Spectral theory of rank one perturbations. selfadjoint operators

of selfadjoint operators Department of Mathematics and Mechanics St. Petersburg State University (joint work with Dmitry Yakubovich, Yurii Belov and Alexander Borichev) The Sixth St. Petersburg Conference in Spectral Theory 3 8 July 2014

Eigenfunction expansions Riesz bases of eigenfunctions, bases with brackets, linear summation methods, hereditary completeness, completeness. Keldyš Theorem, 1951 Suppose A is a selfadjoint Hilbert space operator that belongs to a Schatten ideal S p, 0 < p <, and satisfies Ker A = 0. Let L = A(I + S), where S is compact and Ker(I + S) = 0. Then the operators L and L are complete. Matsaev completeness theorem, 1961 (weak perturbations) Let L = A(I + S), where A, S are compact operators on a Hilbert space, A is selfadjoint, S S ω (i.e., n=1 n 1 s n < ) and ker(i + S) = 0. Then L and L are complete. M. Malamud L. Oridoroga, A. Shkalikov, B. Mityagin J. Aducci, F. Gesztesy V. Tkachenko.

Let A be a compact selfadjoint operator in a Hilbert space H. Moreover, let its point spectrum σ p (A) = {t n } be simple and Ker A = {0}. Rank one perturbations L = L(A, a, b) = A + a b, a, b H, Lf = Af + (f, b)a, f H. Some questions When L(A, a, b) has a complete system of eigenvectors or root vectors (L is complete)? When does completeness of L imply completeness of L? When does L admit spectral synthesis? For which A there exists a rank one perturbation L(A, a, b) which is a Volterra operator (i.e. Ker L = {0}, σ(l) = {0})?

Two counterexamples Completeness of L and L Let L = L(A, a, b) be complete. A trivial obstacle for completeness of L is that Ker L is nontrivial, while Ker L is trivial. First example of a compact operator L such that Ker L = Ker L = {0}, L is complete, L is not H.L. Hamburger (1951, Mathematische Nachrichten, Über die Zerlegung des Hilbertschen Raumes durch vollstetige lineare Transformationen) A simpler example Deckard, Foias, Pearcy (1979), a Hilbert Schmidt class operator. Theorem For any compact selfadjoint operator A there exists a rank one perturbation L = A + (, b)a of A such that Ker L = Ker L = {0} and L is complete while L is not.

Two counterexamples Spectral synthesis We say that an operator T L(H) admits the spectral synthesis if for any invariant subspace E of T the restriction T E is complete (equivalently, the eigenvectors which belong to E, span it). Any normal compact operator admits spectral synthesis (J. Wermer, 1950) First example of a compact operator without the spectral synthesis H.L. Hamburger (1951, Math. Nachr.) Further examples N.K. Nikolski (1969), A.S. Markus (1970). Theorem For any compact selfadjoint operator A there exists a rank one perturbation L = A + (, b)a of A such that both L and L are complete, but L does not admit the spectral synthesis.

Spectral synthesis and geometry of eigenvectors Let {x n } n N be a complete and minimal system in a separable Hilbert space (i.e., Span{x n } = H and Span{x n } n n0 H for any n 0 ). Let {y n } n N be its biorthogonal system, (x m, y n ) = δ mn. x H n N(x, y n )x n. We are interested in the following weak reconstruction property: x Span{(x, y n )x n } for any x H. In this case {x n } n N is said to be a hereditarily complete system or a strong M-basis (M = Markushevich). Equivalent definition: for any partition N = N 1 N 2, N 1 N 2 =, of the index set N, the mixed system {x n } n N1 {y n } n N2 is complete in H. A.S. Marcus (1970): Let L be a compact operator with complete set of eigenvectors {x n } and trivial kernel. Then L admits the spectral synthesis if and only if {x n } is hereditarily complete.

Singular perturbations Let µ = n µ nδ tn be a discrete measure on R and let A be the (unbounded) operator of multiplication by x in L 2 (µ), (Af )(x) = xf (x), f L 2 (µ). Assume that 0 ρ(a). Let the triple (a, b, κ), where a = (a n ), b = (b n ), satisfy and κ R a x, b x L2 (µ); κ C, x 1 a(x)b(x) dµ(x) in the case when a L 2 (µ). The corresponding singular perturbation L = L(A, a, b, κ) of A is defined as follows: D(L) = { y = y 0 + c A 1 a : Ly = Ay 0, c C, y 0 D(A), κ c + y 0, b = 0 } ; y D(L).

Preliminaries on de Branges spaces Let E be an entire function in the Hermite Biehler class, that is E has no zeros in C + R, and E(z) > E (z), z C +, where E (z) = E(z). With any such function we associate the de Branges space H(E) which consists of all entire functions F such that F /E and F /E restricted to C + belong to the Hardy space H 2 = H 2 (C + ). The inner product in H(E) is given by (F, G) E = R F (t)g(t) E(t) 2 H(E) is a reproducing kernel Hilbert space: for any w C there exists a function K w H(E) such that F (w) = (F, K w ) E. Hilbert spaces of entire functions H(E) were introduced by L. de Branges in relation to the inverse spectral problems for differential operators. These spaces are also of a great interest from the function theory point of view. The Paley Wiener space PW a is the de Branges space corresponding to E(z) = exp( iaz), a > 0. dt.

Functional model Let µ = n µ nδ tn, A multiplication by x in L 2 (µ), a = (a n ), b = (b n ), κ C. Theorem Let L = L(A, a, b, κ) be a singular rank one perturbation of A, where b is a cyclic vector for the resolvent of A, i.e., b n 0 for any n. Then there exists a de Branges space H(E) and an entire function G such that G / H(E), G(z) z z 0 H(E) if G(z 0 ) = 0, and L is unitary equivalent to the operator T = T (E, G) which acts on H(E) by the formulas D(T ) := {F H(E) : there exists c = c(f ) C : zf cg H(E)}, TF := zf cg, F D(T ).

Functional model Theorem Conversely, any pair (E, G) where E is an Hermite Biehler function and the entire function G satisfies G / H(E), G(z) z z 0 H(E) if G(z 0 ) = 0, corresponds to some singular rank one perturbation L = L(A, a, b, κ) of the multiplication operator A in some space L 2 (µ) with x 1 a(x), x 1 b(x) L 2 (µ). Functional models: V. Kapustin (rank-one perturbations of unitary operators), S. Naboko, A, Kiselev, V. Ryzhov.

Functional model The functions E and G appearing in the model for L(A, a, b, κ) are related to the data (a, b, κ) by the following formulas. Let E = A E ib E, where A E = (E + E )/2, B E = (E E)/2i. Then B E (z) A E (z) = n ( 1 t n z 1 t n ) b n 2 µ n, (1) and G(z) A E (z) = κ + n ( 1 t n z 1 t n ) a n b n µ n, (2) Note, in particular, that A E vanishes exactly on the set {t n }. The model essentially uses the expansions with respect to the orthogonal basis { A E } z t n of normalized reproducing kernels of H(E).

Spectrum of the model operator σ(a) = Z(A E ) = {t n }; σ(t ) = σ p (T ) = Z(G); TF := zf cg, F D(T ). The eigenspace of T corresponding to an eigenvalue λ, λ Z(G), is spanned by g λ, g λ (z) = G(z) z λ. Suppose T is correctly defined. Then σ(t ) = Z(G) and ker(t λi ) is spanned by the reproducing kernel K λ.

Positive results about the completeness Matsaev completeness theorem, 1961 (weak perturbations) Let L = A(I + S), where A, S are compact operators on a Hilbert space, A is selfadjoint, S S ω (i.e., n=1 n 1 s n < ) and ker(i + S) = 0. Then L and L are complete. Generalized weak perturbations We call L = L(A, a, b.κ) a generalized weak perturbation, if n a n b n µ n t n <, n a n b n µ n t n κ. Theorem If L = L(A, a, b.κ) is a generalized weak perturbations, then L and L are complete.

Positive results about the completeness Theorem (positive perturbations) Let L = L(A, a, b, κ) be a singular rank one perturbation. such that a n b n 0 for all but possibly a finite number of values of n and n t n 1 a n b n µ n =. Then L is correctly defined, and L and L are complete.

Sharpness in Matsaev theorem We call L = L(A, a, b.κ) a generalized weak perturbations, if n a n b n µ n t n <, n a n b n µ n t n κ. A typical example: n a n 2 tn 2α µ n <, n b n 2 tn 2+2α µ n <, α [0, 1]. There exists A S p, p > 1/2, such that for any α 1, α 2 (0, 1) there exist a x α1 L 2 (µ) and b x α2 L 2 (µ) such that the spectrum of the perturbed operator is empty. Thus, there exists a positive compact operator A 0 such that for any α 1, α 2 (0, 1) and there is a rank one perturbation L 0 of A 0 of the form L 0 = A 0 + A α1 0 SAα2 0, where S is a rank one operator and Ker L 0 = Ker L 0 = 0, such that L 0 is a Volterra operator.

Completeness of L and of L Eigenfunctions of L functions G(z) z λ, λ Z(G). Eigenfunctions of L reproducing kernels K λ, λ Z(G). Relation between the completeness of a system of reproducing kernels and of its biorthogonal A.B., Yu. Belov (IMRN, 2011). Theorem Assume that L = L(A, a, b, κ) and L are well defined and a / L 2 (µ). Let L be complete. Then its adjoint L is also complete if any of the following conditions is fulfilled: (i) a n 2 µ n C t n N > 0 for some N > 0; (ii) b n a 1 n C t n N for some N > 0.

Counterexample Theorem For any cyclic selfadjoint operator A with discrete spectrum {t n } such that t n, n, there exists a rank one singular perturbation L of A (with real spectrum and trivial kernel), which is complete, while its adjoint L is correctly defined and incomplete. A counterpart for compact operators For any compact selfadjoint operator A with simple spectrum and trivial kernel there exists a bounded rank one perturbation L of A with real spectrum such that L is complete and ker L = 0, while (L ) is not complete. Moreover, the orthogonal complement to the eigenvectors of (L ) may be infinite-dimensional.

Perturbations without the spectral synthesis Completeness of L completeness of a system of reproducing kernels {K λ } in H(E); Completeness of L completeness of the system biorthogonal to the system of reproducing kernels; Spectral synthesis for L hereditary completeness of {K λ } λ Λ, i.e., for any partition Λ = Λ 1 Λ 2, the system {K λ } λ Λ1 {g λ } λ Λ2 is complete in H(E). Recall that Z(G) = Λ, g λ = G/( λ). Examples of nonhereditarily complete systems of reproducing kernels A.B., Yu. Belov, A. Borichev (Adv. Math., 2013). Theorem For any spectrum {t n } there exists a rank one perturbation L = A + (, b)a of A such that both L and L are complete, but L does not admit the spectral synthesis.

Spectral synthesis for exponential systems Special case: {t n } = Z, µ n 1, b n 1. Then H(E) = PW π. A.B., Yu. Belov, A. Borichev (Adv Math., 2013): there exist nonhereditary complete systems of RK in PW π. However, the orthogonal complement to any mixed system {K λ } λ Λ1 {g λ } λ Λ2 is always at most one-dimensional. Systems of reproducing kernels in PW π are Fourier images of exponential systems in L 2 ( π, π). Let e λ (t) = e iλt. Consider {e λ } λ Λ in L 2 ( π, π). Completeness Levinson (1940), Beurling and Malliavin (1960-s), Makarov, Poltoratskii (2005). Riesz bases of exponentials Paley Wiener (1930s), Kadets (1965), Pavlov, Nikolski, Khruschev (1980). Exponential frames Seip and Ortega Cerda (2002). Hereditary completeness of exponentials B., Belov, Borichev, 2011 (Adv. Math., 2013).

Positive results about synthesis In the example of a rank one perturbation without synthesis, it may fail with an infinite defect, that is, the orthogonal complement to the root vectors in some invariant subspace M will be infinite-dimensional. Theorem Let A 0 be a compact selfadjoint operator with simple spectrum {s n }, s n 0. Assume that {s n } n I is ordered so that s n > 0 and s n decrease for n 0, and s n < 0 and increase for n < 0 and s n+1 s n C 1 s n N1 for some C 1, N 1 > 0. Let L 0 = A 0 + a b be a bounded rank-one perturbation of A 0 such that a, b / xl 2 (µ) and a n 2 µ n C 2 s n N2 for some C 2, N 2 > 0. Assume that operator L 0 is complete and that all its eigenvalues are simple and non-zero. Denote by {f j } j J the eigenvectors of L 0. Then for any L 0 -invariant subspace M we have dim ( M Lin {f j : f j M} ) <, where the upper bound for the dimension depends only N 1 and N 2.

Volterra rank one perturbations Problem Let A 0 be a compact selfadjont operator with the (simple) spectrum {s n }. When (for which spectra {s n }) there exist a rank one perturbation L 0 of A 0 which is a Volterra operator? Equivalent problem for inverse operators: when there is a singular rank one perturbation of an unbounded selfadjoint operator A = A 1 0 which has empty spectrum? Known results: if T = A + ib is a Volterra operator and A S p, p > 1, then B S p (Matsaev) Krein class K 1 An entire function F is in the Krein class K 1, if it is real on R, has only real simple zeros t n and may be represented as 1 F (z) = q + n c n ( 1 t n z 1 t n ), n t 2 n c n <.

Volterra rank one perturbations Theorem Let t n R and t n, n. The following are equivalent: (i) The spectrum {t n } may be removed by a rank one perturbation; (ii) There exists a function F K 1 such that the zero set of F coincides with {t n }. A counterpart for compact operators Let s n R, s n 0, and s n 0, n, and let A 0 be a compact selfadjoint operator with point spectrum {s n }. The following are equivalent: (i) There exists a rank one perturbation L 0 = A 0 + a 0 b 0 such that L 0 is a Volterra operator; (ii) The points t n = sn 1 form the zero set of some function F K 1.

Volterra rank one perturbations Theorem Let t n R and t n, n. The following are equivalent: (i) The spectrum {t n } may be removed by a rank one perturbation; (ii) There exists a function F K 1 such that the zero set of F coincides with {t n }. A rather counterintuitive consequence of this theorem is that adding a finite number of points to the spectrum helps it to become removable, while deleting a finite number of points may make it non-removable. Examples Let t n = n γ, n N. Then the spectrum {t n } is removable for γ 2 and non-removable for γ < 2. γ = 2 is the critical exponent. The spectrum {n 2 } n 1 is removable, but {n 2 } n 2 is not.

Dissipative rank one preturbations (Livshits theorem) Theorem (Livshits, 1946) Let L = A + ib be a Volterra operator (in some Hilbert space H) such that both A and B are selfadjoint and B is of rank one, B = b b. Then the spectrum of A is given by s n = c(n + 1/2) 1, n Z, for some c R, c 0. From this, one may deduce that A is unitary equivalent to the integral operator (having the same spectrum) (Ãf )(x) = i 2πc 0 f (t)sign (x t) dt, f L 2 (0, 2πc), while L is unitary equivalent to the integration operator ( Lf )(x) = 2i x 0 f (t) dt.

Dissipative rank one preturbations (Livshits theorem) Passing to the unbounded inverses we obtain a singular rank one perturbation L = L(A, a, b, κ) of the operator A of multiplication by the independent variable in some space L 2 (µ) where µ = n µ nδ tn, t n = sn 1, κ = 1, a = ib. Let E = A E ib E. Then B E (z) A E (z) = δ + ( 1 n G(z) A E (z) = 1 + i ( 1 n t n z 1 t n t n z 1 t n ) b n 2 µ n, ). b n 2 µ n, Since L (and, thus, the model operator T ) is the inverse to a Volterra operator, Z(G) =, and so G(z) = exp(iπcz) for some real c. Thus, e iπcz = A E (z) + i ( B E (z) δa E (z) ), Hence, A E (z) = cos πcz, t n = c 1( n + 1/2 ).