# Iterative Approximations for Zeros of Sum of Accretive Operators in Banach Spaces

Authors: Yekini Shehu
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

We study the approximation of zero for sum of accretive operators using a modified Mann type forward-backward splitting algorithm and obtain strong convergence of the sequence generated by our scheme to the zero of sum of accretive operators in uniformly convex real Banach spaces which are also uniformly smooth. Our result is new and complements many recent and important results in this direction in the literature.


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## Body

## 1. Introduction and Preliminaries

LetE be a real Banach space. Themodulus of convexity
δ
E
:
[
0,2
]
→
[
0,1
] is defined as (1)
δ
E
ϵ
=
inf
⁡
1
-
x
+
y
2
:
x
=
1
=
y
,
x
-
y
≥
ϵ
.
Eis calleduniformly convex if δ
E
(
ϵ
)
>
0 for any ϵ
∈
(
0,2
] and p
-uniformly convex if there is c
p
>
0 so that δ
E
(
ϵ
)
≥
c
p
ϵ
p for any ϵ
∈
(
0,2
]. Themodulus of smoothness
ρ
E
(
τ
)
:
[
0
,
∞
)
→
[
0
,
∞
) is defined by(2)
ρ
E
τ
=
x
+
τ
y
+
x
-
τ
y
2
-
1
:
x
=
y
=
1
.
E is calleduniformly smooth if lim
n
→
∞
⁡
ρ
E
(
τ
)
/
τ
=
0 and q
-uniformly smooth if there is C
q
>
0 so that ρ
E
(
τ
)
≤
C
q
τ
q for any τ
>
0. Hilbert spaces, L
p
(
o
r
l
p
) spaces, 1
<
p
<
∞, and the Sobolev spaces, W
m
p, 1
<
p
<
∞, are q-uniformly smooth. Hilbert spaces are 2-uniformly smooth while(3)
L
p
or
l
p
or
W
m
p
is
p
-uniformly  smooth
if
1
<
p
≤
2
2
-uniformly  smooth
if
p
≥
2
.It is shown in [1] that there is no Banach space which is q-uniformly smooth with q
>
2. It is obvious that q-uniformly smooth Banach space must be uniformly smooth.The normalized duality mappingJ
:
E
→
2
E
∗ is defined by(4)
J
x
≔
f
∈
E
∗
:
x
,
f
=
x
2
=
f
2
,
x
∈
E
.It is well known that J is single-valued and norm-to-norm uniformly continuous on each bounded subsets of E if E is a real smooth and uniformly convex Banach space (please see [2]). In the sequel, we will denote by j the single-valued normalized duality mapping. If E is reduced to the Hilbert space H, then J
≡
I is the identity mapping.LetT
:
E
→
E be a mapping on E. We will use F
(
T
) to denote its set of fixed points; that is, F
(
T
)
≔
{
x
∈
E
:
T
x
=
x
}. Furthermore, we say that T is afixed contraction if there exists a constant k
∈
(
0,1
) such that, for all x
,
y
∈
E, we have T
x
-
T
y
≤
k
x
-
y. The mapping T is said to benonexpansive if k
=
1. Thus, T is nonexpansive if T
x
-
T
y
≤
x
-
y.LetC be a nonempty, closed, and convex subset of E and let Q be a mapping of E onto C. Then Q is said to be sunny [3] if Q
(
Q
(
x
)
+
t
(
x
-
V
Q
(
x
)
)
)
=
Q
(
x
), for all x
∈
E and t
≥
0. A mapping Q of E into E is said to be a retraction [3] if Q
2
=
Q. If a mapping Q is a retraction, then Q
(
z
)
=
z for every z
∈
R
(
Q
), where R
(
Q
) is the range of Q. In [4], it was shown that if E is uniformly smooth and if D is the fixed point set of a nonexpansive mapping from C into itself, then there is a unique sunny nonexpansive retraction from C onto D. We know that, in a uniformly smooth Banach space, a retraction Q
:
C
→
D is sunny and nonexpansive, if and only if(5)
x
-
Q
x
,
j
z
-
Q
x
≤
0
,
∀
x
∈
C
,
z
∈
D
.A set-valued operator A
:
E
→
2
E, with domain D
(
A
) and range R
(
A
), is said to beaccretive if, for all t
>
0 and every x
,
y
∈
D
(
A
),(6)
x
-
y
≤
x
-
y
+
t
u
-
v
,
u
∈
A
x
,
v
∈
A
y
.Equivalently, by Lemma 1.1 of Kato [5] we know that A is accretive if and only if, for each x
,
y
∈
D
(
A
), there exists j
(
x
-
y
)
∈
J
(
x
-
y
) such that(7)
u
-
v
,
j
x
-
y
≥
0
,
u
∈
A
x
,
v
∈
A
y
.Furthermore, an accretive operator A is said to be m-accretive if the range R
(
I
+
λ
A
)
=
E for all λ
>
0. Given α
>
0 and q
∈
(
1
,
∞
), we say that an accretive operator A is α-inverse strongly accretive (α-isa) of order q if, for each x
,
y
∈
D
(
A
), there exists j
q
(
x
-
y
)
∈
J
q
(
x
-
y
) such that(8)
u
-
v
,
j
q
x
-
y
≥
α
u
-
v
q
,
u
∈
A
x
,
v
∈
A
y
.When q
=
2, we simply say α-isa, instead of α-isa of order 2; that is, A is α-isa if, for each x
,
y
∈
D
(
A
), there exists j
(
x
-
y
)
∈
J
(
x
-
y
) such that(9)
u
-
v
,
j
x
-
y
≥
α
u
-
v
2
,
u
∈
A
x
,
v
∈
A
y
.Let E be a real uniformly convex Banach space which is also uniformly smooth. Suppose A
:
E
→
2
E is an m-accretive operator and B
:
E
→
E an α-inverse strongly accretive. In this paper, we will consider the following inclusion problem (assuming that the solution exists): find x
∈
E such that(10)
0
∈
A
x
+
B
x
.Problem (10) is a very general format for certain concrete problems in machine learning, image processing and linear inverse problem, and many nonlinear problems such as convex programming, variational inequalities, split feasibility problem, and minimization problem.It is well known that a forward-backward splitting method (please see, e.g., [6–9]) is the classical method for solving problem (10). This forward-backward splitting method is given by x
1
∈
E and(11)
x
n
+
1
=
I
+
r
A
-
1
x
n
-
r
B
x
n
,
n
≥
1
,where r
>
0. This method generalizes the proximal point algorithm (please see [10–14]) and the gradient method (see, e.g., [15]). There have been many works concerning the problem of finding zero points of the sum of two monotone operators (in Hilbert spaces) and accretive operators (in Banach spaces). For more details, please, see [16–20] and the references contained therein. Motivated by the results of López et al. [21], Cholamjiak [22] and Wei and Shi [23] have recently extended the results of López et al. [21] from Halpern-type forward-backward method to viscosity-type forward-backward method with fixed contraction in q-uniformly smooth Banach spaces which are also uniformly convex.Remark 1.
We note that forq
=
2 the class of inverse strongly accretive of order q coincides with that of inverse strongly accretive. Thus, in the case of q
≠
2, (8) is not equivalent to (9). Furthermore, for q
<
2, inverse strongly accretive operators of order q  do represent a subclass of inverse strongly accretive operators (see [24]). In light of above comments, the class of inverse strongly accretive operators of order q considered in the results of Cholamjiak [22], López et al. [21], and Wei and Shi [23] is a subclass of inverse strongly accretive operators we will consider in this paper.
Recently, Wei and Duan [25] studied the problem of finding zeros of the sum of finitely many m-accretive operators and finitely many α-inversely strongly accretive operators in a real smooth and uniformly convex Banach space (i.e., find x
∈
E such that 0
∈
∑
i
=
1
N
‍
(
A
i
x
+
B
i
x
)). They proved the following strong convergence result.Theorem 2.
LetE be a real smooth and uniformly convex Banach space and let C be a nonempty, closed, and convex sunny nonexpansive retraction of E, and let Q
C be the sunny nonexpansive retraction of E onto C. Let f
:
E
→
E be a fixed contractive mapping with coefficient k
∈
(
0,1
),  and let  T
:
E
→
E be a strongly positive linear bounded operator with coefficient γ
¯. Suppose that the duality mapping J
:
E
→
E
∗ is weakly sequentially continuous at zero, and 0
<
η
<
γ
¯
/
2
k. Let A
i
:
C
→
2
E be m-accretive operator and let B
i
:
C
→
E be α-inversely strongly accretive operator, where i
=
1,2
,
…
,
N. Suppose that, for r
>
0 and i
=
1,2
,
…
,
N,(12)
B
i
x
-
B
i
y
,
J
I
-
r
B
i
x
-
I
-
r
B
i
y
≥
0
.Let {
x
n
} be generated by(13)
x
0
∈
C
,
y
n
=
Q
C
1
-
α
n
x
n
+
e
n
,
z
n
=
1
-
β
n
x
n
+
β
n
a
0
y
n
+
∑
i
=
1
N
a
i
J
r
n
,
i
A
i
y
n
-
r
n
,
i
B
i
y
n
,
x
n
+
1
=
γ
n
η
f
x
n
+
I
-
γ
n
T
z
n
,
n
≥
0
,
0
<
a
m
<
1, for m
=
0,1
,
…
,
N
,
∑
m
=
0
N
‍
a
m
=
1. Suppose {
e
n
}
⊂
E
,
{
α
n
}
,
{
β
n
}, and {
γ
n
} are three sequences in (
0,1
) and {
r
n
,
i
}
⊂
(
0
,
+
∞
) satisfying the following conditions:(i)
∑
n
=
0
∞
γ
n
=
∞
,
γ
n
→
0
,
γ
n
-
1
/
γ
n
→
0
,
β
n
→
1
,
α
n
→
0, as n
→
∞;
(ii)
∑
n
=
0
∞
|
α
n
+
1
-
α
n
|
<
+
∞
,
∑
n
=
0
∞
|
β
n
+
1
-
β
n
|
<
+
∞
,
∑
n
=
0
∞
(
1
-
γ
n
γ
¯
)
α
n
β
n
<
+
∞;
(iii)
∑
n
=
0
∞
α
n
+
1
,
i
-
α
n
,
i
<
+
∞ and r
n
,
i
≥
ϵ
>
0, for n
≥
0 and i
=
1,2
,
…
,
N;
(iv)
∑
n
=
0
∞
e
n
<
∞.If  ∑
i
=
1
N
(
A
i
+
B
i
)
-
1
(
0
)
≠
∅, then {
x
n
} converges strongly to a point p
0
∈
∑
i
=
1
N
(
A
i
+
B
i
)
-
1
(
0
), which is the unique solution of the following variational inequality: for all z
∈
∑
i
=
1
N
(
A
i
+
B
i
)
-
1
(
0
),(14)
T
-
η
f
p
0
,
J
p
0
-
z
≤
0
.We observe that Theorem2 extends the results of López et al. [21] from q-uniformly smooth Banach spaces which are also uniformly convex to uniformly smooth and uniformly convex Banach spaces under the condition that the normalized duality mapping J is weakly sequentially continuous at zero. We also observe that this result is not applicable in L
p, p
≠
2. This leads to this natural question.Question 1.
Can we obtain strong convergence result for approximation of zero of sum of accretive operators inuniformly smooth Banach spaces which are also uniformly convex without assuming thatthe normalized duality mapping
J
is weakly sequentially continuous at zero or 〈
B
x
-
B
y
,
J
(
(
I
-
r
B
)
x
-
(
I
-
r
B
)
)
y
〉
≥
0
?Our interest in this paper is to answer the above question in the affirmative. That is, we establish strong convergence result for approximation of zero of sum of accretive operators in uniformly smooth Banach spaces which are also uniformly convex without assuming that the normalized duality mappingJ is weakly sequentially continuous at zero or 〈
B
x
-
B
y
,
J
(
(
I
-
r
B
)
x
-
(
I
-
r
B
)
)
y
〉
≥
0. To achieve this, apply generalized forward-backward method which involvesviscosity approximation method with fixed contraction and weakly contractive mappings.In the sequel, we will assume thatA is an m-accretive operator. Then we can define, for each r
>
0, the resolvent J
r
A
:
R
(
I
+
r
A
)
→
D
(
A
) by J
r
A
≔
(
I
+
r
A
)
-
1, and the Yosida approximation A
r
:
R
(
I
+
r
A
)
→
R
(
A
) by A
r
=
I
-
J
r
A
/
r. Denote by A
-
1
(
0
) the zero set of A; that is, A
-
1
(
0
)
≔
{
x
∈
D
(
A
)
:
0
∈
A
x
}. It is well known that J
r
A is single-valued and nonexpansive, F
(
J
r
A
)
=
A
-
1
(
0
) and A
r
x
∈
A
(
J
r
A
x
) for every x
∈
R
(
I
+
r
A
), where F
(
J
r
A
)
≔
{
x
∈
R
(
I
+
r
A
)
:
J
r
A
x
=
x
}.In what follows we will make use of the following lemmas.Lemma 3 (see Maingé [26]).
Let{
s
n
} be a real sequence that does not decrease at infinity, in the sense that there exists a subsequence {
s
n
k
} so that s
n
k
≤
s
n
k
+
1 for all k
≥
1.
For everyn
>
n
0 define an integer sequence {
s
n
k
} as(15)
τ
n
=
max
⁡
n
0
≤
k
≤
n
:
s
k
<
s
k
+
1
.Then τ
(
n
)
→
∞ as n
→
∞ and for all n
>
n
0
(16)
max
⁡
s
τ
n
,
s
n
≤
s
τ
n
+
1
.Lemma 4 (see Xu [27]).
Assume{
a
n
} is a sequence of nonnegative real numbers such that (17)
a
n
+
1
≤
1
-
γ
n
a
n
+
γ
n
δ
n
,
n
≥
0
,where {
γ
n
} is a sequence in (
0,1
) and {
δ
n
} is a sequence in R such that(i)
∑
n
=
0
∞
γ
n
=
∞
,
(ii)
l
i
m
s
u
p
n
→
∞
δ
n
≤
0 or ∑
n
=
0
∞
|
δ
n
γ
n
|
<
∞
.Then l
i
m
n
→
∞
a
n
=
0.Lemma 5 (see Xu [28]).
LetC be a nonempty closed convex subset of a uniformly smooth Banach space E and let T
:
C
→
C be a nonexpansive mapping with a fixed point. Let f
:
E
→
E be a fixed contraction with coefficient k
∈
(
0,1
). If there exists a bounded sequence {
x
n
} such that lim
n
→
∞
⁡
x
n
-
T
x
n
=
0 and p
=
lim
t
→
0
⁡
z
t exists, where {
z
t
} is defined by z
t
=
t
f
(
z
t
)
+
(
1
-
t
)
T
z
t and p solves the variational inequality(18)
I
-
f
p
,
j
p
-
x
≤
0
,
∀
x
∈
F
T
,then(19)
lim
sup
n
→
∞
⁡
f
p
-
p
,
j
x
n
-
p
≤
0
.Lemma 6.
LetE be a real Banach space. Then, for all x
,
y
∈
E, one has(20)
x
+
y
2
≤
x
2
+
2
y
,
j
x
+
y
.Lemma 7 (see López et al. [21]).
LetE be a real Banach space. Let A
:
E
→
2
E be an m-accretive operator and let B
:
E
→
E be an α-inverse strongly accretive mapping on E. Then one has(i)
forr
>
0, F
(
T
r
)
=
(
A
+
B
)
-
1
(
0
),
(ii)
for0
<
s
≤
r and x
∈
E
,
x
-
T
s
x
≤
2
x
-
T
r
x.Lemma 8 (see Suantai and Cholamjiak [29]).
LetE be a real Banach space with Fréchet differentiable norm. For x
∈
E, let β
∗
(
t
) be defined for 0
<
t
<
∞ by(21)
β
∗
t
=
sup
⁡
x
+
t
y
2
-
x
2
t
-
2
y
,
j
x
:
y
=
1
.Then, l
i
m
t
→
0
+
β
∗
(
t
)
=
0 and(22)
x
+
h
2
≤
x
2
+
2
h
,
j
x
+
h
β
∗
hfor all h
∈
E
∖
{
0
}.Remark 9.
In a real Hilbert space, we see thatβ
∗
(
t
)
=
t for t
>
0.In the result of Suantai and Cholamjiak [29], the authors assumed that β
∗
(
t
)
≤
2
t for t
>
0. This naturally leads to this important question.Question 2.
What uniformly smooth Banach spaces (except Hilbert spaces) satisfy the assumptionβ
∗
(
t
)
≤
2
t for t
>
0? In particular, do L
p spaces, 1
<
p
<
∞, satisfy it?InE
=
L
p, 2
≤
p
<
∞, we know that(23)
x
+
y
2
≤
x
2
+
2
y
,
j
x
+
p
-
1
y
2
,
∀
x
,
y
∈
E
.Then β
∗ in (21) is estimated by β
∗
(
t
)
≤
(
p
-
1
)
t for t
>
0.In our more general setting, throughout this paper, we will assume that(24)
β
∗
t
≤
c
t
,
t
>
0
and  for  some
c
>
1
,where β
∗ is the function appearing in (21).By following the same method of proof as contained in the proof of Lemma3.3 of [21], we have the following lemma.Lemma 10.
LetE be a real uniformly convex with Fréchet differentiable norm. Assume that B is a single-valued α-inverse strongly accretive mapping on E. Then, given s
>
0, there exists a continuous, strictly increasing and convex function φ
:
R
+
→
R
+ with φ
(
0
)
=
0 such that, for all x
,
y
∈
E,(25)
T
r
x
-
T
r
y
2
≤
x
-
y
2
-
r
2
α
-
r
c
B
x
-
B
y
2
-
φ
I
-
J
r
A
I
-
r
B
x
-
I
-
J
r
A
I
-
r
B
y
,where T
r
=
J
r
A
(
I
-
r
B
)
=
(
I
+
r
A
)
-
1
(
I
-
r
B
).We will adopt the following notation in this paper:x
n
→
x means that x
n
→
x strongly.


## 2. Main Results

We first prove the following lemma in a real Banach space with Fréchet differentiable norm.Lemma 11.
LetE be a real Banach space with Fréchet differentiable norm. Let A
:
E
→
2
E be an m-accretive operator and let B
:
E
→
E be an α-inverse strongly accretive mapping. Assume that (
A
+
B
)
-
1
(
0
)
≠
∅. Let f
:
E
→
E be a fixed contraction with coefficient k
∈
(
0,1
). Let {
r
n
}
n
=
1
∞ be a sequence of positive real numbers and suppose that {
α
n
}
n
=
1
∞ and {
β
n
}
n
=
1
∞ are sequences in (
0,1
). Let {
x
n
}
n
=
1
∞
⊂
E be a sequence generated by x
1
∈
E,(26)
y
n
=
α
n
f
x
n
+
1
-
α
n
x
n
,
x
n
+
1
=
β
n
x
n
+
1
-
β
n
J
r
n
A
y
n
-
r
n
B
y
n
,
n
≥
1
,where J
r
n
A
=
(
I
+
r
n
A
)
-
1, 0
<
r
n
<
2
α
/
c, n
≥
1. Then {
x
n
}
n
=
1
∞ is bounded.Proof.
For eachn
≥
1, let T
n
≔
J
r
n
A
(
I
-
r
n
B
) and x
∗
∈
(
A
+
B
)
-
1
(
0
). Then, for all x
,
y
∈
E, we have(27)
T
n
x
-
T
n
y
2
≤
J
r
n
A
x
-
r
n
B
x
-
J
r
n
A
y
-
r
n
B
y
2
≤
x
-
y
-
r
n
B
x
-
B
y
2
≤
x
-
y
2
-
2
r
n
B
x
-
B
y
,
j
x
-
y
+
c
r
n
2
B
x
-
B
y
2
≤
x
-
y
2
-
2
r
n
α
B
x
-
B
y
2
+
c
r
n
2
B
x
-
B
y
2
=
x
-
y
2
-
2
α
-
c
r
n
r
n
B
x
-
B
y
2
≤
x
-
y
2
.Thus, T
n is nonexpansive for all n
≥
1. Furthermore,(28)
x
=
T
n
x
⟺
x
=
I
+
r
n
A
-
1
x
-
r
n
B
x
⟺
x
-
r
n
B
x
∈
x
+
r
n
A
x
⟺
0
∈
A
x
+
B
x
.Thus, F
(
T
n
)
=
(
A
+
B
)
-
1
(
0
), ∀
n
≥
1. So,(29)
y
n
-
x
∗
≤
α
n
f
x
n
-
x
∗
+
1
-
α
n
x
n
-
x
∗
≤
α
n
k
x
n
-
x
∗
+
α
n
f
x
∗
-
x
∗
+
1
-
α
n
x
n
-
x
∗
=
1
-
α
n
1
-
k
x
n
-
x
∗
+
α
n
f
x
∗
-
x
∗
.Now, using (29), we have(30)
x
n
+
1
-
x
∗
≤
β
n
x
n
-
x
∗
+
1
-
β
n
T
n
y
n
-
x
∗
≤
β
n
x
n
-
x
∗
+
1
-
β
n
y
n
-
x
∗
≤
β
n
x
n
-
x
∗
+
1
-
α
n
1
-
k
1
-
β
n
x
n
-
x
∗
+
1
-
β
n
α
n
f
x
∗
-
x
∗
=
1
-
1
-
k
α
n
1
-
β
n
x
n
-
x
∗
+
α
n
1
-
β
n
f
x
∗
-
x
∗
≤
max
⁡
x
n
-
x
∗
,
f
x
∗
-
x
∗
1
-
k
≤
⋯
≤
max
⁡
x
1
-
x
∗
,
f
x
∗
-
x
∗
1
-
k
<
∞
.Hence, {
x
n
}
n
=
1
∞ is bounded.Using the idea of the proof partly taken from [18, 26, 30, 31], we state and prove the following strong convergence theorem.Theorem 12.
LetE be a real uniformly convex Banach space which is also uniformly smooth. Let A
:
E
→
2
E be an m-accretive operator and let B
:
E
→
E be an α-inverse strongly accretive. Assume that (
A
+
B
)
-
1
(
0
)
≠
∅. Let f
:
E
→
E be a fixed contraction with coefficient k
∈
(
0,1
). Let {
r
n
}
n
=
1
∞ be a sequence of positive real numbers and suppose that {
α
n
}
n
=
1
∞ and {
β
n
}
n
=
1
∞ are sequences in (
0,1
) satisfying the following conditions:(i)
lim
n
→
∞
⁡
α
n
=
0;
(ii)
∑
n
=
1
∞
‍
α
n
=
∞;
(iii)
0
<
lim
inf
n
→
∞
⁡
r
n
≤
lim
sup
n
→
∞
⁡
r
n
<
2
α
/
c;
(iv)
0
<
lim
inf
n
→
∞
⁡
β
n
≤
lim
sup
n
→
∞
⁡
β
n
<
1.Let the sequence {
x
n
}
n
=
1
∞ be generated by (26). Then {
x
n
}
n
=
1
∞ converges strongly to z
=
Q
(
A
+
B
)
-
1
(
0
)
f
(
z
), where Q
(
A
+
B
)
-
1
(
0
) is the unique sunny nonexpansive retraction of E onto (
A
+
B
)
-
1
(
0
); that is, z solves the variational inequality(31)
I
-
f
z
,
j
z
-
x
≤
0
,
∀
x
∈
A
+
B
-
1
0
.Proof.
Letz
=
Q
(
A
+
B
)
-
1
(
0
)
f
(
z
). Then, using Lemma 6 in (26), we have that(32)
y
n
-
z
2
=
α
n
f
x
n
-
z
+
1
-
α
n
x
n
-
z
2
=
α
n
f
x
n
-
f
z
+
1
-
α
n
x
n
-
z
+
α
n
f
z
-
z
2
≤
α
n
f
x
n
-
f
z
+
1
-
α
n
x
n
-
z
2
+
2
α
n
f
z
-
z
,
j
y
n
-
z
≤
α
n
f
x
n
-
f
z
2
+
1
-
α
n
x
n
-
z
2
+
2
α
n
f
z
-
z
,
j
y
n
-
z
≤
α
n
k
2
x
n
-
z
2
+
1
-
α
n
x
n
-
z
2
+
2
α
n
f
z
-
z
,
j
y
n
-
z
=
1
-
α
n
1
-
k
2
x
n
-
z
2
+
2
α
n
f
z
-
z
,
j
y
n
-
z
.It follows from (32) and Lemma 10 that there exists a continuous, strictly increasing, and convex function φ
:
R
+
→
R
+ with φ
(
0
)
=
0 such that(33)
T
n
y
n
-
z
2
=
T
r
n
y
n
-
z
2
≤
y
n
-
z
2
-
r
n
2
α
-
r
n
c
B
y
n
-
B
z
2
-
φ
I
-
J
r
n
A
I
-
r
n
B
y
n
-
I
-
J
r
n
A
I
-
r
n
B
z
≤
1
-
α
n
1
-
k
2
x
n
-
z
2
+
2
α
n
f
z
-
z
,
j
y
n
-
z
-
r
n
2
α
-
r
n
c
B
y
n
-
B
z
2
-
φ
I
-
J
r
n
A
I
-
r
n
B
y
n
-
I
-
J
r
n
A
I
-
r
n
B
z
.Now, using (26), we have(34)
x
n
+
1
-
z
2
≤
β
n
x
n
-
z
2
+
1
-
β
n
T
n
y
n
-
z
2
≤
β
n
x
n
-
z
2
+
1
-
β
n
1
-
α
n
1
-
k
2
x
n
-
z
2
+
2
α
n
1
-
β
n
f
z
-
z
,
j
y
n
-
z
-
r
n
2
α
-
r
n
c
1
-
β
n
B
y
n
-
B
z
2
-
1
-
β
n
φ
I
-
J
r
n
A
I
-
r
n
B
y
n
-
I
-
J
r
n
A
I
-
r
n
B
z
+
2
α
n
1
-
β
n
f
z
-
z
,
j
y
n
-
z
.Let s
n
=
x
n
-
z
2, c
n
=
r
n
2
α
-
r
n
c
1
-
β
n
B
y
n
-
B
z
2
+
(
1
-
β
n
)
φ
(
y
n
-
r
n
B
y
n
-
T
n
y
n
+
r
n
B
z
), and ϵ
n
=
2
〈
f
(
z
)
-
z
,
j
(
y
n
-
z
)
〉. Then, (34) becomes(35)
s
n
+
1
≤
1
-
α
n
1
-
β
n
1
-
k
2
s
n
+
α
n
1
-
β
n
ϵ
n
-
c
n
,
n
≥
1
.We now show that s
n
→
0, n
→
∞ by considering two possible cases.
Case 1. Suppose that {
s
n
}
n
=
1
∞ is eventually decreasing. That is, there exists N
≥
0 such that {
s
n
}
n
=
1
∞ is decreasing for n
≥
N. This implies that {
s
n
}
n
=
1
∞ is convergent. It then follows from (35) that(36)
0
≤
c
n
≤
s
n
-
s
n
+
1
+
α
n
1
-
β
n
ϵ
n
-
1
-
k
2
s
n
,
n
≥
1
.This implies that(37)
lim
n
→
∞
⁡
B
y
n
-
B
z
2
+
y
n
-
r
n
B
y
n
-
T
n
y
n
+
r
n
B
z
=
0
.Consequently,(38)
lim
n
→
∞
⁡
T
n
y
n
-
y
n
=
0
.Since lim
inf
n
→
∞
⁡
r
n
>
0, there exists ϵ
>
0 such that r
n
≥
ϵ, ∀
n
≥
1
. Then, by Lemma 7, we have(39)
lim
n
→
∞
⁡
T
ϵ
y
n
-
y
n
≤
2
lim
n
→
∞
⁡
T
n
y
n
-
y
n
=
0
.Also, from (26), we obtain(40)
y
n
-
x
n
=
α
n
f
x
n
-
x
n
⟶
0
,
n
⟶
∞
.Since F
(
T
ϵ
)
=
(
A
+
B
)
-
1
(
0
), it follows from Lemma 5 that(41)
lim
sup
n
→
∞
⁡
f
z
-
z
,
j
y
n
-
z
≤
0
.Equivalently, lim
sup
n
→
∞
⁡
ϵ
n
≤
0. We observe further that(42)
x
n
-
T
n
y
n
≤
y
n
-
x
n
+
T
n
y
n
-
y
n
⟶
0
,
n
⟶
∞
.Again from (26), we have that (43)
x
n
+
1
-
T
n
y
n
=
β
n
x
n
-
T
n
y
n
⟶
0
,
n
⟶
∞
.Hence,(44)
x
n
+
1
-
x
n
≤
x
n
-
T
n
y
n
+
x
n
+
1
-
T
n
y
n
⟶
0
,
n
⟶
∞
.From (36), we have that(45)
s
n
≤
s
n
-
s
n
+
1
α
n
1
-
β
n
+
ϵ
n
1
1
-
k
2
.Observe that lim
inf
n
→
∞
⁡
s
n
-
s
n
+
1
/
α
n
(
1
-
β
n
)
=
0 since ∑
n
=
1
∞
α
n
(
1
-
β
n
)
=
∞. Taking l
i
m
i
n
f in (45), we have(46)
lim
n
→
∞
⁡
s
n
≤
1
1
-
k
2
lim
inf
n
→
∞
⁡
s
n
-
s
n
+
1
α
n
1
-
β
n
+
ϵ
n
≤
1
1
-
k
2
lim
inf
n
→
∞
⁡
s
n
-
s
n
+
1
α
n
1
-
β
n
+
lim
sup
n
→
∞
⁡
ϵ
n
≤
0and thus {
s
n
}
n
=
1
∞ must converge to zero.
Case 2. Suppose {
s
n
}
n
=
1
∞ is not eventually decreasing. Hence, we can find a subsequence {
s
n
k
}
k
=
1
∞ of {
s
n
}
n
=
1
∞ such that s
n
k
≤
s
n
k
+
1, ∀
k
≥
1. Then we can define a subsequence {
S
τ
(
n
)
} as in Lemma 3 so that(47)
max
⁡
s
τ
n
,
s
n
≤
s
τ
n
+
1
,
∀
n
>
n
0
.This implies from (36) that(48)
0
≤
c
τ
n
≤
α
τ
n
1
-
β
τ
n
ϵ
τ
n
-
1
-
k
2
s
τ
n
⟶
0
,
n
⟶
∞and thus T
τ
(
n
)
y
τ
(
n
)
-
y
τ
(
n
)
→
0, n
→
∞. In a similar way to Case 1, we have(49)
lim
sup
n
→
∞
⁡
f
z
-
z
,
j
y
τ
n
-
z
≤
0
.Equivalently, lim
sup
n
→
∞
⁡
ϵ
τ
(
n
)
≤
0 which by (48) implies that lim
sup
n
→
∞
⁡
s
τ
(
n
)
≤
0. By repeating the same arguments in Case 1, we can show that(50)
y
n
-
x
n
=
α
n
f
x
n
-
x
n
⟶
0
,
n
⟶
∞
,
x
τ
n
+
1
-
x
τ
n
⟶
0
,
n
⟶
∞
.Using the definition of {
s
n
}, we have(51)
s
τ
n
+
1
-
s
τ
n
=
x
τ
n
+
1
-
z
2
-
x
τ
n
-
z
2
≤
x
τ
n
+
1
-
x
τ
n
x
τ
n
+
1
-
z
+
x
τ
n
-
z
⟶
0
,
n
⟶
∞
.This together with (47) implies that(52)
0
≤
s
n
≤
s
τ
n
+
1
=
s
τ
n
+
s
τ
n
+
1
-
s
τ
n
⟶
0
,
n
⟶
∞
.Consequently, we have that s
n
→
0, n
→
∞ and so x
n
→
z, n
→
∞.If the mappingf maps every point in E to a fixed element, then we have the following result.Corollary 13.
LetE be a real uniformly convex Banach space which is also uniformly smooth. Let A
:
E
→
2
E be an m-accretive operator and let B
:
E
→
E be an α-inverse strongly accretive. Assume that (
A
+
B
)
-
1
(
0
)
≠
∅. Let u be a fixed element in E. Let {
r
n
}
n
=
1
∞ be a sequence of positive real numbers and suppose that {
α
n
}
n
=
1
∞ and {
β
n
}
n
=
1
∞ are sequences in (
0,1
) satisfying the following conditions:(i)
lim
n
→
∞
⁡
α
n
=
0;
(ii)
∑
n
=
1
∞
α
n
=
∞;
(iii)
0
<
lim
inf
n
→
∞
⁡
r
n
≤
lim
sup
n
→
∞
⁡
r
n
<
2
α
/
c;
(iv)
0
<
lim
inf
n
→
∞
⁡
β
n
≤
lim
sup
n
→
∞
⁡
β
n
<
1.For a fixed element u
∈
E, let the sequence {
x
n
}
n
=
1
∞be generated by x
1
∈
E,(53)
y
n
=
α
n
u
+
1
-
α
n
x
n
,
x
n
+
1
=
β
n
x
n
+
1
-
β
n
J
r
n
A
y
n
-
r
n
B
y
n
,
n
≥
1
,where J
r
n
A
=
(
I
+
r
n
A
)
-
1. Then {
x
n
}
n
=
1
∞ converges strongly to z
=
Q
(
A
+
B
)
-
1
(
0
)
u, where Q
(
A
+
B
)
-
1
(
0
) is the unique sunny nonexpansive retraction of E onto (
A
+
B
)
-
1
(
0
); that is, z solves the variational inequality(54)
z
-
u
,
j
z
-
x
≤
0
,
∀
x
∈
A
+
B
-
1
0
.Recall that a mappingh
:
E
→
E is said to beweakly contractive if(55)
h
x
-
h
y
≤
x
-
y
-
ψ
x
-
y
,
∀
x
,
y
∈
E
,where ψ
:
[
0
,
∞
)
→
[
0
,
∞
) is a continuous and strictly increasing function such that ψ is positive on (
0
,
∞
) and ψ
(
0
)
=
0. Let ψ
(
t
)
≔
(
1
-
k
)
t for t
∈
[
0
,
∞
), where k
∈
(
0,1
). Then we see that the weakly contractive mapping h is a contraction with constant k. The next lemma gives existence of unique fixed point for weakly contractive mapping.Lemma 14 (see [32, Theorem 2]).
Let(
X
,
d
) be a complete metric space and h a weakly contractive mapping on X. Then, h has a unique fixed point p in X. Moreover, for x
∈
X
,
{
h
n
x
} converges strongly to p.The following lemma will be used in the next theorem.Lemma 15 (see [33, 34]).
Let{
s
n
} and {
γ
n
} be two sequences of nonnegative real numbers and {
α
n
} a sequence of positive numbers satisfying the following conditions:(i)
∑
n
=
1
∞
α
n
=
∞;
(ii)
lim
n
→
∞
⁡
(
γ
n
/
α
n
)
=
0.Let the recursive inequality (56)
s
n
+
1
≤
s
n
-
α
n
ψ
s
n
+
γ
n
,
n
≥
1
,where ψ
:
0
,
∞
→
0
,
∞ is a continuous and strictly increasing function such that ψ is positive on 0
,
∞ and ψ
(
0
)
=
0. Then, lim
n
→
∞
⁡
s
n
=
0.We now consider the viscosity approximation method with weakly contractive mapping in the following theorem.Theorem 16.
LetE be a real uniformly convex Banach space which is also uniformly smooth. Let A
:
E
→
2
E be an m-accretive operator and let B
:
E
→
E be an α-inverse strongly accretive. Assume that (
A
+
B
)
-
1
(
0
)
≠
∅. Let h
:
E
→
E be a weakly contractive mapping. Let {
r
n
}
n
=
1
∞ be a sequence of positive real numbers and suppose that {
α
n
}
n
=
1
∞ and {
β
n
}
n
=
1
∞ are sequences in (
0,1
) satisfying the following conditions:(i)
lim
n
→
∞
⁡
α
n
=
0;
(ii)
∑
n
=
1
∞
α
n
=
∞;
(iii)
0
<
lim
inf
n
→
∞
⁡
r
n
≤
lim
sup
n
→
∞
⁡
r
n
<
2
α
/
c;
(iv)
0
<
lim
inf
n
→
∞
⁡
β
n
≤
lim
sup
n
→
∞
⁡
β
n
<
1.Let {
x
n
}
n
=
1
∞
⊂
E be a sequence generated by x
1
∈
E,(57)
y
n
=
α
n
h
x
n
+
1
-
α
n
x
n
,
x
n
+
1
=
β
n
x
n
+
1
-
β
n
J
r
n
A
y
n
-
r
n
B
y
n
,
n
≥
1
.Then {
x
n
}
n
=
1
∞ converges strongly to z
=
Q
(
A
+
B
)
-
1
(
0
)
h
(
z
), where Q
(
A
+
B
)
-
1
(
0
) is the unique sunny nonexpansive retraction of E onto (
A
+
B
)
-
1
(
0
); that is, z solves the variational inequality(58)
I
-
h
z
,
j
z
-
x
≤
0
,
∀
x
∈
A
+
B
-
1
0
.Proof.
We first observe thatQ
(
A
+
B
)
-
1
(
0
)
h is a weakly contractive mapping. Indeed, ∀
x
,
y
∈
E,(59)
Q
A
+
B
-
1
0
h
x
-
Q
A
+
B
-
1
0
h
y
≤
h
x
-
h
y
≤
x
-
y
-
ψ
x
-
y
.Then Lemma 14 assures that there exists a unique element z
∈
(
A
+
B
)
-
1
(
0
) such that z
=
Q
(
A
+
B
)
-
1
(
0
)
h
(
z
). Now, define an iterative scheme as follows:(60)
w
n
+
1
=
β
n
w
n
+
1
-
β
n
J
r
n
A
I
-
r
n
B
α
n
h
z
+
1
-
α
n
w
n
,
n
≥
1
.Let {
w
n
} be the sequence generated by (60). Then, Corollary 13 with f
=
h
(
z
) a constant assures that {
w
n
} converges strongly to z
=
Q
(
A
+
B
)
-
1
(
0
)
h
(
z
). Now, for n
≥
1, we obtain(61)
x
n
+
1
-
w
n
+
1
=
β
n
x
n
+
1
-
β
n
J
r
n
A
I
-
r
n
B
α
n
h
x
n
+
1
-
α
n
x
n
-
β
n
w
n
-
1
-
β
n
J
r
n
A
I
-
r
n
B
α
n
h
z
+
1
-
α
n
w
n
≤
β
n
x
n
-
w
n
+
1
-
β
n
α
n
h
x
n
+
1
-
α
n
x
n
-
α
n
h
z
+
1
-
α
n
w
n
≤
β
n
x
n
-
w
n
+
α
n
1
-
β
n
h
x
n
-
h
z
+
1
-
α
n
1
-
β
n
x
n
-
w
n
≤
β
n
x
n
-
w
n
+
α
n
1
-
β
n
h
x
n
-
h
w
n
+
h
w
n
-
h
z
+
1
-
α
n
1
-
β
n
x
n
-
w
n
≤
β
n
x
n
-
w
n
+
α
n
1
-
β
n
x
n
-
w
n
-
ψ
x
n
-
w
n
+
α
n
1
-
β
n
w
n
-
z
-
ψ
w
n
-
z
+
1
-
α
n
1
-
β
n
x
n
-
w
n
=
x
n
-
w
n
-
α
n
1
-
β
n
ψ
x
n
-
w
n
+
α
n
1
-
β
n
w
n
-
z
-
ψ
w
n
-
z
≤
x
n
-
w
n
-
α
n
1
-
β
n
ψ
x
n
-
w
n
+
α
n
1
-
β
n
w
n
-
z
.Take s
n
=
x
n
-
w
n; we have the following recursive inequality:(62)
s
n
+
1
≤
s
n
-
α
n
ψ
s
n
+
α
n
w
n
-
z
.Since w
n
-
z
→
0
,
n
→
∞, it follows from Lemma 15 that x
n
-
w
n
→
0, n
→
∞. Furthermore,(63)
lim
n
→
∞
⁡
x
n
-
z
≤
lim
n
→
∞
⁡
x
n
-
w
n
+
w
n
-
z
=
0
.This completes the proof.Remark 17.
All the results in this paper can still be obtained if error terms are added to our iterative scheme (26). Thus, addition of error terms to our iterative scheme (26) leads to no further generalization of the results obtained in this paper.Remark 18.
We make the following comments which highlight our contributions in this paper (please see also [31]): (1)
The results of Wei and Shi [23] do not hold in L
p spaces with p
≠
2. Here in this paper, our results hold, in particular, in L
p spaces with p
≥
2. Furthermore, the condition ∑
n
=
1
∞
|
r
n
+
1
-
r
n
|
<
∞ assumed in the results of Wei and Shi [23] is dispensed with in our results. Thus, our results improve on the results of Wei and Shi [23].
(2)
The condition∑
n
=
1
∞
r
n
+
1
-
r
n
<
∞ assumed in the results of Wei and Duan [25] is dispensed with in our results. Also, the condition(64)
B
x
-
B
y
,
J
I
-
r
B
x
-
I
-
r
B
y
≥
0
,
for allx
,
y
∈
E and for all r
>
0 assumed in the results of Wei and Duan [25], is dispensed with in our results. Hence, our results improved on the results of Wei and Duan [25].
(3)
Our results extend the results of [21–23, 25] from q-uniformly smooth Banach spaces to uniformly smooth Banach spaces. By Remark 1, our results also extend the results of [21–23, 25] from the class of inverse strongly accretive operators of order q to the more general class of inverse strongly accretive operators.
(4)
In our result in this paper, we consider the viscosity approximation method with weakly contractive mapping for solving (10), which extends all the previous results concerning viscosity approximation method with fixed contraction for solving (10) in q-uniformly smooth Banach spaces to uniformly smooth Banach spaces. Our result in this paper also complements the results of Shehu and Cai [31].Remark 19.
We specifically clarify the differences between the results of this paper and the results obtained in [23, 25]:(1)
We state here that the results obtained in Wei and Duan [25] and Wei and Shi [23] do not hold in L
p spaces with p
≠
2 because the duality mapping J is assumed to be weakly sequentially continuous but here in this paper, our results are applicable, in particular, in L
p spaces with 2
≤
p
<
∞ because the weak sequential continuity of the duality mapping J is dispensed with.
(2)
The choice of sequence{
r
n
} of positive real numbers with(65)
r
n
=
n
-
1
/
2
,
if
n
is  odd
,
n
-
1
/
2
-
1
-
1
,
if
n
is  even
does not satisfy the extra condition∑
n
=
1
∞
r
n
+
1
-
r
n
<
∞ assumed in Wei and Duan [25] and Wei and Shi [23].



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