Spectral theory and applications. An elementary introductory course. Version 2008-2009
Bernard Helffer Universit´eParis-Sud, D´epartement de Math´ematiques, UMR 8628 du CNRS, Bat. 425, F-91405 Orsay Cedex, FRANCE January 7, 2009
Abstract We intend to present in this course the basic tools in spectral analy- sis and to illustrate the theory by presenting examples coming from the Schr¨odinger operator theory and from various branches of physics : sta- tistical mechanics, superconductivity, fluid mechanics.
Contents
1 Introduction 4 1.1 ThefreeLaplacian ...... 4 1.2 Theharmonicoscillator ...... 4 1.3 Theproblemoftheboundary ...... 6 1.3.1 Ill-posedproblems ...... 6 1.3.2 Theperiodicproblem ...... 7 1.3.3 TheDirichletproblem ...... 7 1.3.4 TheNeumannproblem ...... 7 1.3.5 Conclusion ...... 8
2 Unbounded operators, adjoints, Selfadjoint operators 8 2.1 Unboundedoperators ...... 8 2.2 Adjoints...... 11 2.3 Symmetricandselfadjointoperators...... 14
3 Representation theorems 16 3.1 Riesz’sTheorem...... 16 3.2 Lax-Milgram’ssituation...... 16 3.3 An alternative point of view: V, , V ′...... 17 H
1 4 Semi-bounded operators and Friedrichs extension. 20 4.1 Definition ...... 20 4.2 AnalysisoftheCoulombcase...... 20 4.3 Friedrichs’sextension...... 23 4.4 Applications...... 25
5 Compact operators : general properties and examples. 29 5.1 Definitionandproperties...... 29 5.2 Examples ...... 29 5.3 Adjointsandcompactoperators...... 32 5.4 Precompactness...... 33
6 Spectral theory for bounded operators. 35 6.1 Fredholm’salternative ...... 35 6.2 Resolventsetforboundedoperators ...... 36 6.3 Spectraltheoryforcompactoperators ...... 38 6.4 Spectrumofselfadjointoperators...... 40 6.5 Spectral theory for compact selfadjoint operators ...... 41
7 Examples. 43 7.1 Thetransferoperator...... 43 7.1.1 Compactness ...... 43 7.1.2 About the physical origin of the problem...... 43 7.1.3 Krein-Rutman’sTheorem...... 47 7.2 The Dirichlet realization, a model of operator with compact re- solvent...... 49 7.3 Extension to operators with compact resolvent ...... 52 7.4 Operators with compact resolvent : the Schr¨odinger operator in anunboundeddomain...... 52 7.5 The Schr¨odinger operator with magnetic field ...... 55 7.6 Laplace Beltrami operators on a Riemannian compact manifold 56
8 Selfadjoint unbounded operators and spectral theory. 59 8.1 Introduction...... 59 8.2 Spectrum...... 62 8.3 Spectral family and resolution of the identity...... 63 8.4 ThespectraldecompositionTheorem...... 68 8.5 Applicationsofthespectraltheorem: ...... 71 8.6 Examples of functions of a selfadjoint operator ...... 73 8.7 Spectrumandspectralmeasures ...... 74
9 Non-self adjoint operators and ǫ-pseudospectrum 76 9.1 Maindefinitionsandproperties ...... 76 9.2 ǫ-Pseudospectrum : complete analysis of the differentiation op- erator...... 78 9.3 Another example of non selfadjoint operator without spectrum . 82
2 10 Essentially selfadjoint operators 87 10.1Introduction...... 87 10.2Basiccriteria...... 88 10.3 TheKato-Rellichtheorem ...... 90 10.4 Other criteria of selfadjointness for Schr¨odinger operators . . . . 92 10.5 Non-selfadjoint case : Maximal accretivity and application to the Fokker-Planckoperator ...... 94 10.5.1 Accretiveoperators...... 94 10.5.2 Application to the Fokker-Planck operator ...... 95
11 Discrete spectrum, essential spectrum 98 11.1 Discretespectrum ...... 98 11.2 Essentialspectrum ...... 98 11.3 Basicexamples:...... 99 11.4 Weyl’scriterion: ...... 99
12 The max-min principle 103 12.1 Introduction...... 103 12.2Onpositivity ...... 103 12.3 Variational characterization of the discrete spectrum ...... 104 12.4 Max-minprinciple ...... 106 12.5 CLRinequality ...... 110 12.6 Essential spectrum and Persson’sTheorem ...... 111
13 On the Schr¨odinger Operator with constant magnetic field 113 13.1Dimension2...... 113 13.1.1 The case of R2 ...... 113 13.2 Magnetic Schr¨odinger operators in dimension 3 ...... 114 + 13.3 The case of R2 ...... 115 13.3.1 Preliminaries ...... 115 13.3.2 Main results for a 1D-model ...... 115 13.3.3 Regularity...... 117 13.4Notes ...... 120
14 Exercises and Problems 121 14.1 Introduction...... 121 14.2Exercises ...... 121 14.3Problems ...... 125
3 1 Introduction
Our starting point could be the theory of Hermitian matrices, that is of the matrices satisfying : A⋆ = A. If we look for eigenvalues and corresponding eigenvectors of A, that is for pairs (u, λ) with u Ck, u = 0 and λ C such that Au = λu, we know that the eigenvalues are real∈ and6 that one can∈ find an orthonormal basis of eigenvectors associated with real eigenvalues.
In order to extend this theory to the case of spaces with infinite dimension (that is replacing the space Cm by a general Hilbert space ), the first attempt consists in developing the theory of compact selfadjoint opHerators. But it is far to cover all the interesting cases that are present in Quantum Mechanics. So our aim is to present a general theory but it is perhaps good to start by looking at specific operators and to ask naive questions about the existence of pairs (u, λ) with u in some suitable domain, u = 0 and λ C such that Au = λu. We shall discover in particular that the answer6 at these∈ questions may depend strongly on the choice of the domain and on the precise definition of the operator.
1.1 The free Laplacian The Laplacian ∆ has no eigenvectors in L2, but it has for any λ R+ an − m ∈ eigenvector in ′(R ) (actually in L∞) and for any λ C an eigenvector in m S ∈ ′(R ). So what is the right way to extend the theory of Hermitian matrices onD Ck ? On the other hand, it is easy to produce (take for simplicity m = 1) approximate 1 x n2 eigenvectors in the form u (x)= exp ix ξ χ( − ), where χ is a compactly n √n · n supported function of L2-norm equal to 1.
1.2 The harmonic oscillator As we shall see the harmonic oscillator H = d2/dx2 + x2 − plays a central role in the theory of quantum mechanics. When looking for eigenvectors in (R), we obtain that there is a sequence of eigenvalues λn (n N) S ∈ λ = (2n 1) . n − In particular the fundamental level (in other words the lowest eigenvalue) is
λ1 =1 and the splitting between the two first eigenvalues is 2. The first eigenfunction is given by x2 φ (x)= c exp (1.2.1) 1 1 − 2
4 and the other eigenvectors are obtained by applying the creation operator
L+ = d/dx + x . (1.2.2) − We observe indeed that
+ H = L L− + 1 , (1.2.3) · where L− = d/dx + x , (1.2.4) and has the property L−φ1 =0. (1.2.5) Note that if u L2 is a distributional solution of L+u = 0, then u = 0. Note ∈2 also that if u L is a distributional solution of L−u = 0, then u = µφ1 for some µ R. ∈ The nth∈-eigenfunction is then given by
+ n 1 φn = cn(L ) − φ1 .
This can be shown by recursion using the identity
L+(H +2)= HL+ . (1.2.6)
2 It is easy to see that φ (x) = P (x)exp x where P (x) is a polynomial of n n − 2 n order n 1. One can also show that the φn are mutually orthogonal. The proof of− this point is identical to the finite dimensional case, if we observe the following identity (expressing that H is symmetric) :
5 This property is called ”positivity” of the operator. Actually one can show by various ways that
2 u 2 R 2 xu 2 u′ 2 , u (R) . (1.2.11) || ||L ( ) ≤ || ||L || ||L ∀ ∈ S The trick is to observe the identity d d 1= x x . (1.2.12) dx · − · dx The inequality (1.2.10) is simply the consequence of the identity
2 2
Another way is to directly observe the identity
2 2
d2 We mainly consider the operator dx2 and look at various problems that can be asked naively about the existence− of eigenvectors for the problem in L2(]0, 1[).
1.3.1 Ill-posed problems Look first at pairs (u, λ) H1(]0, 1[) C (u = 0) such that ∈ × 6 du/dx = λu, u(0) = 0 . − It is immediate to see that no such pairs exist. Look now at pairs (u, λ) H2(]0, 1[) C (u = 0) such that ∈ × 6 d2u/dx2 = λu . − We can find for any λ two linearly independent solutions.
1 2 Here is a more “physical” version. If u is normalized by u 2 R = 1, u dx defines || ||L ( ) | | a probability measure. One can define x = x u 2dx, mean value of the position and 2 h i | | the variance σx = (x x ) . Similarly, we canR consider: Dx := (Dxu) u¯(x)dx and h −2 h i i h i · σ := (Dx Dx )u . Then (1.2.11) can be extended in the form : R Dx || − h i || 1 σx σD . · x ≥ 4
6 1.3.2 The periodic problem Here we consider pairs (u, λ) H2,per(]0, 1[) C (u = 0) such that ∈ × 6 d2u/dx2 = λu . − Here
2,per 2 H (]0, 1[) = u H (]0, 1[) , u(0) = u(1) and u′(0) = u′(1) . { ∈ } Here we recall that H2(]0, 1[) is included in C1([0, 1]) by the injection theorem of Sobolev. It is an easy exercise to show that the pairs are described by two families
• λ =4π2n2 , u = µ cos2πnx , for n N , µ R 0 , n ∈ ∈ \
• 2 2 λ =4π n , v = µ sin2πnx , for n N∗, µ R 0 . n ∈ ∈ \ One observes that λ = 0 is the lowest eigenvalue and that its multiplicity is one. This means that the corresponding eigenspace is of dimension one (the other eigenspaces are of dimension 2). Moreover an eigenvector in this subspace neither vanishes in ]0, 1[. This is quite evident because u0 = µ = 0. One observes also that one can find an orthonormal basis in L26 (]0, 1[) of eigen- vectors by normalizing the family (cos 2πnx (n N), sin2πnx (n N∗)) or the family exp 2πinx (n Z). ∈ ∈ We are just recovering∈ the L2-theory of the Fourier series.
1.3.3 The Dirichlet problem Here we consider pairs (u, λ) H2,D(]0, 1[) C (u = 0) such that d2u/dx2 = λu. ∈ × 6 − Here H2,D(]0, 1[) = u H2(]0, 1[) , u(0) = u(1) = 0 . { ∈ } It is again an easy exercise to show that the pairs are described by
2 2 λ = π n , v = µ sin πnx , for n N∗, µ R 0 . n ∈ ∈ \ One observes that λ = π2 is the lowest eigenvalue, that its multiplicity is one (Here all the eigenspaces are one-dimensional) and that an eigenvector in this subspace neither vanishes in ]0, 1[.
1.3.4 The Neumann problem Here we consider pairs (u, λ) H2,N (]0, 1[) C (u = 0) such that ∈ × 6 d2u/dx2 = λu . −
7 Here 2,N 2 H (]0, 1[) = u H (]0, 1[) , u′(0) = u′(1) = 0 . { ∈ } It is again an easy exercise to show that the pairs are described by
λ = π2n2 , v = µ cos πnx , for n N, µ R 0 . n ∈ ∈ \ One observes that λ = 0 is the lowest eigenvalue, that its multiplicity is one (Here all the eigenspaces are one-dimensional) and that the corresponding eigenspace is of dimension one and that an eigenvector in this subspace neither vanishes in ]0, 1[.
1.3.5 Conclusion All these examples enter in the so called Sturm-Liouville theory. We emphasize on one property which was always verified in each case: the eigenspace corre- sponding to the lowest eigenvalue is one dimensional and one can find a strictly positive (in ]0, 1[ or in ] , + [ in the case of the harmonic oscillator) cor- responding eigenvector. −We ∞ suggest∞ to the reader to come back at this introduction after have read the course. He will surely realize that the theory has permitted to clarify many badly posed problems.
2 Unbounded operators, adjoints, Selfadjoint op- erators 2.1 Unbounded operators We consider an Hilbert space . The scalar product will be denoted by : u , v or more simply by : uH , v when no confusion is possible. We take theh conventioniH that the scalarh producti is antilinear with respect to the second argument. A linear operator (or more simply an operator) T in is a linear map u Tu defined on a subspace of , denoted by D(T ) andH which is called the domain7→ H0 H of T . We shall also denote by R(T ) (or Im T or Range(T )) the range of 0 by T. We shall say that T is bounded if it is continuous from D(T ) into .H When D(T )= , we recover the notion of linear continuous operators on .H We recall that withH H Tu T ( ) = sup || ||H , (2.1.1) || ||L H u=0 u 6 || ||H ( ) is a Banach space. When D(T ) is not equal to , we shall always assume thatL H H D(T ) is dense in . (2.1.2) H Note that, if T is bounded, then it admits a unique continuous extension to . In this case the generalized notion is not interesting. H We are mainly interested in extensions of this theory and would like to consider
8 unbounded operators. – When using this word, we mean more precisely “non necessarily bounded op- erators”. – The point is to find a natural notion replacing this notion of boundedness. This is the object of the next definition.
The operator is called closed if the graph G(T ) of T is closed in . We recall that H × H G(T ) := (x, y) , x D(T ) , y = T x . (2.1.3) { ∈H×H ∈ } Equivalently, we can say Definition 2.1.1. (Closed operators). Let T be an operator on with (dense) domain D(T ). We say that T is closed if the conditions H u D(T ), • n ∈ u u in , • n → H Tu v in • n → H imply u D(T ), • ∈ v = Tu. • Example 2.1.2.
m 1. T0 = ∆ with D(T0)= C0∞(R ) is not closed. − 2 2 m n For this, it is enough to consider some u in H (R ) and not in C0∞(R ) 2 and to consider a sequence un C0∞ such that un u in H . The ∈ → 2 2 sequence (un, ∆un) is contained in G(T0) and converges in L L to (u, ∆u) which− does not belong to G(T ). × − 0 2 m 2. T1 = ∆ with D(T1)= H (R ) is closed. − 2 2 We observe indeed that if un u in L and ( ∆un) v in L then ∆u = v L2. The last step→ is to observe that− this implies→ that u −H2(Rm) (take∈ the Fourier transform) and (u, ∆u) G(T ). ∈ − ∈ 1 2We recall that the Sobolev space Hs(Rm) is defined as the space
s m ′ m 2 s 2 m H (R ) := u (R ) (1 + ξ ) 2 uˆ L (R ) . { ∈S | | | ∈ } Here ′ is the set of tempered distributions. Hs(Rm) is equipped with the natural Hilbertian normS : 2 2 s 2 u Hs := (1 + ξ ) uˆ(ξ) dξ . || || ZRm | | | | By Hilbertian norm, we mean that the norm is associated to a scalar product. When s N, we can also describe Hs by ∈ Hs(Rm) := u L2(Rm) Dαu L2 , α s. t. α s . { ∈ | x ∈ ∀ | |≤ } The natural norm associated with the second definition is equivalent to the first one.
9 This example suggests another definition.
Definition 2.1.3. The operator T is called closable if the closure of the graph of T is a graph. We can then define the closure T of the operator by a limit procedure via its graph. We observe indeed that we can consider
D(T ) := x y s. t. (x, y) G(T ) . { ∈H|∃ ∈ } For any x D(T ), the assumption that G(T ) is a graph says that y is unique. One can consequently∈ define T by
T x = y .
In a more explicit way, the domain of T is the set of the x such that x x and T x is a Cauchy sequence and for such x’s we∈ define H T by n → ∈ H n
T x = lim T xn . n + → ∞ Example 2.1.4. T = ∆ with D(T )= C∞ is closable and is closure is T . 0 − 0 0 1 2 Let us prove it, as an exercise. Let T0 the closure of T0. Let u L such that 2 ∈ 2 there exists un C0∞ such that un u in L and ∆un v in L . We get by distribution∈ theory that u L2 satisfies→ ∆u =−v L2→. By the ellipticity of the Laplacian (use the Fourier∈ transform),− we get that∈ u H2. We have 2 ∈2 consequently shown that D(T0) H . But C0∞ is dense in H and this gives the inverse inclusion : H2 D(T⊂). We have consequently, ⊂ 0 2 H = D(T1)= D(T0) , and it is then easy to verify that T1 = T0.
These examples lead to a more general question.
Realization of differential operators as unbounded operators. Let n Ω R and let P (x, Dx) be a partial differential operator with C∞ coefficients in⊂ Ω. Then the operator P Ω defined by
Ω Ω D(P )= C∞(Ω) , P u = P (x, D )u, u C∞(Ω) , 0 x ∀ ∈ 0 is closable. Here = L2(Ω). We have indeed H Ω G(P ) G˜ := (u,f) P (x, D )u = f in ′(Ω) . ⊂ Ω { ∈H×H| x D }
10 The proof is then actually a simple exercise in distribution theory. This inclusion shows that G(P Ω) is a graph. Note that the corresponding operator is defined Ω as Pmin with domain
Ω D(Pmin)= 2 2 un u in L (Ω) u L (Ω) a sequence un C0∞(Ω) s.t → 2 . { ∈ | ∃ ∈ P (x, Dx)un converges in L (Ω) } Ω The operator Pmin is then defined for such u by Ω P u = lim P (x, Dx)un . min n + → ∞ Using the theory of distributions, this gives :
Ω Pminu = P (x, Dx)u .
Note that there exists also a natural closed operator whose graph is G˜Ω and extending P Ω : this is the operator P˜Ω, with domain
D˜ Ω := u L2(Ω) , P (x, D )u L2(Ω) , { ∈ x ∈ } and such that P˜Ωu = P (x, D )u , u D˜ Ω , x ∀ ∈ where the last equality is in the distributional sense. Note that P˜Ω is an exten- Ω sion of Pmin in the sense that : P˜Ωu = P Ω u , u D(P Ω ) . min ∀ ∈ min Conclusion. We have associated with a differential operator P (x, Dx) in an open set Ω three natural operators. It would be important to know better the connection between these three operators. Remark 2.1.5. (Link between continuity and closeness). If = , the closed graph Theorem says that a closed operator T is continuous. H0 H 2.2 Adjoints. When we have an operator T in ( ), it is easy to define the Hilbertian adjoint T ⋆ by the identity : L H
T ⋆u, v = u,Tv , u , v . (2.2.1) h iH h iH ∀ ∈ H ∀ ∈ H The map v u,Tv defines a continuous antilinear map on and can be expressed, using7→ h Riesz’siH Theorem, by the scalar product by an elementH which is called T ∗u. The linearity and the continuity of T ∗ is then easily proved using (2.2.1).
Let us now give the definition of the adjoint of an unbounded operator.
11 Definition 2.2.1. (Adjoint ) If T is an unbounded operator on whose domain D(T ) is dense in , we first H H define the domain of T ∗ by
D(T ∗) = u ,D(T ) v u,Tv , can{ be∈ extended H ∋ as an7→ antilinear h i continuous form on . H} Using the Riesz’s Theorem, there exists f such that ∈ H (f, v) = (u,Tv) , u D(T ∗), v D(T ) . ∀ ∈ ∀ ∈ The uniqueness of f is a consequence of the density of D(T ) in and we can H then define T ∗u by T ∗u = f . Remark 2.2.2. When D(T ) = and if T is bounded , then we recover as T ∗ the Hilbertian adjoint. H Example 2.2.3.
T0∗ = T1 . Let us treat in detail this example. We get
2 D(T0∗) = u L the map C0∞ v u, ∆v , can{ be∈ extended| as an antilinear∋ 7→ continuous h − i form on L2 . } We observe that
u, ∆v L2 = u( ∆v)dx = ( ∆u)(¯v) . h − i Rm − − Z The last equality just means that we are considering the distribution ( ∆u) on the test functionv ¯. The condition appearing in the definition is just− that this 2 m distribution is in L (R ). Coming back to the definition of D(T0∗), we get 2 2 D(T ∗)= u L ∆u L . 0 { ∈ | − ∈ } But as already seen, this gives
2 2 D(T ∗)= H ,T ∗u = ∆u , u H . 0 0 − ∀ ∈ Proposition 2.2.4. . T ∗ is a closed operator. Proof. Let (v ) be a sequence in D(T ∗) such that v v in and T ∗v w∗ in n n → H n → H for some pair (v, w∗). We would like to show that (v, w∗) belongs to the graph of T ∗. For all u D(T ), we have : ∈ Tu,v = lim Tu,vn = lim u,T ∗vn = u, w∗ . (2.2.2) n + n + h i → ∞h i → ∞h i h i
12 Coming back to the definition of D(T ∗), we get from (2.2.2) that v D(T ∗) ∈ and T ∗v = w∗. This means that (v, w∗) belongs to the graph of T ∗.
Proposition 2.2.5. . Let T be an operator in with domain D(T ). Then the graph G(T ∗) of T ∗ can be characterized by H G(T ∗)= V ( G(T ) ) ⊥ , (2.2.3) { } where V is the unitary operator defined on by H × H V u, v = v, u . (2.2.4) { } { − } Proof. We just observe that for any u D(T ) and (v, w∗) we have the identity ∈ ∈ H×H V (u,Tu) , (v, w∗) = Tu,v u, w∗ . h iH×H h iH − h iH
The right hand side vanishes for all u D(T ) iff v D(T ∗) and w∗ = T ∗v, that ∈ ∈ is if (v, w∗) belongs to G(T ∗). The left hand side vanishes for all u D(T ) iff ∈ (v, w∗) belongs to V (G(T ))⊥. 1 Standard Hilbertian analysis, using the continuity of V and V − = V , then shows that − V (G(T )) ⊥ = V (G(T )) ⊥ = V (G(T )) ⊥ . { } { } { } End of proof We have not analyzed till now under which condition the do- main of the adjoint is dense in . This is one of the objects of the next theorem. H Theorem 2.2.6. Let T be a closable operator. Then we have
1. D(T ∗) is dense in , H 2. T ∗∗ := (T ∗)∗ = T , where we have denoted by T the operator whose graph is G(T ). Proof. For the first point, let us assume that D(T ∗) is not dense in . Then there exists w = 0 such that w is orthogonal to D(T ). H 6 ∗ We consequently get that for any v D(T ∗), we have ∈ (0, w) , (T ∗v, v) =0 . h − iH×H
This shows that (0, w) is orthogonal to V (G(T ∗)). But the previous proposition gives :
V (G(T )) = G(T ∗)⊥ .
We now apply V to this identity and get, using V 2 = I, − V G(T ∗)⊥ = G(T ) . 13 But, for any closed subspace , we have M ⊂ H
V ( ⊥) = [V ( )]⊥ , M M as a consequence of the identity
V (u, v) , (x, y) = (u, v) , V (x, y) . h iH×H h iH×H We finally obtain that (0, w) belongs to the closure of the graph of T , that is the graph of T because T is closable, and consequently that w = 0. This gives the contradiction. For the second point, we first observe that, D(T ∗) being dense in , we can of H course define (T ∗)∗. Using again the proposition and the closeness of T ∗, we obtain G(T ∗∗)= G(T ) and T ∗∗ = T . This means more explicitly that
D(T ∗∗)= D(T ) ,T ∗∗u = T u , u D(T ) . ∀ ∈ End of the proof.
2.3 Symmetric and selfadjoint operators. Definition 2.3.1. (symmetric operators). We shall say that T : is symmetric if it satisfies H0 7→ H
Tu,v = u,Tv , u, v 0 . h iH h iH ∀ ∈ H Example 2.3.2. . m T = ∆ with D(T )= C∞(R ). − 0 If T is symmetric it is easy to see that
D(T ) D(T ∗) (2.3.1) ⊂ and that Tu = T ∗u , u D(T ) . (2.3.2) ∀ ∈ The two conditions (2.3.1) and (2.3.2) express the property that (T ∗,D(T ∗)) is an extension of (T,D(T )).
Exercise 2.3.3. Show that a symmetric operator is closable. Hint : Show that, if u is a sequence in D(T ) such that, for some ℓ , we have n ∈ H un 0 and Tun ℓ, then ℓ = 0. →For a symmetric→ operator, we have consequently two natural closed exten- sions:
14 The minimal one denoted by Tmin (or previously T ), which is obtained • by taking the operator whose graph is the closure of the graph of T , The maximal one denoted by T the adjoint of T . • max If T sa is a selfadjoint extension of T , then T sa is automatically an extension of 3 Tmin and admits as an extension Tmax. Definition 2.3.4. We shall say that T is selfadjoint if T ∗ = T , i. e.
D(T )= D(T ∗) , and Tu = T ∗u , u D(T ) . ∀ ∈ Starting of a symmetric operator, it is a natural question to ask for the ex- istence and the uniqueness of a selfadjoint extension. We shall see later that a natural way is to prove the equality between Tmin and Tmax.
Exercise 2.3.5. Analysis of differential operators. Give simple criteria in the case of operators with constant coefficients for ob- m taining symmetric operators on C0∞(R ). In particular, verify that the operator 1 Dxj = i ∂xj is symmetric. Proposition 2.3.6. A selfadjoint operator is closed.
This is immediate because T ∗ is closed. Proposition 2.3.7. 1 Let T be a selfadjoint operator which is invertible. Then T − is also selfadjoint.
1 By invertible, we mean here that T admits an inverse T − from R(T ) into D(T ). Let us first show that R(T ) is dense in . Let w such that
3Use Proposition 2.2.5.
15 3 Representation theorems
We assume that the reader knows about this material but recall it for complete- ness4.
3.1 Riesz’s Theorem. Theorem 3.1.1. (Riesz’s Theorem) Let u F (u) a linear continuous form on . Then there exists a unique w such that7→ H ∈ H F (u)= u, w , u . (3.1.1) h iH ∀ ∈ H There is a similar version with antilinear maps : Theorem 3.1.2. Let u F (u) a antilinear continuous form on . Then there exists a unique w 7→such that H ∈ H F (u)=
a(u, v) C u v , u, v V . (3.2.1) | |≤ || ||V · || ||V ∀ ∈ It is immediate to associate, using the Riesz Theorem, a linear map A (V ) such that ∈ L a(u, v)=
a(u,u) α u 2 , u V . (3.2.3) | |≥ || ||V ∀ ∈ Theorem 3.2.2. (Lax-Milgram’s Theorem) Let a be a continuous sesquilinear form on V V . If a is V -elliptic, then A is an isomorphism from V onto V . × The proof is in three steps.
Step 1. A is injective. We get indeed from (3.2.3)
Au,u α u 2 , u V . (3.2.4) |h iV |≥ || ||V ∀ ∈ 4Here we follow, almost verbatim, the book of D. Huet [Hu].
16 Using Cauchy-Schwarz in the left hand side, we first get
Au u α u 2 , u V , || ||V · || ||V ≥ || ||V ∀ ∈ and consequently Au α u , u V . (3.2.5) || ||V ≥ || ||V ∀ ∈ This gives clearly the injectivity but actually more.
Step 2. A(V ) is dense in V . Let us consider u V such that < Av,u >V = O , v V . In particular, we can take v = u. This∈ gives a(u,u) = 0 and u = 0 using∀ (3.2.3).∈
Step 3. R(A) := A(V ) is closed in V . Let vn a Cauchy sequence in A(V ) and un the sequence such that Aun = vn. But using (3.2.5), we get that un is a Cauchy sequence which is consequently convergent to some u V . But the sequence Aun tends to Au by continuity and this shows that v ∈ v = Au and v R(A). n → ∈ Step 4. 1 The three previous steps show that A is bijective. The continuity of A− is a consequence of (3.2.5) or of the Banach Theorem.
Remark 3.2.3. . Let us suppose for simplicity that V is a real Hilbert space. Using the isomor- phism between V and V ′ given by the Riesz Theorem, one gets also a natural I operator from V onto V ′ such that A a(u, v) = ( u)(v) , v V . (3.2.6) A ∀ ∈ We have = A . A I◦
3.3 An alternative point of view: V, ,V ′. H We now consider two Hilbert spaces V and such that H V . (3.3.1) ⊂ H By this notation of inclusion, we mean also that the injection of V into is continuous or equivalently that there exists a constant C > 0 such that, forH any u V , we have ∈ u C u V . || ||H ≤ || || We also assume that V is dense in . (3.3.2) H
17 In this case, there exists a natural injection from into the space V ′ which is defined as the space of continuous linear forms onHV . We observe indeed that if h then V u < u,h> is continuous on V . So there exists ℓh V ′ such∈ that H ∋ 7→ H ∈ ℓh(u)=
D(S)= u V v a(u, v) is continuous on V for the topology induced by . { ∈ | 7→ (3.3.3)H} Using again the Riesz Theorem and assumption (3.3.2), this defines Su in by H a(u, v)=
2 2 α u C α u V C a(u,u) = C < Su,u> C Su u , u D(S) , || ||H ≤ || || ≤ | | | H |≤ || ||H · || ||H ∀ ∈ which leads to α u C Su , u D(S) . (3.3.5) || ||H ≤ || ||H ∀ ∈ We get directly the surjectivity in the following way. If h and if w V is chosen such that ∈ H ∈ h, v = w, v V , v V , h iH h i ∀ ∈ 1 (which follows from Riesz’s Theorem), we can take u = A− w in V , which is a solution of a(u, v)=
18 By the surjectivity of S, there exists v s. t. : Sv = h . We get
We now consider an hermitian sesquilinear form, that is satisfying a(u, v)= a(v,u) , u, v V . (3.3.6) ∀ ∈ This property is transmitted to S in the following way Theorem 3.3.2. If a is hermitian and V -elliptic, we have 1. S is closed;
2. S = S∗; 3. D(S) is dense in V . Proof of 2. We first observe that the assumption of Hermiticity gives
D(S) D(S∗) . (3.3.8) ⊂ Let v D(S∗). Using the surjectivity of S, there exists v D(S) such that ∈ 0 ∈ Sv0 = S∗v . For all u D(S), we get that ∈ < Su,v0 > =< u,Sv0 > =< u,S∗v> =
Proof of 3. Let h V such that ∈
0=
19 4 Semi-bounded operators and Friedrichs exten- sion. 4.1 Definition Definition 4.1.1. Let T0 be a symmetric unbounded operator of domain D(T0). We say that T0 is semibounded (from below) if there exists a constant C such that
2
< P (x, Dx)u,u > = Rm ( ∆u + Vu) u¯ dx H − 2 · 2 = m u(x) dx + m V (x) u(x) dx (4.1.4) RR R C |∇u 2 . | | | ≥−R || ||H R Exercise 4.1.3. Let us consider on L2(R2; C2) the operator 2 α D with domain (R2; C2). j=1 j xj S Here the αj are 2 2 Hermitian matrices such that : × P α α + α α =2δ . j · k k · j jk Show that this operator symmetric but not semi-bounded. This operator is called the Dirac operator. Its domain is H1(R2 , C2) and its square is the Laplacian : 2 2 ( α D ) = ( ∆) IC2 . j xj − ⊗ j=1 X 4.2 Analysis of the Coulomb case. There are two important inequalities which are useful when considering the Coulomb case which plays an important role in atomic Physics. By Coulomb case, we mean the analysis on R3 of the operator Z S := ∆ , Z − − r
20 or of the Klein-Gordon operator Z K := √ ∆+1 . Z − − r The operator √ ∆ + 1 can easily defined on (R3), using the Fourier trans- form , by − S F (√ ∆+1 u)(p)= p2 + 1( u)(p) . F − F The first one is the Hardy Inequality (whichp can be found for example in the book of Kato ([Ka], p. 305-307)) :
2 2 2 2 x − u(x) dx 4 p uˆ(p) dp (4.2.1) R3 | | | | ≤ R3 | | | | Z Z and the second one is due to Kato and says that
1 2 π 2 x − u(x) dx p uˆ(p) dp . (4.2.2) R3 | | | | ≤ 2 R3 | | | | Z Z One proof of the Hardy inequality consists in writing that, for any γ R and 3 ∈ any u C∞(R ; R), we have : ∈ 0
x 2 u + γ 2 u dx 0 . (4.2.3) R3 |∇ x | ≥ Z | | This leads to :
2 2 1 2 x u + γ 2 u dx 2γ u 2 u dx . R3 |∇ | x | ≥− R3 ∇ · x Z | | Z | | But an integration by part gives :
x 2 x 2 R3 u x 2 u dx = R3 u(x) (div ( x 2 )) dx − ∇ · | | | |2 1 | | = R3 u(x) ( x 2 ) dx . R R | | | | 1 R Optimizing over γ leads to γ = 2 and gives then the result. Remark 4.2.1. . The same idea works for N 3 but fails for N = 2. So a good exercise5 is to look for substitutes in this case≥ by starting from the inequality :
x 2 u γ(x) 2 u dx 0 . (4.2.4) R3 |∇ − x | ≥ Z | | The function γ(x) can be assumed radial : γ(x)= g( x ) and the question is to find a differential inequality on g leading at a weaker| | Hardy’s type inequality. One can for example try g(r) = ln(r).
5We thank M.J. Esteban for explaining to us the trick.
21 For the proof of Kato’s inequality, there is a another tricky nice estimate which, as far as we know, goes back to Hardy and Littlewood. In the case of the Coulomb potential, we can write, using the explicit computation for the 1 Fourier-transform of x x : 7→ | | ′ 1 h(p) h(p ) 1 R3 R3 u(p) p p′ 2 u(p′)dp dp′ = R3 R3 u(p) h(p′) h(p) p p′ 2 u(p′)dp dp′ | − | · · | − ′ | · × 1 h(p) h(p ) 1 R R = RR3 R3 p p′ u(p) h(p′) h(p) p p′ u(p′)dp dp′ , b b × b| − | · |b− | · R where h is a strictly positive measurable functionb to be determined later.b We then use Cauchy-Schwarz in the last equality in order to get
1 u(p) p p′ 2 u(p′)dp dp′ | · | 1 1 | − | ′ R R 2 h(p) 2 1 2 2 h(p ) 2 1 2 u(p) h(p′) p p′ 2 dp′ dp u(p′) h(p) p p′ 2 dp′ dp ≤ b | | | b | | − | × | | | | | − | 2 h(p) 2 1 = Ru(p) ( h(p′) p p′ 2 dp′) dp R | b 2| | 2 | |1− 2| 1 b = h(p) u(p) ( ′ ′ 2 dp′) dp . R | R | | h(p ) | p p b | − | R R 1 2 1 We now write pb′ = ω′ p in the integral h(p′) p p′ 2 dp′. We then take | | | | | | | − | h(p)=Rp . | | The integral becomes
1 2 1 1 1 dp′ = p − dω′ , |h(p )| p p 2 | | ω 2 ω ω 2 Z ′ | − ′| Z | ′| | − ′| with p = ω p . This is clearly| | a convergent integral. Moreover, observing the invariance by rotation, one can show that the integral is independent of ω S2. Hence we can compute it with ω = (1, 0, 0). ∈ We finally obtain the existence of an explicit constant C such that
1 2 u(p) 2 u(p′) dp.dp′ C p u(p) dp . | p p |≤ R3 | ||| | Z Z | − ′| Z π The optimization ofb the trick leadsb to C = 2 . b
Let us now show how one can use these inequalities. If we use (4.2.1), we get the semi-boundedness for any Z > 0 for the Schr¨odinger Coulomb operator (using Cauchy-Schwarz Inequality).
1 1 1 2 2 2 u dx ( 2 u dx) u . R3 r | | ≤ r | | · || || Z Z But we can rewrite the Hardy Inequality in the form
1 2 2 R3 2 u dx 4 < ∆u,u >L ( ) . R3 r | | ≤ − Z
22 So we get, for any ǫ> 0,
1 2 1 2 u dx ǫ< ∆u,u >L2 + u . (4.2.5) R3 r | | ≤ − ǫ || || Z This leads to :
Z 2 2 (1 ǫZ) < ∆u,u > u . Z L ≥ − − − ǫ || ||
1 Taking ǫ = Z , we have finally shown that
2 2 2 Z u . (4.2.6) Z L ≥− || || Here we are probably not optimal6 but there is another way to see that the behavior with respect to Z is correct. We just observe some invariance of the model. Let us suppose that we have proved the inequality for Z = 1. In order to treat the general case, we make a change of variable x = ρy. The operator SZ becomes in the new coordinates :
2 Z S˜ = ρ− ( ∆ ) . Z − y − ρy
1 Taking ρ = Z− , we obtain
1 S˜ = Z2 ∆ . Z − y − y The other inequality (4.2.2) is with this respect much better and quite im- portant for the analysis of the relativistic case. Let us see what we obtain in the case of Klein-Gordon using Kato’s Inequality. We have π
Here the nature of the result is different. The proof gives only that KZ is semibounded if Z 2 . This is actually more than a technical problem ! ≤ π 4.3 Friedrichs’s extension Theorem 4.3.1. A symmetric semibounded operator T0 on (with D(T0) dense in ) admits a selfadjoint extension. H H
6It can be proven (see any standard book in quantum mechanics) that the negative spec- trum of this operator is discrete and is described by a sequence of eigenvalues tending to 0 : 2 Z with n N∗. An eigenfunction related to the lowest eigenvalue 1 (Z = 1) is given by − 4n2 ∈ − 4 x exp 1 x . 7→ − 2 | |
23 The extension constructed in the proof is the so-called Friedrichs extension. The proof can be seen as a variant of Lax-Milgram’s Lemma. We can assume indeed by possibly replacing T0 by T0 + λ0Id that T0 satisfies
2
(u, v) a0(u, v) := T0u , v . (4.3.2) 7→ h iH The inequality (4.3.1) says that
2 a0(u,u) u , u D(T0) . (4.3.3) ≥ || ||H ∀ ∈ We introduce V as the completion in of D(T ) for the norm H 0 u p (u)= a (u,u) . 7→ 0 0 More concretely u belongs to V ,p if there exists u D(T ) such that ∈ H n ∈ 0 un u in and un is a Cauchy sequence for the norm p0. As→ a naturalH norm for V , we get as a candidate :
u V = lim p0(un) , (4.3.4) n + || || → ∞ where un is a Cauchy sequence for p0 tending to u in . Let us show that the definition does not depend on theH Cauchy sequence. This is the object of the Lemma 4.3.2. Let xn a Cauchy sequence in D(T0) for p0 such that xn 0 in . Then p (x ) 0. → H 0 n → Proof of the lemma. + First observe that p0(xn) is a Cauchy sequence in R and consequently conver- gent in R+. Let us suppose by contradiction that
p (x ) α> 0 . (4.3.5) 0 n → We first observe that
a (x , x )= a (x , x )+ a (x , x x ) , 0 n m 0 n n 0 n m − n and that a Cauchy-Schwarz inequality is satisfied :
a (x , x x ) a (x , x ) a (x x , x x ) . | 0 n m − n |≤ 0 n n · 0 m − n m − n p p Using also that xn is a Cauchy sequence for p0, we obtain that
ǫ> 0, N s. t. n N, m N, a (x , x ) α2 ǫ . (4.3.6) ∀ ∃ ∀ ≥ ∀ ≥ | 0 n m − |≤
24 α2 We take ǫ = 2 and consider the corresponding N given by (4.3.6). Coming back to the definition of a0 we obtain, 1
But as m + , the left hand side in (4.3.7) tends to 0 because xm 0 and this gives a→ contradiction.∞ → ##. We now observe that u V u , (4.3.8) || || ≥ || ||H as a consequence of (4.3.3) and (4.3.4). This means that the injection of V in is continuous. Note also that V , which contains D(T ), is dense in , by densityH of D(T ) in . Moreover, we get a 0 H 0 H natural scalar product on V by extension of a0 :
Remark 4.3.3. (Friedrichs extension starting from a sesquilinear form) One can also start directly from a semi-bounded sesquilinear form a0 defined on a dense subspace of . As we shall see below,H this is actually the right way to proceed for the Neumann realization of the Laplacian, where we consider on C∞(Ω) the sequilinear form
(u, v) u , v . 7→ h∇ ∇ i 4.4 Applications Application 1: The Dirichlet realization. m Let Ω be an open set in R such that Ω is compact and let T0 be the unbounded operator defined by D(T )= C∞(Ω) ,T = ∆ . 0 0 0 − 2 The involved Hilbert space is = L (Ω). It is clear that T0 is symmetric and positive7 (hence semi-bounded).H Following the previous general construction, we prefer to consider : T˜0 := T0 + Id.
7We shall in fact see later that it is strictly positive.
25 8 1 It is easy to see that V is the closure in H (Ω) of C0∞(Ω). This means, at least 1 if Ω is regular, the space H0 (Ω). The domain of S is then described as D(S) := u H1(Ω) ∆u L2(Ω) . { ∈ 0 | − ∈ } S is then the operator ( ∆ + 1) acting in the sense of distributions. When Ω is regular, a standard− regularity theorem (see [Lio2], [LiMa]) permits to show that D(S)= H2(Ω) H1(Ω) . (4.4.1) ∩ 0 So we have shown the following theorem Theorem 4.4.1. The operator T1 defined by D(T )= H2(Ω) H1(Ω) ,T = ∆ , 1 ∩ 0 1 − is selfadjoint and called the Dirichlet realization of ∆ in Ω. − We have just to observe that T = S 1 and to use Remark 2.3.8. 1 − Note that T1 is a selfadjoint extension of T0. Note that by the technique developed in Subsection ??, we have also constructed another selfadjoint extension of T0. So we have constructed, when Ω is relatively compact, two different selfadjoint realizations of T0. We say in this case that T0 is not essentially selfadjoint.
Application 2: The harmonic oscillator. We can start from H = ∆+ x 2 +1 , 0 − | | with domain m D(H0)= C0∞(R ) . Following the scheme of the construction of the Friedrichs extension, we first get that
V = B1(Rm) := u H1(Rm) x u L2(Rm) , j [1, ,m] . { ∈ | j ∈ ∀ ∈ · · · } One can indeed first show that V B1(Rm) and then get the equality by proving m ⊂ 1 m the property that C0∞(R ) is dense in B (R ). One can then determine the domain of S as
D(S)= u B1(Rm) ( ∆+ x 2 + 1)u L2(Rm) . { ∈ | − | | ∈ } 8 1 We recall that there are two ways for describing H0 (Ω). In the first definition we just ∞ 1 take the closure of C0 (Ω) in H (Ω). 1 1 In the second definition, we describe 0(Ω) as the subspace in H (Ω) of the distributions whose trace is zero at the boundary. ThisH supposes that the boundary Γ = ∂Ω is regular. In 1 1 this case, there exists a unique application γ0 continuous from H (Ω) onto H 2 (Γ) extending ∞ the map C (Ω)¯ u uΓ. It is a standard result (cf Br´ezis [Br] or Lions-Magenes [LiMa]) that 1(Ω) = H1∋(Ω),7→ when the boundary is regular. H0 0
26 By a regularity theorem (differential quotients method [LiMa]), one can show that D(S)= B2(Rm) := u H2(Rm) xαu L2(Rm) , α s. t. α 2 . { ∈ | ∈ ∀ | |≤ } Application 3 : Schr¨odinger operator with Coulomb potential We consequently start from
3 1 D(T )= C∞(R ) ,T = ∆ . (4.4.2) 0 0 0 − − r
We have seen that T0 is semibounded and replacing T0 by T0+2, the assumptions of the proof of Friedrich’s extension theorem are satisfied. We now claim that V = H1(R3) .
3 1 3 Having in mind that C0∞(R ) is dense in H (R ), we have just to verify that 3 the norm p0 and the norm H1(R3) are equivalent on C0∞(R ). This results immediately of (4.2.5). || · || With a little more effort, one gets that D(S)= H2(R3).
Application 4 : Neumann problem. The reader should have some minimal knowledge of Sobolev Spaces and traces of distributions for reading this subsection (See for example Br´ezis [Br]). Let Ω be a bounded domain with regular boundary in Rm. Take = L2(Ω). Let us consider the sesquilinear form : H
a (u, v)= u , v Cm dx + uvdx¯ , 0 h∇ ∇ i ZΩ ZΩ on C∞(Ω). 1 Using Remark 4.3.3 and the density of C∞(Ω) in H (Ω), we can extend the sesquilinear form to V = H1(Ω) According to the definition of the domain of S, we observe that, for u D(S), then it should exist some f L2(Ω) such that, for all v H1(Ω) : ∈ ∈ ∈ a(u, v)= f(x)v(x)dx . (4.4.3) ZΩ Then, one can find first show (by taking in (4.4.3) v C0∞(Ω)) that, in the sense of distribution, ∈ ∆u + u = f , (4.4.4) − and consequently that : D(S) W (Ω) := u H1(Ω) ∆u L2(Ω) . (4.4.5) ⊂ { ∈ | − ∈ } But this is not enough for characterizing the domain. We refer to [Lio1], [Lio2], [LiMa] or better [DaLi], Vol. 4 (p. 1222-1225)) for a detailed explanation. We first recall the Green-Riemann Formula :
u v = ( ∆u) vdx¯ + (∂u/∂ν)¯v dµ , (4.4.6) h∇ | ∇ i − · ∂Ω ZΩ ZΩ Z∂Ω
27 where dµ∂Ω is the induced measure on the boundary, which is clearly true for u H2(Ω) (or for u C1(Ω)) and v H1(Ω). We unfortunately do not know that∈ W (Ω) H2(Ω)∈ and the inclusion∈ in H1(Ω) is not sufficient for defining the normal trace.⊂ But this formula can be extended in the following way. We first observe that, for v H1(Ω) and u W , we have : ∈ 0 ∈
u v = ( ∆u) vdx¯ . (4.4.7) h∇ | ∇ i − · ZΩ ZΩ This shows that the expression
Φ (v) := u v ( ∆u) vdx¯ , u h∇ | ∇ i− − · ZΩ ZΩ which is well defined for u W and v H1(Ω) depends only of the restriction of v to ∂Ω. ∈ ∈ If v0 C∞(∂Ω), we can then extend v0 inside Ω as a function v = Rv0 in ∈ 9 C∞(Ω) and the distribution is defined as the map v0 Φu(Rv0). One observes also that, when u C1(Ω) or u H2(Ω),7→ the Green-Riemann formula shows that : ∈ ∈
Φu(v0)= (∂u/∂ν)¯v0 dµ∂Ω . Z∂Ω So we have found a natural way to extend the notion of trace of the normal derivative for u W and we write : ∈
γ1u =Φu .
We then conclude (using (4.4.4) and again (4.4.3) this time in full generality) that : D(S)= u W (Ω) γ u =0 , { ∈ | 1 } and that S = ∆+1 . − The operator S is called the Neumann realization of the Laplacian in L2(Ω). Remark 4.4.2. Another “standard” regularity theorem shows that
D(S)= u H2(Ω) γ u =0 , { ∈ | 1 } and the notion of normal trace u γ u for u H2(Ω) is more standard10. 7→ 1 ∈
9 1 1 By chosing a more specific R continuous from H 2 (∂Ω) into H (Ω), we get that Φu can 1 be extended as a linear form on H 2 (∂Ω). 1 10The trace is in H 2 (∂Ω).
28 5 Compact operators : general properties and examples. 5.1 Definition and properties. We just recall here very briefly the basic properties of compact operators and their spectral theory. We will emphasize on examples. We refer to the book by H. Br´ezis [Br] (Chap. VI). Let us recall that an operator T from a Banach E into a Banach F is compact if the range of the unit ball in E by T is relatively compact in F . We denote by (E, F ) the space of compact operators which is a closed subspace in (E, F ). ThereK is an alternative equivalent definition in the case when E andL F are Hilbert spaces by using sequences. The operator is compact if and only if, for any sequence xn which converges weakly in E, T xn is a strongly convergent sequence in F . Here we recall that a sequence is said to be weakly convergent in , if, for any y , the sequence < xn,y> is convergent. Such a sequence is boundedH (Banach-Steinhaus’s∈ H Theorem) andH we recall that, in this case, there exists a unique x , such that < xn,y >
Let us recall that when one composes a continuous operator and a compact operator (in any order) one gets a compact operator. This could be one way to prove the compactness. Another efficient way for proving compactness of an operator T is to show that it is the limit (for the norm convergence) of a sequence of continuous operators with finite rank (that is whose range is a finite dimensional space). We observe indeed that a continuous operator with finite rank is clearly a compact operator (in a finite dimensional space the closed bounded sets are compact).
5.2 Examples Continuous kernels . The first example of this type is the operator TK associated to the continuous kernel K on [0, 1] [0, 1]. × By this we mean that the operator TK is defined by 1 E u (T u)(x)= K(x, y)u(y)dy . (5.2.1) ∋ 7→ K Z0 Here E could be the Banach space C0([0, 1]) (with the Sup norm) or L2(]0, 1[). Proposition 5.2.1. If the kernel K is continuous, then the operator TK is compact from E into E. There are two standard proofs for this proposition. The first one is based on Ascoli’s Theorem giving a criterion relating equicontinuity of a subset of con- tinuous functions on a compact and relatively compact sets in C0([0, 1]).
29 The other one is based on the Stone-Weierstrass Theorem permitting to recover the operator as the limit of a sequence of finite rank operators TKn associated jn to kernels Kn of the form Kn(x, y)= j=1 fj,n(x)gj,n(y).
Let us study three other examples.P The first example comes from statistical mechanics, the second one from the spectral theory for the Dirichlet realization of the Laplacian and the third one from Quantum Mechanics.
The transfer operator. The transfer operator is the operator associated with the kernel V (x) V (y) K (x, y) = exp exp t x y 2 exp , (5.2.2) t − 2 − | − | − 2 where t> 0, and V is a C∞(R) function such that
exp V (x)dx < + . R − ∞ Z The L2- boundedness of operators with integral kernel is proven very often through the Lemma 5.2.2. Schur’s Lemma Let K an operator associated with an integral kernel K, that is a function (x, y) K(x, y) on Rm Rm, satisfying 7→ ×
M1 := supx Rm Rm K(x, y) dy < + , ∈ | | ∞ (5.2.3) M2 := supy Rm Rm K(x, y) dx < + . ∈ R | | ∞ m Then K, initially defined for u CR∞(R ) by ∈ 0 (Ku)(x)= K(x, y)u(y)dy , Rm Z can be extended as a continuous linear operator in (L2(Rm)) (still denoted by L K or TK ), whose norm satisfies
K M M . (5.2.4) || || ≤ 1 2 Proof: p By the Cauchy-Schwarz inequality, we have
Ku(x) 2 K(x, y) u(y) 2dy K(x, y) dy M K(x, y) u(y) 2dy . | | ≤ | || | | | ≤ 1 | || | Z Z Z Integrating with respect to x and using Fubini’s Theorem, we then obtain the result.
30 In our case, the operator TK is actually an Hilbert-Schmidt operator, i.e. an operator whose integral kernel is in L2(Rm Rm) (with m = 1). One can indeed prove, using Cauchy-Schwarz’s inequality,× show that :
Ku(x) 2 u(y) 2dy K(x, y) 2dy , | | ≤ | | | | Z Z and one obtains : TK K L2(Rm Rm) . (5.2.5) || || ≤ || || × It is then easy to show that TK is a compact operator. Its kernel K is indeed 2 the limit in L of a sequence Kn such that TKn is of finite rank. If φj is 2 m an orthonormal basis in L (R ), one first shows that the basis φk φℓ is an orthonormal basis in L2(Rm Rm). Here we have by definition : ⊗ × (φ φ )(x, y) := φ (x)φ (y) . k ⊗ ℓ k ℓ We then obtain K(x, y)= ck,ℓφk(x)φℓ(y) . Xk,ℓ We now introduce
Kn(x, y) := ck,ℓφk(x)φℓ(y) . k+ℓ n X≤
It is then easy to see that TKn is of finite rank because its range is included in the linear space generated by the φ ’s (k =1, ,n). Moreover, we have k · · ·
lim K Kn L2(Rm Rm) =0 . n + × → ∞ || − || Coming back to the corresponding operators, we get
m lim TK TKn (L2(R )) =0 . n + L → ∞ || − || The inverse of the Dirichlet operator .
We come back to the operator S, which was introduced in the study of the Dirichlet realization. One can show the following Proposition 5.2.3. 1 The operator S− is compact. Proof. 1 The operator S− can indeed be considered as the composition of a continuous 2 1 operator from L into V = H0 (Ω) and of the continuous injection of V into L2(Ω). If Ω is relatively compact, we know (cf [Br]) that we have compact injection from H1(Ω) into L2(Ω). For the continuity result, we observe that, for all u D(S) : ∈ 2 Su u Su u = a(u,u) α u V α u V u . || ||H|| ||H ≥ h | i ≥ || || ≥ || || || ||H
31 This gives, for all u D(S), the inequality : ∈
Su α u V . (5.2.6) || ||H ≥ || || Using the surjectivity of S, we get :
1 1 S− ( ,V ) . (5.2.7) || ||L H ≤ α Note that in our example α = 1 but that this part of the proof is completely general.
The inverse of the harmonic oscillator.
1 The analysis is analogous. We have seen that the Sobolev space H0 (R) has to be replaced by the space
B1(R) := u L2(R) , xu L2 and du/dx L2 . { ∈ ∈ ∈ } We can then prove, using a standard precompactness criterion, that B1(R) has compact injection in L2(R). One has in particular to use the inequality :
2 1 2 u(x) dx 2 u B1(R) . (5.2.8) x R | | ≤ R || || Z| |≥ It is very important to realize that the space H1(R) is not compactly injected in 2 L . To understand this point, it is enough to consider the sequence un = χ(x n) 2 − where χ is a function in C0∞(R) with norm in L equal to 1. It is a bounded sequence in H1, which converges weakly in H1 to 0 and is not convergent in L2(R).
5.3 Adjoints and compact operators We recall11 that the adjoint of a bounded operator in the Hilbertian case is bounded. When E and F are different Hilbert spaces , the Hilbertian adjoint is defined through the identity :
11We are mainly following Br´ezis’s exposition [Br].
32 In an Hilbert space, we have
M ⊥ = M ⊥ .
In particular, if M is a closed subspace, we have already used the property
(M ⊥)⊥ = M . (5.3.2)
In the case of bounded operators (T (E, F )), one gets easily the proper- ties ∈ L N(T ∗)= R(T )⊥ , (5.3.3) and R(T ) = (N(T ∗))⊥ . (5.3.4) Let us also recall the proposition Proposition 5.3.3. The adjoint of a compact operator is compact.
5.4 Precompactness We assume that the reader is aware of the basic results concerning compact sets in metric spaces. We in particular recall that in a complete metric space E, an efficient way to show that a subset M is relatively compact is to show, that for any ǫ> 0, one can recover M by a finite family of balls of radius ǫ in E. The second standard point is to remember Ascoli’s Theorem, giving a crite- rion for a bounded subset in C0(K) (K compact in Rm) to be relatively compact in term of uniform equicontinuity. Ascoli’s Theorem gives in particular : the compact injection of C1(K) into C0(K) • the compact injection of Hm(Ω) into C0(K), for m> n and with K = Ω. • 2 Let us recall finally a general proposition permitting to show that a subset in L2 is relatively compact. Proposition 5.4.1. Let A L2(Rm). Let us assume that : ⊂ 1. A is bounded in L2(Rm), that is there exists M > 0 such that :
u 2 M , u A . || ||L ≤ ∀ ∈ 2. The expression ǫ(u, R) := u(x) 2dx tends to zero as R + x R | | → ∞ uniformly with respect to u |A|≥. ∈R 3. For h Rm, let τ defined on L2 by : (τ u)(x) = u(x h). Then the ∈ h h − expression δ(u,h) := τhu u L2 tends to zero as h 0, uniformly with respect to A. || − || → Then A is relatively compact in L2.
33 This proposition can be applied for showing : the compact injection of H1(Ω) in L2(Ω) when Ω is regular and bounded, • 0 the compact injection of H1(Ω) in L2(Ω) when Ω is regular and bounded, • the compact injection of B1(Rm) in L2(Rm). •
34 6 Spectral theory for bounded operators. 6.1 Fredholm’s alternative Let us first recall Riesz’s Theorem. Theorem 6.1.1. Let E be a normed linear space such that BE is compact then E is finite dimen- sional. Let us now describe Fredholm’s alternative. Theorem 6.1.2. Let T (E). Then (i) N(∈I KT ) is finite dimensional. (ii) R(I − T ) is closed (of finite codimension). (iii) R(I− T )= E if and only if N(I T )= 0 . − − { } We shall only use this theorem in the Hilbertian framework, so E = , and H we shall prove it for simplicity under the additional assumption that T = T ∗. Proof. We divide the proof in successive steps. Step 1. (i) is a consequence of Riesz’s Theorem. Step 2. Let us show that R(I T ) is closed. − Let yn a sequence in R(I T ) such that yn y in . We would like to show that y R(I T ). − → H ∈ − Let x in N(I T )⊥ such that y = x T x . n − n n − n
Step 2a. Let us first show the weaker property that the sequence xn is bounded.
Let us indeed suppose that there exists a subsequence xnj such that xnj + . Considering u = x / x , we observe that || || → ∞ nj nj || nj || ( ) u Tu 0 . ∗ nj − nj → The sequence being bounded, we observe that (after possibly extracting a sub- sequence) one can consider that the sequence unj is weakly convergent. This implies that Tu is convergent (T is compact). Using now ( ), we get the nj ∗ convergence of unj to u :
u u, Tu = u , u =1 . nj → || ||
But u N(I T )⊥, hence we get u = 0 and a contradiction. ∈ −
Step 2b. We have consequently obtained that the sequence xn is bounded.
One can consequently extract a subsequence xnj that weakly converges to x ∞ in . Using the compactness of T , we get T xnj converges strongly to T x . H ∞
35 Hence the sequence xnj tends strongly to y + T x . We have finally ∞ y + T x = x , ∞ ∞ and consequently proved that y = x T x . ∞ − ∞ Step 3. If N(I T ) = 0 , then N(I T ∗) = 0 (here we use for simplification our additional− assumption){ } and R(I− T ) being closed, we get −
R(I T )= N(I T ∗)⊥ = . − − H
The converse is also immediate as T = T ∗.
Step 4. We have R(I T )⊥ = N(I T ∗)= N(I T ) . − − − This shows, according to (i) that R(I T ) is of finite codimension (second state- ment of (ii)). −
This ends the proof of Fredholm’s alternative in the particular case that T is selfadjoint. Remark 6.1.3. Under the same asumptions, it is possible to show that
dim N(I T )= dim N(I T ∗) . − − 6.2 Resolvent set for bounded operators In this subsection, E could be a Banach on R or C, but we will essentially need the Hilbertian case in the applications treated here.
Definition 6.2.1. (Resolvent set) For T (E), the resolvent set is defined by ∈ L ̺(T )= λ C ; (T λI) is bijective from E on E . (6.2.1) { ∈ − } 1 Note that in this case (T λI)− is continuous (Banach’s Theorem). It is easy to see that ̺(T ) is an open− set in C. If λ ̺(T ), we oberve that 0 ∈ 1 (T λ) = (T λ ) Id + (λ λ )(T λ )− − − 0 − 0 − 0 1 1 Hence (T λ) is invertible if λ λ0 < (T λ0)− − . We also get the following identity− for all λ, λ | ̺−(T ) :| || − || 0 ∈ 1 1 1 1 (T λ)− (T λ )− = (λ λ )(T λ)− (T λ )− . (6.2.2) − − − 0 − 0 − − 0
36 Definition 6.2.2. (Spectrum) The spectrum of T , σ(T ), is the complementary set of ̺(T ) in C. Note that σ(T ) is a closed set in C. This is typically the case when T is a compact injective operator in a Banach space of infinite dimension.
We say that λ is an eigenvalue if N(T λI) = 0. N(T λI) is called the eigenspace associated with λ. − 6 − Definition 6.2.3. (Point spectrum) The point spectrum σp(T ) of T is defined as the set of the eigenvalues of T . Example 6.2.4. (Basic example) 2 0 Let = L (]0, 1[) and f C ([0, 1]). Let Tf be the operator of multiplication by fH. Then one has : ∈ σ(T )= Im f =: λ C x [0, 1] with f(x)= λ . f { ∈ | ∃ ∈ }
1 σ (T )= Sta(f) =: λ C meas(f − (λ)) > 0 . p f { ∈ | } For the first assertion, it is first immediate to see that if λ Im f, then 6∈ T(f λ)−1 is a continuous inverse of Tf λ. On the other side, if λ = f(x0) for − − 1 x x some x ]0, 1[ then we have (T λ)u 0 and u =1 for u = χ( − 0 ), 0 ∈ f − n → || n|| n √n n where χ is a C0∞ function such that χ =1. This shows that f(]0, 1[) σ(Tf ) and we can conclude by considering|| the|| closure. ⊂ Note that the point spectrum is not necessarily closed. Note also that one can have a strict inclusion of the point spectrum in the spectrum as can be observed in the following example : Example 6.2.5. Let us consider E = ℓ2(N) and let T be the shift operator defined by :
(Tu)0 =0 , (Tu)n = un 1 , n> 0 , − 2 where u = (u0, ,un, ) ℓ (N). Then it is easy to see that T est injective (so 0 is not an eigenvalue)· · · · · · and∈ is not surjective (so 0 is in the spectrum of T ). As another interesting example, one can consider : Example 6.2.6. Let E = ℓ2(Z, C) and let T be the operator defined by : 1 (Tu)n = (un 1 + un+1) , n Z , 2 − ∈ for u ℓ2(Z). Then it is easy to see (by using expansion in Fourier series 12 ) that T∈has no eigenvalues and its spectrum is [ 1, +1]. − 12Using the isomorphism between ℓ2(Z; C) and L2(S1 ; C), which associates to the sequence (un)n∈Z the fonction n∈Z un exp inθ, one has to analyze the operator of multiplication by cos θ : P f f = cos θf . 7→ T It is then analyzed as in example 6.2.4. One concludes that the spectrum of is [ 1, +1]. T −
37 Exercise 6.2.7. Let α [0, 1]. Let p and q integers which are mutually prime. Analyze the ∈ 2 spectrum Σα of the operator Hα defined on ℓ (Z) by 1 p (Hαu)n = (un 1 + un+1)+cos2π( n + α) un , n Z . 2 − q ∈ In order to make the analysis easier, one can admit (particular case of the so- called Floquet theory), that one has Σ = θ [0,1]Σθ, where Σθ is the spectrum ∩ ∈ of Hα reduced to the space of the u’s in ℓ∞ such that un+q = exp2iπθ un, for n Z. This∈ operator plays an important role in Solid State Physics and is called the Harper’s operator. p We now replace the rational q by an irrational number β. So we consider the 1 2 operator Hβ,α := 2 (τ+1 + τ 1)+cos2π(β + α) on ℓ (Z, C), where τk (k Z) −2 · ∈ is the operator defined on ℓ (Z, C) by (τku)n = un k. Show that, if β Q, then the spectrum of H is− independent of α. For this, 6∈ β,α one can first prove that Hβ,α is unitary equivalent with Hβ,α+kβ for any k Z. Secondly, one can use the density of the set α + βZ + Z in R. Finally, one∈ can use the inequality { } H H ′ 2π α α′ . || β,α − β,α || ≤ | − | Proposition 6.2.8. The spectrum σ(T ) is a compact set included in the ball B(0, T ). || || T This proposition is immediate if we observe that (I λ ) is invertible if λ > T . − | | || || 6.3 Spectral theory for compact operators In the case of a compact operator, one has a more precise description of the spectrum. Theorem 6.3.1. Let T (E) where E is an infinite dimensional Banach space. Then ∈ K 1. 0 σ(T ) . ∈ 2. σ(T ) 0 = σ (T ) 0 . \{ } p \{ } 3. We are in one (and only one) of the following cases either σ(T )= 0 , • { } either σ(T ) 0 is finite, • \{ } or σ(T ) 0 can be described as a sequence of distincts points tending • to 0. \{ }
38 Proof. 1 a) If 0 σ(T ), then T admits a continuous inverse T − and the identity, which can be6∈ considered as : 1 I = T T − , ◦ is a compact operator, as the composition of the compact operator T and the 1 continuous operator T − . Using Riesz’s Theorem we get a contradiction in the case E is supposed of infinite dimension. b) The fact that, if λ = 0 and λ σ(T ), then λ is an eigenvalue, comes directly 6 ∈ T from the Fredholm’s alternative applied to (I λ ). c) The last step comes essentially from the following− lemma : Lemma 6.3.2. Let (λn)n 1 a sequence of distincts points λn λ and λn σ(T ) 0 , for all n> 0. ≥ → ∈ \{ } Then λ =0. Proof. We just give the proof in the Hilbertian case. For all n > 0, let en be a normalized eigenvector such that (T λn)en = 0 and let En be the vectorial space spanned by e ,e , ,e . Let− us show that we have a strict inclusion { 1 2 · · · n} of En in En+1. We prove this point by recursion. Let us assume the result up to order n 1 and let us show it at order n. If E = E , then e E and we can write− n+1 n n+1 ∈ n n en+1 = αj ej . j=1 X Let us apply T to this relation. Using the property that en+1 is an eigenvector, we obtain n n λ α e = α λ e . n+1 j j j j j j=1 j=1 X X Using the recursion assumption, e , ,e is a basis of E and the λ be- { 1 · · · n} n j ing distincts, we obtain αj = 0 for all j = 1, ,n and a contradiction with e = 1. · · · || n+1|| So we can find a sequence un such that un En En⊥ 1 and un = 1. T being ∈ ∩ − || || compact, one would extract a convergent subsequence (still denoted by Tun) from the sequence (Tun) and, if λn λ = 0, one would also have the conver- gence of this subsequence ( 1 Tu ) and→ consequently6 the Cauchy property. λn n Let us show that this leads to a contradiction. We remark that
(T λn)En En 1 . − ⊂ − Let n>m 2. We have ≥ T un T um 2 (T λn)un (T λm)um 2 = − − + un um || λn − λm || || λn − λm − || (T λn)un (T λm)um 2 2 = − − um + un || λn − λm − || || || u 2 =1 . ≥ || n||
39 We can consequently not extract a Cauchy subsequence from the sequence 1 Tu . This is in contradiction with the assumption λ = 0. λn n This ends the proof of the lemma and of the theorem. 6
We shall now consider the Hilbertian case and see which new properties can be obtained by using the additional assumption that T is selfadjoint.
6.4 Spectrum of selfadjoint operators.
As T = T ∗, the spectrum is real. If Im λ = 0, one immediately verifies that : 6 Im λ u 2 Im (T λ)u,u (T λ)u u . | | || || ≤ | h − i ≤ || − || · || || This shows immediately that the map (T λ) is injective and with close range. But the orthogonal of the range of (T −λ) is the kernel of (T λ¯) which is reduced to 0. So (T λ) is bijective. − − This was actually, a− consequence of the Lax-Milgram theorem (in the simple case when V = H), once we have observed the inequality
(T λ)u,u Im λ u 2 . |h − i| ≥ | ||| || Using again Lax-Milgram’s theorem, we can show Theorem 6.4.1. Let T ( ) be a selfadjoint operator. Then the spectrum of T is contained in [m,M]∈with L Hm = inf Tu,u / u 2 and M = sup Tu,u / u 2. Moreover m and M belong to the spectrumh ofi ||T .|| h i || || Proof: We have already mentioned that the spectrum is real. Now if λ>M, we can apply the Lax-Milgram to the sequilinear-form (u, v) λ u, v Tu,v . Let us now show that M σ(T ). 7→ h i − h i ∈ We observe that, by Cauchy-Schwarz applied to the scalar product (u, v) M u, v Tu,v , we have : 7→ h i − h i 1 1 Mu Tu,v Mu Tu,u 2 Mv Tv,v 2 . |h − i| ≤ h − i h − i In particular, we get :
1 1 2 2 Mu Tu M T ( ) Mu Tu,u . (6.4.1) || − ||H ≤ || − ||L H h − i
Let (un)n Z be a sequence such that un = 1 and Tun,un M as n + . ∈ || || h i → → ∞ By (6.4.1), we get that (T M)un tends to 0 as n + . This implies that − →1 ∞ M σ(T ). If not, we would get that u = (M T )− ((M T )u ) tends to 0 ∈ n − − n in contradiction with un = 1. This theorem admits|| the|| following important corollary
40 Corollary 6.4.2. Let T ( ) be a selfadjoint operator such that σ(T )= 0 . Then T =0. ∈ L H { } We first indeed show that m = M = 0 and consequently that
1 1 Tu,v Tu,u 2 Tv,v 2 . |h i| ≤ h i h i This implies, using Riesz’s Theorem :
1 1 Tu T 2 ( Tu,u ) 2 . || || ≤ || || h i Coming back to the definition of T , we obtain : || || 1 1 T T 2 M 2 , || || ≤ || || and the inequality : T M . || || ≤ But it is immediate that :
Tu,u T u 2 . h i ≤ || || || || This gives the converse inequality and the proposition.
6.5 Spectral theory for compact selfadjoint operators We have a very precise description of the selfadjoint compact operators. Theorem 6.5.1. Let be a separable Hilbert13 space and T a compact selfadjoint operator. Then admitsH an Hilbertian basis consisting of eigenvectors of T . H Proof. Let (λn)n 1 be the sequence of disjoint eigenvalues of T , except 0. Their exis- tence comes≥ from Theorem 6.3.1 and we also observe that the eigenvalues are real. Let us define λ0 = 0.
13that is having a countable dense set.
41 We define E0 = N(T ) and En = N(T λnI); We know (from Riesz’s Theorem) that − 0 < dim E < + . n ∞ Let us show that is the Hilbertian sum of the (En)n 0. H ≥
(i) The spaces (En) are mutually orthogonal. If u Em and v En with m = n, we have ∈ ∈ 6
+ ∞ u = u , with u E . n n ∈ n n=0 X This permits to write + ∞ Tu = λnun . n=1 X If, for k N∗, we define T by ∈ k k Tku = λnun , n=1 X we easily see that Tk is of finite rank and that
T Tk sup λn . || − || ≤ n k+1 | | ≥
Hence the operator T appears as the limit in ( ) of the sequence Tk as k + . L H → ∞
42 7 Examples.
We go back to our previous examples and analyze their properties.
7.1 The transfer operator. 7.1.1 Compactness The transfer operator (which was introduced in (??) is compact and admits consequently a sequence of eigenvalues λn tending to 0 as n + . Let us show the → ∞ Lemma 7.1.1. The transfer operator is positive and injective. Proof of the lemma. Let u L2(R). We can write14, with φ(x) = exp V (x) u(x) ∈ − 2
2
2 ξ 2
The spectrum is consequently the union of a sequence of positive eigenvalues and of its limit 0. TK can be diagonalized in an orthonormal basis of eigenvec- tors associated with strictly positive eigenvalues. We emphasize that 0 is in the spectrum but not an eigenvalue. Theorem 6.4.1 says also that TK is the largest eigenvalue and is isolated. A natural question is then to discuss|| || the multiplicity of each eigenvalue, i. e. the dimension of each associated eigenspace. We shall see later (Krein-Rutman’s Theorem) that the largest eigenvalue is of multiplicity 1.
7.1.2 About the physical origin of the problem. Our initial problem was to find a rather general approach for the estimate of the decay of correlations attached to ”gaussian like” measures of the type
exp Φ(X) dX (7.1.1) − on Rn with Φ. One proof of this estimate (in the case when Φ has a particular structure) is based on the analysis of the transfer matrix method, originally
14If we make only the weak assumption that exp V L1(R), it is better to start for the ∞ R − ∈ proof of the positivity by considering u’s in C0 ( ) and then to treat the general case by using ∞ 2 the density of C0 in L .
43 due to Kramers-Wannier, that we have already seen for the study of the Ising model. We present here briefly the technique for our toy model. We shall only consider the case when d = 1 and treat the periodic case, that is the case when 1, ,n is a representation of Λper = Z/nZ. { ·We · · consider} the particular potential Φ
n 1 x x 2 Φ(n)(X) Φ(X) V (x )+ | j − j+1| (7.1.2) ≡ ≡ h j 4 j=1 X where we take the convention that xn+1 = x1 and where h is possibly a semiclas- sical parameter which is sometimes chosen equal to one if we are not interested in the “semiclassical” aspects. More generally, we could more generally consider examples of the form:
n 1 Φ (X)= (V (x )+ I(x , x )) (7.1.3) h h j j j+1 j=1 X where I is a symmetric ”interaction” potential on R R. Let us mention how- ever that the example (7.1.2) appears naturally in quantum× field theory when the so called ”lattice approximation” is introduced. For this special class of potentials, we shall demonstrate that the informations given by the transfer operator method are complementary to the other approachs we have explained before. We shall present the ”dictionary” between the properties of the measure n n h− 2 exp Φ (X) dX on R and the spectral properties of the transfer operator · − h KV (which is also called Kac operator for some particular models introduced by M. Kac) whose integral kernel is real and given on R R by × 2 1 V (x) x y V (y) K (x, y)= h− 2 exp exp | − | exp . (7.1.4) V − 2h · − 4h · − 2h 15 2 By integral kernel (or distribution kernel), we mean a distribution in ′(R ) D such that the operator is defined from C∞(R) into ′(R) by the formula 0 D
(KVu)(x)v(x) dx = KV (x, y) u(x) v(y) dx dy , u , v C0∞(R) . R R R ∀ ∈ Z Z × (7.1.5) This dictionary permits to obtain interesting connections between estimates for the quotient µ2/µ1 of the two first largest eigenvalues of the transfer operator and corresponding estimates controlling the speed of convergence of thermody- namic quantities. In particular this speed of convergence is exponentially rapid as n + . We→ know∞ that when the operator K is compact, symmetric and injective, then there exists a decreasing (in modulus) sequence µj of eigenvalues tending
15Here we shall always consider much more regular kernels which are in particular continu- ous. So the notation can be interpreted in the usual sense. In general, this means that the distribution kernel KR is applied to the test function (x,y) u(x) v(y). V 7→
44 to 0 and a corresponding sequence of eigenfunctions uj which can be normalized in order to get an orthonormal basis of L2(R). Moreover, the operator becomes the limit in norm of the family of operators K(N) whose corresponding kernel are defined by N (N) K (x, y)= µj uj(x) uj (y) . (7.1.6) j=1 X We recall indeed that (N) K K sup µj . (7.1.7) || − || ≤ j>N | | The symmetric operators16 are called trace class if we have in addition the property that K := µ < + (7.1.8) || ||tr | j | ∞ j X In this case, we get that
K K(N) µ . (7.1.9) || − ||tr ≤ | j | j>NX For a trace class symmetric operator, we can in particular define the trace as
Tr K = µj , (7.1.10) j X and this operation is continuous on the space of the trace class operators
Tr K K . (7.1.11) | | ≤ || ||tr This of course extends the usual notion of trace for matrices. We can actually compute directly the trace of a trace-class operator in the following way. We first observe that Tr K = lim Tr K(N) . (7.1.12) N + → ∞ Then we observe that
N (N) (N) Tr K = µj = K (x, x)dx . (7.1.13) R j=1 X Z We consequently obtain that
Tr K = lim K(N)(x, x)dx . (7.1.14) N + R → ∞ Z 16When K is not symmetric, trace class operators can still be defined by considering √K∗K. Note that when K is trace class, one can compute the trace by considering any orthonormal basis e : Tr K = Ke ,e . i ih i ii P
45 If we observe that x µ u (x) 2 is in L1(R), then it is natural17 to hope 7→ j | j || j | (but this is not trivial!) that x K(x, x) is in L1 and that P 7→
Tr K = K(x, x) dx . (7.1.15) R Z Note that it is only when K is positive that the finiteness of the right hand side in (7.1.15) will imply the trace class property. In the case the operator is Hilbert-Schmidt, that is with a kernel K in L2, then the operator is also compact and we have the identity
2 2 µj = K(x, y) dx dy . (7.1.16) R R | | j X Z × This gives an easy criterion for verifying the compactness of the operator. Let us first look at the thermodynamic limit. This means that we are interested in 1 the limit limn + ln Rn exp Φ(X) dX . We start from the decomposition: → ∞ n − R exp Φ(X)= KV (x1, x2) KV (x2, x3) KV (xn 1, xn) KV (xn, x1) (7.1.17) − · · · · − · and we observe that
exp Φ(X) dX = KV,n(y,y) dy (7.1.18) Rn − R Z Z n where KV,n(x, y) is the distribution kernel of (KV) . Our assumption on V n 18 permits to see that (KV ) is trace class and we rewrite (7.1.18) in the form:
n n exp Φ(X) dX = Tr [(KV) ]= µj , (7.1.19) Rn − j Z X where the µj are introduced in (7.1.24). We note also for future use that
n KV,n(x, y)= µj uj (x) uj (y) . (7.1.20) j X 17Let us sketch a proof of (7.1.15) in our particular case. K being positive and with an V (x) 2 V (y) explicit kernel : exp 2 exp x y exp 2 , one can find an Hilbert-Schmidt operator L satisfying−L⋆L = K.−J| The− kernel| of L−is given, for a suitable θ, by L(x,y) = c exp θ x y 2 exp V (y) . We note indeed that c exp θ x y 2 is the distribution kernel θ − | − | − 2 θ − | − | of exp tθ∆ for a suitable tθ > 0. Then we observe that L is Hilbert-Schmidt and that L 2 = µ = Tr K. Using the || ||H.S j j previously mentioned formula for the Hilbert-Schmidt norm and theP property that
K(x,x)= L⋆(x,z) L(z,x) dz = L⋆(z,x) L(z,x) dz , Z Z one obtains (7.1.15). 18See for example [Ro].
46 In particular we get for the thermodynamic limit:
ln Rn exp Φ(X) dX lim − = ln µ1 . (7.1.21) n n →∞ R Moreover the speed of the convergence is easily estimated by:
ln Rn exp Φ(X) dX µ2 ln − ln µ1 = n ln( ) ln k2 + ln n + (exp δ2n) , − | R n − | − µ1 − O − (7.1.22) where k2 is the multiplicity of µ2.
7.1.3 Krein-Rutman’s Theorem. We observe now that the kernel (x, y) K (x, y) satisfies the condition 7→ V K (x, y) > 0 , x, y R . (7.1.23) V ∀ ∈ In particular it satisfies the assumptions of the extended Perron-Frobenius The- orem also called Krein-Rutman’s Theorem and our positive operator KV admits consequently a largest eigenvalue µ1 equal to KV which is simple and corre- sponds to a unique strictly positive normalized|| eigenfunct|| ion which we denote by u1. Let µj the sequence of eigenvalues that we order as a decreasing sequence tending to 0: 0 µ µ .... µ < µ . (7.1.24) ≤ j+1 ≤ j ≤ ≤ 2 1 We shall denote by uj a corresponding orthonormal basis of eigenvectors with K u = µ u , u =1 . (7.1.25) V j j j || j ||
Let us present the statements: Definition 7.1.2. Let A be a bounded positive operator on a Hilbert space = L2(X, dν) where (X, ν) is a measured space. Then we say that A has a strictlyH positive kernel if, for each choice of a non negative function θ ( θ =0), we have ∈ H || || 6 0 < Aθ almost everywhere. It is immediate to see that the transfer operator satisfies this condition. The theorem generalizing the Perron-Frobenius Theorem is then the following: Theorem 7.1.3. Let A be a bounded positive compact symmetric operator on having a strictly positive kernel and let19 A = λ be the largest eigenvalueH of A. Then λ has || || multiplicity 1 and the corresponding eigenvector uλ can be chosen to be a strictly positive function. 19See Proposition 6.4.3
47 Proof: Since A maps real functions into real functions, we may assume that uλ is real. We now prove that Au ,u A u , u . h λ λi ≤ h | λ| | λ|i This is an immediate consequence of the strict positivity of the kernel. We just write: + u = u u− λ λ − λ and + u = u + u− | λ| λ λ and the above inequality is then a consequence of + Au ,u− 0 , h λ λ i≥ and + u , Au− 0 . h λ λ i≥ We then obtain λ u 2 = Au ,u A u , u A u 2 = λ u 2 . || λ|| h λ λi ≤ h | λ| | λ|i ≤ || || || λ|| || λ|| This implies Au ,u = A u , u . h λ λi h | λ| | λ|i This equality means + + u , Au− + u−, Au =0 . h λ λ i h λ λ i + We then get a contradiction unless uλ = 0 or uλ− = 0. We can then assume u 0 and the assumption gives again λ ≥ 0 < θ,Au = λ θ,u , h λi h λi for any positive θ. This gives u 0 a.e . λ ≥ But 1 uλ = λ− Auλ and this gives uλ > 0 a.e .
Finally if there are two linearly independent eigenvectors vλ and uλ correspond- ing to λ, we would obtain the same property for vλ by considering as new Hilbert space the orthogonal of uλ in . But it is impossible to have two orthogonal vectors which are strictly positive.H q.e.d. Remark 7.1.4. In the case t =0, we keep the positivity but lose the injectivity ! The spectrum is easy to determine. We are indeed dealing with the orthonormal projector asso- V (x) ciated to the function x exp 2 . The real number 1 is a simple eigenvalue and 0 is an eigenvalue whose7→ corresponding− eigenspace is infinite dimensional.
48 7.2 The Dirichlet realization, a model of operator with compact resolvent.
1 We can apply Theorem 6.5.1 to the operator ( ∆D + Id)− . We have seen that this operator is compact and it is clearly injective− (by construction). It was also seen as a selfadjoint and positive. Moreover, the norm of this operator is less or equal to 1. There exists consequently a sequence of distinct eigenvalues µn tending to 0 (with 0 < µn 1) and a corresponding orthonormal basis of eigenvectors such ≤1 that (∆D +I)− is diagonalized. If φn,j is a corresponding eigenvector associated with µn, that is, if 1 ( ∆ + I)− φ = µ φ , − D n,j n n,j 1 2 we first observe that φn,j D( ∆D + I); hence φn,j H0 (Ω) H (Ω) (if Ω is relatively compact with regular∈ − boundary) and ∈ ∩ 1 ∆Dφn,j = ( 1)φn,j . − µn −
The function φn,j is consequently an eigenvector of ∆D associated with the 1 − eigenvalue λn = ( 1). µn − Let us show, as easily imagined, that this basis φn,j permits effectively the diagonalization of ∆D. Let us indeed consider− u = u φ in the domain of ∆ . Let us consider n,j n,j n,j − D the scalar product < ∆Du, φm,ℓ > . Using the selfadjoint character of ∆D, we get − P H − < ∆Du, φm,ℓ > =< u, ∆Dφm,ℓ > = λmum,ℓ . − H − H 1 Observing, that D(∆D) = R(S− ), one obtains that the domain of ∆D is characterized by −
D( ∆ )= u L2 λ2 u 2 < + . − D { ∈ | n| n,j| ∞} n,j X Here we have used the property that for any N, we have the identity :
un,jφn,j = S( λnun,j φn,j ) . n N n N X≤ X≤ We have consequently given a meaning to the following diagonalization formula
∆ = λ Π , (7.2.1) − D n En n X where ΠEn is the orthogonal projector on the eigenspace En associated with the eigenvalue λn. Let us remark that it results from the property that the sequence µn tends to 0 the property that λn tend to + . Let us also prove the ∞
49 Lemma 7.2.1. The lowest eigenvalue of the Dirichlet realization of the Laplacian in a relatively compact domain Ω is strictly positive :
λ1 > 0 . (7.2.2) Proof. We know that λ 0; the Dirichlet realization of the Laplacian is indeed posi- 1 ≥ tive. If λ1 = 0, a corresponding normalized eigenvector φ1 would satisfy < ∆φ , φ >=0 , − 1 1 and consequently φ =0 , dans Ω . ∇ 1 This leads first φ1 = Cste in each connected component of Ω but because the 1 trace of φ1 on ∂Ω vanishes (φ1 H0 (Ω)) implies that φ1 = 0. So we get a contradiction. ∈
Finally it results from standard regularity theorems (See [Br], [LiMa]) that the eigenvectors belong (if Ω is regular) to C∞(Ω).
We now show the Proposition 7.2.2. The lowest eigenvalue is simple and one can choose the first eigenfunction to be strictly positive.
1 The natural idea is to apply Krein-Rutman’s Theorem to ( ∆D + 1)− . One has to show that this operator is positivity improving. This− is indeed the case if the domain is connected but the proof will not be given. We will only 1 show that ( ∆D + I)− is positivity preserving and observe that this implies (following the− proof of Krein Rutman) that if Ω is an eigenfunction then Ω is an eigenfunction. | | Lemma 7.2.3. 1 The operator ( ∆ + I)− is positivity preserving. − D The proof is a consequence of the Maximum principle. It is enough to show that ∆u + u = f,γ u = 0 and f 0 a.e , − 0 ≥ implies that u 0 almost everywhere. ≥ We introduce u+ = max u, 0 and u = inf(u, 0). A standard proposition (see − − 1 below) (see Lieb-Loss) shows that u+ and u belong to H0 (Ω). Multiplying by u , we obtain − − u 2 u 2 = fu dx 0 , −||∇ −|| − || −|| − ≥ ZΩ which implies u = 0. −
50 Proposition 7.2.4. 1 n 1 n 1 n Suppose that f Lloc(R ) with f Lloc(R ). Then also f Lloc(R ) and with the notation∈ ∇ ∈ ∇| | ∈ z z , z =0 signz = | | 6 (7.2.3) (0, z =0, we have f (x)= Re sign(f(x)) f(x) almost everywhere . (7.2.4) ∇| | { ∇ } In particular, f f , ∇| | ≤ |∇ | almost everywhere.
Proof of Proposition 7.2.4. n 2 2 Suppose first that u C∞(R ) and define z ǫ = z + ǫ ǫ, for z C and ǫ> 0. We observe that∈ | | | | − ∈ p 0 z ǫ z[ and lim = z . ǫ 0 ≤ | | ≤ | → | | n Then u C∞(R ) and | |ǫ ∈ Re (u u) u ǫ = ∇ . (7.2.5) ∇| | u 2 + ǫ2 | | Let now f be as in the proposition andp define fδ as the convolution f = f ρ δ ∗ δ with ρδ being a standard approximation of the unity for convolution. Explicitly, n we take a ρ C∞(R ) with ∈ 0 ρ 0, ρ(x) dx =1, ≥ Rn Z n n and define ρδ(x) := δ− ρ(x/δ), for x R and δ > 0. ∈ 1 n Then fδ f, fδ f and fδ f in Lloc(R ) as δ 0. → | | → | | ∇ →n ∇ → Take a test function φ C0∞(R ). We may extract a subsequence δk k N ∈ { } ∈ (with δk 0 for k ) such that fδk (x) f(x) for almost every x supp φ. We restrict→ our attention→ ∞ to this subsequence.→ For simplicity of notation∈ we omit the k from the notation and write limδ 0 instead of limk . We now calculate, using dominated convergence→ and (7.2.5),→∞
( φ) f dx = lim ( φ) f ǫ dx ∇ | | ǫ 0 ∇ | | Z → Z = lim lim ( φ) fδ ǫ dx ǫ 0 δ 0 ∇ | | → → Z Re (f f ) = lim lim φ δ∇ δ dx − ǫ 0 δ 0 f 2 + ǫ2 → → Z | δ| p 51 Using the pointwise convergence of fδ(x) and fδ f L1(supp φ) 0, we can take the limit δ 0 and get k∇ − ∇ k → → Re (f f) ( φ) f dx = lim φ ∇ dx. (7.2.6) ∇ | | − ǫ 0 f 2 + ǫ2 Z → Z | | p Now, φ f L1(Rn) and f(x) signf(x) as ǫ 0, so we get (7.2.4) from √ f 2+ǫ2 ∇ ∈ | | → → (7.2.6) by dominated convergence.
We can now look at the property of a positive eigenfunction. The eigenfunc- tion is positive,in C∞(Ω) by a regularity theorem and satisfies
∆u = λu 0 . − ≥ Hence u is superharmonic and (see Lieb-Loss) satisfies the mean value property : For all y Ω , for all R> 0 such that B(y, R) Ω, then ∈ ∈ 1 u(y) u(z) dz . ≥ B(y, R ZB(y,R) Moreover, we know that inf u = 0. Applying this mean value property with y (if any) such that u(y) = 0, we obtain that u = 0 in B(y, R). Using in addition a connectedness argument, we obtain that in a connected open set u is either identically 0 or strictly positive.
+ Let us come back to what appears in the proof of Krein-Rutamn. uλ and uλ− + are either 0 or strictly positive eigenfunctions and we have also uλ , uλ− = 0. + h i Hence uλ or uλ− should vanish.
7.3 Extension to operators with compact resolvent What we have done for the analysis of the Dirichlet realization is indeed quite general. It can be applied to selfadjoint operators, which are bounded from below and with compact resolvent. We show that in this case, there exists an infinite sequence (if the Hilbert space is infinite dimensional) of real eigenvalues λn tending to + such that the cor- responding eigenspaces are mutually orthogonal, of finite∞ dimension and such that their corresponding Hilbertian sum is equal to . Typically, one can apply the method to the NeumannH realization of the Lapla- cian in a relatively compact domain Ω or to the harmonic oscillator in Rm.
7.4 Operators with compact resolvent : the Schr¨odinger operator in an unbounded domain. We just recall some criteria of compactness for the resolvent of the Schr¨odinger operator P = ∆+ V in Rm in connection with the precompactness criterion. −
52 m In the case of the Schr¨odinger equation on R and if V is C∞ and bounded from below, the domain of the selfadjoint extension is always contained in
1 m 1 m 1 2 m Q(P ) := H (R )= u H (R ) (V + C) 2 u L (R ) . V { ∈ | ∈ } Q(P ) is usually called the form domain of the form
u u 2dx + V (x) u(x) 2 dx . 7→ Rm |∇ | Rm | | Z Z 1 2 It is then easy to see that, if V tends to , then the injection of HV in L is compact (using a criterion of precompactness).∞ We then obtain, observing 1 2 1 1 that (P + 1)− is continuous from L into HV , that the resolvent (P + λ)− is compact for λ σ(P ). In the case of a6∈ compact manifold M and if we consider the Laplace-Beltrami operator on M, then the compactness of the resolvent is obtain without addi- tional assumption on V . The domain of the operator is H2(M) and we have compact injection from H2(M) into L2(M). The condition that V as x is not a necessary condition. We can indeed replace it by the→ weaker ∞ | sufficient| → ∞ condition Proposition 7.4.1. 1 m 2 m Let us assume that the injection of HV (R ) into L (R ) is compact then P is with compact resolvent. More concretely, the way to verify this criterion is to show the existence of a continuous function x ρ(x) tending to as x + such that 7→ ∞ | | → ∞ H1 (Rm) L2(Rm) . (7.4.1) V ⊂ ρ Of course, the preceding case corresponds to ρ = V , but, as typical example of this strategy, we shall show in exercise 7.4.3 that the Schr¨odinger operator on R2, ∆+ x2 y2 + 1, is with compact inverse. On the− other· hand, the criterion that V + as x + is not not too far from optimality. → ∞ | | → ∞ We can indeed prove Lemma 7.4.2. Suppose that V 0 and that there exists r > 0 and a sequence σn such that σ + and≥ such that | n| → ∞ sup sup V (x) < + . (7.4.2) n N x B(σn,r) ∞ ∈ ∈ Then ∆+ V +1 is not with compact inverse. − Proof. Let us consider the sequence
φ (x)= ψ(x σ ) . (7.4.3) n − n
53 Here ψ is a compactly supported function of L2 norm 1 and with support in B(0, r). We observe that the φn are an orthogonal sequence (after possibly extracting a subsequence for obtaining that the supports of φn and φn′ are disjoint for n = n′) which satisfies for some constant C 6 2 2 φ 2 + V φ 2 C . (7.4.4) || n||H || n||L ≤ In particular, there exists C such that :
( ∆+ V + 1)φ 2 C , n N . || − n||L ≤ ∀ ∈ But we can not extract from this sequence a strongly convergent sequence in 2 L , because φn is weakly convergent to 0 and φn = 1. So the operator 1 || || ( ∆+ V + I)− can not be a compact operator. −
Exercise 7.4.3. Show that the unbounded operator on L2(R2)
d2 d2 P := + x2y2 , −dx2 − dy2 is with compact resolvent. Hint. One can introduce 1 1 X = ∂ , X = ∂ , X = xy , 1 i x 2 i y 3 and show, for j =1, 2 and for a suitable constant C, the following inequality :
2 2 1 2 2 (x + y + 1)− 4 [X ,X ]u C
i[X1,X3]= y, i[X2,X3]= x , and then
2 2 1 2 2 2 1 2 2 (x + y + 1)− 4 [X1,X3]u + (x + y + 1)− 4 [X2,X3]u + u || 2 2 1 2 || || || || || (x + y + 1) 4 u . ≥ || || 2 2 1 2 For the control of (x + y + 1)− 4 [X ,X ]u , one can remark that || 1 3 || 1 y 2 2 4 2 (x + y + 1)− [X1,X3]u = 1 u (X1 X3 X3 X1)u , || || h(x2 + y2 + 1) 4 | − i perform an integration by parts, and control a commutator.
54 7.5 The Schr¨odinger operator with magnetic field We can consider on Rm the so-called Schr¨odinger operator with magnetic field :
P := ∆ + V , (7.5.1) A,V − A where m 1 ∆ := ( ∂ A (x))2 . (7.5.2) − A i xj − j j=1 X m Here x A~ = (A1(x), , An(x)) is a vector field on R called the “magnetic potential”7→ and V is called· · · the electric potential. It is easy to see that, when m V is semi-bounded, the operator is symmetric and semi-bounded on C0∞(R ). We can therefore consider the Friedrichs extension and analyze the property of this selfadjoint extension. A general question arises if one can get operator of the type PA,V which are with compact resolvent if V = 0. This is the problem which is called the problem of the magnetic bottle. The “heuristical” idea is that the module of the magnetic field can play in some sense the role of the electric potential if it does not oscillate too rapidly (m 2). For defining the magnetic field it is probably easier to consider the magnetic≥ potential as a one-form n σA = Aj (x)dxj . j=1 X The magnetic field is then defined as the two form
ω = dσ = (∂ A ∂ A )dx dx . B A xj k − xk j j ∧ k Xj ω = Bdx dx , B 1 ∧ 2 20 and we can identify ωB with the function x B(x) =curl (A~)(x). The proof is particularly simple in the case when7→ B(x) has a constant sign (say B(x) 0). In this case, we immediately have the inequality ≥ 2 B(x) u(x) dx < ∆Au,u >L2 . (7.5.3) R2 | | ≤ − Z We observe indeed the following identities between operators 1 B(x)= [X ,X ] , ∆ = X2 + X2 . (7.5.4) i 1 2 − A 1 2 Here 1 1 X = ∂ A (x) , X = ∂ A (x) . 1 i x1 − 1 2 i x2 − 2 20Here, for french readers, curl denotes the rotational (in french “rotationnel”) rot . 55 Note also that : ∆ u, u = X u 2 + X u 2 , h− A i || 1 || || 2 || 2 for all u C0∞(R ). To obtain∈ (7.5.3) is then easy through an integration by parts. One can also use, introducing Z = X1 + iX2, the positivity of Z∗Z or ZZ∗. We then easily obtain as in the previous example that the operator is with compact resolvent if B(x) + . As a simple example, one can→ think∞ of A~ = ( x2x , +x x2) , − 1 2 1 2 which gives 2 2 B(x)= x1 + x2 . Note that the case m = 2 is rather particular and it is more difficult to treat m> 2. We have indeed to introduce partition of unity. 7.6 Laplace Beltrami operators on a Riemannian compact manifold If M is a compact riemannian manifold, it is well known that in this case one can canonically define a measure dµM on M and consequently the Hilbertian space L2(M). We have also a canonical definition of the gradient. At each point x of M, we have indeed a scalar product on TxM giving an isomorphism between TxM and Tx∗M. Using this family of isomorphisms we have a natural identification between the C∞-vector fields on M and the 1-forms on M. In this identification, the vector field grad u associated to a C∞ function on M corresponds to the 1-form du. Considering on C∞(M) C∞(M) the sesquilinear form × (u, v) a (u, v) := < grad u(x) , grad v(x) > dµ . 7→ 0 TxM M ZM There is a natural differential operator ∆M called the Laplace-Beltrami oper- ator on M such that − a (u, v)=< ∆ u, v> 2 . 0 − M L (M) In this context, it is not diffficult to define the Friedrichs extension and to get a 2 selfadjoint extension of ∆M as a selfadjoint operator on L (M). The domain is easily characterized as− being H2(M), the Sobolev space naturally associated to L2(M) and one can show that the injection of H2(M) into L2(M) is com- pact because M is compact. The selfadjoint extension of ∆M is with compact resolvent and the general theory can be applied to this examp− le. 56 The case on the circle S1 The simplest model is the operator d2/dθ2 on the circle of radius one whose spectrum is n2 , n N . For n> 0 the− multiplicity is 2. An orthonormal basis { ∈ } 1 is given by the functions θ (2π)− 2 exp inθ for n Z. Here the form domain of the operator is H1,per(S1)7→ and the domain of the operator∈ is H2,per(S1). These spaces have two descriptions. One is to describe these operators as H1,per := u H1(]0, 2π[ u(0) = u(2π) and H2,per := u H2(]0, 2π[ u(0) = { ∈ | } { ∈ | u(2π) , u′(0) = u′(2π) . The other way is to consider} the Fourier coefficients of u. The Fourier coefficients of u Hk,per are in hk. Here ∈ hk := u ℓ2(Z) nku ℓ2(Z) . { n ∈ | n ∈ } It is then easy to prove the compact injection from H1,per in L2(S1) or equiva- lently from h1 into ℓ2. More generally, elliptic symmetric positive operators of order m > 0 admit a selfadjoint extension with compact resolvent. We refer to the book by Berger- Gauduchon-Mazet [BGM] for this central subject in Riemannian geometry. The Laplacian on S2 One can also consider the Laplacian on S2. We describe as usual S2 by the spherical coordinates, with x = cos φ sin θ,y = sin φ sin θ,z = cos θ , with φ [ π, π[ , θ ]0, π[ , (7.6.1) ∈ − ∈ and we add the two poles “North” and “South”, corresponding to the two points (0, 0, 1) and (0, 0, 1). We are looking for− eigenfunctions of the Fiedrichs extension of 1 ∂2 1 ∂ ∂ L2 = sin θ (7.6.2) −sin2 θ ∂φ2 − sin θ ∂θ ∂θ in L2(sin θdθ dφ), satisfying 2 L Yℓm = ℓ(ℓ + 1)Yℓm . (7.6.3) The standard spherical harmonics, corresponding to ℓ 0 and for an integer m ℓ,...,ℓ , are defined by ≥ ∈ {− } 1 1 d ℓ m 2ℓ Y (θ, φ)= c exp imφ ( ) − sin θ , (7.6.4) ℓm ℓ,m sinm θ −sin θ dθ where cℓ,m is an explicit normalization constant. For future extensions, we prefer to take this as a definition for m 0 and then to observe that ≥ Yℓ, m =ˆcℓ,mYℓ,m . (7.6.5) − 57 For ℓ = 0, we get m = 0 and the constant. For ℓ = 1, we obtain, for m = 1, the function (θ, φ) sin θ exp iφ and for m = 1, the function sin θ exp iφ and for m = 0 the function7→ cos θ, which shows that− the multiplicity is 3 for− the eigenvalue 2. To show the completeness it is enough to show that, for given m 0, the ≥ orthogonal family (indexed by ℓ m + N ) of functions θ ψℓ,m(θ) := 1 1 d ℓ m 2ℓ ∈2 { } 7→ m ( ) − sin θ span all L (]0, π[, sin θdθ). sin θ − sin θ dθ For this, we consider χ C∞(]0, π[) and assume that ∈ 0 π χ(θ)ψ (θ) sin θdθ =0 , ℓ m + N . ℓ,m ∀ ∈{ } Z0 We would like to deduce that this implies χ = 0. After a change of variable t = cos θ and an integration by parts, we obtain that this problem is equivalent to the problem to show that, if 1 2 ℓ (ℓ m) ψ(t)((1 t ) ) − (t)dt =0 , ℓ m + N , 1 − ∀ ∈{ } Z− then ψ = 0. 2 m 2 ℓ (ℓ m) Observing that the space spanned by the functions (1 t )− ((1 t ) ) − (which are actually polynomials of exact order ℓ) is the− space of all− polynomials we can conclude the completeness. 58 8 Selfadjoint unbounded operators and spectral theory. 8.1 Introduction We assume that is an Hilbert space. Once we have a selfadjoint operator we can apply the basicH spectral decomposition, which we shall now describe without to give complete proofs. Before to explain the general case, let us come back to the spectral theorem for compact operators T or operators with compact resolvent. This will permit us to introduce a new vocabulary. We have seen that one can obtain a decomposition of in the form H = k NVk , (8.1.1) H ⊕ ∈ such that Tu = λ u , if u V . (8.1.2) k k k k ∈ k Hence we have decomposed into a direct sum of orthogonal subspaces V in H k which the selfadjoint operator T is reduced to multiplication by λk. If Pk denotes the orthogonal projection operator onto Vk, we can write I = Pk , (8.1.3) Xk (the limit is in the strong-convergence sense) and Tu = λ P u , u D(T ) . (8.1.4) k k ∀ ∈ Xk Here we recall the definition : Definition 8.1.1. 2 An operator P ( ) is called an orthogonal projection if P = P ∗ and P = P . ∈ L H If we assume that T is semibounded (with compact resolvent21), we can introduce for any λ R ∈ Gλ = λk λVk , (8.1.5) ⊕ ≤ and Eλ is the orthogonal projection onto Gλ : Eλ = Pk . (8.1.6) λk λ X≤ It is easy to see that the function λ Eλ has values in ( ) and satisfy the following properties : 7→ L H E = E∗; • λ λ E E = E ; • λ · µ inf(λ,µ) 21 −1 Note that if (T λ0) is compact for some λ0 ρ(T ), then it is true for any λ ρ(T ). − ∈ ∈ 59 for all λ, E = E ; • λ+0 λ limλ Eλ =0 , limλ + Eλ = Id . • →−∞ → ∞ E 0 • λ ≥ All the limits above are in the sense of the strong convergence. We also observe that Eλk Eλk 0 = Pk . − − Then in the sense of vectorvalued distributions, we have dE = δ P , (8.1.7) λ λk ⊗ k Xk where δλk is the Dirac measure at the point λk. Hence, in the sense of Stieltjes integrals (this will be explained in more detail below), one can write + ∞ x = dEλ(x) , Z−∞ and + ∞ T = λdEλ . Z−∞ This is in this form that we shall generalize the previous formulas to the case of any selfadjoint operator T . Functional calculus for operators with compact resolvent : If f is a continuous (or piecewise continuous function) one can also define f(T ) as f(T )= f(λ ) P , k · k Xk as an unbounded operator whose domain is D(f(T )) = x f(λ ) 2 x 2 < + , { ∈ H | | k | || k|| ∞} Xk where xk = Pkx. We can also write f(T ) in the form : 2 D(f(T )) = x f(λ) d < Eλx,x > < + . { ∈ H | R | | H ∞} Z 60 Remark 8.1.2. There are, for semibounded operators with compact resolvent, two possible con- ventions for the notation of the eigenvalues. The first one is to classify them into an increasing sequence µ µ j ≤ j+1 counting each eigenvalue according to its multiplicity. The second one is to describe them as a strictly increasing sequence λk with λk eigenvalue of multi- plicity mk. We now present a list of properties which are easy to verify in this particular case and which will be still true in the general case. 1. If f and g coincide on σ(T ), then f(T )= g(T ). For any (x, y) , the support of the measure d E x, y is contained22 in σ(T ). ∈H×H h λ i 2. If f and g are functions on R, f(T )g(T ) = (f g)(T ) . · In particular, if (T z) is invertible, the inverse is given by f(T ) where f − 1 is a continuous function such that f(λ) = (λ z)− , λ σ(T ). − ∀ ∈ 3. If f is bounded, then f(T ) is bounded and we have f(T ) sup f(λ) . (8.1.8) || || ≤ λ σ(T ) | | ∈ 4. The function f may be complex. Note that, in this case, we get f(T )⋆ = f(T ) . (8.1.9) 1 An interesting case is, for z C R, the function λ (λ z)− . Then we get from (8.1.8) ∈ \ 7→ − 1 1 (T z)− ( ) Im z − . (8.1.10) || − ||L H ≤ | | 5. More generally, this works also for z R σ(T ). We then obtain in this case the spectral theorem ∈ \ 1 1 (T z)− ( ) d(z, σ(T ))− . (8.1.11) || − ||L H ≤ 6. If f C∞(R), we have : ∈ 0 ˜ 1 ∂f 1 f(T )= lim ( )(T z)− dx.dy . (8.1.12) + π ǫ 0 Imz >ǫ ∂z¯ − → Z| | 22At the moment, we have not defined σ(T ) when T is unbounded. Think here simply of the set of the eigenvalues ! 61 Here f˜ is defined by : f˜(x, y) = (f(x)+ iyf ′(x))χ(y) , where χ(y) is equal to 1 in a neighborhood of 0 and with compact support. This formula can be proven using the Green-Formula (first prove it with T replaced by the scalar λ) or using that : 1 ∂ = πδ , z¯ z λ (λ,0) − where δ is the Dirac measure at (λ, 0) R2 and ∂ = 1 (∂ + i∂ ) . (λ,0) ∈ z¯ 2 x y One should observe that f˜ is not holomorphic but “almost” holomorphic in the sense that : ∂f/∂˜ z¯ = (y) , O as y 0. → Exercise 8.1.3. Show that, one can also find, for any N 1, f˜N such that in addition ≥ ∂f/∂˜ z¯ = (yN ) , O as y 0. → 8.2 Spectrum. We now come back to the notion of spectrum that we have only met for bounded operators. Definition 8.2.1. The resolvent set of a closed operator T is the set of the λ in C such that the range of (T λ) is equal to and such that (T λ) admits a continuous operator denoted by −R(λ) whose rangeH is included in D−(T ) such that : R(λ)(T λ)= I , − D(T ) and (T λ)R(λ)= I . − H As in the bounded case, we observe that the resolvent set is open. Note also that the continuity of R(λ) is actually a consequence of the property that the graph of R(λ) is closed (using that T is closed) and that R(λ) is defined on . H Definition 8.2.2. The spectrum of a closed operator T is defined as the complementary set in C of the resolvent set. 62 It is then rather easy to show that the spectrum σ(T ) is closed in C. The proof of the fact that the spectrum is contained in R if T is selfadjoint is very close to the bounded case. The spectrum σ(T ) is not empty if T is selfadjoint. 1 The proof is by contradiction. If T has empty spectrum T − is a bounded selfadjoint operator with spectrum equal to 0 . We observe indeed that, for 1 {1 } 1 1 1 λ = 0, the inverse of T − λ is given, if λ− ρ(T ), by λ− T (T λ− )− . 6 1 − ∈ 1 − Hence T − should be the 0 operator, which contradicts T T − = I . This is no longer true in the non selfadjoint case. At the end of the chap◦ ter, we give some example appearing naturally in various questions in Fluid Mechanics. 8.3 Spectral family and resolution of the identity. Definition 8.3.1. A family of orthogonal projectors E(λ) (or Eλ), <λ< in an Hilbert space is called a resolution of the identity (or−∞ spectral family)∞ if it satisfies the followingH conditions : • E(λ)E(µ)= E(min(λ, µ)) , (8.3.1) • E( )=0 , E(+ )= I , (8.3.2) −∞ ∞ where E( ) is defined 23 by ±∞ E( )x = lim E(λ)x (8.3.3) ±∞ λ →±∞ for all x in , H • E(λ +0)= E(λ) , (8.3.4) where E(λ + 0) is defined by E(λ + 0)x = lim E(µ)x . (8.3.5) µ λ , µ>λ → Remark 8.3.2. We have shown an example of such a family in the previous subsection. Proposition 8.3.3. Let E(λ) be a resolution of identity (=spectral family); then for all x, y , the function ∈ H λ < E(λ)x,y > (8.3.6) 7→ is a function of bounded variation whose total variation24 satisfies V (x, y) x y , x, y . (8.3.7) ≤ || || · || || ∀ ∈ H 23(8.3.1) gives the existence of the limit (cf also Lemma 8.3.4). The limit in (8.3.3) is taken in . We observe indeed that λ E(λ)x,x = E(λ)x 2 is monotonically increasing. 24HSee the definition in (8.3.9) 7→ h i || || 63 Proof. Let λ < λ < < λ . We first get from the assumption (8.3.1) that 1 2 · · · n E = E E ]α,β] β − α is an orthogonal projection. From the Cauchy-Schwarz inequality we have n n j=2 < E]λj−1,λj ]x,y > = j=2 < E]λj−1 ,λj ]x, E]λj−1 ,λj ]y > | | n | | j=2 E]λj−1 ,λj ]x E]λj−1 ,λj ]y] P ≤ P || || || 1 || 1 n 2 2 n 2 2 P j=2 E]λj−1 ,λj ]x j=2 E]λj−1 ,λj ]y ≤ || 1 || 1 || || 2 2 2 2 = PE]λ ,λ ]x E]λ ,λ ]Py . || 1 n || || 1 n || But for m>n, we get m 1 − x 2 E x 2 = E x 2 . (8.3.8) || || ≥ || ]λn,λm] || || ]λi,λi+1] || i=n X We finally obtain that, for any finite sequence λ < λ < < λ , we have 1 2 · · · n n < E x,y > x y . | ]λj−1 ,λj ] | ≤ || || · || || j=2 X This shows the bounded variation and the estimate of the total variation defined as n V (x, y) := sup < E x,y > . (8.3.9) | ]λj−1,λj ] | λ1, ,λn j=2 ··· X Hence we have shown that, for all x and y in , the function λ E(λ)x, y is with bounded variation and we can then showH the existence of 7→E(λ h + 0) andi E(λ 0). This is the object of − Lemma 8.3.4. If E(λ) is a family of projectors satisfying (8.3.1) and (8.3.2), then, for all λ R, the operators ∈ Eλ+0 = lim E(µ) , Eλ 0 = lim E(µ) , (8.3.10) µ λ µ>λ − µ λ µ<λ → → are well defined when considering the limit for the strong convergence topology. Proof. Let us show the existence of the left limit. From (8.3.8), we get that, for any ǫ> 0, there exists λ < λ such that, λ′, λ′′ [λ , λ[, such that λ′ < λ′′ 0 ∀ ∀ ∈ 0 2 E ′ ′′ x ǫ . || ]λ ,λ ] || ≤ It is then easy to show that E 1 x is a Cauchy sequence converging to a limit λ n and that this limit does not depend− on the choice of the sequence tending to λ. 64 The proof of the existence of the limit from the right is the same. This ends the proof of the lemma. It is then classical (Stieltjes integrals) that one can define for any continuous b 25 complex valued function λ f(λ) the integrals a f(λ)d E(λ)x, y as a limit of Riemann sums. 7→ h i R Proposition 8.3.5. Let f be a continuous function on R with complex values and let x . Then it is possible to define for α<β, the integral ∈ H β f(λ) dEλx Zα as the strong limit in of the Riemann sum : H f(λ′ )(E E )x , (8.3.11) j λj+1 − λj j X where α = λ < λ < < λ = β , 1 2 · · · n and λ′ ]λ , λ ] , j ∈ j j+1 when max λ λ 0. j | j+1 − j | → The proof is easy using the uniform continuity of f. Note also that the no- tation could be misleading. May be ]α,β] f(λ)dEλx is less ambiguous. We now arrive like in the standardR theory to the generalized integral. Definition 8.3.6. For any given x and any continuous function f on R, the integral : ∈ H + ∞ f(λ)dEλx Z−∞ β is defined as the strong limit in , if it exists of α f(λ)dEλx when α and β + . H → −∞ → ∞ R Remark 8.3.7. The theory works more generally for any borelian function (cf Reed-Simon, Vol. 1 [RS-I]). This can be important, because we are in particular interested in the case when f(t)=1] ,λ](t). One possibility for the reader−∞ who wants to understand how this can be made is to look at Rudin’s book [Ru1], which gives the following theorem (Theorem 8.14, p. 173) 25The best is to first consider the case when x = y and then use a depolarisation formula, in the same way that, when we have an Hilbertian norm, we can recover the scalar product from the norm. 65 Theorem 8.3.8. . 1. If µ is a complex Borel measure on R and if (⋆) f(x)= µ(] , x]) , x R , − ∞ ∀ ∈ then f is a normalized function with bounded variation (NBV). By NBV we mean, with bounded variation, but also continuous from the right and such that limx f(x)=0. →−∞ 2. Conversely, to every f NBV , there corresponds a unique complex Borel measure µ such that (⋆∈) is satisfied. Theorem 8.3.9. For x given in and if f is a complex valued continuous function on R, the following conditionsH are equivalent + • ∞ f(λ)dEλx exists ; (8.3.12) Z−∞ + • ∞ f(λ) 2d E x 2 < + ; (8.3.13) | | || λ || ∞ Z−∞ + • ∞ y f(λ)d(< Eλy,x > ) (8.3.14) 7→ H Z−∞ is a continuous linear form. Hint for the proof. a) (8.3.12) implies (8.3.14) essentially by using repeatedly the Banach-Steinhaus Theorem (also called Uniform Boundedness Theorem) and the definition of the integral. b) Let us prove that (8.3.14) implies (8.3.13). Let F be the linear form appearing in (8.3.14). If we introduce β y = f(λ)dEλx , Zα then we first observe (coming back to the Riemann integrals) that y = E]α,β]y . 66 It is then not too difficult to show that + F (y) = ∞ f(λ)d < Eλx,y > −∞+ = ∞ f(λ)d < Eλx, E]α,β]y > +R−∞ ∞ f(λ)d < E]α,β]Eλx,y > −∞R = β f(λ)d < E x,y > R α λ = y 2 . ||R || Using (8.3.14), we get y 2 F y and consequently || || ≤ || || · || || y F . (8.3.15) || || ≤ || || Here we observe that the r.h.s. is independent of α and β. On the other hand, coming back to Riemann sums, we get β y 2 = f(λ) 2d E x 2 . || || | | || λ || Zα We finally obtain β f(λ) 2d E x 2 F 2 . (8.3.16) | | || λ || ≤ || || Zα Hence, taking the limits α and β + , we obtain (8.3.13). → −∞ → ∞ c) For the last implication, it is enough to observe that, for α′ <α<β<β′, we have β′ β α β′ 2 2 2 2 2 f(λ)dEλx f(λ)dEλx = f(λ) d Eλx + f(λ) d Eλx . || ′ − || ′ | | || || | | || || Zα Zα Zα Zβ Theorem 8.3.10. Let λ f(λ) be a real-valued continuous function. Let 7→ + ∞ D := x , f(λ) 2 d E(λ)x, x < , . f { ∈ H | | h i ∞ } Z−∞ Then D is dense in and we can define T whose domain is defined by f H f D(Tf )= Df , and + ∞ T x, y = f(λ)d E(λ)x, y h f i h i Z−∞ for all x in D(T ) and y in . f H The operator Tf is selfadjoint. Finally, we have Tf Eλ is an extension of EλTf . (8.3.17) 67 Proof of the theorem. From property (8.3.2), we obtain that, for any y in , there exists a sequence H (αn,βn) such that E]αn,βn]y y as n + . But E y belongs to D , for→ any α, β→, and∞ this shows the density of D in . ]α,β] f f H We now observe that f being real and Eλ being symmetric the symmetry is clear. We observe that, for f0 = 1, we have Tf0 = I and for f1(λ)= λ, we obtain a selfadjoint operator Tf1 := T . In this case we say that + ∞ T = λ dE(λ) Z−∞ is a spectral decomposition of T and we shall note that + + ∞ ∞ T x 2 = λ2d E(λ)x, x = λ2d E(λ)x 2 , || || h i || || Z−∞ Z−∞ for x D(T ). More∈ generally + + ∞ ∞ T x 2 = f(λ) 2d((E(λ)x, x)) = f(λ) 2d( E(λ)x 2) || f || | | | | || || Z−∞ Z−∞ for x D(T ). ∈ f Conclusion. We have consequently seen in this subsection how one can as- sociate to a spectral family of projectors a selfadjoint operator. We have seen in the introduction that the converse was true for a compact operator or an operator with compact resolvent. It remains to prove that this is true in the general case. 8.4 The spectral decomposition Theorem. The spectral decomposition Theorem makes explicit that the preceding situation is actually the general one. Theorem 8.4.1. Any selfadjoint operator T in an Hilbert space admits a spectral decomposition such that H 68 the book of Davies [Da]. Step 1. It is rather natural to imagine that it is essentially enough to treat the case when T is a bounded selfadjoint operator (or at least a normal bounded operator, that is satisfying T ∗T = TT ∗). If A is indeed a general semibounded selfadjoint op- 1 erator, one can come back to the bounded case by considering (A + λ0)− , with 26 λ0 real, which is bounded and selfadjoint. In the general case , one can con- 1 sider (A + i)− . Step 2. We analyze first the spectrum of P (T ) where P is a polynomial. Lemma 8.4.2. . If P is a polynomial, then σ(P (T )) = P (λ) λ σ(T ) (8.4.2) { | ∈ } Proof. We start from the identity P (x) P (λ) = (x λ)Q (x) and from the corre- − − λ sponding identity between bounded operators P (T ) P (λ) = (T λ)Qλ(T ). This permits to construct the inverse of (T λ) if− one knows the− inverse of P (T ) P (λ). − − Conversely, we observe that, if z C and if λj (z) are the roots of λ (P (λ) z), then we can write : ∈ 7→ − (P (T ) z)= c (T λ (z)) . − − j j Y This permits to construct the inverse of (P (T ) z) if one has the inverses of (T λ (z)) (for all j). − − j Lemma 8.4.3. Let T be a bounded self-adjoint operator. Then P (T ) = sup P (λ) . (8.4.3) || || λ σ(T ) | | ∈ We first observe that 2 P (T ) = P (T )∗P (T ) || || || || This is the consequence of the general property for bounded linear operators that : 2 A∗A = A . || || || || 26Here we recall that an example of operator which is not semibounded is given in Exer- cise 4.1.3 69 We recall that the proof is obtained by observing first that : A∗A = sup x 1, y 1 A∗Ax, y || || || ||≤ || ||≤ |h i| = supx,y x 1, y 1 Ax, Ay || ||≤ || ||≤ |h i A 2 , ≤ || || and secondly that : 2 A = sup Ax, Ax = sup A∗Ax, x A∗A . || || x 1h i x 1h i ≤ || || || ||≤ || ||≤ We then observe that : P (T ) 2 = (P¯ P )(T ) || || || || = supµ σ(P¯ P )(T ) µ (using Theorem 6.4.1) ∈ ¯ | | = supλ σ(T ) (P P )(λ) (using Lemma 8.4.2) ∈ | | 2 = supλ σ(T ) P (λ) . ∈ | | Step 3. We have defined a map Φ from the set of polynomials into ( ) defined by L H P Φ(P )= P (T ) , (8.4.4) 7→ which is continuous Φ(P ) ( ) = sup P (λ) . (8.4.5) || ||L H λ σ(T ) | | ∈ The set σ(T ) is a compact in R and using the Stone-Weierstrass theorem (which states the density of the polynomials in C0(σ(T ))), the map Φ can be uniquely extended to C0(σ(T )). We still denote by Φ this extension. The properties of Φ are described in the following theorem Theorem 8.4.4. Let T be as selfadjoint continuous operator on . Then there exists a unique map Φ from C0(σ(T )) into ( ) with the followingH properties : L H 1. Φ(f + g)=Φ(f)+Φ(g) , Φ(λf)= λΦ(f) ; Φ(1) = Id , Φ(f¯)=Φ(f)∗ ; Φ(fg)=Φ(f) Φ(g) . ◦ 2. Φ(f) = sup f(λ) . || || λ σ(T ) | | ∈ 3. If f is defined by f(λ)= λ, then Φ(f)= T . 4. σ(Φ(f)) = f(λ) λ σ(T ) . { | ∈ } 5. If ψ satisfies T ψ = λψ, then Φ(f)ψ = f(λ)ψ. 70 6. If f 0, then Φ(f) 0. ≥ ≥ All these properties are obtained by showing first the properties for polyno- mials P and then extending the properties by continuity to continuous functions. For the last item, note that : Φ(f)=Φ( f) Φ( f)=Φ( f)∗ Φ( f) . · · Step 4. p p p p We are now ready to introduce the measures. Let ψ . Then ∈ H f < ψ,f(T )ψ> =< ψ, Φ(f)ψ> 7→ H H is a positive linear functional on C0(σ(T )). By measure theory (Riesz Theorem) (cf Rudin [Ru1]), there exists a unique measure µψ on σ(T ), such that < ψ,f(T )ψ> = f(λ)dµψ(λ) . (8.4.6) H Zσ(T ) This measure is called the spectral measure associated with the vector ψ . This measure is a Borel measure. This means that we can extend the map∈ H Φ and (8.4.6) to Borelian functions. Using the standard Hilbert calculus (that is the link between sesquilinear form and quadratic forms) we can also construct for any x and y in a complex H measure dµx,y such that < x, Φ(f)y > = f(λ)dµx,y(λ) . (8.4.7) H Zσ(T ) Using the Riesz representation Theorem (Theorem 3.1.1), this gives as, when f is bounded, an operator f(T ). If f =1] ,µ], we recover the operator Eµ = f(T ) which permits to construct indeed the−∞ spectral family announced in Theorem 8.4.1. Remark 8.4.5. Modulo some care concerning the domains of the operator, the properties men- tioned at the first subsection of this section for operators with compact resolvent are preserved in the case of an unbounded selfadjoint operator. 8.5 Applications of the spectral theorem: One of the first applications of the spectral theorem (Property 2.) is the follow- ing property : Proposition 8.5.1. d(λ, σ(T )) x (T λ)x , (8.5.1) || || ≤ || − || for all x in D(T ). 71 This proposition is frequently used in the following context. Except very special cases like the harmonic oscillator, it is usually difficult to get explicitely the values of the eigenvalues of an operator. One consequently tries to localize these eigenvalues by using approximations. Let us suppose for example that one has found λ0 and y in D(L) such that (T λ )y ǫ (8.5.2) || − 0 || ≤ and y = 1 then we deduce the existence of λ in the spectrum of L such that || || λ λ0 ǫ. |Standard− |≤ examples are the case of hermitian matrices or the case of the anhar- 2 d2 2 4 monic oscillator T := h dx2 +x +x . In the second case the first eigenfunction − 2 d2 2 of the harmonic oscillator h dx2 + x can be used as approximate eigenfunc- − 2 tion y in (8.5.2) with λ0 = h. We then find (8.5.2) with ǫ = (h ). Another application is, using the property that the spectrumO is real, the follow- ing inequality Im λ x (T λ)x (8.5.3) | | || || ≤ || − || 1 and this gives an upper bound on the norm of (T λ)− in ( ) by 1/ Im λ . 2 −2 2 L4H | | One can also consider the operator Tǫ = d /dx + x + ǫx . One can show that near each eigenvalue of the harmonic− oscillator (2n + 1), then there exists, when ǫ> 0 is small enough, an eigenvalue λn(ǫ) of Tǫ. Another good example to analyze is the construction of a sequence of approx- imate eigenvectors considered in Subsection 1.1. From the construction of un such that, with T = ∆, − 2 1 (T ξ )u 2 Rm = ( ) , || − n||L ( ) O n one obtains that C d(σ(T ), ξ2) , n N . ≤ n ∀ ∈ As n + , we get ξ2 σ(T ). It is then→ easy∞ to show that∈ σ(P )=[0, + [. ∞ It is enough to prove indeed, using the Fourier transform, that, for any b > 0, 1 2 2 ( ∆+ b) has an inverse ( ∆+ b)− sending L onto H . − Here we have followed− in a particular case the proof of the following general theorem Theorem 8.5.2. Let T be a selfadjoint operator. Then λ σ(P ) if and only if there exists a ∈ sequence (un)n N, un D(T ) such that un = 1 and (T λ)un 0 as n + . ∈ ∈ || || || − || → → ∞ Exercise 8.5.3. Show the “only if”, after reading of the proof of Proposition 8.5.4. 72 Note also the following characterization. Proposition 8.5.4. σ(T )= λ R, s.t. ǫ> 0, E(]λ ǫ, λ + ǫ[) =0 (8.5.4) { ∈ ∀ − 6 } The proof uses in one direction the explicit construction of (T λ) through − Proposition 8.5.1. If λ and ǫ0 > 0 are such E]λ ǫ0,λ+ǫ0[ = 0. Then there exists − 1 a continous function f on R, such that f(t) = (t λ)− on the support of the − measure dEλ. This permits to construct the inverse and to show that λ is in the resolvent set of T . Conversely, let λ in the set defined by the r.h.s of (8.5.4) later denoted byσ ˜(T ). 1 1 For any n N∗, let us take x such that x = 1 and E(]λ , λ+ [)x = x . ∈ n || n|| − n n n n Using property 2., with the function fn(t) = (t λ)1]λ 1 ,λ+ 1 [(t), we get : − − n n 1 1 1 1 (T λ)x = (T λ)E(]λ , λ + [)x x = . || − n|| || − − n n n|| ≤ n|| n|| n Applying Proposition 8.5.1, we obtain : σ˜(T ) σ(T ) . ⊂ 8.6 Examples of functions of a selfadjoint operator We shall for example meet usually in spectral theory the functions 1. f is the characteristic function of ] , λ], Y] ,λ] ; Φ(f)= f(T ) is then Φ(f)= E(λ). − ∞ −∞ 2. f is the characteristic function of ] , λ[, Y] ,λ[ ; f(T ) is then Φ(f)= E(λ 0). − ∞ −∞ − 3. f is a compactly supported continuous function. f(T ) will be an operator whose spectrum is localized in the support of f. 4. ft(λ) = exp(itλ) with t real. ft(T ) is then a solution of the functional equation (∂t iT )(f(t,T )) =0 − f(0,T ) = Id We note here that, for all real t, ft(T ) = exp(itT ) is a bounded unitary operator . 5. gt(λ) = exp( tλ) with t real positive. gt(T ) is then a solution of the functional equation− (∂t + T )(g(t,T )) =0 , for t 0 g(0,T ) = Id . ≥ 73 We have discussed in the introduction the case of an operator with compact resolvent. The other case to understand for the beginner is of course the case of the free Laplacian ∆ on Rn. Using the Fourier transform , we get as unbounded operator the− operator of multiplication by ξ2. It isF not difficult to define directly the functional calculus which simply becomes for a borelian function φ : 1 2 φ( ∆) = − φ(ξ ) . (8.6.1) − F F 1 One possibility is to start from ( ∆+1)− , for which this formula is true and to then follow what was our construction− of the functional calculus. Another possibility is to use the Formula (8.1.12) and to use that (8.6.1) is satisfied for 1 ( ∆+ z)− , with z C R. − The spectral family∈ is\ then defined by ˆ < E(λ)f,g >L2(Rm)= f(ξ) gˆ(ξ) dξ . ξ2 λ · Z ≤ 8.7 Spectrum and spectral measures Another interest of the spectral theorem is to permit the study of the different properties of the spectrum according to the nature of the spectral measure and this leads to the definition of the continuous spectrum and of the pure point spectrum. Let us briefly discuss (without proof) these notions. Starting of a selfadjoint operator T , one defines pp (pure point subspace) as the set defined as H pp = ψ f µ(X)= µ( x ) . (8.7.2) { } x X X∈ One can verify that H is a closed subspace of and that the corresponding pp H orthogonal projection Π pp satisfies H Π pp D(T ) D(T ) . H ⊂ In this case T/ pp is naturally defined as unbounded operator on pp and one defines the pureH point spectrum of T by H σpp(T )= σ(T/ pp ) . (8.7.3) H We can similarly define (continuous subspace) as the set defined as Hc c = ψ f µ( x ))=0 , x X . (8.7.5) { } ∀ ∈ 74 One can verify that H is a closed subspace of and that the corresponding c H orthogonal projection Π c satisfies H Π c D(T ) D(T ) . H ⊂ Moreover, it can be shown (See [RS-I]), that = . (8.7.6) H Hpp ⊕ Hc In this case T/ c is naturally defined as unbounded operator on c and one defines the continuousH spectrum of T by H σc(T )= σ(T/ c ) . (8.7.7) H Example 8.7.1. The spectrum of ∆ is continuous. − We observe indeed that lim fˆ(ξ) 2dξ =0 , f L2(Rm) . ǫ 0 , ǫ>0 ξ 2 λ ǫ | | ∀ ∈ → Z|| | − |≤ One can still refine this discussion by using the natural decomposition of the measure given by the Radon-Nikodym Theorem. This leads to the notion of absolutely continuous spectrum and of singularly continuous spectrum. 75 9 Non-self adjoint operators and ǫ-pseudospectrum When the operators are not selfadjoint, one should think that the spectrum is not the right object because it becomes very unstable by perturbation. It has been realized in the recent years that a family of sets (parametrized by ǫ > 0) in C called the ǫ-pseudospectrum is the right object for getting this stability. 9.1 Main definitions and properties Here we follow Chapter 4 in the book by L.N. Trefethen and M. Embree [TrEm]. Definition 9.1.1. If A is a closed operator with dense domain D(A) in an Hilbert space , the H ǫ-pseudospectrum σǫ(A) of A is defined by 1 1 σ (A) := z C (zI A)− > . ǫ { ∈ | || − || ǫ } Remark 9.1.2. In one part of the literature, > is replaced by in the above definition. We have chosen the definition which leads to the simplest≥ equivalent definitions. 1 We take the convention that (zI A)− = + if z σ(A), where σ(A) denotes the spectrum of A, so it|| is clear− that|| we always∞ have∈ : σ(A) σ (A) . ⊂ ǫ When A is selfadjoint (or more generally normal), σǫ(A) is, by the Spectral Theorem, given by, σ (A)= z C d(z, σ(A)) <ǫ . ǫ { ∈ | } So this is only in the case of non self-adjoint operators that this new concept (first appearing in numerical analysis, see Trefethen [Tr1, Tr2]) becomes interesting. Although formulated in a rather abstract way, the following (easy versionof a) result by Roch-Silbermann [RoSi] explains rather well to what corresponds the ǫ-pseudospectrum : Proposition 9.1.3. σǫ(A)= σ(A + δA) . δA ( ) s. t. δA L H <ǫ { ∈L H [ || || ( ) } In other words, z is in the ǫ-pseudospectrum of A if z is in the spectrum of some perturbation A + δA of A with δA <ǫ. This is indeed a natural notion thinking of the fact that the models we|| are|| analyzing are only approximations of the real problem and of the fact that the numerical analysis of the model goes through the analysis of explicitly computable approximated problems. Numer- ical examples are treated in [Tr2]. 76 Proof Let us show the easy part of this characterization of the ǫ-pseudospectrum. If 1 1 (z A)− ǫ , it is clear that for any δA such that δA <ǫ, (A + δA z) is|| invertible.− || ≤ Its inverse is obtined by observing that || || − 1 1 (z A)− (z A δA)= I (z A)− δA . − − − − − But the left hand side is invertible because 1 1 (z A)− δA (z A)− δA < 1 . || − || ≤ || − || || || The inverse is consequently given by 1 1 j 1 (z A δA)− = ((z A)− δA) (z A)− . − − − − j X 1 1 The converse is not very difficult. If (z A)− > ǫ , by definition of the norm, there exists u such that u =|| 1 and− || ∈ H || || 1 1 (z A)− u = µ> . || − || ǫ 1 Let v = (z A)− u. Let (δA) the linear bounded operator such that − 2 (δA)x = µ− u v x , x . − h | i ∀ ∈ H It is clear that v is an eigenvector of A + δA associated to z and that δA = 1 || || µ <ǫ. Hence we have found a perturbation A + δA of A such that z σ(A + δA) and δA <ǫ. ∈ || Another|| presentation for defining the ǫ-pseudospectrum is to say that z ∈ σǫ(A) iff z σ(A) or if there exists an ǫ-pseudoeigenvector that is u D(A) such that u∈ = 1 and (z A)u <ǫ. ∈ || || || − || Theorem 9.1.4 (ǫ-Pseudospectrum of the adjoint). For any closed densely defined operator A and any ǫ> 0, we have σǫ(A∗)= σǫ(A) , where for a subset Σ in C we denote by Σ the set Σ= z C z¯ Σ . { ∈ | ∈ } Proof This is immediate using that, if z σ(A), 6∈ 1 1 (z A)− = (¯z A∗)− . k − k k − k 77 9.2 ǫ-Pseudospectrum : complete analysis of the differen- tiation operator. We consider the operator A defined on L2(]0, 1[) by D(A)= u H1(]0, 1[) , u(1) = 0 , { ∈ } and Au = u′ , u D(A) . ∀ ∈ This is clearly a closed operator with dense domain. The adjoint of A is defined on L2(]0, 1[) by 1 D(A∗)= u H (]0, 1[) , u(0) = 0 , { ∈ } and Au = u′ , u D(A∗) . − ∀ ∈ Lemma 9.2.1. σ(A)= and A has compact resolvent. ∅ First we can observe that (A z) is injective on D(A) for any z C. Then one easily verifies that for any z − C, the inverse is given by ∈ ∈ 1 1 (z A)− f = exp z(x s) f(s) ds . − − Zx It is also clear that this operator is compact. To analyze the ǫ-pseudospectrum is more interesting. For this we need to estimate 1 ψ(z) := (z A)− . k − k The first remark is that ψ depends only on Re z. For this we can observe that the map u exp iαxu is a unitary transform on L2(]0, 1[), which maps D(A) onto D(A).7→ The main result is the following Theorem 9.2.2. The function ψ is a subharmonic function which depends only of Re z and satisfies 1 ψ(z) , for Re z > 0 , (9.2.1) ≤ Re z and exp Re z 1 ψ(z)= − + ( ) , for Re z < 0 . (9.2.2) − 2 Re z O Re z | | This implies for the ǫ-pseudospectrum of A : 78 Corollary 9.2.3. For ǫ> 0, the ǫ-pseudospectrum of A is an hyperplane of the form σ (A)= z C Rez < c (9.2.3) ǫ { ∈ | ǫ} with (ln ǫ) as ǫ , 0 cǫ → (9.2.4) ∼ ǫ as ǫ + . → ∞ Rough estimate from below for Re z < 0 A first (non optimal) step is to think27 “semi-classical”. Let us take z real and let us consider 1 x φ (x) := 2z 2 exp zx . 7→ z | | This function is not in Ker (z A) because it does not satisfies the boundary condition. But when z −, the boundary condition at x = 1 is “almost → −∞ satisfied”. Actually φz lives very close to 0. Moreover, for z < 0, φ 2 =1 exp2z k zk − Hence the norm tends to 1 as z . → −∞ Let us indeed consider for η > 0 a cut-off function χη such that χη = 1 on [0, 1 η] and χη =0 on [1 η , 1] and let us introduce − − 2 φz,η(x)= χηφz . We now observe that φ D(A) and that z,η ∈ (z A)φ (x)= χ′ φ . − z,η − η z The L2 norm of the right hand side is exponentially small like exp(1 η)z. − This shows that, for any η> 0 there exists zη < 0 such that 1 exp (1 η)z ψ(z) , for z This is not as good as in the statement of the theorem but this suggests a rather general point of view. Rough estimate from above for Re z > 0 Here we will try to estimate ψ(z) from above by using an a priori estimate for (A z) (with z real). For−u D(A), we have ∈ 1 2 (A z)u u = z u + u′(t)¯u(t)dt . h − | i − || || Z0 27The semi-classical parameter is h = 1 Re z 79 But by an integration by parts, we observe that 1 1 2 u′(t)¯u(t) dt = u(t)¯u′(t)dt u(0) . − − | | Z0 Z0 Coming back to the previous equality and taking the real part, we obtain 1 Re (A z)u u = z u 2 + u(0) 2 z u 2 . − h − | i k k 2| | ≥ k k Then we obtain (A z)u z u , u D(A) , (9.2.6) || − || ≥ || || ∀ ∈ which implies (9.2.1). Schur’s Lemma 1 The operator (A z)− being an operator associated to an integral kernel, one can analyze what− is given by Schur’s lemma or by the Hilbert-Schmidt criterion. The kernel is defined by 0 for s < x K(x, s)= exp z(x s) for x According to Schur’s Lemma, we have to consider supx K(x, s)ds and sups K(x, s)dx. If we are interested with the Hilbert-Schmidt norm, we have to compute K(x, s)2dxds. All these computations can be done rather explicitelyR ! R For z = 0, we have R R 6 1 sup K(x, s)ds = (1 exp z) , z − − x Z and 1 sup K(x, s)dx = (1 exp z) z − − s Z This gives 1 ψ(z) (1 exp z) . (9.2.7) ≤ z − − This is actually an improved version of (9.2.1) for z > 0 and for z < 0, it is better to write it in the form 1 ψ(z) − (exp z 1) , (9.2.8) ≤ z − − and to compare it to the lower bound obtained in (9.2.5). 80 A more accurate estimate for z < 0 1 We can rewrite (z A)− in the form − 1 (z A)− = R R , − 1 − 2 with 1 R v(x) := exp z(x s) v(s), ds 1 − Z0 and x R v(x) := exp z(x s) v(s), ds 2 − Z0 Observing that R2∗ can be treated as for the proof of (9.2.7), we first obtain 1 R . || 2||≤− Re z It remains to control R1 . This norm can be computed explicitly. We have indeed || || 1 R v = exp zx exp zs v(s) ds || 1 || || ||| − | Z0 Hence we have just to compute the norm of the linear form 1 v exp zs v(s) ds 7→ − Z0 which is the L2-norm of s exp zs. This gives 7→ − 1 2z 2z 1 z 2z R = exp zx exp zx = (1 e )(e 1) = e− (1 e ) . || 1|| || ||·|| − || −2z − − − −2z − p Combining the estimates of R and R leads to the upper bound for (9.2.2). || 1|| || 2|| Remark 9.2.4. One can discretize the preceding problem by considering, for n N∗, the matrix A = nA with A = I + J where J is the n n matrix such that∈ J = δ . n 1 1 × i,j i+1,j One can observe that the spectrum of An is n. It is also interesting to analyze the pseudospectrum. − Remark 9.2.5. There is a semi-classical version of the pseudospectrum for families Ah. One can then relate the ǫ appearing in the definition of the ǫ-pseudospectrum with the parameter h (which could typically be in ]0,h0]). For example, we can consider ǫ(h)= hN . Exercise 9.2.6. d Analyze the pseudospectrum of dθ + λg(θ) on the circle. 81 9.3 Another example of non selfadjoint operator without spectrum We consider the spectrum of the operator d A = + x2 , dx on the line. We can take as domain D(A) the space of the u L2(R) such that Au L2(R). ∈ ∈ We note that C0∞(R) is dense for the graph norm in D(A). Hence A is the d 2 closed extension of the differential operator dx + x with domain C0∞(R). We note that the operator is not selfadjoint. The adjoint is d + x2. − dx The two following inequalities can be useful. Re Au u xu 2 0 . (9.3.1) h | i ≥ || || ≥ This inequality is first proved for u C∞(R) and then extended to u D(A) ∈ 0 ∈ using the density of C0∞(R). This implies that (D(A) has continuous injection 2 in the weighted space Lρ with ρ(x)= x . Together with the fact that D(A) 1 | | ⊂ Hloc(R), this implies that (A + I) is invertible and that the inverse is compact. The second inequality is 2 2 2 2 2 x u + u′ C Au + u . (9.3.2) || || || || ≤ || || || || To prove it, we observe that 2 2 2 2 2 Au = u′ + x u 2 Re u′ , x u . || || || || || || − h i Using an integration by parts, we get 2 2 Re u′ , x u =2 xu, u . h i h i Hence we get (using Cauchy-Schwarz) 2 2 2 2 Au u′ + x u 2 xu u . || || ≥ || || || || − || |||| || One can then use the consequence of our first inequality (9.3.1) 1 xu 2 Au 2 + u 2 || || ≤ 2 || || || || to get the conclusion. (9.3.2) permits to obtain that D(A)= u H1(R) , x2u L2(R) , (9.3.3) { ∈ ∈ } which is not obvious at all. 82 One will show that A has an inverse, that the inverse is compact but that its spectrum is empty. We consider on R, the differential equation 2 u′ + x u = f . (9.3.4) For all f L2(R), let us show that there exists a unique solution in L2(R) de (9.3.4). An∈ elementary calculus gives 1 x 1 u(x) = exp x3 exp y3f(y) dy . −3 3 Z−∞ One has to work a little for showing that u L2 (this is easier if f is compactly supported). If we denote by K the operator∈ which associates to f the solution u, the distribution kernel of K is given by 0 if y x K(x, y)= . exp 1 (y3 x3) if y <≥ x 3 − We note that if there exists an eigenvalue of K λ = 0 and if u is a corresponding 6 λ eigenfunction, then uλ satisfies 2 1 u′ + x u = u . (9.3.5) λ λ λ λ From this we deduce that K has no non zero eigenvalue. One can indeed solve explicitly (9.3.5) : x3 1 u (x)= C exp exp x . λ − 3 λ It is then easy to see that noone can be in L2(R) when C = 0. To show that K is compact, we can actually show that K6 is Hilbert-Schmidt. That is, we will show that K(x, y) is in L2(R2). We have + x 2 ∞ 2 exp (y3 x3) dxdy = exp (y3 x3) dy dx . y 83 Here we observe that 2 2 exp (y3 x3) = exp (y x)(y2 + yx + x2) , 3 − 3 − and that 1 1 (y2 + yx + x2) (y2 + x2) x2 . ≥ 2 ≥ 2 This leads, as y x, to the upper bound la majoration ≤ 2 1 exp (y3 x3) exp (y x)x2 . 3 − ≤ 3 − and to x 2 3 3 1 exp (y x ) dy dx 3 2 dx < + . x >1 3 − ≤ x >1 x ∞ Z| | Z−∞ Z| | This implies that the spectrum of K is 0. Non self-adjoint effects. We can also try to estimate the “solution” operator Kλ corresponding to the equation 2 u′ + x u = λf . (9.3.6) It is easy, to see that we can reduce the computation to the case when λ r´eel. We have indeed K = exp i Im λ x K exp i Im λ x . λ − Re λ For λ< 0, we easily find, observing that u S(R) and ∈ 2 2 2 Re u′ + x u λ u 2 = x u λ u λ u , h − | iL h − | i≥− || || the estimate 1 Kλ (L2) . || ||L ≤−λ The case λ = 0 has been treated before. Without to much effort, we get the same result for Re λ 1. So we have to consider| |≤ the case when λ 1 and control the estimate as λ + . Proceeding as in the case λ = 0, we first≥ obtain → ∞ 0 si y x K (x, y)= . λ exp 1 (y3 x3) λ(y x) si y <≥ x 3 − − − 2 2 Again, we see that Kλ(x, y) is in L (R ) : + x 2 ∞ 2 3 3 K 2 R2 = exp (y x ) 2λ(y x) dy dx . || λ||L ( ) 3 − − − Z−∞ Z−∞ 84 The proof is similar to the case λ = 0. We cut the domain of integration. 1 x − 2 exp (y3 x3) 2λ(y x) dy dx , 1 4√λ 3 − − − Z− − Z−∞ is bounded after a new partition of the domain of integration by 1 1 4√λ 1 2 (1+4λ 2 )3 − − 2 3 2 (1+2λ 2 )3 4√λe 3 exp y 2λy dy 3√λe 3 , 3 − ≤ Z−∞ and by 1 x − 2 3 3 1 4√λ 1 4√λ exp 3 (y x ) 2λ(y x) dy dx − − − − − − − √ x R 4 λ supR x ] 1 4√λ, 1[ 1 4√λ exp 2λ(y x) dy ≤ 3∈ − − − − − − − 1 8λ 2 2λ− 2 e . R ≤ Then we consider +1 x 2 exp (y3 x3) 2λ(y x) dy dx , 1 3 − − − Z− Z−∞ which is bounded by 2 1 4√λ 2 3 +2λ − − 3 2e exp( 3 y 2λy) dy 2 +2−∞λ − 2e 3 R ≤ 2 +2λ Ce 3 . ≤ and by +1 x 2 3 3 1 1 4√λ exp 3 (y x ) 2λ(y x) dy dx − − − − − − +1 x R 1 R 1 4√λ exp 2λ(y x) dy dx − − − − − 1 exp(4λ(1+2√λ)) R≤ λ R We look now at 1+4√λ x 2 exp (y3 x3) 2λ(y x) dy dx , 1 3 − − − Z Z−∞ which is similarly controlled by 1 4√λ − − 2 4√λ exp2λ(1+4√λ) exp y3 2λy dy 3 − Z−∞ and by 1+4√λ x exp 2λ(y x) dy dx . 1 1 4√λ − − Z Z− − 85 Finally, we have to estimate x 2 exp (y3 x3) 2λ(y x) dy dx , x 1+4√λ 3 − − − Z| |≥ Z−∞ Here we observe that 2 2 exp (y3 x3) = exp (y x)(y2 + yx + x2) , 3 − 3 − and that 1 1 (y2 + yx + x2) (y2 + x2) x2 . ≥ 2 ≥ 2 This leads to the upper bound 2 1 exp (y3 x3)exp 2λ(y x) exp (y x)(x2 6λ) . 3 − − − ≤ 3 − − By simple integration, we get : x 2 3 3 x >1+4√λ exp 3 (y x ) 2λ(y x) dy dx | | −∞ 1 − − − R 3 x >1+4R √λ x2 6 λ dx < + . ≤ | | − ∞ The last bound canR be controlled independently of Re λ. Hence, we have finally proved the existence of C > 0 such that, for Re λ 1, we have ≥ C 3 K C λ exp C Re λ 2 . || λ||HS ≤ | | Remark 9.3.1. Using the Laplace integral method, one can get the asymptotics of K . || λ||HS Note that Kλ HS Kλ (L2). || || ≥ || ||L One can also find a lower bound of Kλ (L2) using quasimodes. || ||L d 2 Note that Kλ is the resolvent of the unbounded operator dx +x . This operator is not selfadjoint. He is with compact resolvent and− has empty spectrum. Moreover, the norm of the resolvent depends only of Re λ. This example shows that when the operators are not selfadjoint many things 1 can occur. In particular the fact that if z ρ(T ), then (T z)− is controlled 1 ∈ || − 1 || 1 by and that in particular that if z R then (T z)− can d(λ,σ(T ) 6∈ || − || ≤ Im z become completely wrong. | | In this context, the notion of pseudo-spectrum can be helpful. We leave this as an exercise to the reader. Remark 9.3.2. Other interesting examples to analyze in the same spirit are for example the d2 2 complex harmonic oscillator ( dx2 + ix ) with c R (See Davies (1999), [Tr2], − 6∈ d2 [Zw] and references therein). The spectrum can be seen as the spectrum of dx2 π − rotated by 4 in the complex plane. 86 10 Essentially selfadjoint operators 10.1 Introduction In most of the examples which were presented, the abstract operators are as- sociated with differential operators. These differential operators are naturally defined on C0∞(Ω) or ′(Ω). Most of the time (for suitable potentials increas- D m m m ing slowly at ) they are also defined (when Ω = R ) on (R ) or ′(R ). It is important∞ to understand how the abstract point of viewS can beS related to the PDE point of view. The theory of the essential selfadjointness gives a clear understanding of the problem. The question is to decide if, starting from a symmetric operator T , whose domain D(T ) = 0 is dense in , there exists a unique selfadjoint extension T ext of T . We recallH that it meansH that D(T ) D(T ext) and T extu = T u , u D(T ). This leads to ⊂ ∀ ∈ Definition 10.1.1. A symmetric operator T with domain 0 is called essentially selfadjoint if its closure is selfadjoint. H Proposition 10.1.2. If T is essentially selfadjoint, then its selfadjoint extension28 is unique. Indeed suppose that S is a selfadjoint extension of T . Then S is closed and being an extension of T , is also an extension of its smallest extension T . We recall from Theorem 2.2.6 that T = T ∗∗. Thus, S = S∗ (T ∗∗)∗ = T ∗∗, and so ⊂ S = T ∗∗. Example 10.1.3. Here we give a list of examples and counter-examples which will be analyzed later. m 1. The differential operator ∆ with domain 0 = C0∞(R ) is essentially selfadjoint (see later). − H 2 m 2. The differential operators ∆ + x with domain 0 = C0∞(R ) or m − | | H H1 = (R ) are essentially selfadjoint and admit consequently a unique selfadjointS extension. This extension is the same for the two operators. The domain can be explicitely described as B2(Rm)= u L2(Rm) xαDβu L2(Rm) , α, β Nm with α + β 2 . { ∈ | x ∈ ∀ ∈ | | | |≤ } 3. The differential operator ∆ with domain C0∞(Ω) (where Ω is an open bounded set with smooth− boundary) is not essentially selfadjoint. There exists a lot of selfadjoint extensions related to the choice of a boundary problem. As we have seen before, we have already met two such exten- sions : 28Although, it will not help in this course, note that the converse is true. See for example the book [Ro], in which the essential selfadjointness is defined differently. 87 the Dirichlet realization whose domain is the set H1(Ω) H2(Ω), • 0 ∩ the Neumann realization whose domain is the set : • u H2(Ω) (∂u/∂ν) =0 { ∈ | /∂Ω } 4. The Laplace Beltrami operator on a compact manifold M with domain 2 C∞(M) is essentially selfadjoint on L (M). The domain of the selfadjoint extension can be described as H2(M) (which can be described using local charts). 10.2 Basic criteria. We now give some criteria in order to verify that an operator is essentially selfadjoint. As already mentioned, one can prove essential selfadjointness by proving that the minimal closed extension Tmin := T coincides with T ∗. One easily verifies (see Proposition 2.2.5 in conjonction with the definition of T ) that T ∗ = T ∗ and we recall that T ∗∗ = T . We now observe the Proposition 10.2.1. Any closed symmetric extension of T is a restriction of T ∗. Proof. Let S is a closed symmetric extension of T . We have indeed T S S∗ and, observing that S∗ T ∗, we consequently get ⊂ ⊂ ⊂ S T ∗. ⊂ In particular, if T is selfadjoint then T is the unique selfadjoint extension of T . We can characterize the selfadjointness through the following general crite- rion Theorem 10.2.2. Let T be a closed symmetric operator. Then the following statements are equiv- alent : 1. T is selfadjoint. 2. Ker (T ⋆ i)= 0 ; ± { } 3. Range (T i)= . ± H Proof. 1. implies 2. This property was already observed (because T = T ∗ and i σ(T )). 2. implies 3. . ± 6∈ We first observe that the property that ker (T ⋆ + i)= 0 implies that R(T i) is dense in . Note that the converse is also true. For{ } getting 3., it remai−ns H 88 to show that R(T i) is closed. But, for all φ in D(T ), we have (using the symmetry of T ) − (T i)φ 2 = T φ 2 + φ 2 . (10.2.1) || − || || || || || If φn is a sequence in D(T ) such that (T + i)φn converges to some ψ . The ∞ previous identity shows that φn is a Cauchy sequence, so there exists φ such ∞ that φn φ in . But T φn = (T + i)φn iφn is convergent and using that the graph→ is∞ closed,H we obtain that φ D(−T ) and T φ = ψ . 3. implies 1. . ∞ ∈ ∞ ∞ Let φ D(T ∗). Let η D(T ) such that (T i)η = (T ∗ i)φ. T being ∈ ∈ − − symmetric, we have also (T ∗ i)(η φ) = 0. But, if (T + i) is surjective, then − − (T ∗ i) is injective and we get φ = η. This proves that φ D(T ). − ∈ Remark 10.2.3. Here we have used and proved during the proof of the assertion “2. implies 3.” the following lemma Lemma 10.2.4. If T is closed and symmetric, then R(T i) is closed. ± This theorem gives as a corollary a criterion for essential selfadjointness in the form Corollary 10.2.5. Let A with domain D(A) be a symmetric operator. Then the following are equivalent 1. A is essentially selfadjoint. 2. Ker(A∗ i)= 0 . ± { } 3. The two spaces R(A i) are dense in . ± H We have indeed essentially to apply the previous theorem to A, oberving in addition that A is symmetric and that Lemma 10.2.4 is true. Let us here emphasize that in this case, to specify the operator A, it is not necessary to give the exact domain of A but a core for A that is a subspace D such that the closure of A/D is A. In the same spirit, we have in the semibounded case the following Theorem 10.2.6. Let T be a positive, symmetric operator. Then the following statements are equivalent : 1. T is essentially selfadjoint. 2. Ker (T ∗ + b)= 0 for some b> 0 . { } 3. Range (T + b) is dense for some b> 0. 89 The proof essentially the same as for the previous corollary, if one observes that if T is positive then the following trivial estimate is a good substitute for (10.2.1) : (T + b)u,u b u 2 . h i≥ || || Example 10.2.7. (The free Laplacian) The operator ∆ with domain C0∞ is essentially selfadjoint. Its selfadjoint extension is −∆ with domain H2. − 10.3 The Kato -Rellich theorem We would like to consider the case when P = ∆+ V when V is regular and tends to 0 as x + . Here V can be considered− as a perturbation of the Laplacian. One| | can → then∞ apply a general theorem due to Kato-Rellich. Theorem 10.3.1. Let A be a selfadjoint operator, B be a symmetric operator whose domain con- tains D(A). Let us assume the existence of a and b such that 0 a < 1 and b 0 such that ≤ ≥ Bu a Au + b u (10.3.1) || || ≤ || || || || for all u D(A). Then A + B is selfadjoint on D(A). If A is essentially∈ selfadjoint on29 D D(A), then A+B has the same property. ⊂ Proof. Step 1. We start from the following identity which only uses that (A + B) with domain D(A) is symmetric. (A + B iλ)u 2 = (A + B)u 2 + λ2 u 2 , u D(A) . (10.3.2) || ± || || || || || ∀ ∈ By the triangle inequality and the symmetry of A + B, we get for a real λ> 0, and for any u D(A) : ∈ √2 (A + B iλ)u (A + B)u + λ u || − || ≥ ||Au Bu|| +||λ ||u (10.3.3) ≥(1 || ||a) −Au || ||+ (λ || b||) u . ≥ − || || − || || We now choose λ>b. Step 2. Let us show that (A+B) with domain D(A) is closed. If we start indeed from a pair (u ,f ) with u D(A) and f = (A+B)u such that (u ,f ) (u,f) in n n n ∈ n n n n → . From (10.3.3), we get that Aun is a Cauchy sequence in . A being closed, Hwe get u D(A) and the existence of g such that Au g =H Au. ∈ n → Now from (10.3.2) and (10.3.1), we get also that Bun is a Cauchy sequence and 29 By this we mean that the closure of A/D is A. 90 there exists v such that Bun v in . We claim that Bu = v. We have→ indeed,H for any h D(A), ∈ Step 3. In order to apply Theorem 10.2.2, we have to show that (A+B iλ) is surjective. The main element in the proof is the following ± Lemma 10.3.2. . For λ> 0 large enough, we have 1 B(A iλ)− < 1 . (10.3.4) || ± || Proof. We observe that, for u D(A), ∈ (A iλ)u 2 = Au 2 + λ2 u 2 . (10.3.5) || ± || || || || || For u D(A), we have, using two times (10.3.5) and then (10.3.1) ∈ Bu a Au + b u || || ≤ || || || || a (A + iλ)u + b (A + iλ)u (10.3.6) ≤ || || λ || || (a + b ) (A + iλ)u . ≤ λ || || It is then enough to choose λ> 0 large enough such that b (a + ) < 1 . λ Writing 1 A + B iλ = [I + B(A iλ)− ](A iλ) , (10.3.7) − − − it is easy to deduce the surjectivity using the lemma and the surjectivity of (A iλ). − Application. As an application, let us treat the case of the Schr¨odinger operator with Coulomb potential. Proposition 10.3.3. 1 3 The operator ∆ x with domain C0∞(R ) is essentially selfadjoint. − − | | 1 2 3 We recall that the operator is well defined because r belongs to Lloc(R ). We first observe a Sobolev type inequality. 91 Lemma 10.3.4. : There exists a constant C such that for all u H2(R3), all a> 0 and all x R3, we have ∈ ∈ 3 u(x) C(a ∆u + a− u ) . | |≤ || ||0 || ||0 (Prove first the inequality30 for all u with x = 0 and a = 1, then use translation and dilation.) 1 In the second step, we show that the potential V = r is a perturbation of the Laplacian. − There exists indeed a constant C such that for all u H2(R3) and all b> 0, we have ∈ 3 Vu C(b ∆u + b− u ) . || ||0 ≤ || ||0 || ||0 For this proof we observe that, for any R> 0, V (x)2 u(x) 2dx = V (x)2 u(x) 2dx + V (x)2 u(x) 2dx , | | x R | | x R | | Z Z| |≤ Z| |≥ and treat the first term of the right hand side using the Sobolev’s type inequality and the second term by the trivial estimate V (x) 1 . | |≤ R Remark 10.3.5. We note that the same proof shows that ∆+V is essentially selfadjoint starting m 2 − 2 from C∞(R ) if V L +L∞, that is if V = V +V with V L and V L∞. 0 ∈ 1 2 1 ∈ 2 ∈ 10.4 Other criteria of selfadjointness for Schr¨odinger op- erators We present in this subsection two criteria which are specific of the Schr¨odinger case. The first one seems due to Rellich (See [Sima]) and we present it in the easy case when the potential is regular. The second one permits to treat singular potentials and is due to Kato (cf [HiSi] or [Ro]). The first theorem is adapted to operators which are already know to be m positive on C0∞(R ). Theorem 10.4.1. A Schr¨odinger operator T = ∆+ V on Rn associated with a C0 potential V , − n which is semibounded on C0∞(R ), is essentially selfadjoint. In other words, the m Friedrichs extension is the unique selfadjoint extension starting from C0∞(R ). This theorem is complementary to the second theorem (Theorem 10.4.4) which will be stated at the end of this subsection because we do not have to assume the positivity of the potential but only the semi-boundedness of the operator T . 30 2 R3 0 R3 In this case, this is just the Sobolev’s injection theorem of H ( ) into Cb ( ), where 0 R3 Cb ( ) is the space of the continuous bounded functions. 92 Proof. Let T be our operator. Possibly by adding a constant, we can assume that 2 m 2 2 2 m u + V (x) u(x) dx u , u C0∞(R ) . ||∇ || Rm | | ≥ || || ∀ ∈ Z 1 m In this form, the inequality can be extended to the elements of Hcomp(R ), corresponding to the distributions of H1(Rm) with compact support : 2 2 2 1 m u + V (x) u(x) dx u , u Hcomp(R ) . (10.4.2) ||∇ || Rm | | ≥ || || ∀ ∈ Z According to the general criterion of essential selfadjointness (cf Theorem 10.2.6), it is enough to verify that R(T ) is dense. Let us show this property. Let f L2(Rm), such that ∈ m 2 Rm (ζkf) (ζku)dx + ζk(x) V (x)u(x) f(x) dx =∇ ( ζ ·)( ∇x) 2u(x)f(x) dx + m (f(∂ u) u(∂ f)) (x)ζ (x)(∂ ζ )(x) dx , R | ∇ k | R i=1 i − i k i k (10.4.6) R m P R for all u C∞(R ). ∈ 1 This formula can be extended to functions u Hloc. In particular, we can take u = f. ∈ We obtain < (ζ f) , (ζ f) > + ζ2V (x) f(x) 2 dx = ζ 2 f(x) 2dx . (10.4.7) ∇ k ∇ k k | | |∇ k| | | Z Z 31This is the property that f L2 (Rm), ∆f L2 (Rm) implies that f H2 (Rm). ∈ loc ∈ loc ∈ loc 93 Using (10.4.1), (10.4.7) and taking the limit k + , we get → ∞ 2 2 f = limk + ζkf || || → ∞ || || 2 2 lim supk + < (ζkf) , (ζkf) > + ζk V (x) f(x) dx ≤ → ∞ ∇ 2 ∇ 2 | | = lim supk + f(x) ( ζk)(x) dx =0 . → ∞ | ∇ | R (10.4.8) R This proves the theorem. Remark 10.4.2. When V is C∞, we get, in the previous proof, that f C∞ and we immediately can prove (10.4.7) without going through the previous∈ discussion. Example 10.4.3. If V 0 and C∞, then the Schr¨odinger operator ∆+ V with domain • ≥m − C∞(R ) is essentially selfadjoint. The operator ∆+V is indeed positive. 0 − m 2 If φ is C∞ on R , then the operators ∆+ φ ∆φ are essentially • − |∇m | ± selfadjoint. They are indeed positive on C0∞(R ). They can actually be written in the form j Zj∗Zj with Zj = ∂xj ∂xj φ. These operators appear naturally in statistical mechanics. ∓ P Let us now mention, without proof, a quite general theorem due to Kato (See for example [Ro]). Theorem 10.4.4. 2 m m Let V in Lloc(R ) such that V 0 almost everywhere on R . Then ∆+ V m ≥ − with domain C0∞(R ) is essentially selfadjoint. This theorem is based on the so called Kato’s inequality. Remark 10.4.5. This last theorem may be extended to the case of the Schr¨odinger operator with magnetic regular potential A (See Subsection 7.5). 10.5 Non-selfadjoint case : Maximal accretivity and ap- plication to the Fokker-Planck operator 10.5.1 Accretive operators We collect here some material on accretive operators. The references could be the books by Dautray-Lions (Vol. 5, Chapter XVII), Reed-Simon or the book of B. Davies. Let be a complex (or real) Hilbert space. H Definition 10.5.1. Let A be an unbounded operator in with domain D(A). We say that A is accretive if H Re Ax x 0 , x D(A) . (10.5.1) h | iH ≥ ∀ ∈ 94 Definition 10.5.2. An accretive operator A is maximally accretive if it does not exist an accretive extension A˜ with strict inclusion of D(A) in D(A˜). Proposition 10.5.3. Let A be an accretive operator with domain D(A) dense in . Then A is closable and its closed extension A is accretive. H For the analysis of the Fokker-Planck operator, the following criterion, which extends the standard criterion of essential self-adjointness, will be the most suitable Theorem 10.5.4. For an accretive operator A, the following conditions are equivalent 1. A is maximally accretive. 2. There exists λ0 > 0 such that A∗ + λ0I is injective. 3. There exists λ > 0 such that the range of A + λ I is dense in . 1 1 H Note that in this case A is the infinitesimal generator of a contraction semi-group. − 10.5.2 Application to the Fokker-Planck operator We would like to show Proposition 10.5.5. n Let V be a C∞ potential on R , then the closure K of the Fokker-Planck operator 2n defined on C0∞(R ) by 1 n K := ∆ + v 2 + X , (10.5.2) − v 4| | − 2 0 where X := V (x) ∂ + v ∂ (10.5.3) 0 −∇ · v · x is maximally accretive. Moreover K∗ is also maximally accretive. The idea is to adapt the proof that a semi-bounded Schr¨odinger operator with regular potential is essentially self-adjoint on L2(Rn). Proof: We apply the abstract criterion taking = L2(Rn Rn) and A = K. The operators being real, we can consider everywhereH real funct× ions. The accretivity 2n on C0∞(R ) is clear. We can then consider the closure K. 95 n Changing K in T := K + ( 2 + 1)I, we would like to show that its range is dense. Let f L2(Rm), with m =2n, such that ∈ m 2 m ( ∆ + v /4+1 X )f =0 , in ′(R ) . − v − 0 D The standard hypoellipticity theorem for the H¨ormander operators32 of type 2 m implies that f C∞(R ). ∈ We now introduce a family of cut-off functions ζk := ζk1,k2 by ζ (x, v) := ζ(x/k )ζ(v/k ) , k N2 , (10.5.5) k1,k2 1 2 ∀ ∈ where ζ is a C∞ function satisfying 0 ζ 1, ζ =1 on B(0, 1) and supp ζ B(0, 2). ≤ ≤ ⊂ For any u C∞, we have the identity ∈ 0 2 2 v(ζkf) v(ζku) dxdv + ζk(x, v) (v /4+1)u(x, v) f(x, v) dx dv ∇ · ∇ 2 + f(x, v)(X0(ζk u))(x, v) dx dv R 2 R = ( vζk)(x, v) u(x, v)f(x, v) dx dv R | ∇m | + i=1 (f(∂vi u) u(∂vi f)) (x, v)ζk(x, v)(∂vi ζk)(x, v) dx dv +R f(x, v) T ζ2u .− h R | k i P (10.5.6) When f satisfies (10.5.4), we get : 2 2 Rm v(ζkf) v(ζku) dxdv + ζk (v /4+1)u(x, v) f(x, v) dx dv ∇ · ∇ 2 + f(x, v)(X0(ζk u))(x, v)dx dv R 2 R (10.5.7) = ( yζk)(x) u(x)f(x, v) dx dv R +| ∇m (f(|∂ u) u(∂ f)) (x, v)ζ (x, v)(∂ ζ )(x, v) dx dv , R i=1 vi − vi k vi k m for all u PC∞(RR ). In particular, we can take u = f. We obtain∈ < (ζ f) (ζ f) > + ζ2(v2/4+1) f(x, v) 2 dx dv ∇v k | ∇v k k | | + f(x, v)(X (ζ2f))(x, v) dx dv (10.5.8) 0 k R = ζ 2 f(x, v) 2 dx dv . R |∇v k| | | With an additionalR integration by part, we get 2 2 2 < v(ζkf) v(ζkf) > + ζk (v /4+1) f(x, v) dx dv ∇ | ∇2 | | + ζkf(x, v) (X0ζk)(x, v) dx dv (10.5.9) = ζ 2 f(x, v) 2 dx dvR . R |∇v k| | | 32 k 2 These operatorsR are in the form P = X + X0 + a(x), where the X are real − j=1 j j vectorfield. If the Xj together with the bracketsP [Xℓ, Xm] span at each point x the whole tangent space, then one can show that the corresponding operator is hypoelliptic. P is said hypoelliptic if for any u ′ and any open set ω, Pu C∞(ω) implies that u C∞(ω). ∈D ∈ ∈ 96 This leads to the existence of a constant C such that, for all k, 2 1 2 ζkf + 4 ζkvf || ||1 2|| 1|| 1 (10.5.10) C k2 f + C k vζkf f + C k V (x)ζkf f . ≤ 2 || || 1 || || || || 2 ||∇ || || || (The constant C will possibly be changed from line to line). This leads to 2 1 2 1 1 2 1 ζk f + ζk vf C( 2 + 2 ) f + C(k1) ζkf f , (10.5.11) || || 8|| || ≤ k2 k1 || || k2 || || || || where C(k1) = sup xV (x) x 2k |∇ | | |≤ 1 This implies ˜ 2 C(k1) 1 2 ζk f C( 2 + 2 ) f . (10.5.12) || || ≤ k2 k1 || || This finally leads to f = 0. For example, one can take first the limit k + , which leads to 2 → ∞ x 2 C 2 ζ( )f 2 f , || k1 || ≤ k1 || || and then the limit k + . 1 → ∞ 97 11 Discrete spectrum, essential spectrum 11.1 Discrete spectrum We have already recalled in Proposition 8.5.4 a characterization of the spectrum. Let us now complete this characterization by introducing different spectra. Definition 11.1.1. If T is a selfadjoint operator, we shall call discrete spectrum of T the set σ (T )= λ σ(T ) s. t. ǫ> 0, dim range (E(]λ ǫ, λ + ǫ[)) < + . disc { ∈ ∃ − ∞} With this new definition, we can say that, for a selfadjoint operator with compact resolvent, the spectrum is reduced to the discrete spectrum. For a compact selfadjoint operator, the spectrum is discrete outside 0. We see in this case that the discrete spectrum is not closed. Equivalently, let us observe now give another characterization : Proposition 11.1.2. Let T be a selfadjoint operator. A real λ is in the discrete spectrum if and only if : λ is an isolated point in σ(T ) and if λ is an eigenvalue of finite multiplicity. Proof. If λ σ (T ), we immediately see that there exists ǫ such that, ǫ such ∈ disc 0 ∀ that 0 <ǫ<ǫ0, E]λ ǫ,λ+ǫ[ becomes a projector independent of ǫ with finite − range. This is actually the projector Πλ = 1 λ (T ) and we observe moreover { } E]λ,λ+ǫ[ = 0 and E]λ ǫ,λ[ = 0. This shows that λ is an isolated point in σ(T ). − Using the spectral representation of T , one immediately get that, if x = Πλx (x = 0), then x is an eigenvector of T . Moreover, one easily obtains that (T λ) 6 − is invertible on R(I Πλ). One can indeed find a continuous bounded f such that f(T )(T λ)(I −Π ) = (I Π ). − − λ − λ Conversely, let λ be isolated. The previous proof as already shown that in this case the range of Πλ is an eigenspace. The assumption of finite multiplicity permits then to conclude. Remark 11.1.3. The discrete spectrum is not a closed set ! If we consider in R3, the Schr¨odinger operator with coulomb potential, the discrete spectrum is a sequence of eigenval- ues tending to 0 but 0 does not belong to the discrete spectrum. 11.2 Essential spectrum Definition 11.2.1. The essential spectrum is the complementary in the spectrum of the discrete spectrum. Intuitively, a point of the essential spectrum corresponds 98 either to a point in the continuous spectrum, • or to a limit point of a sequence of eigenvalues with finite multiplicity, • or to an eigenvalue of infinite multiplicity. • The discrete spectrum being composed of isolated points, we get Proposition 11.2.2. The essential spectrum of a selfadjoint operator T is closed in R. 11.3 Basic examples: 1. The essential spectrum of a compact selfadjoint operator is reduced to 0. 2. The essential spectrum of an operator with compact resolvent is empty. 3. The Laplacian on Rn ∆ is a selfadjoint operator on L2(Rn) whose domain is the Sobolev space −H2(Rn). The spectrum is continuous and equal to R+. The essential spectrum is also R+ and the operator has no discrete spectrum. 4. The Schr¨odinger operator with constant magnetic field (B = 0) in R2: 6 Bx Bx S := (D 2 )2 + (D + 1 )2 , (11.3.1) B x1 − 2 x2 2 with 1 1 ∂ Dxj := ∂xj = . i i ∂xj The spectrum is formed with eigenvalues (2k + 1) B but the spectrum is not discrete because each eigenvalue is with infinite| | multiplicity. This example will be analyzed later. 11.4 Weyl’s criterion: We have already mentioned that the essential spectrum is a closed set. In order to determine the essential spectrum it is useful to have theorems proving the invariance by perturbation. The following characterization is in this spirit quite useful. Theorem 11.4.1. Let T be a selfadjoint operator. Then λ belongs to the essential spectrum iff 33 there exists a sequence un in D(T ) with un =1 such that un tends weakly to 0 and such that (T λ)u 0 as n|| || . || − n|| → → ∞ 33 We recall that we say that a sequence un in a separable Hilbert space is weakly con- H vergent if, for any g in , 99 Let us give a partial proof of the theorem and we refer to [Ro], [HiSi] or [RS-IV] for a complete proof. The sequence appearing in the theorem is called a Weyl sequence. A point λ such that there exists an associated Weyl sequence is said to belong to the Weyl spectrum W (T ). Let us show the inclusion W (T ) σ (T ) . (11.4.1) ⊂ ess We have already seen that W (T ) σ(T ) . (11.4.2) ⊂ Let us suppose by contradiction that λ σdisc(T ). Let Πλ := E λ be the ∈ { } associated spectral projector. We first observe that, Πλ being finite range, hence compact, we have : Π u 0 . (11.4.3) λ n → ∈ H Let us define w = (I Π )u . n − λ n We get w +1 and (T λ)w = (I Π )(T λ)u 0. || n|| → − n − λ − n → But (T λ) is invertible on R(I Πλ), so we get wn 0 and the contradiction. This shows− the announced inclusion− (11.4.1). → Corollary 11.4.2. The operator h2∆+ V with V a continuous function tending to 0 as x (x RD) has−R+ as essential spectrum. | | → ∞ ∈ For proving the inclusion of R+ in the essential spectrum, we can indeed consider the sequence (D/2) u (x) = exp(ix ξ)n− χ((x R )/n) n · · − n 1 with χ 0 and supported in the ball B(0, 1) and equal to one on say B(0, 2 ). The sequence≥ R is chosen such that R (for example R = n2) tends to n | n| | n| ∞ and such that the support of the un are disjoints. This is a particular case of a Weyl sequence (called in [HiSi] a Zhislin sequence). The converse can be obtained by abstract analysis and the fact that we know that the essential spectrum of ∆ is [0, + [. This idea is formalized through the notion of relative compactness.− ∞ Definition 11.4.3. If T is a closed operator with a dense domain DT , we shall say that the operator V is relatively compact with respect to T or T -compact if DT DV and if the ⊂ 2 2 image by V of a closed ball in DT (for the graph-norm u u + Tu ) is relatively compact in . 7→ || || || || H p In other words, we shall say that V is T -compact, if, from each sequence un in D bounded in and such that Tu is bounded in , one can extract a T H n H subsequence uni such that Vuni is convergent in . Here we recall (exercise) that when T is closed, then D(T ) equipped with theH graph norm is an Hilbert space. 100 Example 11.4.4. If V is the multiplication operator by a continuous function V tending to 0 then V is ( ∆)-compact. This is a consequence of Proposition 5.4.1 and of the uniform continuity− of V on each compact. Weyl’s Theorem says Theorem 11.4.5. . Let T be a selfadjoint operator, and V be symmetric and T -compact, then T + V is selfadjoint and the essential spectrum of T + V is equal to the essential spectrum of T . The first part can be deduced from what was discussed in the previous subsection (See Theorem 10.3.1). We observe indeed the following variant of Lions’s Lemma : Lemma 11.4.6. If V is T -compact and closable, then, for any a> 0, there exists b> 0 such that Vu a Tu + b u , u D(T ) . (11.4.4) || || ≤ || || || || ∀ ∈ Proof of the lemma. The proof is by contradiction. If (11.4.4) is not true, then there exists a > 0 such that, n N∗, there exists u D(T ) such that ∀ ∈ n ∈ a Tu + n u < Vu . (11.4.5) || n|| || n|| || n|| Observing that Vu = 0 and that the inequation is homogeneous, we can in || n|| 6 addition assume that un satisfies the condition : Vu =1 . (11.4.6) || n|| From these two properties we get that the sequence Tun is bounded and that u 0. n → On the other hand, by T -compactness, we can extract a subsequence unk such that Vunk is convergent to v with v = 1. But (0, v) is in the closure of the graph of V , hence in the graph of|| the|| closure of V and consequently v = 0 (contradiction). For the second part we can use the Theorem 11.4.1. If we take a Weyl’s sequence u such that u 0 (weakly) and (T λ)u 0 strongly, let us n n → − n → consider (T + V λ)un. We have simply to show that one can extract a subse- quence u such− that (T + V λ)u 0. nk − nk → But Tun is a bounded sequence. By the T -compactness, we can extract a sub- sequence such that Vunk converges strongly to some v in . It remains to show that v = 0. But here we can observe that for any f D(HT ), we have ∈ 101 Here we have used the symmetry of V and the weak convergence of un to 0. Using the density of D(T ) in , we obtain v = 0. This shows that a Weyl sequenceH for T is a Weyl sequence for T + V . For the converse, one can intertwine the roles of T and T + V , once we have shown that V is (T + V )-compact. For this, we can use Lemma 11.4.6, and observe that the following inequality is true : 1 Tu ( (T + V )u + b u ) . (11.4.7) || || ≤ 1 a || || || || − This shows that if un is a sequence such that ( un + (T + V )un ) is bounded, then this sequence has also the property that|| ( u|| +|| Tu ) is|| bounded. || n|| || n|| 102 12 The max-min principle 12.1 Introduction The max-min principle is an alternative way for describing the lowest part of the spectrum when it is discrete. It gives also an efficient way to localize these eigenvalues or to follow their dependence on various parameters. 12.2 On positivity We first recall the following definition Definition 12.2.1. Let A be a symmetric operator. We say that A is positive (and we write A 0), if ≥ < Au,u > 0 , u D(A) . (12.2.1) ≥ ∀ ∈ The following proposition relates the positivity with the spectrum Proposition 12.2.2. Let A be a selfadjoint operator. Then A 0 if and only if σ(A) [0, + [. ≥ ⊂ ∞ Proof. It is clear that if the spectrum is in R+, then the operator is positive. This can be seen for example through the spectral representation : 2 < Au,u >= λ d Eλu . λ σ(A) || || Z ∈ Now, if A 0, then, for any a> 0, A + a is invertible. We have indeed ≥ a u 2 < (A + a)u,u > (A + a)u u , || || ≤ ≤ || || || || which leads to a u (A + a)u , u D(A) . (12.2.2) || || ≤ || || ∀ ∈ This inequality gives the closed range and the injectivity. A being selfadjoint, we get also from the injectivity, the density of the image of (A + a). This shows that a is not in the spectrum of A. − Example 12.2.3. Let us consider the Schr¨odinger operator P := ∆+ V , with V C∞ and semi-bounded, then − ∈ σ(P ) [inf V, + [ . (12.2.3) ⊂ ∞ 103 12.3 Variational characterization of the discrete spectrum Theorem 12.3.1. Let A be a selfadjoint semibounded operator. Let Σ := inf σess(A) and let us con- sider σ(A) ] , Σ[, described as a sequence (finite or infinite) of eigenvalues that we write∩ − in ∞ the form λ < λ < < λ . 1 2 · · · n · · · Then we have 2 λ1 = inf φ − < Aφ, φ > , (12.3.1) φ D(A),φ=0 || || ∈ 6 2 λ2 = inf φ − < Aφ, φ > , (12.3.2) φ D(A) K⊥,φ=0 || || ∈ ∩ 1 6 and, for n 2, ≥ 2 λn = inf φ − < Aφ, φ > , (12.3.3) φ D(A) K⊥ ,φ=0 || || ∈ ∩ n−1 6 where Kj = i j Ker (A λi) . ⊕ ≤ − Proof. Step 1. Let us start, with the lowest eigenvalue. Let us define µ1(A) by 2 µ1(A) := inf φ − < Aφ, φ > . (12.3.4) φ D(A),φ=0 || || ∈ 6 If φ1 is an eigenvector associated to λ1, we get immediately the inequality µ (A) λ (A) . (12.3.5) 1 ≤ 1 Let us prove the converse inequality. Using the spectral theorem, one get im- mediately that A inf σ(A). So we get ≥ inf σ(A) µ (A) . (12.3.6) ≤ 1 Now, if the spectrum below Σ is not empty, we get λ (A) µ (A) . 1 ≤ 1 We have consequently the equality. We have actually a little more. We have indeed proved that, if µ1(A) < Σ, then, the spectrum below Σ is not empty, and the lowest eigenvalue is µ1(A). 104 Step 2. The proof is by recursion, applying Step 1 to A/D(A) K⊥ . ∩ n−1 This ends the proof of Theorem 12.3.1. Example 12.3.2. (Payne-Polya-Weinberger Inequality.) Let P = ∆+ V with V C∞ positive and V + as x + . Let us assume− that V is even∈ → ∞ | | → ∞ V (x)= V ( x) . (12.3.7) − Then λ2 satisfies λ2 inf P (x u )= λ x u 2∂ u . j 1 1 j 1 − j 1 Taking the scalar product with xj u1, we then obtain 2 (λ2 λ1) xj u1 2 < ∂j u1, xj u1 > − || || ≤− 2 u1 (12.3.9) ≤1 || . || ≤ We now use the uncertainty principle (1.2.11) and get : (λ λ ) 4 ∂ u 2 . (12.3.10) 2 − 1 ≤ || j 1|| On the other hand, 2 2 u1 + V (x) u1(x) dx = λ1 , (12.3.11) ||∇ || Rm | | Z and this gives u 2 λ . (12.3.12) ||∇ 1|| ≤ 1 Putting the inequalities (12.3.9) and (12.3.12), we get, summing over j, 4 λ λ λ . (12.3.13) 2 − 1 ≤ m 1 This inequality is not optimal, in the sense that for m = 1 and V (x) = x2, we have λ λ =2λ . 2 − 1 1 105 Example 12.3.3. 2 m Let us consider S := h ∆+ V on R where V is a C∞ potential tending to h − 0 at and such that infx Rm V (x) < 0. ∞ ∈ Then if h > 0 is small enough, there exists at least one eigenvalue for Sh. We note that the essential spectrum is [0, + [. The proof of the existence of this ∞ eigenvalue is elementary. If xmin is one point such that V (xmin) = infx V (x), it λ 2 µn(A) = sup inf Aφ φ . (12.4.1) ψ ,ψ ,...,ψ − h | iH 1 2 n 1 8 φ [span(ψ1,...,ψn 1)]⊥; 9 < φ ∈ D(A) and φ =1− = : ∈ || || ; Then either (a) µn(A) is the n-th eigenvalue when ordering the eigenvalues in increas- ing order (and counting the multiplicity) and A has a discrete spectrum in ] , µn(A)] or− ∞ (b) µn(A) corresponds to the bottom of the essential spectrum. In this case, we have µ (A)= µ (A) for all j n. j n ≥ Remark 12.4.2. In the case when the operator is with compact resolvent, case (b) does not occur and the supremum in (12.4.1) is a maximum. Similarly the infimum is a minimum. This explains the traditional terminology “ Max-Min principle” for this theorem. Proof. If Ω is a borelian, let EΩ be the projection-valued measure for A (see Remark 8.3.7). We first prove that dim RangeE] ,a[ 106 Notons que la conjonction de (12.4.2) et (12.4.3) montre que µn(A) est dans le spectre de A. Step 1: Proof of (12.4.2). Let a and n be given such that a < µn(A). Let us prove (12.4.2) by contra- diction. If it was false, then we would have dim Range(E] ,a[) n and −∞ ≥ we could find an n-dimensional space V Range(E] ,a[). Note now, that A ⊂ −∞ being bounded from below, Range(E] ,a[) is included in D(A). So we can find an n-dimensional space−∞V D(A), such that ⊂ φ V , But then given any ψ1, , ψn 1 in , we can find φ V ψ1, , ψn 1 ⊥ such that φ = 1 and·< · · Aφ,φ− > Ha. Coming back∈ to∩{ the definition,· · · − this} || || ≤ shows that µn(A) a and a contradiction. Note that we have≤ proved in this step the following proposition Proposition 12.4.3. Suppose that there exists a and an n-dimensional subspace V D(A) such that (12.4.4) is satisfied. Then we have the inequality : ⊂ µ (A) a . (12.4.5) n ≤ Modulo the complete proof of the theorem, we obtain Corollary 12.4.4. Under the same assumption as in Proposition 12.4.3, if a is below the bottom of the essential spectrum of A, then A has at least n eigenvalues (counted with multiplicity). Exercise 12.4.5. In continuation of Example 12.3.3, show that for any ǫ > 0 and any N, there exists h0 > 0 such that for h ]0,h0], Sh has at least N eigenvalues in [inf V, inf V +ǫ]. One can treat∈ first the case when V has a unique non degenerate minimum at 0. Step 2 : Proof of (12.4.3). Suppose that (12.4.3) is false. Then dim ( Range (E] ,a[) n 1, so we −∞ ≤ − can find (n 1) generators ψ1, , ψn 1 of this space. Then any φ D(A) − · · · − ∈ ∩ span ψ1, , ψn 1 ⊥ is in Range (E[a,+ [), so { · · · − } ∞ Therefore, coming back to the definition of µn(A), we get µn(A) a in contra- diction with our initial assumption. ≥ Before to continue the proof, let us emphasize on one point. 107 Remark 12.4.6. In the definition of µn(A), ψ1, , ψn 1 are only assumed to belong to the Hilbert space . · · · − H Step 3 : µn(A) < + . First the semi-boundedness∞ from below of A gives a uniform lower bound. Secondly, if µ (A)=+ , this would mean by (12.4.2) that : n ∞ dim(Range(E] ,a[)) As the statement of the theorem suggests, there are two cases to consider and this will be the object of the two next steps. Step 4. Let us first assume (with µn = µn(A)) that dim (Range(E] ,µn+ǫ[))=+ , ǫ> 0 . (12.4.6) −∞ ∞ ∀ We claim that, in this case, we are in the second alternative in the theorem. Using (12.4.2) and (12.4.6), we get indeed dim (Range(E]µn ǫ,µn+ǫ[))=+ , ǫ> 0 . (12.4.7) − ∞ ∀ This shows that µn(A) σess(A). On the other hand, using∈ again (12.4.2), we immediately get that ] , µ (A)[ − ∞ n does not contain any point in the essential spectrum. Thus µn(A) = inf λ λ σ (A) . { | ∈ ess } Let us show now that µn+1 = µn in this case. From the definition of the µk(A), it is clear that µn+1 µn, since one can take ψn = ψn 1. ≥ − But if µn+1 > µn, (12.4.2) would also be satisfied for µn+1, and this is in contradiction with (12.4.6). Step 5. Let us now assume that dim (Range(E] ,µn+ǫ [)) < , for some ǫ0 > 0 . (12.4.8) −∞ 0 ∞ Then it is clear, that the spectrum is discrete in ] , µ + ǫ [. Therefore µ − ∞ n 0 n is an eigenvalue. Then, for ǫ1 > 0 small enough, Range(E] ,µn])= Range(E] ,µn+ǫ [) , −∞ −∞ 1 and by (12.4.3) dim (RangeE] ,µn]) n . (12.4.9) −∞ ≥ So there are at least n eigenvalues E E E µ for A. If E were 1 ≤ 2 ≤···≤ n ≤ n n strictly less than µn, dim(RangeE] ,En]) would equal n in contradiction with −∞ 108 (12.4.2). This ends the proof of Theorem 12.4.1. A first natural extension of Theorem 12.4.1 is obtained by Theorem 12.4.7. Let A be a selfadjoint semibounded operator and Q(A) its form domain 35 . Then µn(A) = sup inf Aφ φ . (12.4.10) ψ ,ψ ,...,ψ − h | iH 1 2 n 1 8 φ [span(ψ1,...,ψn 1)]⊥; 9 < φ ∈ Q(A) and φ =1− = : ∈ || || ; Proof. Letµ ˜n be the right hand side of (12.4.10). By imitating the proof of the previous theorem, we get that eachµ ˜n obeys one of the two conditions. These conditions determine µn and consequently µn =µ ˜n. One can also note (see Subsection 3.3) that, when constructing the Friedrichs extension, one has shown that the domain of the Friedrichs extension is dense in the form domain. Applications It is very often useful to apply the max-min principle by taking the mini- • mum over a dense set in Q(A). The max-min principle permits to control the continuity of the eigenvalues • with respect to parameters. For example the lowest eigenvalue λ1(ǫ) of d2 2 4 dx2 + x + ǫx increases with respect to ǫ. Show that ǫ λ1(ǫ) is right continuous− on [0, + [. (The reader can admit that the7→ corresponding eigenfunction is in ∞(R) for ǫ 0). S ≥ The max-min principle permits to give an upperbound on the bottom of • the spectrum and the comparison between the spectrum of two operators. If A B in the sense that, Q(B) Q(A) and36 ≤ ⊂ < Au,u > 109 Example 12.4.8. (Comparison between Dirichlet and Neumann) Let Ω be a bounded regular connected open set in Rm. Then the N-th eigenvalue of the Neumann realization of ∆+ V is less or equal to the N-th eigenvalue of the Dirichlet realization. It is− indeed enough to observe the inclusion of the form domains. Example 12.4.9. (monotonicity with respect to the domain) Let Ω Ω Rm two bounded regular open sets. Then the n th eigenvalue 1 ⊂ 2 ⊂ − of the Dirichlet realization of the Schr¨odinger operator in Ω2 is less or equal to the n-th eigenvalue of the Dirichlet realization of the Schr¨odinger operator in 1 1 Ω1. We observe that we can indeed identify H0 (Ω1) with a subspace of H0 (Ω2) by just an extension by 0 in Ω2 Ω1. We then have \ 2 2 λn(Ω2) = sup ψ1, ,ψn−1 L (Ω2) inf 1 φ L2(Ω ) { ··· ∈ } 8 φ H0 (Ω2) 9 ||∇ || 2 < ∈ = < φ, ψj > 2 L (Ω2) and φ =1 : || || ; 2 2 sup ψ1, ,ψn−1 L (Ω2) inf 1 φ L2(Ω ) ≤ { ··· ∈ } 8 φ H0 (Ω1) 9 ||∇ || 2 < ∈ = < φ, ψj > 2 L (Ω2) and φ =1 : || || ; 2 2 = sup ψ1, ,ψn−1 L (Ω2) inf 1 φ L2(Ω ) { ··· ∈ } 8 φ H0 (Ω1) 9 ||∇ || 1 < ∈ = < φ, ψj > 2 L (Ω1) and φ =1 : || || ; 2 2 = sup ψ1, ,ψn−1 L (Ω1) inf 1 φ L2(Ω ) { ··· ∈ } 8 φ H0 (Ω1) 9 ||∇ || 1 < ∈ = < φ, ψj > 2 L (Ω1) and φ =1 : || || ; = λn(Ω1) . Note that this argument is not valid for the Neumann realization. 12.5 CLR inequality In order to complete the picture, let us mention (confer [RS-IV], p. 101) that, if m 3, then the following theorem due to Cwickel-Lieb-Rozenbljum is true : ≥ Theorem 12.5.1. m There exists a constant Lm, such that, for any V such that V L 2 , and − ∈ if m 3, the number of strictly negative eigenvalues of S1 N is finite and bounded≥ by − m N Lm ( V ) 2 dx . (12.5.1) − ≤ V (x) 0 − Z ≤ This shows that when m 3, we could have examples of negative potentials V (which are not identically≥ zero) and such that the corresponding Schr¨odinger operator S1 has no eigenvalues. A sufficient condition is indeed m Lm ( V ) 2 dx < 1 . V 0 − Z ≤ In the other direction, we have37 the following results. 37These counterexamples come back (when m = 1 to Avron-Herbst-Simon [AHS] and when m = 2 to Blanchard-Stubbe [BS]). 110 Proposition 12.5.2. 1 m Let V be in L (R ) (m =1, 2). Then S1 = ∆+ V has a negative eigenvalue if the following condition holds : − V (x) dx < 0 . (12.5.2) Rm Z Proof. m We just treat the case when V C0∞(R ). We first observe that the the essential∈ spectrum is [0, + [. For the proof of the proposition, it is then enough to find ψ D(S ) such that∞ ∈ 1 2 ψa′ (x) dx = a , R | | Z and 2 lim V (x) ψa(x) dx = V (x) dx < 0 , a 0 R | | R → Z Z by the dominated convergence Theorem. When m = 2, we can take ψ (x) = exp 1 x a, a> 0, then a − 2 | | 2 π ψa(x) dx = a , R2 ||∇ || 2 Z and 2 lim V (x) ψa(x) dx = V (x)dx < 0 . a 0 R2 | | R2 → Z Z 12.6 Essential spectrum and Persson’s Theorem We refer to Agmon’s book [Ag] for details. Theorem 12.6.1. Let V be a real-valued potential in the Kato-Rellich class38, and let H = ∆+V be the corresponding self-adjoint, semibounded Schr¨odinger operator with− do- main H2(Rm). Then, the bottom of the essential spectrum is given by inf σess(H)=Σ(H) , (12.6.1) where m Σ(H) := sup inf <φ,Hφ> φ C0∞(R ) , (12.6.2) Rm φ =1{ | ∈ \ K } K⊂ || || where the supremum is over all compact subset Rm. K⊂ 38Cf Theorem 10.3.1. 111 Essentially this is a corollary of Weyl’s Theorem 11.4.5. We will indeed play with the fact that Lemma 12.6.2. σess(H)= σess(H + W ) , for any regular potential W with compact support. 112 13 On the Schr¨odinger Operator with constant magnetic field The Schr¨odinger operator with magnetic field has been briefly introduced in Subsection 7.5. We have in particularly given examples where this operator was with compact resolvent. In this section, we analyze more in detail the properties of the Schr¨odinger operator with constant magnetic field in dimension 2 and 3. This appears to play an important role in Superconductivity theory. 13.1 Dimension 2 We would like to analyze the spectrum of : B B S := (D x )2 + (D + x )2 . (13.1.1) B x1 − 2 2 x2 2 1 13.1.1 The case of R2 We first look at the selfadjoint realization in R2. Let us show briefly, how one can analyze its spectrum. We leave as an exercise to show that the spectrum (or the discrete spectrum) of two selfadjoints operators S and T are the same 1 1 1 if there exists a unitary operator U such that U(S i)− U − = (T i)− . We note that this implies that U sends the domain of S±onto the domain± of T . In order to determine the spectrum of the operator SB, we perform a succession of unitary conjugations. The first one is called a gauge transformation. We introduce U on L2(R2) defined, for f L2 by 1 ∈ x x U f = exp iB 1 2 f . (13.1.2) 1 2 It satisfies S U f = U S1 f , f (R2) , (13.1.3) B 1 1 B ∀ ∈ S with 1 2 2 SB := (Dx1 ) + (Dx2 + Bx1) . (13.1.4) Remark 13.1.1. . U1 is a very special case of what is called a gauge transform. More generally, 2 we can consider U = exp iφ where φ is C∞. If∆A := j (Dxj Aj ) is a general Schr¨odinger operator associated with the magnetic potential− A, then 1 ˜ P U − ∆AU = ∆A˜ where A = A + grad φ. Here we observe that B := rot A = rot A˜. The associated magnetic field is unchanged in a gauge transformation. We are discussing in our example the very special case (but important!) when the magnetic potential is constant. 1 We have now to analyze the spectrum of SB. Observing that the operator is with constant coefficients with respect to the x2- variable, we perform a partial Fourier transform with respect to the x2 variable U2 = x ξ , (13.1.5) F 27→ 2 113 and get by conjugation, on L2(R2 ), x1,ξ2 2 2 2 SB := (Dx1 ) + (ξ2 + Bx1) . (13.1.6) We now introduce a third unitary transform U3 ξ (U f)(y , ξ )= f(x , ξ ) , with y = x + 2 , (13.1.7) 3 1 2 1 2 1 1 B and we obtain the operator 3 2 2 2 SB := Dy + B y , (13.1.8) operating on L2(R2 ). y,ξ2 The operator depends only on the y variable. It is easy to find for this operator an orthonormal basis of eigenvectors. We observe indeed that if f L2(R ), ∈ ξ2 and if φn is the (n + 1)-th eigenfunction of the harmonic oscillator, then 1 1 (x, ξ ) B 4 f(ξ ) φ ( B 2 y) 2 7→ | | 2 · n | | is an eigenvector corresponding to the eigenvalue (2n+1) B . So each eigenspace has an infinite dimension. An orthonormal basis of this eigen| | value can be given 1 1 2 by vectors ej (ξ2) B 4 f(ξ2) φn( B 2 y) where ej (j N) is a basis of L (R). We have consequently| | an empty| discrete| spectrum.∈ The eigenvalues are usually called Landau levels. 13.2 Magnetic Schr¨odinger operators in dimension 3 We only consider the Schr¨odinger operator with constant magnetic field in R3. After some rotation in R3, we arrive to the model : 2 2 2 P (h,~b)= D + (Dx bx1) + D , (13.2.1) x1 2 − x3 with b = B . || || This time, we can take the Partial Fourier transform, with respect to x2 and x3 in order to get the operator 2 2 2 D + (ξ2 bx1) + ξ . x1 − 3 When b = 0, we can translate in the x variable and get the operator on L2(R3) 6 1 D2 + ( b y )2 + ξ2 . y1 | | 1 3 It is then easy to see that the spectrum is [ b , + [. | | ∞ 114 + 13.3 The case of R2 . 13.3.1 Preliminaries For the analysis of the spectrum of the Neumann realization of the Schr¨odinger 2,+ 2 operator with constant magnetic field SB in R , we start like in the case of R till (13.1.6). Then we can no more perform the translation and hence we have 2 2 to look at the family (parametrized by ξ) Dx + (x + ξ) on the half line, with Neumann condition and more precisely to the groundstate energy µ(ξ). The bottom of the spectrum is effectively given by : R2,+ inf σ(SN, )= B inf µ(ξ)=Θ B . (13.3.1) B | | 0| | Similarly, for the Dirichlet realization, we find : R2,+ inf σ(SD, )= B inf λ(ξ)= B . (13.3.2) B | | | | 13.3.2 Main results for a 1D-model Let us begin with the analysis of a a family of ordinary differential operators, whose study will play an important role in the analysis of various examples. We consider the Neumann realization HN,ξ in L2(R+) associated to the operator 2 2 Dx + (x ξ) . It is easy to see that the operator is with compact resolvent and that− the lowest eigenvalue µ(ξ) of HN,ξ is simple . For the second point, the following simple argument can be used. Suppose by contradiction that the eigenspace is of dimension 2. Then, we can find in this eigenspace an eigenvector such that u such that u(0) = u′(0) = 0. But then it should be identically 0 by Cauchy uniqueness. We denote by ϕξ the corresponding strictly positive normal eigenvector. The minimax shows that µ(ξ) is continuous as a function of ξ. It is a little more work (admitted (see Kato [Ka]) to show that the function is C∞ and actually analytic (see in the next Subsubsection). It is immediate to show that µ(ξ) + as 2 2 →2 ∞ ξ . We can indeed compare by monotonicity with Dx + x + ξ . →The −∞ second remark is that µ(0) = 1. For this, we use the fact that the lowest 2 2 + eigenvalue of the Neumann realization of Dt +t in R is the same as the lowest 2 2 eigenvalue of Dt + t in R, but restricted to the even functions, which is also 2 2 the same as the lowest eigenvalue of Dt + t in R. Moreover the derivative of µ at 0 is strictly negative. It is a little more difficult to show that lim µ(ξ)=1 . ξ + → ∞ The proof can be done in the following way. For the upper bound, we ob- serve that µ(ξ) λ(ξ) where λ(ξ) is the eigenvalue of the Dirichlet realization. By monotonicity≤ of λ(ξ), it is easy to see that λ(ξ) is larger than one and tend to 1 as ξ + . Another way is to use the function exp 1 (x ξ)2 as a test → ∞ − 2 − 115 function. For the converse, we start from the eigenfunction φξ(x), show some uniform decay of φξ(x) near 0 as ξ + and use x χ(x + ξ)φξ(x + ξ) as a test function for the harmonic oscillator→ ∞ in R. 7→ N,ξ All these remarks lead to the observation that the infimum infξ R inf Sp (H ) is actually a minimum [DaHe] and stricly less than 1. Moreover one∈ can see that µ(ξ) > 0, for any ξ, so the minimum is strictly positive. To be more precise on the variation of µ, let us first establish 2 2 µ′(ξ)= [µ(ξ) ξ ]ϕ (0) . (13.3.3) − − ξ To get (13.3.3), we observe that, if τ > 0, then 2 2 0= R [Dt ϕξ(t) + (t ξ) ϕξ(t)]ϕξ+τ (t + τ)dt + − = ϕξ(0)ϕ′ (τ) + (µ(ξ + τ) µ(ξ)) R ϕξ(t)ϕξ+τ (t + τ) dt . R− ξ+τ − + We then take the limit τ 0 to get the formula.R → + From (13.3.3), it comes that, for any critical point ξc of µ in R 2 µ′′(ξc)=2ξcϕξc (0) > 0 . (13.3.4) So the critical points are necessarily non degenerate local minima. It is then easy to deduce that there exists a unique ξ0 > 0 such that µ(ξ) continues to decay monotonically till some value Θ0 < 1 and is then increasing monotonically and tending to 1 at + . Moreover ∞ 2 Θ0 = ξ0 . (13.3.5) Finally, it is easy to see that ϕ (x) decays exponentially at . ξ ∞ Let us show additional remarks on the properties of µ and ϕξ(x) which are related to the Feynman-Heilmann formula. We admit again (See Kato [Ka] and the next subsection) that we can “freely” differentiate with respect to ξ. Let us start39 from : HN (ξ)ϕ( ; ξ)= µ(ξ)ϕ( ; ξ) . (13.3.6) · · Differentiating with respect to ξ, we obtain : N N (∂ H (ξ) µ′(ξ))ϕ( ; ξ) + (H (ξ)ϕ( ; ξ) µ(ξ))(∂ ϕ)( ; ξ)=0 . (13.3.7) ξ − · · − ξ · 2 + Taking the scalar product with ϕξ in L (R ), we obtain the socalled Feynman- Heilmann Formula + N ∞ 2 µ′(ξ)= (∂ H (ξ)ϕ ϕ = 2 (t ξ) ϕ (t) dt . (13.3.8) h ξ ξ | ξi − − | ξ | Z0 39 N,ξ N We change a little the notations for H (this becomes H (ξ) ) and ϕξ in order to have an easier way for the differentiation. 116 Taking the scalar product with ∂ ϕ)( ; ξ), we obtain the identity : ξ · N N (∂ξH (ξ) µ′(ξ))ϕ( ; ξ) (∂ξϕ)( ; ξ) + (H (ξ)ϕ( ; ξ) µ(ξ))(∂ξ ϕ)( ; ξ) (∂ξϕ)( ; ξ) =0 . h − · | · i h · − · |(13.3.9)· i In particular, we obtain for ξ = ξ0 that : N N (∂ξH (ξ0)ϕ( ; ξ0) ∂ξϕ)( ; ξ0) + (H (ξ)ϕ( ; ξ0) µ(ξ0))(∂ξϕ)( ; ξ0) (∂ξϕ)( ; ξ0) =0 . h · | · i h · − · | (13.3.10)· i We observe that the second term is positive (and with some extra work coming back to (13.3.7) strictly positive) : (∂ HN (ξ ))ϕ( ; ξ ) ∂ ϕ)( ; ξ ) < 0 . (13.3.11) h ξ 0 · 0 | ξ · 0 i Let us differentiate one more (13.3.7) with respect to ξ. N 2(∂ξH (ξ) µ′(ξ))∂ξϕ( ; ξ) N − 2 · +(H (ξ) µ(ξ))(∂ξ ϕ)( ; ξ) (13.3.12) 2 N − · +(∂ H (ξ) µ′′(ξ))ϕ( ; ξ)=0 . ξ − · Taking the scalar product with ϕξ and ξ = ξ0, we obtain N µ′′(ξ )=2+ (∂ H (ξ )ϕ( ; ξ ) ∂ ϕ)( ; ξ ) < 2 . (13.3.13) 0 h ξ 0 · 0 | ξ · 0 i 13.3.3 Regularity Proposition 13.3.1. The eigenvalue µ(ξ) and the corresponding eigenvector φξ are of class C∞ with respect to ξ. Proof : Step 1 : N Let µ(ξ0) the lowest eigenvalue of H (ξ0). We recall that it is simple. Let ϕξ0 the normalized eigenvector attached to µ(ξ0). Let us denote by ϕξ∗0 the orthogonal projection on varphiξ0 . The domain of the operator can be seen as N 2 + D(H (ξ)) = u B (R ) u′(0) = 0 , { ∈ | and we observe that it is independent of ξ. We will use a variant of the so called Grushin’s method. Let us introduce the unbounded operator on L2(R+) C × N H (ξ0) µ(ξ0) ϕξ0 M0 : = − ϕ∗ 0 ξ0 N with domain D(H (ξ0)). Let us show its invertibility. By elementary algebra, we get that the inverse is + E0 E0 R0 = + , E− E − 0 0 117 with : N 1 E0 = ((H (ξ0) µ(ξ0)) ϕ⊥ )− (13.3.14) − | ξ0 + E0 = ϕξ0 (13.3.15) E0− = ϕξ∗0 (13.3.16) + E0 − = 0. (13.3.17) Step 2 : Let us now introduce HN (ξ) µ ϕ M(ξ, µ)= − ξ0 . ϕ∗ 0 ξ0 Let us show that the inversibility is stable when ξ remains near ξ0, and µ remains near µ(ξ0). We observe that : (H(ξ) H(ξ )) (µ µ(ξ )) 0 M(ξ, µ) = M + 0 0 0 − 0− − 0 (HN (ξ) HN (ξ )) (µ µ(ξ )) 0 = M Id + R 0 0 . 0 0 − 0− − 0 N N 2 But the map ξ H (ξ) is continuous from C into (D(H (ξ0),L ). So the result is clear and7→ the inverse can be given by the convergentL Neumann series : j 1 j (H(ξ) H(ξ0)) (µ µ(ξ0)) 0 M(ξ, µ)− = ( 1) R0 − − − R0. − 0 0 j N X∈ (13.3.18) Let us denote by E(ξ, µ) E+(ξ, µ) R(ξ, µ)= + E−(ξ, µ) E −(ξ, µ) the inverse of M(ξ, µ). The following result is standard : Lemma 13.3.2. The inverse of M(ξ, µ) is a C∞ map in a neighborhood of (ξ0, µ(ξ0)) with value in D(HN (ξ )) C). 0 × Proof of Lemma 13.3.2 : It is clear that (H(ξ) H(ξ )) (µ µ(ξ )) 0 T (ξ, µ)= R 0 0 (13.3.19) 0 0− − − 0 is C∞. Let us also observe that : 1 j j M(ξ, µ)− = ( 1) T (ξ, µ)R . (13.3.20) − 0 j N X∈ 118 Consider r(ξ, µ) = (H(ξ) H(ξ )) (µ µ(ξ )) . − 0 − − 0 The derivatives of r(ξ, µ) are given by : ∂ r(ξ, µ) = ∂ H(ξ)= 2(t ξ) ξ ξ (13.3.21) ∂ r(ξ, µ) = 1. − − µ − In view of (13.3.20), we will estimate T (ξ, µ)j : j j r(ξ, µ) 0 E0 ϕξ0 r(ξ, µ) 0 R0 = 0 0 ϕξ∗0 0 0 0 j (13.3.22) (E0r(ξ, µ)) 0 = j 1 . (ϕ∗ r(ξ, µ))(E0r(ξ, µ)) − 0 ξ0 So j 1 j (E0r(ξ, µ)) 0 M(ξ, µ)− = j N( 1) j 1 R0. (13.3.23) − (ϕ∗ r(ξ, µ))(E0r(ξ, µ)) − 0 ∈ ξ0 P It is then easy to show the C∞ property. Lemma 13.3.3. N + µ is an eigenvalue of H (ξ) if and only if E −(ξ, µ)=0. Moreover, if µ is an eigenvalue, E+(ξ, µ) is the associated eigenvector in a neighborhood of (ξ0, µ(ξ0)). Proof of 13.3.3 : Again this is simple linear algebra. Expressing that R(ξ, µ) is the inverse of M(ξ, µ) gives : + E(ξ, µ)(H(ξ) µ)+ E (ξ, µ)ϕ∗ = Id (13.3.24) − ξ0 E−(ξ, µ)ϕξ0 = 1 (13.3.25) E(ξ, µ)ϕξ0 = 0 (13.3.26) + E−(ξ, µ)(H(ξ) µ)+ E −(ξ, µ)ϕ∗ = 0 (13.3.27) − ξ0 (H(ξ) µ)E(ξ, µ)+ ϕξ0 E−(ξ, µ) = Id (13.3.28) − + ϕξ∗0 Eξ,µ = 1 (13.3.29) + + (H(ξ) µ)E (ξ, µ)+ ϕ E −(ξ, µ) = 0 (13.3.30) − ξ0 ϕξ∗0 E(ξ, µ) = 0. (13.3.31) Taking the composition of E+(ξ, µ) with (13.3.27) on the left, we get + + + E (ξ, µ)E−(ξ, µ)(H(ξ) µ)+ E −(ξ, µ)E (ξ, µ)ϕ∗ =0, − ξ0 so + + E (ξ, µ)E−(ξ, µ)(H(ξ) µ) E (ξ, µ)ϕ∗ = . (13.3.32) ξ0 + − − E −(ξ, µ) 119 + This quantity (13.3.32) is well defined if Eξ,µ− = 0. So using (13.3.24) et(13.3.32) we get, 6 + 1 E (ξ, µ)E−(ξ, µ) (H(ξ) µ)− = E(ξ, µ) + . (13.3.33) − − E −(ξ, µ) + So we have shown that if E − = 0, then H(ξ) µ is invertible. ξ,µ 6 − + + Conversely, let us assume that Eξ,µ− = 0. Then using (13.3.30), E (ξ, µ) is an + + + eigenvector as soon that E (ξ, µ) is different from 0. But E = E = ϕξ ξ0,µ(ξ0) 0 0 is non zero, so by continuityit is also true for E+(ξ, µ) . + Step 3 : Analysis of the equation E −(ξ, µ)=0. We have just to compute the implicit function theorem in the neighborhood of (ξ0, µ(ξ0). But by elementary computations, we have + j j 1 E −(ξ, µ)= ( 1) (ϕ∗ rξ,µ)(E0r(ξ, µ)) − ϕξ . (13.3.34) − ξ0 0 j 1 X≥ The derivatives at (ξ0, µ(ξ0) are easily computed as : + ∂ξE(ξ, µ) −(ξ0, µ(ξ0)) = ϕ∗ ∂ξr(ξ0, µ(ξ0))ϕξ (13.3.35) − ξ0 0 2 = 2 (t ξ0)ϕξ0 (t) dt (13.3.36) R+ − + Z ∂ E −(ξ, µ(ξ )) = 1. (13.3.37) µ ξ,µ 0 − + In particular ∂ E −(ξ , µ(ξ )) = 0 and the implicit function theorem leads to µ ξ,µ 0 0 6 Lemma 13.3.4. There exists η> 0 and a C∞ map µ˜ on ]ξ η, ξ + η[ such that 0 − 0 + ξ ]ξ η, ξ + η[, µ ]µ(ξ ) η, µ(ξ )+ η[, E − =0 µ =µ ˜(ξ). ∀ ∈ 0 − 0 ∀ ∈ 0 − 0 ξ,µ ⇐⇒ We then obtain a C∞ function ξ µ˜(ξ) such thatµ ˜(ξ) is an eigenvalue of N 7→ H (ξ) and which is equal to µ(ξ) at ξ0. By uniqueness, we get that for ξ ξ0 small enoughµ ˜(ξ)= µ(ξ). | − | 13.4 Notes Old results are due to Kato [Ka] and Avron-Herbst-Simon [AHS], but we have also added more recent results of Dauge-Helffer [DaHe]. 120 14 Exercises and Problems 14.1 Introduction We present some exercises or problems proposed in the last years. They some- times strongly intersect with the course. 14.2 Exercises Exercise 14.2.1. (After Effros, Avron-Seiler-Simon). Let P and Q two selfadjoint projectors in a Hilbert space . i) Let us assume that A = P Q is compact. Show that if λ = 1 is in theH spectrum, then λ is in the spectrum− with the same multiplicity.6 For± this, one can first show that− with B = I P Q, − − A2 + B2 = I, AB + BA =0 . ii) Assume now that A is in addition trace class, that is that the series µ , | j | where µj are the non zero eigenvalues of A, counted with multiplicity. Compute Tr A := j µj and show that it is an integer. ExerciseP 14.2.2. (Temple’s inequality). Let A be a selfadjoint operator on an Hilbert space and ψ D(A) such that ψ =1. ∈ Suppose|| || that λ is the unique eigenvalue of A in some interval ]α, β[ and that η = ψ Aψ belongs to the interval ]α, β[. Then show that : h | i ǫ2 ǫ2 η λ η + . − β η ≤ ≤ η α − − with : ǫ2 = (A η)ψ 2 . || − || As a preliminary result, one can show that (A α)(A λ) and (A β)(A λ) are positive operators. Then apply the inequalities− with−ψ. − − Show that this inequality is an improvment if ǫ2 (β η)(η α). ≤ − − Exercise 14.2.3. . 2 Let A(x1, x2) = (A1(x1, x2) , A2(x1, x2)) be a C∞ vector field on R . Let V be 2 a C∞ positive function on R . 2 2 Let P := (Dx1 A1(x1, x2)) + (Dx2 A2(x1, x2)) + V (x) the differential − 2 − operator defined on C0∞(R ). a) Show that P admits a selfadjoint extension in L2(R2). b) Show that P is essentially selfadjoint. Exercise 14.2.4. . We admit the results of Exercise 14.2.3. Show that the selfadjoint extension in L2(R2) of 2 d 2 2 d 2 T := ( ix2x1) 2 + x2 , − dx1 − − dx2 is with compact resolvent. 121 Exercise 14.2.5. . 2 Let V be a C∞ positive potential in R . Let us consider, with B R 0 , the operator ∈ \{ } 2 2 P = Dx1 + (Dx2 + B x1) + V (x) . a) Recall briefly the spectrum of its selfadjoint extension in the case V =0. b) We assume that V tends to 0 as x + . Determine the essential spectrum of P¯. | | → ∞ Exercise 14.2.6. . Let K be a kernel in (R2) which is strictly positive and symmetric. a) Show that the associatedS operator which is defined on (R) by K S ( u)(x)= K(x, y)u(y)dy , K R Z can be extended as a compact operator on L2(R). b) Let I be an open interval in R and let us denote by the operator on L2(I) defined by KI ( u)(x)= K(x, y)u(y)dy . KI ZI Let λ1 be the largest eigenvalue of . Show that I KI 1 1 λ λR . I ≤ Show that we have strict inequality when I is not R. 1 1 c) Let u be a normalized eigenfunction of associated with λR. Using its restriction to I, show the inequality : K 1 1 1 1 λR λ (1 u 2 C )− . ≤ I − || ||L (I ) 1 1 d) Let In = [ n,n]. Show that λIn converges rapidly to λR as n + . More precisely, show− that, for all j N, there exists a constant C such→ that∞ : ∈ j 1 1 j λR λ C n− , n N∗ . | − In |≤ j ∀ ∈ Exercise 14.2.7. . Let us consider in Ω =]0, 1[ R, a positive C∞ function V and let S be the × 0 Schr¨odinger operator S0 = ∆+ V defined on C0∞(Ω). − 2 (a) Show that S0 admits a selfadjoint extension on L (Ω). Let S this extension. (b) Determine if S is with compact resolvent in the following cases : 1. V (x)=0, 2 2 2. V (x)= x1 + x2, 2 3. V (x)= x1, 2 4. V (x)= x2 122 5. V (x) = (x x )2. 1 − 2 Determine the spectrum in the cases (1) and (4). One can first determine the spectrum of the Dirichlet realization (or of Neumann) of d2/dx2 on ]0, 1[. − Exercise 14.2.8. . We consider in R2 the operator defined on (R2 ; C2) by S = α D + α D + α . D0 1 x1 2 x2 3 Here the matrices α are hermitian 2 2 matrices such that : j × αiαj + αj αi =2δij , 1 ∂ and we recall that Dxj = for j =1, 2. i ∂xj a) Is 0 symmetric? semi-bounded ? It is suggested to use the Fourier transform.D 2 b) Compute 0. D 2 2 2 c) Show that 0 admits a selfadjoint extension 1 in L (R ; C ), and de- termine its domain.D D d) Determine the spectrum of 1. e) We suppose that, for all x DR2, V (x) is a 2 2 hermitian matrix, with ∈ × bounded C∞ coefficients. Show that V = 0 +V admits a selfadjoint extension and determine its domain. D D Exercise 14.2.9. . Let H be the Dirichlet realization of d2/dx2 + x2 in ] a, +a[. Show that the a − − lowest eigenvalue λ1(a) of Ha is strictly positive, monotonically decreasing as a + and tend exponentially fast to 1 as a + . Give an estimate as fine as→ possible∞ of λ (a) 1 . → ∞ | 1 − | Exercise 14.2.10. . 2 We consider on C0∞(R ), the operator P := (D x )2 + (D x )2 . 0 x1 − 2 x2 − 1 1 We recall that Dxj = i ∂xj . Show that its natural selfadjoint extension P is unitary equivalent to the op- erator ∆ (of domain H2). Determine its spectrum and its essential spectrum. − Exercise 14.2.11. . Show that one can associate to the differential operator on C∞(R ]0, 1[) : 0 × T := (D x x2)2 + (D )2 , 0 x1 − 2 1 x2 an unbounded selfadjoint operator T on L2(R ]0, 1[) whose spectrum is with compact resolvent. × 123 Exercise 14.2.12. . Let φ be a C2- function on Rm such that φ(x) + as x + and with uniformly bounded second derivatives. Let|∇ us consider| → ∞ the d|ifferential| → ∞ operator 2 on C0∞(R ) ∆+2 φ . We consider this operator as an unbounded operator on = L2(−Rm, exp∇ 2·φ ∇ dx). Show that it admits a selfadjoint extension and thatH its spectrum is discrete.− We assume in addition that : Rm exp 2φdx < + . Show that its lowest eigenvalue is simple and determine a corresponding− eigenve∞ ctor. R Exercise 14.2.13. . 3 1 We consider in R the differential operator S0 := ∆ r , a priori defined on 3 − − C0∞(R ). a) Show that the operator admits a selfadjoint extension S. b) Show the continuous injection of H2(R3) into the space of the H¨older func- s 3 1 tions C (R ), with s ]0, 2 [, and the compact injection for all compact K of Cs(K) into C0(K). ∈ c) Determine the essential spectrum of S. One possibility is to start with the χ analysis of Sχ = ∆ r where χ is C∞ with compact support. d) Show using the− minimax-principle− that S has at least one eigenvalue. One 2 can try to minimize over a u 124 Exercise 14.2.18. . Let V (Rd) be positive and let us consider the operator : ∈ S 1 1 T := ( ∆+1)− 2 V ( ∆+1)− 2 . − − 1 2 d a) Explain how to define ( ∆+1)− 2 , as an operator on L (R ). b) Show that T is a bounded,− selfadjoint, positive, compact operator on L2(Rd). c) Discuss the injectivity in function of V ? d) Establish a link with the research of pairs (u, µ) in H2(Rd) R+ such that : × ( ∆+1 µV )u =0 . − − Exercise 14.2.19. . Let δ R. ∈ 2 (a) Show that the operator Pδ defined on C0∞(R ) by 2 2 2 4 4 2 Pδ := Dx + Dy + x y + x y + δ(x + y) is semibounded. One can first show the inequality : P u, u xu 2 + yu 2 . h 0 i ≥ || || || || (b) Show that there exists a natural selfadjoint extension of Pδ. (c) What is the corresponding form domain ? (d) Show that the selfadjoint extension is with compact resolvent. Exercise 14.2.20. . Let us consider in R+, the Neumann realization in R+ of P (ξ) := D2 +(t ξ)2, 0 t − where ξ is a parameter in R. We would like to find an uppr bound for Θ0 = infξ µ(ξ) where µ(ξ) is the smallest eigenvalue of P0(ξ). Following the physicist Kittel, one can proceed by minimizing P0(ξ)φ( ; ρ) φ( ; ρ) over the normalized 2 h · | · i functions φ(t; ρ) := cρ exp ρt (ρ > 0). For which value of ξ is this quantity minimal ?? Deduce the inequality− : 2 Θ0 < 1 . r − π 14.3 Problems Problem 14.3.1. . Let us consider in the disk of R2 Ω := D(0, R) the Dirichlet realization of the Schr¨odinger operator 1 S(h) := ∆+ V (x) , (14.3.1) − h2 where V is a C∞ potential on Ω¯ satisfying : V (x) 0 . (14.3.2) ≥ 125 Here h> 0 is a parameter. a) Show that this operator is with compact resolvent. b) Let λ1(h) be the lowest eigenvalue of S(h). We would like to analyze the behavior of λ1(h) as h 0. Show that h λ1(h) is monotonically increasing. c) Let us assume that V→ > 0 on Ω¯; show→ that there exists ǫ> 0 such that h2λ (h) ǫ . (14.3.3) 1 ≥ d) We assume now that V =0 in an open set ω in Ω. Show that there exists a constant C > 0 such that, for any h> 0, λ (h) C . (14.3.4) 1 ≤ One can use the study of the Dirichlet realization of ∆ in ω. e) Let us assume that : − V > 0 almost everywhere in Ω . (14.3.5) Show that, under this assumption : lim λ1(h)=+ . (14.3.6) h 0 ∞ → One could proceed by contradiction supposing that there exists C such that λ (h) C , h such that 1 h> 0 . (14.3.7) 1 ≤ ∀ ≥ and establishing the following properties. For h> 0, let us denote by x u (h)(x) an L2-normalized eigenfunction • 7→ 1 associted with λ1(h). Show that the family u1(h) (0 < h 1) is bounded in H1(Ω). ≤ Show the existence of a sequence hn (n N) tending to 0 as n + and • u L2(Ω) such that ∈ → ∞ ∞ ∈ lim u1(hn)= u n + ∞ → ∞ in L2(Ω). Deduce that : • V (x) u (x)2 dx =0 . ∞ ZΩ Deduce that u =0 and make explicite the contradiction. • ∞ f) Let us assume that V (0) = 0; show that there exists a constant C, such that : C λ (h) . 1 ≤ h g) Let us assume that V (x)= ( x 4) pr`es de 0. Show that in this case : O | | C λ1(h) 2 . ≤ h 3 126 h) We assume that V (x) x 2 near 0; discuss if one can hope a lower bound in the form ∼ | | 1 λ (h) . 1 ≥ Ch Justify the answer by illustrating the arguments by examples and counterexam- ples. Problem 14.3.2. . 1 2 2 We consider on R and for ǫ I := [ 4 , + [ the operator Hǫ = d /dx + x2 + ǫ x . ∈ − ∞ − | | a) Determine the form domain of Hǫ and show that it is independent of ǫ. b) What is the nature of the spectrum of the associated selfadjoint operator? c) Let λ1(ǫ) the smallest eigenvalue. Give rough estimates permitting to estimate from above or below λ1(ǫ) independently of ǫ on every compact interval of I. d) Show that, for any compact sub-interval J of I, there exists a constant CJ such that, for all ǫ J, any L2-normalized eigenfunction u of H associated ∈ ǫ ǫ with λ1(ǫ) satisfies : u 1 R C . || ǫ||B ( ) ≤ J For this, on can play with : Hǫuǫ,uǫ L2(R). e) Show that the lowest eigenvalueh isi a monotonically increasing sequence of ǫ I. f)∈ Show that the lowest eigenvalue is a locally Lipschitzian function of ǫ I. On utilisera de nouveau le principe du max-min. ∈ g) Show that λ(ǫ) + , as ǫ + and estimate the asymptotic behavior. h) Discuss the same→ questions∞ for→ the∞ case H = d2/dx2 +x2 +ǫx4 (with ǫ 0). ǫ − ≥ Problem 14.3.3. . The aim of this problem is to analyze the spectrum ΣD(P ) of the Dirichlet real- ization of the operator P := (D 1 x )2 + (D + 1 x )2 in R+ R. x1 − 2 2 x2 2 1 × 1. Show that one can a priori compare the infimum of the spectrum of P in R2and the infimum of ΣD(P ). 2. Compare ΣD(P ) with the spectrum ΣD(Q) of the Dirichlet realization of 2 2 + Q := D + (y1 y2) in R R. y1 − × 3. We first consider the following family of Dirichlet problems associated with the family of differential operators : α H(α) defined on ]0, + [ by : 7→ ∞ H(α)= D2 + (t α)2 . t − Compare with the Dirichlet realization of the harmonic oscillator in ] α, + [. − ∞ 4. Show that the lowest eigenvalue λ(α) of H(α) is a monotonic function of α R. ∈ 127 5. Show that α λ(α) is a continuous function on R. 7→ 6. Analyze the limit of λ(α) as α . → −∞ 7. Analyze the limit of λ(α) as α + . → ∞ 8. Compute λ(0). For this, one can compare the spectrum of H(0) with the spectrum of the harmonic oscillator restricted to the odd functions. 9. Let t u(t; α) the positive L2-normalized eigenfunction associated with λ(α).7→ Let us admit that this is the restriction to R+ of a function in (R). + S Let, for α R, T be the distribution in ′(R R) d´efined by ∈ α D × + ∞ φ T (φ)= φ(y , α)u (y )dy . 7→ α 1 α 1 1 Z0 Compute QTα. 2 10. By constructing starting from Tα a suitable sequence of L -functions tend- ing to T , show that λ(α) ΣD(Q). α ∈ 11. Determine ΣD(P ). Problem 14.3.4. . 2 2 2 Let Ha be the Dirichlet realization of d /dx + x in ] a, +a[. (a) Briefly recall the results concerning− the case a =+ −. ∞ (b) Show that the lowest eigenvalue λ1(a) of Ha is decreasing for a ]0, + [ and larger than 1. ∈ ∞ (c) Show that λ1(a) tends exponentially fast to 1 as a + . One can use a suitable construction of approximate eigenvectors. → ∞ (d) What is the behavior of λ1(a) as a 0. 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