Systems of reproducing kernels and their biorthogonal: completeness or incompleteness?
aa r X i v : . [ m a t h . C V ] J un SYSTEMS OF REPRODUCING KERNELS AND THEIRBIORTHOGONAL: COMPLETENESS OR INCOMPLETENESS?
ANTON BARANOV AND YURII BELOV
Abstract.
Let { v n } be a complete minimal system in a Hilbert space H and let { w m } beits biorthogonal system. It is well known that { w m } is not necessarily complete. Howeverthe situation may change if we consider systems of reproducing kernels in a reproducingkernel Hilbert space H of analytic functions. We study the completeness problem for aclass of spaces with a Riesz basis of reproducing kernels and for model subspaces K Θ of the Hardy space. We find a class of spaces where systems biorthogonal to completesystems of reproducing kernels are always complete, and show that in general this is nottrue. In particular we answer the question posed by N.K. Nikolski and construct a modelsubspace with an incomplete biorthogonal system. Introduction and main results
Statement of the problem.
Let H be a separable Hilbert space. A sequence ofvectors { v n } is said to be complete if Span { v n } = H . If, moreover, the system { v n } is minimal , i.e. it fails to be complete when we remove any vector, then we say that thesystem is exact . For every exact system of vectors { v n } there exists a unique biorthogonalsystem { w m } such that h v n , w m i = δ mn .Suppose that H is a space of entire functions with reproducing kernels. Namely, for each λ ∈ C there is an element k λ ∈ H such that h f, k λ i = f ( λ ) for all f ∈ H . We are lookingfor an answer to the following question: Question.
Let { k λ n } be an exact system of reproducing kernels in H . Is it true that itsbiorthogonal system is also complete in H ? Of course, for an arbitrary sequence of vectors its biorthogonal system may be incom-plete. If { e n } ∞ n =1 is an orthonormal basis, then the system { e n + e } ∞ n =2 is complete, but Mathematics Subject Classification.
Primary: 46E22; secondary: 30D15, 30D50, 42C30.
Key words and phrases.
Reproducing kernel Hilbert spaces, de Branges spaces, biorthogonal systems,completeness.This work was supported by the Chebyshev Laboratory (Department of Mathematics and Mechanics,St. Petersburg State University) under RF government grant 11.G34.31.0026, by grant MK 7656.2010.1(Russia), and by the Research Council of Norway, grant 185359/V30. the biorthogonal system { e n } ∞ n =2 is incomplete. On the other hand, it is well known thatif we restrict ourselves to systems of reproducing kernels , then the answer may be positive.R.M. Young [21] proved the completeness of systems biorthogonal to the systems of repro-ducing kernels in the Paley–Wiener spaces or, equivalently, for biorthogonals to systemsof exponentials in L on an interval. E. Fricain [10] extended this result to a class of deBranges spaces of entire functions (see discussion below).Our aim is to exhibit some classes of spaces for which we know the answer (positive ornegative). In particular, we answer the question posed by N.K. Nikolski and construct anexample of a model (shift-coinvariant) subspace of the Hardy space H with an incompletebiorthogonal system. In what follows we consider only systems biorthogonal to systems ofreproducing kernels; therefore, sometimes we write simply the biorthogonal system in placeof the system biorthogonal to an exact system of reproducing kernels. To make the problem more realistic we need some additional structure on H , namelythe existence of a Riesz basis . Recall that a system of vectors { v n } is said to be a Rieszbasis if { v n } is an image of an orthonormal basis under a bounded and invertible operatorin H . We consider the class R of spaces of entire functions satisfying three axioms:(A1) H has a reproducing kernel k λ at every point λ ∈ C ;(A2) If a function f is in H and f ( w ) = 0, then the function f ( z ) z − w is also in H ;(A3) There exists a sequence of distinct points T = { t n } ⊂ C such that the sequence ofnormalized reproducing kernels (cid:8) k t n / k k t n k H (cid:9) is a Riesz basis for H .First example of such spaces is the Paley–Wiener space P W π which is the space ofentire functions of exponential type at most π that are in L ( R ). In this case the sequence { sin( π ( z − n )) π ( z − n ) } n ∈ Z is an orthonormal basis of reproducing kernels and ( A
3) is satisfied. Axioms( A , ( A
2) follow immediately.More general examples are de Branges spaces. We say that an entire function E belongsto the Hermite–Biehler class if it has no real zeros and | E ( z ) | > | E ( z ) | for any z in theupper half-plane C + . The de Branges space H ( E ) consists of all entire functions f suchthat f /E and f ∗ /E belong to the Hardy space H in C + ; here f ∗ ( z ) = f ( z ). The norm in H ( E ) is given by k f k H ( E ) = Z R | f ( x ) | | E ( x ) | dx. As in the Paley–Wiener space, in de Branges spaces there exist orthonormal bases ofreproducing kernels (see [8]). So de Branges spaces form a subclass of our class R .1.2. Parametrization of the class R . We will use an explicit parametrization of theclass R from [6]. Let us briefly remind the description from [6]. YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 3
The Riesz basis { k t n / k k t n k H } has a biorthogonal basis, which we will call { f n } . Usingaxiom (A2) we conclude that f n ( z ) ( z − t n )( z − t m ) ∈ H , m = n . This function vanishes at thepoints t l , l = m , and so it equals f m up to a multiplicative constant. Hence, the function c m f m ( z )( z − t m ) does not depend on m for suitable coefficients c m . We call this func-tion the generating function of the sequence { t n } and denote it by F . Note that usingthis construction we may define the generating function for an arbitrary exact system ofreproducing kernels, not just for a Riesz basis.The sequence f n is also a Riesz basis for H , and therefore any vector h in H can bewritten as(1.1) h ( z ) = X n h ( t n ) F ( z ) F ′ ( t n )( z − t n ) , where the sum converges with respect to the norm of H and A X n | h ( t n ) | k k t n k H ≤ k h k H ≤ B X n | h ( t n ) | k k t n k H for some constants A, B > h . Since point evaluation at every point z is abounded linear functional, the series in (1.1) also converges uniformly on compact subsetsof C \ T . By the assumption that h
7→ { h ( t n ) / k k t n k H } is a bijective map from H to ℓ , weget(1.2) X n k k t n k H | F ′ ( t n ) | | z − t n | < + ∞ whenever z is in C \ T . Therefore (1.2) implies that(1.3) X n b n | t n | < + ∞ , b n := k k t n k H | F ′ ( t n ) | . It follows from (1.1) that we can associate with the space
H ∈ R a space of meromorphicfunctions with prescribed poles . Namely, given a sequence of distinct complex numbers T = { t n } and a weight sequence b = { b n } which satisfy the admissibility condition (1.3),we introduce the space H ( T, b ) consisting of all functions of the form(1.4) f ( z ) = ∞ X n =1 a n b / n z − t n such that k f k H ( T,b ) := ∞ X n =1 | a n | < + ∞ . ANTON BARANOV AND YURII BELOV
The map f F f is an isomorphism of H ( T, b ) onto H which maps reproducing ker-nels to reproducing kernels. So, for our approach, we can consider the pairs ( T, b ) as aparametrization of all spaces from R .The space H ∈ R is a de Branges space H ( E ) for some Hermite–Biehler function E if and only if there exists T such that T ⊂ R . In this case we may choose E so that F = E + E ∗ . Moreover, as was shown in [5], de Branges spaces are the only spaces of theclass R where there exist two different orthogonal bases of reproducing kernels. Thesespaces are the prime example for us. Note, however, that there are many spaces in theclass R which are not isomorphic to a de Branges space. E.g., let T = { u n } ∪ { iw n } , where u n , w n are arbitrary sequences of real points satisfying P n | u n | − = P n | w n | − = ∞ . Thenthe space F H ( T, b ) is not a de Branges space (in any half-plane) since for a Riesz sequenceof normalized kernels { k λ n / k k λ n k H } in a de Branges space H , the sequences { λ n } ∩ C + and { λ n } ∩ C − should satisfy the Carleson interpolation condition [19, Part D, Lemma 4.4.2].1.3. Main theorems.
Now we are ready to state our main results.
Theorem 1.1. If H ∈ R and P n b n < + ∞ , then there exists an exact system of repro-ducing kernels such that its biorthogonal system is not complete. A converse result says that if b n have no more than a power decay, then the biorthogonalsystems are (almost) complete. Theorem 1.2.
If there exists
N > such that inf m b m (1 + | t m | ) N > , then the orthogonal complement to a system biorthogonal to an exact system of reproducingkernels is finite-dimensional.If, moreover, P n b n = + ∞ , then any system biorthogonal to an exact system of repro-ducing kernels is complete in H . The restriction on the decay of b n in Theorem 1.2 is essential. Example 1.3.
There exists a space
H ∈ R such that P n b n = + ∞ , but there exists anexact system of reproducing kernels such that its biorthogonal is not complete.Young’s result about the Paley–Wiener space corresponds to the situation when t n = n , n ∈ Z , and b n = 1, and follows from Theorem 1.2. E. Fricain have proved the completenessof biorthogonal system in de Branges spaces under the assumption that sup x ∈ R ϕ ′ ( x ) < + ∞ , ϕ being the phase function for E , that is, a smooth branch of the argument of E on R . We show now that this assumption implies a lower estimate on b n . We will use the de YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 5
Branges decomposition E = A − iB , where A and B are entire functions real on the realaxis. As known, reproducing kernels in the de Branges space H ( E ) are given by(1.5) k w ( z ) = 1 π · B ( z ) · A ( w ) − A ( z ) B ( w ) z − w . Without loss of generality we may assume that { k t n / k k t n k} is an orthonormal basis, where t n ∈ R are all solutions of the equation A ( z ) = 0 [8, Theorem 22]. Of course, A is thecorresponding generating function. In this notation(1.6) b n = k k t n k | A ′ ( t n ) | = k t n ( t n ) | A ′ ( t n ) | = B ( t n ) πA ′ ( t n ) = 1 πϕ ′ ( t n ) . Hence, inf n b n > Size of the orthogonal complement.
Now we turn to the question of ”size” of theorthogonal complement to a biorthogonal system in the case when it is not complete. Aprecise definition of the ”size” and the main results are given in §
3. Here we only emphasizethe following informal principle:
The size of the orthogonal complement to a biorthogonal system depends on smallness ofthe sequence { b n } . The orthogonal complement becomes bigger if b n tend to zero faster. If,however, { b n } are extremely small, then the orthogonal complement is finite-dimensional. Now we give a precise formulation of the latter property. Put ℓ ( T, b ) := (cid:8) f : T → C : X n | f ( t n ) | b n < + ∞ (cid:9) . The following result relates the size of the biorthogonal system to the density of polynomialson discrete subsets of the real line. Assume that b n are so small that the polynomials belongto ℓ ( T, b ) and are dense there, that is, there is no non-trivial sequence { c n } ∈ ℓ ( T, b ),such that P n c n n k = 0 for any k ∈ N . Theorem 1.4.
Let T ⊂ R and assume that the polynomials are dense in ℓ ( T, b ) . If H is the Hilbert space of the class R corresponding to H ( T, b ) , then the closed linear spanof the system biorthogonal to an exact system of reproducing kernels always has finitecodimension. Density of polynomials in the spaces of the form ℓ ( T, b ) is closely connected to thequasianalyticity phenomena (see [13, 7]). We illustrate this by the following example(further examples are given in Section 3).
ANTON BARANOV AND YURII BELOV
Example 1.5.
Let t n = n , n ∈ Z , and let b n = exp( −| n | ). Then any biorthogonal systemhas finite codimension. More generally, let w = exp( − Ω) be an even function such thatΩ( e t ) is a convex function of t ( w is a so-called normal majorant). Let b n = w ( n ). If P n | log b n | n +1 = + ∞ , then any biorthogonal system has finite codimension.Finally, let us summarize: ∗ If b n has at most power decay, then any biorthogonal system has finite codimension(Theorem 1.2); ∗ If the polynomials belong to ℓ ( T, b ) and are not dense there (”non-quasianalytic case”),then the codimension may be infinite (see Proposition 3.4); ∗ If the polynomials are dense in ℓ ( T, b ) (”quasianalytic case”), then the codimensionis finite (Theorem 1.4).1.5.
Model subspaces.
Our next result is about general model or star-invariant subspaces K Θ of the Hardy space H in the upper half-plane C + . Let Θ be an inner function in C + ,that is, a bounded analytic function such that lim y → +0 | Θ( x + iy ) | = 1 for almost all x ∈ R .With each Θ we associate the subspace K Θ = H ⊖ Θ H . These subspaces, as well as their vector-valued generalizations play an outstanding roleboth in function theory and operator theory. For their numerous applications we refer to[17, 18, 19]. It is well known that if Θ has a meromorphic continuation to the whole plane,then Θ = E ∗ /E for a function E in the Hermite–Biehler class and the mapping f Ef isa unitary operator from K Θ onto H ( E ), which maps reproducing kernels onto reproducingkernels.The reproducing kernels of the space K Θ are of the form(1.7) k λ ( z ) = i π · − Θ( λ )Θ( z ) z − λ , λ ∈ C + . Reproducing kernels of the model spaces have a rich and subtle structure and their geomet-ric properties (such as completeness, Bessel sequences, Riesz basic sequences) are still notcompletely understood (see, e.g., [12, 15, 2] and [19, Part D, Chapter 4]). In particular, itis an open problem whether any model subspace has a Riesz basis of reproducing kernels.A special case of the completeness problem for reproducing kernels is the completeness ofexponential systems in L ( − a, a ) (corresponding to Θ( z ) = e iaz ) settled by the classicalBeurling–Malliavin theory. A recent breakthrough in the completeness problem for repro-ducing kernels in model subspaces is due to N. Makarov and A. Poltoratski [15, 16] who YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 7 suggested a new approach based on singular integrals and extended the Beurling–Malliavintheory to some classes of model subspaces and de Branges spaces.The problem whether the system biorthogonal to an exact system of reproducing kernelsis complete in K Θ was posed by N.K. Nikolski; it was studied by E. Fricain in [10] wherefor the class of inner functions with bounded derivatives a positive answer was obtained.Theorem 1.1, which applies to the case of meromorphic inner functions, already shows thatthe answer in general is negative. However, we are able to prove an analog of Theorem 1.1for general model spaces.Let σ (Θ) be the spectrum of the inner function Θ, that is, the set of all ζ ∈ R ∪ ∞ such that lim z → ζ inf | Θ( z ) | = 0. Note that σ (Θ) is closed and Θ (and any f ∈ K Θ )has analytic continuation across any interval of the set R \ σ (Θ). A point ζ ∈ R issaid to be a Carath´eodory point for Θ if Θ has an angular derivative at ζ , that is, thereexists the nontangential limit Θ( ζ ) with | Θ( ζ ) | = 1, as well as the nontangential limitΘ ′ ( ζ ) = lim z → ζ Θ( z ) − Θ( ζ ) z − ζ . Theorem 1.6.
Let Θ be an inner function in C + such that there exists ζ ∈ σ (Θ) which isa Carath´eodory point for Θ . Then there exists an exact system of reproducing kernels suchthat the biorthogonal system is not complete. If σ (Θ) = {∞} (i.e. Θ is meromorphic in C ) the existence of the ”angular derivative at ∞ ” (appropriately defined) is equivalent to the condition P n b n < + ∞ for some orthogonalbasis of reproducing kernels and we arrive at Theorem 1.1 for de Branges spaces. Wemention also that a result analogous to Theorem 1.6 holds for the model spaces in the unitdisc (see Theorem 4.1).In Section 5, Theorem 5.3, we obtain a condition sufficient for the completeness of abiorthogonal system in a general model space K Θ in terms of the generating function G .Throughout this paper, the notation U ( z ) . V ( z ) (or equivalently V ( z ) & U ( z )) meansthat there is a constant C > U ( z ) ≤ CV ( z ) holds for all suitable z . We write U ( z ) ≍ V ( z ) if U ( z ) . V ( z ) and V ( z ) . U ( z ).2. Proof of Theorems 1.1 and 1.2
Let { k λ n } be an exact system in H . Without loss of generality we can assume that { λ n } ∩ { t m } = ∅ , since we always can move slightly the points { t m } so that { k t m / k k t m k H } remains a Riesz basis. Suppose that F and G are generating functions of systems { k t m } and { k λ n } respectively. Then the system G ( z ) G ′ ( λ n )( z − λ n ) is biorthogonal to { k λ n } . ANTON BARANOV AND YURII BELOV
For any h ∈ H we have a representation with respect to the Riesz basis { k t m / k k t m k H } :(2.1) h = X m a m k t m k k t m k H , { a m } ∈ ℓ . The last series converges in the norm and pointwise.The function h is orthogonal to G ( z ) z − λ n for all n if and only if for any n D Gz − λ n , h E = X m a m k k t m k · G ( t m ) t m − λ n = 0 . Consider the meromorphic function L ( z ) := X m a m k k t m k H · G ( t m ) t m − z . The series converges uniformly on compact subsets of C \ T , since Gz − λ n ∈ H and so (cid:8) G ( t m ) t m k k tm k H (cid:9) ∈ ℓ . The function LF is entire and vanishes at the points { λ n } . Hence, S := LF/G is an entire function, and(2.2) G ( z ) S ( z ) F ( z ) = X m a m k k t m k H · G ( t m ) t m − z . It follows that a m = S ( t m ) k k tm k H F ′ ( t m ) = | F ′ ( t m ) | F ′ ( t m ) S ( t m ) b / m . Hence,(2.3) X m | S ( t m ) | b m < + ∞ . We can consider functions S from (2.2) which satisfy (2.3) as parametrization of all func-tions h orthogonal to a given biorthogonal system { G ( z ) z − λ n } . We denote the space of all suchfunctions S by S . It is a Hilbert space with respect to the norm given as the square rootof the left-hand side of (2.3). Moreover, the mapping S X m | F ′ ( t m ) | F ′ ( t m ) · S ( t m ) b / m · k t m k k t m k H is a unitary operator from S onto the orthogonal complement of the system n G ( z ) z − λ n o .Now we turn to the proof of Theorem 1.1. From (1.4) we get Proposition 2.1.
Function M is in H if and only if (2.4) M ( z ) F ( z ) = X n c n z − t n , X n | c n | b n < + ∞ . Here the series converges uniformly on compact sets in C \ T , while the series P n c n F ( z ) z − t n converges in the norm of the space H . YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 9
Proof. (of Theorem 1.1). We want to construct a generating function G of an exact systemof reproducing kernels such that G ( z ) F ( z ) = P n d n z − t n , where the series converges uniformly oncompact subsets of C \ T . If such a function G is constructed, we can take S ≡ h = X n b / n · | F ′ ( t n ) | F ′ ( t n ) · k t n k k t n k H is orthogonal to all G ( z ) z − λ n .We choose coefficients c n so that c n t n > X n | c n | b n < + ∞ ; (2) X n ( c n t n ) b n = + ∞ , and put G ( z ) = F ( z ) P n c n t n z − t n . It follows from (2.5) that G / ∈ H , but G ( z ) z − λ n ∈ H where λ n are zeros of G . Without loss of generality we can assume that G has no multiple rootsbecause it is enough to change one coefficient c a little. Indeed, we can write G ( z ) = c F ( z ) z − t + H ( z ) = c F ( z ) + H ( z ) , and the functions F and H have no common zeros. So, G and G ′ have common zeros onlyat the points z where F ′ ( z ) H ( z ) − H ′ ( z ) F ( z ) = 0 and c F ( z ) + H ( z ) = 0, but this ispossible only for a countable set of coefficients c .To prove the completeness of { k λ n } we need to show that there is no entire function T such that T G ∈ H . To prove this, we introduce some additional requirements on c n . Choose a subsequence of indexes { n k } so rare that | t n k +1 | > | t n k | and the disks D k = { z : | z − t n k | ≤ t nk } are pairwise disjoint. For other indices we choose a sequence ofpositive numbers h n with P n h n < D n = {| z − t n | ≤ h n } , n / ∈ { n k } ,are also pairwise disjoint. Now assume that additionally to (2.5) we have X k c n k t n k < + ∞ . This will be achieved if we assume P k | t n k | − < + ∞ , P k b / n k < + ∞ and put c n k :=( b n k ) / t − n k . Now we make c n for n
6∈ { n k } so small that X n/ ∈{ n k } | c n t n | h n < X k c n k t n k , X n/ ∈{ n k } c n t n < X k c n k t n k . Assume that
T G ∈ H . Since
T G/F should be represented as in (2.4), we have(2.6) T ( z ) G ( z ) F ( z ) = X n c n t n T ( t n ) z − t n , X n | c n t n T ( t n ) | b n < ∞ . We can estimate G ( z ) /F ( z ) for z / ∈ ( ∪ D k ) ∪ ( ∪ D n ): | G ( z ) || F ( z ) | = (cid:12)(cid:12)(cid:12) X n c n t n z − t n (cid:12)(cid:12)(cid:12) ≥ | z | X n c n t n − X n/ ∈{ n k } | c n t n || z ( z − t n ) | − X n ∈{ n k } | c n t n || z ( z − t n ) |≥ | z | X n c n t n − X n/ ∈{ n k } | c n t n || z | h n − X n ∈{ n k } , | t n | < | z | / c n t n | z | − X n ∈{ n k } , | t n |≥| z | / c n t n | z | . Note that the last sum is o ( | z | − ) as | z | → ∞ . Hence, for sufficiently large | z || G ( z ) || F ( z ) | ≥ | z | X k c n k t n k & | z | , z / ∈ ( ∪ D k ) ∪ ( ∪ D n ) . Using analogous estimates we can show that | T ( z ) G ( z ) || F ( z ) | . , z / ∈ ( ∪ D k ) ∪ ( ∪ D n ) . Hence, | T ( z ) | . | z | for z / ∈ ( ∪ D k ) ∪ ( ∪ D n ). By the choice of t n k and h n there existcircles Γ n = { z : | z | = r n } with r n → ∞ , such that Γ n ∩ (cid:0) ( ∪ D k ) ∪ ( ∪ D n ) (cid:1) = ∅ . Therefore, T ( z ) = az + b . But this contradicts (2.6) and (2) in (2.5) unless T ≡ (cid:3) Remark 2.2.
Note that, by the choice of the coeficients d n = c n t n >
0, all zeros of thefunction G constructed in the proof of Theorem 1.1 are real.In the proof of Theorem 1.2 we will use the following lemma. Lemma 2.3. If S ∈ S , then S ( z ) − S ( w ) z − w ∈ S for any w ∈ C . In particular, if S is of finitedimension n + 1 , then S coincides with the set P n of all polynomials of degree at most n .Proof. Let λ be a zero of G . Then G ( z ) z − λ ∈ H and(2.7) G ( z )( z − λ ) F ( z ) = X m G ( t m )( t m − λ ) F ′ ( t m )( z − t m ) . We have the identity G ( z )( S ( z ) − S ( w )) F ( z )( z − w ) = 1 z − w (cid:18) G ( z ) S ( z ) F ( z ) − G ( w ) S ( w ) F ( w ) (cid:19) + ( w − λ ) S ( w ) z − w (cid:18) G ( w )( w − λ ) F ( w ) − G ( z )( z − λ ) F ( z ) (cid:19) − S ( w ) G ( z )( z − λ ) F ( z ) . By (2.2) and (2.7),1 z − w (cid:18) G ( z ) S ( z ) F ( z ) − G ( w ) S ( w ) F ( w ) (cid:19) = X m a m ( t m − w ) k k t m k H · G ( t m ) t m − z , YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 11 z − w (cid:18) G ( z )( z − λ ) F ( z ) − G ( w )( w − λ ) F ( w ) (cid:19) = X m a m ( t m − λ )( t m − w ) F ′ ( t m ) · G ( t m ) t m − z . Thus, we have shown that the function G ( z ) F ( z ) · S ( z ) − S ( w ) z − w can be represented as a series in(2.2). Condition (2.3) for the function S ( z ) − S ( w ) z − w follows from (1.3). (cid:3) Proof. (of Theorem 1.2) Assume that the system { G ( z ) z − λ n } is not complete. Fix h as in (2.1)which is orthogonal to all functions G ( z ) z − λ n and consider the corresponding space S . If S isan infinite dimensional space then we can find a function S ∈ S with at least N + 2 zeros w , ..., w N +1 different from the points { λ n } and { t n } . By Lemma 2.3 T ( z ) := S ( z ) Q N +1 l =1 ( z − w l ) ∈S . Moreover the corresponding coefficients from (2.2) for T are equal to a m Q N +1 l =1 ( t m − w l ) .Recall that f m ( z ) := k k t m k H F ′ ( t m ) · F ( z ) z − t m = b / m · | F ′ ( t m ) | F ′ ( t m ) · F ( z ) z − t m is the biorthogonal system to the Riesz basis { k t m / k k t m k H } and thus also is a Riesz basis.We have G ( z ) T ( z ) = F ( z ) X m a m Q N +1 l =1 ( t m − w l ) · G ( t m ) k k t m k H ( z − t m ) == X m a m b / m Q Nl =1 ( t m − w l ) · G ( t m )( t m − w N +1 ) k k t m k H · b / m F ( z ) z − t m =: X m d m f m ( z ) . Note that inf m | b / m Q Nl =1 ( t m − w l ) | > λ of G , then (cid:12)(cid:12)(cid:12) G ( t m ) t m − λ (cid:12)(cid:12)(cid:12) ≤ (cid:13)(cid:13)(cid:13) G ( z ) z − λ (cid:13)(cid:13)(cid:13) H · k k t m k H . So the coefficients { d m } are in ℓ and GT ∈ H . However, this contradicts the completenessof the system { k λ n } .Finally, assume that P n b n = + ∞ . If S is a finite-dimensional space, then it followsfrom Lemma 2.3 that S is the space of polynomials P n for some n , which can not be truesince P n | S ( t n ) | b n = + ∞ for any S ∈ S . Thus, the biorthogonal system is complete. (cid:3) At the end of the section we prove Example 1.3.
Proof.
We will construct a space of the form H ( Z , b ), that is, t n = n , n ∈ Z . The corre-sponding generating function is F ( z ) = sin( πz ). Put S ( z ) = ∞ Y k =1 (cid:18) − z (2 k + 1 / (cid:19) , G ( z ) = cos( πz ) /S ( z ) . Then(2.8) G ( z )sin( πz ) = + ∞ X n = −∞ S − ( n ) π ( z − n ) . Indeed, the difference between the left-hand side and the right-hand side in (2.8) should bean entire function of exponential type. Since lim | z |→∞ G ( z )sin πz = 0 along any non-horizontalray in C + or C − , this difference is zero.Put b n = | S ( n ) | − for n = ± k and b n = 1 otherwise. We consider the space H ∈ R corresponding to H ( Z , b ). First of all we want to show that G is the generating functionof an exact system in H . From (2.8) we conclude that G ( z ) z − λ ∈ H for any zero λ of G . Weneed to show that there is no nonzero entire T such that T G ∈ H ( Z , b ). If it exists then(2.9) G ( z ) T ( z )sin( πz ) = ∞ X n = −∞ S − ( n ) T ( n ) π ( z − n ) , X n | S − ( n ) T ( n ) | b n < + ∞ . It follows from (2.9) that T is of zero exponential type, and, at the same time, P k | T (2 k +1) | < + ∞ , which implies T ≡ P n b n = + ∞ and now it remains to show thatthere exists a nonzero S ∈ S . Let S ( z ) = ∞ Y k =2 (cid:18) − z (2 k + δ k ) (cid:19) and choose δ k ∈ (0 ,
1) so small that P ∞ k =1 | S (2 k ) | < + ∞ . By a straightforward estimationwe get | S ( n ) | . | S ( n ) | . So, P n | S ( n ) | b n < + ∞ . On the other hand, lim y →±∞ | S ( iy ) || S ( iy ) | = 0and we have representation (2.2) with condition (2.3) for S .Since S is not a polynomial, we see that S is infinitely-dimensional. Thus, in this casethe orthogonal complement to the biorthogonal system is infinitely-dimensional. (cid:3) Size of the orthogonal complement to a biorthogonal system
In this section we will study in more detail the space S whose elements parameterizefunctions orthogonal to the biorthogonal system via formula (2.2). First of all note thatfor any k ∈ N it is possible that S coincides with the set P k of polynomials of degree atmost k . On the other hand, as we have seen in the proof of Example 1.3, it is possible that S is an infinite-dimensional space. We introduce therefore the following notion of the sizeof the orthogonal complement to the biorthogonal system. Definition 3.1.
Let { k λ } be an exact system of reproducing kernels in a space H ∈ R with the generating function G , and let S be the corresponding space parametrizing the YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 13 orthogonal complement. Let M be a positive increasing function on R + . We say that theorthogonal complement to the biorthogonal system has size M if there exists S ∈ S suchthat, for some y > , log | S ( iy ) | ≥ M ( | y | ) , | y | > y . From now on we will consider only the situation when T = { t n } ⊂ R . As we havementioned in Introduction, this case corresponds to de Branges spaces. Thus we assumethat H = H ( E ) and the generating function F of the sequence T is given by F = E + E ∗ .Our first theorem shows that under some mild restrictions on t n any function in S is ofzero exponential type, and so, for any ε >
0, the orthogonal complement can not have thesize M ( r ) = εr . Theorem 3.2.
Let t n ∈ R . If G is the generating function of an exact system of repro-ducing kernels and S is the corresponding space, then any S ∈ S is of zero exponentialtype. From this we have an immediate corollary.
Corollary 3.3.
Let t n ∈ R . Then for any ε > the orthogonal complement of a biorthog-onal system can not have the size M ( r ) = εr . In what follows we will use essentially the inner-outer factorization of H functions (see,e.g., [13, 17]). Recall that a function f is said to be in the Smirnov class if f = g/h , where g, h are bounded analytic functions in C + (or functions in H p ) and h is outer. Proof. (of Theorem 3.2). If λ is a zero of G , then Gz − λ ∈ H ( E ). Hence, h := G ( z − λ ) E ∈ H .Let us show that h has no singular inner factor of the form e iaz , a > h hasno other singular factors since it is analytic on R ). Indeed, if e − iaz h ∈ H , then put H ( z ) = e − iaz sin azz G ( z ) . Then
H/E ∈ H and also H ∗ /E ∈ H , so H ∈ H ( E ), which contradicts the fact that { λ : G ( λ ) = 0 } is a uniqueness set for H ( E ).Consider the inner function Θ = E ∗ /E . Then 2 F = E (1 + Θ) and we have G ( z ) S ( z ) E ( z ) = X m a m k k t m k H · z ) t m − z G ( t m ) =: f. We show that the right-hand side function f is in the Smirnov class in C + . First of allnote that if v m ≥ { v m } ∈ ℓ , thenIm X m v m t m − z > , z = x + iy ∈ C + , and so this sum is in the Smirnov class. The same is obviously true also for an arbitrarysequence { v m } ∈ ℓ .Since Gz − λ ∈ H ( E ) we have {k k t m k − H t − m G ( t m ) } ∈ ℓ . Hence, { v m } ∈ ℓ where v m = a m k k t m k H · G ( t m ) t m , and we have f ( z )1 + Θ( z ) = X m v m t m t m − z = X m v m + z X m v m t m − z . Hence f is in the Smirnov class.Thus S = f (cid:16) GE (cid:17) − is a ratio of two functions of bounded type, and so is a function ofbounded type (zeros of Blaschke products cancel). Moreover, since S is analytic on R itis in the Smirnov class unless it has a factor of the form e − iaz in its canonical inner-outerfactorization. However, it can not happen, since, as we have seen, G/E has no singularinner factor.By completely identical arguments, S is in the Smirnov class in the lower half-plane C − .Since S is in the Smirnov class both in C + and in C − , it is of zero exponential type byM.G. Krein’s theorem (see, e.g., [11, Chapter I, Section 6]). (cid:3) As we have seen in Corollary 3.3, the linear growth of the function M (which determinesthe size of the orthogonal complement) is not possible. The following proposition, whichapplies to the case H ( Z , b ) (that is, t n = n , n ∈ Z ), provides a converse result: for anyslower growth, the size M for the orthogonal complement may be achieved for some choiceof b n . Proposition 3.4.
Let M ( r ) /r be a decreasing function which tends to zero when r → + ∞ . Then there exist a sequence b n and an exact system { k λ } in the space of entirefunctions H ( corresponding to the space H ( Z , b )) such that the orthogonal complement tothe biorthogonal system has size M .Proof. First of all, we ”atomize” function M . There exists an increasing integer-valuedfunction µ with jumps at some half-integer points such that M − µ ∈ L ∞ . We will assumethat µ (0) = 0, µ (1) = 1. Let S ( z ) = Q t ∈ supp dµ (cid:0) − z t (cid:1) . We want to estimate | S ( iy ) | forlarge | y | : log | S ( iy ) | = Z ∞ log (cid:18) y t (cid:19) dµ ( t ) = 2 y Z ∞ µ ( t ) dtt ( y + t ) ≥ y (cid:16) inf t ∈ [1 / ,y ] µ ( t ) t (cid:17) · Z y / yy + t dt & µ ( y ) , | y | → ∞ . YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 15
Now we put G ( z ) = cos( πz ) /S ( z ), b n = S − ( n ). The function G has exponential type π and we can verify that G is the generating function of some exact system using the samearguments as in the proof of Example 1.3. Indeed, if T G ∈ H , then T is of zero exponentialtype and P n | T ( n ) | < + ∞ , so T ≡
0. It only remains to note that ( z − / − S ( z ) ∈S . (cid:3) As we have seen in the proof of Proposition 3.4 the size of orthogonal complement to abiorthogonal system corresponds to the speed of decrease of the coefficients b n . It becomesbigger when b n decrease faster. Nevertheless from Theorem 1.4 we see that for extremelysmall b n biorthogonal system has finite codimension. This corresponds to polynomial size M ( r ) = n log r . Proof. (of Theorem 1.4). Suppose G is the generating function of some exact system, andlet λ be a zero of G . Hence, G ( z )( z − λ ) F ( z ) = X n c n z − t n , X n | c n | b n < + ∞ . In particular, { c n /b n } ∈ ℓ ( T, b ). Since the polynomials are dense in ℓ ( T, b ), it followsthat there exists N ∈ N ∪ { } such that h{ c n /b n } , t Nn i ℓ ( T,b ) = P n c n t Nn = 0. We take thesmallest N with this property. Let us estimate G/F from below on the imaginary axis.We have X n c n iy − t n = X | t n |≤| y | / c n iy − t n + X | t n | > | y | / c n iy − t n . Since P n | c n | · | t n | k < ∞ for any k , we have (cid:12)(cid:12)(cid:12) X | t n | > | y | / c n iy − t n (cid:12)(cid:12)(cid:12) = O (cid:16) y N +2 (cid:17) , | y | → + ∞ . For | t n | ≤ | y | /
2, we have( iy − t n ) − = X k ≥ t kn ( iy ) k +1 = N X k =0 t kn ( iy ) k +1 + r n ( y ) t N +1 n ( iy ) N +2 , where | r n ( y ) | ≤
2. Hence, X | t n |≤| y | / c n iy − t n = N X k =0 iy ) k +1 X n c n t kn − N X k =0 iy ) k +1 X | t n | > | y | / c n t kn + 1( iy ) N +2 X n c n r n ( y ) t N +1 n = 1( iy ) N +1 X n c n t Nn + O (cid:16) y N +2 (cid:17) . We conclude that | G ( iy ) | / | F ( iy ) | ≥ c | y | − N − , | y | → + ∞ . Now let S ∈ S and so, for some { a m } ∈ ℓ , S ( z ) G ( z ) F ( z ) = X m a m k k t m k H · G ( t m ) t m − z . We have (cid:12)(cid:12)(cid:12) X m a m k k t m k H · G ( t m ) t m − iy (cid:12)(cid:12)(cid:12) ≤ X m | a m | | G ( t m ) |k k t m k H | t m | =: A < + ∞ , since {k k t m k − H | t m | − G ( t m ) } ∈ ℓ . Hence, | S ( iy ) | ≤ A | F ( iy ) || G ( iy ) | ≤ C | y | N +1 , | y | → ∞ . By Theorem 3.2, S is of zero exponential type, and thus, a polynomial of degree at most N + 1. Hence, S has finite dimension. (cid:3) Using the known results on density of polynomials we can give more examples of thesituation where all biorthogonal systems have finite codimension. These examples dealwith one-sided sequences with power growth.
Example 3.5.
1. Let t n = n /β , n ∈ N , and let b n = exp( − At αn ), where A, α >
0. If β ≥ /
2, then the polynomials are dense in ℓ ( T, b ) if and only if α ≥ /
2. If β < /
2, thenthe polynomials are dense in the space ℓ ( T, b ) for α > β and are not dense for α < β ; if α = β , then the polynomials are dense if and only if A ≥ π cot πβ (see [7] for details).2. The situation changes when we add a sparse sequence on the negative semiaxis. Let t n = n /β , n > , n ∈ Z , − | n | , n < , n ∈ Z . Let b n = exp( − t αn ), α >
0. Using the results of [3] one can show that for 1 < β < /
2, thepolynomials are dense in ℓ ( T, b ) for α > / α < /
2. For 3 / ≤ β < ℓ ( T, b ) for α > β − α < β − Incomplete biorthogonal systems in model subspaces
We start with an analogue of Theorem 1.6 for the unit disc D . Let Θ be an inner functionin D and, as in the half-plane case, let K Θ = H ⊖ Θ H . Denote by σ (Θ) the boundaryspectrum of Θ. A point ζ ∈ R is said to be a Carath´eodory point if Θ has an angularderivative at ζ , that is, there exist the nontangential limit Θ( ζ ) with | Θ( ζ ) | = 1, as wellas the nontangential limit Θ ′ ( ζ ) = lim z → ζ Θ( z ) − Θ( ζ ) z − ζ . By the Ahern–Clark theorem [1], thisis equivalent to the fact that the reproducing kernel k ζ belongs to K Θ and any element in K Θ has finite nontangential boundary value at ζ . Finally, if Θ = BI ν is a factorization of YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 17
Θ into a Blaschke product and a singular inner function, then ζ is a Carath´eodory pointif and only if | Θ ′ ( ζ ) | = X n − | z n | | ζ − z n | + Z T dν ( τ ) | ζ − τ | < + ∞ . Here z n are zeros of B and ν is a singular measure on the circle T . Theorem 4.1.
Let Θ be an inner function in D such that there exists ζ ∈ σ (Θ) which isa Carath´eodory point. Then there exists an exact system of reproducing kernels such thatits biorthogonal system is not complete. Without loss of generality we may assume that Θ is a Blaschke product. Indeed, wecan always pass to a Frostman shift B = Θ − γ − γ Θ , | γ | <
1, which is a Blaschke product, andthe map f (1 − | γ | ) / f − γ Θ is a unitary operator from K Θ onto K B , which maps thekernels to the kernels (up to constant bounded factors).We also will move to the upper half-plane so that ζ goes to ∞ . Recall that for a Blaschkeproduct B , the property to have an ”angular derivative at ∞ ” is equivalent to any of thefollowing:a) there exists α ∈ C with | α | = 1 such thatRe α + Θ( z ) α − Θ( z ) = p Im z + Im zπ Z R dµ ( t ) | t − z | , z ∈ C + , for a singular measure µ and p > α such that α − B ∈ K B ;c) there exist a unimodular α and q > − αB ( iy ) = qy + o (cid:16) y (cid:17) , y → + ∞ . In this case(4.1) q = 2 p − = 2 X n y n , where z n = x n + iy n are zeros of B (and, in particular, the series P n y n converges). Ofcourse we will assume α = 1, so 1 − B ∈ K B . Note also that for any g ∈ K B there existsa finite limit lim y →∞ yg ( iy ), and( g, − B ) = 2 π lim y →∞ yg ( iy ) . In what follows we again use essentially the inner-outer factorization of H functions. If m ≥ m ∈ L (cid:16) dtt +1 (cid:17) we denote by O m the outer function with the modulus m on R . If we identify the functions in K Θ and their boundary values on R , then an equivalentdefinition of K Θ is K Θ = H ∩ Θ H . Thus, we have a criterion for the inclusion f ∈ K Θ which we will repeatedly use:(4.2) f ∈ K Θ ⇐⇒ f ∈ H and Θ f ∈ H . In the following lemmas we always assume that Θ is an inner function such that ∞ is aCarath´eodory point and 1 − Θ ∈ K Θ .Our first lemma shows that the zeros of K Θ function may be concentrated in the upperhalf-plane. Lemma 4.2.
Let f = O m BI ∈ K Θ , where B is a Blaschke product and I is some innerfunction. Then there exists a function g = O ˜ m ˜ B ∈ K Θ such that Θ g = O ˜ m is outer, B divides the Blaschke product ˜ B , and | O ˜ m | ≍ | O m | .Moreover, if lim y →∞ yf ( iy ) = 0 , we can choose g so that lim y →∞ yg ( iy ) = 0 .Proof. By the criterion (4.2), we have Θ f ∈ H , and hence, Θ f = O m J for some innerfunction J . Then, again by (4.2), the function f = O m BIJ is in K Θ and Θ f = O m . If IJ is a Blaschke product we are done. Assume now that f = O m B K , where K is a singularinner function. Replace K by its Frostman shift K = K − γ − γK , | γ | <
1, which is a Blaschkeproduct. Put g = O m (1 − γK ) K B = O m ( K − γ ) B , O ˜ m = O m (1 − γK ) . Then Θ g = Θ O m ( K − γ ) B = O m (1 − γK ) since Θ O m B = KO m . Thus, g ∈ K Θ and | O ˜ m | ≍ | O m | .Finally, note that if lim y →∞ yf ( iy ) = 0, that is, f is orthogonal to 1 − Θ, then the sameis true for f . Hence,0 = ( f , − Θ) = (Θ f , Θ −
1) = (Θ − , Θ f ) = (Θ − , O m ) . Thus, lim y →∞ yO m ( iy ) = 0, and the same is true for g , since | O ˜ m | ≍ | O m | . (cid:3) The next lemma shows that one can get rid of real zeros of a function in K Θ . Here wewill use a lemma due to N. Makarov and A. Poltoratski [15]. Lemma 4.3.
Let f be a function in K Θ , let a n ∈ R , a n < a n +1 , | a n | → ∞ , n → ∞ , andassume that there exist nonnegative integers m n such that ( t − a n ) − m n f ∈ L ( a n − δ n , a n + δ n ) for a small δ n > , but ( t − a n ) − m n − f / ∈ L ( a n − δ n , a n + δ n ) . Let { w j } be a sequence ofpoints in C + with | w j | → ∞ . Then there exists a function h ∈ K Θ such that f /h is locally YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 19 bounded on each open interval ( a n , a n +1 ) , for any n we have hz − a n / ∈ L ( a n − δ n , a n + δ n ) ,and | h ( w j ) | & | f ( w j ) | . Moreover, if lim y →∞ yf ( iy ) = 0 , we can take h so that lim y →∞ yh ( iy ) = 0 .Proof. Assume for a moment that all m n = 1. Then we divide f by a function of the form1 − J where J is a meromorphic Blaschke product and J = 1 exactly at { a n } . We alsowant to do this so that f / (1 − J ) is still in H (and hence in K B ). Such a choice of J ispossible by [15, Lemma 3.15]. The function J should be constructed as1 + J ( z )1 − J ( z ) = 1 i X n ν n a n − z , where ν n > t − a n ) − m n f in L ( a n − δ n , a n + δ n )). Since the right-hand side has a positive real part, J is a meromorphic innerfunction, and also, 2 J ( z ) J ( z ) − i X n ν n a n − z . Then we have h = 2 JJ − f ∈ H . Also, Θ h = Θ f · − J ∈ H , and so h ∈ K Θ by (4.2). Obviously, f /h is locally bounded oneach interval ( a n , a n +1 ), Since lim y →∞ J ( iy )1 − J ( iy ) = −
1, we have lim yh ( iy ) = 0. Finally, notethat by choosing ν n sufficiently small we can make | J ( w j ) || − J ( w j ) | as close to 1 as we want.In general case when m n = 1 we repeat the procedure and construct a sequence ofmeromorphic inner functions J k so that a n is in the set { t : J k ( t ) = 1 } exactly for m n ofthe functions J k . Put h = g Y k J k J k − . Chosing the masses ν kn in the definition of J k sufficiently small we may achieve that theproduct converges to h ∈ K Θ , lim y →∞ yh ( iy ) = 0 and | h ( w j ) | ≥ c | f ( w j ) | for a positiveconstant c . We omit the technicalities. (cid:3) Lemma 4.4.
Let f be a Smirnov class function in C + such that f restricted to R has aSmirnov class extension to C − . Assume also that ∆ ⊂ R is an open interval such that f is in L (∆) . Then f has an analytic extension through ∆ . Proof.
Let [ x − r, x + r ] be a subinterval of ∆. Since f is in the Smirnov class and in L (∆)we have f ( · + iε ) → f in L ([ x − r, x + r ]) when ε → +0, and also when ε → −
0. Considerthe contours γ + = [ x − r, x + r ] ∪ { z ∈ C + : | z − x | = r } and γ − = [ x − r, x + r ] ∪ { z ∈ C − : | z − x | = r } with the standard orientations. If z ∈ C + , | z − x | < r , we have f ( z ) = 12 πi Z γ + S ( z ) z − z dz, πi Z γ − S ( z ) z − z dz (step a bit from R , apply the Cauchy formula and then pass to the limit). Taking the sumwe conclude that f ( z ) = 12 πi Z | z − x | = r S ( z ) z − z dz, | z − x | < r, z / ∈ R , and, hence, f has an analytic extension in the whole disc | z − x | < r . (cid:3) Proof. (of Theorems 1.6 and 4.1). As we have seen, it suffices to prove the theorem for thecase where Θ = B is a Blaschke product such that ∞ ∈ σ ( B ), ∞ is a Carath´eodory pointand 1 − B ∈ K B . The symbol q has the same meaning as in (4.1).The idea of the proof is to construct a function g in K B such that(i) g = B Λ O m and Bg = O m , O m is an outer function in K B ;(ii) Λ ′ = Λ ∪ { λ } is a uniqueness set for K B for any λ ∈ C + \ Λ;(iii) g is orthogonal to 1 − B which means that lim y →∞ y | g ( iy ) | = 0.If such g is constructed, then the system ( z − λ ) g ( z ) z − λ , λ ∈ Λ ′ , is biorthogonal to acomplete system { k λ } λ ∈ Λ ′ , but it is not itself complete, since it is orthogonal to 1 − B .We will consider separately two cases with slightly different proofs. Let us start with aneasier case. Case 1.
Assume that(4.3) lim sup y → + ∞ y (cid:12)(cid:12)(cid:12)(cid:12) − B ( iy ) − qy (cid:12)(cid:12)(cid:12)(cid:12) = + ∞ . Consider the function f ( z ) = 1 − B ( z ) − iqz − z , where z = x + iy is some zero of B . Obviously, ( f, − B ) = lim y →∞ y | f ( iy ) | = 0. Alsonote that for t ∈ R , f ( t ) = t − x + iy − iqt − x + iy − B ( t ) . Since q > y we have for almost all t (4.4) | f ( t ) | ≥ (cid:12)(cid:12)(cid:12)(cid:12) t − x + iy − iqt − x + iy (cid:12)(cid:12)(cid:12)(cid:12) − ≥ Ct . YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 21
Now applying Lemma 4.2 to f we obtain a function g = O m B Λ ∈ K B such that | g | ≍ | f | on R and Bg = O m .Let us show that Λ ′ = Λ ∪ { λ } is a uniqueness set for K B for any λ ∈ C + \ Λ. Assumethe converse. Then there exists a function F ∈ K B which vanishes on Λ ′ . We can write F = Sg for a function S which is analytic in C + . Then S = ( F/B Λ ) /O m is in the Smirnovclass in C + . Also, since F ∈ K B we have BF = BgS = O m S ∈ K B . So S ∗ ( z ) = S ( z ) has a Smirnov class extension to the upper half-plane, or S itself is aSmirnov class function in C − . Also, in view of (4.4), S is locally in L on R and hence, byLemma 4.4, S is entire. Since S is in the Smirnov class both in C + and in C − , it is of zeroexponential type by M.G. Krein’s theorem (see, e.g., [11, Chapter I, Section 6]).On the other hand, applying (4.4) once again we conclude that S ∈ ( t + i ) L ( R ), andso S is a polynomial of degree at most 1. On the other hand, we have S ∗ O m ∈ K B . Notethat | O m ( iy ) | ≥ | f ( iy ) | = (cid:12)(cid:12)(cid:12)(cid:12) − B ( iy ) − qy + O (cid:16) y (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) . If S is not a constant, then, by (4.3), lim sup | yS ∗ ( iy ) O m ( iy ) | = + ∞ , but this is impossible,since this limit is finite for any function in K B . Case 2.
Now assume that (4.3) is not satisfied, that is, there exists
M > y (cid:12)(cid:12)(cid:12)(cid:12) − B ( iy ) − qy (cid:12)(cid:12)(cid:12)(cid:12) ≤ M for sufficiently large y . The proof for Case 1 does not work, since if (4.5) is satisfied wecan not claim that S ∗ O m is not in K B when S is a polynomial of degree 1; we need to usethe fact that ∞ is a point of the spectrum (i.e., a limit point for the zeros of B ), sincefor finite Blaschke products the argument should fail. However, we will again construct g satisfying (i)–(iii). Step 1.
Take a very sparse subsequence z n = x n + iy n of zeros of B so that | z n | → ∞ and { z n } is a Carleson interpolating sequence. Thus, the functions √ y n ( z − z n ) − forma Riesz sequence in K B (see, e.g., [17, Lecture VIII]). Also we may assume that y n < x n > M/q for all n , and | x n − x k | > x n / n, k , n = k .Now take a sequence ( c n ) ∈ ℓ , c n >
0, such thata) P n c n x n = + ∞ ;b) P n √ y n c n = 1 (recall that P n y n < + ∞ );c) qc n > √ y n . It is easy to see that these conditions may be achieved, if y n tends to zero sufficiently fast(e.g., if P n √ y n = m < c n = 1 /m ). It follows from b) and the fact that x n > M/q ,thatd) q P √ y n c n x n > M (may be + ∞ ).Put f ( z ) = 1 − B ( z ) − iq X √ y n c n z − z n . Obviously, this is a function in K B which is orthogonal to 1 − B . Also, for sufficiently large y , y | Im f ( iy ) | = y (cid:12)(cid:12)(cid:12)(cid:12) Im (cid:0) − B ( iy ) (cid:1) − q X √ y n c n x n x n + ( y + y n ) (cid:12)(cid:12)(cid:12)(cid:12) ≥ y q X √ y n c n x n x n + ( y + y n ) − M ≥ C > y | Im (1 − B ( iy )) | ≤ M .We will see that f ( z n ) = − qc n √ y n + O (1) (since x n tends to infinity very rapidly), so f hasa good growth along z n and we will see later that zf can not be in H . Step 2.
Applying Lemma 4.2 we would obtain from f a new function g ∈ K B such that Bg is outer. However, f may have real zeros and we shall divide them out first. Considerthe function R ( t ) = (cid:12)(cid:12)(cid:12) − iq X √ y n c n t − z n (cid:12)(cid:12)(cid:12) − (cid:16) − iq X √ y n c n t − z n (cid:17)(cid:16) iq X √ y n c n t − z n (cid:17) − . It is analytic on R which implies that zeros a n of R are of finite multiplicities and a n → ∞ .Hence, we can represent R = ∪ [ a n , a n +1 ] where a n < a n +1 → ∞ and R is locally separatedfrom zero on the open intervals ( a n , a n +1 ). Then | f ( t ) | ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − iq X √ y n c n t − z n (cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12) is locally separated from 0 almost everywhere on each of the intervals ( a n , a n +1 ).By Lemma 4.3 there exists a function h ∈ K B such that hz − a n / ∈ L ( a n − δ, a n + δ ) for any δ > | h/f | is locally separated from 0 on any interval ( a n , a n +1 ) and | h ( z n ) | & | f ( z n ) | .Now we apply Lemma 4.4 to the function h and obtain a function g = O m B Λ ∈ K B suchthat Bg = O m is outer, g is separated from zero locally on the intervals ( a n , a n +1 ), andlim y →∞ yg ( iy ) = 0. Also we have | O m ( z n ) | ≍ | O | h | ( z n ) | ≥ | h ( z n ) | & | f ( z n ) | . Step 3.
To complete the proof we need to show that Λ ∪ { λ } is a uniqueness set for K B . Assume the converse. Then F = Sg ∈ K B for some function S analytic in C + which YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 23 vanishes at λ . As in the proof of Case 1, S is in the Smirnov class both in C + and in C − .Also S = F/g is locally in L on ( a n , a n +1 ). By Lemma 4.4 S is meromorphic with possiblepoles in a n . Since R has zeros of finite multiplicities, there exist m n such that near a n wehave | f ( t ) | ≥ C | R ( t ) | ≥ C | t − a n | m n for some C >
0, and an analogous estimate holds for g . Hence, S may have only poles atthe points a n . However, ( t − a n ) − g / ∈ L ( a n − δ, a n + δ ), and we conclude that S is anentire function which is of zero exponential type by Krein’s theorem.We have | O m ( iy ) | & | f ( iy ) | ≥ Cy − , y → + ∞ . So, | S ( iy ) | ≤ (cid:12)(cid:12)(cid:12) F ( iy ) B Λ ( iy ) (cid:12)(cid:12)(cid:12) | O m ( iy ) | − ≤ Cy / and | S ∗ ( iy ) | ≤ | ( BF )( iy ) || O m ( iy ) | ≤ Cy / , y → + ∞ . We conclude that S is a polynomial of degree at most 1. Step 4.
To finish the proof of completeness of { k λ } λ ∈ Λ ∪{ λ } , we need to show that S is a constant and, thus, zero. Note that we have S ∗ O m ∈ K B . Let S be a polynomial ofdegree 1. Then we have zO m ∈ H , and so, since z n is a Carleson sequence, we have(4.6) X n y n | z n | | f ( z n ) | ≤ C X n y n | z n | | O m ( z n ) | < + ∞ . On the other hand,Re f ( z n ) = 1 − qc n √ y n − q X k = n ( y n + y k ) √ y k c k | z n − z k | = 1 − qc n √ y n − d n . Using the properties of z n we get | d n | = q (cid:12)(cid:12)(cid:12)(cid:12) X k = n √ y k c k ( y n + y k ) | z n − z k | (cid:12)(cid:12)(cid:12)(cid:12) ≤ qx n X k = n √ y k c k , and hence P n y n | z n | | d n | < + ∞ . Recall also that by c) we have qc n √ y n >
2. Thus | z n | · | Re f ( z n ) | + | z n d n | ≥ qc n | z n | √ y n , but, by the choice of c n , X n | z n | | c n | = + ∞ , so P n y n | z n | | Re f ( z n ) | = + ∞ , a contradiction to (4.6). Thus S ≡ const and, since S ( λ ) = 0, we finally conclude that S ≡ (cid:3) Sufficient conditions for the completeness
In this section we discuss conditions sufficient for the completeness of the system biorthog-onal to an exact system of reproducing kernels. Since the model spaces generated bymeromorphic inner functions essentially coincide with de Branges spaces we may obtaincompleteness for a class of model subspaces as a corollary of Theorem 1.2. However, itseems that this method does not work for the general model spaces.On the other hand, even if we can not say that any system biorthogonal to an exactsystem of reproducing kernels is complete, one may look for criteria for completeness ofthe biorthogonal system in terms of the generating function. Results of this type may haveapplications in the spectral theory, since systems biorthogonal to systems of reproducingkernels appear, e.g., as eigenfunctions of certain rank one perturbations of unboundedselfadjoint operators [4].The crucial property of the model spaces which makes it possible to apply the ideasfrom the proof of Theorem 1.2 is the representation of the functions in K Θ via the so-called Clark measures [9], which is similar to (1.4). For any α ∈ C , | α | = 1, the function( α + Θ) / ( α − Θ) has positive real part in the upper half-plane. Hence, there exist p α ≥ µ α with R (1 + t ) − dµ α ( t ) < + ∞ such thatRe α + Θ( z ) α − Θ( z ) = p α Im z + Im zπ Z R dσ α ( t ) | t − z | , z ∈ C + . The Clark theorem states that if µ α is purely atomic (that is, µ α = P n c n δ t n , where δ x denotes the Dirac measure at the point x ) and p α = 0, then the system { k t n } of reproducingkernels is an orthogonal basis in K Θ . In the general case, if p α = 0, then the mapping(5.1) ( C µ g )( z ) = ( α − Θ( z )) Z g ( t ) t − z dµ α ( t )is a unitary map from L ( µ ) onto K Θ . By a theorem of A. Poltoratski [20], any function f ∈ K Θ has nontangential boundary values µ α -a.e. for any α .Note also that the following are equivalent:a) p α > α ;b) α − Θ ∈ H ( R ) (i.e., ∞ is a Carath´eodory point for Θ).c) µ β ( R ) < + ∞ for some β ∈ C , | β | = 1.To see the equivalence of a) and c) note that if µ β ( R ) < + ∞ and p β = 0, then p − β > tempered inner functions ,that is inner functions such that | (arg Θ) ′ ( t ) | . | t | N for some N > YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 25
Theorem 5.1.
Let Θ be a meromorphic inner function. Write Θ = exp(2 iϕ ) on R , where ϕ is a smooth increasing function on R . Let { t n } = { t ∈ R : Θ( t ) = − } . If α − Θ / ∈ L ( R ) for any α ∈ C , | α | = 1 , and for some N > and C > , ϕ ′ ( t n ) ≤ C ( | t n | + 1) N , then any system biorthogonal to an exact system of reproducing kernels is complete in K Θ .Proof. We may write Θ = E ∗ /E for a Hermite–Biehler function E . Then the mapping f Ef is a unitary map of K Θ onto H ( E ). The function A = E + E ∗ = E (1+Θ)2 is thegenerating function for t n , and the functions k t n ( z ) = E ( t n ) πi · A ( z ) z − t n form an orthogonal basisof reproducing kernels in H ( E ) (see formula (1.5)). By (1.6), b n = ( πϕ ′ ( t n )) − . Also notethat X n b n δ t n = X n B ( t n ) πA ′ ( t n ) δ t n is the Clark measure µ − for Θ. Since α − Θ / ∈ L ( R ) for any α we conclude that P n b n =+ ∞ , and the result follows from Theorem 1.2. (cid:3) Remark 5.2.
Theorem 5.1 may be extended to the case when the spectrum of Θ is a finiteset and ϕ ′ has at most power growth near each of these points, that is, ϕ ′ ( t ) . | t − a | − N near a ∈ σ (Θ). However, it is not clear how to deal with the case when σ (Θ) has nonemptyinterior or when Clark measures have singular continuous parts.The following theorem applies to general inner functions. We now impose the restrictionson the generating function G in place of Θ. Recall that G ∈ ( t + i ) H . The followingresult shows that if G is sufficiently regular, that is, has at most power growth, then thebiorthogonal system is complete. Theorem 5.3.
Let Θ be an inner function such that − Θ / ∈ L ( R ) and µ ( R ) = + ∞ ( these conditions are fulfilled, e.g., if ∞ is not a Carath´eodory point for Θ) . Let G be thegenerating function of an exact system of reproducing kernels { k λ n } , λ n ∈ C + . If for some N > and C > | G ( t ) | ≤ C | t + i | N µ − a . e ., then the system G ( z ) z − λ n is complete. Lemma 5.4.
Let G be the generating function of an exact system of reproducing kernels { k λ n } , λ n ∈ C + . Then Θ G is an outer function and Gz − x / ∈ L ( R ) for any x ∈ R .Proof. Fix a zero λ of G . Then we can write G = ( z − λ ) g where g ∈ K Θ and g vanisheson { λ n } n =0 . Then Θ G = ( t − λ )Θ g on R . Since g ∈ K Θ we have Θ g ∈ K Θ and we may write Θ g = IO m . As in the proof of Lemma 4.2, if I is not a Blaschke product we mayreplace it by its Frostman shift I = I − γ − γI , | γ | <
1, which is a Blaschke product. Then h = g (1 − γI ) I is in K Θ . If I = 1, then I is a nontrivial Blaschke product, and so h vanishes on { λ n } n =0 and on the zero set of I . This contradicts the completeness of { k λ n } .If for some x ∈ R , Gz − x ∈ L ( R ), then by (4.2) Gz − x is in K Θ and vanishes on { λ n } . (cid:3) Proof. (of Theorem 5.3). Let µ = µ . Since 1 − Θ / ∈ L ( R ), we have representation (5.1)with α = 1 for the elements of K Θ . Let h ∈ K Θ be orthogonal to all functions G ( z ) z − λ n . Then,for any n ,(5.2) D Gz − λ n , h E H = D Gz − λ n , h E L ( µ ) = Z G ( t ) h ( t ) t − λ n dµ ( t ) = 0 . Consider the function(5.3) L ( z ) = (1 − Θ( z )) Z G ( t ) h ( t ) t − z dµ ( t );it is analytic in C + and vanishes at λ n . Hence we may write L = SG for a function S analytic in C + ,(5.4) S ( z ) G ( z ) = (1 − Θ( z )) Z G ( t ) h ( t ) t − z dµ ( t ) . We denote by S the linear space of all functions S for which (5.4) holds with some h ∈ L ( µ ).Note that S has nontangential boundary values µ -a.e. and G ( t ) S ( t ) = G ( t ) h ( t ) µ -a.e.Indeed, G is locally bounded, and so Ghχ ( − r,r ) ∈ L ( µ ) for any r > χ E we denotethe characteristic function of a set E ). Write(5.5) L ( z ) = (1 − Θ( z )) Z ( − r,r ) G ( t ) h ( t ) t − z dµ ( t ) + (1 − Θ( z )) Z R \ ( − r,r ) G ( t ) h ( t ) t − z dµ ( t ) . Then the first term is in K Θ and has nontangential boundary values G ( x ) h ( x ) for | x | < rµ -a.e. The second term is analytic for | z | < r and tends to zero for a fixed z when r → ∞ .Hence, S ( t ) G ( t ) = G ( t ) h ( t ) µ -a.e.By the arguments similar to Lemma 2.3 it is easy to show that for any w ∈ C + and S ∈ S we have S ( z ) − S ( w ) z − w ∈ S (just replace the sum by the integral with respect to µ ).Assume first that S is infinite dimensional. Then, as in the proof of Theorem 1.2, thereexists S ∈ S with at least N zeros w , ..., w N ∈ C + different from the points { λ n } , and YSTEMS OF REPRODUCING KERNELS AND THEIR BIORTHOGONAL 27 T ( z ) := S ( z ) Q Nl =1 ( z − w l ) ∈ S . For some h ∈ L ( µ ) we have G ( z ) S ( z ) = (1 − Θ( z )) Z G ( t ) h ( t ) t − z dµ ( t ) , and so, G ( z ) T ( z ) = (1 − Θ( z )) Z G ( t ) Q Nl =1 ( t − w l ) · h ( t ) t − z dµ ( t ) . Since | G ( t ) | . | t + i | N µ -a.e., we conclude that G ( t ) Q Nl =1 ( t − w l ) · h ( t ) ∈ L ( µ ). Thus, for thefunction GT we have a representation of the form (5.1), and so GT ∈ K Θ . This contradictsthe completeness of { k λ n } .Now assume that S is finite dimensional. Since the transform ( D w S )( z ) = S ( z ) − S ( w ) z − w preserves the class S , the functions S, D w S, D w S, . . . , D Nw S are linearly dependent for large N . We conclude that S consists of rational functions. Write S = P/Q , where
P, Q arepolynomials without common zeros. Since S is analytic in C + , Q has no zeros in C + . Itfollows from (5.5) that L is locally in L on R . If S has a pole x on R it follows that G ( z ) z − x is in L ( x − δ, x + δ ) which contradicts Lemma 5.4. Assume, finally, that Q has a zero in C − . We have Θ( z ) G ( z ) S ( z ) = (Θ( z ) − Z G ( t ) h ( t ) t − z dµ ( t ) . The right-hand side is analytic in C + whereas, by Lemma 5.4, Θ( z ) G ( z ) is an outer functionin C + . Hence S ( z ) is analytic in C + and, thus, S has no poles in C − . We conclude that S is a polynomial.Without loss of generality we may assume that S ≡
1. To complete the proof recall that S ( t ) G ( t ) = G ( t ) h ( t ) µ -a.e. Put E := { t : G ( t ) = 0 } , E := R \ E . We have h ( t ) = 1 µ -a.e. on E . Since h ∈ L ( µ ), but µ ( R ) = + ∞ , we conclude that µ ( E ) < + ∞ and µ ( E ) = + ∞ . Then the functions h ∈ K Θ such that h = 0 µ -a.e. on E form an infinitedimensional subspace X of K Θ . For any h ∈ X we have Gh = 0 µ -a.e. and hence (5.2)holds. Since X is infinite dimensional and contains h ( z ) z − λ whenever λ ∈ C + , h ( λ ) = 0, thereexists a nonzero function h ∈ X such that ( t + i ) N h ∈ L ( µ ). Then for L defined by(5.3) with h in place of h we have L ( z ) = (1 − Θ( z )) Z G ( t ) h ( t ) t − z dµ ( t ) = (1 − Θ( z )) Z G ( t )( t + i ) N ( t + i ) N h ( t ) t − z dµ ( t ) . The function L vanishes at λ n and belongs to K Θ since it has a representation of the form(5.1). Hence, L ≡ h = 0 µ -a.e. This contradiction proves the theorem. (cid:3) Remark 5.5. If G ∈ L ∞ ( R ), then the conditions 1 − Θ / ∈ L ( R ) and µ ( R ) = + ∞ maybe omitted. Indeed, p α = 0 for all unimodular α except at most one, and the functions in K Θ admit representation (5.1). Then any function L defined by (5.3) is in K Θ (note that Gh ∈ L ( µ )) and vanishes at λ n . Hence, h ≡ References [1] P.R. Ahern, D.N. Clark,
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