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.
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