Contents

**Characteristic Inequalities in Banach Spaces and Applications**

**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 Non-expansive Retraction 23

3.1 Construction of sunny non-expansive retraction in Banach spaces . . . . . . . 23

4 An Application of Sunny Non-expansive Retraction 33

Bibliography 37

**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 hand side is 1. Now

**CHAPTER TWO**

**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 :

**CHAPTER FOUR**

**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.

**Cite this article:**

*Characteristic Inequalities in Banach Spaces and Applications*. Project Topics. (2021). Retrieved September 28, 2021, from https://www.projecttopics.org/characteristic-inequalities-in-banach-spaces-and-applications.html.

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