# Fixed Point Theorems for an Elastic Nonlinear Mapping in Banach Spaces

Authors: Hiroko Manaka
Journal: Abstract and Applied Analysis
 (2015)

---

## Abstract

LetE be a smooth Banach space with a norm ·. Let V
(
x
,
y
)
=
x
2
+
y
2
-
2
x
,
J
y for any x
,
y
∈
E, where ·
,
· stands for the duality pair and J is the normalized duality mapping. We define a V-strongly nonexpansive mapping by V
(
·
,
·
). This nonlinear mapping is nonexpansive in a Hilbert space. However, we show that there exists a V-strongly nonexpansive mapping with fixed points which is not nonexpansive in a Banach space. In this paper, we show a weak convergence theorem and strong convergence theorems for fixed points of this elastic nonlinear mapping and give the existence theorem.


---

## Body

## 1. Introduction

LetE be a smooth Banach space with a norm · and let E
* be the dual space of E. We denote by ·
,
· a duality pair on E
×
E
* and let J be the normalized duality mapping on E. It is well known that J is a continuous single-valued mapping in a smooth Banach space and a one-to-one mapping in a strictly convex Banach space (cf. [1]). We define a mapping V
:
E
×
E
→
R by V
(
x
,
y
)
=
x
2
+
y
2
-
2
x
,
J
y for all x
,
y
∈
E, where R is a set of real numbers. It is obvious that V
(
x
,
y
)
≥
(
x
-
y
)
2
≥
0. Let any y
∈
E be fixed, and then V
(
·
,
y
) is a convex function because of convexity of ·
2. Many nonlinear mappings which are defined by using V
(
·
,
·
) are studied (see [2–4]). We also defined a nonlinear mapping which is called a V-strongly nonexpansive mapping in [5] as follows.Definition 1.
LetC be a nonempty subset of a smooth Banach space E. A mapping T
:
C
→
E is called V-strongly nonexpansive if there exists a constant λ
>
0 such that for all x
,
y
∈
C
(1)
V
(
T
x
,
T
y
)
≤
V
(
x
,
y
)
-
λ
V
I
-
T
x
,
I
-
T
y
,where I is the identity mapping on E.From this definition, it is obvious that the identity mappingI is also a V-strongly nonexpansive mapping. In a Hilbert space, it is trivial that this mapping is nonexpansive since V
(
x
,
y
)
=
x
-
y
2 and that any firmly nonexpansive mapping is a V-strongly nonexpansive mapping with λ
=
1 (see [5]). Moreover, we showed that if there exists a fixed point of a V-strongly nonexpansive mapping T, then T is strongly nonexpansive with a Bregman distance in [5]. However, in Banach spaces, as we give an example in the later section, we find that there exists a V-strongly nonexpansive mapping with fixed points which is not nonexpansive. We should point out that a guarantee of continuity of the V-strongly nonexpansive mappings has not been given in a generalized Banach space yet.In this paper, we prove a weak convergence theorem and strong convergence theorems for finding fixed points of aV-strongly nonexpansive mapping in Banach spaces and show the existence theorem of fixed point with a dissipative property.


## 2. Preliminaries

In this section, at first we show the relationship between aV-strongly nonexpansive mapping and other nonlinear mappings, in a Hilbert space. Secondly, we state some properties of V-strongly nonexpansive mappings in a Banach space and give an example of a V-strongly nonexpansive mapping which is not a quasinonexpansive mapping in a Banach space although T has fixed points. We finally show some lemmas which are necessary in order to prove our theorems.LetC be a subset of a Banach space E and let T
:
C
→
E be a mapping. Then a point p in the closure of C is said to be an asymptotically fixed point of T if C contains a sequence {
x
n
} which converges weakly to p and the sequence {
x
n
-
T
x
n
} converges strongly to 0. F
^
(
T
) denotes the set of asymptotically fixed points of T. In [6], Reich introduced a strongly nonexpansive mapping which is defined by using the Bregman distance D
(
·
,
·
).Definition 2.
LetE be a Banach space. The Bregman distance corresponding to a function f
:
E
→
R is defined by (2)
D
x
,
y
=
f
x
-
f
y
-
f
′
y
x
-
y
,where f is Gâteaux differentiable and f
′
(
x
) stands for the derivative of f at the point x. Let C be a nonempty subset of E. We say that the mapping T
:
C
→
E is strongly nonexpansive if F
^
(
T
)
≠
∅ and (3)
D
p
,
T
x
≤
D
p
,
x
∀
p
∈
F
^
(
T
)
x
∈
C
,and if it holds that lim
n
→
∞
D
(
T
x
n
,
x
n
)
=
0 for a bounded sequence {
x
n
} such that lim
n
→
∞
(
D
(
p
,
x
n
)
-
D
(
p
,
T
x
n
)
)
=
0 for any p
∈
F
^
(
T
).Taking the function·
2 as the convex, continuous, and Gâteaux differentiable function f, we obtain the fact that the Bregman distance D
(
·
,
·
) coincides with V
(
·
,
·
). In particular, in a Hilbert space, it is trivial that D
(
x
,
y
)
=
V
(
x
,
y
)
=
x
-
y
2.Proposition 3 (see [5]).
In a Hilbert space, aV-strongly nonexpansive mapping with F
^
(
T
)
≠
∅ is strongly nonexpansive.Next we recall two mappings of other nonlinear mappings (cf. [6–9]). A firmly nonexpansive mapping and an α-inverse strongly monotone mapping are defined as follows.Definition 4.
LetC be a nonempty, closed, and convex subset of a Banach space E. A mapping T
:
C
→
E is said to be firmly nonexpansive if (4)
T
x
-
T
y
2
≤
x
-
y
,
jfor all x
,
y
∈
C and some j
∈
J
(
T
x
-
T
y
).It is trivial that a firmly nonexpansive mapping is nonexpansive.Definition 5.
LetH be a Hilbert space. A mapping T
:
C
→
H is said to be α-inverse strongly monotone if (5)
α
T
x
-
T
y
2
≤
x
-
y
,
T
x
-
T
yfor all x
,
y
∈
C.The relation among firmly nonexpansive mappings,α-inverse strongly monotone mappings and V-strongly nonexpansive mappings is shown in the following proposition.Proposition 6 (see [5]).
In a Hilbert space, the following hold.(a)
A firmly nonexpansive mapping isV-strongly nonexpansive with λ
=
1.
(b)
LetA be an α-inverse strongly monotone mapping for α
>
1
/
2; then S
=
(
I
-
A
) is V-strongly nonexpansive with (
2
α
-
1
).The above (b) is obvious by showing that, for allx
,
y
∈
H,(6)
S
x
-
S
y
,
x
-
y
≤
x
-
y
2
-
α
(
I
-
S
)
x
-
I
-
S
y
2
.We will introduce some properties of V-strongly nonexpansive mappings in [5].Proposition 7 (see [5]).
In a smooth Banach spaceE, the following hold.(a)
Forc
∈
(
-
1,1
], T
=
c
I is V-strongly nonexpansive. For c
=
1, T
=
I is V-strongly nonexpansive for any λ
>
0. For c
∈
(
-
1,1
), T
=
c
I is V-strongly nonexpansive for any λ
∈
(
0
,
(
1
+
c
)
/
(
1
-
c
)
].
(b)
IfT is V-strongly nonexpansive with λ, then, for any α
∈
[
-
1,1
] with α
≠
0, α
T is also V-strongly nonexpansive with α
2
λ.
(c)
IfT is V-strongly nonexpansive with λ
≥
1, then A
=
I
-
T is V-strongly nonexpansive with λ
-
1.
(d)
Suppose thatT is V-strongly nonexpansive with λ and that α
∈
[
-
1,1
] satisfies α
2
λ
≥
1. Then (
I
-
α
T
) is V-strongly nonexpansive with (
α
2
λ
)
-
1. Moreover, if  T
α
=
I
-
α
T, then (7)
V
(
T
α
x
,
T
α
y
)
≤
V
(
x
,
y
)
-
λ
-
1
V
(
T
x
,
T
y
)
.Now we give an example of aV-strongly nonexpansive mapping in a Banach space.Example 8 (see [10]).
Let1
<
p
,
q
<
∞ such that 1
/
p
+
1
/
q
=
1. Let E
=
R
×
R be a real Banach space with a norm ·
p defined by (8)
x
p
=
x
1
p
+
x
2
p
1
/
p
∀
x
=
(
x
1
,
x
2
)
∈
E
.Then E is smooth, and the normalized duality mapping J is single-valued. J is given by (9)
J
x
=
x
p
2
-
p
x
1
x
1
p
-
2
,
x
2
x
2
p
-
2
∈
l
q
R
×
R
∀
x
=
x
1
,
x
2
∈
E
.Hence, we have for x
,
y
∈
E that (10)
V
x
,
y
=
x
p
2
+
y
p
2
-
2
x
,
J
y
=
x
p
2
+
y
p
2
-
2
y
p
2
-
p
·
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
.We define a mapping T
:
E
→
E as follows: (11)
T
x
=
x
if
x
p
≤
1
,
1
x
p
x
if
x
p
>
1
.In a case of p
=
1, we have shown that the mapping T defined by (11) is a V-strongly nonexpansive mapping (see [5]). We will show that T is V-strongly nonexpansive with any λ
≤
1, for p
>
1.Proposition 9.
Suppose thatT is defined by the formula (11) under the above situation. Then, T is a V-strongly nonexpansive mapping with any λ
≤
1.Proof.
Case (a): suppose thatx
,
y
∈
E with x
p
≤
1 and y
p
>
1.
SinceT
y
=
(
T
y
)
1
,
(
T
y
)
2
=
y
1
y
p
-
1
,
y
2
y
p
-
1, we have that (12)
V
T
x
,
T
y
=
V
x
,
T
y
=
x
p
2
+
T
y
p
2
-
2
T
y
p
2
-
p
·
x
1
T
y
1
T
y
1
p
-
2
+
x
2
T
y
2
T
y
2
p
-
2
=
x
p
2
+
1
-
2
y
p
1
-
p
·
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
.Since (13)
y
-
T
y
=
y
p
-
1
y
p
y
1
,
y
p
-
1
y
p
y
2
,we have that (14)
V
x
-
T
x
,
y
-
T
y
=
V
(
0
,
y
-
T
y
)
=
y
-
T
y
p
2
=
(
y
p
-
1
)
y
p
y
p
2
=
y
p
-
1
2
.Hence, we obtain that (15)
V
(
x
,
y
)
-
V
(
T
x
,
T
y
)
-
λ
V
(
x
-
T
x
,
y
-
T
y
)
=
x
p
2
+
y
p
2
-
2
y
p
2
-
p
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
-
x
p
2
-
1
+
2
y
p
1
-
p
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
-
λ
y
p
-
1
2
=
y
p
2
-
1
-
2
y
p
1
-
p
(
y
p
-
1
)
·
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
-
λ
y
p
-
1
2
≥
(
y
p
-
1
)
y
p
+
1
-
2
y
p
1
-
p
m
m
m
m
m
m
m
m
m
·
(
x
1
y
1
p
-
1
+
x
2
y
2
p
-
1
)
m
m
m
m
m
m
m
m
m
y
p
1
-
p
-
λ
(
y
p
-
1
)
.Hölder’s inequality implies that (16)
x
1
y
1
p
-
1
+
x
2
y
2
p
-
1
≤
x
p
y
1
p
-
1
q
+
y
2
p
-
1
q
1
/
q
=
x
p
y
1
p
+
y
2
p
1
/
q
=
x
p
y
p
p
-
1
.Therefore, we obtain that (17)
V
(
x
,
y
)
-
V
(
T
x
,
T
y
)
-
λ
V
(
x
-
T
x
,
y
-
T
y
)
≥
(
y
p
-
1
)
·
y
p
+
1
-
2
y
p
1
-
p
x
p
y
p
p
-
1
-
λ
y
p
+
λ
=
(
y
p
-
1
)
y
p
+
1
-
2
x
p
-
λ
y
p
+
λ
≥
(
y
p
-
1
)
(
1
-
λ
)
y
p
+
1
-
2
+
λ
=
(
y
p
-
1
)
(
1
-
λ
)
(
y
p
-
1
)
=
(
1
-
λ
)
y
p
-
1
2
≥
0
,
for
any
λ
∈
[
0,1
]
.
That is, the inequality (1) holds.
Case (b): suppose thatx
,
y
∈
E with x
p
≥
1 and y
p
≤
1.
Then we have that(18)
V
T
x
,
T
y
=
V
T
x
,
y
=
1
+
y
p
2
-
2
x
p
-
1
y
p
2
-
p
·
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
,
(19)
V
(
x
-
T
x
,
y
-
T
y
)
=
V
(
x
p
-
1
)
x
p
x
,
0
=
x
p
-
1
2
.Hence, we have that (20)
V
(
x
,
y
)
-
V
(
T
x
,
T
y
)
-
λ
V
(
x
-
T
x
,
y
-
T
y
)
=
x
p
2
+
y
p
2
-
2
y
p
2
-
p
·
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
-
1
-
y
p
2
+
2
y
p
2
-
p
x
p
-
1
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
-
λ
x
p
-
1
2
≥
x
p
2
-
1
-
2
y
p
2
-
p
·
x
1
y
1
p
-
1
+
x
2
y
2
p
-
1
1
-
x
p
-
1
-
λ
1
-
x
p
2
.As (a), we obtain from Hölder’s inequality that (21)
V
(
x
,
y
)
-
V
(
T
x
,
T
y
)
-
λ
V
(
x
-
T
x
,
y
-
T
y
)
≥
x
p
2
-
1
-
2
x
p
y
p
2
-
p
y
p
p
-
1
·
1
-
x
p
-
1
-
λ
x
p
-
1
2
=
x
p
-
1
x
p
+
1
-
2
y
p
x
p
-
1
-
λ
x
p
-
1
2
=
x
p
-
1
x
p
+
1
-
2
y
p
-
λ
x
p
+
λ
≥
x
p
-
1
(
1
-
λ
)
x
p
-
1
=
(
1
-
λ
)
x
p
-
1
2
≥
0
,
for
any
λ
∈
[
0,1
]
.That is, the inequality (1) holds.
Case (c): suppose thatx
,
y
∈
E with x
p
,
y
p
≥
1.
Then we have that(22)
V
T
x
,
T
y
=
1
+
1
-
2
x
p
-
1
x
1
,
x
2
,
m
m
m
m
i
m
m
m
m
y
p
1
-
p
y
1
y
1
p
-
2
,
y
2
y
2
p
-
2
=
2
-
2
x
p
-
1
y
p
1
-
p
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
,
V
(
x
-
T
x
,
y
-
T
y
)
=
V
x
p
-
1
x
p
x
,
y
p
-
1
y
p
y
=
x
p
-
1
2
+
y
p
-
1
2
-
2
(
x
p
-
1
)
(
y
p
-
1
)
x
p
-
1
y
p
-
1
y
p
2
-
p
·
(
x
1
,
x
2
)
,
y
1
p
-
2
y
1
,
y
2
p
-
2
y
2
=
x
p
-
1
2
+
y
p
-
1
2
-
2
x
p
-
1
y
p
-
1
x
p
-
1
y
p
1
-
p
·
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
.Hence, we have that (23)
V
(
x
,
y
)
-
V
(
T
x
,
T
y
)
-
λ
V
(
x
-
T
x
,
y
-
T
y
)
=
x
p
2
+
y
p
2
-
2
y
p
2
-
p
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
-
2
+
2
x
p
-
1
y
p
1
-
p
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
-
λ
x
p
-
1
2
-
λ
y
p
-
1
2
+
2
λ
x
p
-
1
y
p
-
1
x
p
-
1
y
p
1
-
p
·
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
=
x
p
2
+
y
p
2
-
2
-
λ
x
p
-
1
2
-
λ
y
p
-
1
2
-
2
x
p
-
1
y
p
1
-
p
x
1
y
1
y
1
p
-
2
+
x
2
y
2
y
2
p
-
2
·
x
p
y
p
-
1
-
λ
x
p
-
1
y
p
-
1
.It is obvious that (24)
x
p
y
p
-
1
-
λ
x
p
-
1
y
p
-
1
≥
0for any λ
∈
[
0,1
] and x
p
,
y
p
≥
1. Thus, we have from Hölder’s inequality that (25)
V
(
x
,
y
)
-
V
(
T
x
,
T
y
)
-
λ
V
(
x
-
T
x
,
y
-
T
y
)
≥
x
p
2
+
y
p
2
-
2
-
λ
x
p
-
1
2
-
λ
y
p
-
1
2
-
2
x
p
-
1
y
p
1
-
p
x
p
y
p
p
-
1
·
x
p
y
p
-
1
-
λ
x
p
-
1
y
p
-
1
=
x
p
2
+
y
p
2
-
2
-
λ
x
p
-
1
2
-
λ
y
p
-
1
2
-
2
x
p
y
p
-
1
-
λ
x
p
-
1
y
p
-
1
=
x
p
2
+
y
p
2
-
2
-
λ
·
x
p
2
-
2
x
p
+
1
+
y
p
2
-
2
y
p
+
1
-
2
x
p
y
p
+
2
+
2
λ
x
p
y
p
-
x
p
-
y
p
+
1
=
x
p
-
y
p
2
-
λ
x
p
-
y
p
2
=
(
1
-
λ
)
x
p
-
y
p
2
≥
0
,
for
any
λ
∈
[
0,1
]
.That is, the inequality (1) holds.
It is clear that ifx
p
,
y
p
≤
1 then inequality (1) holds. Therefore, from Cases (a), (b), and (c), we obtain the conclusion that T is V-strongly nonexpansive for any λ
∈
(
0,1
].Remark 10.
Whenp
=
1, we have given the result in [5]. When p
=
2, we already know that E is a Hilbert space and a V-strongly nonexpansive mapping T is nonexpansive.Theorem 11.
There exists aV-strongly nonexpansive mapping T with a nonempty subset of fixed points such that T is not nonexpansive for some Banach space.Proof.
It is enough to show that theV-strongly nonexpansive mapping which is given in the previous proposition is not nonexpansive.
Letx
=
(
0,1
)
∈
E. Suppose that y
=
(
y
1
,
y
2
) satisfies that y
p
p
=
|
y
1
|
p
+
|
y
2
|
p
>
1 and 0
<
y
1
,
y
2
<
1. Then T
y
=
y
p
-
1
y. Let h
=
y
2
/
y
1 and t
=
y
p
-
1
y
1
-
y
1. We have that t
<
0 and y
p
-
1
y
2
-
y
2
=
h
t
<
0. Then we obtain that T
y
=
(
y
p
-
1
y
1
,
y
p
-
1
h
y
1
). Then, we have that (26)
T
x
-
T
y
p
p
=
-
y
p
-
1
y
1
,
1
-
y
p
-
1
h
y
1
p
=
-
y
p
-
1
y
1
p
+
1
-
y
p
-
1
h
y
1
p
=
y
p
-
1
y
1
p
+
1
-
y
p
-
1
h
y
1
p
=
y
1
+
t
p
+
1
-
h
y
1
+
t
p
,and since x
-
y
p
p
=
y
1
p
+
(
1
-
h
y
1
)
p, we have that (27)
T
x
-
T
y
p
p
-
x
-
y
p
p
=
y
1
+
t
p
-
y
1
p
+
1
-
h
y
1
+
t
p
-
1
-
h
y
1
p
.Therefore, we will show that (28)
T
x
-
T
y
p
p
-
x
-
y
p
p
>
0
⟺
y
1
+
t
p
-
y
1
p
+
1
-
h
y
1
+
t
p
-
1
-
h
y
1
p
>
0
⟺
{
y
1
+
t
p
-
y
1
p
}
t
-
1
+
1
-
h
y
1
+
t
p
-
1
-
h
y
1
p
t
-
1
<
0
,since t
<
0. Let h be fixed. As y
p
p
=
y
1
p
+
(
h
y
1
)
p
→
1, t
=
y
p
-
1
y
1
-
y
1
→
0. Thus, we have for a sufficiently small |
t
| that (29)
y
1
+
t
p
-
y
1
p
t
-
1
+
{
1
-
h
y
1
+
t
p
-
1
-
h
y
1
p
}
t
-
1
<
0
⟺
p
y
1
p
-
1
-
p
h
1
-
h
y
1
p
-
1
<
0
.It is trivial that (30)
p
y
1
p
-
1
-
p
h
1
-
h
y
1
p
-
1
<
0
⟺
y
1
p
-
1
<
h
1
-
h
y
1
p
-
1
⟺
y
1
p
<
y
2
1
-
y
2
p
-
1
.Let p
=
3
/
2. For y
=
(
0.2,0.95
), we have that (31)
y
1
p
=
0.2
3
/
2
<
0.95
0.05
1
/
2
=
y
2
1
-
y
2
p
-
1
.We obtain that y
p
p
=
(
0.2
)
3
/
2
+
(
0.95
)
3
/
2
>
1 and that (32)
T
x
-
T
y
p
p
=
y
p
-
p
{
0.2
3
/
2
+
y
p
-
0.95
3
/
2
}
>
0.2
3
/
2
+
0.05
3
/
2
=
x
-
y
p
p
.Therefore, we obtain the conclusion.We remark that the symbolsx
n
→
u and x
n
⇀
u mean that {
x
n
} converges strongly and weakly to u, respectively. We will introduce the following important lemmas for proofs of our theorems.Lemma 12.
(a) For allx
,
y
,
z
∈
E, (33)
V
x
,
y
≤
V
x
,
y
+
V
y
,
z
=
V
(
x
,
z
)
-
2
x
-
y
,
J
y
-
J
z
.(b) Let {
x
n
} be a sequence in E such that there exists lim
n
→
∞
V
(
x
n
,
p
)
<
∞ for some p
∈
E; then {
x
n
} is bounded.Lemma 13 (see [3]).
LetE be a smooth and uniformly convex Banach space and C a nonempty, convex, and closed subset of E. Suppose that T
:
C
→
E satisfies (34)
V
T
x
,
T
y
≤
V
x
,
y
∀
x
,
y
∈
C
.If a weakly convergent sequence {
z
n
}
n
≥
1
⊂
C satisfies that lim
n
→
∞
V
(
T
z
n
,
z
n
)
=
0, it holds that z
n
⇀
z
∈
F
(
T
).Theorem 14 (see [1, 11]).
LetY be a compact subset of a topological vector space E and let X be a convex subset of   Y. Let A
:
X
→
2
Y be an operator such that, for each y
∈
Y, A
-
1
y is convex. Suppose that B
:
X
→
2
Y satisfies the following:(1)
B
x
⊂
A
x for each x
∈
X,
(2)
B
-
1
y
≠
∅ for each y
∈
Y,
(3)
B
x is open for each x
∈
X.
Then there exists a pointx
0
∈
X such that x
0
∈
A
x
0.Lemma 15 (see [12]).
Lets
>
0 and let E be a Banach space. Then E is uniformly convex if and only if there exists a continuous, strictly increasing, and convex function g
:
[
0
,
∞
)
→
[
0
,
∞
), g
(
0
)
=
0, such that (35)
x
+
y
2
≥
x
2
+
2
y
,
j
+
g
(
y
)for all x
,
y
∈
{
z
∈
E
:
z
≤
s
} and j
∈
J
x.Lemma 16 (see [13]).
LetE be a smooth and uniformly convex Banach space. Then, there exists a continuous, strictly increasing, and convex function g
:
[
0
,
∞
)
→
[
0
,
∞
) such that g
(
0
)
=
0 and, for each real number r
>
0, (36)
0
≤
g
(
x
-
y
)
≤
V
(
x
,
y
)for all x
,
y
∈
B
r
=
{
z
∈
E
:
z
≤
r
}.Lemma 17 (see [13]).
LetE be a smooth and uniformly convex Banach space and {
y
n
} and {
z
n
} in E. If   lim
n
→
∞
V
(
y
n
,
z
n
)
=
0 and either {
y
n
} or {
z
n
} is bounded, then {
y
n
-
z
n
}
→
0.


## 3. Main Results

In this section, we prove a weak convergence theorem and strong convergence theorems for finding fixed points of aV-strongly nonexpansive mapping T in Banach spaces, and then we show the existence theorem for fixed points of T with a dissipative property (cf. [10]).Theorem 18.
LetE be a smooth and uniformly convex Banach space and C a nonempty, closed, and convex subset of E. Suppose that a mapping T
:
C
→
C is V-strongly nonexpansive with λ and that F
(
T
)
≠
∅. One defines a Mann iterative sequence {
x
n
} as follows: for any x
1
∈
C and n
≥
1, (37)
x
n
+
1
=
β
n
x
n
+
(
1
-
β
n
)
T
x
n
,where {
β
n
}
⊂
(
0,1
) and lim
n
→
∞
β
n
=
0. Then x
n
⇀
p
0 for some p
0
∈
F
(
T
).Proof.
Suppose thatp
∈
F
(
T
). Then we have from the convexity of V that (38)
V
x
n
+
1
,
p
=
V
(
β
n
x
n
+
(
1
-
β
n
)
T
x
n
,
p
)
≤
β
n
V
(
x
n
,
p
)
+
(
1
-
β
n
)
V
(
T
x
n
,
p
)
=
β
n
V
(
x
n
,
p
)
+
(
1
-
β
n
)
V
(
T
x
n
,
T
p
)
.Since T is V-strongly nonexpansive with λ, we have that (39)
V
(
x
n
+
1
,
p
)
≤
β
n
V
(
x
n
,
p
)
+
(
1
-
β
n
)
·
{
V
(
x
n
,
p
)
-
λ
V
(
(
I
-
T
)
x
n
,
(
I
-
T
)
p
)
}
=
V
(
x
n
,
p
)
-
(
1
-
β
n
)
λ
V
(
x
n
-
T
x
n
,
0
)
≤
V
(
x
n
,
p
)
.Hence, we have lim
n
→
∞
V
(
x
n
,
p
)
=
α
<
∞. From Lemma 12 (b), {
x
n
} is bounded. Furthermore, we have that (40)
(
1
-
β
n
)
λ
V
(
x
n
-
T
x
n
,
0
)
≤
V
(
x
n
,
p
)
-
V
(
x
n
+
1
,
p
)
.Since lim
n
→
∞
β
n
=
lim
n
→
∞
{
V
(
x
n
,
p
)
-
V
(
x
n
+
1
,
p
)
}
=
0, we obtain that (41)
lim
n
→
∞
V
(
x
n
-
T
x
n
,
0
)
=
lim
n
→
∞
x
n
-
T
x
n
2
=
0
.This means that {
x
n
-
T
x
n
} converges strongly to 0. Hence, {
T
x
n
} is also bounded, and there exists M
>
0 such that x
n
,
T
x
n
≤
M
-
p for all n
≥
1.
On the other hand, we have from Lemma12 (a) that (42)
0
≤
V
(
x
n
,
T
x
n
)
=
V
(
x
n
,
p
)
-
V
(
T
x
n
,
p
)
-
2
x
n
-
T
x
n
,
J
T
x
n
-
J
p
≤
V
(
x
n
,
p
)
-
V
(
T
x
n
,
p
)
+
2
x
n
-
T
x
n
(
T
x
n
+
p
)
≤
V
(
x
n
,
p
)
-
V
(
T
x
n
,
p
)
+
2
M
x
n
-
T
x
n
=
x
n
2
-
T
x
n
2
-
2
x
n
-
T
x
n
,
J
p
+
2
M
x
n
-
T
x
n
=
(
x
n
-
T
x
n
)
(
x
n
+
T
x
n
)
-
2
x
n
-
T
x
n
,
J
p
+
2
M
x
n
-
T
x
n
≤
x
n
-
T
x
n
(
x
n
+
T
x
n
+
2
M
)
-
2
x
n
-
T
x
n
,
J
p
.Hence, we obtain that lim
n
→
∞
V
(
x
n
,
T
x
n
)
=
lim
n
→
∞
V
(
T
x
n
,
x
n
)
=
0. From Lemma 13, there exists a point p
0
∈
F
(
T
) such that x
n
⇀
p
0 and T
x
n
⇀
p
0.The duality mappingJ of a Banach space E with Gâteaux differentiable norm is said to be weakly sequentially continuous if x
n
⇀
x in E implies that {
J
x
n
} converges weak star to J
x in E
* (cf. [14]). This happens, for example, if E is a Hilbert space, or finite-dimensional and smooth, or l
p if 1
<
p
<
∞ (cf. [15]). Next we prove a strong convergence theorem.Theorem 19.
LetE be a reflexive, smooth, and strictly convex Banach space. Suppose that the duality mapping J of E is weakly sequentially continuous. Suppose that C is a nonempty, closed, and convex subset of E, T
:
C
→
C is V-strongly nonexpansive with λ, and F
(
T
)
≠
∅. One defines a Mann iterative sequence {
x
n
} as follows: for any x
1
∈
C and n
≥
1,(43)
x
n
+
1
=
β
n
x
n
+
(
1
-
β
n
)
T
x
n
,where {
β
n
}
⊂
(
0,1
) and lim
n
→
∞
β
n
=
0. If T satisfies that (44)
x
,
J
T
x
≤
0
∀
x
∈
C
,then x
n
→
p
0 and T
x
n
→
p
0 for some p
0
∈
F
(
T
).Proof.
As in the proof of Theorem18, we obtain that lim
n
→
∞
V
(
x
n
,
T
x
n
)
=
0 and x
n
⇀
p
0 and T
x
n
⇀
p
0 for some p
0
∈
F
(
T
). Furthermore, from Lemma 12 (a), we have that (45)
0
≤
V
(
x
n
,
p
0
)
+
V
(
p
0
,
T
x
n
)
=
V
(
x
n
,
T
x
n
)
-
2
x
n
-
p
0
,
J
p
0
-
J
T
x
n
=
V
(
x
n
,
T
x
n
)
-
2
x
n
-
p
0
,
J
p
0
+
2
x
n
,
J
T
x
n
-
2
p
0
,
J
T
x
n
.Hence, the assumptions imply that (46)
V
x
n
,
p
0
⟶
0
,
V
p
0
,
T
x
n
⟶
0
as
n
⟶
∞
.From Lemma 17, we have the conclusion that x
n
→
p
0 and T
x
n
→
p
0.Condition (44) is a definition of a linear dissipative mapping T (cf. [16]). Moreover, we give a definition of a J-dissipative mapping for nonlinear mappings in a Banach space.Definition 20.
LetJ be a single-valued duality mapping on E and let C be a nonempty subset of E. Then a mapping T
:
C
→
E is called J-dissipative if it holds that (47)
x
-
y
,
J
T
x
-
J
T
y
≤
0for all x
,
y
∈
C.In a Hilbert space, such a mappingT is called dissipative. In Banach spaces, we remark that the J-dissipative mapping is not equal to the dissipative mapping (cf. [17]). Next we give a characterization of J-dissipative mappings by using V
(
·
,
·
).Theorem 21.
LetE be a smooth Banach space, C a nonempty subset of E, and T
:
C
→
E a mapping. Then, the following are equivalent.(a)
T is J-dissipative.
(b)
For allx
,
y
∈
C, (48)
V
(
x
,
T
y
)
+
V
(
y
,
T
x
)
≤
V
(
x
,
T
x
)
+
V
(
y
,
T
y
)
.Proof.
For anyx
,
y
∈
C, (49)
x
-
y
,
J
T
x
-
J
T
y
≤
0is equal to (50)
-
2
x
,
J
T
y
-
2
y
,
J
T
x
≤
-
2
x
,
J
T
x
-
2
y
,
J
T
y
,
-
2
x
,
J
T
y
-
2
y
,
J
T
x
+
x
2
+
T
y
2
+
y
2
+
T
x
2
≤
-
2
x
,
J
T
x
-
2
y
,
J
T
y
+
x
2
+
T
x
2
+
y
2
+
T
y
2
.From the definition of V, this inequality is equivalent to (51)
V
(
x
,
T
y
)
+
V
(
y
,
T
x
)
≤
V
(
x
,
T
x
)
+
V
(
y
,
T
y
)
.Furthermore, we have the following result by this theorem.Lemma 22.
Suppose thatE is a smooth and strictly convex Banach space and that C
⊂
E is a nonempty convex subset. Assume that a mapping T
:
C
→
E is J-dissipative. If there are fixed points of T, then F
(
T
) is singleton.Proof.
Assume that there existp
0 and q
0 such that T
p
0
=
p
0 and T
q
0
=
q
0. Since T is J-dissipative, we have by Theorem 21 that (52)
0
≤
V
p
0
,
T
q
0
+
V
q
0
,
T
p
0
≤
V
(
p
0
,
T
p
0
)
+
V
(
q
0
,
T
q
0
)
=
V
(
p
0
,
p
0
)
+
V
(
q
0
,
q
0
)
=
0
.Thus, we have that V
(
p
0
,
q
0
)
=
V
(
q
0
,
p
0
)
=
0. This implies that (53)
0
≤
p
0
-
q
0
2
≤
V
(
p
0
,
q
0
)
=
0
,
p
0
=
q
0
.
Furthermore, we have(54)
V
p
0
,
q
0
=
p
0
2
+
q
0
2
-
2
p
0
,
J
q
0
=
p
0
2
+
p
0
2
-
2
p
0
,
J
q
0
=
0
,and we have p
0
2
=
p
0
,
J
q
0. Since E is strictly convex and J is one-to-one, we obtain that p
0
=
q
0.We give a result before proving an existence theorem for fixed points.Theorem 23 (see [10]).
LetE be a smooth and uniformly convex Banach space, and let T
:
E
→
E be a V-strongly nonexpansive mapping with λ. Then, one has that (55)
lim
x
-
y
→
0
T
x
-
T
y
=
0
,for x
,
y
,
T
x
,
T
y
≤
r, where r
>
0.Proof.
SinceT is a V-strongly nonexpansive with λ, we have (56)
0
≤
V
T
x
,
T
y
+
λ
V
x
-
T
x
,
y
-
T
y
≤
V
(
x
,
y
)
=
x
2
+
y
2
-
2
x
,
J
y
=
x
2
-
y
2
-
2
x
-
y
,
J
y
≤
x
-
y
(
x
+
y
+
2
y
)
=
x
-
y
(
x
+
3
y
)
,
for
any
x
,
y
∈
E
.Thus, we obtain, for x
,
y with x, y
≤
r, (57)
V
(
T
x
,
T
y
)
⟶
0
,
V
(
x
-
T
x
,
y
-
T
y
)
⟶
0
as
x
-
y
⟶
0
.From Lemma 16, we have that (58)
0
≤
g
T
x
-
T
y
≤
V
T
x
,
T
y
.Therefore, we have from (57) that lim
x
-
y
→
0
g
(
T
x
-
T
y
)
=
0. From the definition of g, we obtain that (59)
lim
x
-
y
→
0
T
x
-
T
y
=
0
.Remark 24.
Ifx
∈
E satisfies that T
x
<
r
0 for r
0
>0, the (57) implies that T
y
<
r
0
+
1 for y in the neighborhood of x.We will prove the following existence theorem by using Theorem14.Theorem 25.
LetE be a reflexive, strictly convex, and smooth Banach space and C a nonempty, bounded, closed, and convex subset of E. Suppose T
:
C
→
C is a V-strongly nonexpansive and J-dissipative mapping. Then, there exists a unique fixed point of T.Proof.
At first, we will show that there existsy
0
∈
C such that (60)
{
x
∈
C
:
V
(
x
,
T
x
)
<
V
(
y
0
,
T
x
)
}
=
∅
.Assume that, for all y
∈
C, (61)
{
x
∈
C
:
V
(
x
,
T
x
)
<
V
(
y
,
T
x
)
}
≠
∅
.Let A
x
=
{
y
∈
C
:
V
(
x
,
T
y
)
<
V
(
y
,
T
y
)
} and B
x
=
{
y
∈
C
:
V
(
x
,
T
x
)
<
V
(
y
,
T
x
)
} for all x
∈
C. Then, from the assumption, B
-
1
y is nonempty for all y
∈
C. Since T is J-dissipative, Theorem 21 implies that (62)
V
(
x
,
T
y
)
-
V
(
y
,
T
y
)
≤
V
(
x
,
T
x
)
-
V
(
y
,
T
x
)for all y
∈
B
x. This means that B
x
⊂
A
x for any x
∈
C. For any y
∈
C, let v
j
∈
A
-
1
y with j
∈
{
1,2
,
…
,
n
}, and suppose that v
=
∑
j
=
1
n
‍
α
j
v
j and ∑
j
=
1
n
‍
α
j
=
1 with α
j
>
0. From the convexity of V, we have (63)
V
v
,
T
y
=
V
∑
j
=
1
n
‍
α
j
v
j
,
T
y
≤
∑
j
=
1
n
‍
α
j
V
v
j
,
T
y
≤
∑
j
=
1
n
‍
α
j
V
(
y
,
T
y
)
=
V
(
y
,
T
y
)
.Thus, we obtain that A
-
1
y is convex for all y
∈
C. Since it is obvious that B
x is open for each x
∈
C, Theorem 14 implies that there exists a point x
0
∈
C such that x
0
∈
A
x
0. This means that (64)
V
(
x
0
,
T
x
0
)
<
V
(
x
0
,
T
x
0
)
.This is a contradiction. Thus, we have for some y
0
∈
C that (65)
{
x
∈
C
:
V
(
x
,
T
x
)
<
V
(
y
0
,
T
x
)
}
=
∅
.This means that there exists y
0
∈
C such that (66)
V
(
y
0
,
T
x
)
≤
V
(
x
,
T
x
)for all x
∈
C.
Furthermore, we will showV
(
y
0
,
T
y
0
)
≤
V
(
x
,
T
y
0
) for all x
∈
C if y
0 satisfies (66). Let y
t
=
(
1
-
t
)
y
0
+
t
x for any t
∈
(
0,1
) and x
∈
C. Since C is convex, then y
t
∈
C. Thus, we obtain that (67)
V
y
0
,
T
y
t
≤
V
(
y
t
,
T
y
t
)
=
V
(
(
1
-
t
)
y
0
+
t
x
,
T
y
t
)
.From the convexity of V
(
·
,
y
) for y
∈
C, (68)
V
(
y
0
,
T
y
t
)
≤
(
1
-
t
)
V
(
y
0
,
T
y
t
)
+
t
V
(
x
,
T
y
t
)and we have V
(
y
0
,
T
y
t
)
≤
V
(
x
,
T
y
t
). From the definition of V
(
·
,
·
), we have that (69)
V
x
,
T
y
t
-
V
x
,
T
y
0
=
T
y
t
2
-
T
y
0
2
-
2
x
,
J
T
y
t
-
J
T
y
0
≤
(
T
y
t
+
T
y
0
)
T
y
t
-
T
y
0
+
2
x
J
T
y
t
-
J
T
y
0
.Therefore, we have, by Theorem 23 and the continuity of J on a smooth Banach space, that lim
t
→
0
+
V
(
x
,
T
y
t
)
=
V
(
x
,
T
y
0
) and (70)
V
y
0
,
T
y
0
=
lim
t
→
0
+
V
y
0
,
T
y
t
≤
lim
t
→
0
+
V
(
x
,
T
y
t
)
=
V
(
x
,
T
y
0
)for all x
∈
C. Letting x
=
T
y
0, we have that (71)
V
(
y
0
,
T
y
0
)
≤
V
(
T
y
0
,
T
y
0
)
=
0
.Hence, V
(
y
0
,
T
y
0
)
=
0. This implies that (72)
y
0
2
+
T
y
0
2
=
2
y
0
,
J
T
y
0
≤
2
y
0
T
y
0
,and then we obtain that (73)
y
0
-
T
y
0
2
≤
0
.Thus, we have y
0
=
T
y
0 and we have by (72) that y
0
2
=
y
0
,
J
T
y
0. Since J is one-to-one on a strictly convex Banach space, J
T
y
0
=
J
y
0 implies that T
y
0
=
y
0. Therefore, we have the conclusion.Finally, we will prove a strong convergence theorem for finding fixed points of aV-strongly nonexpansive mapping T in a Banach space, without the assumption that F
(
T
)
≠
∅.Theorem 26.
LetE be a smooth and uniformly convex Banach space, and let C be a nonempty, compact, and convex subset of E. Suppose that T
:
C
→
C is J-dissipative and V-strongly nonexpansive with λ. One defines a Mann iterative sequence {
x
n
} as follows: for any x
1
∈
C and n
≥
1,(74)
x
n
+
1
=
β
n
x
n
+
(
1
-
β
n
)
T
x
n
,where {
β
n
}
⊂
(
0,1
) and lim
n
→
∞
β
n
=
0. Then, there exists a unique fixed point p
0
∈
C such that x
n
→
p
0 and T
x
n
→
p
0.Proof.
From Theorem25, we have that F
(
T
)
≠
∅. As in the proof of Theorem 18, we obtain that lim
n
→
∞
V
(
x
n
,
T
x
n
)
=
0 and that there exists a point p
0
∈
F
(
T
) such that x
n
⇀
p
0 and T
x
n
⇀
p
0. Since T is J-dissipative, Theorem 21 implies that (75)
0
≤
V
(
x
n
,
T
p
0
)
+
V
(
p
0
,
T
x
n
)
≤
V
(
x
n
,
T
x
n
)
+
V
(
p
0
,
T
p
0
)
.From T
p
0
=
p
0, we have for n
≥
1 that (76)
0
≤
V
x
n
,
p
0
+
V
p
0
,
T
x
n
≤
V
(
x
n
,
T
x
n
)
+
V
(
p
0
,
p
0
)
=
V
(
x
n
,
T
x
n
)
.Since lim
n
→
∞
V
(
x
n
,
T
x
n
)
=
0, we have that (77)
lim
n
→
∞
V
(
x
n
,
p
0
)
=
lim
n
→
∞
V
(
p
0
,
T
x
n
)
=
0
.By Lemma 17, we obtain that x
n
→
p
0 and T
x
n
→
p
0. We have the conclusion.



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*Source: 760671-2015-07-26.xml*