TABLE OF CONTENTS

Dedication iii
Preface v
Acknowledgement vii
1 Preliminaries 1
1.1 Basic notions of functional analysis . . . . . . . . . . . . . . . . . . . . . . . 1
1.1.1 Differentiability in Banach spaces . . . . . . . . . . . . . . . . . . . . 3
1.1.2 Duality mapping in Banach spaces . . . . . . . . . . . . . . . . . . . 5
1.1.3 The signum function . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.1.4 Convex functions and sub-differentials . . . . . . . . . . . . . . . . . 8
2 Characteristic Inequalities 11
2.1 Uniformly convex spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1.1 Strictly convex spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.1.2 Inequalities in uniformly convex spaces . . . . . . . . . . . . . . . . . 14
2.2 Uniformly smooth spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.2.1 Inequalities in uniformly smooth spaces . . . . . . . . . . . . . . . . . 19
2.2.2 Characterization of uniformly smooth spaces by the duality maps . . 21
3 Sunny Nonexpansive Retraction 23
3.1 Construction of sunny nonexpansive retraction in Banach spaces . . . . . . . 23
4 An Application of Sunny Nonexpansive Retraction 33
Bibliography 37
ix

CHAPTER ONE

Preliminaries
1.1 Basic notions of functional analysis
In this chapter, we recall some definitions and results from linear functional analysis.
Proposition 1.1.1 (The Parallelogram Law) Let X be an inner product space. Then for
arbitrary x; y 2 X,
kx + yk2 + kx ? yk2 = 2
?
kxk2 + kyk2
:
Theorem 1.1.1 (The Riesz Representation Theorem) Let H be a Hilbert space and let f be
a bounded linear functional on H. Then there exists a unique vector y0 2 H such that
f(x) = hx; y0i for each x 2 H and ky0k = kfk:
Theorem 1.1.2 Let X be a reflexive and strictly convex Banach space, K be a nonempty,
closed, and convex subset of X. Then for any fixed x 2 X there exists a unique m 2 K
such that
kx ? mk = inf
k2K
kx ? kk:
Proof. Let x 2 X be fixed, and define Px : X ! R [ f1g by
Px(k) =

kx ? kk; if k 2 K,
1; if k =2 K.
Clearly Px is convex. Indeed, let 2 (0; 1) and k1; k2 2 X. If any of k1 or k2 is not in K,
then Px(k1 + (1 ? )k2) Px(k1) + (1 ? )Px(k2) since the right handside is 1. Now
1
2 CHAPTER 1. PRELIMINARIES
suppose both elements are in K. Then
Px(k1 + (1 ? )k2) = k(k1 ? x) + (1 ? )(k2 ? xk
k(k1 ? x)k + k(1 ? )(k2 ? xk
= Px(k1) + (1 ? )Px(k2):
We next show that Px is lower semicontinuous. By the continuity of the map
k 7! kx ? kk; k 2 K;
we have Px is lower semicontinuous on K. We now show Px is lower semicontinuous on
Kc. Let x0 2 Kc and 2 R such that < Px(x0). Since K is closed, Kc is an open
neighbourhood of x0 and < Px(y) 8y 2 Kc . Hence Px is lower semicontinuous on Kc and
therefore on the whole X. Obviously Px is proper. Next, we show that Px is coercive. Let
y 2 X. Then
Px(y) kyk ? kxk
kyk ?
kyk
2
=
kyk
2
provided kyk 2kxk:
This implies that Px(y) ! 1 as kyk ! 1. Thus, Px is lower semicontinuous, convex,
proper, and coersive. Hence there exists m 2 X such that
Px(m) Px(m) 8m 2 X:
Since Px(y) = 1 for all y 2 Kc and K 6= ;, we must have m 2 K. Furthermore kx?mk =
Px(m) Px(k) = kx ? kk; 8k 2 K. This completes the proof. We now show that m 2 K
is unique. Indeed, if x 2 K then m = x and hence it is unique. Suppose x 2 Kc and m 6= n
such that kx?mk = kx?nk kx?kk 8k 2 K, then 1
kx ? mk
k
1
2
((x?m)+(x?n))k < 1.
This implies that kx ? 1
2 (m + n)k < kx ? mk and this contradict the fact that m is a
minimizing vector in K. Therefore m 2 K is unique.
Corollary 1.1.1 Let X be a uniformly convex Banach space and K be any nonempty, closed
and convex subset of X. Then for arbitrary x 2 X there exists a unique k 2 K such that
kx ? kk = inf
k2K
kx ? kk:
Remark If H is a real Hilbert space and M is any nonempty, closed, and convex subset of
H then in view of the above corollary, then there exists a unique map PM : H ! M defined
by x 7! PMx; where kx ? PMxk = inf
m2M
kx ? mk. This map is called the projection map.
The following properties of projection map PM of H onto M are well known.
(1) z = PMx , hx ? z;m ? zi 0 8m 2 M.
(2) kPMx?PMyk2 hx?y; PMx?PMyi 8x; y 2 H, which implies that kPMx?PMyk
kx ? yk 8x; y 2 H, i.e., PM is nonexpansive.
(3) PM(PMx + t(x ? PMx)) = PMx 8t 0.
1.1. BASIC NOTIONS OF FUNCTIONAL ANALYSIS 3
1.1.1 Differentiability in Banach spaces
Let X and Y be two real normed linear spaces and U be a nonempty open subset of X.
Definition 1.1.1 (Directional Differentiability) Let f : U ?! Y be a map. Let x0 2 U and
v 2 Xnf0g. We say that f has directional derivative at x0 in the direction of v if
lim
t!0
f(x0 + tv) ? f(x0)
t
;
exists in the normed linear space Y . We denote by f0(x0; v) to be the directional derivative
of f at x0 in the direction of v.
Example Let f be the function defined from R2 into R by
f(x1; x2) =
(
x1x22
x21
+x22
; if (x1; x2) 6= (0; 0)
0; otherwise.
Then f has directional derivative at (0; 0) in any direction.
To see this, let v = (v1; v2) 2 R2nf0; 0g, t 6= 0; then
f(0 + tv) ? f(0)
t
=
f(tv)
t
=
v1v2
2
v2
1 + v2
2
:
Thus,
lim
t!0
f(0 + tv) ? f(0)
t
=
v1v2
2
v2
1 + v2
2
= f
0
(0; v):
So f has directional derivative at (0; 0) in any direction.
Example Let f be the function defined from R2 into R by
f(x1; x2) =
x1x2
x21
+x22
; if (x1; x2) 6= (0; 0)
0; otherwise.
Then, this function f has no directional derivative at (0; 0) in any direction.
To see this, let v = (v1; v2) 2 R2nf0; 0g; and t 6= 0; then
f(0 + tv) ? f(0)
t
=
f(tv)
t
=
v1v2
t(v2
1 + v2
2)
and so the limit does not exists in R. Therefore, the directional derivative of the function f
does not exists at (0; 0) in any direction.
Definition 1.1.2 (Gateaux Differentiability) Let f : U ?! Y be a map . Let x0 2 U. The
function f is said to be Gâteaux Differentiable at x0 if :
4 CHAPTER 1. PRELIMINARIES
1. f has directional derivative at x0 in every direction v 2 X n f0g and
2. there exists a bounded linear map A 2 B(X; Y ) (depending on x0) such that f0(x0; v) =
A(v) for all v element of X n f0g.
In this case the map f0(x0; ? is called the Gâteaux differential of f at x0 and is denoted by
DGf(x0) or f0G
(x0).
In other words, f is Gâteaux differentiable at x0 if there exists a bounded linear map A 2
B(X; Y ) such that
lim
t!0
f(x0 + tv) ? f(x0)
t
= A(v); 8v 2 X n f0g
Remark From the above definition, it is obvious that if a function is Gâteaux differentiable
at a point, then it has a directional derivative in all directions at that point.
However, the converse is not true in general. We refer to the above example, it is clear
that the directional derivative of the function f exists at (0,0) in any direction but f0(0; )
is not linear.
Definition 1.1.3 A Banach space X is said to have Gateaux differentiable norm if the limit
lim
t!0
kx + tyk ? kxk
t
exists for each x; y 2 X with kxk = kyk = 1:
Definition 1.1.4 A Banach space X is said to have uniformly Gateaux differentiable norm
if for each y 2 X with kyk = 1; the limit
lim
t!0
kx + tyk ? kxk
t
is attained uniformly in x 2 X with kxk = 1.
Definition 1.1.5 Let (M; ) be a metric space. A mapping T : M ! M is called a contraction
if there exists k 2 [0; 1) such that (Tx; Ty) k(x; y) for all x; y 2 M. If k = 1, then
T is called non-expansive.
Theorem 1.1.3 (Banach Contraction Mapping Principle). Let (M; ) be a complete metric
space and T : M ! M be a contraction. Then T has a unique fixed point, i.e., there exists
a unique x 2 M such that Tx = x.
Definition 1.1.6 A mapping L : l1 ! R is called a Banach limit if
(a) L is linear and continuous.
(b) L(x) 0 if x 0, where x = (xn)n with xn 0 8n:
(c) L(x) = L(x); where denotes the shift operator defined by
(x1; x2; x3; :::) = (x2; x3; x4; :::) i:e: (xn) = (xn+1):
(d) L(1) = 1
1.1. BASIC NOTIONS OF FUNCTIONAL ANALYSIS 5
1.1.2 Duality mapping in Banach spaces
In order to find the analogue of the identities (1) and (2) in Banach spaces, we need to have
a suitable replacement for the inner product. In this section, we give the definition and
properties of duality mapping in an arbitrary normed space.
Definition 1.1.7 A continuous and strictly increasing function : R+ ?! R+ such that
(0) = 0 and lim
t!1
(t) = 1 is called a gauge function.
Definition 1.1.8 Given a gauge function , the mapping J : X ?! 2X defined by
J(x) = fx 2 X : hx; xi = kxkkxk; kxk = (kxk)g
is called the duality map with gauge function where X is any normed space. The normalized
duality map J is a particular case of J where (t) = t; t 2 R+.
Lemma 1.1.1 Let be a gauge function and
‘(t) =
Z t
0
(s)ds;
then ‘ is a convex function on R+.
Proposition 1.1.2 In a normed linear space X, for every gauge function , J(x) is not
empty for any x 2 X.
Proof: If x = 0 then trivially we have J(x) 6= ; by taking x = 0.
For x 6= 0 in X, Hahn-Banach theorem guarantee the existence of f 2 X such that
kfk = 1 and hx; fi = kxk. Take x = (kxk)f then clearly kxk = (kxk) and hx; xi =
(kxk)hx; fi = (kxk)kxk. Hence the proof.
Proposition 1.1.3 In a real Hilbert space H, the normalized duality map is the identity map.
Proof: Since H is a Hilbert space, we identify H = H. Let x 2 H, since hx; xi = kxk2,
then x 2 J(x). If y 2 J(x), then hx; yi = kxkkyk and kxk = kyk.
So that kx ? yk2 = hx ? y; x ? yi = hx; xi ? hy; xi ? hx; yi + hy; yi = 0. Therefore y = x.
Hence the proof.
Proposition 1.1.4 In a Banach space X, let J be a duality map of gauge function . Then
for every x in X, x 6= 0 and every in R, we have
J(x) = sign()
(jjkxk)
(kxk)
J(x):
Proof: Let x 2 X such that x 6= 0 and 2 R. We need to show that
J(x) sign()
(jjkxk)
kxk
J(x) and
6 CHAPTER 1. PRELIMINARIES
sign()
(jjkxk)
(kxk)
J(x) J(x):
Now, let u 2 J(x). Then

x; sign()
(jjkxk)
(kxk)
u

= jj
(jjkxk)
(kxk)
hx; ui
= jjkxk(jjkxk)
= kxk(kxk) and
ksign()(kxk)
(kxk) uk = (kxk). Thus we have, sign()(kxk)
(kxk) J(x) J(x).
To see the other inclusion, we take y = x, = 1
and use the above result, we get
sign()(kyk)
(kyk) J(y) J(y). This implies sign( 1
) (kxk)
(jjkxk)J(x) J(x), ie., J(x)
sign()jjkxk
kxk J(x). Hence, we conclude that J(x) = sign()(jjkxk)
(kxk) J(x).
Corollary 1.1.2 Let X be a real Banach space and J be the normalized duality map on X.
Then J(x) = J(x), 8 2 R, 8x 2 X.
Proof: For the normalized duality map, (t) = t, t 2 R+. It follows from the above proposition
that
J(x) = J(x) = sign() jjkxk
kxk J(x) = J(x) = J(x).
Lemma 1.1.2 Let X be a normed linear space and J be a normalized duality map on X.
Then for any x 2 X
J(x) = fx 2 X : hx; xi = kxk2; kxk kxkg:
Proof: For x = 0, the results holds trivially. For x 6= 0, it is immediate that J(x) fx 2
X : hx; xi = kxk2; kxk kxkg.
Suppose that u 2 X such that hx; ui = kxk2 and kuk kxk. We need to show
kxk kuk.
kuk = sup
kxk6=0
hx;ui
kxk which implies that kxk = hx;ui
kxk kuk. Hence kuk = kxk and so
u 2 J(x).
Proposition 1.1.5 Let X be a real Banach space and J be the duality map on X, then
(a) For every x 2 X, the set J(x) is convex and weak closed in X.
(b) J is monotone in the sense that hx?y; x?yi 0 8x; y 2 X and x 2 J(x); y 2 J(y).
Proof:(a) Let x 2 X; x; y 2 J(x) and 2 (0; 1). We need to show that x +(1?)y 2
J(x).
1.1. BASIC NOTIONS OF FUNCTIONAL ANALYSIS 7
hx; x + (1 ? )yi = hx; xi + (1 ? )hx; yi
= kxk2 + (1 ? )kxk2
= kxk2:
Also,kx +(1?)yk kxk+(1?)kyk = kxk. Therfore by lemma (1.1.2) x +(1?
)y 2 J(x). Hence J(x) is convex.
To show J(x) is weak closed, define for each x 2 X, a map x : X ?! R by x(f) = hx; fi.
Then J(x) = ?1
x (kxk2) B(0; kxk) and so is weak closed in X since x is continuous if
we put the weak topology on X.
(b) Let x; y 2 X and x 2 J(x); y 2 J(y). Then
hx ? y; x ? yi = hx; xi ? hx; yi ? hy; xi + hy; yi
= kxk2 + kyk2 ? hx; yi ? hy; xi
kxk2 ? 2kxkkyk + kyk2
= (kxk + kyk)2 0:
Hence, J is monotone.
1.1.3 The signum function
Definition 1.1.9 The signum function denoted by sgn, is the function sgn : R ! R defined
as
sgn(x) =
8<
:
1 if x > 0
0 if x = 0
?1 if x < 0
We now state and prove the following properties:
(a) For all x 2 R, sgn(?x) = ?sgn(x).
(b) For all x 2 R, jxj = sgn(x)x.
(c) For all x 2 R, x 6= 0 d
dx jxj = sgn(x).
Proof.
(a) Let x 2 R.
Case 1. x > 0. In this case sgn(x) = 1, therefore sgn(?x) = ?1 = ?sgn(x).
Case 2. x < 0. In this case sgn(x) = ?1, therefore sgn(?x) = 1 = ?(?1) = ?sgn(x).
Case 3. x = 0, then sgn(?x) = 0 = ?sgn(x).
(b) Let x 2 R.
If x > 0, then jxj = x and sgn(x) = 1 so that jxj = x = sgn(x)x.
If x < 0, then jxj = ?x and sgn(x) = ?1 so that we obtain jxj = ?x = sgn(x)x.
If x = 0. Then jxj = 0 and sgn(x) = 0, so we have jxj = 0 = sgn(x)x.
8 CHAPTER 1. PRELIMINARIES
(c) Let x 2 R such that x 6= 0.
Case 1. x > 0. In this case jxj = x and sgn(x) = 1. Thus d
dx jxj = 1 = sgn(x).
Case 2. x < 0. In this case jxj = ?x and sgn(x) = ?1. So d
dx jxj = ?1 = sgn(x).
Definition 1.1.10 Let C X be nonempty subset of a Banach space X and D C be
non-empty. A retraction Q from C to D is a mapping Q : C ! D such that Qx = x for
x 2 D. Q is nonexpansive if kQx ? Qyk kx ? yk; x; y 2 C.
Definition 1.1.11 A retraction Q from C to D is sunny if Q satisfies the property:
Q(Qx + t(x ? Qx)) = Qx for x 2 C and t > 0 whenever Qx + t(x ? Qx) 2 C. A retraction
Q from C to D is sunny nonexpansive if Q is both sunny and nonexpansive.
Lemma 1.1.3 [8] Let C be a nonempty closed convex subset of a smooth Banach space E,
D C nonempty, j : E ! E the normalized duality mapping of E, and Q : C ! D a
retraction. Then the following are equivalent:
(i) hx ? Qx; j(y ? Qx)i 0 for all x 2 C and y 2 D;
(ii) Q is both sunny and nonexpansive.
1.1.4 Convex functions and sub-differentials
In this section, we present the basic notion of convex functions and sub-differential of a
convex function.
Definition 1.1.12 Let D be a non-empty subset of a normed linear space X. The set D is
called convex if for each x; y 2 D and for any 2 (0; 1) we have x + (1 ? )y 2 D.
Definition 1.1.13 Let f : X ?! R be a map. Then D(f) = fx 2 X : f(x) < +1g is called
the effective domain of f. The function f is proper if D(f) 6= ; i.e 9×0 2 X : f(x0) 2 R.
Definition 1.1.14 Let D be a non-empty convex subset of X. Let f : D ?! R [ f+1g,
then f is said to be convex if for any 2 (0; 1) and for all x; y 2 D we have
f(x + (1 ? )y) f(x) + (1 ? )f(y):
Definition 1.1.15 A convex function f on a convex domain D X is said to be uniformly
convex on D if there exists a function : R+ ?! R+ with (t) = 0 , t = 0 such that
f(x + (1 ? )y) f(x) + (1 ? )f(y) ? (1 ? )(kx ? yk); 8 2 [0; 1]:
If whenever = 1
2 then f is uniformly convex at centre on D.
Definition 1.1.16 The sub-differential of a convex function f is a map @ : X ?! 2X
defined by
@f(x) = fx 2 X : f(y) ? f(x) hy ? x; xi; 8y 2 Xg:
1.1. BASIC NOTIONS OF FUNCTIONAL ANALYSIS 9
Definition 1.1.17 For p > 1, let (t) = tp?1 be the gauge function. Then we define the
generalized duality map Jp : X ?! 2X by
Jp(x) = fx 2 X : hx; xi = kxkkxk; kxk = (kxk) = kxkp?1g:
Observe that for p = 2, we have Jp = J2 = J which is the normalized duality map defined
in the previous section.
Proposition 1.1.6 For every x 6= 0 in a Banach space X,
@kxk = fu 2 X : hx; ui = kxk = kuk; kuk = 1g:
Proof. We note that
@kxk = fu 2 X : hy ? x; ui kyk ? kxk 8y 2 Xg:
Now, let u 2 X : hx; ui = kxk = kuk; kuk = 1: Then, for arbitrary y 2 X we have
hy ? x; ui = hy; ui ? hx; ui kyk ? kxk:
This implies that u 2 @kxk:
Conversely, if u is in @kxk then
hy; ui = h(y + x) ? x; ui kx + yk ? kxk kyk:
From this we get that kuk 1 and with y = 0 in the definition of @kxk; we have kxk
hx; ui kxkkuk. Therefore kuk = 1; kxk = hx; ui and the result holds.
Lemma 1.1.4 J(x) = @ (kxk) for each x in a Banach space X, where (kxk) =
R kxk
0 (s)ds.
Theorem 1.1.4 For p > 1, JP is the sub-differential of the functional 1
p
kxkp.
Proof From the definition of Jp, we note that the gauge function is given by (t) = tp?1; p >
1. By the theorem above, we get
Jp = J(t)(x) = @
Z kxk
0
(t)dt = @
Z kxk
0
tp?1dt = @

1
p
kxkp

;
completing the proof.

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