**TABLE OF CONTENTS**

1 Preliminaries 8

1.1 Linear maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.1.1 A basic result concerning linear maps . . . . . . . . . . . 8

1.1.2 Bounded Linear Maps . . . . . . . . . . . . . . . . . . . . 9

1.2 Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.4 Dierential Calculus in Banach spaces . . . . . . . . . . . . . . . 13

1.5 Convex sets and convex functions . . . . . . . . . . . . . . . . . . 14

1.5.1 Notation and Further denitions . . . . . . . . . . . . . . 17

1.6 Lower Semi-Continuous Functions . . . . . . . . . . . . . . . . . 18

1.7 Existence Result . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.8 Optimality condition: . . . . . . . . . . . . . . . . . . . . . . . . 21

1.9 Optimization with equality constrains . . . . . . . . . . . . . . . 23

2 Pontryagin minimum method principle 26

2.0.1 Towards the principle of pontryagin . . . . . . . . . . . . 28

3 Minimum Principle of Pontryagin: Linear Quadratic Case 30

3.1 Existence and uniqueness of the optimal control . . . . . . . . . . 31

3.2 characterization of the optimal control . . . . . . . . . . . . . . . 35

3.2.1 Riccati equation . . . . . . . . . . . . . . . . . . . . . . . 39

3.3 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

2

**CHAPTER ONE**

Preliminaries

1.1 Linear maps

In this part we dene linear map and present some basic results concerning

them.

Denition :Let X and linear spaces over a scalar eld K.A mapping T : X !

Y is said to be a linear map if:

T(x + y) = T(x) + T(y) (1.1)

for arbitrary x; y 2 X and arbitrary scalars ; 2 K.Some authors use the term

linear operator or linear transformation instead of linear map.Condition (1.1) is

equivalent to the following two conditions:

(i)T(x + y) = T(x) + T(y)8x; y 2 X

(ii)T(x) = T(x)8x 2 X and for each scalar, .

1.1.1 A basic result concerning linear maps

We remark rst that since linear functionals are special forms of linear maps,any

result proved for linear map holds for linear functionals.

Proposition 1.1.1 Let X and Y be two linear spaces over a scalar eld, K, and

let T : X ! Y be a linear map.Then

1. T(0) = 0

2. The rang of T, R(T) = fy 2 Y : T(x) = y for some x 2 Xg is a linear

subspace of Y

3. T is one to one if and only if T(x) = 0 implies that x = 0

4. If T is one to one , then T1 exist on R(T) and T1 : R(T) ! X is also

a linear map.

8

Proof. (1) Since T is linear,we have, T(x) = T(x) for each x 2 X and each

scalar . Take = 0 and (1) follows immediately.

(2)We need to show that for y1; y2 2 R(T) and ; scalars, y1 + y2 2 R(T).

Now, y1; y2 2 R(T) implies that there exists x1; x2 2 X such that T(x1) =

y1; T(x2) = y2. Moreover, x1 + x2 2 X(since X is a linear space). Further-

more, by the linearity of T,we have

T(x1 + x2) = T(x1) + T(x2)

Hence y1 + y2 2 R(T), and so R(T) is a linear subspace of Y .

(3)())Assume that T is one to one. Clearly T(x) = 0 ) T(x) = T(0) since T

is linear (and so T(0) = 0). But T is one-to-one.So,x = 0.

(()Assume that whenever T(u) = 0 then u must be 0. We want to prove that

T is one-to-one.So,let T(x) = T(y). Then T(x) T(y) = 0 and by the linearity

of T; T(x y) = 0.By hypothesis, x y = 0 which implies x = y. Hence T is

one to-one.

(4)Assume that T is one-to-one,since the restriction of T on R(T) is always

onto, then T is bijective from X into R(T).So T1 exists on R(T). For the

linearity, let y1; y2 2 R(T) and a scalar. Then there exists x1; x2 2 X such

that y1 = T(x1); y2 = T(x2). so

T1(y1 + y2) = T1(T(x1) + T(x2)

which is equivalent to:

T1(y1 + y2) = T1(T(x1 + x2)) by using the linearity of T;

which is also equivalent to

T1(y1 + y2) = x1 + x2 = T1(y1) + T1(y2)

Therefore T1 is linear

1.1.2 Bounded Linear Maps

Denition : Let X and Y be normed linear spaces over a scalar eld K, and

let T : X ! Y be a linear map. Then T is said to be bounded if there exists

some constant K 0 such that for each x 2 X,

kT(x)k Kkxk;

the constant K is called a bound for T and in this case, T is called a bounded

linear map. We denote by B(X; Y ), the family of all bounded linear maps from

X into Y .

We now turn our attention to linear maps that are continuous. The notion of

continuity can be state, for linear maps in several useful equivalent forms. We

state these equivalent forms in the following theorem.

Theorem 1.1.1 Let X and Y be normed linear space and let T : X ! Y be a

linear map.Then the following are equivalent:

1. T is continuous;

9

2. T is continuous at the origin(in the sense that if fxng is a sequence of X

such that xn ! 0 as n ! +1,then T(xn) ! 0 as n ! +1)

3. T is Lipschitz, ie , there exists a constant K 0 such that,for each x 2 X,

kT(x)k Kkxk

;

4. If D = fx 2 X : kxk 1g is the closed unit disc in X, then T(D)

is bounded(in the sense that there exists a constant M 0 such that

kT(x)k M for all x 2 D)

1.2 Banach spaces

Denition Let X be a real linear space, and k:kX a norm on X , and dX the

corresponding metric dened by dX(x; y) = kxykX8x; y 2 X. The norm linear

space (X; k:kX) is a real Banach space if the metric space (X; dX) is complete,

ie if any Cauchy sequence of elements of space (X; k:kX) converges in (X; k:kX:)

That is every sequence satisfying the the following Cauchy criterion:

8 > 0; 9n0 2 N : p; q > n0 ) dX(xp; xq) <

Denition Given any vector space V over a vector eld F (where F = R or C),

the topological dual space(or simply) dual space of V is the linear space of all

bounded linear functionals. We shall denote it by V .

Remark 1.2.1 1. The dual space V has a canonical norm dened by

kfk = sup

x2V;kxk6=0

jf(x)j

kxk

; 8f 2 V : (1.2)

2. The dual of every real normed linear space, endowed with its canonical

norm is a Banach space.

In order to dene other useful topologies on dual spaces, we recall the following:

Denition (Initial topology) Let X be an nonempty set, fYigi2I be a family

of topological spaces(where I is an arbitrary index set)and i : X ! Y ; i 2 I

a family of maps.

The smallest topology on X such that the map i; i 2 I are continuous is called

the initial topology.

Next, we dene the weak topology of a normed vector space X and the weak

star topology of its dual space X which are special initial topologies.

Denition (weak topology) Let X be a real normed linear space, and let us

associate to each f 2 X the map f : X ! R given by f (x) = f(x)8x 2 X

The weak topology on X is the smallest topology on X for which all the f are

continuous.

We write !-topology for the weak topology.

10

Denition (weak star topology) Let X be a real normed linear space and X

its dual. Let us associate to each x 2 X the map x : X ! R given by

x(f) = f(x)8f 2 X.

The weak star topology on X is the smallest topology on X for which all the

x are continuous. We write !-topology for the weak star topology

Proposition 1.2.1 Let X be a real normed linear space and X its dual space.

Then, there exists on X three standard topologies, the strong topology given

by the canonical norm k:kx , the weak topology(!-topology) and the weak star

topology ((!-topology) such that:

(X; !) ,! (X; !) ,! (X; k:kX ):

The following part of this section is devoted to reexive spaces. For any normed

real linear space X; the space X of all bounded linear functionals on X is a

Banach space and as a linear space, it has its own corresponding . Let X be

a Banach space, there exists a natural mapping J : X ! X of X into X

dened, for each x 2 X by J(x) = x where x : X ! R is given by

x(f) = hf; xi, for each f 2 X. Thus,

hJ(x); fi = hf; xi for each ; f 2 X

We verify the following properties of J:

(i)J is linear(which is trivial);

(ii)kJ(x)k = kxk for all x 2 X;ie J is an isometry.In fact, for each f 2

X; kJ(x)k = sup

kfk=1;f2X

jhf; xij = kxk.

In general, the map J need not to be onto. Consequently, we always identify X

as a subspace of X. Since an isometry is always injective, it follows that J is

an isomorphism onto J(X) X. The mapping J dened above is called the

canonical map(or canonical embedding) of X into X, and the space X is said

to be embedded in X. This leads to the following denition.

Denition Let X be a norm linear space and let J be the canonical embedding

of X into X. If J is onto, then X is called reexive. Thus, a reexive Banach

space is one in which the canonical embedding is onto.

We now state the following important theorem.

Theorem 1.2.1 (Eberlein-Smul’yan theorem)

A real Banach space X is reexive if and only if every ( norm ) bounded sequence

in X has a subsequence which converges weakly to an element of X.

1.3 Hilbert Spaces

Denition 1. A map : E E ! C is Sesquilinear if

(i)(x + y; z + w) = (x; z) + (x;w) + (y; z) + (y;w)

(ii)(ax; by) = ab(x; y) , where the “bar” indicates the complex conjuga-

tion

11

for all x; y; z;w 2 E and for all a; b 2 C.

2. AHermitian form is a sesquilinear form : E E ! C such that

(x; y) = (y; x)

3. A positive Hermitian form is a Hermitian form such that (x; x) 0

for all x 2 E.

4. A denite Hermitian form is a Hermitian form such that (x; x) =

0 ) x = 0

5. An inner product on E is a positive denite Hermitian form and

will be denoted h:; :i := (:; :). The pair(E; h; i) is called an inner prod-

uct space.

We shall simply write E for the inner product space (E, h; i) when the

inner product h; i) is known.

In the case where we are using more than one inner product spaces, spec-

ication will be made by writing h; h; iE when talking about the inner

product space (E; h; i)

Denition Two vectors x and y in an inner product space E are said to be

orthogonal if hx; yi = 0. For a subset F of E, we have x is orthogonal to F if x

is orthogonal to y for all y in F.

Proposition 1.3.1 Let E be an inner product space andx; y 2 E: Then

khx; yik (hx; xi)

1

2 (hy; yi)

1

2

For an inner product space (E; h; i),the function k:k : E ! R dened by

kxkE =

p

hx; xiE

is a norm on E

Thus, (E; k:kE ) is a normed vector space, hence a metric space endowed with

the distance

dE : E E ! R dened by dE(x; y) = kx ykE.

Denition (Hilbert Space). An inner product space E is called a Hilbert

space if it is complete.

Remark 1.3.1 1. Hilbert spaces are thus a special class of Banach spaces.

2. Every nite dimension inner product space is complete and simply called

Euclidian Space.

Proposition 1.3.2 : Let H be a Hilbert space. Then,

for all u 2 H; Tu(v) := hu; vi denes a bounded linear functional,ie Tu 2 H.

Furthermore kukH = kTukH

Theorem 1.3.1 (Riesz Representation theorem) Let H be a Hilbert space and

let f be a bounded linear functional on H: Then,

12

1. There exists a unique vector y0 2 H such that

f(x) = hx; y0i for each ; x 2 H

2. Moreover, kfk = ky0k.

Remark 1.3.2 The map T : H ! H dened by T(u) = Tu is linear,(anti-

linear in the complex case) and isometric. Therefore the canonical embedding

is an isometry showing that “any Hilbert space is reexive”.

At the end of this part, we state this important proposition which is just a

corollary of Eberlein-Smul’yan theorem.

Proposition 1.3.3 Let H be a Hilbert space, then any bounded sequence in H

has a subsequence which converges weakly to an element of H

1.4 Dierential Calculus in Banach spaces

In this section we dene the derivative of a map dened between real Banach

spaces.

Denition ( Directional Dierentiability) Let f be a function dened from a

real linear space X into a real normed linear space Y and let x0 in X and v in

X/{0}. The function f is said to be dierentiable at x0 in the direction v if the

function t 7! f(x0 + tv) is dierentiable at t = 0. i.e.

t 7!

f(x0 + tv) f(x0)

t

; t 6= 0

has a limit in Y when t tends to 0.This,when it exists is denoted f0(xo; v) or

@f

@v (x0)

Denition ( Gateau Dierentiability) A function f dened from a real linear

space X into a real normed linear space Y is Gateaux Dierentiable at a point

x0 in X if :

1. f is dierentiable at x0 in every direction v

2. there exists a bounded linear map A : X ! Y such that f0(x0; v) =

A(v);in others words,the mapv 7! f0(x0; v) is a bounded linear map

from X into Y

In this case the map f0(x0; 🙂 is called the Gateaux dierential of f at x0

and is denoted by DGf(x0; 🙂 or f0G

(x0).

Denition (Frechet Dierentiability) A map f : U X ! Y whose domain

U is an open set of a real Banach space X and whose range is a real Banach

space Y is (Frechet) dierentiable at x 2 U if there exists a bounded linear map

A : X ! Y such that

lim

kuk!0

k

f(x + u) f(x) Au

kuk

k = 0

or equivalently to

f(x + u) f(x) Au = o(kuk)

13

Proposition 1.4.1 : If f : U X ! Y is Frechet Dierentiable, then f is

Gateaux Dierentiable.

Proof. Indeed by taking u = tv; in the denition of Frechet Dierentiability we

have:f(x + tv) f(x) = A(tv) + o(ktvk). Simplifying by t and using the fact

that A is linear,we have:

f(x + tv) F(x)

t

= (A(v) +

(oktvk)

t

)

by the Frechet Dierentiability of f And since as t ! 0; u ! 0 so we have

lim

t!0

f(x + tv) f(x)

t

= Av

Proposition 1.4.2 Let X be a real Banach space and Y be a real normed linear

space. Then

1. The set of Gateaux dierentiable mappings from X into Y is a linear sub-

space of the linear space of all the mappings dened from X into Y space

is contained in B(X; Y );

2. The set of Frechet Dierentiable mappings from X into Y is also a sub-

space of B(X; Y )

1.5 Convex sets and convex functions

Let us recall rst the following basic notions of optimization.

Denition Let V be a norm linear space, and let F : V ! R [ f+1g.

1. A point u 2 K is a minimizer of F on K if

F(u) F(v)8v 2 K

2. A point u 2 K is a local minimizer of F on K if there exists r > 0 such that

F(u) F(v)8v 2 K \ B(u; r)

3. A point u 2 K is a strict minimizer of F on K if

F(u) < F(v)8v 2 K; v 6= u

4. A point u 2 K is a strict local minimizer of F on K if there exists r > 0 such

that

F(u) < F(v)8v 2 K \ B(u; r); v 6= u

Denition Let X be a norm linear space.A set C X is convex if and only if:

8x; y 2 C; 8 2 (0; 1), we have x + (1 )y 2 C

ie [x; y] C ; 8x; y 2 C

14

Denition Let C be a convex subset of a norm linear space X. A function

f : X ! R [ f+1g is convex on C if 8x; y 2 C; 8 2 (0; 1)

f(x + (1 )y) f(x) + (1 )f(y) (1.3)

If (1.3) is strict for x; y 2 C, with x 6= y and f(x); f(y) nite then f is strictly

convex.

f linear functional implies that f is convex and concave

Theorem 1.5.1 (Slope Inequality) Let I be an interval of R and let f : I !

R [ f+1g convex. Let r1; r2; r3 2 I : r1 < r2 < r3;with f(r1); f(r2) nite then:

f(r1) f(r2)

r2 r1

f(r3) f(r1)

r3 r1

f(r3) f(r2)

r3 r2

Proof. Set =

r2 r1

r3 r1

2 (0; 1). We have 1 =

r3 r2

r3 r1

2 (0; 1) and

r2 = (1 )r1 + r3. Using the convexity of f, it follows that f(r2)

f(r3) + (1 )f(r1) =

r3 r2

r3 r1

f(r1) +

r2 r1

r3 r1

f(r3). Adding f(r1) in both

sides,we get f(r2) f(r1)

r1 r2

r3 r1

f(r1) +

r2 r1

r3 r1

f(r3). Which implies that

f(r2) f(r1)

r2 r1

f(r3) f(r1)

r3 r1

. This prove rst inequality. Replacing f(r1)

by f(r3) in this rst inequality and using the same argument, we get the

second inequality.

Corollary 1.5.1 Let f : I ! R derivable on I, then The following are equiva-

lent:

1. f is convex

2. f0 is increasing

3. f(y) f(x) + f0(x)(y x)8x; y 2 I

If f is twice derivable on I the f is convex if and only if f0(x) 08x 2 I

Proof.

(1) ) (2). Let r < t.By the rst inequality of the slope inequality we have

f(s) f(r)

s r

f(t) f(r)

t r

, for r < s < t. So we have, f0(r) = lim

s&r

f(s) f(r)

s r

f(t) f(r)

t r

. By taking s % t, and by using the second inequality of the slope

inequality, we have

f(t) f(r)

t r

lim

s%t

f(s) f(t)

s t

= f0(t). Therefore we have

f0(r) f0(t)8r < t. Thus f0 is increasing.

(2)) (3)

Let x 2 I, let us dene g(y) = f(y) f(x) f0(x)(y x). We have that

g is dierentiable and g0(y) = f0(y) f0(x). this give us:g0(y) 0 if and

only if f0(y) f0(x), if and only if y x since f0 is increasing. And also

we have,g0(y) 0if and only if y x. So by studding the function g, we

have that x is a minimizer of this function. So we have g(y) g(x)8y. Since

g(x) = 0,this implies that g(y) 0 which implies f(y)f(x)f0(x)(yx) 0,

15

ie f(y) f(x) + f0(x)(y x), ie (3)

(3)) (1)

We have by assumption that f(y) f(x) + f0(x)(y x)8x 2 I, which implies

that f(y) sup

x2I

[f(x) + f0(x)(y x)],which is equivalent to f(y) sup

x2I

hx(y),

where hx(y) = f(x)+f0(x)(yx) = f0(x):y+f(x)f0(x):x. Since for x = y we

have equality on the previous inequality,then we have: f(y) = sup

x2I

hx(y), which

give us that f is convex.

Theorem 1.5.2 Let X a norm linear space, U X; is open, convex, nonempty.

Let f : U ! R dierentiable, then the following are equivalent:

1. f is convex

2. f0 : U ! X is monotone increasing, ie hf0(y)f0(x); yxi 08x; y 2 U

3. f(y) f(x) + hf0(x); y xi8x; y 2 U

If f is twice dierentiable then f is convex if and only if f”(x) is positive

semi-denite ie:

hf0(x):y; yi 08y 2 X; 8x 2 U

Proof.

Let x; y 2 U, dene I = fs 2 R : x + s(y x) 2 Ug

Claim:I is an interval of R such that 0; 1 2 I

Let s1; s2 2 I; t 2 (0; 1), then we have: x + s1(y x) 2 U and x + s2(y x) 2

U. Our aim is to show that x + (ts1 + (1 t)s2)(y x) 2 U to get that

ts1 + (1 t)s2(y x) 2 I, which is equivalent to say that I is an interval. For

that, we have x + (ts1 + (1 t)s2)(y x) = x + ts1(y x) + (1 t)s2(y x)

= tx + (1 t)x + ts1(y x) + (1 t)s2(y x)

= t(x + s1(y x)) + (1 t)(x + s2(y x))

Since x + s1(y x) 2 U and x + s2(y x) 2 U and U is convex, t 2 (0; 1), we

have = t(x+s1(y x))+(1t)(x+s2(y x)) 2 U. Which is equivalent to say

x + (ts1 + (1 t)s2)(y x) 2 U, ie I is an interval.

Assume that (1) is true, we want to show (2), ie f0 is monotone. Let us dene

h : I ! R by h(s) = f(x + s(y x))

We have that:

i)h is derivable on I

ii)h is convex if and only if f is convex on U

f convex implies that h is convex. so according to the previous theorem, we

have h0 is increasing, which give us that h0(1) h0(0)

Buy dierentiating h, we have h0(s) = hf0(x + s(y x)); y xi. So h0(0) =

hf0(x); y xi and h0(1) = hf0(y); y xi. So using our inequality h0(1) h0(0),

we have hf0(y); y xi hf0(x); y xi, which is equivalent to f0 is monotone.

we assume (2), ie a dire f is monotone, we have to prove rst that f0 monotone

implies that h0 is increasing. For that, let us evaluate the dierence (h0(s2)

h0(s1))(s2 s1). We have that

(h0(s2)h0(s1))(s2 s1) = (hf0(x+s2(yx)f0(y+s1(yx); yxi)(s2 s1)

= hf0(x + s2(y x) f0(y + s1(y x); y x; (s2 s1)(y x)i

16

= hf0(z2) f0(z1); z2 z1i

By using the monotony of f0, we have that hf0(z2) f0(z1); z2 z1i 0,which

implies that (h0(s2) h0(s1))(s2 s1) 0, which means that h0 is increasing.

So buy using our previous theorem, we have:h(s) h(r)+h0(r)(sr)8s; r 2 I.

So h(1) h(0)+h0(0), which implies that f(y) f(x)+hf0(x); yxi8x; y 2 U,

ie (3)

we assume (3) and we have to (1), ie f is convex. We have by hypothesis that

) f(y) f(x) + hf0(x); y xi, which implies that

f(y) = sup

x2U

(f(x) + hf0(x); y xi) = sup

x2U

fx(y)

where fx(y) = f(x) + hf0(x); y xi.

Therefore f is convex.

Denition (Domain of a Function) Let F : X ! R [ f+1g be a map. The

domain of F is the set dened by

dom(F) := fx 2 X : F(x) < +1g

Domain of F is sometimes called the eective domain of F. The map F is called

Proper if D(F) 6= ;.Recall that this means there exists at least one x0 2 D(F)

such that F(x0) 2 R or F is not identically +1

1.5.1 Notation and Further denitions

Let X be a real normed space and D X is a convex subset of X. Let

f : D ! R be a convex function. We dene the convex extension of f,

F : X ! R [ f+1g by

F(x) =

f(x); if x 2 D

+1; if x 2 XnD

(1.4)

We observe that f is convex on D if and only if F is convex on X. Moreover

dom(f) = dom(F).

Denition The epigraph of F is the subset of X] R denoted by epi(F) and

dened by

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

Denition Let 2 R. We have the following denitions:

SF; = fx 2 X : F(x) g = fx 2 D(F) : F(x) g

Proposition 1.5.1 F is convex if and only if epi(F) is convex

Proof.

))Assume that F is convex. Let (x1; 1) 2 epi(F); (x2; 2) 2 epi(F); t 2 (0; 1).

Now F(x1) 1; F(x2) 2. Hence F(tx1+(1t)x2) tF (x1)+(1t)F(x2)

t1 + (1 t)2:

17

Thus (tx1 + (1 t)x2; t1 + (1 t)2) 2 epi(F). But

(tx1 + (1 t)x2; t1 + (1 t)2) = t(x1; 1) + (1 t)(x2; 2). Hence, epi(F) is

convex.

(() Assume that epi(F) is convex. We show that F is convex. Let x1; x2 2

D(F); t 2 (0; 1). Then x1; x2 2 D(F) implies that F(x1) 2 R; F(x2) 2 R.

Thus (x1; F(x1)) 2 epi(F) and (x2; F(x2) 2 epi(F). But epi(F) is convex,thus

t(x1; F(x1) + (1 t)(x2; F(x2)) 2 epi(F) if and only if F(tx1 + (1 t)x2)

tF (x1) + (1 t)F(x2) if and only if f is convex.

1.6 Lower Semi-Continuous Functions

Let V be a real norm linear space, let F : V ! R [ f+1g be a map.

Denition : We says that F is lower semi-continuous(lsc)at x0 2 Dom(F) if:

Given; > 0; 9 > 0 : kx x0k < ) F(x0) < F(x):

Denition One says that F is lsc if and only if epi(F) is closed in X R

Proposition 1.6.1 : Let F : V ! R [ f+1g. F is lower semi-continuous if

and only if for all sequence (xn) of V converging to x 2 V , we have F(x)

lim inf F(xn:)

Proof. Assume that F is lsc and xn ! x in V . We take a subsequence (xnk)

of (xn) such that lim

k

F(xnk) = lim inf

n

F(xn). It follows that(xnk; F(xnk) is a

sequence of epi(F) converging to (x; lim inf

n

F(xn)). Since epi(F) is closed, then

(x; lim inf

n

F(xn)) 2 epi(F). Thus

F(x) lim inf

n

F(xn)

Reversely suppose that(xn ! xin V ) ) F(x) lim inf

n

F(xn). Let (an; xn) be

a sequence of epi(F) converging to (x; a) in V R. Then xn ! x and an ! a.

So by hypothesis we have F(x) lim inf

n

F(xn) lim inf an = limn an = a

therefore (x; a) 2 epi(F)

Proposition 1.6.2 : A function F : V ! R [ f+1g is lsc if and only if for

all a 2 R; Sa;F = fx 2 V : F(x) ag is closed.

Proof. Assume that F is lsc. Let a 2 R and let (xn) be a sequence of Sa;F

converges to x 2 V .

Since F(xn) a8n,then lim inf

n

F(xn) a. So according to proposition 1,

F(x) lim inf

n

F(xn) a.

Thus x 2 Ca, which means that Ca is closed.

Reversely assume that Sa;F is closed for each a 2 R. Let xn be a sequence of V

converging to x 2 V . There exists a subsequence xnk of xn such that

lim

k

F(xnk) = lim inf

n

F(xn)

Assume that F(x) > lim inf

n

F(xn). Then there exists a 2 R such that

lim inf

n

F(xn) < a < F(x): (1.5)

18

So there exists N 2 IN such that F(xnk < a for all k N. So xnk 2 Sa;F 8k

N. Since xnk ! x and Sa;F is closed then x 2 Sa;F or equivalent to say that

F(x) a which contradict (2.1) Thus F(x) lim inf

n

F(xn)

Proposition 1.6.3 Let F : V ! R [ f+1g be a l.s.c. function on a compact

set K. Then there exists x 2 K such that:

F(x) = inf

x2K

F(x)

.

Proof. Let fxng be a minimizing sequence of F on K. Since K is compact,there

exists a subsequence (xnk) of (xn) such that xnk ! x 2 K. From Proposition 2

we have:

F(x) lim inf

k

F(xnk) = lim

k

F(xnk) = inf

x2K

F(x):

Therefore F(x) = inf

x2K

F(x).

1.7 Existence Result

A concept that will be needed in the proof of one of our fundamental theorems

of optimization is that of coercivity. We introduce this concept now

Denition : Let X be a topological space. Let K be a subset of X. The set

K is said to be sequentially compact if every sequence in K has a subsequence

which converges to an element of K. The subset K is called Weakly sequentially

compact if every sequence in K has a subsequence which converges weakly to

an element of K.

Denition : Let X be a topological space. A function F : X ! R[f+1g is

called (weakly, respectively) sequentially coercive if the closure of every section

S;F is (weakly, respectively) sequentially compact in X

Denition :let X be a real reexive.A function F : X ! R [ f+1g is called

coercive if:

lim

kxk!+1

F(x) = +1

In fact, we have the following Proposition:

Proposition 1.7.1 : Let X be a real reexive space. Then F : X ! R[f+1g

is weakly sequentially coercive if and only if lim

kxk!+1

F(x) = +1

Proof. ()) Let F be weakly sequentially coercive. We prove that

limkxk!+1F(x) = +1 i.e. , 8M 2 R; 9N0 2 R :if kxk > N0; then F(x) > M.

Assume by contradiction that this is not the case. Then there exists a sequence

fxng such that lim

n!+1

kxnk = +1and fF(xn)g is bounded. Let 2 R be such

that jF(xn)j 8n 1:Since F is weakly sequentially coercive, the section

19

S;F X is weakly sequentially compact. Hence the sequence fxng which is

in the section S;F = fx 2 X : F(x) g, has a subsequence fxnjg which

converges weakly to an element of S;F ; F(xnj) whereas lim kxnjk = +1.

This subsequence is therefore bounded(Since every weakly convergent sequence

is bounded), which is a contradiction.

(()We assume that lim

kxk!+1

F(x) = +1

we want to prove that F is weakly sequentially coercive, ie for arbitrary 2

R; S;F is weakly sequentially compact, or that any sequence in S;F has a

weakly convergent subsequence. So let fxng be an arbitrary sequence in S;F

for arbitrary 2 R.

Then fxng is bounded. Since X is reexive, fxng has a subsequence fxnkg

converges weakly to an element of S;F . Thus F is weakly sequentially coercive.

Remark 1.7.1 If X is reexive and if lim

kxk!+1

F(x) = +1, we simply say that

F is coercive. This condition actually implies that

8A > 0; 9B > 0 : 8x 2 X; kxk > B ) F(x) > A

Equivalently,this means that F(x) A ) kxk B, i.e. ,if the range of F is

bounded, then the domain of F is also bounded.

Theorem 1.7.1 (Existence of minimum in nite dimension)

Let K be a nonempty closed subset of Rn and F a continuous real value appli-

cation on K satisfying the following property:

8(un) K; lim

n!1

kunk = +1 ) lim

n!1

F(un) = +1 (1.6)

Then there exists a least one minimum point of F over K. Moreover for any

minimizing sequence of F over K,one can take a subsequence converging to a

minimum point.

Proof.

Let (un) be a minimizing sequence of F over K. The condition (1:3) implies

that (un) is bounded since (F(un)) is a bounded real sequence. So by Bonzano-

weistrass there exists a subsequence (unk) of(un) converging to a point u of Rn.

But u is in K since K is closed,and F(unk) converges to F(u) by the continuity

of F, thus F(U) = inf

v2K

F(v)

Theorem 1.7.2 (Existence in innite dimension)

Let K be a closed,convex and nonempty subset of a reexive real Banach space

V . Let F : V ! R [ f+1g convex and lsc.

If K is bounded or if F is coercive, then there exists u 2 K such that F(u) =

min

v2K

F(v)

Moreover if F is strictly convex, then u is unique

Proof. Assume that F 6= +1, ie a there exists at least x0 2 V : F(x0) 6= +1.

In the case where F = +1, we obviously have inf

v2K

F(v) = +1

K is nonempty,let (vn) be a minimizing sequence of F on K. K bounded or

20

F coercive implies that (vn) is bounded. Indeed, if K is bounded,then vn is

bounded as a sequence of K, if F is coercive,assume by contradiction that vn is

not bounded,then vn has a subsequence vnk : vnk ! +1 as k ! +1. Since F

is coercive,then lim

k!+1

F(vnk) = +1. But since (vn) is a minimizing sequence of

F, we have that lim

k!+1

F(vnk) = inf

v2K

F(v) which implies that inf

v2K

F(v) = +1.

Therefore we have F(x) = +1 which is a contradiction,thus vn is bounded.

V reexive implies that there exists (vnk) subsequence of vn such that vnk

converges weakly to u in V . K convex and closed implies that K is weakly

closed,which implies that u 2 K So in one hand we have lim

k!+1

F(vnk) =

inf

v2K

F(v),and in a other hand, we have F convex and lsc implies that epi(f) is

convex and closed, which implies that F is weakly lsc.Or vnk converges weakly

to u,so we have F(u) lim inf F(vnk) = inf

v2K

F(v). Thus F(u) = inf

v2K

F(v).

but we have u 2 K,which implies that F(u) = min

v2K

F(v) For the uniqueness,

assume that F is strictly convex. Let u and w two minimum of F over K

Assume by contradiction that u 6= w. K convex implies that 1

2 u + 1

2 w 2 K.

So we have min

v2K

F(v) F( 1

2 u + 1

2 w). Using the strict convexity of F,we have

min

v2K

F(v) < 1

2F(u) + 1

2F( w), which is equivalent that

min

v2K

F(v) < min

v2K

F(v)

which is a contradiction, therefore u = w, ie the minimum is unique

Corollary 1.7.1 Let A be an (n n) positif dened matrix. Then there exists

> 0 such that

hAx; xi kxk28x 2 Rn

Proof. Consider the following problem

8<

:

min J(x) := hAx; xi

kxk = 1

(1.7)

-J is continuous

-dim(Rn) < 1

-Dene K = fx 2 Rn : kxk = 1g is compact

Then by Weistrass theorem there exists x 2 K : x = arg min J(x).

ie 9x 2 K : J(x) J( x

kxk )8x 6= 0. Taking = J(x) > 0,we have

1

kxk2 hAx; xi ,

which implies that hAx; xi kxk2

1.8 Optimality condition:

-Convex case

In this section, we will try to get the necessary condition,and sometimes su-

cient, for minimizer. Our aim is now in one hand more practical than the last

section, since these condition will more often be useful for computing a mini-

mizer.

21

These conditions will be achieved by using the rst derivative(necessary condi-

tions of rst order) or second(necessary conditions of second order)

Theorem 1.8.1 : Let K be a nonempty convex set of norm linear space V . Let

F : V ! R derivable (in the sens of Gateaux).

If u is a local minimum of F over K then

F0(u)(v u) 08v 2 K (1.8)

Moreover if u 2 Int(K) then (1.8) become

F0(u) = 0 EULER EQUATION ;

ie u is a critical point.

Proof.

u minimum local implies that there exists r 0 such that F(u) F(v); 8v 2

K \ B(u; r). Let v 2 K; v 6= u, let t 2]0; 1[, let wt = u + t(v u) =

(1 t)u + tv 2 K. For wt to be in B(u; r), it suces to have kwt uk < r,

ie 0 < t <

r

kv uk

. let 0 < t < minf

r

kv uk

; 1g = , for all t 2]0; [, we

have wt 2 K \ B(u; r), which implies that F(u) F(u + t(v u)); which

gives that

F(u) F(u + t(v u))

t

08t 2]0; [. Letting t goes to 0, we get

F0(u)(v u) 0

Assume u 2 int(K), so 9r > 0;B(u; r) K. we already know that 8u 2

K; F0(u)(v u) 0, let h 6= 0. Consider for all t 2 R; u + th. So for all

tsuch that jtj <

r

krk

,we have u + th 2 K,which implies that for all t such that

jtj <

r

krk

, we have tF 0(u):h 0. this give us tF 0(u):(h) = 0, which implies

that F0(u):h = 08h 6= 0, which means that F0(u) = 0

-General case

Denition : Let K be a nonempty subset of a norm linear space V . Let u 2 K,

let d 2 V; d 6= 0.

One says that d is an admissible direction of K on U if it exists > 0 such

that u + td 2 K, for all t 2]o; [. We denote by Dad(u) the set of admissible

direction on the point u.

Theorem 1.8.2 Consider

MinF (v)

v 2 K 6= ;

(1.9)

If F is dierentiable and if F is a local minimum of F over K, then F0(u):d

08d 2 Dad(u)

In particular if u 2 int(K), then F0(u) = 0

Proof.

u minimum local of F implies that there existsr 0 : F(u) F(v), for all

v 2 K \ B(u; r). Let d 2 Dad(u), it follows that : u + td 2 K; 8t 2]0; [.So for

22

u+td to be on B(u; r),it suces to have 0 < t <

r

kdk

. Now let displaystyle =

minf; r

kdkg, so for all t 2]0; [, we have u + td 2 K \ B(u; r), therefore F(u +

td) F(u); 8t 2]0; [ implies that

F(u + td) F(u)

t

0; 8t 2]0; [, letting t

goes to 0, we have F0(u):d 0

If u 2 int(K),then Dad = V=f0g implies that F0(u):d 0; 8d 2 V=f0g, which

implies that F0(u):d 0; 8d 2 V=f0g, which implies that F0(u) = 0

Theorem 1.8.3 (necessary and sucient condition of second order)

One assume that K is nonempty, F is of class C2.

If u 2 K is a local minimum of F on K then:

1) F0(u):d 08d 2 Dad(U)

2) If d 2 Dad(u) and F0(u):d = 0 then F00(u):d:d 0

3) If F0(u):d = 0 and F00(u):d:d kdk2; 8d 6= 0; > 0, then u is strict local

minimum of F over K

Proof.

(1)Let d 2 Dad(u) : F0(u):d = 0, let > 0 : 8t 2]0; [; u + td 2 K \ B(u; r). We

have by Taylor expansion that:

F(u + td) = F(u) + tF 0(u):d +

t2

2

F00(u):d:d + t2kdk2″(t)

where “(t) ! 0 as t ! 0. This implies that t2[ 1

2F00(u):d:d + kdk2″(t) 0. By

simplifying the last inequality by t2 and bu letting t goes to 0 we have:F0(u):d

0

(2) Let v 2 V; v 6= u,we have also by the Taylor expansion of F that

F(v) F(u) = F0(u)(v u) +

1

2

:(v u)(v u) + kv uk2″(v)

where “(v) ! 0 as v ! 0. This implies that F(v) F(u) kv uk2[

2

+ “(v)]

But

2 > 0 and varepsilon(v) ! 0 as v ! 0 implies that there exists r > 0 :

kv uk < r implies that “(v)j <

2 . This implies that for kv uk < r,we have

2

< “(v) <

2

,which implies that for all v 2 B(u; r),we have

2

+ “(v) > 0,

which give us: for all v 2 B(u; r); v 6= u, we have F(v) > F(u)

Therefore u is a strict local minimum If F is convex, then F0(u):(v u) 0

is a (Necessary and sucient condition)

1.9 Optimization with equality constrains

In this part we are interested with optimization problem with equality constrains

in Rn. Given

an open set of Rn; F and F1; cdots; Fp functions dened on

taking value on R. One consider the following problem:

inf

v2K

F(v) (1.10)

with

fv 2

; gi(v) = 0; 1 i mg (1.11)

23

The function F is called objective or cost.The function gi dene the equality

constrains,the elements of K are called the admissible elements. Obviously the

rst question we have to ask is the one on the existence of solution of (1.10), for

that we use our last results. Indeed they have been obtained in spaces which

are more general, they are applicable in particularly in Rn. It is preferable in

the following to denote in a more synthetic form the constrains. For that, one

pose.

G(x) = (g1(x); ; gm(x))8x 2

Hence the set of the constrains is written K = G1f0g.

Consider the following optimization problem

8>>>><

>>>>:

min F(x)

gi(x) = 0; i = 1; :::p

x 2 RN(p N)

(1.12)

where F; gi : Rn ! R

Set C = fx 2 RN : gi(x) = 0; 8i = 1; ; pg

So our problem (1.4) become:

8<

:

min F(x)

x 2 C

(1.13)

Denition A point x satises the qualication condition (QC) or we can say

x is regular if rgi(x); i = 1; :::p are linearly independant

If p = 1; x is regular if rg1(x) 6= 0

G : RN ! Rp

x 7! G(x) = (rg1(x); ;rgp(x))

C = fx 2 RN=G(x) = 0g

x is regular if and only if JG(x) has rank p

Theorem 1.9.1 (Lagrange)

Assume that x 2 C is regular. If x is an extremum then:

91; ; p 2 R :

8>>><

>>>:

rF(x) +

Xp

i=1

irgi(x) = 0

gi(x) = 0; i = 1; ; p

(1.14)

The real 1; ; p are called the Lagrange multipliers associated with the

constrains of the problem and the minimum local point. The uniqueness of these

reals is assured by the qualication condition(QC)

24

Proposition 1.9.1 Let

be an open convex subset of Rn. Let F :

! R be a

convex dierentiable function on

and gi :

! R be ane. Let x 2 K such

that there exists 1; 2; ; p 2 R such that

rF(x) +

Xp

i=1

irgi(x) = 0: (1.15)

Then, x is a minimizer of F on K.

Proof. Let x 2 K,by convexity of F we have:

F(x) F(x) rF(x)(x x) (1.16)

Using (1.15) it follows that:

F(x) F(x)

Xp

i=1

irgi(x)(x x) (1.17)

Since gi are ane and gi(x) = gi(x) = 0,we have also the identities

gi(x) gi(x) = rgi(x):(x x) = 0

By replacing in (1.17),we obtain F(x) F(x)

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