The Singularly Continuous Spectrum and Non-Closed Invariant Subspaces

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1 Operator Theory: Advances and Applications, Vol. 160, c 2005 Birkhäuser Verlag Basel/Switzerland The Singularly Continuous Spectrum and Non-Closed Invariant Subspaces Vadim Kostrykin and Konstantin A. Makarov Dedicated to Israel Gohberg on the occasion of his 75th birthday Abstract. Let A be a bounded self-adjoint operator on a separable Hilbert space H and H 0 H aclosedinvariantsubspaceofa. AssumingthatH 0 is of codimension 1, we study the variation of the invariant subspace H 0 under bounded self-adjoint perturbations V of A that are off-diagonal with respect to the decomposition H = H 0 H 1.Inparticular,weprovetheexistenceofa one-parameter family of dense non-closed invariant subspaces of the operator A + V provided that this operator has a nonempty singularly continuous spectrum. We show that such subspaces are related to non-closable densely defined solutions of the operator Riccati equation associated with generalized eigenfunctions corresponding to the singularly continuous spectrum of B. Mathematics Subject Classification (2000). Primary 47A55, 47A15; Secondary 47B15. Keywords. Invariant subspaces, operator Riccati equation, singular spectrum. 1. Introduction In the present article we address the problem of a perturbation of invariant subspaces of self-adjoint operators on a separable Hilbert space H and related questions on the existence of solutions to the operator Riccati equation. Given a self-adjoint operator A and a closed invariant subspace H 0 H of A we set A i = A Hi, i =0, 1, with H 1 = H H 0.AssumingthattheperturbationV is off-diagonal with respect to the orthogonal decomposition H = H 0 H 1 consider the self-adjoint operator B = A + V = A0 V V A 1 with V = 0 V V, 0

2 300 V. Kostrykin and K.A. Makarov where V is a linear operator from H 1 to H 0.Itiswellknown(see,e.g.,[7])that the Riccati equation A 1 X XA 0 XVX + V =0 (1) has a closed (possibly unbounded) solution X : H 0 H 1 if and only if its graph G(H 0,X) := {x H x = x 0 Xx 0,x 0 Dom(X) H 0 } (2) is an invariant closed subspace for the operator B. Sufficient conditions guaranteeing the existence of a solution to equation (1) require in general the assumption that the spectra of the operators A 0 and A 1 are separated, d := dist(spec(a 0 ), spec(a 1 )) > 0, (3) and hence H 0 and H 1 are necessarily spectral invariant subspaces of the operator A. Inparticular(see[9]),if V <c π d with c π = 3π π π 2 4 = , (4) then the Riccati equation (1) has a bounded solution X satisfying the bound X π V < 1 1+ X 2 2 d δ V with 1 2 V δ V = V tan arctan. 2 d It is plausible to conjecture that condition (4) can be relaxed by the weaker requirement V < 3d/2 (see[9]fordetails).however,noproofofthatisavailable as yet. In general, without additional assumptions, neither condition (3) nor a smallness assumption like (4) on the magnitude of the perturbation V can be dropped. However, if the spectra of A 0 and A 1 are subordinated in the sense that sup spec(a 0 ) inf spec(a 1 ), then for any V with arbitrary large norm the Riccati equation (1) has a contractive solution [8] (see also [1]). Note that in this case the invariant subspaces H 0 and H 1 are not necessarily supposed to be spectral invariant subspaces of A. In the present work we prove new existence results for the Riccati equation under the assumption that the subspace H 1 is one-dimensional. In particular, these results imply the existence of a one-parameter family of non-closed invariant subspaces of the self-adjoint operator B, providedthatb has nonempty singularly continuous spectrum. The main result of our paper is presented by the following theorem. Theorem 1. Assume that dim H 1 =1and suppose that H 0 is a cyclic subspace for the operator A 0 generated by the one-dimensional subspace Ran V.LetS pp denote the set of all eigenvalues of the operator B.

3 Non-Closed Invariant Subspaces 301 Then there exists a minimal support S s of the singular part of the spectral measure of the operator B such that: (i) For any λ S sc = S s \ S pp the subspace Ψ(λ) =G(H 0,X λ ) H is a dense non-closed graph subspace with X λ : H 0 H 1 anon-closeddenselydefined operator solving the Riccati equation (1) in the sense of Definition 2.3 below. (ii) For any λ S pp S s the subspace Ψ(λ) =G(H 0,X λ ) H is a closed graph subspace of codimension 1 with X λ : H 0 H 1 aboundedoperatorsolving the Riccati equation (1). Moreover,theoperatorX λ is an isolated point (in the operator norm topology) of the set of all bounded solutions to the Riccati equation. The mapping Ψ from S s to the set M(B) of all (not necessarily closed) subspaces of H invariant with respect to the operator B is injective. The article is organized as follows. In Section 2 we establish a link between non-closable densely defined solutions to the Riccati equation (1) and the associated non-closed invariant subspaces of the operator B.InSection3accommodating the Simon-Wolff theory [10] to rank two off-diagonal perturbations we perform the spectral analysis of this operator under the assumption that dim H 1 = 1. The main result of this section is Theorem 3.4. Theorem 1 will be proven in Section 4. Throughout the whole work the Hilbert space H will assumed to be separable. The notation B(M, N) isusedforthesetofboundedlinearoperatorsfromthe Hilbert space M to the Hilbert space N. We will write B(N) insteadofb(n, N). 2. Non-closed graph subspaces Let H 0 be a closed subspace of a Hilbert space H and X adenselydefined(possibly unbounded and not necessarily closed) operator from H 0 to H 1 = H 0 := H H 0 with domain Dom(X). A linear subspace G(H 0,X) := {x H x = x 0 Xx 0,x 0 Dom(X) H 0 } is called the graph subspace of H associated with the pair (H 0,X)or,inshort,the graph of X. Recalling general facts on densely defined closable operators (see, e.g., [6]) we mention the following: If X : H 0 H 1 is a densely defined non-closable operator, then G(H 0,X)isanon-closedsubspaceofH. Itsclosureisnotagraphsubspace, i.e., there is no closed operator Y such that G(H 0,X)=G(H 0,Y). Proposition 2.1. Let X : H 0 H 1 be a densely defined non-closable operator. Then the closed subspace G(H 0,X) contains an element orthogonal to H 0. Proof. First, for X : H 0 H 1 being a densely defined non-closable operator we prove the following alternative: either the closed subspace G(H 0,X)contains an element orthogonal to H 0 or the subspace H 0 contains an element orthogonal

4 302 V. Kostrykin and K.A. Makarov to G(H 0,X). Indeed, assume on the contrary that neither the closed subspace G(H 0,X)containsanelementorthogonaltoH 0 nor the subspace H 0 contains an element orthogonal to G(H 0,X). Then by Theorem 3.2 in [7] there is a closed densely defined operator Y : H 0 H 1 such that G(H 0,X)=G(H 0,Y), which is a contradiction. Now assume that the subspace H 0 contains an element x 0 orthogonal to G(H 0,X). Obviously, this element is orthogonal to G(H 0,X), that is, x 0 0,x 0 Xx 0 =0,andhencex 0 = 0. Then, by the alternative proven above the subspace G(H 0,X)containsanelementorthogonaltoH 0,completingtheproof. For notational setup assume the following hypothesis. Hypothesis 2.2. Let B be a self-adjoint operator represented with respect to the decomposition H = H 0 H 1 as a 2 2 operator block matrix A0 V B = V, (5) A 1 where A i B(H i ), i =0, 1, areboundedself-adjointoperatorsinh i while V B(H 1, H 0 ) is a bounded operator from H 1 to H 0.Moreexplicitly,B = A + V, where A is the bounded diagonal self-adjoint operator, A0 0 A =, (6) 0 A 1 and the operator V = V is an off-diagonal bounded operator 0 V V = V. (7) 0 Definition 2.3. Adenselydefined(possiblyunboundedandnotnecessarilyclosable) operator X from H 0 to H 1 with domain Dom(X) is called a strong solution to the Riccati equation A 1 X XA 0 XVX + V =0 (8) if and Ran(A 0 + VX) Dom(X) Dom(X) A 1 Xx X(A 0 + VX)x + V x =0 for any x Dom(X). Theorem 2.4. Assume Hypothesis 2.2. Adenselydefined(possiblyunboundedand not necessarily closed) operator X from H 0 to H 1 with domain Dom(X) is a strong solution to the Riccati equation (8) if and only if the graph subspace G(H 0,X) is invariant for the operator B. Proof. First, assume that G(H 0,X)isinvariantforB. Then B(x Xx)=(A 0 x + VXx) (A 1 Xx + V x) G(H 0,X)

5 Non-Closed Invariant Subspaces 303 for any x Dom(X). In particular, A 0 x + VXx Dom(X) and A 1 Xx + V x = X(A 0 x + VXx)forallx Dom(X), which proves that X is a strong solution to the Riccati equation (21). To prove the converse statement assume that X is a strong solution to the Riccati equation (8), that is, A 0 x + VXx Dom(X) and A 1 Xx + V x = X(A 0 x + VXx), x Dom(X), which proves that the graph subspace G(H 0,X)isB-invariant. Remark 2.5. By Lemma 4.3 in [7] acloseddenselydefinedoperatorx : H 0 H 1 is a strong solution to the Riccati equation (8) if and only if it is a weak solution to (8). 3. The singular spectrum of the operator B Assume the following hypothesis. Hypothesis 3.1. Assume Hypothesis 2.2. AssumeinadditionthattheHilbertspace H 1 is one-dimensional, H 1 = C, and the Hilbert space H 0 is the cyclic subspace generated by Ran V. Note that under Hypothesis 3.1 the Hilbert space H 0 can be realized as a space of square integrable functions with respect to a Borel probability measure m with compact support, H 0 = L 2 (R; m) such that the bounded operator A 0 acts on L 2 (R,m)asthemultiplicationoperator (A 0 x 0 )(λ) =λx 0 (λ), x 0 L 2 (R,m), A 1 is the multiplication by a real number a 1 and, finally, the linear bounded map is given by for some v H 0. V : H 0 H 1 V x 0 = v, x 0 H0, x 0 H 0 Lemma 3.2. Assume Hypothesis 3.1. Thentheelement0 1 H = H 0 H 1 is cyclic for the operator B given by (5) (7) and, hence, B has a simple spectrum. Proof. By hypothesis (in the above notations) the element v H 0 is cyclic for the operator A 0. Therefore, the cyclic subspace with respect to the operator B generated by the elements v 0 H and 0 1 H is the whole H. Withoutloss of generality we may assume that a 1 =0.ObservingthatB(0 1) = v 0proves the claim.

6 304 V. Kostrykin and K.A. Makarov Theorem 3.3. Assume Hypothesis 3.1. ThentheHerglotzfunction φ(z) = 1+(a 1 z) v, (A 0 z) 1 v H0 (a 1 z) v, (A 0 z) 1 v H0 (9) admits the representation dω(λ) φ(z) = λ z, where ω is a probability measure on R with compact support. Moreover, the operator B is unitarily equivalent to the multiplication operator by the independent variable on L 2 (R,ω). Proof. Introduce the Borel measure Ω with values in the set of non-negative operators on H 1 H 1 by V 0 V 0 Ω(δ) = E 0 1 B (δ), 0 1 V 0 where is the linear map from H H 1 to H 0 H 1 and let ω(δ) =trω(δ), δ R aborelset. Clearly, the measure ω vanishes on all Borel sets δ such that E B (δ) =0.Infact, these measures have the same families of Borel sets, on which they vanish. Indeed, assuming ω(δ) =0yields and, hence, in particular, v 0, E B (δ) v 0 H + 0 1, E B (δ)0 1 H =0 0 1, E B (δ)0 1 H =0, (10) which implies E B (δ) =0. Introducing the B(H 1 H 1 )-valued Herglotz function V 0 M(z) = (B z) 1 V one concludes that the Herglotz function M(z) admitstherepresentation dω(λ) M(z) = λ z, and hence R trm(z) = R dω(λ) λ z. Straightforward computations show that the operator-valued function (11) with respect to the orthogonal decomposition H = H 0 H 1 can be represented as the 2 2matrix M00 (z) M M(z) = 01 (z) M 10 (z) M 11 (z) (11)

7 Non-Closed Invariant Subspaces 305 with the entries given by M 00 (z) =(a 1 z) v, (A 0 z) 1 v [a 1 z v, (A 0 z) 1 v ] 1, M 11 (z) =[a 1 z v, (A 0 z) 1 v ] 1, M 01 (z) = (a 1 z) 1 M 00 (z), M 10 (z) = (a 1 z) 1 M 00 (z). Taking the trace of M(z) yieldsrepresentation(9). Since by Lemma 3.2 the element 0 1iscyclicandthemeasureω and the spectral measure E B have the same families of Borel sets, on which they vanish, one concludes (see, e.g., [3]) that the operator B is unitarily equivalent to the multiplication operator by the independent variable on L 2 (R,ω), completing the proof. Recall that a measurable not necessarily closed set S R is a support of a measure ν if ν(r \ S) =0.AsupportS is said to be minimal if any measurable subset S S with ν(s )=0hasLebesguemeasurezero. Theorem 3.4. The sets { S s := λ R a 1 λ = and { S sc := λ R a 1 λ = v(µ) 2 dm(µ) µ λ i0, v(µ) 2 } dm(µ) µ λ i0 v(µ) 2 } dm(µ) µ λ 2 = are minimal supports of the singular part ω s and the singularly continuous part ω sc of the measure ω, respectively.theset { v(µ) 2 dm(µ) v(µ) 2 } dm(µ) S pp := λ R a 1 λ =, µ λ µ λ 2 < (14) coincides with the set of all atoms of the measure ω. Proof. The fact that (12) is a minimal support of ω s follows from Lemma 3.5 in [4], where one sets m + a (z) =(a 1 z) and v(µ) m + b (z) = v, (A 0 z) 1 2 dm(µ) v H0 =, Im z 0. µ z It is not hard to see (cf., e.g., Example 1 in [2]) that the set S pp coincides with the set of all eigenvalues of the operator B. Hence,byTheorem3.3oneproves that S pp coincides with the set of all atoms of the measure ω. Therefore, to prove that (13) is a minimal support of ω sc it suffices to check the inclusion Assume that λ S pp,thatis, v(µ) 2 dm(µ) a 1 λ = µ λ (12) (13) S pp S s. (15) (16)

8 306 V. Kostrykin and K.A. Makarov and v(µ) 2 dm(µ) µ λ 2 <. Since v(µ) 2 ( dm(µ) v(µ) 2 ) 1/2 dm(µ) µ λ µ λ 2 v L2(R;m), the dominated convergence theorem yields v(µ) 2 dm(µ) v(µ) 2 dm(µ) v(µ) 2 dm(µ) lim =, µ λ i0 ε +0 µ λ iε µ λ which together with (16) proves inclusion (15). The proof is complete. Remark 3.5. By Lemma 5 in [5] from Theorem 3.3 it follows that there exist minimal supports of the absolutely continuous part ω ac, the singular part ω s,and the singularly continuous part ω sc of the measure ω such that their closures coincide with the absolute continuous part spec ac (B), the singular part spec s (B), andthe singularly continuous part spec sc (B) of the spectrum, respectively. 4. Riccati equation Given λ R, introducetheoperator(linearfunctional) on by X λ : L 2 (R; m) H 1 = C { Dom(X λ )= ϕ L 2 (R; m) lim ε +0 X λ ϕ = lim ε +0 } v(µ)ϕ(µ) dm(µ) exists finitely µ λ iε v(µ)ϕ(µ) µ λ iε dm(µ), ϕ Dom(X λ). (17) Lemma 4.1. If λ S s,thentheoperatorx λ is densely defined. Proof. Since the element v L 2 (R; m) isgeneratingfortheoperatora 0,theset D = {ϕ ϕ(µ) = v(µ)ψ(µ), ψ is continuously differentiable on R} is dense in L 2 (R; m). For ϕ D and ε>0oneobtains v(µ)ϕ(µ) v(µ) 2 dm(µ) =ψ(λ) dm(µ) (18) µ λ iε µ λ iε v(µ) 2 (ψ(µ) ψ(λ)) + dm(µ). (19) µ λ iε Since λ S s,bytheorem3.4thelimit v(µ) 2 dm(µ) lim ε +0 µ λ iε = v(µ) 2 dm(µ) µ λ i0

9 Non-Closed Invariant Subspaces 307 exists finitely. The integral (19) also has a limit as ε +0 since ψ is a continuously differentiable which proves that the left-hand side of (18) has a finite limit as ε +0. Therefore, D Dom(X λ ), that is, X λ is densely defined. Remark 4.2. Note that by the Riesz representation theorem X λ is bounded whenever the condition v(µ) 2 dm(µ) < (20) λ µ 2 holds true. The converse is also true: If X λ is bounded, then (20) holds. Indeed, by the uniform boundedness principle from definition (17) it follows that sup ε (0,1] v(µ) 2 (µ λ) 2 dm(µ) <, + ε2 proving (20) by the monotone convergence theorem. Theorem 4.3. Let λ S s.thentheoperatorx λ is a strong solution to the Riccati equation A 1 X XA 0 XVX + V =0. (21) Moreover, if λ S pp,thesolutionx λ is bounded and if λ S sc = S s \ S pp,the operator X λ is non-closable. Proof. Note that A 0 Dom(X λ ) Dom(X λ ). If λ S s, then by Theorem 3.4 v(µ) 2 dm(µ) a 1 λ = µ λ i0. In particular, v Dom(X λ )and v(µ) 2 X λ VX λ ϕ = µ λ i0 dm(µ) X λϕ v(µ)ϕ(µ) =(a 1 λ) µ λ i0 dm(µ), ϕ Dom(X λ). Therefore, for an arbitrary ϕ Dom(X λ )onegets A 1 X λ ϕ X λ A 0 ϕ X λ VX λ ϕ v(µ)ϕ(µ)(a1 µ) v(µ)ϕ(µ) = dm(µ) (a 1 λ) µ λ i0 µ λ i0 dm(µ) v(µ)ϕ(µ)(λ µ) = dm(µ) = v(µ)ϕ(µ)dm(µ) = V ϕ, µ λ i0 which proves that the operator X λ is a strong solution to the Riccati equation (21). If λ S pp,then(20)holds,inwhichcasex λ is bounded. If λ S sc = S s \S pp, then X λ is an unbounded densely defined operator (functional) (cf. Remark 4.2). Since every closed finite-rank operator is bounded [6], it follows that for λ S sc the unbounded solution X λ is non-closable.

10 308 V. Kostrykin and K.A. Makarov Proof of Theorem 1. Introduce the mapping Ψ(λ) =G(H 0,X λ ), λ S s, (22) where X λ is the strong solution to the Riccati equation referred to in Theorem 4.3. By Theorem 2.4 the subspace Ψ(λ), λ S s is invariant with respect to B. To prove the injectivity of the mapping Ψ, assume that Ψ(λ 1 )=Ψ(λ 2 )forsome λ 1,λ 2 S s.dueto(22),x λ1 = X λ2 which by (17) implies λ 1 = λ 2. (i) Let λ S sc.bytheorem4.3thefunctionalx λ is non-closable. Since X λ is densely defined, the closure G(H 0,X λ )ofthesubspaceg(h 0,X λ )containsthe subspace H 0. By Proposition 2.1, the subspace G(H 0,X λ )containsanelementorthogonal to H 0.SinceH 0 H is of codimension 1, one concludes that G(H 0,X λ )= H 0 H 1 = H. (ii) Let λ S pp.bytheorem5.3in[7]thesolutionx λ is an isolated point (in the operator norm topology) of the set of all bounded solutions to the Riccati equation (21) if and only if the subspace G(H 0,X λ )isspectral,thatis,thereisaborelset R such that G(H 0,X λ )=RanE B ( ). Observe that the one-dimensional graph subspace G(H 1, Xλ )isinvariant with respect to the operator B. This subspace is spectral since by Lemma 3.2 λ is asimpleeigenvalueoftheoperatorb. Thus,G(H 0,X λ )=G(H 1, Xλ ) is also a spectral subspace of the operator B. Acknowledgments The authors are grateful to C. van der Mee for useful suggestions. K.A. Makarov is indebted to the Graduiertenkolleg Hierarchie und Symmetrie in mathematischen Modellen for its kind hospitality during his stay at the RWTH Aachen in the Summer of References [1] V. Adamyan, H. Langer, and C. Tretter, Existence and uniqueness of contractive solutions of some Riccati equations, J.Funct.Anal.179 (2001), [2] S. Albeverio, K.A. Makarov, and A.K. Motovilov, Graph subspaces and the spectral shift function, Canad.J.Math.55 (2003), arxiv:math.sp/ [3] M.S. Birman and M.Z. Solomyak, Spectral Theory of Self-Adjoint Operators in Hilbert Space, D.Reidel,Dordrecht,1987. [4] D.J. Gilbert, On subordinacy and analysis of the spectrum of Schrödinger operators with two singular endpoints, Proc.RoyalSoc.Edinburgh112A (1989), [5] D.J. Gilbert and D.B. Pearson, On subordinacy and analysis of the spectrum of onedimensional Schrödinger operators, J.Math.Anal.Appl.128 (1987), [6] T. Kato, Perturbation Theory for Linear Operators, Springer-Verlag, Berlin, 1966.

11 Non-Closed Invariant Subspaces 309 [7] V. Kostrykin, K.A. Makarov, and A.K. Motovilov, Existence and uniqueness of solutions to the operator Riccati equation. A geometric approach, inyu.karpeshina, G. Stolz, R. Weikard, Y. Zeng (Eds.), Advances in Differential Equations and Mathematical Physics, ContemporaryMathematics327, Amer.Math.Soc.,2003,pp arxiv:math.sp/ [8] V. Kostrykin, K.A. Makarov, and A.K. Motovilov, Ageneralizationofthetan 2Θ theorem, inj.a.ball,m.klaus,j.w.helton,andl.rodman(eds.),current Trends in Operator Theory and Its Applications. OperatorTheory:AdvancesandApplications 149, Birkhäuser, Basel, 2004, pp arxiv:math.sp/ [9] V. Kostrykin, K.A. Makarov, and A.K. Motovilov, Perturbation of spectra and spectral subspaces, Trans.Amer.Math.Soc.(toappear),arXiv:math.SP/ [10] B. Simon and T. Wolff, Singular continuous spectrum under rank one perturbations and localization for random Hamiltonians, Comm.PureAppl.Math.39 (1986), Vadim Kostrykin Fraunhofer-Institut für Lasertechnik Steinbachstraße 15 D Aachen, Germany kostrykin@ilt.fraunhofer.de, kostrykin@t-online.de URL: Konstantin A. Makarov Department of Mathematics University of Missouri Columbia, MO 65211, USA makarov@math.missouri.edu URL: