Elements of spectral theory Nicolas Raymond To cite this version: Nicolas Raymond. Elements of spectral theory. Master. France. 2017. cel-01587623v2 HAL Id: cel-01587623 https://hal.archives-ouvertes.fr/cel-01587623v2 Submitted on 11 Apr 2018 (v2), last revised 9 Mar 2020 (v3) HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. ELEMENTS OF SPECTRAL THEORY N. RAYMOND CONTENTS Prolegomena 3 1. Preliminary considerations 4 1.1. A question 4 1.2. An answer 4 1.3. Some density results 6 2. Unbounded operators 8 2.1. Definitions 8 2.2. Adjoint and closedness 9 2.3. Self-adjoint operators and essentially self-adjoint operators 13 2.4. Lax-Milgram theorems 16 2.5. Examples 18 3. Spectrum 19 3.1. Definitions and basic properties 19 3.2. Spectral radius and resolvent bound in the self-adjoint case 23 3.3. About the Riesz projections 24 4. Compact operators 27 4.1. Reminders 27 4.2. A compactness criterion 28 4.3. Operators with compact resolvent 29 5. Fredholm theory 30 5.1. Grushin formalism 30 5.2. On the spectrum of compact operators 32 5.3. On the index of Fredholm operators 32 5.4. Toeplitz operators on the circle 33 6. Spectrum of self-adjoint operators 34 6.1. Compact normal operators 34 6.2. About the harmonic oscillator 36 6.3. Characterization of the different spectra 39 6.4. Min-max principle 42 7. Examples 44 7.1. Sturm-Liouville’s oscillation theorem 44 7.2. Weyl’s law in one dimension 44 8. Hille-Yosida’s theorem 48 8.1. Semi-groups 48 8.2. Hille-Yosida’s theorem 49 8.3. Stone’s theorem 50 9. About the spectral measure 51 9.1. A functional calculus based on the Fourier transform 51 9.2. Where the spectral measure comes into play 53 1 9.3. Spectral projections 55 References 61 2 PROLEGOMENA These notes are freely inspired by many books [LB03, Dav07, Hel13, Bre83, Rud80, RS, Kat95, Ray17, Zwo12]. They also owe very much to some lecture notes by Z. Ammari, C. Gerard´ and S. Vu˜ Ngo.c. The Reader is warned that these notes are devoted to teaching. They only exist in order to refresh the memory of the Author and, consequently, some important comments do not appear in this electronic document. Hopefully, the students will also help correcting my mistakes, won’t they? 3 1. PRELIMINARY CONSIDERATIONS This section is here to help the reader revising some notions that he encountered in the past. 1.1. A question. We endow the space L2(I) with the usual scalar product Z hu; viL2(I) = u v dx : I We define H1(I) = f 2 L2(I): 0 2 L2(I)g ; and we endow it with the following Hermitian form 0 0 hu; viH1(I) = hu; viL2(I) + hu ; v iL2(I) : 1 Lemma 1.1. (H (I); h·; ·iH1(I)) is a Hilbert space. We define H1 1 1 H0(I) = C0 (I) : 1 Lemma 1.2. (H0(I); h·; ·iH1(I)) is a Hilbert space. For J = (a; b), we let R 0 2 J j j dx (1.1) λ1 = inf R ; 2H1(J) 2 0 J j j dx 6=0 and we would like to give an explicit value of λ1. 1.2. An answer. Lemma 1.3 (Sobolev embedding). The following assertions hold. (i) We have H1(R) ⊂ C 0(R) and, for all 2 H1(R), 1 8x 2 R ; j (x)j ≤ p k kH1( ) : 2 R 1 0 1 (ii) We have H0(J) ⊂ C (J) and, for all 2 H0(J), (a) = (b) = 0 and 1 0 8x 2 J; j (x)j ≤ jJj 2 k kL2(J) : 1 (iii) For all 2 H0(J), we have, for all x; y 2 J, p 0 j (x) − (y)j ≤ jx − yjk kL2(J) : Proof. Let us deal with (i). We use the (unitary) Fourier transform to get, for all 2 H1(R), Z 2 2 2 k k 1 = (1 + ξ )j (ξ)j dξ : H (R) b R In particular, we deduce, by Cauchy-Schwarz, that b 2 L1(R). By using the inverse Fourier transform, we get Z 1 ixξ 8x 2 R ; (x) = p b(ξ)e dξ : 2π R By dominated convergence, we see that is continuous. Moreover, it goes to 0 by the Riemann- Lebesgue lemma. In addition, 1 1 − 2 − 2 −1 8x 2 [0; 1] ; j (x)j ≤ (2π) k bkL1(R) ≤ (2π) khξi kL2(R)khξi bkL2(R) : 4 1 Let us now consider (ii). Consider 2 H0(J). Let us extend by zero outside J and denote 2 1 1 by this extension. We have 2 L (R). Since C0 (J) is dense in H0(J), we can consider a 1 1 sequence ( n) ⊂ C0 (J) converging to in H -norm. Note that, for all n; p 2 N, (1.2) k n − mkH1(R) = k n − mkH1(J) : 1 2 Thus, ( n) converges in H (R) to some v. Since ( n) converges in L (R) to , we get v = 2 H1(R). By (i), we deduce that is continuous on J. Coming back to (1.2) and using again (i), we get that ( n) uniformly converges to . In particular, (a) = (b) = 0. Then, we write, for all x 2 J, Z x 0 n(x) = n(t) dt ; a so that 1 2 0 j n(x)j ≤ jJj k nkL2(J) : Let us prove (iii). We use again the sequence ( n) and we write, for all x; y 2 J, Z y 0 n(x) − n(y) = n(t) dt ; x and we use the Cauchy-Schwarz inequality. Lemma 1.4. The infinimum (1.1) is a minimum. 0 Proof. Let ( n) be a minimizing sequence such that k nkL2(J) = 1. In particular ( n) is 2 bounded in L (J). Thus, ( n) is equicontinuous on [a; b] and pointwise bounded. We can apply the Ascoli theorem and we may assume that ( n) uniformly converges to on [a; b] and 2 1 thus in L (J). We get k kL2(J) = 1. Since ( n) is bounded in H0(J), we can assume that it 1 0 0 is weakly convergent (to φ) in H0(J), and thus (why?) in D (J). We must have φ = . We deduce 0 0 lim inf k nkL2(J) ≥ k kL2(J) n!+1 0 2 0 since ( n) weakly converges in L (J) to (why?). We get 0 2 λ1 ≥ k kL2(J) ; 1 where 2 H0(J) and k kL2(J) = 1: It follows that 0 2 k kL2(J) = λ1 : Lemma 1.5. Let be a minimum. Then, satisfies, in the sense of distributions on J, 00 (1.3) − = λ1 : 1 Proof. Let ' 2 C0 (J) and > 0. We define R j( + ϕ)0j2 dx f() = J : R 2 J j + ϕj dx We get f 0(0) = 0 and then Z Z 0 0 ' dx = λ1 ' dx : J J Lemma 1.6. The λ1 for which there are non trivial solutions to (1.3) being 0 at a and at b are exactly the numbers (b−a)−2n2π2 where n 2 N∗. The corresponding solutions are proportional to sin(nπ(b − a)−1·). 5 1.3. Some density results. The Reader is invited to read [Bre83, Section IV. 4]. Let us con- R sider a sequence of smooth non-negative functions (ρn)n2 ∗ such that d ρn(x) dx = 1 with N R 1 suppρn = B 0; n . Consider a smooth function with compact support 0 ≤ χ ≤ 1 equal to 1 in −1 a neighborhood of 0, and define χn(·) = χ(n ·). p d Lemma 1.7. Let p 2 [1; +1). Let f 2 L (R ). Then, ρn ? f and χn(ρn ? f) converges to f in p d 1 d p L (R ). In particular, C0 (R ) is dense in (L (R); k · kLp(Rd)). 0 d Proof. Let " > 0 and f 2 C0 (R ) such that kf − f0kLp(Rd) ≤ ". We have Z ρn ? f0(x) − f0(x) = ρn(y)(f0(x − y) − f0(x)) dy ; Rd and, by the Holder¨ inequality (with measure ρn dy), Z Z p p kρn ? f0 − f0k p d ≤ ρn(y)jf0(x − y) − f0(x)j dy dx : L (R ) Rd Rd By using the uniform continuity of f0 and the support of ρn, we see that ρn ? f0 converges to f0 in Lp(Rd). It remains to notice that kρn ? (f − f0)kLp(Rd) ≤ kf − f0kLp(Rd) ; p d to see that ρn ? f converges to f in L (R ). Then, we consider Z Z p p p k(1 − χn)ρn ? fk p d ≤ (1 − χn(x)) ρn(y)jf(x − y)j dy dx ; L (R ) Rd Rd and we get Z Z p p p k(1 − χn)ρn ? fk p d ≤ (1 − χn(x + y)) ρn(y)jf(x)j dy dx L (R ) jx|≥n−1 Rd Z ≤ jf(x)jp dx ; jx|≥n−1 p d and the conclusion follows since f 2 L (R ).