Characteristic Inequalities in Banach Spaces and Applications

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.