1 Introduction
Let E be a Banach space, and let C be a nonempty closed convex subset of E. A mapping T: C → E is nonexpansive if ||Tx - Ty|| ≤ ||x - y|| for every x, y ∈ C. Let F(T): = {x ∈ C: x = Tx} denote the set of fixed points of T. Besides, a mapping T: C → E is quasinonexpansive if
F
(
T
)
≠
∅
and ||Tx - y|| ≤ ||x - y|| for all x ∈ C and y ∈ F(T).
In 2008, Suzuki
1
introduced the following generalized nonexpansive mapping on Banach spaces. A mapping T: C → E is said to satisfy condition (C) if for all x, y ∈ C,
1
2
|
|
x
-
T
x
|
|
≤
|
|
x
-
y
|
|
⇒
|
|
T
x
-
T
y
|
|
≤
|
|
x
-
y
|
|
.
In fact, every nonexpansive mapping satisfies condition (C), but the converse may be false [1, Example 1]. Besides, if T: C → E satisfies condition (C) and
F
(
T
)
≠
∅
, then T is a quasinonexpansive mapping. However, the converse may be false [1, Example 2].
Construction of approximating fixed points of nonlinear mappings is an important subject in the theory of nonlinear mappings and its applications in a number of applied areas.
Let C be a nonempty closed convex subset of a real Hilbert space H, and let T: C → C be a mapping. In 1953, Mann
2
gave an iteration process:
x
n
+
1
=
α
n
x
n
+
(
1
-
α
n
)
T
x
n
,
n
≥
0
,
where x
0 is taken in C arbitrarily, and {α
n
} is a sequence in [0,1].
In 2001, Soltuz
3
introduced the following Mann-type implicit process for a nonexpansive mapping T: C → C:
x
n
=
α
n
x
n
-
1
+
(
1
-
α
n
)
T
x
n
,
n
∈
ℕ
,
where x
0 is taken in C arbitrarily, and {t
n
} is a sequence in [0,1].
In 2001, Xu and Ori
4
have introduced an implicit iteration process for a finite family of nonexpansive mappings. Let T
1, T
2, ..., T
N
be N self-mappings of C and suppose that
F
:
=
∩
i
=
1
N
F
(
T
i
)
≠
∅
, the set of common fixed points of T
i
, i = 1, 2, ..., N. Let I: = {1, 2, ..., N}. Xu and Ori
4
gave an implicit iteration process for a finite family of nonexpansive mappings:
x
n
=
t
n
x
n
-
1
+
(
1
-
t
n
)
T
n
x
n
,
n
∈
ℕ
,
where x
0 is taken in C arbitrarily, {t
n
} is a sequence in [0,1], and T
k
= T
k mod N
. (Here the mod N function takes values in I.) And they proved the weak convergence of process (1.3) to a common fixed point in the setting of a Hilbert space.
In 2010, Khan et al.
5
presented an implicit iterative process for two nonexpansive mappings in Banach spaces. Let E be a Banach space, and let C be a nonempty closed convex subset of E, and let T, S: C → C be two nonexpansive mappings. Khan et al.
5
considered the following implicit iterative process:
x
n
=
α
n
x
n
-
1
+
β
n
S
x
n
+
γ
n
T
x
n
,
n
∈
ℕ
,
where {α
n
}, {β
n
}, and {γ
n
} are sequences in [0,1] with α
n
+ β
n
+ γ
n
= 1.
Motivated by the above works in
5
, we want to consider the following implicit iterative process. Let E be a Banach space, C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings, and let S: C → C be a mapping which satisfy condition (C). We first consider the weak convergence theorems for the following implicit iterative process:
x
0
∈
C
chosen arbitrary
,
x
n
=
a
n
x
n
-
1
+
b
n
S
x
n
-
1
+
c
n
T
1
x
n
+
d
n
T
2
x
n
,
where {a
n
}, {b
n
}, {c
n
}, and {d
n
} are sequences in [0,1] with a
n
+ b
n
+ c
n
+ d
n
= 1.
Next, we also consider weak convergence theorems for another implicit iterative process:
x
0
∈
C
chosen arbitrary
,
y
n
=
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
,
x
n
=
d
n
y
n
+
(
1
-
d
n
)
S
y
n
,
where {a
n
}, {b
n
}, {c
n
}, and {d
n
} are sequences in [0,1] with a
n
+ b
n
+ c
n
= 1.
In fact, for the above implicit iterative processes, most researchers always considered weak convergence theorems, and few researchers considered strong convergence theorem under suitable conditions. For example, one can see
5
6
7
. However, some conditions are not natural. For this reason, we consider the following shrinking-implicit iterative processes and study the strong convergence theorem. Let {x
n
} be defined by
x
0
∈
C
chosen arbitrary and
C
0
=
D
0
=
C
,
y
n
=
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
,
z
n
=
J
-
1
(
d
n
J
y
n
+
(
1
-
d
n
)
J
S
y
n
)
,
C
n
=
{
z
∈
C
n
-
1
:
ϕ
(
z
,
z
n
)
≤
ϕ
(
z
,
y
n
)
}
,
D
n
=
{
z
∈
D
n
-
1
:
|
|
y
n
-
z
|
|
≤
|
|
x
n
-
1
-
z
|
|
}
,
x
n
=
Π
C
n
∩
D
n
x
0
,
where {a
n
}, {b
n
}, {c
n
}, and {d
n
} are sequences in (0, 1) with a
n
+ b
n
+ c
n
= 1.
In this article, we first consider weak convergence theorems of implicit iterative processes for two nonexpansive mappings and a mapping which satisfy condition (C). And we generalize Khan et al.'s result
5
as special case. Next, we consider strong convergence theorem of an implicit-shrinking iterative process for two non-expansive mappings and a relative nonexpansive mapping on Banach spaces. Note that the conditions of strong convergence theorem are different from the strong convergence theorems for the implicit iterative processes in the literatures. Finally, we discuss a strong convergence theorem concerning two nonexpansive mappings and the resolvent of a maximal monotone operator in a Banach space.
2 Preliminaries
Throughout this article, let ℕ and ℝ be the sets of all positive integers and real numbers, respectively. Let E be a Banach space and let E* be the dual space of E. For a sequence {x
n
} of E and a point x ∈ E, the weak convergence of {x
n
} to x and the strong convergence of {x
n
} to x are denoted by x
n
⇀ x and x
n
→ x, respectively.
A Banach space E is said to satisfy Opial's condition if {x
n
} is a sequence in E with x
n
⇀ x, then
limsup
n
→
∞
|
|
x
n
-
x
|
|
<
limsup
n
→
∞
|
|
x
n
-
y
|
|
,
∀
y
∈
E
,
y
≠
x
.
Let E be a Banach space. Then, the duality mapping
J
:
E
⊸
E
*
is defined by
J
x
:
x
*
∈
E
*
:
x
,
x
*
=
|
|
x
|
|
2
=
|
|
x
*
|
|
2
,
∀
x
∈
E
.
Let S(E) be the unit sphere centered at the origin of E. Then, the space E is said to be smooth if the limit
lim
t
→
0
|
|
x
+
t
y
|
|
-
|
|
x
|
|
t
exists for all x, y ∈ S(E). It is also said to be uniformly smooth if the limit exists uniformly in x, y ∈ S(E). A Banach space E is said to be strictly convex if
x
+
y
2
<
1
whenever x, y ∈ S(E) and x ≠ y. It is said to be uniformly convex if for each ε ∈ (0, 2], there exists δ > 0 such that
x
+
y
2
<
1
-
δ
whenever x, y ∈ S(E) and ||x - y|| ≥ ε. Furthermore, we know that
8
(i) if E in smooth, then J is single-valued;
(ii) if E is reflexive, then J is onto;
(iii) if E is strictly convex, then J is one-to-one;
(iv) if E is strictly convex, then J is strictly monotone;
(v) if E is uniformly smooth, then J is uniformly norm-to-norm continuous on each bounded subset of E.
A Banach space E is said to have Kadec-Klee property if a sequence {x
n
} of E satisfying that x
n
⇀ x and ||x
n
|| → ||x||, then x
n
→ x. It is known that if E uniformly convex, then E has the Kadec-Klee property
8
.
Let E be a smooth, strictly convex and reflexive Banach space and let C be a nonempty closed convex subset of E. Throughout this article, define the function ϕ: C × C → ℝ by
ϕ
(
x
,
y
)
:
=
|
|
x
|
|
2
-
2
x
,
J
y
+
|
|
y
|
|
2
,
∀
x
,
y
∈
E
.
Observe that, in a Hilbert space H, ϕ(x, y) = ||x - y||2 for all x, y ∈ H. Furthermore, for each x, y, z, w ∈ E, we know that:
(1) (||x|| - ||y||)2 ≤ ϕ(x, y) ≤ (||x|| + ||y||)2;
(2) ϕ(x, y) ≥ 0;
(3) ϕ(x, y) = ϕ(x, z) + ϕ(z, y) + 2〈x - z, Jz - Jy〉;
(4) 2〈x - y, Jz - Jw〉 = ϕ(x, w) + ϕ(y, z) - ϕ(x, z) - ϕ(y, w);
(5) if E is additionally assumed to be strictly convex, then
ϕ
(
x
,
y
)
=
0
if
and
only
if
x
=
y
;
(6) ϕ(x, J
-1(λJy + (1 - λ)Jz)) ≤ λϕ(x, y) + (1 - λ)ϕ(x, z).
Lemma 2.1.
9
Let E be a uniformly convex Banach space and let r > 0. Then, there exists a strictly increasing, continuous, and convex function g: [0, 2r] → [0, ∞) such that g(0) = 0 and
|
|
a
x
+
b
y
+
c
z
+
d
w
|
|
2
≤
a
|
|
x
|
|
2
+
b
|
|
y
|
|
2
+
c
|
|
z
|
|
2
+
d
|
|
w
|
|
2
-
a
b
g
(
|
|
x
-
y
|
|
)
for all x, y, z, w ∈ B
r
and a, b, c, d ∈ [0,1] with a + b + c + d = 1, where B
r
: = {z ∈ E: ||z|| ≤ r}.
Lemma 2.2.
10
Let E be a uniformly convex Banach space and let r > 0. Then, there exists a strictly increasing, continuous, and convex function g: [0, 2r] → [0, ∞) such that g(0) = 0 and
ϕ
(
x
,
J
-
1
(
λ
J
y
+
(
1
-
λ
)
J
z
)
)
≤
λ
ϕ
(
x
,
y
)
+
(
1
-
λ
)
ϕ
(
x
,
z
)
-
λ
(
1
-
λ
)
g
(
|
|
J
y
-
J
z
|
|
)
for all x, y, z ∈ B
r
and λ ∈ [0,1], where B
r
: = {z ∈ E: ||z|| ≤ r}.
Lemma 2.3.
11
Let E be a uniformly convex Banach space, let {α
n
} be a sequence of real numbers such that 0 < b ≤ α
n
≤ c < 1 for all n ∈ ℕ, and let {x
n
} and {y
n
} be sequences of E such that lim sup
n→∞ ||x
n
|| ≤ a, lim sup
n→∞ ||y
n
|| ≤ a, and lim
n→∞ ||α
n
x
n
+ (1 - α
n
)y
n
|| = a for some a ≥ 0. Then, lim
n→∞ ||x
n
- y
n
|| = 0.
Lemma 2.4.
12
Let E be a smooth and uniformly convex Banach space, and let {x
n
} and {y
n
} be sequences in E such that either {x
n
} or {y
n
} is bounded. If lim
n→∞
ϕ(x
n
, y
n
) = 0, then lim
n→∞ ||x
n
- y
n
|| = 0.
Remark 2.1.
13
Let E be a uniformly convex and uniformly smooth Banach space. If {x
n
} and {y
n
} are bounded sequences in E, then
lim
n
→
∞
ϕ
(
x
n
,
y
n
)
=
0
⇔
lim
n
→
∞
|
|
x
n
-
y
n
|
|
=
0
⇔
lim
n
→
∞
|
|
J
x
n
-
J
y
n
|
|
=
0
.
Let C be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space E. For an arbitrary point x of E, the set
z
∈
C
:
ϕ
(
z
,
x
)
=
min
y
∈
C
ϕ
(
y
,
x
)
is always nonempty and a singleton
14
. Let us define the mapping Π
C
from E onto C by Π
C
x = z, that is,
ϕ
(
Π
C
x
,
x
)
=
min
y
∈
C
ϕ
(
y
,
x
)
for every x ∈ E. Such Π
C
is called the generalized projection from E onto C
14
.
Lemma 2.5.
14
15
Let C be a nonempty closed convex subset of a smooth, strictly convex, and reflexive Banach space E, and let (x, z) ∈ E × C. Then:
(i) z = Π
C
x if and only if 〈y - z, Jx - Jz〉 ≤ 0 for all y ∈ C;
(ii) ϕ(z, Π
C
x) + ϕ(Π
C
x, x) ≤ ϕ(z, x).
Lemma 2.6.
16
Let E be a uniformly convex Banach space, C be a nonempty closed convex subset of E and T: C → C is a nonexpansive mapping. Let {x
n
} be a sequence in C with x
n
⇀ x ∈ C and lim
n→∞ ||x
n
- Tx
n
|| = 0. Then, Tx = x.
Lemma 2.7.
1
Let C be a nonempty subset of a Banach space E with the Opial property. Assume that T: C → E satisfies condition (C). Let {x
n
} be a sequence in C with x
n
⇀ x ∈ C and lim
n→∞ ||x
n
- Tx
n
|| = 0. Then, Tx = x.
Lemma 2.8.
1
Let T be a mapping on a closed subset C of a Banach space E. Assume that T satisfies condition (C). Then, F(T) is a closed set. Moreover, if E is strictly convex and C is convex, then F(T) is also convex.
Lemma 2.9.
17
Let C be a nonempty closed convex subset of a strictly convex Banach space E, and T: C → C be a nonexpansive mapping. Then, F(T) is a closed convex subset of C.
3 Weak convergence theorems
Lemma 3.1. Let E be a uniformly convex Banach space, C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings, and let S: C → C be a mapping with condition (C). Let {a
n
}, {b
n
}, {c
n
}, and {d
n
} be sequences with 0 < a ≤ a
n
, b
n
, c
n
, d
n
≤ b < 1 and a
n
+ b
n
+ c
n
+ d
n
= 1. Suppose that
Ω
:
=
F
(
S
)
∩
F
(
T
1
)
∩
F
(
T
2
)
≠
∅
. Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary,
x
n
=
a
n
x
n
-
1
+
b
n
S
x
n
-
1
+
c
n
T
1
x
n
+
d
n
T
2
x
n
.
Then, we have:
(i)
lim
n
→
∞
|
|
x
n
-
p
|
|
exists for each p ∈ Ω.
(ii)
lim
n
→
∞
|
|
x
n
-
S
x
n
|
|
=
lim
n
→
∞
|
|
x
n
-
T
1
x
n
|
|
=
lim
n
→
∞
|
|
x
n
-
T
2
x
n
|
|
=
0
.
Proof. First, we show that {x
n
} is well-defined. Now, let f(x): = a
1
x
0+b
1
Sx
0+c
1
T
1
x+d
1
T
2
x. Then,
|
|
f
(
x
)
-
f
(
y
)
|
|
≤
c
1
|
|
T
1
x
-
T
1
y
|
|
+
d
1
|
|
T
2
x
-
T
2
y
|
|
≤
(
c
1
+
d
1
)
|
|
x
-
y
|
|
≤
(
1
-
2
a
)
|
|
x
-
y
|
|
.
By Banach contraction principle, the existence of x
1 is established. Similarly, the existence of {x
n
} is well-defined.
(i) For each p ∈ Ω and n ∈ ℕ, we have:
|
|
x
n
-
p
|
|
≤
a
n
|
|
x
n
-
1
-
p
|
|
+
b
n
|
|
S
x
n
-
1
-
p
|
|
+
c
n
|
|
T
1
x
n
-
p
|
|
+
d
n
|
|
T
2
x
n
-
p
|
|
≤
a
n
|
|
x
n
-
1
-
p
|
|
+
b
n
|
|
x
n
-
1
-
p
|
|
+
(
c
n
+
d
n
)
|
|
x
n
-
p
|
|
.
This implies that (1 - c
n
- d
n
)||x
n
- p|| ≤ (a
n
+ b
n
)||x
n-1-p||. Hence, ||x
n
-p||≤ ||x
n-1-p||, lim
n →∞ ||x
n
-p|| exists, and {x
n
} is a bounded sequence.
(ii) Take any p ∈ Ω and let p be fixed. Suppose that
lim
n
→
∞
|
|
x
n
-
p
|
|
=
d
. Clearly,
limsup
n
→
∞
|
|
T
2
x
n
-
p
|
|
≤
d
, and we have:
lim
n
→
∞
|
|
x
n
-
p
|
|
=
lim
n
→
∞
|
|
a
n
x
n
-
1
+
b
n
S
x
n
-
1
+
c
n
T
1
x
n
+
d
n
T
2
x
n
-
p
|
|
=
lim
n
→
∞
(
1
-
d
n
)
a
n
1
-
d
n
(
x
n
-
1
-
p
)
+
b
n
1
-
d
n
(
S
x
n
-
1
-
p
)
+
c
n
1
-
d
n
(
T
1
x
n
-
p
)
+
d
n
(
T
2
x
n
-
p
)
.
Besides,
limsup
n
→
∞
a
n
1
-
d
n
(
x
n
-
1
-
p
)
+
b
n
1
-
d
n
(
S
x
n
-
1
-
p
)
+
c
n
1
-
d
n
(
T
1
x
n
-
p
)
≤
limsup
n
→
∞
a
n
1
-
d
n
|
|
x
n
-
1
-
p
|
|
+
b
n
1
-
d
n
|
|
S
x
n
-
1
-
p
|
|
+
c
n
1
-
d
n
|
|
T
1
x
n
-
p
|
|
≤
limsup
n
→
∞
a
n
1
-
d
n
|
|
x
n
-
1
-
p
|
|
+
b
n
1
-
d
n
|
|
S
x
n
-
1
-
p
|
|
+
c
n
1
-
d
n
|
|
T
1
x
n
-
p
|
|
≤
limsup
n
→
∞
a
n
+
b
n
1
-
d
n
|
|
x
n
-
1
-
p
|
|
+
c
n
1
-
d
n
|
|
x
n
-
p
|
|
≤
limsup
n
→
∞
a
n
+
b
n
+
c
n
1
-
d
n
|
|
x
n
-
1
-
p
|
|
=
d
.
By Lemma 2.3,
lim
n
→
∞
a
n
1
-
d
n
(
x
n
-
1
-
p
)
+
b
n
1
-
d
n
(
S
x
n
-
1
-
p
)
+
c
n
1
-
d
n
(
T
1
x
n
-
p
)
-
(
T
2
x
n
-
p
)
=
0
.
This implies that lim
n→∞ ||x
n
- T
2
x
n
|| = 0. Similarly, lim
n→∞ ||x
n
- T
1
x
n
|| = 0.
Since {x
n
} is bounded, there exists r > 0 such that 2 sup{||x
n
-p||:n ∈ ℕ}≤ r.
By Lemma 2.1, there exists a strictly increasing, continuous, and convex function g: [0, 2r] → [0, ∞) such that g(0) = 0 and
|
|
x
n
-
p
|
|
2
≤
a
n
|
|
x
n
-
1
-
p
|
|
2
+
b
n
|
|
S
x
n
-
1
-
p
|
|
2
+
c
n
|
|
T
1
x
n
-
p
|
|
2
+
d
n
|
|
T
2
x
n
-
p
|
|
2
-
a
n
b
n
g
(
|
|
x
n
-
1
-
S
x
n
-
1
|
|
)
≤
(
a
n
+
b
n
)
|
|
x
n
-
1
-
p
|
|
2
+
(
c
n
+
d
n
)
|
|
x
n
-
p
|
|
2
-
a
n
b
n
g
(
|
|
x
n
-
1
-
S
x
n
-
1
|
|
)
.
This implies that
a
n
b
n
g
(
|
|
x
n
-
1
-
S
x
n
-
1
|
|
)
≤
(
a
n
+
b
2
)
(
|
|
x
n
-
1
-
p
|
|
2
-
|
|
x
n
-
p
|
|
2
)
.
By the properties of g and lim
n→∞ ||x
n
- p|| = d, we get lim
n→∞ ||x
n
- Sx
n
|| = 0.
Theorem 3.1. Let E be a uniformly convex Banach space with Opial's condition, C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings, and let S: C → C be a mapping with condition (C). Let {a
n
}, {b
n
}, {c
n
}, and {d
n
} be sequences with 0 < a ≤ a
n
, b
n
, c
n
, d
n
≤ b < 1 and a
n
+ b
n
+ c
n
+ d
n
= 1. Suppose that
Ω
:
=
F
(
S
)
∩
F
(
T
1
)
∩
F
(
T
2
)
≠
∅
. Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary
,
x
n
=
a
n
x
n
-
1
+
b
n
S
x
n
-
1
+
c
n
T
1
x
n
+
d
n
T
2
x
n
.
Then, x
n
⇀ z for some z ∈ Ω.
Proof. By Lemma 3.1, {x
n
} is a bounded sequence. Then, there exists a subsequence
{
x
n
k
}
of {x
n
} and z ∈ C such that
x
n
k
⇀
z
. By Lemmas 2.6, 2.7, and 3.1, we know that z ∈ Ω. Since E has Opial's condition, it is easy to see that x
n
⇀ z.
Hence, the proof is completed.
Remark 3.1. The conclusion of Theorem 3.1 is still true if S: C → C is a quasi-nonexpansive mapping, and I - S is demiclosed at zero, that is, x
n
⇀ x and (I - S)x
n
⇀ 0 implies that (I - S)x = 0.
In Theorem 3.1, if S = I, then we get the following result. Hence, Theorem 3.1 generalizes Theorem 4 in
5
.
Corollary 3.1.
5
Let E be a uniformly convex Banach space with Opial's condition, C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings. Let {a
n
}, {b
n
}, and {c
n
} be sequences with 0 < a ≤ a
n
, b
n
, c
n
≤ b < 1 and a
n
+ b
n
+ c
n
= 1. Suppose that
Ω
:
=
F
(
T
1
)
∩
F
(
T
2
)
≠
∅
.
Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary
,
x
n
=
a
n
x
n
-
1
+
b
n
T
1
x
n
+
c
n
T
2
x
n
.
Then, x
n
⇀ z for some z ∈ Ω.
Besides, it is easy to get the following result from Theorem 3.1.
Corollary 3.2. Let E be a uniformly convex Banach space with Opial's condition, C be a nonempty closed convex subset of E, and let S: C → C be a mapping with condition (C). Let {a
n
} be a sequence with 0 < a ≤a
n
≤b < 1. Suppose that
F
(
S
)
≠
0̸
. Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary
,
x
n
=
a
n
x
n
-
1
+
(
1
-
a
n
)
S
x
n
-
1
.
Then, x
n
⇀ z for some z ∈ F(S).
Proof. Let T
1 = T
2 = I, where I is the identity mapping. For each n ∈ ℕ, we know that
x
n
=
a
n
2
x
n
-
1
+
1
-
a
n
2
S
x
n
-
1
+
1
4
T
1
x
n
+
1
4
T
2
x
n
.
By Theorem 3.1, it is easy to get the conclusion.
Theorem 3.2. Let E be a uniformly convex Banach space with Opial's condition, C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings, and let S: C → C be a mapping with condition (C). Let {a
n
}, {b
n
}, {c
n
}, and {d
n
} be sequences with 0 < a ≤ a
n
, b
n
, c
n
, d
n
≤ b < 1 and a
n
+ b
n
+ c
n
= 1. Suppose that
Ω
:
=
F
(
S
)
∩
F
(
T
1
)
∩
F
(
T
2
)
≠
∅
. Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary,
y
n
=
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
,
x
n
=
d
n
y
n
+
(
1
-
d
n
)
S
y
n
.
Then, x
n
⇀ z for some z ∈ Ω.
Proof. Following the same argument as in Lemma 3.1, we know that {y
n
} is well-defined. Take any w ∈ Ω and let w be fixed. Then, for each n ∈ ℕ, we have
|
|
y
n
-
w
|
|
=
|
|
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
-
w
|
|
≤
a
n
|
|
x
n
-
1
-
w
|
|
+
b
n
|
|
T
1
y
n
-
w
|
|
+
c
n
|
|
T
2
y
n
-
w
|
|
≤
a
n
|
|
x
n
-
1
-
w
|
|
+
(
b
n
+
c
n
)
|
|
y
n
-
w
|
|
.
This implies that ||y
n
- w|| ≤ ||x
n-1 - w|| for each n ∈ ℕ. Besides, we also have
|
|
x
n
-
w
|
|
=
|
|
d
n
y
n
+
(
1
-
d
n
)
S
y
n
-
w
|
|
≤
d
n
|
|
y
n
-
w
|
|
+
(
1
-
d
n
)
|
|
S
y
n
-
w
|
|
≤
|
|
y
n
-
w
|
|
.
Hence, ||x
n
- w|| ≤ ||y
n
- w|| ≤ ||x
n-1 - w|| for each n ∈ ℕ. So, lim
n→∞ ||x
n
- w|| and lim
n→∞||y
n
- w|| exist, and {x
n
}, {y
n
} are bounded sequences.
Suppose that lim
n→∞ ||x
n
-w|| = lim
n→∞||y
n
-w|| = d. Clearly, lim sup
n→∞ ||T
2
y
n
-w|| ≤ d, and we have
lim
n
→
∞
|
|
y
n
-
w
|
|
=
lim
n
→
∞
|
|
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
-
w
|
|
=
lim
n
→
∞
(
1
-
c
n
)
a
n
1
-
c
n
(
x
n
-
1
-
w
)
+
b
n
1
-
c
n
(
T
1
y
n
-
w
)
+
c
n
(
T
2
y
n
-
w
)
.
Besides,
limsup
n
→
∞
a
n
1
-
c
n
(
x
n
-
1
-
w
)
+
b
n
1
-
c
n
(
T
1
y
n
-
w
)
≤
limsup
n
→
∞
a
n
1
-
c
n
|
|
x
n
-
1
-
w
|
|
+
b
n
1
-
c
n
|
|
T
1
y
n
-
w
|
|
≤
limsup
n
→
∞
a
n
1
-
c
n
|
|
x
n
-
1
-
w
|
|
+
b
n
1
-
c
n
|
|
y
n
-
w
|
|
≤
limsup
n
→
∞
|
|
x
n
-
1
-
w
|
|
=
d
.
By Lemma 2.3,
lim
n
→
∞
a
n
1
-
c
n
(
x
n
-
1
-
w
)
+
b
n
1
-
c
n
(
T
1
y
n
-
w
)
-
(
T
2
y
n
-
w
)
=
0
.
This implies that lim
n→∞ ||y
n
- T
2
y
n
|| = 0. Similarly, lim
n→∞ ||y
n
- T
1
y
n
|| = 0.
Since {x
n
} and {y
n
} are bounded sequences, there exists r > 0 such that
2
sup
{
|
|
x
n
|
|
,
|
|
y
n
|
|
,
|
|
x
n
-
w
|
|
,
|
|
y
n
-
w
|
|
:
n
∈
ℕ
}
≤
r
.
By Lemma 2.1, there exists a strictly increasing, continuous, and convex function g: [0, 2r] → [0, ∞) such that g(0) = 0 and
|
|
d
n
y
n
+
(
1
-
d
n
)
S
y
n
-
w
|
|
2
≤
d
n
|
|
y
n
-
w
|
|
2
+
(
1
-
d
n
)
|
|
S
y
n
-
w
|
|
2
-
d
n
(
1
-
d
n
)
g
(
|
|
y
n
-
S
y
n
|
|
)
.
This implies that
d
n
(
1
-
d
n
)
g
(
|
|
y
n
-
S
y
n
|
|
)
≤
|
|
y
n
-
w
|
|
2
-
|
|
x
n
-
w
|
|
2
.
Since lim
n→∞ ||x
n
- w|| = lim
n→∞ ||y
n
- w|| = d, and the properties of g, we get lim
n→∞ ||y
n
- Sy
n
|| = 0. Besides,
|
|
x
n
-
y
n
|
|
=
|
|
d
n
y
n
+
(
1
-
d
n
)
S
y
n
-
y
n
|
|
=
(
1
-
d
n
)
|
|
y
n
-
S
y
n
|
|
.
Hence, lim
n→∞ ||x
n
-y
n
|| = 0. Finally, following the same argument as in the proof of Theorem 3.1, we know that x
n
⇀ z for some z ∈ Ω.
Next, we give the following examples for Theorems 3.1 and 3.2.
Example 3.1. Let E = ℝ, C: = [0,3], T
1
x = T
2
x = x, and let S: C → C be the same as in
1
:
S
x
:
=
0
if
x
≠
3
,
1
if
x
=
3
.
For each n, let
a
n
=
b
n
=
c
n
=
d
n
=
1
4
. Let x
0 = 1. Then, for the sequence {x
n
}, in Theorem 3.1, we know that
x
n
=
1
2
n
for all n ∈ ℕ, and x
n
→ 0, and 0 is a common fixed point of S, T
1, and T
2.
Example 3.2. Let E, C, T
1
, T
2, S be the same as in Example 3.1. For each n, let
a
n
=
b
n
=
c
n
=
1
3
, and
d
n
=
1
2
. Let x
0 = 1. Then, for the sequence {x
n
} in Theorem 3.1, we know that
x
n
=
1
2
n
for all n ∈ ℕ, and x
n
→ 0, and 0 is a common fixed point of S, T
1, and T
2.
Example 3.3. Let E, C, {a
n
}, {b
n
}, {c
n
}, {d
n
}, and let S: C → C be the same as in Example 3.1. Let T
1
x = T
2
x = 0 for each x ∈ C. Then, for the sequence {x
n
} in Theorem 3.1, we know that
x
n
=
1
4
n
for all n ∈ ℕ.
Example 3.4. Let E, C, {a
n
}, {b
n
}, {c
n
}, {d
n
}, and let S: C → C be the same as in Example 3.2. Let T
1
x = T
2
x = 0 for each x ∈ C. Then, for the sequence {x
n
} in Theorem 3.2, we know that
x
n
=
1
6
n
for all n∈ ℕ.
Remark 3.2.
(i) For the rate of convergence, by Examples 3.3 and 3.4, we know that the iteration process in Theorem 3.2 may be faster than the iteration process in Theorem 3.1. But, the times of iteration process for Theorem 3.2 is much than ones in Theorem 3.1.
(ii) The conclusion of Theorem 3.2 is still true if S: C → C is a quasi-nonexpansive mapping, and I - S is demiclosed at zero, that is, x
n
⇀ x and (I - S)x
n
→ 0 implies that (I - S)x = 0.
(iii) Corollaries 3.1 and 3.2 are special cases of Theorem 3.2.
Definition 3.1.
18
Let C be a nonempty subset of a Banach space E. A mapping T:C → E satisfy condition (E) if there exists μ ≥ 1 such that for all x, y ∈ C,
|
|
x
-
T
y
|
|
≤
μ
|
|
x
-
T
x
|
|
+
|
|
x
-
y
|
|
.
By Lemma 7 in
1
, we know that if T satisfies condition (C), then T satisfies condition (E). But, the converse may be false [18, Example 1]. Furthermore, we also observe the following result.
Lemma 3.2.
18
Let C be a nonempty subset of a Banach space E. Let T: C → E be a mapping. Assume that:
(i)
lim
n
→
∞
|
|
x
n
-
T
x
n
|
|
=
0
and x
n
⇀ x;
(ii) T satisfies condition (E);
(iii) E has Opial condition.
Then, Tx = x.
By Lemma 3.2, if S satisfies condition (E), then the conclusions of Theorems 3.1 and 3.2 are still true. Hence, we can use the following condition to replace condition (C) in Theorems 3.1 and 3.2 by Proposition 19 in
19
.
Definition 3.2.
19
Let T be a mapping on a subset C of a Banach space E.
Then, T is said to satisfy (SKC)-condition if
1
2
|
|
x
-
T
x
|
|
≤
|
|
x
-
y
|
|
⇒
|
|
T
x
-
T
y
|
|
≤
N
(
x
,
y
)
,
where
N
(
x
,
y
)
:
=
max
{
|
|
x
-
y
|
|
,
1
2
(
|
|
x
-
T
x
|
|
+
|
|
T
y
-
y
|
|
)
,
1
2
(
|
|
T
x
-
y
|
|
+
|
|
x
-
T
y
|
|
)
}
for all x, y ∈ C.
4 Strong convergence theorems (I)
Let C be a nonempty closed convex subset of a Banach space E. A point p in C is said to be an asymptotic fixed point of a mapping T: C → C if C contains a sequence {x
n
} which converges weakly to p such that lim
n→∞, ||x
n
- Tx
n
|| = 0. The set of asymptotic fixed points of T will be denoted by
F
^
(
T
)
. A mapping T: C → C is called relatively nonexpansive
20
if
F
(
T
)
≠
0̸
,
F
^
(
T
)
=
F
(
T
)
, and ϕ(p,Tx) ≤ ϕ(p,x) for all x ∈ C and p ∈ F(T). Note that every identity mapping is a relatively nonexpansive mapping.
Lemma 4.1.
21
Let E be a strictly convex and smooth Banach space, let C be a closed convex subset of E, and let T: C → C be a relatively nonexpansive mapping. Then, F(T) is a closed and convex subset of C.
The following property is motivated by the property (Q
4) in
22
.
Definition 4.1. Let E be a Banach space. Then, we say that E satisfies condition (Q) if for each x, y, z
1
, z
2 ∈ E and t ∈ [0,1],
{
x
0
∈
C
chosen
arbitrary
and
C
0
=
D
0
=
C
,
y
n
=
a
n
x
n
−
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
,
z
n
=
J
−
1
(
d
n
J
y
n
+
(
1
−
d
n
)
J
S
y
n
)
,
C
n
=
{
z
∈
C
n
−
1
:
ϕ
(
z
,
z
n
)
≤
ϕ
(
z
,
y
n
)
}
,
D
n
=
{
z
∈
D
n
−
1
:
|
|
y
n
−
z
|
|
≤
|
|
x
n
−
1
−
z
|
|
}
,
x
n
=
Π
C
n
∩
D
n
x
0
.
Remark 4.1. If H is a Hilbert space, then H satisfies condition (Q).
Theorem 4.1. Let E be a uniformly convex and uniformly smooth Banach space with condition (Q), and let C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings, and let S: C → C be a relatively nonexpansive mapping. Let {a
n
}, {b
n
}, {c
n
}, and {d
n
} be sequences in (0,1) with and a
n
+ b
n
+ c
n
= 1. Suppose that
Ω
:
=
F
(
S
)
∩
F
(
T
1
)
∩
F
(
T
2
)
≠
∅
. Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary
and
C
0
=
D
0
=
C
,
y
n
=
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
,
z
n
=
J
-
1
(
d
n
J
y
n
+
(
1
-
d
n
)
J
S
y
n
)
,
C
n
=
{
z
∈
C
n
-
1
:
ϕ
(
z
,
z
n
)
≤
ϕ
(
z
,
y
n
)
}
,
D
n
=
{
z
∈
D
n
-
1
:
|
|
y
n
-
z
|
|
≤
|
|
x
n
-
1
-
z
|
|
}
,
x
n
=
Π
C
n
∩
D
n
x
0
.
Assume that lim inf
n→∞
b
n
> 0, lim inf
n→∞
c
n
> 0, and lim inf
n→∞
d
n
(1 - d
n
) > 0.
Then, lim
n→∞
x
n
= lim
n→∞
y
n
= lim
n→∞
z
n
= ΠΩ
x
0.
Proof. Following the same argument as in Lemma 3.1, we know that {y
n
} is well-defined.
Clearly, C
0 and D
0 are nonempty closed convex subsets of C, and C
n
is a closed subset of C for every n ∈ ℕ. Since ϕ(z, z
n
) ≤ ϕ(z, y
n
) is equivalent to
2
z
,
J
y
n
-
J
z
n
≤
|
|
y
n
|
|
2
-
|
|
z
n
|
|
2
,
it is easy to see that C
n
is a convex set for each n ∈ ℕ. Besides, by condition (Q), it is easy to see that D
n
is a nonempty closed convex subset of C.
Next, we want to show that Ω ⊆ C
n
⋂ D
n
for each n ∈ ℕ ∪ {0}. Clearly, Ω ⊆ C
0. Suppose that Ω ⊆ C
n-1. Let w ∈ Ω. Then, w ∈ F(S) and
ϕ
(
w
,
z
n
)
=
ϕ
(
w
,
J
-
1
(
d
n
J
y
n
+
(
1
-
d
n
)
J
S
y
n
)
)
≤
d
n
ϕ
(
w
,
y
n
)
+
(
1
-
d
n
)
ϕ
(
w
,
S
y
n
)
≤
d
n
ϕ
(
w
,
y
n
)
+
(
1
-
d
n
)
ϕ
(
w
,
y
n
)
=
ϕ
(
w
,
y
n
)
.
So, Ω ⊆ C
n
. By induction, Ω ⊆ C
n
for each n ∈ ℕ ∪ {0}.
Clearly, Ω ⊆ D
0. Suppose that Ω ⊆ D
n-1. Let w ∈ Ω. Then, w ∈ F(T
1) ⋂ F(T
2) and
|
|
y
n
-
w
|
|
≤
a
n
|
|
x
n
-
1
-
w
|
|
+
b
n
|
|
T
1
y
n
-
w
|
|
+
c
n
|
|
T
2
y
n
-
w
|
|
≤
a
n
|
|
x
n
-
1
-
w
|
|
+
b
n
|
|
y
n
-
w
|
|
+
c
n
|
|
y
n
-
w
|
|
.
This implies that ||y
n
- w|| ≤ ||x
n-1 - w|| and w ∈ D
n
. By induction, Ω ⊆ D
n
for each n ∈ ℕ ∪ {0}. So, Ω ⊆ C
n
⋂ D
n
for each n ∈ ℕ ∪ {0}.
Since
x
n
=
Π
C
n
∩
D
n
x
0
,
ϕ
(
x
n
,
x
0
)
≤
ϕ
(
w
,
x
0
)
-
ϕ
(
w
,
x
n
)
≤
ϕ
(
w
,
x
0
)
for each w ∈ Ω. Therefore, {ϕ(x
n
, x
0
)} is a bounded sequence. Furthermore, {x
n
} is a bounded sequence.
By Lemma 2.5,
x
n
=
Π
C
n
∩
D
n
x
0
, and
x
n
+
1
=
Π
C
n
+
1
∩
D
n
+
1
x
0
,
ϕ
(
x
n
+
1
,
x
n
)
=
ϕ
(
x
n
+
1
,
Π
C
n
∩
D
n
x
0
)
≤
ϕ
(
x
n
+
1
,
x
0
)
-
ϕ
(
x
n
,
x
0
)
.
Hence, ϕ(x
n
, x
0) ≤ ϕ(x
n+1, x
0), lim
n→∞
ϕ(x
n
, x
0) exists, and lim
n→∞
ϕ(x
n+1, x
n
) = 0. By Lemma 2.4, lim
n→∞, ||x
n+1 - x
n
|| = 0. Since x
n
∈ D
n
, we know that ||y
n
-x
n
|| ≤ ||x
n-1-x
n
|| and lim
n→∞ ||x
n
-y
n
|| = 0. Furthermore, lim
n→∞
ϕ (x
n
, y
n
) = 0. Since x
n
∈ C
n
, it is easy to see that lim
n→∞
ϕ(x
n
, z
n
) = 0. Hence, lim
n→∞ ||x
n
-z
n
|| = 0.
Take any w ∈ Ω and let w be fixed. Let r: = 2sup{||x
n
||, ||x
n
- w||, ||y
n
||, ||y
n
-w||: n ∈ ℕ}. By Lemma 2.1, there exists a strictly increasing, continuous, and convex function g: [0, 2r] → [0, ∞) such that g(0) = 0 and
|
|
y
n
-
w
|
|
2
≤
a
n
|
|
x
n
-
1
-
w
|
|
2
+
b
n
|
|
T
1
y
n
-
w
|
|
2
+
c
n
|
|
T
2
y
n
-
w
|
|
2
-
a
n
b
n
g
(
|
|
x
n
-
1
-
T
1
y
n
|
|
)
≤
a
n
|
|
x
n
-
1
-
w
|
|
2
+
b
n
|
|
y
n
-
w
|
|
2
+
c
n
|
|
y
n
-
w
|
|
2
-
a
n
b
n
g
(
|
|
x
n
-
1
-
T
1
y
n
|
|
)
.
This implies that
b
n
g
(
|
|
x
n
-
1
-
T
1
y
n
|
|
)
≤
|
|
x
n
-
1
-
y
n
|
|
(
|
|
x
n
-
1
-
w
|
|
+
|
|
y
n
-
w
|
|
)
.
So, lim
n→∞
b
n
g(||x
n-1 - T
1
y
n
||) = 0. By (ii), lim
n→∞ ||x
n-1 - T
1
y
n
|| = 0. Furthermore, lim
n→∞ ||y
n
- T
1
y
n
|| = 0. Similarly, lim
n→∞ ||y
n
- T
2
y
n
|| = 0.
By Lemma 2.2, there exists a strictly increasing, continuous, and convex function g': [0, 2r] → [0, ∞) such that g'(0) = 0 and
ϕ
(
w
,
z
n
)
≤
d
n
ϕ
(
w
,
y
n
)
+
(
1
-
d
n
)
ϕ
(
w
,
S
y
n
)
-
d
n
(
1
-
d
n
)
g
′
(
|
|
J
y
n
-
J
S
y
n
|
|
)
≤
ϕ
(
w
,
y
n
)
-
d
n
(
1
-
d
n
)
g
′
(
|
|
J
y
n
-
J
S
y
n
|
|
)
.
Hence,
d
n
(
1
-
d
n
)
g
′
(
|
|
J
y
n
-
J
S
y
n
|
|
)
≤
ϕ
(
w
,
y
n
)
-
ϕ
(
w
,
z
n
)
=
(
|
|
w
|
|
2
+
|
|
y
n
|
|
2
-
2
w
,
J
y
n
)
-
(
|
|
w
|
|
2
+
|
|
z
n
|
|
2
-
2
w
,
J
z
n
)
=
|
|
y
n
|
|
2
-
|
|
z
n
|
|
2
+
2
w
,
J
z
n
-
J
y
n
=
|
|
y
n
-
z
n
|
|
(
|
|
y
n
|
|
+
|
|
z
n
|
|
)
+
2
|
|
w
|
|
⋅
|
|
J
z
n
-
J
y
n
|
|
.
By Remark 2.1, lim
n→∞
d
n
(1 - d
n
)g(||Jy
n
- JSy
n
||) = 0. By assumptions and the properties of g, lim
n→∞ ||Jy
n
- JSy
n
|| = 0. Furthermore, lim
n→∞ ||y
n
- Sy
n
|| = 0.
Since {y
n
} is a bounded sequence, there exists a subsequence
{
y
n
k
}
of {y
n
} and
x
̄
∈
C
such that
y
n
k
⇀
x
̄
. By Lemma 2.6,
x
̄
∈
F
(
T
1
)
∩
F
(
T
2
)
. Besides, since S is a relatively nonexpansive mapping,
x
̄
∈
F
^
(
S
)
=
F
(
S
)
. So,
x
̄
∈
Ω
.
Finally, we want to show that y
n
→ ΠΩ
x
0. Let q = ΠΩ
x
0. Then, q ∈ Ω ⊆ C
n
⋂ D
n
for each n ∈ ℕ. So,
ϕ
(
x
n
,
x
0
)
=
min
y
∈
C
n
∩
D
n
ϕ
(
y
,
x
0
)
≤
ϕ
(
q
,
x
0
)
.
On the other hand, from weakly lower semicontinuity of the norm and lim
n→∞ ||x
n
-y
n
|| = 0, we have
ϕ
(
x
̄
,
x
0
)
=
|
|
x
̄
|
|
2
-
2
x
̄
,
J
x
0
+
|
|
x
0
|
|
2
≤
liminf
n
→
∞
(
|
|
y
n
k
|
|
2
-
2
y
n
k
,
J
x
0
+
|
|
x
0
|
|
2
)
=
liminf
n
→
∞
(
|
|
x
n
k
|
|
2
-
2
x
n
k
,
J
x
0
+
|
|
x
0
|
|
2
)
≤
liminf
n
→
∞
ϕ
(
x
n
k
,
x
0
)
≤
limsup
n
→
∞
ϕ
(
x
n
k
,
x
0
)
≤
ϕ
(
q
,
x
0
)
.
Since
q
=
Π
Ω
x
0
,
x
̄
=
q
. Hence,
lim
n
→
∞
ϕ
(
x
n
k
,
x
0
)
=
ϕ
(
x
̄
,
x
0
)
. So, we have
lim
n
→
∞
|
|
x
n
k
|
|
=
|
|
x
̄
|
|
. Using the Kadec-Klee property of E, we obtain that
lim
k
→
∞
x
n
k
=
q
=
Π
Ω
x
0
.
Furthermore, for each weakly convergence subsequence
{
x
n
m
}
of {x
n
}, we know that
lim
m
→
∞
x
n
m
=
q
=
Π
Ω
y
1
by following the same argument as the above conclusion. Therefore,
lim
n
→
∞
x
n
=
lim
n
→
∞
y
n
=
lim
n
→
∞
z
n
=
Π
Ω
x
0
.
Hence, the proof is completed.
Remark 4.2. Since nonspreading mappings with fixed points in a strictly convex Banach space with a uniformly Gateaux differentiable norm are relatively nonex-pansive mappings [
23
, Theorem 3.3], we know that the conclusion of Theorem 4.1 is still true if S is replaced by a nonspreading mapping.
Next, we give an easy example for Theorem 4.1.
Example 4.1. Let E = ℝ, C: = [0,3], T
1
x = T
2
x = x, and let S: C → C be the as in
1
:
S
x
:
=
0
if
x
≠
3
,
1
if
x
=
3
.
For each n, let
a
n
=
b
n
=
c
n
=
1
3
and
d
n
=
1
2
. Let x
0 = 1. Hence, we have
(a) y
n
= x
n-1 for each n ∈ ℕ;
(b)
z
n
=
1
2
y
n
for each n ∈ ℕ;
(c)
C
n
:
=
{
z
∈
C
n
-
1
:
|
z
-
z
n
|
≤
|
z
-
y
n
|
}
=
0
0
,
y
n
+
z
n
2
;
(d) D
n
: = {z ∈ D
n-1 : |z - y
n
| ≤ |z - x
n-1|} = [0,3];
(e)
x
n
=
1
2
(
y
n
+
z
n
)
=
1
2
x
n
-
1
+
1
2
x
n
-
1
=
3
4
x
n
-
1
.
By (e) and x
0 = 1, we know that
x
n
=
3
4
n
for each n ∈ ℕ ∪ {0}, lim
n→∞
x
n
= 0, and 0 is a common fixed point of S, T
1, and T
2.
The following results are special cases of Theorem 4.1.
Corollary 4.1. Let E be a uniformly convex and uniformly smooth Banach space with condition (Q), and let C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings. Let {a
n
}, {b
n
}, {c
n
} be sequences in (0,1) with and a
n
+ b
n
+ c
n
= 1. Suppose that
Ω
:
=
F
(
T
1
)
∩
F
(
T
2
)
≠
∅
. Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary
and
D
0
=
C
,
y
n
=
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
,
D
n
=
{
z
∈
D
n
-
1
:
|
|
y
n
-
z
|
|
≤
|
|
x
n
-
1
-
z
|
|
}
,
x
n
=
Π
D
n
x
0
.
Assume that lim inf
n→∞
b
n
> 0, lim inf
n→∞
c
n
> 0. Then, lim
n→∞
x
n
= lim
n→∞, y
n
= ∏Ω
x
0.
Corollary 4.2. Let E be a uniformly convex and uniformly smooth Banach space, and let C be a nonempty closed convex subset of E, and let S: C → C be a relatively nonexpansive mapping. Let {d
n
} be a sequence in (0,1). Suppose that
F
(
S
)
≠
0̸
. Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary
and
C
0
=
C
,
z
n
=
J
-
1
(
d
n
J
x
n
-
1
+
(
1
-
d
n
)
J
S
x
n
-
1
)
,
C
n
=
{
z
∈
C
n
-
1
:
ϕ
(
z
,
z
n
)
≤
ϕ
(
z
,
x
n
-
1
)
}
,
x
n
=
Π
C
n
x
0
.
Assume that lim inf
n→∞
d
n
(1 - d
n
) > 0. Then, lim
n→∞
x
n
= lim
n→∞
z
n
= Π
F(s)
x
0.
Remark 4.3. Corollary 4.2 is a generalization of Theorem 4.1 in
24
. But, it is a special case of Theorem 3.1 in
25
.
5 Strong convergence theorems (II)
In this section, we need the following important lemmas.
Lemma 5.1.
26
Let E be a reflexive Banach space and f: E → ℝ ∪ {+∞} be a convex and lower semicontinuous function. Let C be a nonempty bounded and closed convex subset of E. Then, the function f attains its minimum on C. That is, there exists x* ∈ C such that f(x*) ≤ f(x) for all x ∈ C.
Lemma 5.2. In a Banach space E, there holds the inequality
|
|
x
+
y
|
|
2
≤
|
|
x
|
|
2
+
2
y
,
j
(
x
+
y
)
,
x
,
y
∈
E
,
where j(x+y) ∈ J(x+y).
Lemma 5.3.
27
Let C be a nonempty closed convex subset of a Banach space E with a uniformly Gâteaux differentiable norm. Let {x
n
} be a bounded sequence of E and let μ
n
be a Banach limit and z ∈ C. Then,
μ
n
|
|
x
n
-
z
|
|
2
=
min
y
∈
C
μ
n
|
|
x
n
-
y
|
|
2
⇔
μ
n
y
-
z
,
J
(
x
n
-
z
)
≤
0
for
all
y
∈
C
.
Lemma 5.4.
28
Let α be a real number and (x
0, x
1,...) ∈ ℓ
2 such that μ
n
x
n
≤ α for all Banach μ
n
. If lim sup
n→∞ (x
n+1 - x
n
) ≤ 0, then lim sup
n→∞
x
n
≤ α.
Lemma 5.5.
29
Assume that {a
n
}
n∈ℕ is a sequence of nonnegative real numbers such that a
n+1 < (1 - γ
n
)a
n
+ δ
n
, n ∈ ℕ, where {γ
n
} ⊆ (0,1) and δ
n
is a sequence in ℝ such that (i)
∑
n
=
1
∞
γ
n
=
∞
; (ii)
lim
sup
n
→
∞
δ
n
γ
n
≤
0
or
∑
n
=
1
∞
|
δ
n
|
<
∞
. Then, lim
n→∞
a
n
= 0.
Theorem 5.1. Let E be a uniformly convex and uniformly smooth Banach space with Opial's condition, C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings, and let S: C → C be a mapping with condition (C). Let {a
n
}, {b
n
}, and {c
n
} be sequences in (a, b) for some 0 < a, b < 1 with a
n
+ b
n
+ c
n
= 1. Let {d
n
} be a sequence in [0,1]. Suppose that
Ω
:
=
F
(
S
)
∩
F
(
T
1
)
∩
F
(
T
2
)
≠
∅
. Define a sequence {x
n
} by
x
0
∈
C
chosen
arbitrary
,
y
n
=
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
,
x
n
=
d
n
x
0
+
(
1
-
d
n
)
S
y
n
.
Assume that:
(i)
lim
n
→
∞
d
n
=
0
,
∑
n
=
1
∞
d
n
=
∞
, and
lim
n
→
∞
|
d
n
+
1
-
d
n
|
d
n
=
0
;
(ii)
lim
n
→
∞
(
a
n
+
1
-
a
n
)
=
lim
n
→
∞
(
b
n
+
1
-
b
n
)
=
lim
n
→
∞
(
c
n
+
1
-
c
n
)
=
0
.
Then,
lim
n
→
∞
x
n
=
lim
n
→
∞
y
n
=
x
̄
for some
x
̄
∈
Ω
.
Proof. Following the same argument as in Lemma 3.1, we know that {y
n
} is well-defined. Take any w ∈ Ω: = F(S) ⋂ F(T
1) ⋂ F(T
2) and let w be fixed. Then, for each n ∈ ℕ, we have
y
n
-
w
≤
a
n
x
n
-
1
-
w
+
b
n
T
1
y
n
-
w
+
c
n
T
2
y
n
-
w
≤
a
n
x
n
-
1
-
w
+
(
b
n
+
c
n
)
y
n
-
w
.
This implies that ||y
n
- w|| ≤ ||x
n-1 - w|| for each n ∈ ℕ. Next, we have
|
|
x
n
-
w
|
|
≤
d
n
|
|
x
0
-
w
|
|
+
(
1
-
d
n
)
|
|
S
y
n
-
w
|
|
≤
d
n
|
|
x
0
-
w
|
|
+
(
1
-
d
n
)
|
|
y
n
-
w
|
|
≤
d
n
|
|
x
0
-
w
|
|
+
(
1
-
d
n
)
|
|
x
n
-
1
-
w
|
|
⋮
≤
max
{
|
|
x
0
-
w
|
|
,
|
|
x
1
-
w
|
|
}
.
Then, {x
n
} is a bounded sequence. Furthermore, {y
n
}, {Sy
n
}, {T
1
y
n
}, {T
2
y
n
} are bounded sequences. Define M as
M
:
=
sup
{
|
|
x
n
|
|
,
|
|
y
n
|
|
,
|
|
T
1
y
n
|
|
,
|
|
T
2
y
n
|
|
,
|
|
S
y
n
|
|
,
|
|
x
n
-
w
|
|
,
|
|
y
n
-
w
|
|
:
n
∈
ℕ
}
.
Besides, we know that
limsup
n
→
∞
|
|
x
n
-
w
|
|
≤
limsup
n
→
∞
(
d
n
|
|
x
0
-
w
|
|
+
(
1
-
d
n
)
|
|
y
n
-
w
|
|
)
≤
limsup
n
→
∞
d
n
|
|
x
0
-
w
|
|
+
limsup
n
→
∞
|
|
y
n
-
w
|
|
≤
limsup
n
←
∞
|
|
x
n
-
1
-
w
|
|
.
This implies that
limsup
n
→
∞
|
|
x
n
-
w
|
|
=
limsup
n
→
∞
|
|
y
n
-
w
|
|
.
By Lemma 2.1, there exists a strictly increasing, continuous, and convex function g: [0, 2M] → ℝ such that
|
|
y
n
-
w
|
|
2
≤
a
n
|
|
x
n
-
1
-
w
|
|
2
+
b
n
|
|
T
1
y
n
-
w
|
|
2
+
c
n
|
|
T
2
y
n
-
w
|
|
2
-
a
n
b
n
|
|
x
n
-
1
-
T
1
y
n
|
|
2
≤
a
n
|
|
x
n
-
1
-
w
|
|
2
+
b
n
|
|
y
n
-
w
|
|
2
+
c
n
|
|
y
n
-
w
|
|
2
-
a
n
b
n
|
|
x
n
-
1
-
T
1
y
n
|
|
2
.
Then,
|
|
y
n
-
w
|
|
2
≤
|
|
y
n
-
w
|
|
2
+
a
|
|
x
n
-
1
-
T
1
y
n
|
|
2
≤
|
|
y
n
-
w
|
|
2
+
b
n
|
|
x
n
-
1
-
T
1
y
n
|
|
2
≤
|
|
x
n
-
1
-
w
|
|
2
.
This implies that
lim
n
→
∞
|
|
x
n
-
1
-
T
1
y
n
|
|
=
0
.
Similar, we have
lim
n
→
∞
|
|
x
n
-
1
-
T
2
y
n
|
|
=
0
.
By (i),
lim
n
→
∞
|
|
x
n
-
S
y
n
|
|
=
lim
n
→
∞
d
n
|
|
x
0
-
S
y
n
|
|
=
0
and
|
|
x
n
+
1
-
x
n
|
|
=
|
|
d
n
+
1
x
0
+
(
1
-
d
n
+
1
)
S
y
n
+
1
-
d
n
x
0
-
(
1
-
d
n
)
S
y
n
|
|
≤
|
|
d
n
+
1
x
0
+
(
1
-
d
n
+
1
)
S
y
n
+
1
-
d
n
x
0
-
(
1
-
d
n
)
S
y
n
+
1
|
|
+
|
|
d
n
x
0
+
(
1
-
d
n
)
S
y
n
+
1
-
d
n
x
0
-
(
1
-
d
n
)
S
y
n
|
|
≤
|
d
n
+
1
-
d
n
|
⋅
|
|
x
0
|
|
+
|
d
n
+
1
-
d
n
|
⋅
|
|
S
y
n
+
1
|
|
+
(
1
-
d
n
)
⋅
|
|
S
y
n
+
1
-
S
y
n
|
|
≤
|
d
n
+
1
-
d
n
|
⋅
|
|
x
0
|
|
+
|
d
n
+
1
-
d
n
|
⋅
|
|
S
y
n
+
1
|
|
+
(
1
-
d
n
)
⋅
|
|
x
n
+
1
-
d
n
+
1
x
0
-
x
n
+
d
n
x
0
|
|
≤
2
M
⋅
|
d
n
+
1
-
d
n
|
+
(
1
-
d
n
)
⋅
(
|
|
x
n
+
1
-
x
n
|
|
+
|
d
n
+
1
-
d
n
|
⋅
|
|
x
0
|
|
)
.
So,
|
|
x
n
+
1
-
x
n
|
|
≤
3
M
⋅
|
d
n
+
1
-
d
n
|
d
n
.
By (i),
lim
n
→
∞
|
|
x
n
+
1
-
x
n
|
|
=
0
.
Furthermore,
lim
n
→
∞
|
|
S
y
n
+
1
-
S
y
n
|
|
=
0
.
Next, we have
|
|
y
n
+
1
−
y
n
|
|
=
|
|
(
a
n
+
1
x
n
+
b
n
+
1
T
1
y
n
+
1
+
c
n
+
1
T
2
y
n
+
1
)
−
(
a
n
x
n
−
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
)
|
|
≤
|
|
(
a
n
+
1
x
n
+
b
n
+
1
T
1
y
n
+
1
+
c
n
+
1
T
2
y
n
+
1
)
−
(
a
n
x
n
+
b
n
T
1
y
n
+
1
+
c
n
T
2
y
n
+
1
)
|
|
+
|
|
(
a
n
x
n
+
b
n
T
1
y
n
+
1
+
c
n
T
2
y
n
+
1
)
−
(
a
n
x
n
−
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
)
|
|
≤
|
a
n
+
1
−
a
n
|
·
|
|
x
n
|
|
+
|
b
n
+
1
−
b
n
|
·
|
|
T
1
y
n
+
1
|
|
+
|
c
n
+
1
−
c
n
|
·
|
|
T
2
y
n
+
1
)
|
|
+
a
n
|
|
x
n
−
x
n
−
1
|
|
+
b
n
|
|
T
1
y
n
+
1
-
T
1
y
n
|
|
+
c
n
|
|
T
2
y
n
+
1
-
T
2
y
n
|
|
≤
M
·
(
|
a
n
+
1
−
a
n
|
+
|
b
n
+
1
−
b
n
|
+
|
c
n
+
1
−
c
n
|
)
+
a
n
|
|
x
n
−
x
n
−
1
|
|
+
b
n
|
|
y
n
+
1
−
y
n
|
|
+
c
n
|
|
y
n
+
1
−
y
n
|
|
.
This implies that
|
|
y
n
+
1
-
y
n
|
|
≤
M
⋅
(
|
a
n
+
1
-
a
n
|
+
|
b
n
+
1
-
b
n
|
+
|
c
n
+
1
-
c
n
|
)
a
n
+
|
|
x
n
-
x
n
-
1
|
|
.
So,
lim
n
→
∞
|
|
y
n
+
1
-
y
n
|
|
=
0
.
Besides,
lim
n
→
∞
|
|
y
n
-
x
n
-
1
|
|
=
lim
n
→
∞
|
|
b
n
(
T
1
y
n
-
x
n
-
1
)
+
c
n
(
T
2
y
n
-
x
n
-
1
)
|
|
=
0
and
lim
n
→
∞
|
|
y
n
-
S
y
n
|
|
=
lim
n
→
∞
|
|
y
n
-
T
1
y
n
|
|
=
lim
n
→
∞
|
|
y
n
-
T
2
y
n
|
|
=
0
.
Let φ: C → ℝ be defined by φ(u): = μ
n
||x
n
- u|| for each u ∈ C. Clearly, φ is convex and continuous. Taking p ∈ Ω and defining a subset D of C by
D
:
=
{
x
∈
C
:
|
|
x
-
p
|
|
≤
r
}
,
where r: = max{||x
0 - p||, ||x
1 - p||}. Then, D is a nonempty closed bounded convex subset of C and {x
n
} ⊆ D. By Lemma 5.1,
C
min
:
=
{
z
∈
D
:
φ
(
z
)
:
=
min
y
∈
D
φ
(
y
)
}
≠
0̸
.
Obviously, C
min is a bounded closed convex subset. Following the property of Banach limit μ
n
, for all z ∈ C
min, we have
φ
(
S
z
)
=
μ
n
|
|
x
n
-
S
z
|
|
2
≤
μ
n
(
|
|
x
n
-
y
n
|
|
+
|
|
y
n
-
S
z
|
|
)
2
≤
μ
n
(
|
|
x
n
-
y
n
|
|
+
3
|
|
y
n
-
S
y
n
|
|
+
|
|
y
n
-
z
|
|
)
2
=
μ
n
|
|
y
n
-
z
|
|
2
≤
μ
n
(
|
|
y
n
-
x
n
|
|
+
|
|
x
n
-
z
|
|
)
2
≤
μ
n
|
|
x
n
-
z
|
|
2
.
Then, Sz ∈ C
min. By Theorem 4 in
1
, there exists
x
̄
∈
C
min
such that
S
x
̄
=
x
̄
. By Lemma 5.3,
μ
n
⟨
y
-
x
̄
,
J
(
x
n
-
x
̄
)
⟩
≤
0
for
all
y
∈
C
.
Take any y ∈ C and let y be fixed. Since lim
n→∞ ||x
n+1 - x
n
|| = 0, then it follows from the norm-weak* uniformly continuity of the duality mapping J that
lim
n
→
∞
(
⟨
y
-
x
̄
,
J
(
x
n
+
1
-
x
̄
)
⟩
-
⟨
y
-
x
̄
,
J
(
x
n
-
x
̄
)
⟩
)
=
0
.
By Lemma 5.4,
lim
n
→
∞
⟨
y
-
x
̄
,
J
(
x
n
-
x
̄
)
⟩
≤
0
for
all
y
∈
C
.
By Lemma 5.2,
|
|
x
n
-
x
̄
|
|
2
=
|
|
d
n
(
x
0
-
x
̄
)
+
(
1
-
d
n
)
(
S
y
n
-
x
̄
)
|
|
2
≤
(
1
-
d
n
)
2
|
|
S
y
n
-
x
̄
|
|
2
+
2
d
n
⟨
x
0
-
x
̄
,
J
(
x
n
-
x
̄
)
⟩
≤
(
1
-
d
n
)
2
|
|
y
n
-
x
̄
|
|
2
+
2
d
n
⟨
x
0
-
x
̄
,
J
(
x
n
-
x
̄
)
⟩
≤
(
1
-
d
n
)
|
|
x
n
-
1
-
x
̄
|
|
2
+
2
d
n
⟨
x
0
-
x
̄
,
J
(
x
n
-
x
̄
)
⟩
.
By Lemma 5.5,
lim
n
→
∞
|
|
x
n
-
x
̄
|
|
=
0
. Furthermore, since T
1 and T
2 are nonexpan-sive mappings, we know that
x
̄
is also a fixed point of T
1 and T
2. Therefore, the proof is completed.
The following is a special case of Theorem 5.1 when T
1 and T
2 are identity mappings.
Theorem 5.2. Let E be a uniformly convex and uniformly smooth Banach space with Opial's condition, C be a nonempty closed convex subset of E, and let S: C → C be a mapping with condition (C). Let {d
n
} be a sequence in (0,1). Suppose that
F
(
S
)
≠
0̸
. Define a sequence {x
n
} by
x
0
∈
C
chosen arbitrary
,
x
n
=
d
n
x
0
+
(
1
-
d
n
)
S
x
n
-
1
,
n
∈
ℕ
.
Assume that
lim
n
→
∞
d
n
=
0
,
∑
n
=
1
∞
d
n
=
∞
, and
lim
n
→
∞
|
d
n
+
1
-
d
n
|
d
n
=
0
.
. Then,
lim
n
→
∞
x
n
=
lim
n
→
∞
y
n
=
x
̄
for some
x
̄
∈
F
(
S
)
.
6 Application
Let E be a reflexive, strictly convex, and smooth Banach space and let A ⊆ E × E* be a set-valued mapping with range R(A): = {x* ∈ E*: x* ∈ Ax} and domain
D
(
A
)
=
{
x
∈
E
:
A
x
≠
0̸
}
. Then, the mapping A is said to be monotone if 〈x -y,x* - y*〉 ≥ 0 whenever (x, x*), (y, y*) ∈ A. It is also said to be maximal monotone if A is monotone and there is no monotone operator from E into E* whose graph properly contains the graph of A. It is known that if A ⊆ E × E* is maximal monotone, then A
-10 is closed and convex.
Lemma 6.1.
30
Let E be a reflexive, strictly convex, and smooth Banach space and let A ⊆ E × E* be a monotone operator. Then, A is maximal monotone if and only if R(J + rA) = E* for all r > 0.
By Lemma 6.1, for every r > 0 and x ∈ E, there exists a unique x
r
∈ D(A) such that Jx ∈ Jx
r
+ rAx
r
. Hence, define a single valued mapping J
r
: E → D(A) by J
r
x = x
r
, that is, J
r
= (J + rA)-1
J and such J
r
is called the relative resolvent of A. We know that A
-10 = F(J
r
) for all r > 0
8
.
Lemma 6.2.
21
Let E be a uniformly convex and uniformly smooth Banach space and let A ⊆ E × E* be a maximal monotone operator. Let J
r
be the relative resolvent of A, where r > 0. If A
-10 is nonempty, then J
r
is a relatively nonexpansive mapping on E.
By Theorem 4.1 and Lemma 6.2, it is easy to get the following result.
Theorem 6.1. Let E be a uniformly convex and uniformly smooth Banach space with property (Q), and let C be a nonempty closed convex subset of E, and let T
1, T
2 : C → C be two nonexpansive mappings, and let A ⊆ E × E* be a maximal monotone operator. Let {a
n
}, {b
n
}, {c
n
}, and {d
n
} be sequences in (0,1) with and a
n
+ b
n
+ c
n
= 1. Suppose that
Ω
:
=
A
-
1
0
∩
F
(
T
1
)
∩
F
(
T
2
)
≠
0̸
. Define a sequence {x
n
} by
x
0
∈
C
chosen arbitrary and
C
0
=
D
0
=
C
,
y
n
=
a
n
x
n
-
1
+
b
n
T
1
y
n
+
c
n
T
2
y
n
,
z
n
=
J
-
1
(
d
n
J
y
n
+
(
1
-
d
n
)
J
J
r
y
n
)
,
C
n
=
{
z
∈
C
n
-
1
:
ϕ
(
z
,
z
n
)
≤
ϕ
(
z
,
y
n
)
}
,
D
n
=
{
z
∈
D
n
-
1
:
|
|
y
n
-
z
|
|
≤
|
|
x
n
-
1
-
z
|
|
}
,
x
n
=
Π
C
n
∩
D
n
x
0
.
Assume that lim inf
n→∞
b
n
> 0, lim inf
n→∞
c
n
> 0, and lim inf
n→∞
d
n
(1 - d
n
) > 0.
Then, lim
n→∞
x
n
= lim
n→∞
y
n
= lim
n→∞
z
n
= ΠΩ
x
0.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
L-JL responsible for problem resign, coordinator, discussion, revise the important part, and submit. C-SC is responsible for the important results of this article, discuss, and draft. Z-TY is responsible for discussion and the applications. All authors read and approved the final manuscript.