## 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

0<t<1

j’0(t)j ;

that is

kg f(x1) g f(x2)k sup

0<t<1

j’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

0<t<1

kDGf(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

Xn1

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

Xn1

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

0<t<1

j’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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