The mountain pass theorem and applications.
Table Of Contents
- <p> Epigraph ii<br>Dedication iii<br>Preface iv<br>Acknowledgement v<br>Introduction 1<br>1 Dierentiability in Banach Spaces 6<br>
- 1.1Gâteaux Derivative . . . . . . . . . . . . . . . . . . . 6<br>
- 1.2Fréchet Dierentiability . . . . . . . . . . . . . . . . 10<br>
- 1.3Second order derivative . . . . . . . . . . . . . . . . . 16<br>2 Nemytskii Operators 20<br>
- 2.1Denition of a Nemytskii Operator . . . . . . . . . . 20<br>
- 2.2Carathéodory condition . . . . . . . . . . . . . . . . . 21<br>
- 2.3Continuity and Dierentiability of Nemytskii Operator 23<br>3 Variational Principles and Minimization 28<br>
- 3.1Lower Semicontinuous Functions . . . . . . . . . . . . 28<br>
- 3.2Ekeland Theorem in Complete metric space . . . . . 31<br>
- 3.3PalaisSmale Conditions and Minimization . . . . . . 35<br>
- 3.4Deformation Theorem and Palais-Smale Conditon . . 37<br>
- 3.5Mountain-Pass Theorem . . . . . . . . . . . . . . . . 41<br>4 Application: The lane Emden Equation 44<br>vi<br>Bibliography 54 <br></p>
Project Abstract
The mountain pass theorem is a powerful tool in the field of mathematical analysis and nonlinear functional analysis. This theorem provides a way to prove the existence of critical points for functionals defined on Banach spaces. By constructing a path connecting two critical points with a specific energy level, the mountain pass theorem guarantees the existence of a critical point with energy level higher than the given threshold. This abstract setting allows for the study of a wide range of problems in various disciplines, including differential equations, mathematical physics, and optimization. The applications of the mountain pass theorem are vast and diverse. In the context of partial differential equations, this theorem has been used to study the existence of solutions for elliptic and parabolic equations. By utilizing the mountain pass approach, researchers have been able to establish the existence of nontrivial solutions for a variety of boundary value problems. Furthermore, the mountain pass theorem has been applied to prove the existence of multiple solutions for nonlinear problems, leading to a deeper understanding of the behavior of solutions in nonlinear systems. Moreover, the mountain pass theorem has found applications in the field of mathematical physics, particularly in the study of variational problems and critical points of energy functionals. By applying the mountain pass approach, researchers have been able to identify critical points that correspond to stable or unstable equilibrium states in physical systems. This has implications for understanding the stability and dynamics of physical phenomena governed by nonlinear equations. In the realm of optimization, the mountain pass theorem provides a theoretical foundation for finding global minima of functionals subject to certain constraints. By constructing suitable mountain pass paths, researchers can establish the existence of critical points that minimize the energy functional over a given domain. This has practical implications for optimizing complex systems in engineering, physics, and economics. Overall, the mountain pass theorem serves as a fundamental tool in nonlinear functional analysis and has broad applications across various scientific disciplines. Its ability to guarantee the existence of critical points with specific energy levels makes it a valuable technique for studying a wide range of problems in mathematics, physics, and engineering.
Project Overview
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</p><p>Dierentiability in Banach Spaces<br>We will dene here two types of dierentiability in Banach spaces<br>as generalizations of the concept of dierentiability in R.<br>1.1 Gâteaux Derivative<br>Let us denote by B(X; Y ) the space of all bounded linear maps from<br>X to Y where X; Y are Banach spaces.<br>Recall that a bounded linear map means a continuous linear map.<br>Denition1.1<br>Let f : U 7! Y be a mapping and x 2 U ; where U X open. We<br>say that f is Gâteaux dierentiable at xo if there exists A 2 B(X; Y ),<br>such that<br>8 h 2 X n f0g, the map t 7! f(x0+th)ô€€€f(x0)<br>t has a limit as t ! 0<br>equal to A(h); that is,<br>lim<br>t!0<br>f(x0 + th) ô€€€ f(x0)<br>t<br>= A(h) (1:2)<br>or equivalently<br>f(x0 + th) ô€€€ f(x0) = tA(h) + o(t) 8 h 2 X<br>where o(t) holds for the remainder r(t) = f(x0+th)ô€€€f(x0)ô€€€tA(h)<br>6<br>satisfying<br>lim<br>t!0<br>kr(t)k<br>t<br>= 0 :<br>For simplicity we will write Ah instead of A(h):<br>Ah is called the Gâteaux derivative of f at x0 in the direction of<br>h denoted @f<br>@h (x0).<br>The bounded linear operator A, depending on x0, is denoted by<br>DGf(xo) or f0G<br>(xo) and called the Gâteaux dierential.<br>Remarks.<br>– In Denition 1.1, one can simply require that t ! 0+.<br>– Whenever h 6= 0 and the ratio f(x0+th)ô€€€f(x0)<br>t has a limit in Y as<br>t ! 0, we say that f is dierentiable in the direction of h at xo,<br>and we call lim<br>t!0<br>f(x0 + th) ô€€€ f(x0)<br>t<br>the directional dérivative of<br>f at xo in the direction h.<br>Example 1.1:<br>The function f : R2 7! R dened by<br>f(x; y) = x2 + y2<br>is Gateaux dierentiable at every point (x0; y0) 2 R2.<br>Indeed: Let u0 = (x0; y0) and h = (h1; h2). Then<br>f(u0 + th) ô€€€ f(u0) = 2t(x0h1 + y0h2) + t2(h21<br>+ h22<br>); 8t 2 R:<br>It follows that<br>lim<br>t!0<br>f(u0 + th) ô€€€ f(u0)<br>t<br>= 2(x0h1 + y0h2) = 2hu0; hi<br>et since the map h 7! 2hu0; hi is linear and continuous from R2<br>to R, we conclude that f is Gâteaux dierentiable and<br>DGf(u0)(h) = 2hu0; hi 8 hR2:<br>Moreover by regarding R2 as a euclidean space, we can derive the<br>gradient of f at uo as<br>rf(uo) = 2uo :<br>which is actually linear and bounded with respect to h as the inner<br>product( since RN is an inner product space).<br>7<br>Theorem 1.1: (Euler necessary condition for extrema)<br>Let X and Y be real Banach spaces, f : U 7! Y be a mapping and<br>x 2 U where U X is open. If f is Gâteaux dierentiable at an extremum<br>point x0 (maximum or minimum point), then DGf(x0) = 0<br>Proof : Under the hypothesis of this theorem, suppose without loss<br>of generality that x0 is a minimum point (otherwise consider the<br>function ô€€€f instead of f).<br>Since xo 2 U and U is open, there exists a positive real number<br>r such that the open ball Br(xo) is contained in U. Now let<br>h 2 X n f0g. Then for every t such that jtj < r=jjhjj, we have,<br>f(xo + th) f(xo) and so by the Gâteaux dierentiability of f at<br>xo, we have<br>DGf(xo)(h) = lim<br>t!0+<br>f(u0 + th) ô€€€ f(u0)<br>t<br>0<br>It also follows that DGf(xo)(ô€€€h) 0, i.e. DGf(xo)(h) 0 by linearity<br>of DGf(xo). Therefore DGf(xo)(h) = 0 for all h 2 X indeed.<br>Thus DGf(xo) = 0.<br>Theorem 1.2: (Mean Value Theorem in Banach Spaces)<br>Let X and Y be Banach spaces, U X be open and let f : U ! Y<br>be Gâteaux dierentiable. Then for all x1 x2 2 X, we have<br>kf(x1) ô€€€ f(x2)k sup<br>t2[0;1]<br>kDGf(x1 + t(x2 ô€€€ x1)k kx1 ô€€€ x2k<br>provided that the sup<br>t2[0;1]<br>kDGf(x1 + t(x2 ô€€€ x1)k is nite.<br>Proof. Suppose that the assumptions of Theorem 1.2 hold. Let<br>g 2 Y (the dual of Y ) such that jjgjj 1. Then the real-valued<br>function ‘ : [0; 1] ô€€€! R dened by<br>‘(t) = g f(x1 + th) where h = x2 ô€€€ x1<br>is dierentiable on [0; 1 in the usual sense. Moreover we see that<br>‘0(t) = gô€€€<br>DGf(x1 + th)(h)</p><p>; 8 t 2 (0; 1) :<br>It follows from the classical mean valued theorem that<br>j'(1) ô€€€ ‘(0)j sup<br>0j’0(t)j ;<br>that is<br>kg f(x1) ô€€€ g f(x2)k sup<br>0j’0(t)j :<br>8<br>Moreover for all t 2 (0; 1), we have<br>j’0(t)j =</p><p>g<br>ô€€€<br>DGf(x1 + th)(h)</p><p>jjgjj kDGf(x1 + th)k khk<br>kDGf(x1 + th)k khk:<br>And so<br>kgô€€€<br>f(x1)ô€€€f(x2)</p><p>k = kgof(x1)ô€€€gf(x2)kj</p><p>sup<br>0kDGf(x1 + th)k</p><p>khk :<br>But it is well known as a consequence of the Hahn-Banach theorem<br>that<br>kyk = supfu(y) ; u 2 Y ; kuk 1 g:<br>Therefore we nally have<br>kf(x1) ô€€€ f(x2)k sup<br>t2[0;1]<br>kDGf(x1 + t(x2 ô€€€ x1)k kx1 ô€€€ x2k :<br>Remark. If f satises the assumptions of Theorem 1.2 and has a<br>continuous Gâteaux dierential, then one can prove the conclusion<br>of Theorem 1.2 by using the notion of Riemann integration in Banach<br>spaces following the next lemma.<br>Lemma. (Cf. ) Let X be a Banach space and ‘ : [a; b] ! X<br>be continuous, where ô€€€1 < a < b < +1. Then the sequence of<br>partial sums<br>b ô€€€ a<br>n<br>Xnô€€€1<br>k=0<br>‘</p><p>a + k<br>b ô€€€ a<br>n</p><p>converges as n ! 1;<br>and its limit is called the Riemann integral of f over [a; b] and is<br>denoted by Z b<br>a<br>‘(t) dt :<br>That is<br>Z b<br>a<br>‘(t)dt = lim<br>n!1<br>b ô€€€ a<br>n<br>Xnô€€€1<br>k=0<br>‘</p><p>a + k<br>b ô€€€ a<br>n</p><p>:<br>It is easily seen that</p><p>Z b<br>a<br>‘(t) dt</p><p>Z b<br>a<br>k'(t)k dt:<br>9<br>Furthermore if ‘ is continuously (Gâteaux) diérentiable on [a; b] ,<br>then<br>‘(b) = ‘(a) +<br>Z b<br>a<br>‘0(t) dt :<br>Whenener the convergence in (1.2) is uniform for h, there arises<br>an interesting stronger type of dierentiability called the Fréchet<br>dierentiability.<br>1.2 Fréchet Dierentiability<br>It is a rened notion dierentiabilty of which concept is implicitly<br>closer to that of the standard notion of dierentiability known in R<br>Recall: A function f : R ô€€€! R is said to be dierentiable at<br>x0 2 R if and only if the mapping dened on R n f0g as<br>h 7!<br>f(x0 + h) ô€€€ f(x0)<br>h<br>has a limit a 2 R as h ! 0; that is,<br>lim<br>h!0<br>f(x0 + h) ô€€€ f(x0)<br>h<br>= a 2 R:<br>Obeserve that this condition is equivalent to the existence of a real<br>number a 2 R such that<br>f(x0 + h) = f(x0) + ah + o(h) :<br>Now, how can we extend this notion to operators dened between<br>Banach spaces? The answer is in the following denition.<br>Denition 1.2:<br>A function f : U ô€€€! Y ; where Xand Y are Banach spaces and U<br>open in X, is said to be Fréchet dierentiable at a point x0 2 U, if<br>there exists a bounded linear map A : X ! Y such that:<br>lim<br>khk!0<br>kf(x0 + h) ô€€€ f(x0) ô€€€ Ahk<br>khk<br>= 0 ; (1:1)<br>or equivalently<br>f(x0 + h) ô€€€ f(x0) = Ah + o(khk) ; (1:2)<br>10<br>where<br>r(h) := f(x0 + h) ô€€€ f(x0) ô€€€ Ah = o(h)<br>in the sense that<br>lim<br>khk!0<br>kr(h)k<br>khk<br>= 0 :<br>Such an operator A is unique and called the Fréchet dierential<br>of f at x0 and is denoted by Df(x0) or f0(x0) (somtimes it is also<br>denoted by df(xo)).<br>The function f is said to be Fréchet dierentiable (or simply differentiable)<br>on U, if it is Fréchet dierentiable at every point of U:<br>When there is no ambiguity about the domain of f, we just say that<br>f is dierentiable.<br>Denition 1.3:<br>Let X and Y be Banach spaces, U open in X and let f : U ! Y<br>be Fréchet dierentiable on U. The Fréchet dierential of f on U is<br>the mapping<br>Df : U ! B(X; Y )<br>x 7! Df(x) :<br>We say that f is continuously dierentiable on U or a mapping of<br>class C1 (or simply a C1-mapping) if Df is continuous as a mapping<br>from U into B(X; Y ).<br>Examples (Fréchet dierentiable functions).<br>Let H be a real Hilbert space. Then the function F : H ô€€€! R<br>dened by<br>F(x) =<br>1<br>2<br>kxk2<br>is Fréchet dierentiable on H and its Fréchet dierential is dened<br>by:<br>DF(x)(h) = hx; hi = hh; xi:<br>Thanks to the Riesz representation we can write<br>rF(x) = x :<br>11<br>Indeed let us x xo 2 H arbitrarily. Then for every h 2 H, we have<br>F(xo + h) ô€€€ F(xo) = 1<br>2 (kxo + hk2 ô€€€ kxok2)<br>= 1<br>2 (hxo + h; xo + hi ô€€€ hxo; xoi)<br>= 1<br>2 (hxo; xoi + 2hxo; hi + hh; hi ô€€€ hxo; xoi)<br>= hxo; hi + hh;hi<br>2<br>= hh; xoi + jjhjj2<br>2 :<br>Now dene the operator A : H ! H by A(h) = hh; xoi. Then<br>A is linear (since the real inner product is bilinear) and bounded<br>(according to Cauchy-Schwarz inequality). Moreover it is clear that<br>lim<br>khk!0<br>kF(x0 + h) ô€€€ F(x0) ô€€€ A(h)k<br>khk<br>= lim<br>khk!0<br>khk=2 = 0 :<br>Next we present some properties of the Fréchet dierential.<br>Proposition 1.1: Let X and Y be Banach spaces and U X<br>open.<br>1. If F : U ! Y is Fréchet dierentiable at some point x0 2 U,<br>then F is continuous at x0.<br>2. If F : U ! Y is Fréchet dierentiable according to a norm in<br>X,<br>then it is also Fréchet dierentiable according to any norm<br>equivalent to the rst norm.<br>3. (linéarity)<br>If F;G : U ! Y are Fréchet dierentiable at some point xo 2<br>U,<br>then for any a; b 2 R, aF + bG is Fréchet dierentiable at xo<br>and<br>D(aF + bG)(xo) = aDF(xo) + bDG(xo) :<br>4. (Chain rule).<br>Let also V be an open set of a Banach space Z and consider<br>two mappings F : U ô€€€! Y and G : V ô€€€! Z such that<br>F(U) V . If F is Frechet dierentiable at some point xo 2 U<br>12<br>and G : V ô€€€! Z is Frechet dierentiable at yo = F(xo) 2 V ,<br>then G F is Fréchet dierentiable at xo and<br>D(G F)(xo) = DG(yo) DF(xo) :<br>Proof:<br>1. Suppose that f is Fréchet dierentiable at x0 2 U. Then there<br>exists a bounded linear map A : X ! Y such that:<br>f(x0 + h) ô€€€ f(x0) = Ah + o(khk) :<br>It follows from the continuity of A and the denition of o(h)<br>that<br>lim<br>jjhjj!0<br>jjf(x0 + h) ô€€€ f(x0)jj = 0 ;<br>that is<br>lim<br>h!0<br>ô€€€<br>f(x0 + h) ô€€€ f(x0)</p><p>= 0 in Y<br>or simply lim<br>h!0<br>f(x0 + h) = f(x0).<br>2. Let k:k1 and k:k2 be two equivalent norm in X. Then there<br>exist constant > 0 and > 0 such that<br>kxk1 kxk2 kxk1 ; 8 x 2 X :<br>There a mapping g is dened from an open neighbourhood of<br>0 in (X; k:k1) into Y if and only if is dened from an open<br>neighbourhood of 0 in (X; k:k2) into Y . Moreover for any h 6= 0<br>in the domain of g, we have<br>kg(h)k<br>khk1</p><p>kg(h)k<br>khk2</p><p>kg(h)k<br>khk1</p><p>kg(h)k<br>khk2<br>which implies that<br>lim<br>khk1!0<br>kg(h)k<br>khk1<br>= 0 () lim<br>khk2!0<br>kg(h)k<br>khk2<br>= 0 :<br>3. Let ” > 0. Then by the Fréchet dierentiability of the two<br>functions F and G at xo 2 U, we get (indeed) the existence of<br>> 0 such that for every h 2 X satisfying jjhjj < , we have<br>kF(xo + h) ô€€€ F(xo) ô€€€ DF(xo)(h)k<br>”<br>2(jaj + 1)<br>khk<br>13<br>and<br>kG(xo + h) ô€€€ G(xo) ô€€€ DG(xo)(h)k<br>”<br>2(jbj + 1)<br>khk :<br>Thus we have<br>k(aF+bG)(xo+h)ô€€€(aF+bG)(xo)ô€€€aDF(x)(h)ô€€€bDG(x)(h)k “jjhjj :<br>4. We know that<br>F(xo + h) ô€€€ F(xo) = DF(xo)(h) + o(h) (i)<br>and<br>G(yo + h) ô€€€ G(yo) = DG(yo)(h) + o(h) (ii):<br>Therefore<br>G F)(xo + h) ô€€€ (G F)(xo) = G(F(xo + h)) ô€€€ G(F(xo))<br>= G<br>ô€€€<br>F(xo) + DF(xo)(h) + o(khk)</p><p>ô€€€ G(F(xo))<br>= G(F(xo)) + DG(yo)((DF(xo)(h) + o(h) ô€€€ G(F(xo= DG(yo)DF(xo)(h) + DG(yo)(o(h))<br>= DG(yo)DF(xo)(h) + ~o(h)<br>which gives the result since DG(yo)DF(xo) is a bounded linear<br>map and<br>jj~o(h)jj jjDG(yo)jj jjo(h)jj:<br>Theorem 1.3:<br>Every Fréchet dierentiable function is Gâteaux dierentiable and<br>the dierentials coincide..<br>Proof : Let f : U ô€€€! Y ; where Xand Y are Banach spaces and U<br>open in X, be Fréchet dierentiable at a point x0 2 U. We show<br>that f is Gâteaux dierentiable at xo. Fix any v 2 X n f0g. Then<br>we have f0(xo) 2 B(X; Y ) and<br>lim<br>t!0+</p><p>f(xo+tv)ô€€€f(xo)<br>t ô€€€ f0(xo)(v)</p><p>= lim<br>t!0+<br>jjvjj f(xo+tv)ô€€€f(xo)ô€€€f0(xo)(tv)<br>jjtvjj<br>= lim<br>khk!0<br>jjvjj f(xo+h)ô€€€f(xo)ô€€€f0(xo)(h)<br>jjhjj<br>= 0 :<br>Therefore f is Gâteaux dierentiable at xo and moreover DGf(xo) =<br>f0(xo).<br>14<br>Remark: The converse of Theorem 1.2 is false as it can be seen<br>by the example below.<br>Example 1.2 :<br>The function g : R2 ô€€€! R dened by<br>g(x; y) =<br>8><<br>>:</p><p>x2y<br>x4+y2<br>4<br>if y 6= 0 ;<br>0 if y = 0<br>is Gâteaux dierentiable at (0; 0) but not Fréchet dierentiable at<br>this point.<br>Indeed, let h = (h1; h2) 2 R2. Then for t > 0, we have<br>g(th) ô€€€ g(0; 0)<br>t<br>=<br>8<<br>:<br>t4h41<br>h22<br>(t2h41<br>+h22<br>)4 if h2 6= 0 ;<br>0 if h2 = 0 ;<br>and so<br>lim<br>t!0+<br>g(th) ô€€€ g(0; 0)<br>t<br>= 0 ;<br>yielding the Gâteaux dierentiability of g at (0; 0) with g0(0; 0) 0.<br>But it is seen that g is not Fréchet dierentiable at (0; 0), according<br>to Theorem 1.3, by considering the perturbations H = (h1; h21<br>) as<br>follows :<br>lim<br>h1ô€€€!0+<br>g(h1; h21<br>) ô€€€ g(0; 0)<br>k(h1; h21<br>)k<br>=<br>1<br>16<br>6= 0 :<br>The next theorem gives a useful sucient condition under which<br>Gâteaux dierentiability implies Fréchet dierrentiability.<br>Theorem 1.4:<br>Let X and Y be Banach spaces, U open and nonempty in X and<br>let f : U ! Y . If f has a continuous Gâteaux dierential, then f<br>is Fréchet dierentiable and f 2 C1(U;R).<br>Proof :<br>Let x 2 U and choose > 0 such that B(x; ) U. Let h 2 X such<br>that jjhjj < . Consider the function ‘ : [0; 1] ô€€€! R dened by :<br>‘(t) = f(x + th) ô€€€ f(x) ô€€€ tDGf(x)(h)<br>Since f is Gâteaux dierentiable, it follows that ‘ is dierentiable<br>and<br>‘0(t) = DGf(x + th)(h) ô€€€ DGf(x)(h) :<br>15<br>By applying the Mean Value Theorem to ‘ we have :<br>j'(1) ô€€€ ‘(0)j sup<br>0j’0(t)j ;<br>that is,<br>kf(x + h) ô€€€ f(x) ô€€€ DGf(x)(h)k sup<br>t2(0;1)<br>kDGf(x + th)(h) ô€€€ DGF(x)(h)k<br>sup<br>t2(0;1)<br>kDGf(x + th) ô€€€ DGF(x)k khk:<br>By continuity of the mapping Df : U ! B(X; Y ), we have<br>lim<br>h!0</p><p>sup<br>t2(0;1)<br>kDGf(x + th) ô€€€ DGf(x)k<br>!<br>= 0 ;<br>and so<br>f(x + h) ô€€€ f(x) ô€€€ DGf(x)(h) = o(h)<br>with Df(x) 2 B(X; Y ).<br>Denition 1.2:<br>Let H be a Hilbert space equipped with inner product h:; :i and<br>f : X ô€€€! R be Fréchet dierentiable. Then the mapping<br>rf : H ô€€€! H<br>x 7! rf(x) ;<br>(where rf(x) is the gradient of f at x) is called a potential operator<br>with a potential f : H ô€€€! R.<br>1.3 Second order derivative<br>Let X and Y be real Banach spaces, U open and nonempty in X,<br>and let f : U ! Y be dierentiable. If<br>f0 : U ô€€€! B(X; Y )<br>is dierentiable, then for every x 2 U,<br>(f0)0(x) 2 L(X; B(X; Y ))<br>and is simply denoted by f00(x) or D2f(x). In this case we say that<br>f is twice dierentiable at x and f00(x) is called the second order<br>16<br>dierential of f at x.<br>Observe that in fact<br>f00(x) : X X ô€€€! Y<br>is bilinear and bounded (i.e., continuous) .<br>We recall that a mapping : X X ô€€€! Y is a bounded bilinear<br>map if:<br>1. 8 (x1; x2) 2 X X, 8 y 2 X and 8 ; 2 R,<br>(x1 + x2; y) = (x1; y) + (x2; y)<br>(y; x1 + x2) = (y; x1) + (y; x2)<br>2. 9K 2 (0; 1) such that<br>k(x1: x2)kY Kkx1kXkx2kX :<br>The norm of such a bounded bilinear map is given by:<br>kk = sup fk(x1; x2)kY ; kx1kX 1 and kx2k 1g<br>Note that, more generally, if E1, E2 and E3 are given three normed<br>linear spaces, we can dene a bounded linear map from E1E2 into<br>E3.<br>The space of bounded bilinear maps from X X into Y is isometric<br>to B(X; B(X; Y )). Indeed the map<br>j : B(X2; Y )) ô€€€! B(X; B(X; Y ))<br>A 7! j(A)<br>where j(A) is such that for all x 2 X and for all y 2 Y ,<br>ô€€€<br>j(A)(x)</p><p>(y) = A(x; y) :<br>Moreover jjj(A)jj = jjAjj.<br>Going back to the setting of the denition of the second order<br>dierential, if f : U ! Y is twice dierentiable, then f0 : U !<br>B(X2; Y )). And if f00 is continuous, we say that f is of class C2<br>and we write f 2 C2(U; Y ).<br>17<br>Furthermore we have the following Taylor formula for x 2 U and h<br>suciently small :<br>f(x + h) = f(x) + f0(x)(h) +<br>1<br>2<br>f00(x)(h; h) + o<br>ô€€€<br>khk2<br>X</p><p>(1:5)<br>that can be established by using a notion of Riemann integration in<br>Y such as<br>f(x + h) = f(x) + f0(x)(h) +<br>Z 1<br>0<br>(1 ô€€€ t)f00(x + th)(h; h) dt :<br>These Taylor expansions give the simplest sucient conditions<br>on a critical a C2 functional to be a local extrema.<br>Proposition 1.4<br>Let X and Y be Banach spaces, U open in X and let f : U ! Y<br>be twice continuously dierentiable. Suppose that xo is a critical<br>point of f.<br>1. If there exists a positive real number such that<br>D2f(xo)(h; h) jjhjj2 ; 8 h 2 X ;<br>Then xo is a local minimum point of f.<br>2. If for every x in a neighbourhood of xo, D2f(x) is positive<br>semidenite (in the sense that<br>D2f(x)(h; h) 0 ; 8 h 2 X );<br>then xo is a local minimum point of f over U.<br>3. If U is convex and for every x 2 U, D2f(x) is positive semidefinite<br>(in the sense that<br>D2f(x)(h; h) 0 ; 8 h 2 X );<br>then xo is a minimum point of f over U. Observe in this case<br>that f is convex on U.<br>For instance if H is a real Hilbert space and b 2 H is given then,<br>the critical point xo of the the functional ‘ dened on H by<br>‘(x) =<br>jjxjj2<br>2<br>+ hb; xi ;<br>is the minimum point (i.e., xo = ô€€€b) .<br>18<br>Note.<br>Let H be a Hilbert space and f : H ! R be twice continuously<br>dierentiable. Then for every x 2 H, there exists (according<br>to the Riesz Representation Theorem) a bounded linear operator<br>A : H ! H which is symmetric and satises<br>D2f(x)(h1; h2) = hAh1; h2i :<br>I<br>19</p>
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