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Modified noor iterations for nonexpansive semigroups with generalized
contraction in Banach spaces
Journal of Inequalities and Applications 2012, 2012:6 doi:10.1186/1029-242X-2012-6
Rabian Wangkeeree ()
Pakkapon Preechasilp ()
ISSN 1029-242X
Article type Research
Submission date 1 June 2011
Acceptance date 12 January 2012
Publication date 12 January 2012
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Modified noor iterations for nonexpansive semigroups with
generalized contraction in Banach spaces
Rabian Wangkeeree
∗1,2
and Pakkapon Preechasilp
1
1
Department of Mathematics, Faculty of Science, Naresuan University,
Phitsanulok 65000, Thailand
2

Centre of Excellence in Mathematics, CHE, Si Ayutthaya Road,
Bangkok 10400, Thailand
∗
Corresp onding author:
Email address:
PP:
Abstract
In this article, the modified Noor iterations are considered for the generalized contraction and a
nonexpansive semigroup in the framework of a reflexive Banach space which admits a weakly sequen-
tially continuous duality mapping. The strong convergence theorems are obtained under very mild
conditions imposed the parameters. The results presented in this article improve and extend the
corresponding results announced by Chen and He and Chen et al. and many others.
AMS subject classification: 47H09; 47H10; 47H17.
Keywords: generalized contraction; Meir–Keeler type mapping; nonexpansive semigroup; fixed
point; reflexive Banach space.
1. Introduction and preliminaries
Let E be a real Banach space. A mapping T of E into itself is said to be nonexpansive if Tx−Ty≤
x − y for each x, y ∈ E. We denote by Fix(T ) the set of fixed points of T . A mapping f : E −→ E
is called α-contraction, if there exists a constant 0 <α<1 such that f(x) − f(y)≤αx − y for all
1
x, y ∈ E. Throughout this article, we denote by N and R
+
the sets of positive integers and nonnegative
real numbers, respectively. A mapping ψ : R
+
−→ R
+
is said to be an L-function if ψ(0) = 0,ψ(t) > 0,
for each t>0andforeveryt>0andforeverys>0 there exists u>ssuch that ψ(t) ≤ s, for all
t ∈ [s, u], As a consequence, every L-function ψ satisfies ψ(t) <t,foreacht>0.

Definition 1.1. Let (X, d) be a matric space. A mapping f : X −→ X is said to be :
(i) a (ψ,L)-function if ψ : R
+
−→ R
+
is an L-function and d(f(x),f(y)) <ψ(d(x, y)), for all x, y ∈ X,
with x = y:
(ii) a Meir–Keeler type mapping if for each ε>0 there exists δ = δ(ε) > 0 such that for each x, y ∈ X,
with ε ≤ d(x, y) <ε+ δ we have d(f(x),f(y)) <ε.
If, in Definition 1.1 we consider ψ(t)=αt,foreacht ∈ R
+
, where α ∈ [0, 1), then we get the usual
contraction mapping with coefficient α.
Proposition 1.2. [1] Let (X, d) be a matric space and f : X −→ X be a mapping. The following
assertions are equivalent:
(i) f is a Meir–Keeler type mapping :
(ii) there exists an L-function ψ : R
+
−→ R
+
such that f is a (ψ,L)-contraction.
Lemma 1.3. [2] Let X be a Banach space and C be a convex subset of it. Let T : C −→ C be a
nonexpansive mapping and f is a (ψ, L)-contraction. Then the following assertions hold:
(i) T ◦ f is a (ψ, L)-contraction on C and has a unique fixed point in C ;
(ii) for each α ∈ (0, 1) the mapping x → αf(x)+(1− α)T (x) is a Meir–Keeler type mapping on
C.
Lemma 1.4. [3, Proposition 2] Let E be a Banach space and C a convex subset of it. Let f : C −→ C
be a Meir–Keeler type mapping. Then for each ε>0 there exists r ∈ (0, 1) such that
for each x, y ∈ C with x − y≥ε we have f (x) − f(y)≤r x − y.
From now on, by a generalized contraction mapping we mean a Meir–Keeler type mapping or (ψ, L)-

contraction. In the rest of the article we suppose that the ψ from the definition of the (ψ, L)-contraction
is continuous, strictly increasing and η(t) is strictly increasing and onto, where η(t):=t −ψ(t),t∈ R
+
.
As a consequence, we have the η is a bijection on R
+
.
A family S = {T (t):0≤ t<∞} of mappings of E into itself is called a nonexpansive semigroup on
E if it satisfies the following conditions:
(i) T (0)x = x for all x ∈ E ;
(ii) T (s + t)=T (s)T(t) for all s, t ≥ 0;
(iii) T (t)x − T (t)y≤x − y for all x, y ∈ E and t ≥ 0;
(iv) for all x ∈ E, the mapping t → T(t)x is continuous.
We denote by Fix(S) the set of all common fixed points of S,thatis,
Fix(S):={x ∈ E : T(t)x = x, 0 ≤ t<∞} = ∩
t≥0
Fix(T (t)).
In [4], Shioji and Takahashi introduced the following implicit iteration in a Hilbert space
x
n
= α
n
x +(1− α
n
)
1
t
n
t
n


0
T (s)x
n
ds, ∀n ∈ N (1.1)
where {α
n
} is a sequence in (0, 1), {t
n
} is a sequence of positive real numbers which diverges to ∞.
Under certain restrictions on the sequence {α
n
}, Shioji and Takahashi [4] proved strong convergence of
the sequence {x
n
} toamemberofF(S). In [5], Shimizu and Takahashi studied the strong convergence
of the sequence {x
n
} defined by
x
n+1
= α
n
x +(1− α
n
)
1
t
n
t

n

0
T (s)x
n
ds, ∀n ∈ N (1.2)
in a real Hilbert space where {T (t):t ≥ 0} is a strongly continuous semigroup of nonexpansive
mappings on a closed convex subset C of a Banach space E and lim
n−→ ∞
t
n
= ∞. Using viscosity
iterative method, Chen and Song [6] studied the strong convergence of the following iterative method
for a nonexpansive semigroup {T (t):t ≥ 0} with Fix(S) = ∅ in a Banach space:
x
n+1
= α
n
f(x)+(1− α
n
)
1
t
n
t
n

0
T (s)x
n

ds, ∀n ∈ N, (1.3)
where f is a contraction. Note however that their iterate x
n
at step n is constructed through the
average of the semigroup over the interval (0,t). Suzuki [7] was the first to introduce again in a Hilbert
space the following implicit iteration process:
x
n
= α
n
u +(1− α
n
)T (t
n
)x
n
, ∀n ∈ N, (1.4)
for the nonexpansive semigroup case. In 2002, Benavides et al. [8] in a uniformly smooth Banach space,
showed that if S satisfies an asymptotic regularity condition and {α
n
} fulfills the control conditions
lim
n−→ ∞
α
n
=0,

∞
n=1
α

n
= ∞, and lim
n−→ ∞
α
n
α
n+1
= 0, then both the implicit iteration process (1.4)
and the explicit iteration process (1.5)
x
n+1
= α
n
u +(1− α
n
)T (t
n
)x
n
, ∀n ∈ N, (1.5)
converge to a same point of F (S). In 2005, Xu [9] studied the strong convergence of the implicit
iteration process (1.1) and (1.4) in a uniformly convex Banach space which admits a weakly sequen-
tially continuous duality mapping. Recently Chen and He [10] introduced the viscosity approximation
methods:
y
n
= α
n
f(y
n

)+(1− α
n
)T (t
n
)y
n
, ∀n ∈ N, (1.6)
and
x
n+1
= α
n
f(x
n
)+(1− α
n
)T (t
n
)x
n
, ∀n ∈ N, (1.7)
where f is a contraction, {α
n
} is a sequence in (0, 1) and a nonexpansive semigroup {T(t):t ≥ 0}.
The strong convergence theorem of {x
n
} is proved in a reflexive Banach space which admits a weakly
sequentially continuous duality mapping. Very recently, motivated by the above results, Chen et al. [11]
proposed the following two modified Mann iterations for nonexpansive semigroups {T(t):0≤ t<∞}
and obtained the strong convergence theorems in a reflexive Banach space E which admits a weakly

sequentially continuous duality mapping:







y
n
= α
n
x
n
+(1− α
n
)T (t
n
)x
n
,
x
n
= β
n
f(x
n
)+(1− β
n
)y

n
,
(1.8)
and















x
0
∈ C,
y
n
= α
n
x
n
+(1− α
n

)T (t
n
)x
n
,
x
n+1
= β
n
f(x
n
)+(1− β
n
)y
n
,
(1.9)
where f : C −→ C is a contraction. They proved that the implicit iterative scheme {x
n
} defined by
(1.8) converges to an element q of Fix(S), which solves the following variation inequality problem:
(f − I)q, j(x − q)≤0 for all x ∈ Fix(S).
Furthermore, Moudafi’s viscosity approximation methods have been recently studies by many au-
thors; see the well known results in [12,13]. However, the involved mapping f is usually considered as a
contraction. Note that Suzuki [14] proved the equivalence between Moudafi’s viscosity approximation
with contractions and Browder-type iterative processes (Halpern-type iterative processes); see [14] for
more details.
In this article, inspired by above result, we introduce and study the explicit viscosity iterative scheme
for the generalized contraction f and a nonexpansive semigroup {T (t):t ≥ 0}:
























x
0
∈ C,
z
n
= γ
n
x

n
+(1− γ
n
)T (t
n
)x
n
,
y
n
= α
n
x
n
+(1− α
n
)T (t
n
)z
n
,
x
n+1
= β
n
f(x
n
)+(1− β
n
)y

n
,n≥ 0.
(1.10)
The iterative schemes (1.10) are called the three-step(modified Noor) iterations which inspired by
three-step(Noor) iterations [15–23]. It is well known that three-step(Noor) iterations, include Mann
and two-step iterative methods as special cases. If γ ≡ 1, then (1.10) reduces to (1.9). Furthermore,
the implicit iteration (1.8) and explicit iteration (1.10) are considered for the generalized contraction
and a nonexpansive semigroup in the framework of a reflexive Banach space which admits a weakly
sequentially continuous duality mapping. The strong convergence theorems are obtained under very
mild conditions imposed the parameters. The results presented in this article improve and extend the
corresponding results announced by Chen and He [10] and Chen et al. [11] and many others.
In order to prove our main results, we need the following lemmas.
Definition 1.5. [24] A Banach space is said to admit a weakly sequentially continuous normalized
duality mapping J from E in E
∗
,ifJ : E −→ E
∗
is single-valued and weak to weak
∗
sequentially
continuous, that is, if x
n
xin E, then J(x
n
) 
∗
J(x)inE
∗
.
A Banach space E is said to satisfy Opial’s condition if for any sequence {x

n
} in E, x
n
x
(n −→ ∞ ) implies
lim sup
n−→ ∞
x
n
− x < lim sup
n−→ ∞
x
n
− y, ∀y ∈ E with x = y. (1.11)
By [25, Theorem 1], it is well known that if E admits a weakly sequentially continuous duality
mapping, then E satisfies Opial’s condition, and E is smooth.
In order to prove our main result, we need the following lemmas.
Lemma 1.6. Let E be a Banach s pace and x, y ∈ E,j(x) ∈ J(x),j(x + y) ∈ J(x + y).Then
x
2
+2y,j(x)≤x + y
2
≤x
2
+2y,j(x + y).
In the following, we also need the following lemma that can be found in the existing literature [13,26].
Lemma 1.7. Let {a
n
} be a sequence of non-negative real numbers satisfying the property
a

n+1
≤ (1 − γ
n
)a
n
+ δ
n
,n≥ 0,
where {γ
n
}⊆(0, 1) and {δ
n
}⊆R such that

∞
n=1
γ
n
= ∞, and either lim sup
n→∞
δ
n
γ
n
≤ 0 or

∞
n=1
|δ
n

| < ∞.
Then lim
n−→ ∞
a
n
=0.
Lemma 1.8. [27] Let {x
n
} and {y
n
} be bounded sequences in a Banach space E and {β
n
} asequence
in [0, 1] with 0 < lim inf
n→∞
β
n
≤ lim sup
n→∞
β
n
< 1. Suppose that x
n+1
=(1− β
n
)y
n
+ β
n
x

n
for all
n ≥ 0 and
lim sup
n→∞
(y
n+1
− y
n
−x
n+1
− x
n
) ≤ 0.
Then lim
n→∞
y
n
− x
n
 =0.
2. Modified Mann iteration for generalized contractions
Now, we are a position to state and prove our main results.
Theorem 2.1. Let E be a reflexive Banach space which admits a weakly sequenctially continuous
duality mapping J from E into E
∗
,supposeC is a nonempty closed convex subset of E.LetS :=
{T (t):t ≥ 0} be a nonexpansive semigroup on C such that Fix(S) = ∅,andf : C −→ C a generalized
contraction on C.Let{α
n

}⊂(0, 1), {β
n
}⊂(0, 1),and{t
n
}⊂(0, ∞) be sequences of real numbers
satisfying lim
n−→ ∞
α
n
= lim
n−→ ∞
t
n
= lim
n−→ ∞
β
n
t
n
=0. Define a sequence {x
n
} in C by







y

n
= α
n
x
n
+(1− α
n
)T (t
n
)x
n
,
x
n
= β
n
f(x
n
)+(1− β
n
)y
n
, for all n ≥ 1.
(2.1)
Then {x
n
} converges strongly to q,asn −→ ∞ ; q is the element of Fix(S) such that q is the unique
solution in Fix(S) to the following variational inequality:
(f − I)q, j(x − q)≤0 for all x ∈ Fix(S). (2.2)
Proof. We first show that {x

n
} is well defined. For any n ≥ 1, we consider a mapping G
n
on C defined
by
G
n
x = β
n
f(x)+(1− β
n
)U
n
x, ∀x ∈ C,
where U
n
:= α
n
I +(1− α
n
)T (t
n
). It follows from nonexpansivity of U
n
and Lemma 1.3 that G
n
is
a Meir–Keeler type contraction. Hence G
n
has a unique fixed point, denoted as x

n
, which uniquely
solves the fixed point equation
x
n
= α
n
f(x
n
)+(1− α
n
)U
n
x
n
, ∀n ≥ 1.
Hence {x
n
} generated in (2.1) is well defined. Now we show that {x
n
} is bounded. Indeed, if we take
a fixed point x ∈ Fix(S), we have
y
n
− x≤α
n
x
n
− x +(1− α
n

)T (t
n
)x
n
− x≤x
n
− x, (2.3)
and so
x
n
− x
2
= β
n
(f(x
n
) − x)+(1− β
n
)(y
n
− x),j(x
n
− x)
= β
n
f(x
n
) − f(x)+f (x) − x, j(x
n
− x) +(1− β

n
)(y
n
− x),j(x
n
− x)
≤ β
n
f(x
n
) − f(x)x
n
− x + β
n
f(x) − x, j(x
n
− x) +(1− β
n
)y
n
− xx
n
− x
≤ β
n
ψ(x
n
− x)x
n
− x + β

n
f(x) − xx
n
− x +(1− β
n
)x
n
− x
2
,
and hence
x
n
− x
2
≤ ψ(x
n
− x)x
n
− x + f(x) − xx
n
− x. (2.4)
Therefore
η(x
n
− x):=x
n
− x−ψ(x
n
− x) ≤f(x) − x,

equivalent to
x
n
− x≤η
−1
(f(x) − x).
Thus {x
n
} is bounded, and so are {T (t
n
)x
n
}, {f (x
n
)},and{y
n
}. Next, we claim that {x
n
} is relatively
sequentially compact. Indeed, By reflexivity of E and boundedness of the sequence {x
n
} there exists
a weakly convergent subsequence {x
n
j
}⊂{x
n
} such that x
n
j

pfor some p ∈ C. Now we show that
p ∈ Fix(S). Put x
j
= x
n
j
,y
j
= y
n
j
,α
j
= α
n
j
,β
j
= β
n
j
and t
j
= t
n
j
for j ∈ N,fixedt>0. Notice that
x
j
− T (t)p≤

[t/t
j
]−1

k=0
T ((k +1)t
j
)x
j
− T (kt
j
)x
j

+T ([t/t
j
]t
j
) x
j
− T ([t/t
j
]t
j
) p + T ([t/t
j
]t
j
) p − T(t)p
≤ [t/t

j
]T (t
j
)x
j
− x
j
 + x
j
− p + T (t − [t/t
j
]t
j
) p − p
=[t/t
j
]
β
j
1 − α
j
x
j
− f(x
j
) + x
j
− p + T (t − [t/t
j
]t

j
) p − p
≤
t
1 − α
j
β
j
t
j
x
j
− f(x
j
) + x
j
− p + max {T (s)p − p :0≤ s ≤ t
j
}.
For all j ∈ N,wehave
lim sup
j−→ ∞
x
j
− T (t)p≤lim sup
j−→ ∞
x
j
− p.
Since Banach space E with a weakly sequentially continuous duality mapping satisfies Opial’s condition,

T (t)p = p. Therefore p ∈ Fix(S). In Equation (2.4), replace p with x to obtain
x
j
− p (x
j
− p−ψ(x
j
− p)) ≤f(p) − p, j(x
j
− p).
Using that the duality map j is single-valued and weakly sequentially continuous from E to E
∗
,weget
that
lim
j−→ ∞
x
j
− p (x
j
− p−ψ(x
j
− p)) ≤ lim
j−→ ∞
f(p) − p, j(x
j
− p) =0.
If lim
j−→ ∞
x

j
− p = 0, then we have done.
If lim
j−→ ∞
(x
j
− p−ψ(x
j
− p)) = 0, then we have lim
j−→ ∞
x
j
− p = lim
j−→ ∞
ψ(x
j
− p).
Since ψ is a continuous function, lim
j−→ ∞
x
j
− p = ψ(lim
j−→ ∞
x
j
− p). By Definition of ψ,
we have lim
j−→ ∞
x
j

− p = 0. Hence {x
n
} is relatively sequentially compact, i.e., there exists a
subsequence {x
n
j
}⊆{x
n
} such that x
n
j
−→ p as j −→ ∞ . Next, we show that p is a solution in
Fix(S) to the variational inequality (2.2). In fact, for any x ∈ Fix(S),
x
n
− x
2
= β
n
f(x
n
)+(1− β
n
)y
n
− x, j(x
n
− x)
= β
n

(f(x
n
) − x
n
+ x
n
− x)+(1− β
n
)(y
n
− x),j(x
n
− x)
= β
n
f(x
n
) − x
n
,j(x
n
− x) + β
n
x
n
− x, j(x
n
− x) +(1− β
n
)y

n
− x, j(x
n
− x)
≤ β
n
f(x
n
) − x
n
,j(x
n
− x) + β
n
x
n
− x
2
+(1− β
n
)y
n
− xx
n
− x
≤ β
n
f(x
n
) − x

n
,j(x
n
− x) + β
n
x
n
− x
2
+(1− β
n
)x
n
− x
2
.
Therefore,
f(x
n
) − x
n
,j(x − x
n
)≤0. (2.5)
Since the sets {x
n
− x} and {x
n
− f (x
n

)} are bounded and the duality mapping j is singled-valued and
weakly sequentially continuous from E into E
∗
, for any fixed x ∈ Fix(S). It follows from (2.5) that
f(p) − p, j(x − p) = lim
j−→ ∞
f(x
n
j
) − x
n
j
,j(x − x
n
j
)≤0, ∀x ∈ Fix(S).
This is, p ∈ Fix(S) is a solution of the variational inequality (2.2).
Finally, we show that p ∈ Fix(S) is the unique solution of the variational inequality (2.2). In fact,
supposing p, q ∈ Fix(S) satisfy the inequality (2.2) with p = q, we get that there exists ε>0 such that
p − q≥ε. By Proposition 1.4 there exists r ∈ (0, 1) such that f(p) − f(q)≤rp − q. We get that
(f − I)p, j(q − p)≤0and(f − I)q,j(p − q)≤0.
Adding the two above inequalities, we have that
0 < (1 − r)ε
2
≤ (1 − r)p − q
2
≤((I − f)p − (I − f)q, j(p − q))≤0,
which is contradiction. We must have p = q, and the uniqueness is proved.
In a similar way, it can be shown that each cluster point of sequence {x
n

} is equal to q. Therefore,
the entire sequence {x
n
} converges to q and the proof is complete. 
Setting f is a contraction on C in Theorem 2.1, we have the following results immediately.
Corollary 2.2. [11, Theorem 3.1] Let E be a reflexive Banach space which admits a weakly sequenc-
tially continuous duality mapping J from E into E
∗
,supposeC is a nonempty closed convex subset of
E.LetS := {T (t):t ≥ 0} be a nonexpansive semigroup on C such that Fix(S) = ∅,andf : C −→ C
a contraction on C.Let{α
n
}⊂(0, 1), {β
n
}⊂(0, 1),and{t
n
}⊂(0, ∞) be sequences of real numbers
satisfying lim
n−→ ∞
α
n
= lim
n−→ ∞
t
n
= lim
n−→ ∞
β
n
t

n
=0. Define a sequence {x
n
} in C by







y
n
= α
n
x
n
+(1− α
n
)T (t
n
)x
n
,
x
n
= β
n
f(x
n

)+(1− β
n
)y
n
, for all n ≥ 1.
Then {x
n
} converges strongly to q,asn −→ ∞ ; q is the element of Fix(S) such that q is the unique
solution in Fix(S) to the following variational inequality:
(f − I)q, j(x − q)≤0 for all x ∈ Fix(S).
Theorem 2.3. Let E be a reflexive Banach space which admits a weakly sequenctially continuous
duality mapping J from E into E
∗
,supposeC is a nonempty closed convex subset of E.Let{T (t):
t ≥ 0}, be a nonexpansive semigroup on C such that Fix(S) = ∅,andf : C −→ C be a generalized
contraction on C.Let{α
n
}⊂(0, 1), {β
n
}⊂(0, 1), {γ
n
}⊂[0, 1],and{t
n
}⊂(0, ∞) be sequences of
real numbers satisfying the conditions:
(C1) lim
n−→ ∞
β
n
=0,


∞
n=0
β
n
= ∞ and lim
n−→ ∞
t
n
=0,
(C2) lim
n−→ ∞
α
n
=0and lim
n−→ ∞
γ
n
=1,
(C3)

∞
n=0
|α
n+1
− α
n
| < ∞,

∞

n=0
|β
n+1
− β
n
| < ∞,and

∞
n=0
|γ
n+1
− γ
n
| < ∞.
Define a sequence {x
n
} in C by
























x
0
∈ C,
z
n
= γ
n
x
n
+(1− γ
n
)T (t
n
)x
n
,
y
n
= α
n

x
n
+(1− α
n
)T (t
n
)z
n
,
x
n+1
= β
n
f(x
n
)+(1− β
n
)y
n
,n≥ 0.
(2.6)
Suppose
∞

n=0
sup
x∈
˜
C
T (t

n
)x − T(t
n−1
)x < ∞, (*)
where
˜
C is any bounded subset of C.Then{x
n
} converges strongly to q,asn −→ ∞ ;whereq is the
unique solution in Fix(S) to the variational inequality (2.2).
Proof. First, we show that {x
n
} is bounded. Indeed, if we take a fixed point x ∈ Fix(S). We will prove
by induction that
x
n
− x≤M for all n ≥ 0,
where M := {x
0
− z,η
−1
(f(x) − x)}. From Definition of (2.8), notice that
z
n
− x≤γ
n
x
n
− x +(1− γ
n

)T (t
n
)x
n
− x≤x
n
− x.
It follows that
y
n
− x≤α
n
x
n
− x +(1− α
n
)T (t
n
)z
n
− x≤α
n
x
n
− x +(1− α
n
)x
n
− x≤x
n

− x.
The case n = 0 is obvious.
Suppose that x
n
− x≤M,wehave
x
n+1
− x≤β
n
f(x
n
) − x +(1− β
n
)y
n
− x
≤ β
n
f(x
n
) − f(x) + β
n
f(x) − x +(1− β
n
)y
n
− x
≤ β
n
ψ(x

n
− x)+β
n
f(x) − x +(1− β
n
)x
n
− x
= β
n
ψ(x
n
− x)+β
n
η(η
−1
(f(x) − x)) + (1 − β
n
)x
n
− x
≤ β
n
ψ(M)+β
n
η(M)+(1− β
n
)M
= β
n

ψ(M)+β
n
(M − ψ(M)) + (1 − β
n
)M = M.
By induction,
x
n
− x≤max

x
0
− x,η
−1
(f(x) − x)

, ∀ n ≥ 0.
Thus {x
n
} is bounded, and so are {T (t
n
)x
n
}, {y
n
}, {z
n
}, {T (t
n
)z

n
},and{f(x
n
)}. As a result, we
obtain by condition (C1),
x
n+1
− y
n
 = β
n
f(x
n
) − y
n
−→0. (2.7)
We next show that
x
n
− T (t
n
)x
n
−→0. (2.8)
It suffices to show that
x
n+1
− x
n
−→0. (2.9)

Indeed, if (2.9) holds, then noting (2.7), we obtain
x
n
− T (t
n
)x
n
≤x
n
− x
n+1
 + x
n+1
− y
n
 + y
n
− T (t
n
)z
n
 + T(t
n
)z
n
− T (t
n
)x
n


≤x
n
− x
n+1
 + x
n+1
− y
n
 + α
n
x
n
− T (t
n
)z
n
 + z
n
− x
n

≤x
n
− x
n+1
 + x
n+1
− y
n
 + α

n
x
n
− T (t
n
)x
n
 + α
n
T (t
n
)x
n
− T (t
n
)z
n

+z
n
− x
n

≤x
n
− x
n+1
 + x
n+1
− y

n
 + α
n
x
n
− T (t
n
)x
n
 + α
n
x
n
− z
n
 + z
n
− x
n

≤x
n
− x
n+1
 + x
n+1
− y
n
 + α
n

x
n
− T (t
n
)x
n
 +(1+α
n
)x
n
− z
n

≤x
n
− x
n+1
 + x
n+1
− y
n
 + α
n
x
n
− T (t
n
)x
n


+(1 + α
n
)(1 − γ
n
)x
n
− T (t
n
)x
n
.
It follows from (C2) that
x
n
− T (t
n
)x
n
≤x
n
− x
n+1
 + x
n+1
− y
n
 + α
n
x
n

− T (t
n
)x
n

+(1 + α
n
)(1 − γ
n
)x
n
− T (t
n
)x
n
−→0asn −→ ∞ .
Suppose that (2.9) is not holds, there exists ε>0 and subsequence x
n
j
+1
− x
n
j
 of x
n+1
− x
n

such that x
n

j
+1
− x
n
j
≥ε for all j ∈ N. By Proposition 1.4, there exists r ∈ (0, 1) such that
f(x
n
j
+1
) − f(x
n
j
)≤rx
n
j
+1
− x
n
j
 for all j ∈ N.Putx
j
= x
n
j
,α
j
= α
n
j

,β
j
= β
n
j
,γ
j
= γ
n
j
and
t
j
= t
n
j
for j ∈ N. We calculate x
j+1
− x
j
. Observing that
x
j+1
= β
j
f(x
j
)+(1− β
j
)y

j
and x
j
= β
j−1
f(x
j−1
)+(1− β
j−1
)y
j−1
,
we get
x
j+1
− x
j
= β
j
f(x
j
)+(1− β
j
)y
j
− β
j−1
f(x
j−1
) − (1 − β

j−1
)y
j−1
= β
j
(f(x
j
) − f(x
j−1
)) + (β
j
− β
j−1
)f(x
j−1
)+(1− β
j
)(y
j
− y
j−1
)
−(β
j
− β
j−1
)y
j−1
. (2.10)
That is

x
j+1
− x
j
≤β
j
rx
j
− x
j−1
 + |β
j
− β
j−1
|f(x
j−1
) +(1− β
j
)y
j
− y
j−1

+|β
j
− β
j−1
|y
j−1
. (2.11)

Noticing that
y
j
= α
j
x
j
+(1− α
j
)T (t
j
)z
j
and y
j−1
= α
j−1
x
j−1
+(1− α
j−1
)T (t
j−1
)z
j−1
.
We obtain that
y
j
− y

j−1
= α
j
x
j
+(1− α
j
)T (t
j
)z
j
− α
j−1
x
j−1
− (1 − α
j−1
)T (t
j−1
)z
j−1
.
= α
j
(x
j
− x
j−1
)+(α
j

− α
j−1
)x
j−1
+(1 − α
j−1
)(T (t
j
)z
j−1
− T (t
j−1
)z
j−1
)+(1− α
j
)(T (t
j
)z
j
− T (t
j
)z
j−1
)
−(α
j
− α
j−1
)T (t

j
)z
j−1
. (2.12)
This implied that
y
j
− y
j−1
≤α
j
x
j
− x
j−1
 + |α
j
− α
j−1
|x
j−1

+(1 − α
j−1
)T (t
j
)z
j−1
− T (t
j−1

)z
j−1
 +(1− α
j
)z
j
− z
j−1

+|α
j
− α
j−1
|T (t
j
)z
j−1
. (2.13)
Again from (2.6) we obtain
z
j
− z
j−1
= γ
j
(x
j
− x
j−1
)+(1− γ

j
)(T (t
j
)x
j
− T (t
j
)x
j−1
)
+(1 − γ
j
)(T (t
j
)x
j−1
− T (t
j−1
)x
j−1
)+(γ
j
− γ
j−1
)(x
j−1
− T (t
j−1
)x
j−1

),
that is,
z
j
− z
j−1
≤γ
j
x
j
− x
j−1
 +(1− γ
j
)T (t
j
)x
j
− T (t
j
)x
j−1

+(1 − γ
j
)T (t
j
)x
j−1
− T (t

j−1
)x
j−1
 + |γ
j
− γ
j−1
|x
j−1
− T (t
j−1
)x
j−1

≤x
j
− x
j−1
 +(1− γ
j
)T (t
j
)x
j−1
− T (t
j−1
)x
j−1

+|γ

j
− γ
j−1
|x
j−1
− T (t
j−1
)x
j−1
. (2.14)
Substituting (2.14) into (2.13),
y
j
− y
j−1
≤α
j
x
j
− x
j−1
 + |α
j
− α
j−1
|x
j−1

+(1 − α
j−1

)T (t
j
)z
j−1
− T (t
j−1
)z
j−1
 + |α
j
− α
j−1
|T (t
j
)z
j−1

+(1 − α
j
)x
j
− x
j−1
 +(1− α
j
)(1 − γ
j
)T (t
j
)x

j−1
− T (t
j−1
)x
j−1

+(1 − α
j
)|γ
j
− γ
j−1
|x
j−1
− T (t
j−1
)x
j−1
]
≤x
j
− x
j−1
 + |α
j
− α
j−1
|x
j−1


+T (t
j
)z
j−1
− T (t
j−1
)z
j−1
 + |α
j
− α
j−1
|T (t
j
)z
j−1

+T (t
j
)x
j−1
− T (t
j−1
)x
j−1

+|γ
j
− γ
j−1

|x
j−1
− T (t
j−1
)x
j−1
. (2.15)
Substituting (2.15) into (2.11),
x
j+1
− x
j
≤β
j
rx
j
− x
j−1
 + |β
j
− β
j−1
|f(x
j−1
) +(1− β
j
)x
j
− x
j−1


+(1 − β
j
)|α
j
− α
j−1
|x
j−1

+(1 − β
j
)T (t
j
)z
j−1
− T (t
j−1
)z
j−1
 +(1− β
j
)|α
j
− α
j−1
|T (t
j
)z
j−1


+(1 − β
j
)T (t
j
)x
j−1
− T (t
j−1
)x
j−1

+(1 − β
j
)|γ
j
− γ
j−1
|x
j−1
− T (t
j−1
)x
j−1

+|β
j
− β
j−1
|y

j−1

≤ (1 − (1 − r)β
j
)x
j
− x
j−1
 + |β
j
− β
j−1
|f(x
j−1
)
+|α
j
− α
j−1
|x
j−1

+T (t
j
)z
j−1
− T (t
j−1
)z
j−1

 + |α
j
− α
j−1
|T (t
j
)z
j−1

+T (t
j
)x
j−1
− T (t
j−1
)x
j−1

+|γ
j
− γ
j−1
|x
j−1
− T (t
j−1
)x
j−1

+|β

j
− β
j−1
|y
j−1
. (2.16)
Hence,
x
j+1
− x
j
≤(1 − (1 − r)β
j
)x
j
− x
j−1
 +(2|α
j
− α
j−1
| +2|β
j
− β
j−1
| + |γ
j
− γ
j−1
|) M

+sup
x∈{x
n
}
T (t
j
)x − T(t
j−1
)x +sup
z∈{z
n
}
T (t
j
)z − T (t
j−1
)z, (2.17)
where M ≥ max {x
j−1
− T (t
j−1
)x
j−1
, y
j−1
, x
j−1
, T (t
j
)z

j−1
, f(x
j−1
)} for all j.
By assumption, we have that

∞
j=1
β
j
= ∞,and

∞
j=1
(2|α
j
− α
j−1
| +2|β
j
− β
j−1
| + |γ
j
− γ
j−1
| +
sup
x∈{x
n

}
T (t
j
)x − T(t
j−1
)x+sup
z∈{z
n
}
T (t
j
)z − T (t
j−1
)z) < ∞. Hence, Lemma 1.7 is applicable
to (2.17) and we obtain x
j+1
− x
j
−→0, which is a contradiction. So (2.9) is proved. Applying
Theorem 2.1, there is a unique solution q ∈ Fix(S) to the following variational inequality:
f(q) − q, j(x − q)≤0 for all x ∈ Fix(S).
Next, we show that
lim sup
n−→ ∞
f(q) − q, j(x
n+1
− q)≤0. (2.18)
Indeed, we can take a subsequence {x
n
i

} of {x
n
} such that
lim sup
n−→ ∞
f(q) − q, j(x
n+1
− q) = lim
i−→ ∞
f(q) − q, j(x
n
i
+1
− q).
By the reflexivity of E and boundedness of the sequence {x
n
}, we may assume, without loss of gener-
ality, that x
n
i
pfor some p ∈ C. Now we show that p ∈ Fix(S). Put x
i
= x
n
i
,α
i
= α
n
i

,β
i
= β
n
i
and t
i
= t
n
i
for i ∈ N,lett
i
≥ 0 be such that
t
i
−→ 0and
T (t
i
)x
i
− x
i

t
i
−→ 0,i−→ ∞ .
Fix t>0. Notice that
x
i
− T (t)p≤

[t/t
i
]−1

k=0
T ((k +1)t
i
)x
i
− T (kt
i
)x
i

+T ([t/t
i
]t
i
) x
i
− T ([t/t
i
]t
i
)p + T([t/t
i
]t
i
)p − T(t)p
≤ [t/t

i
]T (t
i
)x
i
− x
i
 + x
i
− p + T (t − [t/t
i
]t
i
)p − p
≤ t
T (t
i
)x
i
− x
i

t
i
+ x
i
− p + T (t − [t/t
i
]t
i

)p − p
≤ t
T (t
i
)x
i
− x
i

t
i
+ x
i
− p + max {T (s)p − p :0≤ s ≤ t
i
}.
For all i ∈ N,wehave
lim sup
i−→ ∞
x
i
− T (t)p≤lim sup
i−→ ∞
x
i
− p.
Since Banach space E with a weakly sequentially continuous duality mapping satisfies Opial’s condition,
this implies T (t)p = p. Therefore p ∈ Fix(S). In view of the variational inequality (2.2) and the
assumption that duality mapping J is weakly sequentially continuous, we conclude
lim sup

n−→ ∞
f(q) − q, j(x
n+1
− q) = lim
i−→ ∞
f(q) − q, j(x
n
i
+1
− q)
= f(q) − q,j(p − q)≤0.
Then (2.18) is proved. Finally, show that x
n
−→ q, i.e. x
n
− q−→0. Suppose that x
n
− q  0,
then there exists ε>0 and a subsequence {x
n
j
} of {x
n
} such that x
n
j
− q≥ε for all j ∈ N.Put
x
j
= x

n
j
,α
j
= α
n
j
,β
j
= β
n
j
and t
j
= t
n
j
for j ∈ N. By Proposition 1.4, there exists r ∈ (0, 1) such
that f(x
j
) − f(q)≤rx
j
− q for all j ∈ N. As a matter of fact, from Lemma 1.6 we have that
x
j+1
− q
2
= β
j
f(x

j
)+(1− β
j
)(α
j
x
j
+(1− α
j
)T (t
j
)z
j
) − q
2
= (1 − β
j
)(α
j
(x
j
− q)+(1− α
j
(T (t
j
)z
j
− q)) + β
j
(f(x

j
) − q)
2
≤ (1 − β
j
)
2
α
j
(x
j
− q)+(1− α
j
)(T (t
j
)z
j
− q)
2
+2β
j
f(x
j
) − q,j(x
j+1
− q)
≤ (1 − β
j
)
2

(α
j
x
j
− q +(1− α
j
)T (t
j
)z
j
− q)
2
+2β
j
f(x
j
) − f(q),j(x
j+1
− q) +2β
j
f(q) − q, j(x
j+1
− q)
≤ (1 − β
j
)
2
x
j
− q

2
+2β
j
f(x
j
) − f(q)x
j+1
− q +2β
j
f(q) − q, j(x
j+1
− q)
≤ (1 − β
j
)
2
x
j
− q
2
+2β
j
rx
j
− qx
j+1
− q +2β
j
f(q) − q, j(x
j+1

− q)
≤ (1 − β
j
)
2
x
j
− q
2
+ β
j
r

x
j
− q
2
+ x
j+1
− q
2

+2β
j
f(q) − q, j(x
j+1
− q)
≤

(1 − β

j
)
2
+ β
j
r

x
j
− q
2
+ β
j
rx
j+1
− q
2
+2β
j
f(q) − q, j(x
j+1
− q).
It follows that
x
j+1
− q
2
≤

1 − (2 − r)β

j
+ β
2
j

1 − β
j
r
x
j
− q
2
+
2β
j
1 − β
j
r
f(q) − q, j(x
j+1
− q)
=
1 − (2 − r)β
j
1 − β
j
r
x
j
− q

2
+
β
2
j
1 − β
j
r
x
j
− q
2
+
2β
j
1 − β
j
r
f(q) − q, j(x
j+1
− q)
≤
1 − (2 − r)β
j
1 − β
j
r
x
j
− q

2
+
β
2
j
1 − r
x
j
− q
2
+
2β
j
1 − r
f(q) − q, j(x
j+1
− q)
≤
1 − β
j
r − 2(1 − r)β
j
1 − β
j
r
x
j
− q
2
+ β

2
j
M +
2β
j
1 − r
f(q) − q, j(x
j+1
− q)
=

1 −
2(1 − r)β
j
1 − β
j
r

x
j
− q
2
+ β
2
j
M +
2β
j
1 − r
f(q) − q, j(x

j+1
− q)
≤ (1 − 2(1 − r)β
j
)x
j
− q
2
+ β
j

2
1 − r
f(q) − q, j(x
j+1
− q) + β
j
M

,
where M>0 such that M ≥
1
1−r
x
j
− q
2
.Thatis,
x
j+1

− q
2
≤ (1 − γ
j
)x
j
− q
2
+ δ
j
, (2.19)
where γ
j
= 2(1 − r)β
j
and
δ
j
γ
j
=
1
(1−r)
2
f(q) − q, j(x
j+1
− q) +
M
2(1−r)
β

j
.
It follows by condition (B1) that γ
j
−→ 0and

∞
j=1
γ
j
= ∞. From (2.18) we have
lim sup
j−→ ∞
δ
j
γ
j
≤ lim sup
j−→ ∞
1
(1 − r)
2
f(q) − q, j(x
j+1
− q) + lim
j−→ ∞
M
2(1 − r)
β
j

≤ lim sup
j−→ ∞
1
(1 − r)
2
f(q) − q, j(x
j+1
− q)≤0.
Using Lemma 1.7 onto (2.19), we conclude that x
j
−q−→0. This is a contradiction. Hence x
n
−→ q.
The proof is completed. 
If γ
n
≡ 1, then we have the following Corollary.
Corollary 2.4. Let E be a reflexive Banach space which admits a weakly sequenctially continuous
duality mapping J from E into E
∗
,supposeC is a nonempty closed convex subset of E.Let{T (t):
t ≥ 0} be a nonexpansive semigroup on C such that Fix(S) = ∅,andf : C −→ C be a generalized
contraction on C.Let{α
n
}⊂(0, 1), {β
n
}⊂(0, 1),and{t
n
}⊂(0, ∞) be sequences of real numbers
satisfying the conditions:

(C1) lim
n−→ ∞
β
n
=0,

∞
n=0
β
n
= ∞ and lim
n−→ ∞
t
n
=0,
(C2) lim
n−→ ∞
α
n
=0,
(C3)

∞
n=0
|α
n+1
− α
n
| < ∞,


∞
n=0
|β
n+1
− β
n
| < ∞.
Define a sequence {x
n
} in C by















x
0
∈ C,
y
n

= α
n
x
n
+(1− α
n
)T (t
n
)x
n
,
x
n+1
= β
n
f(x
n
)+(1− β
n
)y
n
,n≥ 0.
(2.20)
Suppose

∞
n=0
T (t
n
)x

n−1
− T (t
n−1
)x
n−1
 < ∞.Then{x
n
} converges strongly to q,asn −→ ∞ ;
where q is the unique solution in Fix(S) to the variational inequality (2.2).
Setting f is a contraction on C in Corollary 2.4, we have the following results immediately.
Corollary 2.5. [11, Theorem 3.2] Let E be a reflexive Banach space which admits a weakly sequenc-
tially continuous duality mapping J from E into E
∗
,supposeC is a nonempty closed convex subset of
E.Let{T (t):t ≥ 0} be a nonexpansive semigroup on C such that Fix(S) = ∅,andf : C −→ C be
a contraction on C.Let{α
n
}⊂(0, 1), {β
n
}⊂(0, 1),and{t
n
}⊂(0, ∞) be sequences of real numbers
satisfying the conditions:
(C1) lim
n−→ ∞
β
n
=0,

∞

n=0
β
n
= ∞ and lim
n−→ ∞
t
n
=0,
(C2) lim
n−→ ∞
α
n
=0,
(C3)

∞
n=0
|α
n+1
− α
n
| < ∞,

∞
n=0
|β
n+1
− β
n
| < ∞.

Define a sequence {x
n
} in C by















x
0
∈ C,
y
n
= α
n
x
n
+(1− α
n
)T (t

n
)x
n
,
x
n+1
= β
n
f(x
n
)+(1− β
n
)y
n
,n≥ 0.
(2.21)
Suppose

∞
n=0
T (t
n
)x
n−1
− T (t
n−1
)x
n−1
 < ∞.Then{x
n

} converges strongly to q,asn −→ ∞ ;
where q is the unique solution in Fix(S) to the variational inequality (2.2).
Questions
(i) Could we obtain Theorem 2.3 with other control conditions which are different from (C2) and
(C3)?
(ii) Could we weaken the control condition (*) by the strictly weaker condition (**):
lim
n−→ ∞
sup
x∈
˜
C
T (t
n
)x − T(t
n−1
)x =0?
The following theorem gives the affirmative answers to these question mentioned above.
Theorem 2.6. Let E be a reflexive Banach space which admits a weakly sequenctially continuous
duality mapping J from E into E
∗
,supposeC is a nonempty closed convex subset of E.Let{T (t):
t ≥ 0}, be a nonexpansive semigroup on C such that Fix(S) = ∅,andf : C −→ C be a generalized
contraction on C.Let{α
n
}⊂(0, 1), {β
n
}⊂(0, 1), {γ
n
}⊂[0, 1],and{t

n
}⊂(0, ∞) be sequences of
real numbers satisfying the conditions:
(B1) lim
n−→ ∞
β
n
=0,

∞
n=0
β
n
= ∞ and lim
n−→ ∞
t
n
=0,
(B2) α
n
+(1+α
n
)(1 − γ
n
) ∈ [0,a) for some a ∈ (0, 1),
(B3) 0 < lim inf
n−→ ∞
α
n
≤ lim sup

n−→ ∞
α
n
< 1,
(B4) lim
n−→ ∞
(γ
n+1
− γ
n
)=0.
Define a sequence {x
n
} in C by
























x
0
∈ C,
z
n
= γ
n
x
n
+(1− γ
n
)T (t
n
)x
n
,
y
n
= α
n
x
n
+(1− α

n
)T (t
n
)z
n
,
x
n+1
= β
n
f(x
n
)+(1− β
n
)y
n
,n≥ 0.
(2.22)
Suppose
lim
n−→ ∞
sup
x∈
˜
C
T (t
n
)x − T(t
n−1
)x =0, (**)

where
˜
C is any bounded subset of C.Then{x
n
} converges strongly to q,asn −→ ∞ ;whereq is the
unique solution in Fix(S) to the variational inequality (2.2).
Proof. First, we show that {x
n
} is bounded. Indeed, if we take a fixed point x ∈ Fix(S). We will prove
by induction that
x
n
− x≤M for all n ≥ 0,
where M := {x
0
− z,η
−1
(f(x) − x)}. From Definition of (2.22), notice that
z
n
− x≤γ
n
x
n
− x +(1− γ
n
)T (t
n
)x
n

− x≤x
n
− x.
It follows that
y
n
− x≤α
n
x
n
− x +(1− α
n
)T (t
n
)z
n
− x≤α
n
x
n
− x +(1− α
n
)x
n
− x≤x
n
− x.
The case n = 0 is obvious.
Suppose that x
n

− x≤M,wehave
x
n+1
− x≤β
n
f(x
n
) − x +(1− β
n
)y
n
− x
≤ β
n
f(x
n
) − f(x) + β
n
f(x) − x +(1− β
n
)y
n
− x
≤ β
n
ψ(x
n
− x)+β
n
f(x) − x +(1− β

n
)x
n
− x
= β
n
ψ(x
n
− x)+β
n
η

η
−1
(f(x) − x)

+(1− β
n
)x
n
− x
≤ β
n
ψ(M)+β
n
η(M)+(1− β
n
)M
= β
n

ψ(M)+β
n
(M − ψ(M)) + (1 − β
n
)M = M.
By induction,
x
n
− x≤max

x
0
− x,η
−1
(f(x) − x)

, ∀ n ≥ 0.
Thus {x
n
} is bounded, and so are {T (t
n
)x
n
}, {y
n
}, {z
n
} and {f(x
n
)}. As a result, we obtain by

condition (B1),
x
n+1
− y
n
 = β
n
f(x
n
) − y
n
−→0. (2.23)
We next show that
x
n
− T (t
n
)x
n
−→0. (2.24)
It suffices to show that
x
n+1
− x
n
−→0. (2.25)
Indeed, if (2.25) holds, then noting (2.23), we obtain
x
n
− T (t

n
)x
n
≤x
n
− x
n+1
 + x
n+1
− y
n
 + y
n
− T (t
n
)z
n
 + T(t
n
)z
n
− T (t
n
)x
n

≤x
n
− x
n+1

 + x
n+1
− y
n
 + α
n
x
n
− T (t
n
)z
n
 + z
n
− x
n

≤x
n
− x
n+1
 + x
n+1
− y
n
 + α
n
x
n
− T (t

n
)x
n
 + α
n
T (t
n
)x
n
− T (t
n
)z
n

+z
n
− x
n

≤x
n
− x
n+1
 + x
n+1
− y
n
 + α
n
x

n
− T (t
n
)x
n
 + α
n
x
n
− z
n
 + z
n
− x
n

≤x
n
− x
n+1
 + x
n+1
− y
n
 + α
n
x
n
− T (t
n

)x
n
 +(1+α
n
)x
n
− z
n

≤x
n
− x
n+1
 + x
n+1
− y
n
 + α
n
x
n
− T (t
n
)x
n

+(1 + α
n
)(1 − γ
n

)x
n
− T (t
n
)x
n

and hence (1 − (α
n
+(1+α
n
)(1 − γ
n
)))x
n
− T (t
n
)x
n
≤x
n
− x
n+1
 + x
n+1
− y
n
−→0asn −→ ∞ .
Using (B2), we conclude that (2.24) holds. Define the sequence {u
n

} by
u
n
=
x
n+1
− σ
n
x
n
(1 − σ
n
)
,
where σ
n
=(1− β
n
)α
n
. Then x
n+1
= σ
n
x
n
+(1− σ
n
)u
n

. Next multiplication gives us that
u
n+1
− u
n
=
(x
n+2
− σ
n+1
x
n+1
)
1 − σ
n+1
−
x
n+1
− σ
n
x
n
1 − σ
n
=
β
n+1
f(x
n+1
)+(1− β

n+1
)y
n+1
− σ
n+1
x
n+1
1 − σ
n+1
−
β
n
f(x
n
)+(1− β
n
)y
n
− σ
n
x
n
1 − σ
n
=

β
n+1
f(x
n+1

)
1 − α
n+1
−
β
n
f(x
n
)
1 − α
n

+
(1 − β
n+1
)(α
n+1
x
n+1
+(1− α
n+1
)T (t
n+1
)z
n+1
) − α
n+1
x
n+1
1 − σ

n+1
−
(1 − β
n
)(α
n
x
n
+(1− α
n
)T (t
n
)z
n
) − α
n
x
n
1 − σ
n
=

β
n+1
f(x
n+1
)
1 − σ
n+1
−

β
n
f(x
n
)
1 − σ
n

+

T (t
n+1
)z
n+1
−
α
n+1
1 − σ
n+1
T (t
n+1
)z
n+1

−

T (t
n
)z
n

−
α
n
1 − σ
n
T (t
n
)z
n

=
β
n+1
1 − σ
n+1
(f(x
n+1
) − T(t
n+1
)z
n+1
) −
β
n
1 − σ
n
(f(x
n
) − T(t
n

)z
n
)
+(T (t
n+1
)z
n+1
− T (t
n+1
)z
n
) − (T(t
n+1
)z
n
− T (t
n
)z
n
) .
Then we have
u
n+1
− u
n
≤
β
n+1
1 − σ
n+1

f(x
n+1
) − T(t
n+1
)z
n+1
−
β
n
1 − σ
n
f(x
n
) − T(t
n
)z
n

+ z
n+1
− z
n
 + T(t
n+1
)z
n
− T (t
n
)z
n

 . (2.26)
From (2.22) we have
z
n+1
− z
n
= γ
n+1
x
n+1
+(1− γ
n+1
)T (t
n+1
)x
n+1
− γ
n
x
n
− (1 − γ
n
)T (t
n
)x
n
= γ
n+1
(x
n+1

− x
n
)+(γ
n+1
− γ
n
) x
n
+(1− γ
n+1
)(T (t
n+1
)x
n+1
− T (t
n+1
)x
n
)
+(1 − γ
n+1
)(T (t
n+1
)x
n
− T (t
n
)x
n
) − (γ

n+1
− γ
n
)T (t
n
)x
n
= γ
n+1
(x
n+1
− x
n
)+(γ
n+1
− γ
n
)(x
n
− T (t
n
)x
n
)
+(1 − γ
n+1
)(T (t
n+1
)x
n+1

− T (t
n+1
)x
n
)+(1− γ
n+1
)(T (t
n+1
)x
n
− T (t
n
)x
n
) ,
that is,
z
n+1
− z
n
≤γ
n+1
x
n+1
− x
n
 + |γ
n+1
− γ
n

|x
n
− T (t
n
)x
n
 +(1− γ
n+1
) x
n+1
− x
n

+(1 − γ
n+1
) T (t
n+1
)x
n
− T (t
n
)x
n

≤x
n+1
− x
n
 + |γ
n+1

− γ
n
|x
n
− T (t
n
)x
n
 + T(t
n+1
)x
n
− T (t
n
)x
n
 . (2.27)
Substituting (2.27) into (2.26) that
u
n+1
− u
n
−x
n+1
− x
n
≤
β
n+1
1 − σ

n+1
f(x
n+1
) − T(t
n+1
)z
n+1
−
β
n
1 − σ
n
f(x
n
) − T(t
n
)z
n

+ |γ
n+1
− γ
n
|x
n
− T (t
n
)x
n
 + T(t

n+1
)x
n
− T (t
n
)x
n

+ T (t
n+1
)z
n
− T (t
n
)z
n

≤
β
n+1
1 − σ
n+1
f(x
n+1
) − T(t
n+1
)z
n+1
−
β

n
1 − σ
n
f(x
n
) − T(t
n
)z
n

+ |γ
n+1
− γ
n
|x
n
− T (t
n
)x
n
 +sup
x∈{x
n
}
T (t
n+1
)x − T(t
n
)x
+sup

z∈{z
n
}
T (t
n+1
)z − T (t
n
)z. (2.28)
By (B1), (B4), lim
n−→ ∞
sup
x∈{x
n
}
T (t
n+1
)x − T(t
n
)x =0,
lim
n−→ ∞
sup
z∈{z
n
}
T (t
n+1
)z − T (t
n
)z = 0 and (2.28), we obtain that

lim sup
n−→ ∞
(u
n+1
− u
n
−x
n+1
− x
n
) ≤ 0.
Hence by Lemma 1.8, we have lim
n−→ ∞
u
n
− x
n
 = 0. It follows from (B3) that
lim
n−→ ∞
x
n+1
− x
n
 = lim
n−→ ∞
(1 − σ
n
)u
n

− x
n
 =0.
Hence (2.24) holds. Applying Theorem 2.1, there is a unique solution q ∈ Fix(S) to the following
variational inequality:
f(q) − q, j(x − q)≤0 for all x ∈ Fix(S).
Next, we show that
lim sup
n−→ ∞
f(q) − q, j(x
n+1
− q)≤0. (2.29)
Indeed, we can take a subsequence {x
n
i
} of {x
n
} such that
lim sup
n−→ ∞
f(q) − q, j(x
n+1
− q) = lim
i−→ ∞
f(q) − q, j(x
n
i
+1
− q).
By the reflexivity of E and boundedness of the sequence {x

n
}, we may assume, without loss of gener-
ality, that x
n
i
pfor some p ∈ C. Now we show that p ∈ Fix(S). Put x
i
= x
n
i
,α
i
= α
n
i
,β
i
= β
n
i
and t
i
= t
n
i
for i ∈ N,lett
i
≥ 0 be such that
t
i

−→ 0and
T (t
i
)x
i
− x
i

t
i
−→ 0,i−→ ∞ .
Fix t>0. Notice that
x
i
− T (t)p≤
[t/t
i
]−1

k=0
T ((k +1)t
i
)x
i
− T (kt
i
)x
i

+T ([t/t

i
]t
i
)x
i
− T ([t/t
i
]t
i
)p + T([t/t
i
]t
i
)p − T(t)p
≤ [t/t
i
]T (t
i
)x
i
− x
i
 + x
i
− p + T (t − [t/t
i
]t
i
)p − p
≤ t

T (t
i
)x
i
− x
i

t
i
+ x
i
− p + T (t − [t/t
i
]t
i
)p − p
≤ t
T (t
i
)x
i
− x
i

t
i
+ x
i
− p + max {T (s)p − p :0≤ s ≤ t
i

}.
For all i ∈ N,wehave
lim sup
i−→ ∞
x
i
− T (t)p≤lim sup
i−→ ∞
x
i
− p.
Since Banach space E with a weakly sequentially continuous duality mapping satisfies Opial’s condition,
this implies T (t)p = p. Therefore p ∈ Fix(S). In view of the variational inequality (2.2) and the
assumption that duality mapping J is weakly sequentially continuous, we conclude
lim sup
n−→ ∞
f(q) − q, j(x
n+1
− q) = lim
i−→ ∞
f(q) − q, j(x
n
i
+1
− q)
= f(q) − q,j(p − q)≤0.
Then (2.29) is proved. Finally, show that x
n
−→ q, i.e., x
n

− q−→0. Suppose that x
n
− q  0,
then there exists ε>0 and a subsequence {x
n
j
} of {x
n
} such that x
n
j
− q≥ε for all j ∈ N.Put
x
j
= x
n
j
,α
j
= α
n
j
,β
j
= β
n
j
and t
j
= t

n
j
for j ∈ N. By Proposition 1.4, there exists r ∈ (0, 1) such
that f(x
j
) − f(q)≤rx
j
− q for all j ∈ N. As a matter of fact, from 1.6 we have that
x
j+1
− q
2
= β
j
f(x
j
)+(1− β
j
)(α
j
x
j
+(1− α
j
)T (t
j
)z
j
) − q
2

= (1 − β
j
)(α
j
(x
j
− q)+(1− α
j
)(T (t
j
)z
j
− q)) + β
j
(f(x
j
) − q)
2
≤ (1 − β
j
)
2
α
j
(x
j
− q)+(1− α
j
)(T (t
j

)z
j
− q)
2
+2β
j
f(x
j
) − q,j(x
j+1
− q)
≤ (1 − β
j
)
2
(α
j
x
j
− q +(1− α
j
)T (t
j
)z
j
− q)
2
+2β
j
f(x

j
) − f(q),j(x
j+1
− q) +2β
j
f(q) − q, j(x
j+1
− q)
≤ (1 − β
j
)
2
x
j
− q
2
+2β
j
f(x
j
) − f(q)x
j+1
− q +2β
j
f(q) − q, j(x
j+1
− q)
≤ (1 − β
j
)

2
x
j
− q
2
+2β
j
rx
j
− qx
j+1
− q +2β
j
f(q) − q, j(x
j+1
− q)
≤ (1 − β
j
)
2
x
j
− q
2
+ β
j
r(x
j
− q
2

+ x
j+1
− q
2
)+2β
j
f(q) − q, j(x
j+1
− q)
≤ ((1 − β
j
)
2
+ β
j
r)x
j
− q
2
+ β
j
rx
j+1
− q
2
+2β
j
f(q) − q, j(x
j+1
− q).

It follows that
x
j+1
− q
2
≤

1 − (2 − r)β
j
+ β
2
j

1 − β
j
r
x
j
− q
2
+
2β
j
1 − β
j
r
f(q) − q, j(x
j+1
− q)
=

1 − (2 − r)β
j
1 − β
j
r
x
j
− q
2
+
β
2
j
1 − β
j
r
x
j
− q
2
+
2β
j
1 − β
j
r
f(q) − q, j(x
j+1
− q)
≤

1 − (2 − r)β
j
1 − β
j
r
x
j
− q
2
+
β
2
j
1 − r
x
j
− q
2
+
2β
j
1 − r
f(q) − q, j(x
j+1
− q)
≤
1 − β
j
r − 2(1 − r)β
j

1 − β
j
r
x
j
− q
2
+ β
2
j
M +
2β
j
1 − r
f(q) − q, j(x
j+1
− q)
=

1 −
2(1 − r)β
j
1 − β
j
r

x
j
− q
2

+ β
2
j
M +
2β
j
1 − r
f(q) − q, j(x
j+1
− q)
≤ (1 − 2(1 − r)β
j
)x
j
− q
2
+ β
j

2
1 − r
f(q) − q, j(x
j+1
− q) + β
j
M

,
where M>0 such that M ≥
1

1−r
x
j
− q
2
.Thatis,
x
j+1
− q
2
≤ (1 − γ
j
)x
j
− q
2
+ δ
j
, (2.30)
where γ
j
= 2(1 − r)β
j
and
δ
j
γ
j
=
1

(1−r)
2
f(q) − q, j(x
j+1
− q) +
M
2(1−r)
β
j
.