July 16, 2017

Download Analyse mathématique IV: Intégration et théorie spectrale, by Roger Godement PDF

By Roger Godement

Ce 4?me quantity de l'ouvrage Analyse math?matique initiera le lecteur ? l'analyse fonctionnelle (int?gration, espaces de Hilbert, examine harmonique en th?orie des groupes) et aux m?thodes de l. a. th?orie des fonctions modulaires (s?ries L et theta, fonctions elliptiques, utilization de l'alg?bre de Lie de SL2). Tout comme pour les volumes 1 ? three, on reconna?tra ici encore, le type inimitable de l'auteur et pas seulement par son refus de l'ecriture condens?e en utilization dans de nombreux manuels. Mariant judicieusement les math?matiques dites 'modernes' et' classiques', los angeles premi?re partie (Int?gration) est d'utilit? universelle tandis que l. a. seconde oriente le lecteur vers un domaine de recherche sp?cialis? et tr?s actif, avec de vastes g?n?ralisations possibles.

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Extra resources for Analyse mathématique IV: Intégration et théorie spectrale, analyse harmonique, le jardin des délices modulaires

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Now, we define a mapping T : E → E by t T x(t) = x0 + f (s, x(s))ds for all x(·) ∈ E. 0 It is easy to check that T is continuous. Moreover, T : K → K is a mapping and T K is compact. , x(t) = x0 + 0 f (s, x(s))ds. Therefore, x (t) = f (t, x(t)) for all t ∈ (0, t0 ). 1) has a solution. This completes the proof. In the following, let Ω ⊂ RN be an open bounded subset with smooth boundary and ai , b : Ω×R×RN → R, i = 1, 2, · · · , N , be continuous functions such that (1) ∂ai ∂ηj ξi ξj ≥ |ξ|2 for all (x, z, ξ) ∈ Ω × R × Rn ; (2) |ai (x, z, 0) ≤ g(z), i = 1, 2, · · · , N , where g(·) ∈ Lq (Ω), and q > N ; ∂ai ∂ai 2 i (3) (1 + |ξ|2 )| ∂a ∂ξj | + (1 + |ξ|)(| ∂z | + |ai |) + | ∂xj | + |b| ≤ µ(|z|)(1 + |ξ| ) for i, j = 1, 2 · · · , N , where µ : [0, +∞) → [0, +∞) is a increasing function; (4) −b(x, z, ξ)signz ≤ L(|ξ| + f (x)) for all (x, z, ξ) ∈ Ω × R × RN , where L > 0 is a constant.

1), we deduce that M = R(Tλi ) is closed. For (2) notice that N1 ⊆ N2 ⊆ · · · and R1 ⊇ R2 ⊇ · · · . We cannot have Ni = Ni+1 for all i. 9, there exists a sequence of xi ∈ Ni+1 \ Ni such that xi = 1 and xi − xj ≥ 21 for i = j and thus we have T xi − T xj ≥ 2−1 λ for all j < i, which is impossible since T is compact. Thus Nj = Ni for some i and all j > i and, consequently, Rj = Ri for all j > i. If λ = 0 is not an eigenvalue of T , then Tλ is one to one. For any y ∈ Ri−1 , we have Tλ y ∈ Ri = Ri+1 and thus there exists x such that Tλ y = Tλi+1 x.

The degree theory in this chapter can be established in locally convex spaces (see [197]) or admissible topological vector spaces (see [170], [235], [307]). 6 Exercises 1. Let f (x, y) : R × R → R be a continuous function and T : C[a, b] → s C[a, b] be defined by T x(·)(s) = a f (t, x(t))dt for all x(·) ∈ C[a, b] and s ∈ [a, b]. Show that T is continuous and compact. 2. Let T : C[0, π] → C[0, π] be defined by T x(t) = 2 π π [sin t sin s + c sin 2t sin 2s][2x(s) + x3 (s)]ds 0 for all x(·) ∈ C[0, π].

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