Berezin Symbol and Invertibility of Operators on the Functional Hilbert Spaces

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Journal of Functional Analysis 238 (2006) 181–192 www.elsevier.com/locate/jfa

Berezin symbol and invertibility of operators on the functional Hilbert spaces

Mubariz T. Karaev

Suleyman Demirel University, Isparta (MYO) Vocation School, 32260 Isparta, Turkey Received 7 June 2005; accepted 24 April 2006 Available online 16 June 2006 Communicated by Paul Malliavin

Abstract We give in terms of reproducing kernel and Berezin symbol the sufﬁcient conditions ensuring the invertibility of some linear bounded operators on some functional Hilbert spaces. © 2006 Elsevier Inc. All rights reserved.

Keywords: Hardy space; Bergman space; Toeplitz operator; Berezin symbol; Reproducing kernel

1. Introduction ∞ ∞ Let T be the unit circle T ={ζ ∈ C: |ζ |=1}, ϕ ∈ L = L (T), and let Tϕ be the Toeplitz operator acting in the Hardy space H 2(D) on the unit disc D ={z ∈ C: |z| < 1} by the formula Tϕf = P+ϕf , where P+ is the Riesz projector. Let ϕ˜ denote the harmonic extension of the function ϕ to D. In [5] Douglas posed the following problem: if ϕ is a function in L∞ for which |˜ϕ(z)| δ>0, z ∈ D, then is Tϕ invertible? In [18] Tolokonnikov ﬁrstly gave a positive answer to this question under the condition that δ is near enough to 1, namely, he proved that if 45 1 ϕ(z)˜ δ> ,z∈ D, 46

then Tϕ is invertible and −1 − − −1 Tϕ 1 46(1 δ) .

E-mail address: [email protected].

This assertion was also proved by Wolff [19]. Nikolskii [15] has somewhat improved the result of Tolokonnikov proving invertibility of Tϕ and the estimate −1 − −1/2 Tϕ (24δ 23) under condition δ>23/24. Finally, Wolff [19] has constructed a function ϕ ∈ L∞ such that infD |˜ϕ(z)| > 0 but the corresponding operator Tϕ is not invertible, and thus showed that the answer to the question of Douglas is negative in general. Since ϕ˜ coincides with the Berezin symbol Tϕ of the operator Tϕ (see Lemma 2.1), in this context the following natural problem arises.

Problem 1. Let A be a linear bounded operator acting in the functional Hilbert space H(Ω) of complex-valued functions over the some (non-empty) set Ω, such that |A(z) | δ for all z ∈ Ω and for some δ>0. To ﬁnd the number δ0, which can be (more or less) easily computed from the data of A, and due to which the inequality A(z) δ>δ0,z∈ D, ensures the invertibility of A, where A denotes the Berezin symbol of the operator A.

In particular, the following problem is also interesting, which is closely related with the ﬁnite section method of Böttcher and Silbermann [3].

Problem 2. Let E ⊂ H(Ω) be a closed subspace of the functional Hilbert space H(Ω), and let A be a linear bounded operator acting in H(Ω) such that A(z) δ for all z ∈ Ω and for some δ>0. To ﬁnd a number δ0, such that δ>δ0 ensures the invertibility of operator PEA | E (the compression of the operator A to the subspace E), where PE is an orthogonal projection from H(Ω) onto E.

In this article we solve these problems in some special cases. Our argument uses the concept of reproducing kernel and Berezin symbol.

2. Notations and preliminaries

2.1. Recall that a functional Hilbert space is a Hilbert space H = H(Ω) of complex-valued functions on a (non-empty) set Ω, which has the property that point evaluations are continuous (i.e., for each λ ∈ Ω,themapf → f(λ)is a continuous linear functional on H). Then the Riesz representation theorem ensures that for each λ ∈ Ω there is a unique element kλ of H such that f(λ)=f,kλ for all f ∈ H. The collection {kλ: λ ∈ Ω} is called the reproducing kernel of H. It is well known (see, for instance, [8, Problem 37] that if {en} is an orthonormal basis for a functional Hilbert space H, then the reproducing kernel of H is given by kλ(z) = en(λ)en(z). n M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192 183

For λ ∈ Ω,letkˆ = kλ be the normalized reproducing kernel of H. For a bounded linear λ kλ operator A on H, the function A deﬁned on Ω by

ˆ ˆ A(λ) =Akλ, kλ is the Berezin symbol of A, which ﬁrstly have been introduced by Berezin [1,2]. It is clear that the Berezin symbol A is the bounded function on Ω whose values lies in the numerical range of the operator A, and hence def supA(z) = ber(A) (“Berezin number”) z∈D w(A) (numerical radius).

More typical examples of functional Hilbert spaces are the Hardy and Bergman spaces.

2.2. Let dm2 denote Lebesgue area measure on the unit disk D, normalized so that the D 2 = 2 D measure of equals 1. The Bergman space La La( ) is the Hilbert space consisting of the 2 analytic functions on D that are also in L (D,dm2). For z ∈ D, the Bergman reproducing ker- ∈ 2 = ∈ 2 nel is the function kλ La such that f(λ) f,kλ for every f La. It is well known that 2 1 ˆ kλ 1−|λ| kλ(z) = . The normalized Bergman reproducing kernel kλ is the function = . (1−λz)2 kλ 2 (1−λz)2 2 = 2 D = n The Hardy space H H ( ) is the Hilbert space of analytic functions f(z) n0 anz D ={ ∈ | | } | |2 ∞ deﬁned in the unit disc z C: z < 1 , such that n0 an < . Alternately, it can be identiﬁed with a closed subspace of the Lebesgue space L2 = L2(T) on the unit circle, by associating to each analytic function its radial limit. The algebra of bounded analytic functions on D is denoted by H ∞.Anyϕ ∈ H ∞ acts as a multiplication operator on H 2, that we will denote by Tϕ. Norm and inner product in L2 or H 2 will be denoted by . and ·,·, respectively. Evaluations at points λ ∈ D are bounded functionals on H 2 and the corresponding reproducing kernel is k (z) = 1 ; thus, f(λ)=f,k .Ifϕ ∈ H ∞, then k is an eigenvector for T ∗ and T ∗k = λ 1−λz¯ λ λ ϕ ϕ λ ϕ(λ)kλ. By normalizing kλ we obtain

ˆ kλ 2 kλ = = 1 −|λ| kλ. kλ

2.3. The Berezin symbol have been investigated in detail for the Toeplitz and Hankel operators on the Hardy and Bergman spaces; it is widely applied in the various questions of analysis (see, for instance, [9–14,16,17,20]). In particular, it is known (see [11,20]) the following result which we will use in what follows.

∞ 2 Lemma 2.1. The Berezin symbol Tϕ of the Toeplitz operator Tϕ,ϕ∈ L , on the Hardy space H coincides with the harmonic extension ϕ˜ of the function ϕ into the unit disc D, that is Tϕ(λ) = ϕ(λ)˜ for all λ ∈ D.

Suppose now θ is an inner function. We deﬁne the corresponding model space by the formula 2 2 Kθ = H θH . 184 M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192

2.4. We recall some basic deﬁnitions concerning geometric properties of sequences in a Hilbert space. For most of the deﬁnitions and facts below, one can use [7,15] as a main references (see also [4,6]). Let H be a complex Hilbert space. If {xn}n1 ⊂ H , we denote by span{xn: n 1} the closure def of the linear hull generated by {xn}n1. The sequence X ={xn}n1 is called:

• complete if span{xn: n 1}=H ; • minimal if for all n 1, xn ∈/ span{xm: m = n}; x • uniformly minimal if inf dist( n , span(x : m = n)) > 0; n 1 xn m • a Riesz basis if there exists an isomorphism U mapping X onto an orthonormal family {Uxn: n 1}; • the operator U will be called the orthogonalizer of X.

The expression “a Riesz basis in H ” means a Riesz basis X with the completeness property span(X) = H . It is well known that X is a Riesz basis in its closed linear span if there are positive constants C1, C2 such that 1/2 1/2 2 2 C1 |an| anxn C2 |an| (1) n1 n1 n1 for all ﬁnite complex sequences {an}n1. Note that if U is an orthogonalizer of the family X then def the product r(X) =UU −1 characterizes the deviation of the basis X from an orthonormal one and U−1 and U −1 are the best constants in the inequality (1); r(X) will be referred to as the Riesz constant of the family X. Obviously, r(X) 1. We now recall some well-known facts (see [15]) concerning reproducing kernels in H 2.Let Λ ={λn}n1 be a sequence of distinct points in D. Then we have:

{ } { } (i1) kλn n1 is minimal if and only if λn n1 is Blaschke sequence (which means that −| | ∞ n1(1 λn )< ). As usual, we denote by

| | − = = = λn λn z B BΛ bλn , where bλn (z) . λn 1 − λ z n1 n

{ } { } (i2) If λn n1 is a Blaschke sequence, then kλn n1 is complete in KB . {ˆ } (i3) kλn n1 is a Riesz basis of KB if and only if it is uniformly minimal which is equivalent to {λn}n1 satisﬁes the Carleson condition inf Bn(λn) > 0, n1

= { } ∈ where Bn B/bλn ; we will write in this case λn n1 (C).

3. Results

In this section we partially solve Problems 1 and 2. M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192 185

Theorem 3.1. Let Λ ={λn}n1 be a Carleson sequence of distinct points in D, B the corresponding Blaschke product, and

def={ˆ } X kλn : n 1

2 2 be a corresponding Riesz basis in the model space KB = H BH (see assertion (i3) above), and denote by r(X) =UU −1 the corresponding Riesz constant of the family X. Let A be 2 def a linear bounded operator on the Hardy space H such that A KB ⊂ KB , and denote MA = 2 PB A|KB , where PB is an orthogonal projection from H onto KB . Suppose that: ∞ 1/2 ˆ − ˆ 2 def= Λ +∞ (1) Akλn A(λn)kλn τA < and n=1 ∞ 1/2 ˆ − ˆ 2 def= Λ +∞ (2) A kλn A (λn)kλn τA < , n=1 where A denotes the Berezin symbol of the operator A. If def= def= Λ Λ inf A(z) δ>δ0 r(X) U max τA ,τA , z∈D

then the operator MA is invertible in KB and

−1 − δ M 1 −Uτ Λ . A r(X) A

∈ ={ˆ } Proof. Since Λ (C), the sequence X kλn : n 1 is a Riesz basis in KB (see assertion (i3) above). If U is an orthogonalizer of X , then U−1 and U −1 are corresponding best constants appearing in (1), that is 1/2 1/2 −1 | |2 ˆ −1 | |2 U an ankλn U an (2) n1 n1 n1 for all ﬁnite complex sequences {an}n1. Now it is clear from (2) and the condition |A(z)| δ>0, z ∈ D, of the theorem that N N 1/2 N 1/2 − − a A(λ )kˆ U 1 a A(λ )2 δU 1 |a |2 n n λn n n n n=1 n=1 n=1 δ N δ N a kˆ = a kˆ UU −1 n λn r(X) n λn n=1 n=1 and hence N δ N a A(λ )kˆ a kˆ , (3) n n λn r(X) n λn n=1 n=1 186 M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192

={ˆ } where r(X) is the Riesz constant of the family X kλn : n 1 . Now using condition (1) of the theorem and inequalities (2) and (3), for every ﬁnite N>0 and for arbitrary numbers an ∈ C (n = 1, 2,...,N)we have: N M a kˆ A n λn n=1 N N N = (P A | K ) a kˆ = P A a kˆ = a P Akˆ B B n λn B n λn n B λn n=1 n=1 n=1 N = a P Akˆ − A(λ )kˆ + A(λ )kˆ n B λn n λn n λn n=1 N N a P A(λ )kˆ − a P Akˆ − A(λ )kˆ n B n λn n B λn n λn n=1 n=1 N N a A(λ )kˆ − |a |Akˆ − A(λ )kˆ n n λn n λn n λn n=1 n=1 1/2 1/2 δ N N N a kˆ − |a |2 Akˆ − A(λ )kˆ 2 r(X) n λn n λn n λn n=1 n=1 n=1 ∞ 1/2 δ N N a kˆ −U a kˆ Akˆ − A(λ )kˆ 2 r(X) n λn n λn λn n λn n=1 n=1 n=1 δ N N a kˆ −Uτ Λ a kˆ r(X) n λn A n λn n=1 n=1 δ N = −Uτ Λ a kˆ . r(X) A n λn n=1

Thus N δ N M a kˆ −Uτ Λ a kˆ (4) A n λn r(X) A n λn n=1 n=1 for all ﬁnite N>0 and complex numbers an, n = 1, 2,...,N. Since the Carleson condition implies the Blaschke condition, we have from (4) that

δ M f −Uτ Λ f (5) A r(X) A for all f ∈ KB (see assertion (i2) above). M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192 187

Since A KB ⊂ KB , it is easy to see that

= | = | MA (PB A KB ) A KB .

Analogously, using condition (2) of the theorem, the equality |A(z)|=|A(z) | (z ∈ D) and the inequalities (2) it can be proved that (we omit it) N N ˆ δ Λ ˆ M a k −Uτ a k A n λn r(X) A n λn n=1 n=1 for all ﬁnite N>0 and complex numbers an, n = 1, 2,...,N,which yields that

δ Λ M f −Uτ f (6) A r(X) A for all f ∈ KB . Now by considering that Λ Λ = δ>r(X) U max τA ,τA δ0, we deduce from the estimates (5) and (6) that the operator MA is invertible in KB and

−1 − δ M 1 −Uτ Λ , A r(X) A which completes the proof. 2 ∞ It is necessary to note that when A is an analytic Toeplitz operator (i.e., A = Tϕ,ϕ∈ H )in Theorem 3.1, the invertibility of the operator MA follows only from the condition infA(z) > 0, z∈D because in this case A is invertible. Our next result concerns to the Bergman space operators. We recall that

2 ˆ 1 −|λ| kλ(z) = (1 − λz)2

2 are the normalized reproducing kernels of the Bergman space La. These normalized reproducing 2 kernels are the right building blocks for La. In some sense, they play the role of an orthonormal 2 basis for La, although they are clearly not mutually orthogonal. (This and other properties of Bergman kernel can be found in Zhu [20].) The following key lemma (see [20, Theorem 4.4.6]) gives the so-called atomic decomposition 2 for functions in the Bergman space La.

Lemma 3.2. There exists a sequence Λ ={λn}n1 in D and a constant C>0 with the following properties: 188 M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192

2 (a) For any {an} in l , the function

∞ 2 1 −|λn| f(z)= an (1 − λ z)2 n=1 n

2 { } ; is in La with f C an l2 ∈ 2 { } 2 (b) If f La, then there is an in l , such that

∞ −| |2 = 1 λn { } f(z) an and an 2 C f . (1 − λ z)2 l n=1 n

Theorem 3.3. Let Λ ={λn}n1 and C>0 are the same as in Lemma 3.2. Let A be a linear 2 bounded operator on the Bergman space La satisfying

∞ def (1) Akˆ − A(λ )kˆ 2 = τ Λ < +∞ and n=1 λn n λn A ∞ ˆ − ˆ 2 def= Λ +∞ (2) n=1 A kλn A (λn)kλn τA < ,

| | { 3 Λ 3 Λ } ∈ D where A denotes the Berezin symbol of operator A. If A(z) δ>max C τA ,C τA , z , then A is invertible and

2 − C A 1 . − 3 Λ δ C τA

∈ 2 { } 2 Proof. If f La is an arbitrary function then by Lemma 3.2, there exists an in l such that ∞ = ˆ { } f(z) ankλn (z) and an l2 C f . n=1

| |= { } ∈ 2 Since supz∈D A(z) ber(A) A , we have thatanA(λn) n1 l , and therefore it follows ∞ ˆ 2 from the claim (a) of Lemma 3.2 that the function n=1 anA(λn)kλn is in La with ∞ a A(λ )kˆ C a A(λ ) . (7) n n λn n n l2 n=1

Since ∞ a A(λ ) ber(A){a } C ber(A)f =C ber(A) a kˆ , n n l2 n l2 n λn n=1 the inequality (7) means that the diagonal operator D deﬁned in L2 by the formula {A(λn)} a ∞ ∞ ˆ ˆ D a k = a A(λ )k , {A(λn)} n λn n n λn n=1 n=1 M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192 189 is bounded operator with

2 D C ber(A). {A(λn)} On the other hand, it follows from the condition of theorem that |A(λn)| δ for all n 1, and D 2 D−1 = D therefore {A(λ )} is an invertible in La, { } { 1 }, and n A(λn) A(λn)

2 C D{ 1 } . (8) A(λn) δ

Now using condition (1) of the theorem and inequality (8), we have ∞ ∞ Af = A a kˆ = a Akˆ 2 n λn n λn = = n 1 n 1 ∞ ∞ a A(λ )kˆ − a Akˆ − A(λ )kˆ n n λn n λn n λn n=1 n=1 ∞ ∞ 1/2 ∞ 1/2 −1 ˆ 2 ˆ ˆ 2 D 1 ankλ − |an| Akλ − A(λn)kλ { } n n n A(λn) = = = n 1 n 1 n 1 ∞ δ − CτΛ a kˆ 2 A n λn C = n 1 δ = − CτΛ f . C2 A

Thus

δ Af − CτΛ f (9) 2 C2 A

∈ 2 for all f La. Since |A(z)|=|A(z) |, z ∈ D, by similar arguments we can prove that

δ Λ A f − Cτ f (10) C2 A

∈ 2 for all f La. 2 { Λ Λ } Also, since δ/C >Cmax τA ,τA (see the condition of theorem), inequalities (9) and (10) 2 mean that A is invertible in La and

−1 − δ A 1 − CτΛ , C2 A which proves the theorem. 2 190 M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192

Our more general result is the following theorem.

Theorem 3.4. Let H = H(Ω) be a functional Hilbert space of complex-valued functions on a (non-empty) set Ω with the orthonormal basis {en(z)}n0, and let A be a linear bounded operator on H such that:

(a) |A(z) | δ>0, z ∈ Ω; (b) there exists a sequence Λ ={λn}n0 ⊂ Ω such that: ∞ def (1) ( Ae (z) − A(λ )e (z)2)1/2 = δΛ < +∞; n=0 n n n A ∞ − 2 1/2 def= Λ +∞ (2) ( n=0 A en(z) A (λn)en(z) ) δA < . { Λ Λ } If δ>max δA ,δA , then A is invertible and −1 − Λ −1 A δ δA .

D Proof. Let us consider the diagonal operator {A(λ )} with respect to the orthonormal basis n {e (z)} of the space H, that is D e (z) = A(λ )e (z), n 0. Since δ |A(λ )| A n n 0 {A(λn)} n n n n for all n 0, we have that D =sup A(λ ) A, {A(λn)} n n0

−1 D exists and {A(λn)} D−1 = D D−1 1 { } { 1 } and { } . A(λn) A(λn) A(λn) δ = ∞ ∈ H Then by considering these and the conditions of theorem, for all f(z) n=1 anen(z) we have ∞ ∞ = = Af H A anen(z) anAen(z) n=0 H n=0 H ∞ ∞ − − anA(λn)en(z) an Aen(z) A(λn)en(z) n=0 H n=0 H ∞ ∞ 1/2 ∞ 1/2 1 a e (z) − |a |2 Ae (z) − A(λ )e (z) 2 D−1 n n n n n n = = = {A(λn)} n 0 H n 0 n 0 = 1 − Λ = − Λ f H δA f H inf A(λn) f H δA f H 1 n0 supn0 A(λn) − Λ = − Λ δ f H δA f H δ δA f H. M.T. Karaev / Journal of Functional Analysis 238 (2006) 181–192 191

Analogously, we can show that − Λ ∈ H A f H δ δA f H for all f , and hence A is invertible and −1 − Λ −1 A δ δA .

The theorem is proved. 2

Corollary 3.5. Let ϕ ∈ L∞(T), and denote as before, by ϕ˜ its harmonic extension (by the Poisson 2 formula) into D. Let Tϕ be a corresponding Toeplitz operator on the Hardy space H . Suppose def that δ = infz∈D |˜ϕ(z)| > 0 and there exists a sequence Λ ={λn}n0 ⊂ D such that

∞ |ˆ|2 − ˜ ˆ + ˜ 2 def= Λ +∞ ϕ (0) 2Reϕ(λn)ϕ(0) ϕ(λn) νϕ < . (11) n=0

Λ If δ>νϕ , then Tϕ is invertible and − − T 1 δ − δΛ 1. ϕ Tϕ

Proof. An easy computation shows that n n2 n2 n n 2 Tϕz −˜ϕ(λn)z = Tϕz − 2Reϕ(λ˜ n) Tϕz ,z + ϕ(λ˜ n) n2 − ˜ ˆ + ˜ 2 ϕz L2(T) 2Reϕ(λn)ϕ(0) ϕ(λn) 2 2 = |ˆϕ| (0) − 2Reϕ(λ˜ n)ϕ(ˆ 0) + ϕ(λ˜ n) ,

¯ = | |=| | from which by replacing ϕ with ϕ and by considering that Tϕ Tϕ¯ and Tϕ(z) Tϕ (z) , we also have n − ¯ n2 |ˆ|2 − ˜ ˆ + ˜ 2 Tϕ z ϕ(λn)z ϕ (0) 2Reϕ(λn)ϕ(0) ϕ(λn) .

Λ Now, by considering the conditions δ>νϕ and (11), and by applying Lemma 2.1 and Theo- rem 3.4, we have the desired result. 2

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