# A Theorem of Nehari Type on Weighted Bergman Spaces of the Unit

Authors: Yufeng Lu; Jun Yang
Journal: Abstract and Applied Analysis
 (2008)

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## Abstract

This paper shows that ifS is a bounded linear operator acting on
the weighted Bergman spaces Aα2 on the unit ball in ℂn such that STzi=Tz¯iS(i=1,…,n), where Tzi=zif and
Tz¯i=P(z¯if); and where P is the weighted Bergman projection, then S must be a Hankel operator.


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## Body

## 1. Introduction

LetBn be the  open unit ball in the complex vector
space ℂn.
For z=(z1,…,zn),w=(w1,…,wn)∈ℂn, let 〈z,w〉=z1w¯1+⋯+znw¯n, where w¯k is the complex conjugate of wk, and |z|=〈z,z〉. For a multi-index m=(m1,…,mn) and z=(z1,…,zn)∈ℂn,
we also write(1.1)zm=z1m1…znmn.LetdV be the volume measure on Bn,
normalized so that V(Bn)=1. For α>−1, the weighted Lebesgue measure dVα is defined by(1.2)dVα(z)=cα(1−|z|2)αdV(z),where(1.3)cα=Γ(n+α+1)n!Γ(α+1)is a normalizing constant so
that dVα is a probability measure on Bn.Forp≥1 and α>−1,
the weighted Bergman space Aαp consists of holomorphic functions f in Lp(Bn,dVα),
that is,(1.4)Aαp=Lp(Bn,dVα)∩H(Bn).When α=0,Aαp is the standard (unweighted) Bergman spaces,
which is simply denoted by Ap.The weighted Bergman spaceAαp is a closed subspace of Lp(Bn,dVα) and the set of all polynomials is dense in Aαp.
See, for example, [1].With the norm(1.5)∥f∥p=(∫Bn|f(z)|pdVα(z))1/p,Lp(Bn,dVα) and Aαp become Banach spaces. L2(Bn,dVα) is a Hilbert space whose inner product will be
denoted by 〈⋅,⋅〉α.
Some other properties of Bergman spaces as well as some recent results on the
operators on them, can be found, for example, in [2–13] (see, also the references therein).Forφ∈L∞(Bn),
the Hankel operator Hφ is defined on Aα2 by(1.6)Hφ(f)=P(J(φf)),where J is the unitary operator defined on L2(Bn,dVα) by(1.7)J(f(z))=J(f(z1,…,zn))=f(z¯)=f(z¯1,…,z¯n),and P is the weighted Bergman projection from L2(Bn,dVα) onto Aα2.The Toeplitz operator with the symbolφ∈L∞(Bn) is defined on Aα2 by(1.8)Tφf=P(fφ),f∈Aα2.Toeplitz operators have the following properties: ifa and b are complex numbers, and φ and ψ∈L∞(Bn),
then Taφ+bψ=aTφ+bTψ,Tφ*=Tφ¯; moreover, if φ∈H∞(Bn), then TψTφ=Tφψ and Tφ¯Tψ=Tφ¯ψ.The symbolzi will denote the ith coordinate function (i=1,…,n).It is easy to see thatHφTzi=Tz¯iHφ.
Thus, the Hankel operators Hφ are particular solutions of the operator equation(1.9)STzi=Tz¯iS,i=1,…,n,where S is a bounded linear operator on Aα2.It is well known that on the classical Hardy spaceH2,
Toeplitz operators and Hankel operators are of the same status, and present
different operators classes. The authors of [14] regarded Hankel operators as an essential part of
Toeplitz operator theory, and many authors studied Hankel operators and their
related problems in [14–22].On the Hardy spaceH2,
Nehari [19] proved
that if S is a bounded linear operator such that STz=Tz¯S, then S=Hφ for some φ∈L∞;
moreover, φ can be chosen such that ∥Hφ∥=∥φ∥.
Faour  [20] proved a theorem of Nehari
type on the Bergman spaces of the unit disk.  In [21], the authors gave the
characterization of Hankel operators on the generalized H2 spaces, which is also similar to the Nehari
theorem on the Hardy space.The motivation for this paper is the question whether
solutions of the operator (1.9) must be the Hankel operator on the Bergman
space Aα2.In this paper, we take the weighted Bergman spaceAα2 as our domain and prove a Nehari-type theorem.
While our method is basically adapted from [20, 21], substantial amount of extra work is necessary for
the setting of the weighted Bergman spaces on the unit ball.


## 2. Nehari-Type Theorem

To establish a Nehari-type theorem on the weighted
Bergman spaces on the unit ball, we recall the atomic decomposition of the
weighted Bergman spaceAαp,
which plays an important role in this paper. It is shown that every function in
the weighted Bergman space Aαp can be decomposed into a series of nice
functions called atoms. These atoms are defined in terms of kernel functions
and in some sense act as a basis for Aαp.
The following lemma is Theorem 2.30 in [1].
Lemma 2.1.
Supposep>0,α>−1,
and(2.1)b>nmax(1,1p)+α+1p.Then there exists a sequence {ak} in Bn such that Aαp consists exactly of functions of the form(2.2)f(z)=∑k=1∞ck(1−|ak|2)(pb−n−1−α)/p(1−〈z,ak〉)b,z∈Bn,where {ck} belongs to the sequence space lp and the series converges in the norm topology
of Aαp.Remark 2.2.
By the proof of Theorem 2.30 in [1], it can be seen that the sequence {ak} in Lemma 2.1 is chosen independent of p,α,
and b.Remark 2.3.
The proof of Theorem 2.30 in [1] tells us that for any f∈Aαp,
we can choose a sequence {ck} in Lemma 2.1 so that(2.3)∑k|ck|p≤C∫Bn|f(z)|pdVα(z),where C is a positive constant independent of f.The following lemma follows immediately from Lemma2.1.Lemma 2.4.
Suppose{ak} is a sequence as in Lemma 2.1, α>−1,
and b>n+α+1. Let(2.4)la(z)=(1−|a|2)b−n−1−α(1−〈z,a〉)b.Then, Aα1(Bn) consists exactly of the functions of the form(2.5)f(z)=∑k=1∞cklak,z∈Bn,where {ck} belongs to the sequence space l1 and the series converges in the norm topology
of Aα1.From now on, we assume thatb>2(n+α+1) is fixed and {ak} and la(z) are defined as in Lemma 2.4.The following two lemmas follow immediately from
Theorem 1.12 in [1].
Lemma 2.5.
Letα>−1, 0<r<1,
then for every a∈Bn, one has(2.6)∥la(rz)∥2≤k(r),where k(r) is a constant which only depends on r.Lemma 2.6.
There
exists a constantC such that for every a∈Bn,r∈(0,1),(2.7)∥la(rz)∥1≤C,where C is independent of a and r.Theorem 2.7.
LetS be a bounded linear operator acting on the
weighted Bergman space Aα2 such that STzi=Tz¯iS(i=1,…,n).
Then, there exists φ∈L∞(Bn) such that S=Hφ.Proof.
Define the linear functionalG on Aα2 by G(f)=〈Sf,1〉α.
Clearly, G is a bounded linear functional on Aα2.
Note that Aα2⊂Aα1.
From Lemma 2.4 and Remark 2.3, given f∈Aα2,
there exists {ck} in l1 such that f=∑kcklak converges in Aα1 and ∑|ck|≤β∥f∥1,
where β is a positive constant independent f.
Forf∈Aα2,
let f+(z)=f(z¯)¯∈Aα2.
From (1.9), it is easy to see that STzik=Tz¯ikS(i=1,…,n;k=1,2,…,).
If p=azik,q=bzjm,
then we have(2.8)〈Sp,q〉α=ab¯〈STzik1,Tzjm1〉α=ab¯〈Tz¯jmSTzik1,1〉α=ab¯〈STzjmTzik1,1〉α=ab¯〈S(zjmzik),1〉α=〈S(pq+),1〉α.Hence, we establish that 〈S(pq+),1〉α=〈Sp,q〉α, where p and q are polynomials in z=(z1,…,zn).
Since the set of all polynomials is dense inAα2,
there are sequences of polynomials pn(z) and qn(z) such that(2.9)∥pn−lak1/2∥2→0,∥qn−(lak1/2)+∥2→0(asn→∞).Furthermore, ∥qn+−lak1/2∥2→0.
Since(2.10)〈S(pnqn+),1〉α=〈Spn,qn〉α,by using the boundedness of S and the continuity of the scalar product, it
follows that(2.11)〈Slak1/2,(lak1/2)+〉α=〈Slak,1〉α.
Givenr∈(0,1),
from Lemma 2.5, f(rz)=∑kcklak(rz) converges in Aα2.
Thus, with fr(z)=f(rz),
we see that(2.12)〈Sfr,1〉α=〈S(∑kcklak(rz)),1〉α=∑kck〈S(lak(rz)),1〉α=∑kck〈S(lak1/2(rz)),(lak1/2(rz))+〉α.Note that(2.13)∥lak1/2(rz)∥2=(∫Bn|(lak(rz))|dVα(z))1/2,∥(lak1/2(rz))+∥2=(∫Bn|(lak(rz¯))¯|dVα(z))1/2=∥lak1/2(rz)∥2,and consequently(2.14)∥lak1/2(rz)∥2∥(lak1/2(rz))+∥2=∥lak(rz)∥1.
Therefore,(2.15)|〈Sfr,1〉α|≤∑k|ck|∥S∥⋅supak∥lak(rz)∥1.Consequently, it follows from
Lemma 2.6 that(2.16)|〈Sfr,1〉α|≤Cβ∥S∥∥f∥1;but fr→f in Aα2(Bn).
Thus, by the continuity of G it follows that |G(f)|≤γ∥f∥1 for some constant γ.
Since Aα2 is dense in Aα1,
it follows that G is extended by continuity to an element of (Aα1)*,
and consequently, by the  Hahn-Banach theorem to an
element of (L1(Bn))*=L∞(Bn). Therefore, there exists φ∈L∞(Bn) such that(2.17)〈Sf,1〉α=〈φf,1〉α=〈J(φf),1〉α=〈P(J(φf)),1〉α=〈Hφf,1〉α.Since(2.18)〈Hφ(pq+),1〉α=∫Bnφ(z¯)p(z¯)q(z)¯dVα(z)=〈Hφp,q〉α,and by using the fact that 〈S(pq+),1〉α=〈Sp,q〉α,
where p, q are polynomials in z,
it follows that(2.19)〈Sp,q〉α=〈Hφp,q〉α.Hence, S=Hφ,
finishing the proof of the theorem.



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*Source: 538573-2008-12-22.xml*