## TABLE OF CONTENTS

Preliminaries 7

1.1 Geometry of Banach Spaces . . . . . . . . . . . . . . . . . . . 7

1.1.1 Uniformly Convex Spaces . . . . . . . . . . . . . . . . 7

1.1.2 Strictly Convex Spaces . . . . . . . . . . . . . . . . . . 9

1.1.3 Duality Mappings. . . . . . . . . . . . . . . . . . . . 10

1.1.4 Duality maps of Lp Spaces (p > 1) . . . . . . . . . . . 13

1.2 Convex Functions and Subdierentials . . . . . . . . . . . . . 15

1.2.1 Basic notions of Convex Analysis . . . . . . . . . . . . 15

1.2.2 Subdierential of a Convex function . . . . . . . . . . 19

1.2.3 Jordan Von Neumann Theorem for the Existence of

Saddle point . . . . . . . . . . . . . . . . . . . . . . . 20

2 Monotone operators. Maximal monotone operators. 23

2.1 Maximal monotone operators . . . . . . . . . . . . . . . . . . 23

2.1.1 Denitions, Examples and properties of Monotone

Operators . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.1.2 Rockafellar’s Characterization of Maximal Monotone

Operators . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.3 Topological Conditions for Maximal Monotone Oper-

ators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

2.2 The sum of two maximal monotone operators . . . . . . . . . 37

2.2.1 Resolvent and Yosida Approximations of Maximal Mono-

tone Operators . . . . . . . . . . . . . . . . . . . . . . 37

2.2.2 Basic Properties of Yosida Approximations . . . . . . 38

3 On the Characterization of Maximal Monotone Operators 46

3.1 Rockafellar’s characterization of maximal monotone operators. 46

4 Applications 51

4.1 Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2 Uniformly Monotone Operators . . . . . . . . . . . . . . . . . 52

6

## CHAPTER ONE

Preliminaries

The aim of this chapter is to provide some basic results pertaining

to geometric properties of normed linear spaces and convex functions.

Some of these results, which can be easily found in textbooks are given

without proofs or with a sketch of proof only.

1.1 Geometry of Banach Spaces

Throughout this chapter X denotes a real norm space and X denotes

its corresponding dual. We shall denote by the pairing hx; xi the value

of the function x 2 X at x 2 X. The norm in X is denoted by k k,

while the norm in X is denoted by k k. If there is no danger of

confusion we omit the asterisk from the notation kk and denote both

norm in X and X by the symbol k k.

As usual We shall use the symbol ! and * to indicate strong and

weak convergence in X and X respectively. We shall also use w-lim

to indicate the weak-star convergence in X. The space X endowed

with the weak-star topology is denoted by Xw

1.1.1 Uniformly Convex Spaces

Denition 1.1. Let X be a normed linear space. Then X is said to

be uniformly convex if for any ” 2 (0; 2] there exist a = (“) > 0 such

that for each x; y 2 X with kxk 1, kyk 1, and kx ? yk “, we

have k1

2 (x + y)k 1 ? .

Theorem 1.2. Let X be a uniformly convex space. Then for any

d > 0; ” > 0 and x; y 2 X with kxk d, kyk d, and kx ? yk “,

there exist a = ( ”

d ) > 0 such that k1

2 (x + y)k (1 ? )d.

7

Proof. Let x; y 2 X, set k1 = x

d ; k2 = y

d and ” = ”

d . Then obviously we

see that ” > 0. Moreover , kk1k 1; kk2k 1 and kk1 ? k2k ”

d = “.

Now, by the uniform convexity of X, we have for some ( ”

d ) > 0,

1

2

(k1 + k2)

1 ? (“);

that is,

1

2d

(x + y)

1 ? “();

which implies,

1

2

(x + y)

[1 ? (

”

d

)]d:

Hence we have the result.

Proposition 1.3. Let X be a uniformly convex space, let 2 (0; 1)

and ” > 0, then for any d > 0, x; y 2 X such that kxk d, kyk d,

and kx ? yk ” there exist (“) > 0 independent of x and y such that

kx + (1 ? )yk [1 ? 2(“) minf; 1 ? g]d:

Proof. Without loss of generality we shall assume that 2 (0; 1

2 ], we

also observe that

kx+(1?)yk = k(x + y)+(1?2)yk 2k

1

2

(x+y)k+(1?2)kyk

Thus from the uniform convexity of X we have for some (“) > 0

kx + (1 ? )yk 2

1

2

(x + y)

+ (1 ? 2)kyk

2(1 ? (“))d + (1 ? 2)d

= (1 ? 2(“))d

[1 ? 2(“)minf; 1 ? g]d:

Which completes the proof.

8

1.1.2 Strictly Convex Spaces

Denition 1.4. A normed linear space X is said to be strictly convex

if for all x; y 2 X x 6= y, kxk = kyk = 1, we have

kx + (1 ? )yk < 1 for all 2 (0; 1):

Theorem 1.5. Every uniformly convex space is strictly convex.

Proof. Suppose X is uniformly convex, since x 6= y, set ” = kx?yk >

0 and d = 1. Then in view of proposition (1:3) we see that for each

2 (0; 1); kx + (1 ? )yk < 1, which gives the desired result.

We now give some examples to illustrate uniformly and strictly convex

spaces.

Example 1. Every inner product space H is uniformly convex. In

particular Rn with the euclidean norm is uniformly convex.

To see this we shall apply parallelogram law which is valid in any inner

product space. That is for all x; y;2 H, we have

kx + yk2 + kx ? yk2 = 2(kxk2 + kyk2)

Now let ” 2 (0; 2] be given, let x; y 2 H such that kxk 1; kyk 1,

and kx ? yk “, then from the above identity we have

1

2

(x + y)

2

1

4

2(2) ? kx ? yk2

= 1 ?

1

2

(x ? y)

2

1 ?

1

4

“2

So that

1

2

(x + y)

r

1 ?

1

4

“2

To complete the proof we choose =

q

1 ? 1

4″2 > 0.

Example 2. Rn with kk1 is not strictly convex. To see this we choose

the canonical bases e1; e2 in Rn and take = 1

2 . Clearly ke1k = ke2k =

1; e1 6= e2 and

1

2

e1 +

1

2

e2

=

1

2

ke1 + e2k = 1:

Thus we have Rn with k k1 is not strictly convex.

Example 3. The space C[a; b] of all real valued continuous func-

tions on the compact interval [a; b] endowed with the “sup norm” is

9

not strictly convex. To see this we choose two functions such that

f(t) := 1 for all t 2 C[a; b]; g(t) :=

b ? t

b ? a

for all t 2 C[a; b]:

Take ” = 1

2 . Clearly, f; g 2 C[a; b], kfk = kgk = 1 and kf?gk = 1 > “.

But k1

2 (x + y)k = 1. Thus, C[a; b] is not strictly convex.

Theorem 1.6. Let X be a re exive Banach space with norm k k.

Then there exist an equivalent norm k k0 such that X is strictly con-

vex in this norm and X is strictly convex in the dual norm k k0

.

1.1.3 Duality Mappings.

Denition 1.7. Dene a map J : X ?! 2X by

Jx :=

n

x 2 X : hx; xi = kxkkxk; kxk = kxk

o

:

By Hahn Banach theorem for each x 2 X; x 6= 0, there exist y 2 X

such that kyk = 1, and hx; yi = kxk. So if we set x = kxky, then

we see that for each x 2 X 9 x 2 X such that kxk = kxk and

hx; xi = kxk2. So we see that for each x 2 X; Jx 6= ;. The mapping

J : X ?! 2X is called the duality mapping of the space X. In general

J is multivalued.

Remark. More generally, given an increasing continuous function ‘ :

[0; +1) ! [0; +1) such that ‘(0) = 0 and lim

+1

‘ = +1, one denes

the duality map J’ corresponding to the (normalization) function ‘,

by

J’x :=

n

x 2 X : hx; xi = kxkkxk; kxk = ‘(kxk)

o

:

Proposition 1.8. Let H be a real Hilbert space and identify H with

H, then

Jx = fxg for all x 2 H;

i.e The duality map J is the identity map.

Proof. Let a 2 H. Dene

‘a(x) = ha; xi 8x 2 H:

10

Then ‘a 2 H, k’ak = kak and ‘a(a) = kak2: Therefore ‘a 2 J(a)and

since ‘a is identied with a, via Riesz representation theorem, we can

write a 2 J(a). Conversely, if y 2 Ja then ha; yi = kakkyk and

kak = kyk so that

ka?yk2 = ha ? y; a?yi = kak2 +kyk2 ?2ha; yi = 2kak2 ?2kyk2 = 0:

So we have y = a. Therefore Ja = fag.

Proposition 1.9. Let X be a real Banach and J be the duality mapping

on X, then

J(x) = J(x) 8 2 R 8 x 2 X:

Proof. Let y 2 J(x) and 2 R. For = 0 the result follows trivially.

suppose 6= 0, then we have

hx; yi = 2hx; yi = kxkkyk;we also have kxk = kyk:

Thus we have y 2 J(x), which implies J(x) J(x). From the

preceding inclusion we also obtained that 1

J(x) J(x) which implies

J(x) J(x). Therefore J(x) = J(x) 8 2 R, 8x 2 X.

Denition 1.10. Let f : X ?! Y be a map. Then f is said to

be demi-continuous if it is norm to weak-star continuous, i.e f is continuous

from X (endowed with the strong topology) to Y (endowed

with the weak-star topology).

Proposition 1.11. Let X be a real norm space and J be the dual-

ity mapping on X. Then the following are true.

a) For each x 2 X, Jx is a closed, convex subset of B[0; kxk] in X.

Where B[0; kxk] = fy 2 X : kyk kxk:g

b) If X is strictly convex, then for each x 2 X, Jx is single valued.

Moreover the mapping J is demi-continuous, i.e J is continuous as

a mapping from X with the strong topology to X with the weak-star

topology.

c) If X is uniformly convex, then for each x 2 X, Jx is single valued

and the mapping x 7?! Jx is uniformly continuous on bounded subsets

of X.

Proof. (a) Obviously we have Jx B[0; kxk]. Let fyn

gn1 Jx

such that yn

! y, for each n 1 we have hx; yn

i = kxkkyn

k and

kxk = kyn

k. Letting n ! +1 we see that hx; yi = kxkkyk and

11

kxk = kyk. Hence we have y 2 Jx, which implies that Jx is closed.

For convexity, let x; y 2 Jx and 2 (0; 1), then

hx; x + (1 ? )yi = hx; xi + (1 ? )hx; yi

= kxkkxk + (1 ? )kyk = kxk2

We also have from the triangular inequality that kx + (1 ? )yk

kxk, also,

kxk2 = hx; x + (1 ? )yi

kxkkx + (1 ? )yk;

which implies that kxk kx + (1 ? )yk. Hence we have

kxk = kx + (1 ? )yk; which shows that x + (1 ? )y 2 Jx:

(b) Assume X is strictly convex, and suppose that there exit x; y 2

Jx such that x 6= y, then kxk = kyk = kxk and by the strict

convexity of X we have that for any 2 (0; 1), kx + (1 ? )yk <

kxk. In particular taking = 1

2 , we have k1

2 (x + y)k < kxk, which

contradicts the fact that k1

2 (x + y)k = kxk. (Since Jx is convex)

Let fxngn1 X such that xn ! x. using the fact that kJxnk = kxnk,

i.e fJxngn1 is bounded and the fact that the unit ball is w-compact

in X (Banach Alaoglo Theorem) we see that there exist a limit point

y of fJxngn1. Now let fJxnkgk1 X such that w?limJxnk = y,

then we have lim

k!1

hxnk ; Jxnki = hx; yi. We also have that

lim

k!1

hxnk ; Jxnki = lim

k!1

kxnkk2 = kxk2:

So we get hx; yi = kxk2, which implies kxk kyk. To get the

reverse inequality we use the fact that w ? limJxnk = y implies

kyk lim inf kJxnkk = lim inf kxnkk = kxk: Thus we have kxk = kyk

and hx; yi = kxk2. i.e y = Jx. Therefore J is demicontinuous.

(c) Since a uniformly convex space is also strictly, then by part (b)

above we see that J is single valued.

Assume J is not uniformly continuous on bounded subsets of X, then

there exist a constant M > 0, 0 > 0, and subsequences fung; fvng

X such that

kunk M; kvnk M; n 1;

kun ? vnk ! 0 as n ! 1;

kJun ? Jvnk 0; n 1: (1.1)

12

Let > 0 such that kunk ; kvnk ; for n 1: Such exist, for

if there exist a subsequence funkg X such that unk ! 0 as n ! +1,

then we see that vnk ! 0. From the denition of duality map we

obtained that Junk ! 0 and Jvnk ! 0, and this contradicts (1:1).

Now set

xn =

un

kunk

; yn =

vn

kvnk

un; vn 6= 0: Then we have,

kxn ? ynk =

1

kunkkvnk

kunkvnk ? kunkvnk

1

2 (kvnkkun ? vnk + kkvnk ? kunkkkvnk)

2M

2

kun ? vnk ! 0 as n ! +1

We also have 2 kJxn + Jynk hxn; Jxn + Jyni which together with

hxn; Jxn + Jyni = kxnk2 + kynk2 + hxn ? yn; Jyni

= 2 + hxn ? yn; Jyni 2 ? kxn ? ynk

implies

lim

n!1

kJxn + Jynk = 2 i.e lim

n!1

k

1

2

(Jxn + Jyn)k = 1: (1.2)

Now suppose there exist “0 > 0 and a subsequence fxnkg; fynkg X

such that kJxnk ? Jynkk “0, for n 1. Observing that kJxnkk =

kJynkk = 1 and using the uniform convexity of X we see that there

exist (“0) > 0 such that k1

2 (Jxnk+Jynk)k 1?(“0) which contradicts

(1.2). Therefore we have lim

n!1

kJxn ? Jynk = 0, which implies

kJun ? Jvnk = kJ(kunkxn) ? J(kvnkyn)k

= kkunkJxn ? kvnkJynk

kunkkJxn ? Jynk + kvnkkkunk ? kvnkk

MkJxn ? Jynk + kun ? vnk ! 0 as n ! +1:

This contradicts (1.1). Hence we have the result.

1.1.4 Duality maps of Lp Spaces (p > 1)

Proposition 1.12. The duality map on Lp([0; 1]), p > 1 is given by

J(0) = f0g and for f 6= 0

J(f) = ffg

13

where f 2 (Lp) = Lq is dened by

f(g) =

Z 1

0

f(t)g(t)dt 8g 2 Lp

and

f =

jfjp?1signf

kfkp?2

p

Observe that when p 2, this f has the following expression:

f =

fjfjp?2

jjfjjp?2

p

:

Proof. Now set Af = ffg. By denition of the duality map

J(f) = f 2 (Lp) : (f) = kfkkk; kfk = kkg:

Let 2 J(f), since 2 (Lp), then by Riez representation theorem

there exist a unique f 2 Lq, 1

p + 1

q = 1 p; q > 1 such that

(f) = hf; fi =

Z 1

0

f(t)f(t) dt; kk = kfk:

Setting = f , we have

f(f) = hf; fi =

Z 1

0

f(t)f(t) dt; kfk = kfk:

So we have

kfkkfk =

f (f) =

Z 1

0

f(t)f(t) dt; k

fk = kfk = kfk:

which implies

Z 1

0

f(t)f(t) dt = kfk2; kfkq = kfkp: (1.5)

We now show that f(t) := jf(t)jp?1signf(t)

kfkp?2

p

satises (1.5). But we have

Z 1

0

jf(t)jq dt

1

q

=

Z 1

0

jf(t)jq(p?1)

kfk(p?2)q dt

1

q

=

1

kfk(p?2)

Z 1

0

jf(t)jq dt

1

q

=

kfk

p

q

kfkp?2 =

kfkp?1

kfkp?2 = kfkp:

14

Therefore kfkq = kfkp. Also

Z 1

0

f(t)f(t) dt =

Z 1

0

jf(t)jp?1signf(t)

kfkp?2

p

f(t) dt

=

1

kfkp?2

Z 1

0

jf(t)jp dt

=

kfkp

kfkp?2 = kfkpkfkp = kfkpkfkq = kfkkfk:

Thus J(f) Af . On the other hand for arbitrary h 2 Lp([0; 1]),

f(h) =

Z 1

0

f(t)h(t) dt; kfk = kfk:

In particular

f(f) =

Z 1

0

g(t)f(t) dt

=

Z 1

0

f(t)

jf(t)jp?1signf(t)

kfkp?2

p

dt

=

1

kfkp?2

Z 1

0

jf(t)jp dt = kfk2

p:

So we have f(f) = kfkpkfkp and kfkp = kfkq so that kfk = kfkp:

Thus Af J(f). Therefore Af = J(f).

1.2 Convex Functions and Subdierentials

In this section we present the basic properties of convex functions and

subdierentials as we shall use them in the next chapter.

1.2.1 Basic notions of Convex Analysis

Denition 1.13. Let C be a non empty subset of a real norm linear

space X. The set C is said to be convex if for each x; y 2 C and for

each t 2 (0; 1) we have tx + (1 ? t)y 2 C:

Denition 1.14. Let C be a non empty convex subset of X. Then the

convex hull of C denoted by coC is the intersection of all convex sets

containing C. (Equivalently, convex hull of C is the set of all convex

15

combinations of nite subsets of points of C).

Denition 1.15. Let C be a non empty convex subset of X. Let

f : C ?! R [ f+1g. Then f is said to be convex if for each t 2 (0; 1)

and for all x; y 2 C we have

f(tx + (1 ? t)y) tf (x) + (1 ? t)f(y):

Moreover f is said to be proper if f is not identically +1 (i.e 9 x0 2 C

such that f(x0) 2 R)

Denition 1.16. Let f : C ?! R [ f+1g be a map. The eective

domain of f is dened by

D(f) = fx 2 X : f(x) < +1g:

The set

epi(f) = f(x; ) 2 X R : f(x) g

is called the epigraph of f, while

S = fx 2 X : f(x) g

is called the section of f.

Proposition 1.17. A mapping f : X ?! R[f+1g is convex if and

only if its epigraph is convex.

Proposition 1.18. (Slope Inequality) Let I be an interval of R and

f : I ?! R be a convex function. Assume r1 < r2 < r3 with ri 2 I for

i = 1; 2; 3: and f(r1); f(r2) are nite. Then

f(r2) ? f(r1)

r2 ? r1

f(r3) ? f(r1)

r3 ? r1

f(r3) ? f(r2)

r3 ? r2

:

Proposition 1.19. Suppose f : I ?! R is convex and derivable on I.

Then f0 is increasing.

Proof. Let r < t we show that f0(r) f0(t). Now

f0(r) = lim

s!r+

f(s) ? f(r)

s ? r

f(t) ? f(r)

t ? r

lim

s!r?

f(s) ? f(t)

s ? t

= f0(t)

Denition 1.20. Let f : X ?! R[ f+1g be a map. Let x0 2 D(f),

16

then f is lower semicontinuous at x0 if for each ” > 0 there exist > 0

such that f(x0) ? ” < f(x) for all x 2 B(x0; ).

Proposition 1.21. Let f : X ?! R[f+1g be a map. Let x0 2 D(f),

then f is lower semicontinuous at x0 if and only if

lim inf f(xn) f(x0)

for all fxng X such that xn ! x0.

Proposition 1.22. Let f : X ?! R [ f+1g be a map. Then the

following are equivalent.

(a) f is lower semicontinuous,

(b) epi(f) is closed,

(c) S is closed for each 2 R.

Denition 1.23. Let f : X ?! R [ f+1g be a map. Then f is

said to be coercive if

lim

kxk!1

f(x) = +1:

Proposition 1.24. Let f : X ?! R [ f+1g be a map. Then f is

convex and l.s.c if and only if f is convex and weakly l.s.c.

Proof.

f is convex and l.s.c , epi(f) is convex and closed

, epi(f) is convex and weakly closed

, f is convex and weakly l.s.c.

Theorem 1.25. Assume X is re exive. Let f : X ?! R [ f+1g

be proper, convex, coercive and l.s.c function on X. Then f has a

minimum on X. That is there exist x0 2 X such that

f(x0) = inf

x2X

f(x):

Proof. Let = inf

x2X

f(x). Since f is proper we see that < +1.

Let fxng X such that f(xn) ! < +1, then from the coercivity

condition of f we see that fxng is bounded. Since X is re exive, then

there exist x0 2 X and a subsequence fxnkg such that xnk * x0. In

17

view of proposition (1.24) f is weakly lower semi continuous. So we

have

f(x0) lim inf f(xnk) = lim

k!1

f(xnk) = :

Therefore

f(x0) = = inf

x2X

f(x):

Theorem 1.26. Let f be proper, convex, and lower semi-continuous

on X. Then f is bounded from below by an ane function. i.e There

exist x 2 X and a constant c 2 R such that

f(x) hx; xi + c for all x 2 X:

Proof. Let x0 2 D(f) and 2 R such that f(x0) > . This is

possible since f is proper, i.e D(f) 6= ;: Clearly (x0; ) =2 X R, also

in view of proposition (1.22) epi(f) is closed and convex, so by Hahn

Banach theorem there exists a closed hyperplane

H = f(x; ) 2 X R : hx; x

0i + = g

that separates epi(f) and (x0; ) i.e

hx; x

0i + hx0; x

0i +

Considering the left hand side of the inequality only, it is easy to see

that < 0, otherwise we arrived at contradiction. Therefore we have

+ hx;?

x0

i for all (x; ) 2 epi(f):

Since for each x in X (x; f(x)) 2 epi(f), we see that

f(x) hx; xi + c for all x 2 X; where x = ?

x0

and c =

:

Denition 1.27. Let X be Banach space. Let f : X ?! R [ f+1g

be a function. Let x 2 D(f) and v 2 X, then we say that f has a

directional derivative at x in the direction of v 6= 0 if the limit

lim

t!0+

f(x + tv) ? f(x)

t

exist:

We denote by f0(x; v) the directional derivative of f at x in the direction

of v, and we write

f

0

(x; v) = lim

t!0+

f(x + tv) ? f(x)

t

:

18

The function f : X ?! R is said to be G^ateaux dierentiable at x 2 X

if for all v 2 X f0(x; v) exists in R and the function v 7! f0(x; v) is

linear and continuous. We denote by DGf(x) the G^ateaux dierential

of f at x and

hDGf(x); vi := f

0

(x; v) for all v 2 X:

1.2.2 Subdierential of a Convex function

Denition 1.28. Let f : X ?! R [ f+1g be a proper and convex

function. Let x 2 D(f), then the subdierential @f(x) of f at x is the

set

@f(x) = fx 2 X : hy ? x; xi f(y) ? f(x) 8y 2 Xg:

We remarked that if x is not in D(f) then @f(x) = ;:

Proposition 1.29. Let X be proper and convex function which is

G^ateaux dierentiable at x 2 D(f) then

@f(x) = fDGf(x)g:

Proof. To see this we pick y 2 X. Convexity of f implies that

f(x + t(y ? x)) ? f(x)

t

f(y) ? f(x); 0 < t < 1:

Since f is G^ateaux dierentiable at x we obtained that

hy ? x;DGf(x)i = lim

t!0+

f(x + t(y ? x)) ? f(x)

t

f(y) ? f(x):

Thus DGf(x) 2 @f(x).

Conversely, let w 2 @f(x), then for any y 2 X and t > 0

f(x + ty) ? f(x)

t

hy;wi:

Using the G^ateaux dierentiability of f at x we obtained that

hy;DGf(x)i hy;wi 8y 2 X

which implies DGf(x) = w 2 @f(x). Therefore @f(x) = fDGf(x)g:

Example 1. Dene a function f : X ?! R [ f+1g by

f(x) =

1

2

kxk2 8x 2 X:

19

Then f is proper, convex and continuous. Moreover @f(x) = J(x) for

each x 2 X where J is the duality map on X.

Indeed choose rst x 2 Jx. Then for any y 2 x we have

hy ? x; xi = hy; xi ? kxk2 kykkxk ? kxk2

1

2

kyk2 ?

1

2

kxk2

= f(y) ? f(x):

Thus we have x 2 @f(x). Conversely, for x 2 @f(x) we have

hy ? x; xi f(y) ? f(x) 8y 2 X:

So considering x + ty, t 2 (0; 1) we get

hx; yi

1

2t

(kx + tyk2 ? kxk2) kxkkyk +

t

2

kyk:

As t ! 0+ we have hx; yi kxkkyk; which implies kxk kxk: Also

using the fact that x 2 @f(x) and considering x + tx 2 X we have

2th?x; xi kx ? txk2 ? kxk2 = (t2 ? 2t)kxk2; t > 0:

So we have (2 ? t)kxk2 2hx; xi: Now as t ! 0+ we obtained

kxk2 hx; xi kxkkxk which implies kxk kxk:

Therefore we have kxk = kxk and hx; xi = kxk2: Thus x 2 J(x).

Example 2. Let K be a closed, convex subset of X. Dene a map Ik

on X by

Ik(x) =

(

0 for x 2 K,

+1 for x =2 K.

It is easy to see that Ik is convex and lower semi-continuous (since K

is convex and closed). Futhermore for any x 2 K we get

@Ik(x) = fx 2 X : hy ? x; xi 0; 8y 2 Xg:

1.2.3 Jordan Von Neumann Theorem for the Existence of

Saddle point

We now state and prove Jordan von Neumann Theorem for the existence

of saddle point for an upper semi-continuous function dened on

a compact convex subset of a Banach space. But before that we state

20

Kakutani xed point theorem without proof.

Theorem 1.30. (Kakutani) Let K be a nonempty compact convex

subset of a Banach space and let

T : K ?! 2K

be a mapping having a closed graph (i.e., T is upper semicontinuous)

and such that for every x 2 K, T(x) K is nonempty, closed and

convex. Then there exists at least one x 2 K such that x 2 T(x).

Theorem 1.31. (J.Von Neumann) Let X and Y be real Banach

spaces and let U X and V Y be compact convex subsets of X and

Y , respectively. Let H : U V ?! R be a continuous, convex-concave

function (i.e H(u; v) is convex as a function of u and concave as a

function of v). Then there exists (u0; v0) 2 U V such that

H(u0; v) H(u0; v0) H(u; v0) 8 u 2 U and 8 v 2 V:

Such a point (u0; v0) is called the saddle point of the function H.

Proof. Dene the mappings T1 : U ?! V , T2 : V ?! U and

T : U V ?! 2UV

respectively by

T1(u) =

n

v 2 V : H(u; v) H(u;w) 8 w 2 U

o

;

T2(v) =

n

u 2 U : H(u; v) H(w; v) 8 w 2 V

o

;

T(u; v) = T2(v) T1(u):

Since H is continuous we see that the graph of T;

Graph(T) =

n?

(u; v); (x; y)

2 (U V )2 : (x; y) 2 T(u; v)

o

;

is closed. Also for each (u; v) 2 U V , T(u; v) is convex. To see this it

is enough to show that T1(u) and T2(v) are convex. Let u1; u2 2 T2(v)

and 2 (0; 1). Since H is convex as a function of u we see that

H(u1 + (1 ? )u2; v) H(u1; v) + (1 ? )H(u2; v)

H(w; v) + (1 ? )H(w; v)

= H(w; v) 8w 2 V:

21

which implies u1+(1?)u2 2 T2(v). So we have T2(v) is convex. Similarly

we have T1(u) is convex, and hence T(u; v) is convex. Therefore

by theorem (1:30) there exists (u0; v0) 2 U V such that

(u0; v0) 2 T(u0; v0) = T2(v0) T1(u0)

which implies

H(u0; v) H(u0; v0) H(u; v0) 8 u 2 U and 8 v 2 V:

The proof is complete.

22

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