ITERATION METHODS FOR APPROXIMATION OF SOLUTIONS OF NONLINEAR
EQUATIONS IN BANACH SPACES
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory committee. This
dissertation does not include proprietary or classified information.
Chukwudi Chidume
Certificate of Approval:
Asheber Abebe
Associate Professor
Mathematics and Statistics
Geraldo Soares de Souza, Chair
Professor
Mathematics and Statistics
Narendra Govil
Professor
Mathematics and Statistics
Yonsheng Han
Professor
Mathematics and Statistics
George T. Flowers
Interim Dean, Graduate School
ITERATION METHODS FOR APPROXIMATION OF SOLUTIONS OF NONLINEAR
EQUATIONS IN BANACH SPACES
Chukwudi Chidume
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulfillment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
August 9, 2008
ITERATION METHODS FOR APPROXIMATION OF SOLUTIONS OF NONLINEAR
EQUATIONS IN BANACH SPACES
Chukwudi Chidume
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Dissertation Abstract
ITERATION METHODS FOR APPROXIMATION OF SOLUTIONS OF NONLINEAR
EQUATIONS IN BANACH SPACES
Chukwudi Chidume
Doctor of Philosophy, August 9, 2008
(DICTP, International Center for Theoretical Physics, Trieste, Italy 2004)
(BSBA, University of Alabama in Huntsville, 2002)
(B.S., University of Alabama in Huntsville, 2000)
87 Typed Pages
Directed by Geraldo Soares de Souza
The objective in this manuscript is to study some iterative methods used to approxi-
mate solutions of nonlinear equations in Banach Spaces. In particular, we study a Halpern-
type iterative scheme in relation to nonexpansive and asymptotically nonexpansive map-
pings and prove convergence theorems in both of these cases. We also study a hybrid
steepest descent iterative scheme in relation to the variational inequality problem, and us-
ing this process, we prove convergence theorems for the approximation of the solution of
the variational inequality problem in certain Banach spaces, in particular for Lp spaces.
iv
Acknowledgments
I would like to start by thanking God for making all this possible. I would like to thank
my advisor and major professor Geraldo deSouza for his constant advice and guidance. He
has shared his great knowledge and expertise with me throughout this endeavor. He was
patient and understanding and without his support I would have had a difficult time here at
Auburn University. My sincere appreciation and modest gratitude goes to Professor Abebe,
Professor Govil and Professor Han the members of my PhD academic committee for their
scientific support and direction. I also want to thank Professor Chidume, Professor Li and
Dr. Ali for reading this manuscript countless number of times and offering their insight and
helpful corrections.
This work is dedicated to my father Professor C.E. Chidume. Without his love, influ-
ence and support, I would not be on this career path. Thank you dad. Special thanks to
my mother Dr. Ifeoma Chidume, my sister Ada and my brothers Kene and Okey and all
my family for their continued support.
Of all, I am most thankful to my lovely children Gabriel and Michael and I am extremely
grateful to my wife Tiffani for the unconditional love, care, patience and understanding. I
definitely would not be where I am today without her invaluable support throughout my
graduate studies here at Auburn University.
v
Style manual or journal used Journal of Approximation Theory (together with the style
known as ?aums?). Bibliograpy follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (specifically LATEX)
together with the departmental style-file aums.sty.
vi
Table of Contents
1 Introduction 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2 Iterative Approximation of Fixed Points of Nonexpansive Mappings 18
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.3 Convergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3 A Strong Convergence Theorem for Fixed Points of Asymptotically Nonex-
pansive Mappings in Banach Spaces 27
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.2 Convergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4 Convergence of a Hybrid Steepest Descent Method for Variational Inequal-
ities in Banach Spaces 34
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.3 Convergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
4.4 The case of Lp spaces, 1 < p ? 2. . . . . . . . . . . . . . . . . . . . . . . . . 46
5 Approximation of Fixed Points of Nonexpansive Mappings and Solutions of
Variational Inequalities 53
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.2 Convergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
5.3 The case of Lp spaces, 1 < p ? 2. . . . . . . . . . . . . . . . . . . . . . . . . 66
Bibliography 73
vii
Chapter 1
Introduction
1.1 Introduction
The contributions of this thesis fall within the general area of nonlinear operator theory,
an area with vast amount of applicability in recent years, as such becoming the object of
an increasing amount of study. We devote our attention to an important topic within the
area: iterative methods for approximating fixed points and solutions of variational inequality
problems for nonexpansive and accretive-type nonlinear mappings.
Let K be a nonempty subset of a real normed linear space E and let T : K ? K
be a map. A point x ? K is said to be a fixed point of T if Tx = x. We shall denote
the set of fixed points for an operator T by F(T). Now, consider the differential equation
du
dt + Au(t) = 0 which describes an evolution system where A is an accretive map from a
Banach space to itself. At equilibrium state, dudt = 0 and so a solution of Au = 0 describes
the equilibrium or stable state of the system. This is very desirable in many applications
in, for example, ecology, economics, physics, to name a few. Consequently, considerable
research efforts have been devoted to methods of solving the equation Au = 0 when A is
accretive. Since, in general A is a nonlinear operator, there is no closed form solution to
this equation. The standard technique is to replace A by an operator (I?T) where I is the
identity map on E and T maps E to itself. Such a map T is called a pseudo-contraction
(or is called pseudo-contractive). It is then clear that any zero of A is a fixed point of T. As
a result of this, the study of fixed point theory for pseudo-contractive maps has attracted
the interest of numerous scientists and has become a flourishing area of research, especially
1
within the past 30 years or so, for numerous mathematicians. A very important subclass
of the class of pseudo-contractive mappings is the class of nonexpansive mappings. In this
dissertation, we shall devote attention particularly to this class of mappings. Interest in,
and the importance of this class of mappings will become evident in the sequel.
One of the most important fixed point theorems in applications is the classical contrac-
tion mapping principle, or, in other words, the Banach-Cacciopoli [14] fixed point theorem
which is the following:
Theorem 1.1.1 (Banach Contraction Mapping Principle) Let (X,?) be a complete metric
space and let T : (X,?) ? (X,?) satisfy
?(T(x),T(y)) ? ??(x,y) (1.1)
for some nonnegative constant k < 1 and for each x,y ? X. Then, T has a unique fixed
point in X. Moreover, starting with arbitrary x0 ? X, the sequence {xn} defined by
xn+1 = Txn = Tnx0, n ? 1, (1.2)
converges strongly to the unique fixed point.
The iterative technique of Theorem 1.1.1 is due to Picard [69]. A mapping T satisfying
(1.1) is called a strict contraction. If ? = 1 in the relation (1.1), then T is called nonexpan-
sive. If however, ? is an arbitrary fixed positive constant, then T is called a Lipschitz map
or a ?-Lipschitzian map.
2
For the contractive condition (1.1), it was observed that if the condition ? < 1 on the
operator T is weakened to ? = 1, the operator T may no longer have a fixed point and even
when it does have a fixed point, the sequence {xn} defined by (1.2) may fail to converge to
such a fixed point. This can be seen by considering an anti-clockwise rotation of the unit
disc of R2 about the origin through an angle of say, pi4. This map is nonexpansive with the
origin as the unique fixed point, but the Picard sequence fails to converge with any starting
point x0 negationslash= (0,0). Krasnosel?skii [58], however, showed that in this example, if the Picard
iteration formula is replaced by the following formula,
x0 ? K, xn+1 = 12(xn +Txn),n ? 0, (1.3)
then the iterative sequence converges to the unique fixed point.
In general, if E is a normed linear space and T is a nonexpansive mapping, a generalization
of (1.3) which has proved successful in the approximation of a fixed point of T (when it
exists) was given by Schaefer [74] :
x0 ? K, xn+1 = (1??)xn +?Txn, n ? 0, ? ? (0,1). (1.4)
However, the most general iterative formula for approximation of fixed points of nonex-
pansive mappings, which is called the Mann iteration formula (in the light of [63]), is the
following:
x0 ? K, xn+1 = (1??n)xn +?nTxn, n ? 0, (1.5)
3
where {?n} is a real sequence in the interval (0,1) satisfying the following conditions:
(i) limn???n = 0 and (ii)
?summationdisplay
n=1
?n = ?.
The recursion formula (1.4) is consequently called the Krasnoselskii-Mann (KM) formula
for finding fixed points of nonexpansive mappings. This iterative process has become very
important and applicable as noted below.
? ?Many well-known algorithms in signal processing and image reconstruction are it-
erative in nature .... A wide variety of iterative procedures used in signal processing
and image reconstruction and elsewhere are special cases of the (Krasnoselskii-Mann)
iteration procedure, for particular choices of the (nonexpansive) operator....?
(Charles Byrne , [13]).
Apart from being an obvious generalization of the contraction mappings, nonexpansive maps
are important, as has been observed by Bruck [9], mainly for the following two reasons:
? Nonexpansive maps are intimately connected with the monotonicity methods devel-
oped since the early 1960?s and constitute one of the first classes of nonlinear mappings
for which fixed point theorems were obtained by using the fine geometric properties
of the underlying Banach spaces instead of compactness properties.
? Nonexpansive mappings appear in applications as the transition operators for initial
value problems of differential inclusions of the form 0 ? dudt +T(t)u, where the opera-
tors {T(t)} are, in general, set-valued and are accretive or dissipative and minimally
continuous.
4
Nonexpansive maps have been studied, and are still being studied, extensively by numerous
authors (see e.g., Bauschke [2], Belluce and Kirk [3], Browder [4], Bruck [7, 8], Chidume
[19, 18], Chidume and Ali [21, 26], Chidume and Chidume [27], Chidume et al. [28, 34],
Chidume and Shahzad [33], De Marr [43], G?ohde [47]. Jung and Kim [49], Jung [50], Jung
et al. [51], Khan and Fukharu-ud-din [53], Kirk [56], Lim [59], Matsuhita and Kuroiwa [64],
O?Hara et al. [67], Oka [68], Reich [68], Senter and Dotson [77], Shahzad [78], Shahzad and
Al-dubiban [79], Takahashi and Tamura [85], Takahashi and Kim [86], Tan and Xu [88], Xu
and Yin [94], Zeng and Yao [100] and a host of other authors).
Let E be a real Banach space, K a closed convex subset of E and T : K ? K a non-
expansive mapping. For fixed t ? (0,1] and arbitrary u ? K, define a map Tt : K ? K
by Ttx := tu + (1 ? t)Tx, x ? K. Then Tt is a strict contraction for every fixed constant
t ? (0,1]. Denote the unique fixed point of Tt by zt ? K, and assume F(T) := {x ? K :
Tx = x} negationslash= ?.
In 1967, Browder [5] proved that if E = H, a Hilbert space, then limt?0zt exists and is a fixed
point of T. In 1980, Reich [71] extended this result to uniformly smooth Banach spaces. In
1981, Kirk [57] obtained the same result in arbitrary Banach spaces under the additional
assumption that T has pre-compact range. We have mentioned that every nonexpansive
mapping is a pseudo-contractive mapping. Following this, in 2000, Morales and Jung [65]
proved the same result for T a continuous pseudocontraction in a real reflexive Banach
space with uniformly G?ateaux differentiable norm.
For a sequence {?n} of real numbers in [0,1] and an arbitrary u ? K, let the sequence {xn}
5
in K be iteratively defined by x0 ? K,
xn+1 = ?nu+ (1??n)Txn,n ? 0. (1.6)
The recursion formula (1.6) was first introduced in 1967 by Halpern [48] with u = 0, in the
framework of Hilbert spaces. Under appropriate conditions on the domain of T, and some
restrictions on the parameter {?n} (?n = n?a, a ? (0,1)), he proved strong convergence
of {xn} to a fixed point of T. Iteration formulas of the form (1.6) are now said to be of
the Halpern-type. Lions [62] considered a more general parameter {?n} and improved the
result of Halpern, still in Hilbert spaces. He proved strong convergence of {xn} to a fixed
point of T where the real sequence {?n} satisfies the conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?; C3 : lim |?n ??n?1|?2
n
= 0.
In 1980, Reich [71] proved that the result of Halpern remains true when E is uniformly
smooth. It was observed that both Halpern?s and Lions?s conditions on the real sequence
{?n} excluded the canonical choice ?n = 1n+1. This was overcome in 1992 by Wittmann
[93] who proved, still in Hilbert spaces, the strong convergence of {xn} to a fixed point of
T if {?n} satisfies the conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?; C4 :
summationdisplay
|?n+1 ??n| < ?.
In 1994, Reich [72] extended the result of Wittmann to Banach spaces which are uniformly
smooth and have weakly sequentially continuous duality maps (e.g., lp spaces, 1 < p < ?),
where {?n} satisfies C1 and C2 and is also required to be decreasing (and hence also satisfies
6
C4). These spaces exclude Lp spaces, 1 < p < ?,p negationslash= 2. Shioji and Takahashi [80] extended
Wittmann?s result to real Banach spaces with uniformly G?ateaux differentiable norms and
in which each nonempty closed convex and bounded subset has the fixed point property
for nonexpansive mappings (e.g., Lp spaces, 1 < p < ?). In 2002, Xu [95] (see also [96])
improved the result of Lions twofold. First, he weakened the condition C3 by removing the
square in the denominator so that the canonical choice of ?n = 1n+1 is possible. Secondly,
he proved the strong convergence of the scheme (1.6) in the framework of real uniformly
smooth Banach spaces. Xu also remarked ([95], Remark 3.2) that Halpern [48] observed
that conditions (C1) and (C2) are necessary for the strong convergence of algorithm (1.6)
for all nonexpansive mappings T : K ? K. It is not clear if they are sufficient. This
brought about the following question which has been open for many years:
Question 1: Are the conditions C1 : lim?n = 0 and C2 : summationtext?n = ? sufficient for
the strong convergence of algorithm (1.6) for all nonexpansive mappings T : K ? K?
In Chapter 2 of this dissertation, we modify the recursion formula (1.6) by introducing
an auxiliary operator that has the same set of fixed points as T. With the help of this
operator, we prove that conditions C1 and C2 are sufficient for the modified iteration al-
gorithm to converge strongly to a fixed point of T, even in the more general setting where
E is a real Banach space with uniformly G?ateaux differentiable norm. Consequently, our
theorems in Chapter 2 (also see [27]), provide a partial answer to Question 1. The general
question still remains open.
7
One important class of nonlinear mappings more general than the class of nonexpansive
mappings which has been studied extensively by various authors is the class of asymptotically
nonexpansive mappings. This class of mappings was introduced in 1972 by Goebel and Kirk
[46].
Definition 1.1.2 Let K be a nonempty subset of a normed linear space, a mapping T :
K ? K is called asymptotically nonexpansive if there exists a sequence {kn} ? [1,?) with
limn??kn = 1 such that bardblTnx?Tnybardbl ? knbardblx?ybardbl for all x,y ? K and n = 1,2,....
It was proved in [46] that if K is a nonempty closed, convex and bounded subset of a
uniformly convex real Banach space and T is an asymptotically nonexpansive self-mapping
of K, then T has a fixed point. It is clear that every nonexpansive mapping is asymptotically
nonexpansive. The following example shows that the class of asymptotically nonexpansive
mappings properly contains the class of nonexpansive mappings.
Example 1.1.3 (Goebel and Kirk, [46]) Let B be a unit ball of the real Hilbert space
l2 and let T : B ? B be defined by T({x1,x2,...}) = {0,x21,a2x2, a3x3,...} where {an}
is a sequence of numbers such that 0 < an < 1 and
?productdisplay
n=2
an = 12. Then T is Lipschitzian
and bardblTx ? Tybardbl ? 2bardblx ? ybardbl, for all x,y ? B and moreover, bardblTnx ? Tnybardbl ? knbardblx ? ybardbl,
with kn := 2
nproductdisplay
i=2
ai. Observe that T is not nonexpansive and that limn??kn = 1, so that T is
asymptotically nonexpansive map.
In Chapter 3, we prove a strong convergence theorem for a Halpern-type iteration sequence
for approximation of a fixed point of asymptotically nonexpansive mappings. Our main
theorem in Chapter 3 (also see [35]) is proved in a real Banach space which has a uni-
formly G?ateaux differentiable norm. The main theorem extends some important known
8
results from the class of nonexpansive mappings to the more general class of asymptotically
nonexpansive mappings.
In the second half of this manuscript, we turn attention to the approximation of a solution
of a variational inequality problem.
Definition 1.1.4 Let E be a real normed linear space and E? be its dual space. For some
real number q (1 < q < ?), the generalized duality mapping Jq : E ? 2E? is defined by
Jq(x) = {f? ? E? : ?x,f?? = bardblxbardblq,bardblf?bardbl = bardblxbardblq?1},
where ?.,.? denotes the duality pairing between elements of E and elements of E?. If q = 2,
then J2 is called the normalized duality map on E.
Let K be a nonempty closed convex subset of E and S : E ? E be a nonlinear operator.
The variational inequality problem is formulated as follows: Find a point x? ? K such that
VI(S,K) : ?jq(Sx?),(y ?x?)? ? 0 ?y ? K. (1.7)
If E = H, a real Hilbert space, the variational inequality problem reduces to the following:
Find a point x? ? K such that
VI(S,K) : ?Sx?,y ?x?? ? 0 ?y ? K. (1.8)
9
Definition 1.1.5 A mapping G : E ? E is said to be accretive if ?x,y ? E, there exists
jq(x?y) ? Jq(x?y) such that
?Gx?Gy,jq(x?y)? ? 0. (1.9)
Definition 1.1.6 For some real number ? > 0, G is called ??strongly accretive if ?x,y ? E,
there exists jq(x?y) ? Jq(x?y) such that
?Gx?Gy,jq(x?y)? ? ?bardblx?ybardblq. (1.10)
In Hilbert spaces, accretive (strongly accretive) operators are called monotone (strongly
monotone) where inequalities (1.9) and (1.10) hold with jq replaced by the identity map of
H.
Applications of variational inequalities span as diverse disciplines as differential equations,
time-optimal control, optimization, mathematical programming, mechanics, finance and so
on (see, for example, Kinderlehrer and Stampacchia [55], Noor [66] for more details).
It is known that if S is Lipschitz and strongly accretive, then problem VI(S,K) has a
unique solution. An important problem is how to find a solution of the problem VI(S,K)
whenever it exists. Considerable efforts have been devoted to this problem (see, e.g. Xu
[98] , Yamada [99] and the references contained therein).
10
It is known that in a real Hilbert space, the problem VI(S,K) is equivalent to the fol-
lowing fixed point equation
x? = PK(x? ??Sx?), (1.11)
where ? > 0 is an arbitrary fixed constant and PK is the nearest point projection map from
H onto K, i.e., PKx = y, where bardblx ? ybardbl = inf
u?K
bardblx ? ubardbl for x ? H. Consequently, under
appropriate conditions on S and ?, fixed point methods can be used to find or approximate
a solution of problem VI(S,K). For instance, if S is strongly monotone and Lipschitz then
a mapping G : H ? H defined by Gx = PK(x ? ?Sx), x ? H with ? > 0 sufficiently
small is a strict contraction. Hence, the Picard iteration, x0 ? H, xn+1 = Gxn, n ? 0 of
the classical Banach contraction mapping principle converges to the unique solution of the
problem VI(K,S).
In applications, however, the projection operator PK in the fixed point formulation (1.11)
may make the computation of the iterates difficult due to possible complexity of the convex
set K. In order to reduce the possible difficulty with the use of PK, Yamada [99] recently
introduced a hybrid descent method for solving the problem VI(K,S). Let T : H ? H be a
map and let K := {x ? H : Tx = x} negationslash= ?. Let S be ??strongly monotone and ??Lipschitz
on H. Let ? ? (0, 2??2) be arbitrary but fixed real number and let a sequence {?n} in (0,1)
satisfy the following conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?; and C3 : lim?n ??n+1?2
n
= 0.
11
Starting with an arbitrary initial guess x0 ? H, let a sequence {xn} be generated by the
following algorithm
xn+1 = Txn ??n+1?S(Txn), n ? 0. (1.12)
Then, Yamada [99] proved that {xn} converges strongly to the unique solution of VI(K,S).
In the case that K = r?i=1F(Ti) negationslash= ?, where {Ti}ri=1 is a finite family of nonexpansive map-
pings, Yamada [99] studied the following algorithm,
xn+1 = T[n+1]xn+1 ??n+1?S(T[n+1]xn), n ? 0, (1.13)
where T[k] = Tk mod r, for k ? 1, with the mod function taking values in the set {1,2,...,r},
and where the sequence{?n}satisfies the conditionsC1, C2 andC4 : summationtext|?n??n+N| < ?.
Under these conditions, he proved the strong convergence of {xn} to the unique solution of
the VI(K,S).
Recently, Xu and Kim [98] studied the convergence of the algorithms (1.12) and (1.13),
still in the framework of Hilbert spaces, and proved strong convergence with condition C3
replaced by C5: lim?n??n+1?n+1 = 0 and with condition C4 replaced by C6 : lim?n??n+r?n+r = 0.
These are improvements on the results of Yamada. In particular, the canonical choice
?n := 1n+1 is applicable in the results of Xu and Kim but is not in the result of Yamada
with condition C3.
12
In Chapter 4, we prove theorems that extend the results of Xu and Kim [98] (and con-
sequently those of Wang [91], Xu and Kim [98], Yamada [99], Zheng and Yao [100]) from
real Hilbert spaces to the more general real q?uniformly smooth Banach spaces, q ? 2. In
particular, our theorems are applicable in Lp spaces, 2 ? p < ?. (see e.g., Chidume et al.
[28]).
The condition q > 2, however, excludes the Lp spaces, 1 < p < 2. In Section 4.4, we
employ another tool to prove convergence theorems that extend the results of Xu and Kim
to Lp spaces, 1 < p ? 2 (see Chidume and deSouza [36]). These theorems complement
those of the first part of Chapter 4 to provide convergence theorems valid in all Lp spaces,
1 < p < ?.
In Chapter 5, we continue our interest in fixed points of nonexpansive mappings and so-
lutions of variational inequality problems. In this chapter, we introduce a new recursion
formula and prove strong convergence theorems for the unique solution of the variational
inequality problem VI(K,S) of Chapter 4, requiring only conditions C1 and C2 on the
parameter sequence {?n}. Furthermore, in the case Ti : E ? E, i = 1,2,...,r is a family
of nonexpansive mappings with K = r?i=1F(Ti) negationslash= ?, we prove a convergence theorem where
condition C6 is replaced with limn??bardblT[n+2]xn ? T[n+1]xnbardbl = 0. An example satisfying this
condition is presented by Chidume and Ali in [22]. All our theorems in Chapter 5 (see
also [29]) are proved in q?uniformly smooth Banach spaces, q ? 2. In particular, they are
applicable in Lp spaces, 2 ? p < ?.
13
As in chapter 4, we also use a different tool to extend our theorems to include Lp spaces,
1 < p ? 2. Our theorems in Chapter 5 (see also [30]) still extend the results of Xu and Kim
[98] (and consequently those of Wang [91], Xu and Kim [98], Yamada [99], Zheng and Yao
[100]) from real Hilbert spaces to, in particular, the more general real Lp spaces, 1 < p < ?.
Moreover, in this more general setting, the iteration parameter {?n} is required to satisfy
only conditions C1 and C2.
1.2 Preliminaries
Definition 1.2.1 Let S := {x ? E : ||x|| = 1} denote the unit sphere of the real Banach
space E. The space E is said to have a G?ateaux differentiable norm if the limit
limt?0 ||x+ty||?||x||t
exists for each x,y ? S; and E is said to have a uniformly G?ateaux differentiable norm if
for each y ? S, the limit is attained uniformly for x ? S.
Definition 1.2.2 We shall denote a Banach limit by ?. Recall that ? is an element of (l?)?
such that bardbl?bardbl = 1, liminfn?? an ? ?nan ? limsup
n??
an and ?nan = ?n+1an for all {an}n?0 ? l?
(see e.g. Chidume et al. [31], Chidume [17]).
To motivate the definition of modulus of smoothness which will be used in the sequel,
we begin with the following definition.
Definition 1.2.3 A real Banach space is called smooth if for every x in X with bardblxbardbl = 1,
there exists a unique x? in X? such that bardblx?bardbl = ?x,x?? = 1.
14
Assume now that X is not smooth and take x in X and u?, v? in X? such that bardblxbardbl = bardblu?bardbl =
bardblv?bardbl = ?x,u?? = ?x,v?? = 1 and u? negationslash= v?. Let y in X be such that bardblybardbl = 1, ?y,u?? > 0 and
?y,v?? < 0 . Then for every t > 0 we have
1 +t?y,u?? = ?x+ty,u?? ? bardblx+tybardbl,
1?t?y,v?? = ?x?ty,v?? ? bardblx?tybardbl
which imply
2 < 2 +t(?y,u????y,v??) ? bardblx+tybardbl+bardblx?tybardbl
or equivalently
0 < t(?y,u
????y,v??
2 ) ?
bardblx+tybardbl+bardblx?tybardbl
2 ?1.
With this motivation we introduce the following definition.
Definition 1.2.4 Let E be a normed space with dimE ? 2. The modulus of smoothness
of E is the function ?E : [0,?) ? [0,?) defined by
?E(?) := sup
braceleftbiggbardblx+ybardbl +bardblx?ybardbl
2 ?1 : bardblxbardbl = 1;bardblybardbl = ?
bracerightbigg
.
The space E is called uniformly smooth if and only if lim
t?0+
?E(t)
t = 0.
Definition 1.2.5 For some positive constant q, E is called q?uniformly smooth if there
exists a constant c > 0 such that ?E(t) ? ctq, t > 0.
15
Lp spaces, 1 < p < ? are p?uniformly smooth (see e.g., Lindenstrauss and Tzafriri [61]).
In fact, it is known that
Lp(or lp) spaces are
??
????
?
???
???
2? uniformly smooth, if, 2 ? p < ?,
p? uniformly smooth, if, 1 < p ? 2,
(see e.g., Lindenstrauss and Tzafriri [61]). It is well known that if E is smooth then the
duality mapping is singled-valued, and if E is uniformly smooth then the duality mapping
is norm-to-norm uniformly continuous on bounded subset of E.
Definition 1.2.6 Let E be a real Banach space and K be a nonempty, closed and convex
subset of E. Let P be a mapping of E onto K. Then, P is said to be sunny if P(Px+t(x?
Px)) = Px for all x ? E and t ? 0. A mapping P of E into E is said to be a retraction if
P2 = P.
Definition 1.2.7 A subset K of E is said to be sunny nonexpansive retract of E if there
exists a sunny nonexpansive retraction of E onto K. A retraction P is said to be orthogonal
if for each x, x?P(x) is normal to K in the sense of James [54].
It is well known (see Bruck [10]) that if E is uniformly smooth and there exists a nonexpan-
sive retraction of E onto K, then there exists a nonexpansive projection of E onto K. If E
is a real smooth Banach space, then P is an orthogonal retraction of E onto K if and only
if P(x) ? K and ?P(x)?x,jq(P(x)?y)? ? 0 for all y ? K. It is also known (see e.g., Shioji
and Takahashi [81]) that if K is a convex subset of a uniformly convex Banach space whose
norm is uniformly G?ateaux differentiable and T : K ? K is nonexpansive with F(T) negationslash= ?,
16
then, F(T) is a nonexpansive retract of K.
Let K be a nonempty closed convex and bounded subset of a Banach space E and let
the diameter of K be defined by d(K) := sup{bardblx ? ybardbl : x,y ? K}. For each x ? K,
let r(x,K) := sup{bardblx ? ybardbl : y ? K} and let r(K) := inf{r(x,K) : x ? K} denote the
Chebyshev radius of K relative to itself. The normal structure coefficient N(E) of E (see
e.g. [12]) is defined by N(E) := inf
braceleftBigd(K)
r(K) : K is a closed convex and bounded subset of E
with d(K) > 0
bracerightBig
. A space E such that N(E) > 1 is said to have uniform normal structure.
It is known that all uniformly convex and uniformly smooth Banach spaces have uniform
normal structure (see e.g., [60]).
17
Chapter 2
Iterative Approximation of Fixed Points of Nonexpansive Mappings
2.1 Introduction
Let K be a nonempty closed convex subset of a real Banach space E which has a
uniformly G?ateaux differentiable norm and T : K ? K a nonexpansive mapping with
F(T) negationslash= ?.
In this chapter, we prove that the conditions C1: lim?n = 0 and C2: summationtext?n = ? which
are known to be necessary are, under appropriate conditions, also sufficient for the strong
convergence of a Halpern-type iterative scheme to a fixed point of a nonexpansive mapping
T. Our result gives a partial answer to Question 1 mentioned in the introduction.
We begin with the following well known theorem.
Theorem 2.1.1 (Morales and Jung [65], Reich [71]) Let K be a nonempty closed convex
subset of a Banach space E which has uniformly G?ateaux differentiable norm and T : K ?
K a nonexpansive mapping with F(T) negationslash= ?. Suppose that every nonempty closed convex
bounded subset of K has the fixed point property for nonexpansve mappings. Then there
exists a continuous path t ? zt,0 < t < 1 satisfying zt = tu + (1?t)Tzt, for arbitrary but
fixed u ? K, which converges to a fixed point of T.
Recently, Shioji and Takahashi [80] proved the following theorem.
Theorem 2.1.2 (Shioji and Takahashi [80]) Let E be a real Banach space whose norm is
uniformly G?ateaux differentiable and let K be a closed convex subset of E. Let T : K ? K
be a nonexpansive mapping with F(T) := {x ? K : Tx = x} negationslash= ?. Let {?n} be a sequence
18
which satisfies the following conditions:
(i) 0 ? ?n ? 1, lim?n = 0;
(ii) summationtext?n = ?;
(iii)summationtext?n=0 |?n+1 ??n| < ?.
Let u ? K and let {xn} be defined by x0 ? K,
xn+1 = ?nu+ (1??n)Txn,n ? 0. (2.1)
Assume that {zt} converges strongly to z ? F(T) as t ? 0, where for 0 < t < 1, zt is the
unique element of K which satisfies zt = tu+(1?t)Tzt. Then, {xn} converges strongly to
z.
Xu [96] (see also [95]) proved the following theorem.
Theorem 2.1.3 (Xu [96], Theorem 3.1) Let E be a uniformly smooth real Banach space,
K a closed convex subset of E, and T : K ? K a nonexpansive mapping with a fixed point.
Let u,x0 ? K be given. Assume that {?n} ? [0,1] satisfies the conditions:
(1) lim?n = 0;
(2) summationtext?n = ?;
(3) lim ?n??n?1?n = 0.
Then the sequence {xn} generated by x0 ? K,
xn+1 = ?nu+ (1??n)Txn,n ? 0,
converges strongly to a fixed point of T.
19
It is our purpose in this chapter to prove a significant improvement of Theorem 2.1.2
and Theorem 2.1.3 in the following sense. We prove the strong convergence of the algorithm
(2.1) in the framework of real Banach spaces E with uniformly G?ateaux differentiable norms
and without condition (iii) of Theorem 2.1.2. Our theorem then also extends Theorem 2.1.3
to the more general real Banach spaces with uniformly G?ateaux differentiable norms and
at the same time dispenses with condition (3) of that theorem. Furthermore, our theorem
gives a partial affirmative answer to Question 1 mentioned in Chapter 1.
2.2 Preliminaries
Lemma 2.2.1 Let E be a real normed linear space. Then, the following inequality holds:
||x+y||2 ? ||x||2 + 2?y,j(x+y)? ? x,y ? E, ? j(x+y) ? J(x+y).
In the sequel, we shall also make use of the following lemmas.
Lemma 2.2.2 (Suzuki, [83]) Let {xn} and {yn} be bounded sequences in a Banach space
E and let {?n} be a sequence in [0,1] with 0 < liminf?n ? limsup?n < 1. Suppose
xn+1 = ?nyn+(1??n)xn for all integers n ? 0 and limsup(||yn+1?yn||?||xn+1?xn||) ? 0.
Then, lim||yn ?xn|| = 0.
Lemma 2.2.3 (Xu, [96]). Let {an} be a sequence of nonnegative real numbers satisfying
the relation:
an+1 ? (1??n)an +?n?n +?n,n ? 0,
where,
(i) {?n} ? [0,1], summationtext?n = ?;
20
(ii) limsup ?n ? 0;
(iii) ?n ? 0; (n ? 0), summationtext?n < ?.
Then, an ? 0 as n ? ?.
2.3 Convergence Theorems
Theorem 2.3.1 (C.E.Chidume and C.O.Chidume [27]) Let K be a nonempty closed convex
subset of a real Banach space E which has a uniformly G?ateaux differentiable norm and
T : K ? K be a nonexpansive mapping with F(T) negationslash= ?. For a fixed ? ? (0,1), define
S : K ? K by Sx := (1 ? ?)x + ?Tx ? x ? K. Assume that {zt} converges strongly
to a fixed point z of T as t ? 0, where zt is the unique element of K which satisfies
zt = tu+(1?t)Tzt for arbitrary u ? K. Let {?n} be a real sequence in (0,1) which satisfies
the conditions: C1 : lim?n = 0; C2 : summationtext?n = ?. For arbitrary x0 ? K, let the sequence
{xn} be defined iteratively by
xn+1 = ?nu+ (1??n)Sxn. (2.2)
Then, {xn} converges strongly to a fixed point of T.
Proof. Observe first that S is nonexpansive and has the same set of fixed points as T.
Define
?n := (1??)?n +? ? n ? 0; yn := xn+1 ?xn +?nxn?
n
, n ? 0.
Observe also that ?n ? ? as n ? ?, and that if {xn} is bounded, then {yn} is bounded.
Let x? ? F(T) = F(S). One easily shows by induction that ||xn ? x?|| ? max{||x0 ?
21
x?||,||u?x?||} for all integers n ? 0, and so, {xn},{yn}, {Txn} and {Sxn} are all bounded.
Also,
||xn+1 ?Sxn|| = ?n||u?Sxn|| ? 0, n ? ?. (2.3)
Observe that from the definitions of ?n and S, we obtain that
yn = ?nu+ (1??n)?Txn?
n
,
which implies
||yn+1 ?yn|| ? ||xn+1 ?xn||
?
vextendsinglevextendsingle
vextendsingle?n+1?
n+1
? ?n?
n
vextendsinglevextendsingle
vextendsingle.||u||+ (1??n+1)?
n+1
? ||Txn+1 ?Txn||
+
vextendsinglevextendsingle
vextendsingle1??n+1?
n+1
? 1??n?
n
vextendsinglevextendsingle
vextendsingle ? ||Txn||?||xn+1 ?xn||.
Since {xn} and {Txn} are bounded, we obtain (for some constants M1 > 0, and M2 > 0)
that,
limsup(||yn+1 ?yn|| ? ||xn+1 ?xn||)
? limsup
braceleftBigvextendsinglevextendsingle
vextendsingle?n+1?
n+1
? ?n?
n
vextendsinglevextendsingle
vextendsingle.||u||
+
vextendsinglevextendsingle
vextendsingle(1??n+1)?
n+1
? ?1
vextendsinglevextendsingle
vextendsingle M1
+
vextendsinglevextendsingle
vextendsingle1??n+1?
n+1
? 1??n?
n
vextendsinglevextendsingle
vextendsingle?M2
bracerightBig
? 0.
22
Hence, by Lemma 2.2.2, ||yn ?xn|| ? 0 as n ? ?. Consequently,
lim||xn+1 ?xn|| = lim?n||yn ?xn|| = 0.
Combining this with (2.3) yields that
||xn ?Sxn|| ? 0 as n ? ?.
We now show that
limsup?u?z,j(xn ?z)? ? 0.
For each integer n ? 0, let tn ? (0,1) be such that
tn ? 0, and ||xn ?Sxn||t
n
? 0, n ? ?.
Let ztn ? K be the unique fixed point of the contraction mapping Stn given by
Stnx = tnu+ (1?tn)Sx, x ? K.
Then,
ztn ?xn = tn(u?xn) + (1?tn)(Sztn ?xn).
23
Using the inequality of Lemma 2.2.1, we compute as follows:
||ztn ?xn||2 ? (1?tn)2||Sztn ?xn||2 + 2tn?u?xn,j(ztn ?xn)?
? (1?tn)2(||Sztn ?Sxn||+||Sxn ?xn||)2 + 2tn(||ztn ?xn||2
+ ?u?ztn,j(ztn ?xn)?
? (1 +t2n)||ztn ?xn||2 +||Sxn ?xn||?
(2||ztn ?xn||+||Sxn ?xn||)
+ 2tn?u?ztn,j(ztn ?xn)?,
and hence,
?u?ztn,j(xn ?ztn)? ? tn2 ||ztn ?xn||2 + ||Sxn ?xn||2t
n
?(2||ztn ?xn||+||Sxn ?xn||).
Since {xn},{ztn} and {Sxn} are bounded and ||Sxn?xn||2tn ? 0,n ? ?, it follows from the
last inequality that
limsup?u?ztn,j(xn ?ztn)? ? 0.
Moreover, we have that
?u?ztn,j(xn ?ztn)? = ?u?z,j(xn ?z)?+?u?z,j(xn ?ztn)?j(xn ?z)?
+ ?z ?ztn,j(xn ?ztn)?. (2.4)
24
But, by hypothesis, ztn ? z ? F(S), n ? ?. Thus, using the boundedness of {xn} we
obtain that
?z ?ztn,j(xn ?ztn)? ? 0, n ? ?. (2.5)
Also,
?u?z,j(xn ?ztn)?j(xn ?z)? ? 0, n ? ?,
since j is norm-to-weak? uniformly continuous on bounded subsets of E. Hence, we obtain
from (2.4) and (2.5) that
limsup?u?z,j(xn ?z)? ? 0.
Furthermore, from the recurrence relation (2.2) we get that xn+1 ? z = ?n(u ? z) + (1 ?
?n)(Sxn ?z). It then follows that
||xn+1 ?z||2 ? (1??n)2||Sxn ?z||2 + 2?n?u?z,j(xn+1 ?z)?
? (1??n)||xn ?z||2 +?n?n,
where ?n := 2?u?z,j(xn+1 ?z)?; ?n ? 0 ? n ? 0. Thus, by Lemma 2.2.3, {xn} converges
strongly to a fixed point of T. a50
Remark 2.3.2 We note that every uniformly smooth Banach space has a uniformly G?ateaux
differentiable norm and is such that every nonempty closed convex and bounded subset of E
has the fixed point property for nonexpansive maps (see e.g., [1]).
25
Remark 2.3.3 Theorem 2.3.1 is a significant generalization of Theorem 2.1.2 and of The-
orem 2.1.3 as has been explained in the introduction. Furthermore, our method of proof
which is different from the method of Shioji and Takahashi [80] is of independent interest.
Let Sn(x) := 1n summationtextn?1k=0 Skx. With this definition, Xu also proved the following theorem.
Theorem 2.3.4 (Xu [96], Theorem 3.2) Assume that E is a real uniformly convex and
uniformly smooth Banach space. For given u,x0 ? K, let {xn} be generated by the algorithm:
xn+1 = ?nu+ (1??n)Snxn,n ? 0. (2.6)
Assume that
(i) lim?n = 0;
(i) summationtext?n = ?.
Then, {xn} converges strongly to a fixed point of S : K ? K nonexpansive.
Remark 2.3.5 Theorem 2.3.1 is also a significant improvement of Theorem 2.3.4 in the
sense that the recursion formula (2.2) is simpler and requires less computer time than
the recursion formula (2.6). Moreover, the requirement that E is also uniformly convex
imposed in Theorem 2.3.4 is dispensed with in Theorem 2.3.1. Furthermore, Theorem 2.3.1
is proved in the framework of the more general real Banach spaces with uniformly G?ateaux
differentiable norms.
26
Chapter 3
A Strong Convergence Theorem for Fixed Points of Asymptotically Nonexpansive
Mappings in Banach Spaces
3.1 Introduction
In this chapter, we extend the result of Chapter 2 from the class of nonexpansive mappings
to the class of asymptotically nonexpansive ones.
Recall that a mapping T : K ? K is called asymptotically nonexpansive if there exists a
sequence {kn},kn ? 1, such that limn??kn = 1 and ||Tnx?Tny|| ? kn||x?y|| holds for
each x,y ? K and for each integer n ? 1.
This class of mappings has been studied extensively by various authors (see e.g.,Chidume
and Ali [20],[22], [24], Chidume et al.[32, 33], Chang et al. [15], Falset et al. [45], Kaczor
[52], Oka [68], Schu [75, 76], Wang [92], Qihou [70], Shioji and Takahashi [81], Sun [82],Tan
and Xu [89, 87] and the references contained therein).
Suppose now K is a nonempty closed convex subset of real uniformly smooth Banach space
E and T : K ? K is an asymptotically nonexpansive mapping with sequence kn ? 1 for all
n ? 1. Fix u ? K and define, for each integer n ? 1, the contraction mapping Sn : K ? K
by
Sn(x) =
parenleftBig
1? tnk
n
parenrightBig
u+ tnk
n
Tnx,
27
where {tn} ? [0,1) is any sequence such that tn ? 1. Then, by the Banach contraction
mapping principle, there exists unique xn such that
xn =
parenleftBig
1? tnk
n
parenrightBig
u+ tnk
n
Tnxn.
The question now arises as to whether or not this sequence converges to a fixed point of T.
A partial answer was given in 1994 by Lim and Xu who proved the following theorem:
Theorem 3.1.1 (Lim and Xu [60] ) Suppose E is a real uniformly smooth Banach space
and suppose {tn} is chosen such that limn??parenleftbig kn?1kn?tnparenrightbig = 0. Suppose, in addition, the follow-
ing condition holds:
lim||xn ?Txn|| = 0.
Then, the sequence {xn} defined, for a fixed u ? K, by
xn =
parenleftBig
1? tnk
n
parenrightBig
u+ tnk
n
Tnxn (3.1)
converges strongly to a fixed point of T.
Remark 3.1.2 Observe that equation (4.10) can be re-written as follows:
xn = ?nu+ (1??n)Tnxn
where ?n := 1? tnkn and ?n ? 0 as n ? ?.
It is our purpose in this chapter to extend Theorem 2.3.1 of Chapter 2 to the more
general class of asymptotically nonexpansive mappings.
28
3.2 Convergence Theorems
Theorem 3.2.1 (Chidume and de Souza [35]) Let K be a nonempty closed convex subset of
a real Banach space E which has a uniformly G?ateaux differentiable norm and T : K ? K
be an asymptotically nonexpansive mapping with sequence {kn},kn ? 1 and limkn = 1
such that summationtext(k2n ? 1) < ? and F(T) := {x ? K : Tx = x} negationslash= ?. For a fixed ? ? (0,1),
define Sn : K ? K by Snx := (1 ? ?)x + ?Tnx ? x ? K. Assume that {zt} converges
strongly to a fixed point z of T as t ? 0, where zt is the unique element of K which satisfies
zt = tu + (1 ? t)Tnzt for arbitrary u ? K. Let {?n} be a real sequence in (0,1) which
satisfies the following conditions: C1 : lim?n = 0; C2 : summationtext?n = ?. For arbitrary x0 ? K,
let the sequence {xn} be defined iteratively by
xn+1 = ?nu+ (1??n)Snxn.
Assume {xn} is bounded and bardblxn ? Txnbardbl ? 0 as n ? ?. Then, {xn} converges strongly
to a fixed point of T.
Proof. Observe first that
bardblSnx?Snybardbl ? (1??)bardblx?ybardbl+?bardblTnx?Tnybardbl ? (1?? +kn?)bardblx?ybardbl.
Furthermore, Snx = x if and only if Tnx = x, and hence S is asymptotically nonexpansive
and has the same set of fixed points as T. Define
?n := (1??)?n +? ? n ? 0; yn := xn+1 ?xn +?nxn?
n
, n ? 0. (3.2)
29
Observe that ?n ? ? as n ? ?, and that {xn}, {yn}, {Txn} and {Sxn} are all bounded.
Observe also that from the definitions of ?n and Sn, we obtain that yn = ?nu+(1??n)?Tnxn?n
so that,
||yn+1 ?yn|| ? ||xn+1 ?xn||
?
vextendsinglevextendsingle
vextendsingle?n+1?
n+1
? ?n?
n
vextendsinglevextendsingle
vextendsingle .||u||+ (1??n+1)?
n+1
? ||Tn+1xn+1 ?Tnxn||
+
vextendsinglevextendsingle
vextendsingle
vextendsinglevextendsingle
vextendsingle(1??n+1)?
n+1
?Tn+1xn ? (1??n)?
n
?Tn+1xn
+ (1??n)?
n
?Tn+1xn ? (1??n)?
n
?Tnxn
vextendsinglevextendsingle
vextendsingle
vextendsinglevextendsingle
vextendsingle
?
vextendsinglevextendsingle
vextendsingle?n+1?
n+1
? ?n?
n
vextendsinglevextendsingle
vextendsingle ||u||+ (1??n+1)?
n+1
?kn+1 ||xn+1 ?xn||
+
vextendsinglevextendsingle
vextendsingle
parenleftBig(1??n+1)
?n+1 ?
(1??n)
?n
parenrightBig
?
vextendsinglevextendsingle
vextendsingle ||Tn+1xn||
+
vextendsinglevextendsingle
vextendsingle(1??n+1?
n+1
? 1??n?
n
) ?Kn
vextendsinglevextendsingle
vextendsingle ||xn ?Txn||.
Hence we have for some constant M1 > 0,
||yn+1 ?yn|| ? ||xn+1 ?xn||
?
vextendsinglevextendsingle
vextendsingle?n+1?
n+1
? ?n?
n
vextendsinglevextendsingle
vextendsingle.||u||+
vextendsinglevextendsingle
vextendsingle(1??n+1)?
n+1
?kn+1 ?1
vextendsinglevextendsingle
vextendsingle ||xn+1 ?xn||
+
parenleftBig1??n+1
?n+1 ?
1??n
?n
parenrightBig
?M1 +
parenleftBig1??n
?n
parenrightBig
?kn||xn ?Txn||,
and so,
limsup(||yn+1 ?yn||?||xn+1 ?xn||) ? 0.
30
Hence, by Lemma 2.2.2, ||yn ? xn|| ? 0 as n ? ?. Consequently, lim||xn+1 ? xn|| =
lim?n||yn ? xn|| = 0. Furthermore, ||xn+1 ? Snxn|| = ?n||u ? Snxn|| ? 0 as n ? ?.
Hence, ||xn ?Snxn|| ? ||xn ?xn+1||+||xn+1 ?Snxn|| ? 0 as n ? ?.
Claim: limsup?u?z,j(xn ?z)? ? 0.
For each integer n ? 0, let tn ? (0, ?n1??n) be such that
k2n ?1
tn ? 0, and
||xn ?Sxn||
tn ? 0, n ? ?. (3.3)
Clearly tn ? 0 as n ? ?. Now observe that
ztn = tnu+ (1?tn)Snztn
so that
||ztn ?xn||2 ? (1?tn)2||Snztn ?xn||2 + 2tn?u?ztn,j(ztn ?xn)?
? (1?tn)2
bracketleftBig
||Snztn ?Snxn||+||Snxn ?xn||
bracketrightBig2
+ 2tn||ztn ?xn||2 + 2tn?u?ztn,j(ztn ?xn)?
? (1?tn)2kn2||ztn ?xn||2 + 2||Snztn ?Snxn||||Snxn ?xn||
+ ||Snxn ?xn||2 + 2tn||ztn ?xn||2 + 2tn?u?ztn,j(ztn ?xn)?
31
and because
?u?ztn,j(xn ?ztn)? ? 12
bracketleftBig(1?tn)2k2
n + 2tn ?1
tn
bracketrightBig
||ztn ?xn||2 + ||S
nxn ?xn||M
tn ,
for some constant M > 0, this yields that
limsup?u?ztn,j(xn ?ztn)? ? 0.
Moreover,
?u?ztn,j(xn ?ztn)? = ?u?z,j(xn ?z)?+?u?z,j(xn ?ztn)?j(xn ?z)?
+ ?z ?ztn,j(xn ?ztn)?,
and since j is norm-to-weak? uniformly continuous on bounded sets and ztn ? z, we obtain
that
limsup?u?z,j(xn ?z)? ? 0,
establishing the claim. From xn+1 = ?nu+ (1??n)Snxn we have
||xn+1 ?z||2 ? (1??n)2k2n||xn ?z||2 + 2?n?u?z,j(xn+1 ?z?.
Since tn ? (0, ?n1??n), there exists an integer N0 > 0 such that,
||xn+1 ?z||2 ? (1??n)||xn ?z||2 +?n?n,
32
for all n ? N0, where ?n := 2?u ? z,j(xn+1 ? z)? ? n ? 0. Thus, by Lemma 2.2.3, {xn}
converges strongly to a fixed point of T. a50
Corollary 3.2.2 Let K be a nonempty closed convex subset of a real Banach space E which
has a uniformly G?ateaux differentiable norm. Let T : K ? K be a nonexpansive mapping
with F(T) negationslash= ?. For a fixed ? ? (0,1), define S : K ? K by Sx := (1??)x+?Tx, ? x ? K.
Assume that {zt} converges strongly to a fixed point z of T as t ? 0, where zt is the unique
element of K which satisfies zt = tu + (1 ? t)Tzt for arbitrary u ? K. Let {?n} be a real
sequence in (0,1) which satisfies the conditions: C1 : lim?n = 0; C2 : summationtext?n = ?. For
arbitrary x0 ? K, let the sequence {xn} be defined iteratively by
xn+1 = ?nu+ (1??n)Sxn. (3.4)
Then, {xn} converges strongly to a fixed point of T.
Proof. It is easy to see from equation (3.4) that {xn} is bounded and lim||xn ?Txn|| = 0.
Hence, the result follows from Theorem 3.2.1.
Remark 3.2.3 Theorem 3.2.1 extends Theorem 2.3.1 (and consequently, extends Theorem
2.1.2, Theorem 2.1.3 and Theorem 2.3.4, (see Remarks 3.1, 3.2 and 3.4 of [27]) to the more
general class of asymptotically nonexpansive mappings.
33
Chapter 4
Convergence of a Hybrid Steepest Descent Method for Variational Inequalities in
Banach Spaces
4.1 Introduction
In this chapter, we extend the results of Xu and Kim [98] from real Hilbert spaces to
q?uniformly smooth real Banach spaces which are much more general than Hilbert spaces.
In particular, our theorems will be applicable in Lp spaces, 1 < p < ?.
4.2 Preliminaries
We shall make use of the following lemmas.
Lemma 4.2.1 (Shoiji and Takahashi, [80]) Let (a0,a1,...) ? l? such that ?n(an) ? 0 for
all Banach limit ? and limsup
n??
(an+1 ?an) ? 0. Then, limsup
n??
an ? 0.
Lemma 4.2.2 (Xu, [97]) Let E be a q-uniformly smooth real Banach space for some q > 1,
then there exists some positive constant dq such that
bardblx+ybardblq ? bardblxbardblq +q?y,jq(x)?+dqbardblybardblq ? x,y ? E and jq(x) ? Jq(x).
Lemma 4.2.3 ((Lim and Xu,) [60], Theorem 1) Suppose E is a Banach space with uniform
normal structure, K is a nonempty bounded subset of E, and T : K ? K is uniformly
k?Lipschitzian mapping with k < N(E)12. Suppose also there exists a nonempty bounded
34
closed convex subset C of K with the property (P) :
(P) x ? C implies ?w(x) ? C,
where ?w(x) is the ??limi set of T at x, i.e., the set
{y ? E : y = weak?limj Tnjx for j ? ?}.
Then, T has a fixed point in C.
Lemma 4.2.4 Let X be a real reflexive Banach space and f : X ?? R?{+?} be a convex
proper lower semi-continuous function. Suppose
lim
bardblxbardbl??
f(x) = +?.
Then, ? ?x ? X such that f(?x) ? f(x), x ? X, i.e.,
f(?x) = inf
x?X
f(x).
4.3 Convergence Theorems
Lemma 4.3.1 (Chidume et al. [28]) Let E be a q?uniformly smooth real Banach space
with constant dq, q ? 2. Let T : E ? E be a nonexpansive mapping and G : E ?
E be an ?? strongly accretive and ??Lipschitzian map. For ? ?
parenleftBig
0, 2q(q?1)
parenrightBig
and ? ?
parenleftBig
0,min
braceleftBig
q
4?,(
q?
dq?q)
1
(q?1)
bracerightBigparenrightBig
, define a map T? : E ? E by T?x = Tx ? ??G(Tx), x ? E.
35
Then, T? is a strict contraction. Furthermore,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl ? x,y ? E, (4.1)
where ? := q2 ?
radicalBig
q2
4 ??(q? ??q?1dq?q) ? (0,1).
Proof. For x,y ? E, using Lemma 4.2.2, we have:
bardblT?x?T?ybardblq = bardblTx?Ty ???(G(Tx)?G(Ty))bardblq
? bardblTx?Tybardblq ?q???G(Tx)?G(Ty),jq(Tx?Ty)?
+ dq?q?qbardblG(Tx)?G(Ty)bardblq
? bardblTx?Tybardblq ?q???bardblTx?Tybardblq +dq?q?q?qbardblTx?Tybardblq
?
bracketleftBig
1???parenleftbigq? ?dq?q?1?q?1?qparenrightbig
bracketrightBig
bardblx?ybardblq
?
bracketleftBig
1???parenleftbigq? ?dq?q?1?qparenrightbig
bracketrightBig
bardblx?ybardblq.
Define
f(?) := 1???(q? ?dq?q?1?q) = (1???)q, for some ? ? (0,1), say.
By Taylor development, there exists ? ? (0,?) such that
1???(q? ?dq?q?1?q) = 1?q??+ 12q(q ?1)(1???)q?2?2?2.
Using ? ?
parenleftBig
0, 2q(q?1)
parenrightBig
which implies 12q(q ?1)? < 1, we obtain that
1???(q? ?dq?q?1?q) < 1?q??+ 12q(q ?1)?2?2 < 1?q??+??2,
36
so that
?2 ?q? +?(q? ?dq?q?1?q) > 0.
Solving this quadratic inequality in ?, we obtain, ? < q2 ?
radicalBig
q2
4 ??(q? ?dq?q?1?q).
Now, set
? := q2 ?
radicalbigg
q2
4 ??(q? ?dq?
q?1?q).
Observe that
q2
4 ??(q? ?dq?
q?1?q) =
parenleftBigq2
4 ??q?
parenrightBig
+dq?q?1?q > 0,
since ? < q4?. Moreover, since q ? 2 and ? < 2q(q?1) < 2q, we have
1??? = 1? ?q2 +
radicalbigg
q2?2
4 ??
2?(q? ?dq?q?1?q) ? (0,1).
The proof is complete.
We note that Lp spaces, 2 ? p < ?, are 2?uniformly smooth and the following inequality
holds (see e.g., [97]): For each x,y ? Lp, 2 ? p < ?,
bardblx+ybardbl2 ? bardblxbardbl2 + 2?y,j(x)?+ (p?1)bardblybardbl2.
It then follows that by setting q = 2, dq = (p?1) in Lemma 4.3.1, we obtain the following
corollary.
Corollary 4.3.2 Let E = Lp, 2 ? p < ?. Let T : E ? E, be a nonexpansive map and
G : E ? E be an ??strongly accretive and ??Lipschitzian map. For ? ?
parenleftBig
0,1
parenrightBig
and
? ?
parenleftBig
0,min
braceleftBig
1
2?,
2?
(p?1)?2
bracerightBigparenrightBig
, define a map T? : E ? E by T?x = Tx ? ??G(Tx), x ? E.
37
Then, T? is a strict contraction. In particular,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl, x,y ? E, (4.2)
where ? := 1?radicalbig1??(2? ?(p?1)??2) ? (0,1).
By setting p = 2 in Corollary 4.3.2, we obtain the following corollary.
Corollary 4.3.3 Let H be a real Hilbert space, T : H ? H be a nonexpansive map and
G : H ? H be an ??strongly monotone and ??Lipschitzian map. For ? ?
parenleftBig
0,1
parenrightBig
and
? ?
parenleftBig
0,min
braceleftBig
1
2?,
2?
?2
bracerightBigparenrightBig
, define a map T? : H ? H by T?x = Tx???G(Tx), x ? H. Then,
T? is a strict contraction. In particular,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl, x,y ? H, (4.3)
where ? := 1?radicalbig1??(2? ???2) ? (0,1).
Remark 4.3.4 Corollary 4.3.3 is a result of Yamada [99] and is the main tool used in Wang
[91], Xu and Kim [98], Yamada [99], Zheng and Yao [100]. Lemma 4.3.1 and Corollary
4.3.2 which extend this result to q-uniformly smooth spaces, q ? 2, and Lp spaces, 2 ? p <
?, respectively, are new.
We prove the following theorem for family of nonexpansive maps. In the theorem, dq is the
constant which appears in Lemma 4.2.2.
Theorem 4.3.5 (Chidume et al. [28]) Let E be a q?uniformly smooth real Banach space
with constant dq, q ? 2. Let Ti : E ? E, i = 1,2,...,r be a finite family of nonexpansive
38
mappings with K := r?i=1F(Ti) negationslash= ?. Let G : E ? E be an ??strongly accretive map which is
also ??Lipschitzian. Let {?n} be a real sequence in [0,1] satisfying
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?; C6 : lim?n ??n+r?
n+r
= 0.
For ? ?
parenleftBig
0,min
braceleftBig
q
4?,(
q?
dq?q)
1
(q?1)
bracerightBigparenrightBig
, define a sequence {xn} iteratively in E by x0 ? E,
xn+1 = T?n+1[n+1]xn = T[n+1]xn ???n+1G(T[n+1]xn), n ? 0, (4.4)
where T[n] = Tn mod r. Assume also that
K = F(TrTr?1...T1) = F(T1Tr...T2) = ... = F(Tr?1Tr?2...Tr).
Then, {xn} converges strongly to the unique solution x? of the variational inequality VI(G,K).
Proof. Let x? ? K, then the sequence {xn} satisfies
bardblxn ?x?bardbl ? max
braceleftBig
bardblx0 ?x?bardbl, ??bardblG(x?)bardbl
bracerightBig
, n ? 0.
39
It is obvious that this is true for n = 0. Assume it is true for n = k for some k ? N.
From the recursion formula (4.4) and condition C1, we have
bardblxk+1 ?x?bardbl = bardblT?k+1[k+1]xk ?x?bardbl
? bardblT?k+1[k+1]xk ?T?k+1[k+1]x?bardbl+bardblT?k+1[k+1]x? ?x?bardbl
? (1??k+1?)bardblxk ?x?bardbl+?k+1?bardblG(x?)bardbl
? max
braceleftBig
bardblx0 ?x?bardbl, ??bardblG(x?)bardbl
bracerightBig
,
and the claim follows by induction. Thus the sequence {xn} is bounded and so are
{T[n+1]xn} and {G(T[n+1]xn)}. Using the recursion formula (4.4) we get,
bardblxn+1 ?T[n+1]xnbardbl = ?n+1?bardblG(T[n+1]xn)bardbl ? 0 as n ? ?.
Also,
bardblxn+r ?xnbardbl = bardblT?n+r[n+r]xn+r?1 ?T?n[n] xn?1bardbl
? bardblT?n+r[n+r]xn+r?1 ?T?n+r[n+r]xn?1bardbl+bardblT?n+r[n+r]xn?1 ?T?n[n] xn?1bardbl
? (1??n+r?)bardblxn+r?1 ?xn?1bardbl
+ ??n+r
parenleftBig|?n+r ??n|
??n+r ?bardblG(T[n]xn?1)bardbl
parenrightBig
.
By Lemma 2.2.3 and condition C6, we have
bardblxn+r ?xnbardbl ? 0 as n ? ?. (4.5)
40
In particular,
bardblxn+1 ?xnbardbl ? 0 as n ? ?. (4.6)
Replacing n by n+r ?1 in (4.4) we have,
||xn+r ?Tn+rxn+r?1|| = ??n+r||G(T[n+r]xn+r?1)|| ? 0, n ? ?.
Using the fact that Ti is nonexpansive for each i, we obtain the following finite table:
xn+r ?Tn+rxn+r?1 ? 0 as n ? ?;
Tn+rxn+r?1 ?Tn+rTn+r?1xn+r?2 ? 0 as n ? ?;
...
Tn+rTn+r?1...Tn+2xn+1 ?Tn+rTn+r?1...Tn+2Tn+1xn ? 0 as n ? ?;
and adding up the table yields
xn+r ?Tn+rTn+r?1...Tn+1xn ? 0 as n ? ?.
Using this and (4.5) we get that
limn??||xn ?Tn+rTn+r?1...Tn+1xn|| = 0. (4.7)
Define a map ? : E ? R by ?(y) = ?nbardblxn+1?ybardbl2, where ?n denotes a Banach limit. Then,
? is continuous, convex and ?(y) ? +? as bardblybardbl ? +?. Thus, since E is a reflexive Banach
41
space, there exists y? ? E such that ?(y?) = min
u?E
?(u). So, the set K? := {x ? E : ?(x) =
min
u?E
?(u)} negationslash= ?. We now show Ti has a fixed point in K? for each i = 1,2,...,r. We shall
assume, from equation (4.7), that ?i ,
limn??||xn ?Tixn|| = 0. (4.8)
We shall make use of Lemma 4.2.3. If x is in K? and y := ? ? limj Tmji x, belongs to the
weak ? ? limit set ?w(x) of Ti at x, then, from the w-l.s.c. of ? and equation (4.8), we
have, (since equation (4.8) implies ||xn ? Tmi xn|| ? 0 as n ? ?, this is easily proved by
induction),
?(y) ? liminfj ?
parenleftBig
Tmji x
parenrightBig
? limsup
m
?
parenleftBig
Tmi x
parenrightBig
= limsup
m
parenleftBig
?n||xn ?Tmi x||2
parenrightBig
= limsup
m
parenleftBig
?n||xn ?Tmi xn +Tmi xn ?Tmi x||2
parenrightBig
? limsup
m
parenleftBig
?n||Tmi xn ?Tmi x||2
parenrightBig
? limsup
m
parenleftBig
?n||xn ?x||2
parenrightBig
= ?(x)
= inf
u?E
?(u).
So, y ? K?. By Lemma 4.2.3, Ti has a fixed point in K? ? i and so K? ?K negationslash= ?.
Let x? ? K? ? K and t ? (0,1). It then follows that ?(x?) ? ?(x? ? tG(x?)). Using
the inequality of Lemma 2.2.1, we have that
bardblxn ?x? +tG(x?)bardbl2 ? bardblxn ?x?bardbl2 + 2t?G(x?),j(xn ?x? +tG(x?))?.
42
Thus, taking Banach limits over n ? 1 gives
?nbardblxn ?x? +tG(x?)bardbl2 ? ?nbardblxn ?x?bardbl2
+ 2t?n?G(x?),j(xn ?x? +tG(x?))?.
This implies,
?n??G(x?),j(xn ?x? +tG(x?))? ? ?(x?)??(x? ?tG(x?)) ? 0.
This therefore implies that
?n??G(x?),j(xn ?x? +tG(x?))? ? 0 ? n ? 1.
Since the normalized duality mapping is norm-to-norm uniformly continuous on bounded
subsets of E, we obtain, as t ? 0, that
??G(x?),j(xn ?x?)????G(x?),j(xn ?x? +tG(x?))? ? 0.
Hence, for all ? > 0, there exists ? > 0 such that ?t ? (0,?) and for all n ? 1,
??G(x?),j(xn ?x?)? < ??G(x?),j(xn ?x? +tG(x?))?+?.
Consequently,
?n??G(x?),j(xn ?x?)? ? ?n??G(x?),j(xn ?x? +tG(x?))?+? ? ?.
43
Since ? is arbitrary, we have
?n??G(x?),j(xn ?x?)? ? 0.
Moreover, from the norm-to-norm uniform continuity of j on bounded sets, we obtain, that
limn??
parenleftBig
??G(x?),j(xn+1 ?x?)????G(x?),j(xn ?x?)?
parenrightBig
= 0.
Thus, the sequence {??G(x?),j(xn ?x?)?} satisfies the conditions of Lemma 4.2.1. Hence,
we obtain that
limsup
n??
??G(x?),j(xn ?x?)? ? 0.
Define
?n := max {??G(x?),j(xn+1 ?x?)?,0}.
Then, lim?n = 0, and ??G(x?),j(xn+1 ? x?)? ? ?n. From the recursion formula (4.4),
and Lemma 2.2.1, we have,
bardblxn+1 ?x?bardbl2 = bardblT?n+1[n+1]xn ?T?n+1[n+1]x? +T?n+1[n+1]x? ?x?bardbl2
? bardblT?n+1[n+1]xn ?T?n+1[n+1]x?bardbl2 + 2?n+1???G(x?),j(xn+1 ?x?)?
? (1??n+1?)bardblxn ?x?bardbl2 + 2?n+1???G(x?),j(xn+1 ?x?)?
and by Lemma 2.2.3, we have that xn ? x? as n ? ?. This completes the proof.a50
The following corollaries follow from Theorem 4.3.5.
44
Corollary 4.3.6 Let E = Lp, 2 ? p < ?. Let Ti : E ? E, i = 1,2,...,r be a finite
family of nonexpansive mappings with K = r?i=1F(Ti) negationslash= ?. Let G : E ? E be an ??strongly
accretive map which is also ??Lipschitzian. Let {?n} be a real sequence in [0,1] that satisfies
conditions C1, C2 and C6 as in theorem 4.3.5. For ? ?
parenleftBig
0,min
braceleftBig
1
2?,
2?
(p?1)?2
bracerightBigparenrightBig
, define a
sequence {xn} iteratively in E by (4.4). Then, {xn} converges strongly to the unique solution
x? of the variational inequality VI(G,K).
Corollary 4.3.7 Let H be a real Hilbert space. Let Ti : H ? H, i = 1,2,...,r be a finite
family of nonexpansive mappings with K = r?i=1F(Ti) negationslash= ?. Let G : H ? H be an ??strongly
monotone map which is also ??Lipschitzian. Let {?n} be a real sequence in [0,1] that
satisfies conditions C1, C2 and C6 as in theorem 4.3.6. For ? ?
parenleftBig
0,min
braceleftBig
1
2?,
2?
?2
bracerightBigparenrightBig
, define
a sequence {xn} iteratively in H by (4.4). Then, {xn} converges strongly to the unique
solution x? of the variational inequality VI(G,K).
Theorem 4.3.8 (Chidume et al. [28]) Let E be a real q?uniformly smooth Banach space
with constant dq, q ? 2. Let T : E ? E be a nonexpansive map. Assume that K := F(T) =
{x ? E : Tx = x} negationslash= ?. Let G : E ? E be an ??strongly accretive and ??Lipschitzian map.
Let {?n} be a real sequence in [0,1] satisfying the following conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?; C5 : lim|?n ??n+1|?
n+1
= 0.
For ? ?
parenleftBig
0,min
braceleftBig
q
4?,(
q?
dq?q)
1
(q?1)
bracerightBigparenrightBig
, define a sequence {xn} iteratively in E by x0 ? E,
xn+1 = T?n+1xn = Txn ???n+1G(Txn), n ? 0. (4.9)
Then, {xn} converges strongly to the unique solution x? of the variational inequality VI(G,K).
45
Proof. Take T1 = T2 = ... = Tr = T in Theorem 4.3.5 and the result follows.
The following corollaries follow from Theorem 4.3.8.
Corollary 4.3.9 Let E = Lp, 2 ? p < ?. Let T : E ? E, be a nonexpansive map. Assume
that K := F(T) = {x ? E : Tx = x} negationslash= ?. Let G : E ? E be an ??strongly accretive
and ??Lipschitzian map. Let {?n} be a real sequence in [0,1] that satisfies conditions C1,
C2 and C5 as in theorem 4.3.8. For ? ?
parenleftBig
0,min
braceleftBig
1
2?,
2?
(p?1)?2
bracerightBigparenrightBig
, define a sequence {xn}
iteratively in E by (4.9). Then, {xn} converges strongly to the unique solution x? of the
variational inequality VI(G,K).
Corollary 4.3.10 Let H be a real Hilbert space. Let T : H ? H, be a nonexpansive
map. Assume that K := F(T) = {x ? E : Tx = x} negationslash= ?. Let G : H ? H be an
??strongly monotone and ??Lipschitzian map. Let {?n} be a real sequence in [0,1] that
satisfies conditions C1, C2 and C5 as in theorem 4.3.8. For ? ?
parenleftBig
0,min
braceleftBig
1
2?,
2?
?2
bracerightBigparenrightBig
, define
a sequence {xn} iteratively in H by (4.9). Then, {xn} converges strongly to the unique
solution x? of the variational inequality VI(G,K).
4.4 The case of Lp spaces, 1 < p ? 2.
We begin with the following definition.
Definition 4.4.1 A Banach space E is called a lower weak parallelogram space with con-
stant b ? 0 or, briefly, E is LWP(b), in the terminology of Bynum [12] if
||x+y||2 +b||x?y||2 ? 2(||x||2 +||y||2) (4.10)
46
holds for all x,y ? E.
It is proved in [12] that lp space, 1 < p ? 2, is a lower weak parallelogram space with (p?1)
as the largest number b for which (4.10) holds. Furthermore, if Lp, (1 < p ? 2) has at least
two disjoint sets of positive finite measure, then it is a lower weak parallelogram space with
(p ? 1) as the largest number b for which (4.10) holds. We shall assume, without loss of
generality, that Lp, (1 < p ? 2) has at least two disjoint sets of positive finite measure. In
the sequel, we shall state all our theorems and lemmas only for Lp spaces, 1 < p ? 2, with
the understanding that they also hold for lp spaces, 1 < p ? 2.
In terms of the normalized duality mapping, Bynum [12] proved that a real Banach
space is a lower weak parallelogram space if and only if for each x,y ? E and f ? J(x), the
following inequality holds:
||x+y||2 ? ||x||2 +b||y||2 + 2?y,f?. (4.11)
In particular, for E = Lp, 1 < p ? 2, the following inequality holds:
||x+y||2 ? ||x||2 + (p?1)||y||2 + 2?y,j(x)? ? x,y ? E. (4.12)
We now prove the following lemmas which will be central in the sequel.
Lemma 4.4.2 Let E = Lp,1 < p ? 2. Then, for all x,y ? E, the following inequality
holds:
(p?1)||x+y||2 ? ||x||2 + 2?y,j(x)?+||y||2. (4.13)
47
Proof. Observe first that E is smooth so that the normalized duality map on E is single-
valued. Now, replacing x by (?x) and y by (x+y) in inequality (4.12), we obtain
||y||2 ? ||x||2 + 2?x+y,j(?x)?+ (p?1)||x+y||2,
which implies
(p?1)||x+y||2 ? ||y||2 ?||x||2 + 2?x+y,j(x)?
= ||x||2 + 2?y,j(x)?+||y||2,
establishing inequality (4.13) and completing proof of the lemma.a50
Lemma 4.4.3 (Chidume and de Souza [36]) Let E = Lp, 1 < p ? 2, T : E ? E a non-
expansive mapping and G : E ? E an ??strongly accretive and ?- Lipschitzian mapping.
For,
? ?
parenleftBig
0, 1p?1
parenrightBig
, and ? ?
parenleftBig
0 , min
braceleftBig2?(p?1)
?2 ,
(p?1)2
?
bracerightBigparenrightBig
,
define a map T? : E ? E by: T?x := Tx ? ??G(Tx), x ? E. Then, T? is a strict
contraction. Furthermore,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl ? x,y ? E, (4.14)
where
? := (p?1)?
radicalbig
(p?1)2 ??[2? ??(p?1)?1?2] ? (0,1).
48
Proof. For x,y ? E, using Lemma 4.4.2, we have,
bardblT?x?T?ybardbl2 = bardblTx?Ty ???(G(Tx)?G(Ty))bardbl2
? 1(p?1)
bracketleftBig
bardblTx?Tybardbl2 ?2???G(Tx)?G(Ty),j(Tx?Ty)?
+ ?2?2bardblG(Tx)?G(Ty)bardbl2
bracketrightBig
? 1(p?1)
bracketleftBig
bardblTx?Tybardbl2 ?2???bardblTx?Tybardbl2 +?2?2?2bardblTx?Tybardbl2
bracketrightBig
? 1(p?1)
bracketleftBig
1???[2? ??(p?1)?1?2]
bracketrightBig
bardblx?ybardbl2, since ? < 1(p?1).
Define
f(?) := 1(p?1)
bracketleftBig
1???[2? ??(p?1)?1?2]
bracketrightBig
.
If f(?) = (1 ? ??)2 for some ? ? (0,1) then, ?2 ? 2(p ? 1)? + ? ? 0, where ? :=
?[2? ??(p?1)?1?2]. Thus we obtain that
? ? (p?1)?
radicalbig
(p?1)2 ??[2? ??(p?1)?1?2] ? (0,1).
Now set
? := (p?1)?
radicalbig
(p?1)2 ??[2? ??(p?1)?1?2] ? (0,1),
and the proof is complete. a50
Remark 4.4.4 In Hilbert space, by putting p = 2 and observing that ? can always be as-
sumed to be arbitrarily small, without any loss of generality, we get, min
braceleftBig2?(p?1)
?2 ,
(p?1)2
?
bracerightBig
=
2?
?2.
Thus, we have the following corollary.
49
Corollary 4.4.5 Let H be a real Hilbert space, T : H ? H a nonexpansive mapping, G :
H ? H an ??strongly monotone and ??Lipschitzian map. For ? ? (0,1) and ? ? (0, 2??2),
define a map T? : H ? H by: T?x = Tx ? ??G(Tx), x ? H. Then, T? is a strict
contraction. In particular,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl ? x,y ? H, (4.15)
where ? := 1?radicalbig1??(2? ???2) ? (0,1).
Proof. Set p = 2 in Lemma 4.4.3 and the result follows.
Remark 4.4.6 Corollary 4.4.5 is a result of Yamada [99] and is the main tool used in Wang
[91], Xu and Kim [98], Yamada [99] and Zheng and Yao [100]. Consequently, Lemma 4.4.3
is an important extension of these results to Lp spaces, 1 < p ? 2.
We now prove the following theorems. In the theorem, F(Ti) := {x ? E : Tix = x}.
Theorem 4.4.7 (Chidume and de Souza [36]) Let E = Lp, 1 < p ? 2, T : E ? E
a nonexpansive mapping. Assume K := {x ? E : Tx = x} negationslash= ?. Let G : E ? E be
an ??strongly accretive and ??Lipschitzian mapping. Let {?n} be a real sequence in [0,1]
satisfying the following conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?;C3 : lim?n ??n+1?
n+1
= 0.
For ? ?
parenleftBig
0, min
braceleftBig2?(p?1)
?2 ,
(p?1)2
?
bracerightBigparenrightBig
, define a sequence {xn} iteratively in E by x0 ? E,
xn+1 = T?n+1xn = Txn ???n+1G(Txn), n ? 0. (4.16)
50
Then, {xn} converges strongly to the unique solution x? of the variational inequality problem
VI(G,K).
Proof. This follows using Lemma 4.4.3.
The following corollary follows from Theorem 4.4.7.
Corollary 4.4.8 Let H be a real Hilbert space, T : H ? H a nonexpansive mapping.
Assume K := {x ? E : Tx = x} negationslash= ?. Let G : H ? H be an ??strongly monotone and
??Lipschitzian mapping. Let {?n} be a real sequence in [0,1] that satisfies conditions C1,
C2 and C3 as in Theorem 4.4.7. For ? ? (0, 2??2), define a sequence {xn} iteratively in H by
(4.16). Then, {xn} converges strongly to the unique solution x? of the variational inequality
problem VI(G,K).
Following the method of section 4.3 and using Lemma 4.4.3, the following theorem and
corollary are easily proved.
Theorem 4.4.9 Let E = Lp, 1 < p ? 2, Ti : E ? E,i = 1,2,...,r a finite family of
nonexpansive mappings with K := r?i=1F(Ti) negationslash= ?. Let G : E ? E be an ??strongly accretive
and ??Lipschitzian mapping. Let {?n} be a real sequence in [0,1] satisfying the conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?; C6 : lim?n ??n+r?
n+r
= 0.
For ? ?
parenleftBig
0, min
braceleftBig2?(p?1)
?2 ,
(p?1)2
?
bracerightBigparenrightBig
, define a sequence {xn} iteratively in E by: x0 ? E,
xn+1 = T?n+1[n+1]xn = T[n+1]xn ???nG(T[n+1]xn), n ? 0, (4.17)
51
where T[n] = Tn mod r. Assume also that
K = F(TrTr?1...T1) = F(T1Tr...T2) = ... = F(Tr?1Tr?2...Tr)
and limn??bardblT[n+2]xn ? T[n+1]xnbardbl = 0. Then, {xn} converges strongly to the unique solution
x? of the variational inequality problem VI(G,K).
Corollary 4.4.10 Let H be a real Hilbert space, Ti : H ? H, i = 1,2,...,r a finite family
of nonexpansive mappings with K := r?i=1F(Ti) negationslash= ?. Let G : H ? H be an ??strongly
monotone and ??Lipschitzian mapping. Let {?n} be a real sequence in [0,1] that satisfies
conditions C1, C2 and C6 as in Theorem 4.4.9 and let limn??bardblT[n+2]xn ?T[n+1]xnbardbl = 0. For
? ? (0, 2??2), define a sequence {xn} iteratively in H by (5.9). Then, {xn} converges strongly
to the unique solution x? of the variational inequality problem VI(G,K).
Remark 4.4.11 Our theorems in this chapter which are extensions of the results of Yamada
[99], Wang [91], Xu and Kim [98], Zeng and Yao [100] from real Hilbert spaces to Lp spaces,
1 < p ? 2 complement the theorems earlier in the chapter (see also Chidume et al. [28]) to
provide convergence theorems, for the problems considered here, in all Lp spaces, 1 < p < ?.
52
Chapter 5
Approximation of Fixed Points of Nonexpansive Mappings and Solutions of
Variational Inequalities
5.1 Introduction
In Chapter 4, we extended the results of Xu and Kim [98] to q?uniformly smooth
Banach spaces, q ? 2. In particular, we proved theorems which are applicable in Lp spaces,
1 < p < ? under conditions C1,C2 and C5 or C6 (as in the result of Xu and Kim).
In this chapter, we introduce new recursion formulas and prove strong convergence
theorems for the unique solution of the variational inequality problem VI(K,S), requiring
only conditions C1 and C2 on the parameter sequence {?n}. Furthermore in the case Ti :
E ? E i = 1,2,...,r is a family of nonexpansive mappings withK = r?i=1F(Ti) negationslash= ?,we prove
a convergence theorem where condition C6 is replaced with limn??bardblT[n+2]xn ? T[n+1]xnbardbl =
0. An example satisfying this condition is given in [21]. All our theorems are proved in
q?uniformly smooth Banach spaces, q ? 2. In particular, our theorems are applicable in Lp
spaces, 1 < p < ?.
5.2 Convergence Theorems
We first prove the following lemma which will be central in the sequel.
Lemma 5.2.1 (Chidume et al. [29]) Let E be a q?uniformly smooth real Banach space
with constant dq, q ? 2, T : E ? E a nonexpansive mapping and G : E ? E an
53
??strongly accretive and ??Lipschitzian mapping. For
? ?
parenleftBig
0,min
braceleftBig q
4??,(
q?
dq?q)
1
(q?1)
bracerightBigparenrightBig
, ?,? ? (0,1),
define a mapping T? : E ? E by:
T?x := (1??)x+?[Tx???G(Tx)], x ? E.
Then, T? is a strict contraction. Furthermore,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl x,y ? E, (5.1)
where
? := q2 ?
radicalbigg
q2
4 ???(q? ??
q?1dq?q) ? (0,1).
Proof. For x,y ? E, using the convexity of ||.||q and Lemma 4.2.2, we have,
bardblT?x?T?ybardblq = bardbl(1??)(x?y) +?[Tx?Ty ???(G(Tx)?G(Ty))]bardblq
? (1??)bardblx?ybardblq +?
bracketleftBig
bardblTx?Tybardblq ?q???G(Tx)?G(Ty),jq(Tx?Ty)?
+ dq?q?qbardblG(Tx)?G(Ty)bardblq
bracketrightBig
? (1??)bardblx?ybardblq +?
bracketleftBig
bardblTx?Tybardblq ?q???bardblTx?Tybardblq
+ dq?q?q?qbardblTx?Tybardblq
bracketrightBig
?
bracketleftBig
1????parenleftbigq? ?dq?q?1?q?1?qparenrightbig
bracketrightBig
bardblx?ybardblq
?
bracketleftBig
1????parenleftbigq? ?dq?q?1?qparenrightbig
bracketrightBig
bardblx?ybardblq.
54
Define
f(?) := 1????(q? ?dq?q?1?q) = (1???)q, for some ? ? (0,1), say.
Then, there exists ? ? (0,?) such that
1????(q? ?dq?q?1?q) = 1?q??+ 12q(q ?1)(1???)q?2?2?2,
and since q ? 2, this implies
1????(q? ?dq?q?1?q) ? 1?q??+ 12q(q ?1)?2?2,
which yields,
?2 ?q? +??
parenleftBig
q? ?dq?q?1?q
parenrightBig
> 0,
since ? ?
parenleftBig
0, 2q(q?1)
parenrightBig
. Thus we have,
? ? q2 ?
radicalbigg
q2
4 ???(q? ??
q?1dq?q) ? (0,1).
Set
? := q2 ?
radicalbigg
q2
4 ???(q? ??
q?1dq?q) ? (0,1).
and the proof is complete. a50
We note that in Lp spaces, 2 ? p < ?, the following inequality holds (see e.g., [17]):
55
For each x,y ? Lp, 2 ? p < ?,
bardblx+ybardbl2 ? bardblxbardbl2 + 2?y,j(x)?+ (p?1)bardblybardbl2.
It then follows that by setting q = 2, dq = p?1 in Lemma 5.2.1, the following corollary is
easily proved.
Corollary 5.2.2 Let E = Lp, 2 ? p < ?, T : E ? E a nonexpansive mapping and
G : E ? E an ??strongly accretive and ??Lipschitzian mapping. For ?, ? ? (0,1), and
? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
(p?1)?2
bracerightBigparenrightBig
, define a mapping T? : E ? E by:
T?x := (1??)x+?[Tx???G(Tx)] ? x ? E.
Then, T? is a strict contraction. In particular,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl x,y ? H, (5.2)
where ? := 1?radicalbig1???(2? ?(p?1)??2) ? (0,1).
We also have the following corollary.
Corollary 5.2.3 Let H be a real Hilbert space, T : H ? H a nonexpansive mapping,
G : H ? H an ??strongly monotone and ??Lipschitzian mapping. For ?,? ? (0,1) and
? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
?2
bracerightBigparenrightBig
, define a mapping T? : H ? H by:
T?x = (1??)x+?[Tx???G(Tx)] ? x ? H.
56
Then, T? is a strict contraction. In particular,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl ? x,y ? H, (5.3)
where ? := 1?radicalbig1???(2? ???2) ? (0,1).
Proof. Set p = 2 in Corollary 5.2.2 and the result follows.
We now prove the following convergence theorems.
Theorem 5.2.4 (Chidume et al. [29]) Let E be a q?uniformly smooth real Banach space
with constant dq, q ? 2 and T : E ? E a nonexpansive mapping. Assume K := {x ? E :
Tx = x} negationslash= ?. Let G : E ? E be an ??strongly accretive and ??Lipschitzian mapping. Let
{?n} be a real sequence in [0,1] satisfying the conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?.
For ? ?
parenleftBig
0,min
braceleftBig
q
4??,(
q?
dq?q)
1
(q?1)
bracerightBigparenrightBig
, ? ? (0,1), define a sequence {xn} iteratively in E by
x0 ? E,
xn+1 = T?n+1xn = (1??)xn +?[Txn ???n+1G(Txn)], n ? 0. (5.4)
Then, {xn} converges strongly to the unique solution x? of the variational inequality VI(G,K).
Proof. Let x? ? K := F(T), then the sequence {xn} satisfies
bardblxn ?x?bardbl ? max
braceleftBig
bardblx0 ?x?bardbl, ??bardblG(x?)bardbl
bracerightBig
, n ? 0.
57
It is obvious that this is true for n = 0. Assume it is true for n = k for some k ? N. From
the recursion formula (5.4), we have
bardblxk+1 ?x?bardbl = bardblT?k+1xk ?x?bardbl
? bardblT?k+1xk ?T?k+1x?bardbl+bardblT?k+1x? ?x?bardbl
? (1??k+1?)bardblxk ?x?bardbl+?k+1?bardblG(x?)bardbl
? max
braceleftBig
bardblx0 ?x?bardbl, ??bardblG(x?)bardbl
bracerightBig
,
and the claim follows by induction. Thus the sequence {xn} is bounded and so are the
sequences {Txn} and {G(Txn)}.
Define two sequences {?n} and {yn} by ?n := (1??)?n+1 +? and
yn := xn+1?xn+?nxn?n . Then,
yn = (1??)?n+1xn +?[Txn ??n+1?G(Txn)]?
n
.
Observe that {yn} is bounded and that
bardblyn+1 ?ynbardbl?bardblxn+1 ?xnbardbl ?
vextendsinglevextendsingle
vextendsingle ??
n+1
?1
vextendsinglevextendsingle
vextendsingle bardblxn+1 ?xnbardbl
+
vextendsinglevextendsingle
vextendsingle ??
n+1
? ??
n
vextendsinglevextendsingle
vextendsinglebardblTxnbardbl+ ?n+2(1??)?
n+1
bardblxn+1 ?xnbardbl
+ (1??)
vextendsinglevextendsingle
vextendsingle?n+2?
n+1
? ?n+1?
n
vextendsinglevextendsingle
vextendsinglebardblxnbardbl+ ?n+1???
n
bardblG(Txn)?G(Txn+1)bardbl
+ ??
vextendsinglevextendsingle
vextendsingle?n+1?
n
? ?n+2?
n+1
vextendsinglevextendsingle
vextendsinglebardblG(Txn+1)bardbl.
58
This implies, limsup
n??
(||yn+1 ?yn||?||xn+1 ?xn||) ? 0, and therefore by Lemma 2.2.2,
limn??||yn ?xn|| = 0.
Hence,
||xn+1 ?xn|| = ?n||yn ?xn|| ? 0 as n ? ?. (5.5)
From the recursion formula (5.4), we have that
?bardblxn+1 ?Txnbardbl ? (1??)bardblxn+1 ?xnbardbl+?n+1??bardblG(Txn)bardbl ? 0 as n ? ?.
which implies,
bardblxn+1 ?Txnbardbl ? 0 as n ? ?. (5.6)
From (5.5) and (5.6) we have
bardblxn ?Txnbardbl ? bardblxn ?xn+1bardbl+bardblxn+1 ?Txnbardbl ? 0 as n ? ?. (5.7)
We now prove that
limsup
n??
??G(x?),j(xn+1 ?x?)? ? 0.
Define a map ? : E ? R by
?(x) = ?nbardblxn ?xbardbl2 ? x ? E,
59
where ?n is a Banach limit for each n. Then, ?(x) ? ? as bardblxbardbl ? ?, ? is continuous and
convex, so as E is reflexive, there exists y? ? E such that ?(y?) = min
u?E
?(u). Hence, the set
K? :=
braceleftBig
x ? E : ?(x) = min
u?E
?(u)
bracerightBig
negationslash= ?.
We now show T has a fixed point in K?. We know
limn??||xn ?Tixn|| = 0. (5.8)
We shall make use of Lemma 4.2.3. If x is in K? and y := ? ? limj Tmjx, belongs to the
weak ? ?limit set ?w(x) of T at x, then, from the w-l.s.c. (since ? is l.s.c. and convex) of
? and equation (5.8), we have,
?(y) ? liminfj ?
parenleftBig
Tmjx
parenrightBig
? limsup
m
?
parenleftBig
Tmx
parenrightBig
= limsup
m
parenleftBig
?n||xn ?Tmx||2
parenrightBig
= limsup
m
parenleftBig
?n||xn ?Tmxn +Tmxn ?Tmx||2
parenrightBig
? limsup
m
parenleftBig
?n||Tmxn ?Tmx||2
parenrightBig
? limsup
m
parenleftBig
?n||xn ?x||2
parenrightBig
= ?(x)
= inf
u?E
?(u).
So, y ? K?. By Lemma 4.2.3, T has a fixed point in K? and so K? ?K negationslash= ?.
By Lemma 4.2.3, K? ? K negationslash= ?. Let x? ? K? ? K and let t ? (0,1). Then, it follows
60
that ?(x?) ? ?(x? ?tG(x?)) and using Lemma 2.2.1, we obtain that
bardblxn ?x? +tG(x?)bardbl2 ? bardblxn ?x?bardbl2 + 2t?G(x?),j(xn ?x? +tG(x?))?
which implies,
?n??G(x?),j(xn ?x? +tG(x?))angbracketrightbig? 0.
The rest now follows exactly as in the proof of Theorem 4.3.5 to yield that xn ? x? as n ?
?. This completes the proof. a50
The following corollaries follow from Theorem 5.2.4.
Corollary 5.2.5 Let E = Lp, 2 ? p < ?, T : E ? E a nonexpansive mapping. Assume
K := {x ? E : Tx = x} negationslash= ?. Let G : E ? E be an ??strongly accretive and ??Lipschitzian
mapping. Let {?n} be a real sequence in [0,1] that satisfies conditions C1 and C2 as
in theorem 5.2.4. For ? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
(p?1)?2
bracerightBigparenrightBig
, ? ? (0,1), define a sequence {xn}
iteratively in E by (5.4). Then, {xn} converges strongly to the unique solution x? of the
variational inequality problem VI(G,K).
Corollary 5.2.6 Let H be a real Hilbert space, T : H ? H a nonexpansive mapping.
Assume K := {x ? H : Tx = x} negationslash= ?. Let G : H ? H be an ??strongly monotone
??Lipschitzian mapping. Further, let {?n} be a real sequence in [0,1] that satisfies con-
ditions C1 and C2 as in Theorem 5.2.4. For ? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
?2
bracerightBigparenrightBig
, ? ? (0,1), define
a sequence {xn} iteratively in H by (5.4). Then, {xn} converges strongly to the unique
solution x? of the variational inequality problem VI(G,K).
Finally, we prove the following theorem for a finite family of nonexpansive mappings.
61
Theorem 5.2.7 (Chidume et al. [29]) Let E be a q?uniformly smooth real Banach space
with constant dq, q ? 2, Ti : E ? E, i = 1,2,...,r a finite family of nonexpan-
sive mappings with K := r?i=1F(Ti) negationslash= ?. Let G : E ? E be an ??strongly accretive and
??Lipschitzian mapping, and {?n} a real sequence in [0,1] satisfying the conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?.
For a fixed real number ? ?
parenleftBig
0,min
braceleftBig
q
4??,(
q?
dq?q)
1
(q?1)
bracerightBigparenrightBig
, ? ? (0,1), define a sequence {xn}
iteratively in E by x0 ? E,
xn+1 = T?n+1[n+1]xn = (1??)xn +?[T[n+1]xn ???nG(T[n+1]xn)], n ? 0, (5.9)
where T[n] = Tn mod r. Assume also that
K = F(TrTr?1...T1) = F(T1Tr...T2) = ... = F(Tr?1Tr?2...Tr)
and limn??bardblT[n+2]xn ? T[n+1]xnbardbl = 0. Then, {xn} converges strongly to the unique solution
x? of the variational inequality problem VI(G,K).
Proof. Let x? ? K, then the sequence {xn} satisfies
bardblxn ?x?bardbl ? max
braceleftBig
bardblx0 ?x?bardbl, ??bardblG(x?)bardbl
bracerightBig
, n ? 0.
62
It is obvious that this is true for n = 0. Assume it is true for n = k for some k ? N.
From the recursion formula (5.9), we have
bardblxk+1 ?x?bardbl = bardblT?k+1[k+1]xk ?x?bardbl
? bardblT?k+1[k+1]xk ?T?k+1[k+1]x?bardbl+bardblT?k+1[k+1]x? ?x?bardbl
? (1??k+1?)bardblxk ?x?bardbl+?k+1?bardblG(x?)bardbl
? max
braceleftBig
bardblx0 ?x?bardbl, ??bardblG(x?)bardbl
bracerightBig
,
and the claim follows by induction. Thus the sequence {xn} is bounded and so are {T[n]xn}
and {G(T[n]xn)}.
Define two sequences {?n} and {yn} by ?n := (1??)?n+1 +? and
yn := xn+1?xn+?nxn?n . Then,
yn = (1??)?n+1xn +?[T[n+1]xn ??n+1?G(T[n+1]xn)]?
n
.
Observe that {yn} is bounded and that
bardblyn+1 ?ynbardbl?bardblxn+1 ?xnbardbl ?
vextendsinglevextendsingle
vextendsingle ??
n+1
?1
vextendsinglevextendsingle
vextendsingle bardblxn+1 ?xnbardbl
+ ??
n+1
bardblT[n+2]xn ?T[n+1]xnbardbl+
vextendsinglevextendsingle
vextendsingle ??
n+1
? ??
n
vextendsinglevextendsingle
vextendsinglebardblT[n+1]xnbardbl
+ ?n+2(1??)?
n+1
bardblxn+1 ?xnbardbl+ (1??)
vextendsinglevextendsingle
vextendsingle?n+2?
n+1
? ?n+1?
n
vextendsinglevextendsingle
vextendsinglebardblxnbardbl
+ ?n+1???
n
bardblG(T[n+1]xn)?G(T[n+2]xn+1)bardbl
+ ??
vextendsinglevextendsingle
vextendsingle?n+1?
n
? ?n+2?
n+1
vextendsinglevextendsingle
vextendsinglebardblG(T[n+2]xn+1)bardbl.
63
This implies,
limsup
n??
(||yn+1 ?yn||?||xn+1 ?xn||) ? 0,
and by Lemma 2.2.2,
limn??||yn ?xn|| = 0. Hence,
||xn+1 ?xn|| = ?n||yn ?xn|| ? 0 (5.10)
as n ? ?. From the recursion formula (5.9), we have that
?bardblxn+1 ?T[n+1]xnbardbl ? (1??)bardblxn+1 ?xnbardbl+?n+1??bardblG(T[n+1]xn)bardbl ? 0,
as n ? ?, which implies,
bardblxn+1 ?T[n+1]xnbardbl ? 0 as n ? ?. (5.11)
Note that from (5.10) and (5.11) we have
bardblxn ?T[n+1]xnbardbl ? bardblxn ?xn+1bardbl+bardblxn+1 ?T[n+1]xnbardbl ? 0 as n ? ?. (5.12)
Also,
bardblxn+r ?xnbardbl ? bardblxn+r ?xn+r?1bardbl+bardblxn+r?1 ?xn+r?2bardbl+???+bardblxn+1 ?xnbardbl
and so,
bardblxn+r ?xnbardbl ? 0 as n ? ?. (5.13)
64
Using the fact that Ti is nonexpansive for each i, we obtain the following finite table:
xn+r ?Tn+rxn+r?1 ? 0 as n ? ?;
Tn+rxn+r?1 ?Tn+rTn+r?1xn+r?2 ? 0 as n ? ?;
...
Tn+rTn+r?1 ???Tn+2xn+1 ?Tn+rTn+r?1 ???Tn+2Tn+1xn ? 0 as n ? ?;
and adding up the table yields
xn+r ?Tn+rTn+r?1 ???Tn+1xn ? 0 as n ? ?.
Using this and (5.13) we get that limn??||xn ?Tn+rTn+r?1 ???Tn+1xn|| = 0.
Carrying out similar arguments as in the proof of Theorem 5.2.4, we easily get that
limsup
n??
??G(x?),j(xn+1 ?x?)? ? 0.
From the recursion formula (5.9), and Lemma 2.2.1 we have
bardblxn+1 ?x?bardbl2 = bardblT?n+1[n+1]xn ?T?n+1[n] x? +T?n+1[n+1]x? ?x?bardbl2
? bardblT?n+1[n+1]xn ?T?n+1[n+1]x?bardbl2 + 2?n+1????G(x?),j(xn+1 ?x?)?
? (1??n+1?)bardblxn ?x?bardbl2 + 2?n+1????G(x?),j(xn+1 ?x?)?
65
which by using Lemma 2.2.3, gives that xn ? x? as n ? ?, completing the proof. a50
The following corollaries follow from Theorem 5.2.7.
Corollary 5.2.8 Let E = Lp, 2 ? p < ?, Ti : E ? E, i = 1,2,...,r a finite family
of nonexpansive mappings with K = r?i=1F(Ti) negationslash= ?. Let G : E ? E be an ??strongly
accretive and ??Lipschitzian mapping. Let {?n} be a real sequence in [0,1] that satisfies
conditions C1 and C2 as in Theorem 5.2.7 and let limn??bardblT[n+2]xn ? T[n+1]xnbardbl = 0. For
? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
(p?1)?2
bracerightBigparenrightBig
, ? ? (0,1), define a sequence {xn} iteratively in E by (5.9).
Then, {xn} converges strongly to the unique solution x? of the variational inequality problem
VI(G,K).
Corollary 5.2.9 Let H be a real Hilbert space, Ti : H ? H, i = 1,2,...,r a finite family
of nonexpansive mappings with K = r?i=1F(Ti) negationslash= ?. Let G : H ? H be an ??strongly
monotone and ??Lipschitzian mapping. Let {?n} be a real sequence in [0,1] that satisfies
conditions C1 and C2 as in theorem 5.2.7 and let limn??bardblT[n+2]xn ? T[n+1]xnbardbl = 0. For
? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
?2
bracerightBigparenrightBig
, ? ? (0,1), define a sequence {xn} iteratively in H by (5.9). Then,
{xn} converges strongly to the variational inequality problem VI(G,K).
5.3 The case of Lp spaces, 1 < p ? 2.
We first prove the following lemmas.
We begin with the following definition. A Banach space E is called a lower weak paral-
lelogram space with constant b ? 0 or, briefly, E is LWP(b), in the terminology of Bynum
66
[11] if
||x+y||2 +b||x?y||2 ? 2(||x||2 +||y||2) (5.14)
holds for all x,y ? E. It is proved in [11] that lp space, 1 < p ? 2, is a lower weak
parallelogram space with (p?1) as the largest number b for which (5.14) holds. Furthermore,
if Lp, (1 < p ? 2), has at least two disjoint sets of positive finite measure, then it is a lower
weak parallelogram space with (p?1) as the largest number b for which (5.14) holds. We
shall assume, without loss of generality, that Lp, (1 < p ? 2), has at least two disjoint sets
of positive finite measure. In the sequel, we shall state all our theorems and lemmas only for
Lp spaces, 1 < p ? 2, with the understanding that they also hold for lp spaces, 1 < p ? 2.
In terms of the normalized duality mapping, Bynum [11] proved that a real Banach
space is a lower weak parallelogram space if and only if for each x,y ? E and f ? J(x), the
following inequality holds:
||x+y||2 ? ||x||2 +b||y||2 + 2?y,f?. (5.15)
In particular, for E = Lp, 1 < p ? 2, the following inequality holds:
||x+y||2 ? ||x||2 + (p?1)||y||2 + 2?y,j(x)? ? x,y ? E. (5.16)
We now obtain the following lemmas which will be central in the sequel.
67
Lemma 5.3.1 Let E = Lp,1 < p ? 2. Then, for all x,y ? E, the following inequality
holds:
(p?1)||x+y||2 ? ||x||2 + 2?y,j(x)?+||y||2. (5.17)
Proof. Observe first that E is smooth so that the normalized duality map on E is single-
valued. Now, replacing x by (?x) and y by (x + y) in inequality (5.16), we obtain ||y||2 ?
||x||2 + 2?x+y,j(?x)?+ (p?1)||x+y||2, so that
(p?1)||x+y||2 ? ||y||2 ?||x||2 + 2?x+y,j(x)?
= ||x||2 + 2?y,j(x)?+||y||2,
establishing the lemma. a50
Lemma 5.3.2 (Chidume et al. [30]) Let E = Lp, 1 < p ? 2, T : E ? E be a nonexpansive
mapping and G : E ? E be an ??strongly accretive and ?- Lipschitzian mapping. For,
? ? (0,1), ? ? (0,1), ? ?
parenleftBig
0 , min
braceleftBig2?
?2 ,
(p?1)2
2??
bracerightBigparenrightBig
,
define a map T? : E ? E by
T?x := (1??)x+?
bracketleftBig
Tx???G(Tx)
bracketrightBig
, x ? E.
Then, T? is a strict contraction. Furthermore,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl ? x,y ? E, (5.18)
68
where
? := (p?1)?
radicalbig
(p?1)2 ???(2? ???2) ? (0,1).
Proof. For x,y ? E, using the convexity of ||.||2, and Lemma 5.3.1, we have,
bardblT?x?T?ybardbl2 = bardbl(1??)(x?y) +?[Tx?Ty ???(G(Tx)?G(Ty))]bardbl2
? (1??)bardblx?ybardbl2 + ?(p?1)
bracketleftBig
bardblTx?Tybardbl2
? 2???G(Tx)?G(Ty),j(Tx?Ty)?+?2?2bardblG(Tx)?G(Ty)bardbl2
bracketrightBig
? (1??)bardblx?ybardbl2 + ?(p?1)
bracketleftBig
bardblTx?Tybardbl2 ?2???bardblTx?Tybardbl2
+ ?2?2?2bardblTx?Tybardbl2
bracketrightBig
?
bracketleftBig
1 +?
parenleftBig 1
p?1 ?1
parenrightBig
? 2????(p?1) + ???
2?2
(p?1)
bracketrightBig
bardblx?ybardbl2, (? < 1).
Define
f(?) := 1 +?
parenleftBig 1
p?1 ?1
parenrightBig
? 2????(p?1) + ???
2?2
(p?1) = (1???)
2,
for some ? ? (0,1), say. Since
parenleftBig
1
p?1 ?1
parenrightBig
> 0, and ?(p?1) ? 1, this implies,
? 2???(p?1) + ??
2?2
(p?1) ? ?2? +?
2,
which yields
?2 ?2(p?1)? + 2??? ???2?2 ? 0,
implying that
? ? (p?1)?
radicalbig
(p?1)2 ???(2? ???2) ? (0,1).
69
Now set
? := (p?1)?
radicalbig
(p?1)2 ???(2? ???2) ? (0,1),
and the proof is complete. a50
Remark 5.3.3 In a Hilbert space, by putting p = 2 and observing that ? can always be as-
sumed to be arbitrarily small, without any loss of generality, we get, min
braceleftBig2?(p?1)
?2 ,
(p?1)2
?
bracerightBig
=
2?
?2.
By Remark 5.3.3, we have the following corollary.
Corollary 5.3.4 Let H be a real Hilbert space, T : H ? H be a nonexpansive map-
ping, G : H ? H be an ??strongly ??Lipschitzian mapping. For ? ? (0,1) and ? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
?2
bracerightBigparenrightBig
, ? ? (0,1), define a mapping T? : H ? H by: T?x = (1??)x+?[Tx?
??G(Tx)], x ? H. Then, T? is a strict contraction. In particular,
bardblT?x?T?ybardbl ? (1???)bardblx?ybardbl ? x,y ? H, (5.19)
where ? := 1?radicalbig1???(2? ???2) ? (0,1).
Proof. Set p = 2 in Lemma 5.3.2 and the result follows.
We now prove the following theorem.
Theorem 5.3.5 (Chidume et al. [30]) Let E = Lp, 1 < p ? 2, T : E ? E be a
nonexpansive mapping. Assume K := {x ? E : Tx = x} negationslash= ?. Let G : E ? E be an
??strongly accretive and ??Lipschitzian mapping. Let {?n} be a real sequence in [0,1]
70
satisfying the conditions:
C1 : lim?n = 0; C2 :
summationdisplay
?n = ?.
For ? ? (0,1), and ? ?
parenleftBig
0, min
braceleftBig
2?
?2 ,
(p?1)2
2??
bracerightBigparenrightBig
, define a sequence {xn} iteratively in E by
x0 ? E,
xn+1 = T?n+1xn = (1??)xn +?[Txn ???n+1G(Txn)], n ? 0. (5.20)
Then, {xn} converges strongly to the unique solution x? of the variational inequality problem
VI(G,K).
Proof. This follows exactly as in the proof of theorem 5.2.4, using Lemma 5.3.2.
The following corollary follows from Theorem 5.3.5.
Corollary 5.3.6 Let H be a real Hilbert space, T : H ? H be a nonexpansive mapping.
Assume K := {x ? E : Tx = x} negationslash= ?. Let G : H ? H be an ??strongly monotone
??Lipschitzian mapping. Let {?n} be a real sequence in [0,1] that satisfies conditions C1
and C2 as in theorem 5.2.4. For ? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
?2
bracerightBigparenrightBig
, ? ? (0,1), define a sequence {xn}
iteratively in H by (5.4). Then, {xn} converges strongly to the unique solution x? of the
variational inequality VI(G,K).
Following the method of Section 5.2, the following theorem and corollary are easily proved.
Theorem 5.3.7 Let E = Lp, 1 < p ? 2, and Ti : E ? E,i = 1,2,...,r be a finite family of
nonexpansive mappings with K := r?i=1F(Ti) negationslash= ?. Let G : E ? E be an ??strongly accretive
and ??Lipschitzian mapping. Let {?n} be a real sequence in [0,1] satisfying the conditions:
71
C1 : lim?n = 0; C2 : summationtext?n = ?. For ? ? (0,1), and ? ?
parenleftBig
0, min
braceleftBig
2?
?2 ,
(p?1)2
2??
bracerightBigparenrightBig
, define
a sequence {xn} iteratively in E by x0 ? E,
xn+1 = T?n+1[n+1]xn = (1??)xn +?[T[n+1]xn ???nG(T[n+1]xn)], n ? 0, (5.21)
where T[n] = Tn mod r. Assume also that
K = F(TrTr?1...T1) = F(T1Tr...T2) = ... = F(Tr?1Tr?2...Tr)
and limn??bardblT[n+2]xn ? T[n+1]xnbardbl = 0. Then, {xn} converges strongly to the unique solution
x? of the variational inequality problem VI(G,K).
Corollary 5.3.8 Let H be a real Hilbert space, Ti : H ? H, i = 1,2,...,r be a finite
family of nonexpansive mappings with K := r?i=1F(Ti) negationslash= ?. Let G : H ? H be an ??strongly
monotone and ??Lipschitzian mapping. Let {?n} be a real sequence in [0,1] that satisfies
conditions C1 and C2 as in theorem 5.3.7 and let limn??bardblT[n+2]xn ? T[n+1]xnbardbl = 0. For
? ?
parenleftBig
0,min
braceleftBig
1
2??,
2?
?2
bracerightBigparenrightBig
, ? ? (0,1), define a sequence {xn} iteratively in H by (5.21).
Then, {xn} converges strongly to the unique solution x? of the variational inequality problem
VI(G,K).
Remark 5.3.9 Our theorems in this chapter are extensions of the results of Yamada [99],
Wang [91], Xu and Kim [98], Zeng and Yao [100] from real Hilbert spaces to Lp spaces,
1 < p < ?. Moreover, in this our general setting, the iteration parameter is required to
satisfy only conditions C1 and C2.
72
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