Minimum principle of pontryagin

Project Abstract

The Minimum Principle of Pontryagin is a fundamental concept in the field of optimal control theory. Developed by Russian mathematician Lev Pontryagin in the mid-20th century, this principle provides a powerful framework for determining the optimal control strategies for dynamical systems. The main idea behind the Minimum Principle is to find the control functions that minimize a given cost functional while satisfying the dynamic constraints of the system. The Pontryagin Minimum Principle states that for a system described by a set of differential equations and subject to control constraints, the optimal control trajectory must satisfy a certain set of conditions. These conditions involve the Hamiltonian function, which is a key component in the formulation of the principle. The Hamiltonian represents the total cost associated with the system and the control inputs and plays a crucial role in the optimization process. By formulating the Hamiltonian and deriving the necessary conditions for optimality, the Minimum Principle of Pontryagin provides a systematic approach to solving optimal control problems. The principle allows researchers and engineers to analyze complex systems and determine the control strategies that minimize a given performance criterion. This is particularly useful in fields such as aerospace engineering, robotics, economics, and biology, where optimizing control actions is essential for achieving desired outcomes. One of the key strengths of the Minimum Principle of Pontryagin is its generality and applicability to a wide range of dynamical systems. Whether the system is linear or nonlinear, deterministic or stochastic, the principle can be used to derive optimal control laws that lead to the best possible system performance. This versatility has made the Pontryagin Minimum Principle a cornerstone of modern control theory and a valuable tool for researchers and practitioners alike. In conclusion, the Minimum Principle of Pontryagin is a powerful tool for solving optimal control problems in dynamical systems. By formulating the Hamiltonian and deriving the necessary conditions for optimality, this principle provides a systematic approach to determining the control strategies that minimize a given cost functional. Its generality and applicability make it a valuable tool for researchers and engineers working in a wide range of fields.

Project Overview

<p> </p><p>Preliminaries<br>1.1 Linear maps<br>In this part we dene linear map and present some basic results concerning<br>them.<br>Denition :Let X and linear spaces over a scalar eld K.A mapping T : X ô€€€!<br>Y is said to be a linear map if:<br>T(x + y) = T(x) + T(y) (1.1)<br>for arbitrary x; y 2 X and arbitrary scalars ; 2 K.Some authors use the term<br>linear operator or linear transformation instead of linear map.Condition (1.1) is<br>equivalent to the following two conditions:<br>(i)T(x + y) = T(x) + T(y)8x; y 2 X<br>(ii)T(x) = T(x)8x 2 X and for each scalar, .<br>1.1.1 A basic result concerning linear maps<br>We remark rst that since linear functionals are special forms of linear maps,any<br>result proved for linear map holds for linear functionals.<br>Proposition 1.1.1 Let X and Y be two linear spaces over a scalar eld, K, and<br>let T : X ô€€€! Y be a linear map.Then<br>1. T(0) = 0<br>2. The rang of T, R(T) = fy 2 Y : T(x) = y for some x 2 Xg is a linear<br>subspace of Y<br>3. T is one to one if and only if T(x) = 0 implies that x = 0<br>4. If T is one to one , then Tô€€€1 exist on R(T) and Tô€€€1 : R(T) ô€€€! X is also<br>a linear map.<br>8<br>Proof. (1) Since T is linear,we have, T(x) = T(x) for each x 2 X and each<br>scalar . Take = 0 and (1) follows immediately.<br>(2)We need to show that for y1; y2 2 R(T) and ; scalars, y1 + y2 2 R(T).<br>Now, y1; y2 2 R(T) implies that there exists x1; x2 2 X such that T(x1) =<br>y1; T(x2) = y2. Moreover, x1 + x2 2 X(since X is a linear space). Further-<br>more, by the linearity of T,we have<br>T(x1 + x2) = T(x1) + T(x2)<br>Hence y1 + y2 2 R(T), and so R(T) is a linear subspace of Y .<br>(3)())Assume that T is one to one. Clearly T(x) = 0 ) T(x) = T(0) since T<br>is linear (and so T(0) = 0). But T is one-to-one.So,x = 0.<br>(()Assume that whenever T(u) = 0 then u must be 0. We want to prove that<br>T is one-to-one.So,let T(x) = T(y). Then T(x) ô€€€ T(y) = 0 and by the linearity<br>of T; T(x ô€€€ y) = 0.By hypothesis, x ô€€€ y = 0 which implies x = y. Hence T is<br>one to-one.<br>(4)Assume that T is one-to-one,since the restriction of T on R(T) is always<br>onto, then T is bijective from X into R(T).So Tô€€€1 exists on R(T). For the<br>linearity, let y1; y2 2 R(T) and a scalar. Then there exists x1; x2 2 X such<br>that y1 = T(x1); y2 = T(x2). so<br>Tô€€€1(y1 + y2) = Tô€€€1(T(x1) + T(x2)<br>which is equivalent to:<br>Tô€€€1(y1 + y2) = Tô€€€1(T(x1 + x2)) by using the linearity of T;<br>which is also equivalent to<br>Tô€€€1(y1 + y2) = x1 + x2 = Tô€€€1(y1) + Tô€€€1(y2)<br>Therefore Tô€€€1 is linear<br>1.1.2 Bounded Linear Maps<br>Denition : Let X and Y be normed linear spaces over a scalar eld K, and<br>let T : X ô€€€! Y be a linear map. Then T is said to be bounded if there exists<br>some constant K 0 such that for each x 2 X,<br>kT(x)k Kkxk;<br>the constant K is called a bound for T and in this case, T is called a bounded<br>linear map. We denote by B(X; Y ), the family of all bounded linear maps from<br>X into Y .<br>We now turn our attention to linear maps that are continuous. The notion of<br>continuity can be state, for linear maps in several useful equivalent forms. We<br>state these equivalent forms in the following theorem.<br>Theorem 1.1.1 Let X and Y be normed linear space and let T : X ô€€€! Y be a<br>linear map.Then the following are equivalent:<br>1. T is continuous;<br>9<br>2. T is continuous at the origin(in the sense that if fxng is a sequence of X<br>such that xn ! 0 as n ! +1,then T(xn) ! 0 as n ! +1)<br>3. T is Lipschitz, ie , there exists a constant K 0 such that,for each x 2 X,<br>kT(x)k Kkxk<br>;<br>4. If D = fx 2 X : kxk 1g is the closed unit disc in X, then T(D)<br>is bounded(in the sense that there exists a constant M 0 such that<br>kT(x)k M for all x 2 D)<br>1.2 Banach spaces<br>Denition Let X be a real linear space, and k:kX a norm on X , and dX the<br>corresponding metric dened by dX(x; y) = kxô€€€ykX8x; y 2 X. The norm linear<br>space (X; k:kX) is a real Banach space if the metric space (X; dX) is complete,<br>ie if any Cauchy sequence of elements of space (X; k:kX) converges in (X; k:kX:)<br>That is every sequence satisfying the the following Cauchy criterion:<br>8 &gt; 0; 9n0 2 N : p; q &gt; n0 ) dX(xp; xq) &lt;<br>Denition Given any vector space V over a vector eld F (where F = R or C),<br>the topological dual space(or simply) dual space of V is the linear space of all<br>bounded linear functionals. We shall denote it by V .<br>Remark 1.2.1 1. The dual space V has a canonical norm dened by<br>kfk = sup<br>x2V;kxk6=0<br>jf(x)j<br>kxk<br>; 8f 2 V : (1.2)<br>2. The dual of every real normed linear space, endowed with its canonical<br>norm is a Banach space.<br>In order to dene other useful topologies on dual spaces, we recall the following:<br>Denition (Initial topology) Let X be an nonempty set, fYigi2I be a family<br>of topological spaces(where I is an arbitrary index set)and i : X ô€€€! Y ; i 2 I<br>a family of maps.<br>The smallest topology on X such that the map i; i 2 I are continuous is called<br>the initial topology.<br>Next, we dene the weak topology of a normed vector space X and the weak<br>star topology of its dual space X which are special initial topologies.<br>Denition (weak topology) Let X be a real normed linear space, and let us<br>associate to each f 2 X the map f : X ô€€€! R given by f (x) = f(x)8x 2 X<br>The weak topology on X is the smallest topology on X for which all the f are<br>continuous.<br>We write !-topology for the weak topology.<br>10<br>Denition (weak star topology) Let X be a real normed linear space and X<br>its dual. Let us associate to each x 2 X the map x : X ô€€€! R given by<br>x(f) = f(x)8f 2 X.<br>The weak star topology on X is the smallest topology on X for which all the<br>x are continuous. We write !-topology for the weak star topology<br>Proposition 1.2.1 Let X be a real normed linear space and X its dual space.<br>Then, there exists on X three standard topologies, the strong topology given<br>by the canonical norm k:kx , the weak topology(!-topology) and the weak star<br>topology ((!-topology) such that:<br>(X; !) ,! (X; !) ,! (X; k:kX ):<br>The following part of this section is devoted to reexive spaces. For any normed<br>real linear space X; the space X of all bounded linear functionals on X is a<br>Banach space and as a linear space, it has its own corresponding . Let X be<br>a Banach space, there exists a natural mapping J : X ô€€€! X of X into X<br>dened, for each x 2 X by J(x) = x where x : X ô€€€! R is given by<br>x(f) = hf; xi, for each f 2 X. Thus,<br>hJ(x); fi = hf; xi for each ; f 2 X<br>We verify the following properties of J:<br>(i)J is linear(which is trivial);<br>(ii)kJ(x)k = kxk for all x 2 X;ie J is an isometry.In fact, for each f 2<br>X; kJ(x)k = sup<br>kfk=1;f2X<br>jhf; xij = kxk.<br>In general, the map J need not to be onto. Consequently, we always identify X<br>as a subspace of X. Since an isometry is always injective, it follows that J is<br>an isomorphism onto J(X) X. The mapping J dened above is called the<br>canonical map(or canonical embedding) of X into X, and the space X is said<br>to be embedded in X. This leads to the following denition.<br>Denition Let X be a norm linear space and let J be the canonical embedding<br>of X into X. If J is onto, then X is called reexive. Thus, a reexive Banach<br>space is one in which the canonical embedding is onto.<br>We now state the following important theorem.<br>Theorem 1.2.1 (Eberlein-Smul’yan theorem)<br>A real Banach space X is reexive if and only if every ( norm ) bounded sequence<br>in X has a subsequence which converges weakly to an element of X.<br>1.3 Hilbert Spaces<br>Denition 1. A map : E E ô€€€! C is Sesquilinear if<br>(i)(x + y; z + w) = (x; z) + (x;w) + (y; z) + (y;w)<br>(ii)(ax; by) = ab(x; y) , where the “bar” indicates the complex conjuga-<br>tion<br>11<br>for all x; y; z;w 2 E and for all a; b 2 C.<br>2. AHermitian form is a sesquilinear form : E E ô€€€! C such that<br>(x; y) = (y; x)<br>3. A positive Hermitian form is a Hermitian form such that (x; x) 0<br>for all x 2 E.<br>4. A denite Hermitian form is a Hermitian form such that (x; x) =<br>0 ) x = 0<br>5. An inner product on E is a positive denite Hermitian form and<br>will be denoted h:; :i := (:; :). The pair(E; h; i) is called an inner prod-<br>uct space.<br>We shall simply write E for the inner product space (E, h; i) when the<br>inner product h; i) is known.<br>In the case where we are using more than one inner product spaces, spec-<br>ication will be made by writing h; h; iE when talking about the inner<br>product space (E; h; i)<br>Denition Two vectors x and y in an inner product space E are said to be<br>orthogonal if hx; yi = 0. For a subset F of E, we have x is orthogonal to F if x<br>is orthogonal to y for all y in F.<br>Proposition 1.3.1 Let E be an inner product space andx; y 2 E: Then<br>khx; yik (hx; xi)<br>1<br>2 (hy; yi)<br>1<br>2<br>For an inner product space (E; h; i),the function k:k : E ô€€€! R dened by<br>kxkE =<br>p<br>hx; xiE<br>is a norm on E<br>Thus, (E; k:kE ) is a normed vector space, hence a metric space endowed with<br>the distance<br>dE : E E ô€€€! R dened by dE(x; y) = kx ô€€€ ykE.<br>Denition (Hilbert Space). An inner product space E is called a Hilbert<br>space if it is complete.<br>Remark 1.3.1 1. Hilbert spaces are thus a special class of Banach spaces.<br>2. Every nite dimension inner product space is complete and simply called<br>Euclidian Space.<br>Proposition 1.3.2 : Let H be a Hilbert space. Then,<br>for all u 2 H; Tu(v) := hu; vi denes a bounded linear functional,ie Tu 2 H.<br>Furthermore kukH = kTukH<br>Theorem 1.3.1 (Riesz Representation theorem) Let H be a Hilbert space and<br>let f be a bounded linear functional on H: Then,<br>12<br>1. There exists a unique vector y0 2 H such that<br>f(x) = hx; y0i for each ; x 2 H<br>2. Moreover, kfk = ky0k.<br>Remark 1.3.2 The map T : H ô€€€! H dened by T(u) = Tu is linear,(anti-<br>linear in the complex case) and isometric. Therefore the canonical embedding<br>is an isometry showing that “any Hilbert space is reexive”.<br>At the end of this part, we state this important proposition which is just a<br>corollary of Eberlein-Smul’yan theorem.<br>Proposition 1.3.3 Let H be a Hilbert space, then any bounded sequence in H<br>has a subsequence which converges weakly to an element of H<br>1.4 Dierential Calculus in Banach spaces<br>In this section we dene the derivative of a map dened between real Banach<br>spaces.<br>Denition ( Directional Dierentiability) Let f be a function dened from a<br>real linear space X into a real normed linear space Y and let x0 in X and v in<br>X/{0}. The function f is said to be dierentiable at x0 in the direction v if the<br>function t 7ô€€€! f(x0 + tv) is dierentiable at t = 0. i.e.<br>t 7ô€€€!<br>f(x0 + tv) ô€€€ f(x0)<br>t<br>; t 6= 0<br>has a limit in Y when t tends to 0.This,when it exists is denoted f0(xo; v) or<br>@f<br>@v (x0)<br>Denition ( Gateau Dierentiability) A function f dened from a real linear<br>space X into a real normed linear space Y is Gateaux Dierentiable at a point<br>x0 in X if :<br>1. f is dierentiable at x0 in every direction v<br>2. there exists a bounded linear map A : X ô€€€! Y such that f0(x0; v) =<br>A(v);in others words,the mapv 7ô€€€! f0(x0; v) is a bounded linear map<br>from X into Y<br>In this case the map f0(x0; <img alt="" src="https://s.w.org/images/core/emoji/11/svg/1f642.svg">&nbsp;is called the Gateaux dierential of f at x0<br>and is denoted by DGf(x0; <img alt="" src="https://s.w.org/images/core/emoji/11/svg/1f642.svg">&nbsp;or f0G<br>(x0).<br>Denition (Frechet Dierentiability) A map f : U X ô€€€! Y whose domain<br>U is an open set of a real Banach space X and whose range is a real Banach<br>space Y is (Frechet) dierentiable at x 2 U if there exists a bounded linear map<br>A : X ô€€€! Y such that<br>lim<br>kuk!0<br>k<br>f(x + u) ô€€€ f(x) ô€€€ Au<br>kuk<br>k = 0<br>or equivalently to<br>f(x + u) ô€€€ f(x) ô€€€ Au = o(kuk)<br>13<br>Proposition 1.4.1 : If f : U X ô€€€! Y is Frechet Dierentiable, then f is<br>Gateaux Dierentiable.<br>Proof. Indeed by taking u = tv; in the denition of Frechet Dierentiability we<br>have:f(x + tv) ô€€€ f(x) = A(tv) + o(ktvk). Simplifying by t and using the fact<br>that A is linear,we have:<br>f(x + tv) ô€€€ F(x)<br>t<br>= (A(v) +<br>(oktvk)<br>t<br>)<br>by the Frechet Dierentiability of f And since as t ! 0; u ! 0 so we have<br>lim<br>t!0<br>f(x + tv) ô€€€ f(x)<br>t<br>= Av<br>Proposition 1.4.2 Let X be a real Banach space and Y be a real normed linear<br>space. Then<br>1. The set of Gateaux dierentiable mappings from X into Y is a linear sub-<br>space of the linear space of all the mappings dened from X into Y space<br>is contained in B(X; Y );<br>2. The set of Frechet Dierentiable mappings from X into Y is also a sub-<br>space of B(X; Y )<br>1.5 Convex sets and convex functions<br>Let us recall rst the following basic notions of optimization.<br>Denition Let V be a norm linear space, and let F : V ô€€€! R [ f+1g.<br>1. A point u 2 K is a minimizer of F on K if<br>F(u) F(v)8v 2 K<br>2. A point u 2 K is a local minimizer of F on K if there exists r &gt; 0 such that<br>F(u) F(v)8v 2 K B(u; r)<br>3. A point u 2 K is a strict minimizer of F on K if<br>F(u) &lt; F(v)8v 2 K; v 6= u<br>4. A point u 2 K is a strict local minimizer of F on K if there exists r &gt; 0 such<br>that<br>F(u) &lt; F(v)8v 2 K B(u; r); v 6= u<br>Denition Let X be a norm linear space.A set C X is convex if and only if:<br>8x; y 2 C; 8 2 (0; 1), we have x + (1 ô€€€ )y 2 C<br>ie [x; y] C ; 8x; y 2 C<br>14<br>Denition Let C be a convex subset of a norm linear space X. A function<br>f : X ô€€€! R [ f+1g is convex on C if 8x; y 2 C; 8 2 (0; 1)<br>f(x + (1 ô€€€ )y) f(x) + (1 ô€€€ )f(y) (1.3)<br>If (1.3) is strict for x; y 2 C, with x 6= y and f(x); f(y) nite then f is strictly<br>convex.<br>f linear functional implies that f is convex and concave<br>Theorem 1.5.1 (Slope Inequality) Let I be an interval of R and let f : I ô€€€!<br>R [ f+1g convex. Let r1; r2; r3 2 I : r1 &lt; r2 &lt; r3;with f(r1); f(r2) nite then:<br>f(r1) ô€€€ f(r2)<br>r2 ô€€€ r1</p><p>f(r3) ô€€€ f(r1)<br>r3 ô€€€ r1</p><p>f(r3) ô€€€ f(r2)<br>r3 ô€€€ r2<br>Proof. Set =<br>r2 ô€€€ r1<br>r3 ô€€€ r1<br>2 (0; 1). We have 1 ô€€€ =<br>r3 ô€€€ r2<br>r3 ô€€€ r1<br>2 (0; 1) and<br>r2 = (1 ô€€€ )r1 + r3. Using the convexity of f, it follows that f(r2)<br>f(r3) + (1 ô€€€ )f(r1) =<br>r3 ô€€€ r2<br>r3 ô€€€ r1<br>f(r1) +<br>r2 ô€€€ r1<br>r3 ô€€€ r1<br>f(r3). Adding ô€€€f(r1) in both<br>sides,we get f(r2) ô€€€ f(r1)<br>r1 ô€€€ r2<br>r3 ô€€€ r1<br>f(r1) +<br>r2 ô€€€ r1<br>r3 ô€€€ r1<br>f(r3). Which implies that<br>f(r2) ô€€€ f(r1)<br>r2 ô€€€ r1</p><p>f(r3) ô€€€ f(r1)<br>r3 ô€€€ r1<br>. This prove rst inequality. Replacing ô€€€f(r1)<br>by ô€€€f(r3) in this rst inequality and using the same argument, we get the<br>second inequality.<br>Corollary 1.5.1 Let f : I ô€€€! R derivable on I, then The following are equiva-<br>lent:<br>1. f is convex<br>2. f0 is increasing<br>3. f(y) f(x) + f0(x)(y ô€€€ x)8x; y 2 I<br>If f is twice derivable on I the f is convex if and only if f0(x) 08x 2 I<br>Proof.<br>(1) ) (2). Let r &lt; t.By the rst inequality of the slope inequality we have<br>f(s) ô€€€ f(r)<br>s ô€€€ r</p><p>f(t) ô€€€ f(r)<br>t ô€€€ r<br>, for r &lt; s &lt; t. So we have, f0(r) = lim<br>s&amp;r<br>f(s) ô€€€ f(r)<br>s ô€€€ r</p><p>f(t) ô€€€ f(r)<br>t ô€€€ r<br>. By taking s % t, and by using the second inequality of the slope<br>inequality, we have<br>f(t) ô€€€ f(r)<br>t ô€€€ r<br>lim<br>s%t<br>f(s) ô€€€ f(t)<br>s ô€€€ t<br>= f0(t). Therefore we have<br>f0(r) f0(t)8r &lt; t. Thus f0 is increasing.<br>(2)) (3)<br>Let x 2 I, let us dene g(y) = f(y) ô€€€ f(x) ô€€€ f0(x)(y ô€€€ x). We have that<br>g is dierentiable and g0(y) = f0(y) ô€€€ f0(x). this give us:g0(y) 0 if and<br>only if f0(y) f0(x), if and only if y x since f0 is increasing. And also<br>we have,g0(y) 0if and only if y x. So by studding the function g, we<br>have that x is a minimizer of this function. So we have g(y) g(x)8y. Since<br>g(x) = 0,this implies that g(y) 0 which implies f(y)ô€€€f(x)ô€€€f0(x)(yô€€€x) 0,<br>15<br>ie f(y) f(x) + f0(x)(y ô€€€ x), ie (3)<br>(3)) (1)<br>We have by assumption that f(y) f(x) + f0(x)(y ô€€€ x)8x 2 I, which implies<br>that f(y) sup<br>x2I<br>[f(x) + f0(x)(y ô€€€ x)],which is equivalent to f(y) sup<br>x2I<br>hx(y),<br>where hx(y) = f(x)+f0(x)(yô€€€x) = f0(x):y+f(x)ô€€€f0(x):x. Since for x = y we<br>have equality on the previous inequality,then we have: f(y) = sup<br>x2I<br>hx(y), which<br>give us that f is convex.<br>Theorem 1.5.2 Let X a norm linear space, U X; is open, convex, nonempty.<br>Let f : U ô€€€! R dierentiable, then the following are equivalent:<br>1. f is convex<br>2. f0 : U ô€€€! X is monotone increasing, ie hf0(y)ô€€€f0(x); yô€€€xi 08x; y 2 U<br>3. f(y) f(x) + hf0(x); y ô€€€ xi8x; y 2 U<br>If f is twice dierentiable then f is convex if and only if f”(x) is positive<br>semi-denite ie:<br>hf0(x):y; yi 08y 2 X; 8x 2 U<br>Proof.<br>Let x; y 2 U, dene I = fs 2 R : x + s(y ô€€€ x) 2 Ug<br>Claim:I is an interval of R such that 0; 1 2 I<br>Let s1; s2 2 I; t 2 (0; 1), then we have: x + s1(y ô€€€ x) 2 U and x + s2(y ô€€€ x) 2<br>U. Our aim is to show that x + (ts1 + (1 ô€€€ t)s2)(y ô€€€ x) 2 U to get that<br>ts1 + (1 ô€€€ t)s2(y ô€€€ x) 2 I, which is equivalent to say that I is an interval. For<br>that, we have x + (ts1 + (1 ô€€€ t)s2)(y ô€€€ x) = x + ts1(y ô€€€ x) + (1 ô€€€ t)s2(y ô€€€ x)<br>= tx + (1 ô€€€ t)x + ts1(y ô€€€ x) + (1 ô€€€ t)s2(y ô€€€ x)<br>= t(x + s1(y ô€€€ x)) + (1 ô€€€ t)(x + s2(y ô€€€ x))<br>Since x + s1(y ô€€€ x) 2 U and x + s2(y ô€€€ x) 2 U and U is convex, t 2 (0; 1), we<br>have = t(x+s1(y ô€€€x))+(1ô€€€t)(x+s2(y ô€€€x)) 2 U. Which is equivalent to say<br>x + (ts1 + (1 ô€€€ t)s2)(y ô€€€ x) 2 U, ie I is an interval.<br>Assume that (1) is true, we want to show (2), ie f0 is monotone. Let us dene<br>h : I ô€€€! R by h(s) = f(x + s(y ô€€€ x))<br>We have that:<br>i)h is derivable on I<br>ii)h is convex if and only if f is convex on U<br>f convex implies that h is convex. so according to the previous theorem, we<br>have h0 is increasing, which give us that h0(1) h0(0)<br>Buy dierentiating h, we have h0(s) = hf0(x + s(y ô€€€ x)); y ô€€€ xi. So h0(0) =<br>hf0(x); y ô€€€ xi and h0(1) = hf0(y); y ô€€€ xi. So using our inequality h0(1) h0(0),<br>we have hf0(y); y ô€€€ xi hf0(x); y ô€€€ xi, which is equivalent to f0 is monotone.<br>we assume (2), ie a dire f is monotone, we have to prove rst that f0 monotone<br>implies that h0 is increasing. For that, let us evaluate the dierence (h0(s2) ô€€€<br>h0(s1))(s2 ô€€€ s1). We have that<br>(h0(s2)ô€€€h0(s1))(s2 ô€€€s1) = (hf0(x+s2(yô€€€x)ô€€€f0(y+s1(yô€€€x); yô€€€xi)(s2 ô€€€s1)<br>= hf0(x + s2(y ô€€€ x) ô€€€ f0(y + s1(y ô€€€ x); y ô€€€ x; (s2 ô€€€ s1)(y ô€€€ x)i<br>16<br>= hf0(z2) ô€€€ f0(z1); z2 ô€€€ z1i<br>By using the monotony of f0, we have that hf0(z2) ô€€€ f0(z1); z2 ô€€€ z1i 0,which<br>implies that (h0(s2) ô€€€ h0(s1))(s2 ô€€€ s1) 0, which means that h0 is increasing.<br>So buy using our previous theorem, we have:h(s) h(r)+h0(r)(sô€€€r)8s; r 2 I.<br>So h(1) h(0)+h0(0), which implies that f(y) f(x)+hf0(x); yô€€€xi8x; y 2 U,<br>ie (3)<br>we assume (3) and we have to (1), ie f is convex. We have by hypothesis that<br>) f(y) f(x) + hf0(x); y ô€€€ xi, which implies that<br>f(y) = sup<br>x2U<br>(f(x) + hf0(x); y ô€€€ xi) = sup<br>x2U<br>fx(y)<br>where fx(y) = f(x) + hf0(x); y ô€€€ xi.<br>Therefore f is convex.<br>Denition (Domain of a Function) Let F : X ô€€€! R [ f+1g be a map. The<br>domain of F is the set dened by<br>dom(F) := fx 2 X : F(x) &lt; +1g<br>Domain of F is sometimes called the eective domain of F. The map F is called<br>Proper if D(F) 6= ;.Recall that this means there exists at least one x0 2 D(F)<br>such that F(x0) 2 R or F is not identically +1<br>1.5.1 Notation and Further denitions<br>Let X be a real normed space and D X is a convex subset of X. Let<br>f : D ô€€€! R be a convex function. We dene the convex extension of f,<br>F : X ô€€€! R [ f+1g by<br>F(x) =</p><p>f(x); if x 2 D<br>+1; if x 2 XnD<br>(1.4)<br>We observe that f is convex on D if and only if F is convex on X. Moreover<br>dom(f) = dom(F).<br>Denition The epigraph of F is the subset of X] R denoted by epi(F) and<br>dened by<br>epi(F) = f(x; ) 2 X R : x 2 dom(F) and; F(x) g:<br>Denition Let 2 R. We have the following denitions:<br>SF; = fx 2 X : F(x) g = fx 2 D(F) : F(x) g<br>Proposition 1.5.1 F is convex if and only if epi(F) is convex<br>Proof.<br>))Assume that F is convex. Let (x1; 1) 2 epi(F); (x2; 2) 2 epi(F); t 2 (0; 1).<br>Now F(x1) 1; F(x2) 2. Hence F(tx1+(1ô€€€t)x2) tF (x1)+(1ô€€€t)F(x2)<br>t1 + (1 ô€€€ t)2:<br>17<br>Thus (tx1 + (1 ô€€€ t)x2; t1 + (1 ô€€€ t)2) 2 epi(F). But<br>(tx1 + (1 ô€€€ t)x2; t1 + (1 ô€€€ t)2) = t(x1; 1) + (1 ô€€€ t)(x2; 2). Hence, epi(F) is<br>convex.<br>(() Assume that epi(F) is convex. We show that F is convex. Let x1; x2 2<br>D(F); t 2 (0; 1). Then x1; x2 2 D(F) implies that F(x1) 2 R; F(x2) 2 R.<br>Thus (x1; F(x1)) 2 epi(F) and (x2; F(x2) 2 epi(F). But epi(F) is convex,thus<br>t(x1; F(x1) + (1 ô€€€ t)(x2; F(x2)) 2 epi(F) if and only if F(tx1 + (1 ô€€€ t)x2)<br>tF (x1) + (1 ô€€€ t)F(x2) if and only if f is convex.<br>1.6 Lower Semi-Continuous Functions<br>Let V be a real norm linear space, let F : V ô€€€! R [ f+1g be a map.<br>Denition : We says that F is lower semi-continuous(lsc)at x0 2 Dom(F) if:<br>Given; &gt; 0; 9 &gt; 0 : kx ô€€€ x0k &lt; ) F(x0) ô€€€ &lt; F(x):<br>Denition One says that F is lsc if and only if epi(F) is closed in X R<br>Proposition 1.6.1 : Let F : V ô€€€! R [ f+1g. F is lower semi-continuous if<br>and only if for all sequence (xn) of V converging to x 2 V , we have F(x)<br>lim inf F(xn:)<br>Proof. Assume that F is lsc and xn ! x in V . We take a subsequence (xnk)<br>of (xn) such that lim<br>k<br>F(xnk) = lim inf<br>n<br>F(xn). It follows that(xnk; F(xnk) is a<br>sequence of epi(F) converging to (x; lim inf<br>n<br>F(xn)). Since epi(F) is closed, then<br>(x; lim inf<br>n<br>F(xn)) 2 epi(F). Thus<br>F(x) lim inf<br>n<br>F(xn)<br>Reversely suppose that(xn ! xin V ) ) F(x) lim inf<br>n<br>F(xn). Let (an; xn) be<br>a sequence of epi(F) converging to (x; a) in V R. Then xn ! x and an ! a.<br>So by hypothesis we have F(x) lim inf<br>n<br>F(xn) lim inf an = limn an = a<br>therefore (x; a) 2 epi(F)<br>Proposition 1.6.2 : A function F : V ô€€€! R [ f+1g is lsc if and only if for<br>all a 2 R; Sa;F = fx 2 V : F(x) ag is closed.<br>Proof. Assume that F is lsc. Let a 2 R and let (xn) be a sequence of Sa;F<br>converges to x 2 V .<br>Since F(xn) a8n,then lim inf<br>n<br>F(xn) a. So according to proposition 1,<br>F(x) lim inf<br>n<br>F(xn) a.<br>Thus x 2 Ca, which means that Ca is closed.<br>Reversely assume that Sa;F is closed for each a 2 R. Let xn be a sequence of V<br>converging to x 2 V . There exists a subsequence xnk of xn such that<br>lim<br>k<br>F(xnk) = lim inf<br>n<br>F(xn)<br>Assume that F(x) &gt; lim inf<br>n<br>F(xn). Then there exists a 2 R such that<br>lim inf<br>n<br>F(xn) &lt; a &lt; F(x): (1.5)<br>18<br>So there exists N 2 IN such that F(xnk &lt; a for all k N. So xnk 2 Sa;F 8k<br>N. Since xnk ! x and Sa;F is closed then x 2 Sa;F or equivalent to say that<br>F(x) a which contradict (2.1) Thus F(x) lim inf<br>n<br>F(xn)<br>Proposition 1.6.3 Let F : V ô€€€! R [ f+1g be a l.s.c. function on a compact<br>set K. Then there exists x 2 K such that:<br>F(x) = inf<br>x2K<br>F(x)<br>.<br>Proof. Let fxng be a minimizing sequence of F on K. Since K is compact,there<br>exists a subsequence (xnk) of (xn) such that xnk ! x 2 K. From Proposition 2<br>we have:<br>F(x) lim inf<br>k<br>F(xnk) = lim<br>k<br>F(xnk) = inf<br>x2K<br>F(x):<br>Therefore F(x) = inf<br>x2K<br>F(x).<br>1.7 Existence Result<br>A concept that will be needed in the proof of one of our fundamental theorems<br>of optimization is that of coercivity. We introduce this concept now<br>Denition : Let X be a topological space. Let K be a subset of X. The set<br>K is said to be sequentially compact if every sequence in K has a subsequence<br>which converges to an element of K. The subset K is called Weakly sequentially<br>compact if every sequence in K has a subsequence which converges weakly to<br>an element of K.<br>Denition : Let X be a topological space. A function F : X ô€€€! R[f+1g is<br>called (weakly, respectively) sequentially coercive if the closure of every section<br>S;F is (weakly, respectively) sequentially compact in X<br>Denition :let X be a real reexive.A function F : X ô€€€! R [ f+1g is called<br>coercive if:<br>lim<br>kxk!+1<br>F(x) = +1<br>In fact, we have the following Proposition:<br>Proposition 1.7.1 : Let X be a real reexive space. Then F : X ô€€€! R[f+1g<br>is weakly sequentially coercive if and only if lim<br>kxk!+1<br>F(x) = +1<br>Proof. ()) Let F be weakly sequentially coercive. We prove that<br>limkxk!+1F(x) = +1 i.e. , 8M 2 R; 9N0 2 R :if kxk &gt; N0; then F(x) &gt; M.<br>Assume by contradiction that this is not the case. Then there exists a sequence<br>fxng such that lim<br>n!+1<br>kxnk = +1and fF(xn)g is bounded. Let 2 R be such<br>that jF(xn)j 8n 1:Since F is weakly sequentially coercive, the section<br>19<br>S;F X is weakly sequentially compact. Hence the sequence fxng which is<br>in the section S;F = fx 2 X : F(x) g, has a subsequence fxnjg which<br>converges weakly to an element of S;F ; F(xnj) whereas lim kxnjk = +1.<br>This subsequence is therefore bounded(Since every weakly convergent sequence<br>is bounded), which is a contradiction.<br>(()We assume that lim<br>kxk!+1<br>F(x) = +1<br>we want to prove that F is weakly sequentially coercive, ie for arbitrary 2<br>R; S;F is weakly sequentially compact, or that any sequence in S;F has a<br>weakly convergent subsequence. So let fxng be an arbitrary sequence in S;F<br>for arbitrary 2 R.<br>Then fxng is bounded. Since X is reexive, fxng has a subsequence fxnkg<br>converges weakly to an element of S;F . Thus F is weakly sequentially coercive.<br>Remark 1.7.1 If X is reexive and if lim<br>kxk!+1<br>F(x) = +1, we simply say that<br>F is coercive. This condition actually implies that<br>8A &gt; 0; 9B &gt; 0 : 8x 2 X; kxk &gt; B ) F(x) &gt; A<br>Equivalently,this means that F(x) A ) kxk B, i.e. ,if the range of F is<br>bounded, then the domain of F is also bounded.<br>Theorem 1.7.1 (Existence of minimum in nite dimension)<br>Let K be a nonempty closed subset of Rn and F a continuous real value appli-<br>cation on K satisfying the following property:<br>8(un) K; lim<br>n!1<br>kunk = +1 ) lim<br>n!1<br>F(un) = +1 (1.6)<br>Then there exists a least one minimum point of F over K. Moreover for any<br>minimizing sequence of F over K,one can take a subsequence converging to a<br>minimum point.<br>Proof.<br>Let (un) be a minimizing sequence of F over K. The condition (1:3) implies<br>that (un) is bounded since (F(un)) is a bounded real sequence. So by Bonzano-<br>weistrass there exists a subsequence (unk) of(un) converging to a point u of Rn.<br>But u is in K since K is closed,and F(unk) converges to F(u) by the continuity<br>of F, thus F(U) = inf<br>v2K<br>F(v)<br>Theorem 1.7.2 (Existence in innite dimension)<br>Let K be a closed,convex and nonempty subset of a reexive real Banach space<br>V . Let F : V ô€€€! R [ f+1g convex and lsc.<br>If K is bounded or if F is coercive, then there exists u 2 K such that F(u) =<br>min<br>v2K<br>F(v)<br>Moreover if F is strictly convex, then u is unique<br>Proof. Assume that F 6= +1, ie a there exists at least x0 2 V : F(x0) 6= +1.<br>In the case where F = +1, we obviously have inf<br>v2K<br>F(v) = +1<br>K is nonempty,let (vn) be a minimizing sequence of F on K. K bounded or<br>20<br>F coercive implies that (vn) is bounded. Indeed, if K is bounded,then vn is<br>bounded as a sequence of K, if F is coercive,assume by contradiction that vn is<br>not bounded,then vn has a subsequence vnk : vnk ! +1 as k ! +1. Since F<br>is coercive,then lim<br>k!+1<br>F(vnk) = +1. But since (vn) is a minimizing sequence of<br>F, we have that lim<br>k!+1<br>F(vnk) = inf<br>v2K<br>F(v) which implies that inf<br>v2K<br>F(v) = +1.<br>Therefore we have F(x) = +1 which is a contradiction,thus vn is bounded.<br>V reexive implies that there exists (vnk) subsequence of vn such that vnk<br>converges weakly to u in V . K convex and closed implies that K is weakly<br>closed,which implies that u 2 K So in one hand we have lim<br>k!+1<br>F(vnk) =<br>inf<br>v2K<br>F(v),and in a other hand, we have F convex and lsc implies that epi(f) is<br>convex and closed, which implies that F is weakly lsc.Or vnk converges weakly<br>to u,so we have F(u) lim inf F(vnk) = inf<br>v2K<br>F(v). Thus F(u) = inf<br>v2K<br>F(v).<br>but we have u 2 K,which implies that F(u) = min<br>v2K<br>F(v) For the uniqueness,<br>assume that F is strictly convex. Let u and w two minimum of F over K<br>Assume by contradiction that u 6= w. K convex implies that 1<br>2 u + 1<br>2 w 2 K.<br>So we have min<br>v2K<br>F(v) F( 1<br>2 u + 1<br>2 w). Using the strict convexity of F,we have<br>min<br>v2K<br>F(v) &lt; 1<br>2F(u) + 1<br>2F( w), which is equivalent that<br>min<br>v2K<br>F(v) &lt; min<br>v2K<br>F(v)<br>which is a contradiction, therefore u = w, ie the minimum is unique<br>Corollary 1.7.1 Let A be an (n n) positif dened matrix. Then there exists<br>&gt; 0 such that<br>hAx; xi kxk28x 2 Rn<br>Proof. Consider the following problem<br>8&lt;<br>:<br>min J(x) := hAx; xi<br>kxk = 1<br>(1.7)<br>-J is continuous<br>-dim(Rn) &lt; 1<br>-Dene K = fx 2 Rn : kxk = 1g is compact<br>Then by Weistrass theorem there exists x 2 K : x = arg min J(x).<br>ie 9x 2 K : J(x) J( x<br>kxk )8x 6= 0. Taking = J(x) &gt; 0,we have<br>1<br>kxk2 hAx; xi ,<br>which implies that hAx; xi kxk2<br>1.8 Optimality condition:<br>-Convex case<br>In this section, we will try to get the necessary condition,and sometimes su-<br>cient, for minimizer. Our aim is now in one hand more practical than the last<br>section, since these condition will more often be useful for computing a mini-<br>mizer.<br>21<br>These conditions will be achieved by using the rst derivative(necessary condi-<br>tions of rst order) or second(necessary conditions of second order)<br>Theorem 1.8.1 : Let K be a nonempty convex set of norm linear space V . Let<br>F : V ô€€€! R derivable (in the sens of Gateaux).<br>If u is a local minimum of F over K then<br>F0(u)(v ô€€€ u) 08v 2 K (1.8)<br>Moreover if u 2 Int(K) then (1.8) become<br>F0(u) = 0 EULER EQUATION ;<br>ie u is a critical point.<br>Proof.<br>u minimum local implies that there exists r 0 such that F(u) F(v); 8v 2<br>K B(u; r). Let v 2 K; v 6= u, let t 2]0; 1[, let wt = u + t(v ô€€€ u) =<br>(1 ô€€€ t)u + tv 2 K. For wt to be in B(u; r), it suces to have kwt ô€€€ uk &lt; r,<br>ie 0 &lt; t &lt;<br>r<br>kv ô€€€ uk<br>. let 0 &lt; t &lt; minf<br>r<br>kv ô€€€ uk<br>; 1g = , for all t 2]0; [, we<br>have wt 2 K B(u; r), which implies that F(u) F(u + t(v ô€€€ u)); which<br>gives that<br>F(u) ô€€€ F(u + t(v ô€€€ u))<br>t<br>08t 2]0; [. Letting t goes to 0, we get<br>F0(u)(v ô€€€ u) 0<br>Assume u 2 int(K), so 9r &gt; 0;B(u; r) K. we already know that 8u 2<br>K; F0(u)(v ô€€€ u) 0, let h 6= 0. Consider for all t 2 R; u + th. So for all<br>tsuch that jtj &lt;<br>r<br>krk<br>,we have u + th 2 K,which implies that for all t such that<br>jtj &lt;<br>r<br>krk<br>, we have tF 0(u):h 0. this give us tF 0(u):(ô€€€h) = 0, which implies<br>that F0(u):h = 08h 6= 0, which means that F0(u) = 0<br>-General case<br>Denition : Let K be a nonempty subset of a norm linear space V . Let u 2 K,<br>let d 2 V; d 6= 0.<br>One says that d is an admissible direction of K on U if it exists &gt; 0 such<br>that u + td 2 K, for all t 2]o; [. We denote by Dad(u) the set of admissible<br>direction on the point u.<br>Theorem 1.8.2 Consider</p><p>MinF (v)<br>v 2 K 6= ;<br>(1.9)<br>If F is dierentiable and if F is a local minimum of F over K, then F0(u):d<br>08d 2 Dad(u)<br>In particular if u 2 int(K), then F0(u) = 0<br>Proof.<br>u minimum local of F implies that there existsr 0 : F(u) F(v), for all<br>v 2 K B(u; r). Let d 2 Dad(u), it follows that : u + td 2 K; 8t 2]0; [.So for<br>22<br>u+td to be on B(u; r),it suces to have 0 &lt; t &lt;<br>r<br>kdk<br>. Now let displaystyle =<br>minf; r<br>kdkg, so for all t 2]0; [, we have u + td 2 K B(u; r), therefore F(u +<br>td) F(u); 8t 2]0; [ implies that<br>F(u + td) ô€€€ F(u)<br>t<br>0; 8t 2]0; [, letting t<br>goes to 0, we have F0(u):d 0<br>If u 2 int(K),then Dad = V=f0g implies that F0(u):d 0; 8d 2 V=f0g, which<br>implies that F0(u):d 0; 8d 2 V=f0g, which implies that F0(u) = 0<br>Theorem 1.8.3 (necessary and sucient condition of second order)<br>One assume that K is nonempty, F is of class C2.<br>If u 2 K is a local minimum of F on K then:<br>1) F0(u):d 08d 2 Dad(U)<br>2) If d 2 Dad(u) and F0(u):d = 0 then F00(u):d:d 0<br>3) If F0(u):d = 0 and F00(u):d:d kdk2; 8d 6= 0; &gt; 0, then u is strict local<br>minimum of F over K<br>Proof.<br>(1)Let d 2 Dad(u) : F0(u):d = 0, let &gt; 0 : 8t 2]0; [; u + td 2 K B(u; r). We<br>have by Taylor expansion that:<br>F(u + td) = F(u) + tF 0(u):d +<br>t2<br>2<br>F00(u):d:d + t2kdk2″(t)<br>where “(t) ! 0 as t ! 0. This implies that t2[ 1<br>2F00(u):d:d + kdk2″(t) 0. By<br>simplifying the last inequality by t2 and bu letting t goes to 0 we have:F0(u):d<br>0<br>(2) Let v 2 V; v 6= u,we have also by the Taylor expansion of F that<br>F(v) ô€€€ F(u) = F0(u)(v ô€€€ u) +<br>1<br>2<br>:(v ô€€€ u)(v ô€€€ u) + kv ô€€€ uk2″(v)<br>where “(v) ! 0 as v ! 0. This implies that F(v) ô€€€ F(u) kv ô€€€ uk2[</p><p>2<br>+ “(v)]<br>But<br>2 &gt; 0 and varepsilon(v) ! 0 as v ! 0 implies that there exists r &gt; 0 :<br>kv ô€€€ uk &lt; r implies that “(v)j &lt;<br>2 . This implies that for kv ô€€€ uk &lt; r,we have<br>ô€€€</p><p>2<br>&lt; “(v) &lt;</p><p>2<br>,which implies that for all v 2 B(u; r),we have</p><p>2<br>+ “(v) &gt; 0,<br>which give us: for all v 2 B(u; r); v 6= u, we have F(v) &gt; F(u)<br>Therefore u is a strict local minimum If F is convex, then F0(u):(v ô€€€ u) 0<br>is a (Necessary and sucient condition)<br>1.9 Optimization with equality constrains<br>In this part we are interested with optimization problem with equality constrains<br>in Rn. Given<br>an open set of Rn; F and F1; cdots; Fp functions dened on</p><p>taking value on R. One consider the following problem:<br>inf<br>v2K<br>F(v) (1.10)<br>with<br>fv 2<br>; gi(v) = 0; 1 i mg (1.11)<br>23<br>The function F is called objective or cost.The function gi dene the equality<br>constrains,the elements of K are called the admissible elements. Obviously the<br>rst question we have to ask is the one on the existence of solution of (1.10), for<br>that we use our last results. Indeed they have been obtained in spaces which<br>are more general, they are applicable in particularly in Rn. It is preferable in<br>the following to denote in a more synthetic form the constrains. For that, one<br>pose.<br>G(x) = (g1(x); ; gm(x))8x 2</p><p>Hence the set of the constrains is written K = Gô€€€1f0g.<br>Consider the following optimization problem<br>8&gt;&gt;&gt;&gt;&lt;<br>&gt;&gt;&gt;&gt;:<br>min F(x)<br>gi(x) = 0; i = 1; :::p<br>x 2 RN(p N)<br>(1.12)<br>where F; gi : Rn ô€€€! R<br>Set C = fx 2 RN : gi(x) = 0; 8i = 1; ; pg<br>So our problem (1.4) become:<br>8&lt;<br>:<br>min F(x)<br>x 2 C<br>(1.13)<br>Denition A point x satises the qualication condition (QC) or we can say<br>x is regular if rgi(x); i = 1; :::p are linearly independant<br>If p = 1; x is regular if rg1(x) 6= 0<br>G : RN ô€€€! Rp<br>x 7! G(x) = (rg1(x); ;rgp(x))<br>C = fx 2 RN=G(x) = 0g<br>x is regular if and only if JG(x) has rank p<br>Theorem 1.9.1 (Lagrange)<br>Assume that x 2 C is regular. If x is an extremum then:<br>91; ; p 2 R :<br>8&gt;&gt;&gt;&lt;<br>&gt;&gt;&gt;:<br>rF(x) +<br>Xp<br>i=1<br>irgi(x) = 0<br>gi(x) = 0; i = 1; ; p<br>(1.14)<br>The real 1; ; p are called the Lagrange multipliers associated with the<br>constrains of the problem and the minimum local point. The uniqueness of these<br>reals is assured by the qualication condition(QC)<br>24<br>Proposition 1.9.1 Let<br>be an open convex subset of Rn. Let F :<br>! R be a<br>convex dierentiable function on<br>and gi :<br>! R be ane. Let x 2 K such<br>that there exists 1; 2; ; p 2 R such that<br>rF(x) +<br>Xp<br>i=1<br>irgi(x) = 0: (1.15)<br>Then, x is a minimizer of F on K.<br>Proof. Let x 2 K,by convexity of F we have:<br>F(x) ô€€€ F(x) rF(x)(x ô€€€ x) (1.16)<br>Using (1.15) it follows that:<br>F(x) ô€€€ F(x) ô€€€<br>Xp<br>i=1<br>irgi(x)(x ô€€€ x) (1.17)<br>Since gi are ane and gi(x) = gi(x) = 0,we have also the identities<br>gi(x) ô€€€ gi(x) = rgi(x):(x ô€€€ x) = 0<br>By replacing in (1.17),we obtain F(x) F(x)</p> <br><p></p>