TABLE OF CONTENTS

Epigraph ii
Dedication iii
Preface iv
Acknowledgement v
Introduction 1
1 Dierentiability in Banach Spaces 6
1.1 Gâteaux Derivative . . . . . . . . . . . . . . . . . . . 6
1.2 Fréchet Dierentiability . . . . . . . . . . . . . . . . 10
1.3 Second order derivative . . . . . . . . . . . . . . . . . 16
2 Nemytskii Operators 20
2.1 Denition of a Nemytskii Operator . . . . . . . . . . 20
2.2 Carathéodory condition . . . . . . . . . . . . . . . . . 21
2.3 Continuity and Dierentiability of Nemytskii Operator 23
3 Variational Principles and Minimization 28
3.1 Lower Semicontinuous Functions . . . . . . . . . . . . 28
3.2 Ekeland Theorem in Complete metric space . . . . . 31
3.3 PalaisSmale Conditions and Minimization . . . . . . 35
3.4 Deformation Theorem and Palais-Smale Conditon . . 37
3.5 Mountain-Pass Theorem . . . . . . . . . . . . . . . . 41
4 Application: The lane Emden Equation 44
vi
Bibliography 54

 

 

CHAPTER ONE

 

Dierentiability in Banach Spaces
We will dene here two types of dierentiability in Banach spaces
as generalizations of the concept of dierentiability in R.
1.1 Gâteaux Derivative
Let us denote by B(X; Y ) the space of all bounded linear maps from
X to Y where X; Y are Banach spaces.
Recall that a bounded linear map means a continuous linear map.
Denition1.1
Let f : U 7! Y be a mapping and x 2 U ; where U X open. We
say that f is Gâteaux dierentiable at xo if there exists A 2 B(X; Y ),
such that
8 h 2 X n f0g, the map t 7! f(x0+th)?f(x0)
t has a limit as t ! 0
equal to A(h); that is,
lim
t!0
f(x0 + th) ? f(x0)
t
= A(h) (1:2)
or equivalently
f(x0 + th) ? f(x0) = tA(h) + o(t) 8 h 2 X
where o(t) holds for the remainder r(t) = f(x0+th)?f(x0)?tA(h)
6
satisfying
lim
t!0
kr(t)k
t
= 0 :
For simplicity we will write Ah instead of A(h):
Ah is called the Gâteaux derivative of f at x0 in the direction of
h denoted @f
@h (x0).
The bounded linear operator A, depending on x0, is denoted by
DGf(xo) or f0G
(xo) and called the Gâteaux dierential.
Remarks.
– In Denition 1.1, one can simply require that t ! 0+.
– Whenever h 6= 0 and the ratio f(x0+th)?f(x0)
t has a limit in Y as
t ! 0, we say that f is dierentiable in the direction of h at xo,
and we call lim
t!0
f(x0 + th) ? f(x0)
t
the directional dérivative of
f at xo in the direction h.
Example 1.1:
The function f : R2 7! R dened by
f(x; y) = x2 + y2
is Gateaux dierentiable at every point (x0; y0) 2 R2.
Indeed: Let u0 = (x0; y0) and h = (h1; h2). Then
f(u0 + th) ? f(u0) = 2t(x0h1 + y0h2) + t2(h21
+ h22
); 8t 2 R:
It follows that
lim
t!0
f(u0 + th) ? f(u0)
t
= 2(x0h1 + y0h2) = 2hu0; hi
et since the map h 7! 2hu0; hi is linear and continuous from R2
to R, we conclude that f is Gâteaux dierentiable and
DGf(u0)(h) = 2hu0; hi 8 hR2:
Moreover by regarding R2 as a euclidean space, we can derive the
gradient of f at uo as
rf(uo) = 2uo :
which is actually linear and bounded with respect to h as the inner
product( since RN is an inner product space).
7
Theorem 1.1: (Euler necessary condition for extrema)
Let X and Y be real Banach spaces, f : U 7! Y be a mapping and
x 2 U where U X is open. If f is Gâteaux dierentiable at an extremum
point x0 (maximum or minimum point), then DGf(x0) = 0
Proof : Under the hypothesis of this theorem, suppose without loss
of generality that x0 is a minimum point (otherwise consider the
function ?f instead of f).
Since xo 2 U and U is open, there exists a positive real number
r such that the open ball Br(xo) is contained in U. Now let
h 2 X n f0g. Then for every t such that jtj < r=jjhjj, we have,
f(xo + th) f(xo) and so by the Gâteaux dierentiability of f at
xo, we have
DGf(xo)(h) = lim
t!0+
f(u0 + th) ? f(u0)
t
0
It also follows that DGf(xo)(?h) 0, i.e. DGf(xo)(h) 0 by linearity
of DGf(xo). Therefore DGf(xo)(h) = 0 for all h 2 X indeed.
Thus DGf(xo) = 0.
Theorem 1.2: (Mean Value Theorem in Banach Spaces)
Let X and Y be Banach spaces, U X be open and let f : U ! Y
be Gâteaux dierentiable. Then for all x1 x2 2 X, we have
kf(x1) ? f(x2)k sup
t2[0;1]
kDGf(x1 + t(x2 ? x1)k kx1 ? x2k
provided that the sup
t2[0;1]
kDGf(x1 + t(x2 ? x1)k is nite.
Proof. Suppose that the assumptions of Theorem 1.2 hold. Let
g 2 Y (the dual of Y ) such that jjgjj 1. Then the real-valued
function ‘ : [0; 1] ?! R dened by
‘(t) = g f(x1 + th) where h = x2 ? x1
is dierentiable on [0; 1 in the usual sense. Moreover we see that
‘0(t) = g?
DGf(x1 + th)(h)

; 8 t 2 (0; 1) :
It follows from the classical mean valued theorem that
j'(1) ? ‘(0)j sup
0j’0(t)j ;
that is
kg f(x1) ? g f(x2)k sup
0j’0(t)j :
8
Moreover for all t 2 (0; 1), we have
j’0(t)j =

g
?
DGf(x1 + th)(h)

jjgjj kDGf(x1 + th)k khk
kDGf(x1 + th)k khk:
And so
kg?
f(x1)?f(x2)

k = kgof(x1)?gf(x2)kj

sup
0kDGf(x1 + th)k

khk :
But it is well known as a consequence of the Hahn-Banach theorem
that
kyk = supfu(y) ; u 2 Y ; kuk 1 g:
Therefore we nally have
kf(x1) ? f(x2)k sup
t2[0;1]
kDGf(x1 + t(x2 ? x1)k kx1 ? x2k :
Remark. If f satises the assumptions of Theorem 1.2 and has a
continuous Gâteaux dierential, then one can prove the conclusion
of Theorem 1.2 by using the notion of Riemann integration in Banach
spaces following the next lemma.
Lemma. (Cf. ) Let X be a Banach space and ‘ : [a; b] ! X
be continuous, where ?1 < a < b < +1. Then the sequence of
partial sums
b ? a
n
Xn?1
k=0
‘

a + k
b ? a
n

converges as n ! 1;
and its limit is called the Riemann integral of f over [a; b] and is
denoted by Z b
a
‘(t) dt :
That is
Z b
a
‘(t)dt = lim
n!1
b ? a
n
Xn?1
k=0
‘

a + k
b ? a
n

:
It is easily seen that

Z b
a
‘(t) dt

Z b
a
k'(t)k dt:
9
Furthermore if ‘ is continuously (Gâteaux) diérentiable on [a; b] ,
then
‘(b) = ‘(a) +
Z b
a
‘0(t) dt :
Whenener the convergence in (1.2) is uniform for h, there arises
an interesting stronger type of dierentiability called the Fréchet
dierentiability.
1.2 Fréchet Dierentiability
It is a rened notion dierentiabilty of which concept is implicitly
closer to that of the standard notion of dierentiability known in R
Recall: A function f : R ?! R is said to be dierentiable at
x0 2 R if and only if the mapping dened on R n f0g as
h 7!
f(x0 + h) ? f(x0)
h
has a limit a 2 R as h ! 0; that is,
lim
h!0
f(x0 + h) ? f(x0)
h
= a 2 R:
Obeserve that this condition is equivalent to the existence of a real
number a 2 R such that
f(x0 + h) = f(x0) + ah + o(h) :
Now, how can we extend this notion to operators dened between
Banach spaces? The answer is in the following denition.
Denition 1.2:
A function f : U ?! Y ; where Xand Y are Banach spaces and U
open in X, is said to be Fréchet dierentiable at a point x0 2 U, if
there exists a bounded linear map A : X ! Y such that:
lim
khk!0
kf(x0 + h) ? f(x0) ? Ahk
khk
= 0 ; (1:1)
or equivalently
f(x0 + h) ? f(x0) = Ah + o(khk) ; (1:2)
10
where
r(h) := f(x0 + h) ? f(x0) ? Ah = o(h)
in the sense that
lim
khk!0
kr(h)k
khk
= 0 :
Such an operator A is unique and called the Fréchet dierential
of f at x0 and is denoted by Df(x0) or f0(x0) (somtimes it is also
denoted by df(xo)).
The function f is said to be Fréchet dierentiable (or simply differentiable)
on U, if it is Fréchet dierentiable at every point of U:
When there is no ambiguity about the domain of f, we just say that
f is dierentiable.
Denition 1.3:
Let X and Y be Banach spaces, U open in X and let f : U ! Y
be Fréchet dierentiable on U. The Fréchet dierential of f on U is
the mapping
Df : U ! B(X; Y )
x 7! Df(x) :
We say that f is continuously dierentiable on U or a mapping of
class C1 (or simply a C1-mapping) if Df is continuous as a mapping
from U into B(X; Y ).
Examples (Fréchet dierentiable functions).
Let H be a real Hilbert space. Then the function F : H ?! R
dened by
F(x) =
1
2
kxk2
is Fréchet dierentiable on H and its Fréchet dierential is dened
by:
DF(x)(h) = hx; hi = hh; xi:
Thanks to the Riesz representation we can write
rF(x) = x :
11
Indeed let us x xo 2 H arbitrarily. Then for every h 2 H, we have
F(xo + h) ? F(xo) = 1
2 (kxo + hk2 ? kxok2)
= 1
2 (hxo + h; xo + hi ? hxo; xoi)
= 1
2 (hxo; xoi + 2hxo; hi + hh; hi ? hxo; xoi)
= hxo; hi + hh;hi
2
= hh; xoi + jjhjj2
2 :
Now dene the operator A : H ! H by A(h) = hh; xoi. Then
A is linear (since the real inner product is bilinear) and bounded
(according to Cauchy-Schwarz inequality). Moreover it is clear that
lim
khk!0
kF(x0 + h) ? F(x0) ? A(h)k
khk
= lim
khk!0
khk=2 = 0 :
Next we present some properties of the Fréchet dierential.
Proposition 1.1: Let X and Y be Banach spaces and U X
open.
1. If F : U ! Y is Fréchet dierentiable at some point x0 2 U,
then F is continuous at x0.
2. If F : U ! Y is Fréchet dierentiable according to a norm in
X,
then it is also Fréchet dierentiable according to any norm
equivalent to the rst norm.
3. (linéarity)
If F;G : U ! Y are Fréchet dierentiable at some point xo 2
U,
then for any a; b 2 R, aF + bG is Fréchet dierentiable at xo
and
D(aF + bG)(xo) = aDF(xo) + bDG(xo) :
4. (Chain rule).
Let also V be an open set of a Banach space Z and consider
two mappings F : U ?! Y and G : V ?! Z such that
F(U) V . If F is Frechet dierentiable at some point xo 2 U
12
and G : V ?! Z is Frechet dierentiable at yo = F(xo) 2 V ,
then G F is Fréchet dierentiable at xo and
D(G F)(xo) = DG(yo) DF(xo) :
Proof:
1. Suppose that f is Fréchet dierentiable at x0 2 U. Then there
exists a bounded linear map A : X ! Y such that:
f(x0 + h) ? f(x0) = Ah + o(khk) :
It follows from the continuity of A and the denition of o(h)
that
lim
jjhjj!0
jjf(x0 + h) ? f(x0)jj = 0 ;
that is
lim
h!0
?
f(x0 + h) ? f(x0)

= 0 in Y
or simply lim
h!0
f(x0 + h) = f(x0).
2. Let k:k1 and k:k2 be two equivalent norm in X. Then there
exist constant > 0 and > 0 such that
kxk1 kxk2 kxk1 ; 8 x 2 X :
There a mapping g is dened from an open neighbourhood of
0 in (X; k:k1) into Y if and only if is dened from an open
neighbourhood of 0 in (X; k:k2) into Y . Moreover for any h 6= 0
in the domain of g, we have
kg(h)k
khk1

kg(h)k
khk2

kg(h)k
khk1

kg(h)k
khk2
which implies that
lim
khk1!0
kg(h)k
khk1
= 0 () lim
khk2!0
kg(h)k
khk2
= 0 :
3. Let ” > 0. Then by the Fréchet dierentiability of the two
functions F and G at xo 2 U, we get (indeed) the existence of
> 0 such that for every h 2 X satisfying jjhjj < , we have
kF(xo + h) ? F(xo) ? DF(xo)(h)k
”
2(jaj + 1)
khk
13
and
kG(xo + h) ? G(xo) ? DG(xo)(h)k
”
2(jbj + 1)
khk :
Thus we have
k(aF+bG)(xo+h)?(aF+bG)(xo)?aDF(x)(h)?bDG(x)(h)k “jjhjj :
4. We know that
F(xo + h) ? F(xo) = DF(xo)(h) + o(h) (i)
and
G(yo + h) ? G(yo) = DG(yo)(h) + o(h) (ii):
Therefore
G F)(xo + h) ? (G F)(xo) = G(F(xo + h)) ? G(F(xo))
= G
?
F(xo) + DF(xo)(h) + o(khk)

? G(F(xo))
= G(F(xo)) + DG(yo)((DF(xo)(h) + o(h) ? G(F(xo= DG(yo)DF(xo)(h) + DG(yo)(o(h))
= DG(yo)DF(xo)(h) + ~o(h)
which gives the result since DG(yo)DF(xo) is a bounded linear
map and
jj~o(h)jj jjDG(yo)jj jjo(h)jj:
Theorem 1.3:
Every Fréchet dierentiable function is Gâteaux dierentiable and
the dierentials coincide..
Proof : Let f : U ?! Y ; where Xand Y are Banach spaces and U
open in X, be Fréchet dierentiable at a point x0 2 U. We show
that f is Gâteaux dierentiable at xo. Fix any v 2 X n f0g. Then
we have f0(xo) 2 B(X; Y ) and
lim
t!0+

f(xo+tv)?f(xo)
t ? f0(xo)(v)

= lim
t!0+
jjvjj f(xo+tv)?f(xo)?f0(xo)(tv)
jjtvjj
= lim
khk!0
jjvjj f(xo+h)?f(xo)?f0(xo)(h)
jjhjj
= 0 :
Therefore f is Gâteaux dierentiable at xo and moreover DGf(xo) =
f0(xo).
14
Remark: The converse of Theorem 1.2 is false as it can be seen
by the example below.
Example 1.2 :
The function g : R2 ?! R dened by
g(x; y) =
8><
>:

x2y
x4+y2
4
if y 6= 0 ;
0 if y = 0
is Gâteaux dierentiable at (0; 0) but not Fréchet dierentiable at
this point.
Indeed, let h = (h1; h2) 2 R2. Then for t > 0, we have
g(th) ? g(0; 0)
t
=
8<
:
t4h41
h22
(t2h41
+h22
)4 if h2 6= 0 ;
0 if h2 = 0 ;
and so
lim
t!0+
g(th) ? g(0; 0)
t
= 0 ;
yielding the Gâteaux dierentiability of g at (0; 0) with g0(0; 0) 0.
But it is seen that g is not Fréchet dierentiable at (0; 0), according
to Theorem 1.3, by considering the perturbations H = (h1; h21
) as
follows :
lim
h1?!0+
g(h1; h21
) ? g(0; 0)
k(h1; h21
)k
=
1
16
6= 0 :
The next theorem gives a useful sucient condition under which
Gâteaux dierentiability implies Fréchet dierrentiability.
Theorem 1.4:
Let X and Y be Banach spaces, U open and nonempty in X and
let f : U ! Y . If f has a continuous Gâteaux dierential, then f
is Fréchet dierentiable and f 2 C1(U;R).
Proof :
Let x 2 U and choose > 0 such that B(x; ) U. Let h 2 X such
that jjhjj < . Consider the function ‘ : [0; 1] ?! R dened by :
‘(t) = f(x + th) ? f(x) ? tDGf(x)(h)
Since f is Gâteaux dierentiable, it follows that ‘ is dierentiable
and
‘0(t) = DGf(x + th)(h) ? DGf(x)(h) :
15
By applying the Mean Value Theorem to ‘ we have :
j'(1) ? ‘(0)j sup
0j’0(t)j ;
that is,
kf(x + h) ? f(x) ? DGf(x)(h)k sup
t2(0;1)
kDGf(x + th)(h) ? DGF(x)(h)k
sup
t2(0;1)
kDGf(x + th) ? DGF(x)k khk:
By continuity of the mapping Df : U ! B(X; Y ), we have
lim
h!0

sup
t2(0;1)
kDGf(x + th) ? DGf(x)k
!
= 0 ;
and so
f(x + h) ? f(x) ? DGf(x)(h) = o(h)
with Df(x) 2 B(X; Y ).
Denition 1.2:
Let H be a Hilbert space equipped with inner product h:; :i and
f : X ?! R be Fréchet dierentiable. Then the mapping
rf : H ?! H
x 7! rf(x) ;
(where rf(x) is the gradient of f at x) is called a potential operator
with a potential f : H ?! R.
1.3 Second order derivative
Let X and Y be real Banach spaces, U open and nonempty in X,
and let f : U ! Y be dierentiable. If
f0 : U ?! B(X; Y )
is dierentiable, then for every x 2 U,
(f0)0(x) 2 L(X; B(X; Y ))
and is simply denoted by f00(x) or D2f(x). In this case we say that
f is twice dierentiable at x and f00(x) is called the second order
16
dierential of f at x.
Observe that in fact
f00(x) : X X ?! Y
is bilinear and bounded (i.e., continuous) .
We recall that a mapping : X X ?! Y is a bounded bilinear
map if:
1. 8 (x1; x2) 2 X X, 8 y 2 X and 8 ; 2 R,
(x1 + x2; y) = (x1; y) + (x2; y)
(y; x1 + x2) = (y; x1) + (y; x2)
2. 9K 2 (0; 1) such that
k(x1: x2)kY Kkx1kXkx2kX :
The norm of such a bounded bilinear map is given by:
kk = sup fk(x1; x2)kY ; kx1kX 1 and kx2k 1g
Note that, more generally, if E1, E2 and E3 are given three normed
linear spaces, we can dene a bounded linear map from E1E2 into
E3.
The space of bounded bilinear maps from X X into Y is isometric
to B(X; B(X; Y )). Indeed the map
j : B(X2; Y )) ?! B(X; B(X; Y ))
A 7! j(A)
where j(A) is such that for all x 2 X and for all y 2 Y ,
?
j(A)(x)

(y) = A(x; y) :
Moreover jjj(A)jj = jjAjj.
Going back to the setting of the denition of the second order
dierential, if f : U ! Y is twice dierentiable, then f0 : U !
B(X2; Y )). And if f00 is continuous, we say that f is of class C2
and we write f 2 C2(U; Y ).
17
Furthermore we have the following Taylor formula for x 2 U and h
suciently small :
f(x + h) = f(x) + f0(x)(h) +
1
2
f00(x)(h; h) + o
?
khk2
X

(1:5)
that can be established by using a notion of Riemann integration in
Y such as
f(x + h) = f(x) + f0(x)(h) +
Z 1
0
(1 ? t)f00(x + th)(h; h) dt :
These Taylor expansions give the simplest sucient conditions
on a critical a C2 functional to be a local extrema.
Proposition 1.4
Let X and Y be Banach spaces, U open in X and let f : U ! Y
be twice continuously dierentiable. Suppose that xo is a critical
point of f.
1. If there exists a positive real number such that
D2f(xo)(h; h) jjhjj2 ; 8 h 2 X ;
Then xo is a local minimum point of f.
2. If for every x in a neighbourhood of xo, D2f(x) is positive
semidenite (in the sense that
D2f(x)(h; h) 0 ; 8 h 2 X );
then xo is a local minimum point of f over U.
3. If U is convex and for every x 2 U, D2f(x) is positive semidefinite
(in the sense that
D2f(x)(h; h) 0 ; 8 h 2 X );
then xo is a minimum point of f over U. Observe in this case
that f is convex on U.
For instance if H is a real Hilbert space and b 2 H is given then,
the critical point xo of the the functional ‘ dened on H by
‘(x) =
jjxjj2
2
+ hb; xi ;
is the minimum point (i.e., xo = ?b) .
18
Note.
Let H be a Hilbert space and f : H ! R be twice continuously
dierentiable. Then for every x 2 H, there exists (according
to the Riesz Representation Theorem) a bounded linear operator
A : H ! H which is symmetric and satises
D2f(x)(h1; h2) = hAh1; h2i :
I
19

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