Krein - de Branges theory in spectral analysis Alexei Poltoratski - PowerPoint PPT Presentation

Krein - de Branges theory in spectral analysis Alexei Poltoratski Texas A&M October 23, 2013 Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 1 / 24

Krein’s systems � 0 � 1 Symplectic structure on R 2 : Ω = , { x , y } = (Ω x , y ) . − 1 0 Consider a 2 × 2 differential system with a spectral parameter z : Ω ˙ X = zH ( t ) X − Q ( t ) X , t − < t < t + � u ( t ) � where X ( t ) = . We assume the (real-valued) coefficients to satisfy v ( t ) H , Q ∈ L 1 loc (( t − , t + ) → R 2 × 2 ) . By definition, a solution X = X z ( t ) is a C 2 (( t − , t + ))-function satisfying the equation. Theorem Every IVP has a unique solution on ( t − , t + ) . For each fixed t, this solution presents an entire function u z ( t ) + iv z ( t ) of z of exponential type. Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 2 / 24

Self-adjoint systems ( ∗ ) Ω ˙ X = zH ( t ) X − Q ( t ) X , t − < t < t + . We may further assume that H ( t ) , Q ( t ) are real symmetric locally summable matrix-valued functions and that H ( t ) ≥ 0. The Hilbert space L 2 ( H ) consists of (equivalence classes) of vector-functions with � t + || f || 2 H = { Hf , f } dt < ∞ . t − The system ( ∗ ) is an eigenvalue equation DX = zX for the (formal) differential operator � Ω d � D = H − 1 dt + Q . Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 3 / 24

Schr¨ odinger equations: − ¨ u = zu − qu . u and X = ( u , v ) T to obtain Put v = − ˙ � 1 � � q � 0 0 Ω ˙ X = z X − X . 0 − 1 0 0 Dirac systems: H ≡ I . The general form is � 1 � � q 11 � 0 q 12 Ω ˙ X = z X − X , q 12 = q 21 . 0 1 q 21 q 22 � − q 2 � − q 1 The ”standard form”: Q = . In this case f = q 1 + iq 2 is the − q 1 q 2 potential function. Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 4 / 24

Krein’s Canonical Systems Canonical Systems are self-adjoint systems with Q ≡ 0: Ω ˙ X = zH ( t ) X . A general self-adjoint system can be reduced to canonical form: To reduce Ω ˙ X = zH ( t ) X − Q ( t ) X , ( ∗ ) solve Ω ˙ V = − QV and make a substitution X = VY . Then ( ∗ ) becomes Ω ˙ Y = z [ V ∗ HV ] Y . Example � t � � e − 2 0 f 0 H CS = Dirac system with real potential f : . � t e − 2 0 f 0 Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 5 / 24

de Branges’ spaces of entire functions Hardy space in C + : � H 2 = { f ∈ Hol ( C + ) | || f || 2 | f ( x + iy ) | 2 dx < ∞} . H 2 = sup y > 0 R Notation: if E ( z ) is entire we denote E # ( z ) = ¯ E (¯ z ). Hermite-Biehler entire functions An entire E ( z ) is a Hermit-Biehler function ( E ∈ HB ) if | E # ( z ) | < | E ( z ) | , z ∈ C + . de Branges’ space B ( E ) If E ∈ HB then B ( E ) is defined as the space of entire functions F such that F / E , F # / E ∈ H 2 . Hilbert structure: if F , G ∈ B ( E ) then G ( x ) dx � F ( x )¯ < F , G > B ( E ) = < F / E , G / E > H 2 = | E | 2 . R Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 6 / 24

de Branges’ spaces of entire functions: axiomatic definition Theorem (de Branges) Suppose that H is a Hilbert space of entire functions that satisfies (A1) F ∈ H , F ( λ ) = 0 ⇒ F ( z )( z − ¯ λ ) / ( z − λ ) ∈ H with the same norm (A2) ∀ λ �∈ R , the point evaluation is bounded (A3) F → F # is an isometry Then H = B ( E ) for some E ∈ HB. Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 7 / 24

Examples of dB spaces Example E is a polynomial. E ∈ HB ⇔ all zeros are in ¯ C − . B ( E ) consists of all poynomials of lesser degree. Example E = e − iaz , B ( E ) = PW a (Payley-Wiener space). Example Let µ > 0 be a finite measure on R such that polynomials are incomplete in L 2 ( µ ). Then the closure of polynomials is a de Branges space. Example The same example with E a = { e ict , 0 ≤ c ≤ a } in place of polynomials. (What is E in the last two examples ???!!!) Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 8 / 24

Krein Systems meet de Branges’ spaces Let E be an Hermite-Biehler function. Put A = ( E + E # ) / 2 , B = ( E − E # ) / 2 i . Reproducing kernels for B ( E ): for any λ ∈ C , F ∈ B ( E ), F ( λ ) = < F , K λ > where B ( z )¯ A ( λ ) − A ( z )¯ K λ ( z ) = 1 B ( λ ) . z − ¯ π λ We will consider canonical systems Ω ˙ X ( t ) = zH ( t ) X ( t ) without ”jump intervals”, i.e. intervals where H is a constant matrix of rank 1. Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 9 / 24

Krein Systems meet de Branges’ spaces Solve a canonical system with any real initial condition at t − . Denote the solution by X z ( t ) = ( A t ( z ) , B t ( z )). Theorem For any fixed t, E t ( z ) = A t ( z ) − iB t ( z ) is a Hermit-Biehler entire function. The map W defined as WX z = K t z extends unitarily to ¯ W : L 2 ( H , ( t − , t )) → B ( E t ) (Weyl transform). The formula for W : � t Wf ( z ) = < Hf , X ¯ z > L 2 ( H , ( t − , t )) = < H ( t ) f ( t ) , X ¯ z ( t ) > dt . t − Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 10 / 24

Examples of Weyl transforms Krein- de Branges’ theory: W Canonical System on ( t − , t + ) ← → B ( E t ) , t ∈ [ t − , t + ) Example Orthogonal polynomials satisfy difference equations corresponding to Krein systems with jump intervals. B ( E t ) = B n is the same on each jump interval, B n = P n . Example Free Dirac ( Q = 0): E t = e − 2 π izt , B ( E t ) = PW t as sets. Theorem Let B ( E t ) be the chain of de Branges’ spaces corresponding to a Dirac system with an L 1 loc -potential. Then B ( E t ) = PW t as sets. Gelfand-Levitan theory: a study of systems with B ( E t ) = PW t as sets. Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 11 / 24

Let B ( E t ) be a chain of de Branges’ spaces, t ∈ [ t − , t + ) (the final space B ( E t + ) may or may not exist). There exists a locally finite positive measure µ on R such that || f || B ( E t ) = || f || L 2 ( µ ) for all f ∈ B ( E t ) and all t . µ is the spectral measure for the corresponding Krein’s system. Relation with de Branges’ functions: 1 | E t | 2 → µ as t → t + . (In the limit circle case the limit will produce one of spectral measures.) Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 12 / 24

Consider the Dirac system � 1 � � − q 2 � 0 q 1 Ω ˙ X = z X − X 0 0 − q 1 q 2 with potential q = q 1 + iq 2 . Let B ( E t ) = PW t (as sets) be the corresponding chain of de Branges’ spaces. If K t 0 is the reproducing kernel for B t = B ( E t ) then via the formula for the Weyl transform we get d 0 (0) = d dt K t dt || K t 0 || 2 B t = E 2 t (0) . Recalling that E t = A t − iB t , where X z ( t ) = ( A t ( z ) , B t ( z )) T is a solution to the initial system, we obtain d dt E t (0) = − qE t (0) or q = − 1 dt log E t (0) 2 = − 1 d dt log d d 0 || 2 dt || K t 2 2 (Gelfand-Levitan formula). Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 13 / 24

We denote by ˆ µ the Fourier transform of µ : � e − 2 π izt d µ ( t ) . µ ( z ) = ˆ Theorem (Krein) Let q and µ be the potential and spectral measure of a Dirac system on R + . Then q ∈ C ( R + ) iff ˆ µ = δ 0 + φ , where φ ∈ C ( R ) . Proof of the ’if’ part: For any f ∈ B t ( ∈ PW t ) � t � f = f (0) = < f , K t ˆ f ¯ K t 0 > B t = 0 d µ − t 0 = ψ t then the last equation implies if we put ˆ K t 1 = ψ t ∗ ˆ µ = ψ t + ψ t ∗ φ [ − t , t ] on Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 14 / 24

We obtained that the Fourier transform ψ t = ψ of K t 0 satisfies the Volterra equation ( I + K t ) ψ = 1 , Where K t is an operator on L 2 [ − t , t ], K t f = f ∗ φ . The operator K t is an integral operator with a continuous kernel. Hence K t is compact (approximate the kernel with polynomials). Hence I + K t is Fredholm. Since < ( I + K t ) f , g > L 2 [ − t , t ] = < f , g > L 2 ( µ ) = < f , g > B t , I + K t has a trivial kernel. Therefore, I + K t is invertible and ψ t = ( I + K t ) − 1 1 . By the Fredholm-Hilbert Lemma on solutions of integral equations, ψ t ( x ) is differentiable with respect to t for each fixed x ∈ [ − t , t ] and the derivative d dt ψ t ( x ) is a continuous function of x . Alexei Poltoratski (Texas A&M) Krein - de Branges theory in spectral analysis October 23, 2013 15 / 24