three converge to the same limit.

By (2.6), (2.7), (2.11) and (2.12) we have



16 Binayak S. Choudhury & Subhajit Kundu CUBO
15, 3 (2013)

ωw(xn)=ωw(un)= ωw(zn) =ωw(yn), ωs(xn) = ωs(un)=ωs(zn)=ωs(yn). (2.13)

Now we have, p = PFix(S)∩ EP(F)f(p). We shall prove that lim sup
n→∞

〈f(p) − p, yn − p〉 ≤ 0.

We take a subsequence {yni} of { yn} such that

lim sup
n→∞

〈f(p) − p, yn − p〉 = lim
i→∞

〈f(p) − p, yni − p〉. (2.14)

Since {yni} is bounded and the Hilbert space H is reflexive, by Lemma 1.8, there exists a subse-

quence {yni
k
} of {yni} which converges weakly to x

∗. Then x∗ is also a weak limit of {xn}. Let

v0=PFix(S)∩ EP(F)x0. Since {xn} is a bounded sequence, there exists K such that B(v0, K) contains

{xn}. Moreover, B(v0, K) is T(s)-invariant for every s ≥ 0. Therefore, we can assume that (T(s))s≥ 0
is a nonexpansive semigroup on B(v0, K). So by (2.13), x

∗ ∈ ωw(un)= ωw(zn). Then, from

Lemma 1.4, we have , for every h ≥ 0, lim
n→∞

‖
1

sn

∫sn

0

T(s)ynds−T(h)
1

sn

∫sn

0

T(s)ynds‖= lim
n→∞

‖un−

T(h)un‖ = 0. Therefore from Lemma 1.5 , we have x
∗ ∈ Fix(S). Next we prove that x∗ ∈ EP(F).

Let {xni
k
} be a subsequence of {xni} such that xni

k
⇀ x∗. From (2.11) we can say that zk ⇀ x

∗.

Moreover, by (C2) we obtain

(1/rk)〈y − zk, zk − xk〉 ≥ F(y, zk), for all y ∈ H.

By condition (C4), for fixed x ∈ H, the function F(x, .) is lower semicontinuous and convex and

thus is weakly lower semicontinuous.

Since zk ⇀ x, by (2.11) and the fact that lim inf
n→∞

rn = b > 0, we get (zk − xk)/rk → 0. Letting

k → ∞, we have, F(y, x∗) ≤ lim inf
k→∞

F(y, zk) ≤ 0, for all y ∈ H.

Replacing y by yt where yt = ty + (1 − t)x
∗, t ∈ [0, 1] and using (C1) and (C4), we get

0=F(yt, yt) ≤ tF(yt, y) + (1 − t)F(yt, x
∗) ≤ F(yt, y).

Therefore, F(ty+(1−t)x∗, y) ≥ 0, t ∈ [0, 1], y ∈ H. Letting t → 0+ and using (C3), we conclude

that

F(x∗, y) ≥ 0, y ∈ H. Therefore, x∗ ∈ EP(F).

Since x∗ ∈ Fix(S) ∩ EP(F), from Lemma 1.2, we have, lim
n→∞

〈f(p) − p, yn − p〉

= lim
i→∞

〈f(p) − p, yni − p〉 (by using(2.14))

=〈f(p) − p, x∗ − p〉 ≤ 0.

Now for p ∈ Fix(S) ∩ EP(F), for all n ≥ 0, we have,

‖xn+1 − p‖
2 = ‖βn(xn − p) + (1 − βn)(un − p) ‖

2

≤ βn‖xn − p‖
2 + (1 − βn)‖un − p‖

2

≤ βn‖xn − p‖
2 + (1 − βn)‖yn − p‖

2

≤ βn‖xn − p‖
2 + (1 − βn){(1 − αn)

2‖zn − p‖
2 + 2αn〈f(xn) − p, yn − p〉}

≤ βn‖xn −p‖
2 +(1−βn)(1−2αn +α

2
n)‖xn −p‖

2 +2αn(1−βn)〈f(xn)−p, yn −p〉

≤ (1 − 2(1 − βn)αn)‖xn − p‖
2 + α2n‖xn − p‖

2

+ 2αn(1 − βn){〈f(xn) − f(p), yn − p〉 + 〈f(p) − p, yn − p〉}

≤ (1 − 2(1 − βn)αn)‖xn − p‖
2 + α2nM0 + 2(1 − βn)αn(θ‖ xn − p ‖.‖ yn − p ‖ + ηn)

≤ (1−2(1−βn)αn)‖xn−p‖
2+α2nM0 +(1−βn)θαn(‖xn−p‖

2+‖yn−p‖
2)+2αnηn

where ηn=max{〈f(p) − p, yn − p〉, 0} and M0 =sup
n≥0

{‖xn − p‖
2 + ‖f(xn) − p‖

2}.

Now, for all n ≥ 0, ‖yn − p‖
2 ≤ αn‖f(xn) − p ‖

2 + (1 − αn)‖zn − p ‖
2



CUBO
15, 3 (2013)

Approximating a solution of an equilibrium problem . . . 17

≤ αn‖f(xn) − p ‖
2 + (1 − αn)‖xn − p ‖

2

≤ αnM0 + ‖xn − p ‖
2.

Therefore, for all n ≥ 0, ‖xn+1 − p ‖
2

≤ (1 − 2(1 − βn)αn)‖xn − p ‖
2 + α2nM0 + (1 − βn)θαn(2‖xn − p ‖

2 + αnM0) + 2αnηn

≤ (1 − 2(1 − βn)(1 − θ)αn)‖xn − p‖
2 + α2nM0 + θα

2
nM0 + 2αnηn

≤ (1 − 2(1 − d)(1 − θ)αn)‖xn − p‖
2 + (α2nM0 + θα

2
nM0 + 2αnηn)

Therefore, by Lemma 1.3, we get xn → p as n → ∞.

Received: November 2012. Accepted: September 2013.

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