ABSTRACT

In this thesis, an iterative algorithm for approximating the solutions of a
variational inequality problem for a strongly accretive, L-Lipschitz map and
solutions of a multiple sets split feasibility problem is studied in a uniformly
convex and 2-uniformly smooth real Banach space under the assumption
that the duality map is weakly sequentially continuous. A strong convergence
theorem is proved.

TABLE OF CONTENTS

Acknowledgment i
Certication ii
Approval iii
Abstract v
Dedication vi
1 General Introduction 2
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Statement of the Problem . . . . . . . . . . . . . . . . . . . . 10
1.4 Signicance of the Study . . . . . . . . . . . . . . . . . . . . . 11
1.5 Aim and Objectives . . . . . . . . . . . . . . . . . . . . . . . 11
1.6 Scope and Limitations . . . . . . . . . . . . . . . . . . . . . . 11
1.7 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2 Literature Review 12
2.1 Literature review . . . . . . . . . . . . . . . . . . . . . . . . . 12
3 Strong convergence theorem for solving variational inequal-
ity with multiple set split feasibility problem 16
3.1 introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3.2 Main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
4 Summary and Conclusion 32
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.2 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.4 Recomendation . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1

CHAPTER ONE

General Introduction
In this chapter, we give a brief introduction of the subject matter and denitions
of some basic terms which will be used in our subsequent discussions.
1.1 Introduction
The Multiple sets split feasibility problem is to nd a point contained in the
intersection of a family of closed convex sets in one space so that its image
under a bonded linear transformation is contained in the intersection of a
family of closed convex sets in the image space. It generalizes the convex
feasibility problem and the two sets split feasibility problem. The problem
is formulated as
find x 2
n
i=1
Ci such that A(x) 2
m
t=1
Qt:
where A : X ! Y is a bounded linear operator, Ci X; i = 1; 2; 3; ; n
and Qt Y; t = 1; 2; 3; ;m are nonempty closed convex sets.
When n = m = 1, the problem reduce to the Split feasibiliry problem (SFP)
which is to nd
x 2 C such that A(x) 2 Q:
where C and Q are two nonempty closed convex subsets of X and Y respectively.
In Banach space, the multiple sets split feasibility problem is formulated
as nding an element x 2 X satisfying
x 2
n
i=1
Ci; A(x) 2
m
t=1
Qt:
2
3
where X and Y are two Banach spaces, m; n are two given integers, A :
X ! Y is a bounded linear operator, Ci; i = 1; 2; 3; ; n are closed convex
sets in X, and Qt; t = 1; 2; 3; ;m closed convex sets in Y .
The multiple sets split feasibility problem was rst introduced by Censor
and Elfving [9]. The problem arises in many practical elds such as
signal processing, image reconstruction [11], Intensity modulated radiation
therapy(IMRT)[10] and so on.
1.2 Preliminaries
Denition 1.2.1 A vector space over some eld say F is s nonempty set
E together with two binary operations of addition(+) and scalar multiplica-
tion(.) satisfying the following conditions for any v;w; z 2 E; ; 2 F:
1. v + w = w + v; the commutative law of addition,
2. (v + w) + z = v + (w + z); the associative law for addition,
3. There exists 0 2 E satisfying v + 0 = v; the existence of an additive
identity,
4. 8v 2 E there exists (?v) 2 E such that v+(-v) = 0; the existence of
an additive inverse,
5. (v + w) = v + w;
6. ( + ) v = v + v;
7. ( v) = () v;
8. 1 v = v.
Here, the scalar multiplication v is often written as v: The eld of scalars
will always be assumed to be either R or C and the vector space will be called
real or complex depending on whether the eld is R or C. A vector space is
also called a linear space.
Example 1.2.2 Space Rn. This is the Euclidean space, the underlying set
being the set of all n?tuples of real numbers, written as x = (x1::::; xn), y =
(y1::::; yn), etc., and we now see that this is a real vector space with the two
algebraic operations dened in the usual fashion x+y = (x1+y1; :::; xn+yn)
and ax = (ax1; :::; axn), a 2 R.
Denition 1.2.3 The vectors fx1; x2; x3; g are said to form a basis for
E if they are linearly independent and E = spanfx1; x2; x3; g.
4
Denition 1.2.4 A vector space E is said to be nite dimensional if the
number of vectors in a basis of E is nite.
Note that if E is not nite dimensional, it is said to be innite dimensional.
Example 1.2.5 In analysis, innite dimensional vector spaces are of greater
interest than nite dimensional ones. For instance, C[a; b] and l2 are innite
dimensional, whereas Rn and Ck are nite dimensional for some n; k 2 N.
Denition 1.2.6 A normed space E is a vector space with a norm dened
on it, here a norm on a (real or complex) vector space E is a real-valued
function on E whose value at an x 2 E is denoted by kxk and which satises
the following properties, for x; y 2 E and 2 R
1. kxk 0;
2. kxk = 0 i x = 0;
3. kxk = jjkxk;
4. kx + yk kxk + kyk;
Denition 1.2.7 A sequence fxng in a normed linear space X is
(i) convergent to x 2 X if given > 0, there exists N 2 N such that
kxn ? xk < whenever n N
(ii) said to be Cauchy if given > 0; there exists N 2 N such that
Remark 1.2.8 Every convergent sequence is Cauchy but the converse is not
necessarily true.
Denition 1.2.9 A space X is said to be complete if every Cauchy sequence
in X converges to an element of X.
Denition 1.2.10 A Banach space is a complete normed space (complete
in the metric dened by the norm).
Example 1.2.11 The space lp is a Banach space with norm given by
kxk = (
1X
j=1
jxjp)
1
p
Denition 1.2.12 An inner product space (E; h; i) is a vector space E
together with an inner product h; i : E E ! C such that for all vectors x,
y, z and scalar a we have
1. hx + y; zi = hx; zi + hy; zi;
5
2. hx; yi = hx; yi;
3. hx; yi = hy; xi;
4. hx; xi 0 and hx; xi = 0 i x = 0;
A norm on E can also be dene as
1. kxk2 = hx; xi, 8x 2 E
2. x and y are orthogonal if hx; yi = 0
Inner product space generalizes notion of dot product of nite dimensional
spaces.
Denition 1.2.13 A Hilbert space is a complete inner product space.
In a Banach space E, beside the strong convergence dened by the norm,
i.e., fxng E converges strongly to a if and only if limn!1 kxn ? ak = 0,
we shall consider the weak convergence, corresponding to the weak topology
in E. We say that fxng E converges weakly to a if for any f 2 E
hxn; fi ! ha; fi as n ! 1.
Remark 1.2.14 Any weakly convergent sequence fxng in a Banach space
is bounded.
Denition 1.2.15 Let E be a Banach space. Consider the following map
J : E ! E dened for each x 2 E, by
J(x) = x 2 E
where
x : E ! R
is given by
x(f) = hf; xi; for each f 2 E:
Clearly J is linear, bounded and one-to-one. The mapping J dened above
is called the canonical map(or canonical embedding) of E onto E.
Denition 1.2.16 Let E be a normed linear space and J be the canonical
embedding of E onto E. If J is onto, then E is called re exive.
Proposition 1.2.17 1. In re exive Banach space each bounded sequence
has a weakly convergent subsequence.
2. The spaces Lp and lp, p > 1, are re exive.
6
3. The spaces L1 and l1 are non-re exive.
Denition 1.2.18 A Banach space E is said to be strictly convex if kx+yk
2 <
1 for all x; y 2 U; where U = fz 2 E : kzk = 1g with x 6= y.
Denition 1.2.19 A Banach space E is said to be smooth, if for every 0 6=
x 2 E there exists a unique x 2 E such that kxk = 1 and hx; xi = kxk
i.e., there exists a unique supporting hyperplane for the ball around the origin
with radius kxk at x.
Denition 1.2.20 The modulus of convexity of a normed space E is the
function E : (0; 2] ! [0; 1] dened by
E() = inff1 ? k
1
2
(x + y)k; kxk = kyk = 1; kx ? yk = g:
Denition 1.2.21 The modulus of smoothness of a normed space E is the
fuction E : [0;1) ! [0;1) dened by
E(r) =
1
2
supfkx + yk + kx ? yk ? 2 : kxk = 1; kyk rg:
Denition 1.2.22 A Banach space E is said to be uniformly convex, if for
any 2 (0; 2] there exists a = () > 0; such that for any x; y 2 E with kxk =
kyk = 1 and kx ? yk then kx+y
2 k 1 ? :
Remark 1.2.23 1. Every uniformly convex space is re exive
2. E is uniformly convex i E() > 0:8 2 (0; 2]
Denition 1.2.24 A Banach space E is said to be uniformly smooth, if
lim
r!0
(
E(r)
r
) = 0:
where E(r) is the modulus of smoothness.
Remark 1.2.25 1. E is continuous, convex and nondecreasing with E(0) =
0 and E(r) r
2. The function r 7! E(r)
r is nondecreasing and fullls E(r)
r > 0 for all
r > 0:
Denition 1.2.26 Let q > 1 be a real number. A normed space E is said
to be q-uniformly smooth if there is a constant d > 0 such that
E(r) dq:
When 1 < q 2; E is said to be 2-uniformly smooth.
7
Denition 1.2.27 A mapping A : E1 ! E2 is said to be bounded and linear
if there exists real numbers c; and such that for x; y 2 E1,
kAxk ckxk
and
A(x + y) = Ax + Ay:
Denition 1.2.28 Let E1 and E2 be two re exive, strictly convex and smooth
Banach spaces. The mapping A : E1 ?! E2 is called a bounded linear op-
erator endowered with the operator norm kAk = supkxk1 kAxk. The dual
operator A : E
2 ?! E
1 dened by hAy; xi = hy; Axi8x 2 E1; y 2 E
2 is
called the adjoint operator of A. The adjoint operator A has the property.
kAk = kAk
Denition 1.2.29 A continuous strictly incresing function g : R+ ?! R+
such that g(0) = 0 and limt!1 g(t) = 1 is called a guage function.
Denition 1.2.30 The ganeralized daulity map J : E ?! 2E
with respect
to the guage function is dened by
J(x) = fx 2 E; hx; xi = kxkkxk; kxk = (kxk)g:
For p > 1; if (t) = tp?1; then Jp : E ?! 2E
dened by
Jp(x) = fx 2 E; hx; xi = kxkkxk; kxk = (kxk) = kxkp?1g:
is also called the generalized duality map.
In particular, if p = 2 then
J2x := Jx = ff 2 E : hx; fi = kxk2 = kfk2g
is called the normalized duality mapping
Proposition 1.2.31 The duality map of a Banach space E has the follow-
ing properties;
1. It is homogeneous
2. It is additive i E is a Hilbert space.
3. It is single-valued i E is smooth.
4. It is surjective i E is re exive.
5. It is injective or strictly monotone i E is strictly convex
8
6. It is norm to weak* uniformly continous on bounded subsets of E if E
is smooth
7. If E is Hilbert, J and J?1 are identity.
If E is re exive, strictly convex and smooth, then J is bijective. In this case
the inverse J?1 : E ?! E is given by J?1 = J with J being the duality
mapping of E.
Denition 1.2.32 The duality mapping Jp
E is said to be weakly sequentially
continuous if for each xn ! x weakly, we have Jp
E(xn) ! Jp
E(x) weakly.
Denition 1.2.33 A mapping PC : E ! C is said to be a projection of x
onto C if for all y 2 C there exists a unique element PC(x) 2 C such that
kx ? PC(x)k = miny2C kx ? yk. Moreover, if Jp is the duality mapping of
E, then x0 2 C is the projection of x onto C i
hJp(x0 ? x); y ? x0i 0 8y 2 C:
Denition 1.2.34 A mapping T : C ! C is said to be nonexpansive if
kTx ? Tyk kx ? yk for any x; y 2 C:
Denition 1.2.35 A mapping T : E ! E is said to be accretive if
hTx ? Ty; j(x ? y)i 0 8x; y 2 E:
Denition 1.2.36 A mapping T : E ! E is said to be strongly accretive if
there exists > 0 such that
hTx ? Ty; j(x ? y)i kx ? yk2 8x; y 2 E:
Remark 1.2.37 If E := H a Hilbert space, then a strongly accretive map
T is strongly monotone.
Denition 1.2.38 A mapping T : E ! E is said to be L-Lipschitz contin-
uous on E if there exists L > 0 such that
kTx ? Tyk Lkx ? yk 8 x; y 2 E:
Denition 1.2.39 Let C be a nonempty, closed and convex subset of E.
The problem of nding x 2 C such that
hj(x ? x); Txi 0
is called a variational inequality problem(VI), where T : E ! E is strongly
accretive and L?Lipschitz continuous.
9
Lemma 1.2.40 [8] Let E be q- uniformly smooth Banach space. Then,
there exists a constant dq > 0, such that
kx + ykq kxkq + qhy; jxi + dqkykq:
Lemma 1.2.41 [25] Let E be 2- uniformly smooth Banach space with best
smoothness constant k > 0: Then,
kx + yk2 kxk2 + 2hy; jxi + 2kkyk2:
Lemma 1.2.42 Let C be a nonempty, closed and convex subset of a smooth,
strictly convex and re exive Banach space E and let (x; z) 2 E C. Then ,
z = PCx iff hy ? z; j(x ? z)i 0 8y 2 C:
Lemma 1.2.43 [4] Let E be a normed linear space. Then, the following
inequality hold:
kx + yk2 kxk2 + 2hy; j(x + y)i for x; y 2 E; j(x + y) 2 J(x + y):
Lemma 1.2.44 ([19]see also [3]) Let E be a uniformly convex Banach space
with modulus of convexity () of order q; q 2, then there exists > 0
such that the following inequality hold:
kPCx ? xkq kx ? ykq ? kPCx ? ykq 8 y 2 C:
Lemma 1.2.45 Let E be a uniformly smooth Banach space with best smooth-
ness constant k satisfying 0 < k < p1
2
: Suppose T : E ! E is strongly
accretive and L- Lipschitz continuous on E , 0 < < 1, 0 1 ? and
0 < < 2
L2 : Then,
k(1?)x?Tx?[(1?)y ?Ty]k (1? ? )kx?yk 8 x; y 2 E;
where
= 1 ?
p
1 ? (2 ? L2) 2 (0; 1]:
Proof
By L-Lipschitz contuinity and strongly accretivity of F, we have
k(x ? y) ? (Tx ? Ty)k2 = k(Tx ? Ty) ? (x ? y)k2
2kTx ? Tyk2 ? 2hTx ? Ty; j(x ? y)i
+ 2k2kx ? yk2
2kTx ? Tyk2 ? 2hTx ? Ty; j(x ? y)i
+ kx ? yk2
2L2kx ? yk2 ? 2kx ? yk2 + kx ? yk2
= (1 ? 2 + 2L2)kx ? yk2
10
Thus
k(x ? y) ? (Tx ? Ty)k
p
1 ? 2 + 2L2kx ? yk:j (1.2.1)
Now, using (1.2.1)
k(1 ? )x ? Tx ? [(1 ? )y ? Ty]k
= k(1 ? )(x ? y) ? (Tx ? Ty)k
= k(1 ? )x ? y) ? [(x ? y) ? (Tx ? Ty)]k
(1 ? ? )kx ? y)k + k(x ? y) ? (Tx ? Ty)k
(1 ? ? )kx ? y)k +
p
1 ? 2 + 2L2kx ? yk
= (1 ? ? )kx ? y)k
where
= 1 ?
p
1 ? (2 ? L2):
This completes the proof.
Lemma 1.2.46 [2] Let fang be a sequence of nonnegative real numbers.
Suppose that for any integer m, there exists an integer p such that p m
and ap ap+1. Let n0 be an integer such that an0 an0+1 and dene for
all integer n n0 by
(n) = maxfk 2 N : n0 k n; ak ak+1g:
Then f (n)gnn0 is a nondecreasing sequence satisfying limn!1 (n) = 1
and the following inequalities hold true:
a(n) a(n)+1 an a(n)+1 8n n0:
Lemma 1.2.47 [24] Assume fang is a sequence of nonnegative real num-
bers satisfying the condition
an+1 (1 ? n)an + nn; 8 n n0:
where fng is a sequence in (0; 1) and fng is a sequence in R such that
(4i)
P1
n=0 n = 1;
(ii) lim supn!1 n 0:
Then limn!1 an = 0:
1.3 Statement of the Problem
The split feasiblity problem (SFP) in nite-dimensional Hilbert spaces was
rst introduced by Censor et al.[9] for modeling inverse problems which
arise from phase retrievals and in medical image reconstruction. Since then,
11
various algorithms have been introduced and studied for solving SFP and
MSSFP
Recently, Anh[1] proposed a parallel method for solving the variational
inequality with the MSSFP and proved a strong convergence of the iterative
process in frame work of a Hilbert space. Now the problem is to establish
new convergence theorems that hold in more general Banach space than
Hilbert space.
1.4 Signicance of the Study
When a multiple set split feasibility problem exists in a setting of a Banach
space more general than Hilbert, the result of Anh[1] cannot be used to get
solutions. While our result in this thesis can be applied to such problem to
some extent.
1.5 Aim and Objectives
The aim of this work is to present a new theorem for solution of some nonlinear
operator problem. The aim is achieved through the following objective
To establish some existance results of solutions and the convergence of an
iterative process for solving variational inequalities with multiple sets split
feasibility problem in a uniformly convex and 2-uniformly smooth Banach
spaces
1.6 Scope and Limitations
The theorems considered in this research work hold in a uniformly convex
and 2-uniformly smooth Banach space whose duality mapping is weakly
sequentially continuous. It involves solution of a variational inequality problem.
The work is limited to Banach spaces with weakly sequentially continuous
duality maps which excludes some Banach spaces such as Lp spaces and
sobolev spaces.
1.7 Methodology
Various well known computational techniques, theorems and results in the
literature related to approximations of solutions of the multiple sets split
feasibility problems are used.

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