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**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 kxmk =

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 kxmk = kxnk kxkk 8k 2 K, then 1

kx mk

k

1

2

((xm)+(xn))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) kPMxPMyk2 hxy; PMxPMyi 8x; y 2 H, which implies that kPMxPMyk

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 hxy; xyi 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) = tp1 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) = kxkp1g:

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) = tp1; p >

1. By the theorem above, we get

Jp = J(t)(x) = @

Z kxk

0

(t)dt = @

Z kxk

0

tp1dt = @

1

p

kxkp

;

completing the proof.