# Hyponormal Toeplitz Operators on the Dirichlet Spaces

Authors: Puyu Cui; Yufeng Lu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We completely characterize the hyponormality of bounded Toeplitz operators with Sobolev symbols on the Dirichlet space and the harmonic Dirichlet space.


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## Body

## 1. Introduction

Let𝔻 be the open unit disk in the complex plane ℂ and dA be the normalized Lebesgue area measure on 𝔻. L∞(𝔻,dA) and L2(𝔻,dA) denote the essential bounded measurable function space and the space of square integral functions on 𝔻 with respect to dA, respectively. The Bergman space La2 consists of all analytic functions in L2(𝔻,dA). The Sobolev space W1,2(𝔻) is the space of functions f:𝔻→ℂ with the following norm:
(1)∥f∥=[|∫𝔻fdA|2+∫𝔻|∂f∂z|2+|∂f∂z¯|2dA]1/2<∞.W1,2(𝔻) is a Hilbert space with the inner product
(2)〈f,g〉=∫𝔻fdA∫𝔻g¯dA+〈∂f∂z,∂g∂z〉L2(𝔻,dA)+〈∂f∂z¯,∂g∂z¯〉L2(𝔻,dA).
The Dirichlet space 𝒟 consists of all analytic functions h in W1,2(𝔻) with h(0)=0. The Sobolev space W1,∞(𝔻) is defined by
(3)W1,∞(𝔻)={u∈W1,2(𝔻):u,∂u∂z,∂u∂z¯∈L∞(𝔻,dA)},
with the norm
(4)∥u∥1,∞=max{∥u∥∞,∥∂u∂z∥∞,∥∂u∂z¯∥∞}.
Let P be the orthogonal projection of W1,2(𝔻) onto 𝒟. P is an integral operator represented by
(5)P(f)(z)=∫𝔻∂f∂w(∂Kz∂w)-dA,
where Kz(w)=∑k=1∞(z¯kwk/k) is the reproducing kernel of 𝒟. For u∈W1,∞(𝔻), the Toeplitz operator Tu with symbol u is defined by
(6)Tuh=P(uh)h∈𝒟.Tu is a bounded operator for u∈W1,∞(𝔻) on 𝒟.Yu gave a decomposition of the Sobolev spaceW1,2(𝔻) in [1]. Let 𝒫0 be the set of all the following polynomials:
(7)∑j≥-l∑l≥0al+j,lzl+jz¯l,
where j and l run over a finite subset of ℤ and ∑l≥0‍al+j,l=0. Let 𝒜0 denote the closure of 𝒫0 in W1,2(𝔻), and let 𝒜 denote 𝒜0+ℂ. Since the set of all polynomials in z and z¯ is dense in W1,2(𝔻), there is the following decomposition:
(8)W1,2(𝔻)=𝒜⊕𝒟⊕𝒟¯.SinceW1,∞(𝔻)⊆W1,2(𝔻) and by the above decomposition, it follows that, if u∈W1,∞(𝔻), then u=u0+f+g¯, where u0∈𝒜, f, g∈H(𝔻) (the space of the analytic functions on 𝔻) with f(0)=g(0)=0.For the space𝒜0, there is the following proposition.Proposition 1 (see [1]).
Letϕ∈W1,∞(𝔻). Then ϕ𝒜0⊂𝒜0.A bounded linear operatorA on a Hilbert space is called hyponormal if A*A-AA* is a positive operator. There is an extensive literature on hyponormal Toeplitz operators on H2(𝕋) (the Hardy space on 𝕋) [2–4]. The corresponding problems for the Toeplitz operators on the Bergman space have been characterized in [5–9]. In the case of the Dirichlet space and the harmonic Dirichlet space, Lu and Yu proved that there are no nonconstant hyponormal Toeplitz operators with certain symbols [10]. In this paper, we completely characterize the Toeplitz operators Tu with u∈W1,∞(𝔻) on Dirichlet space 𝒟 and harmonic Dirichlet space 𝒟h.


## 2. Case on the Dirichlet Space

In this section, the hyponormality ofTu with u∈W1,∞(𝔻) on 𝒟 will be discussed.Theorem 2.
Letu=u0+c+f+g¯∈W1,∞(𝔻) with u0+c∈𝒜, f,g∈H(𝔻), and f(0)=g(0)=0. Then Tu is hyponormal on 𝒟 if and only if u∈𝒜.Proof.
By Proposition1, we only need to prove the necessity with u(z)=f(z)+g(z)¯=∑k=1∞‍fkzk+∑k=1∞‍g¯kz¯k.
Leth(z)=∑k=1∞‍hkzk∈𝒟. Simple calculations imply that
(9)Tf+g¯h(z)=Tfh(z)+Tg¯h(z)=f(z)h(z)+P(g¯h)(z)=f(z)h(z)+〈g¯h,Kz〉=∑m=2∞(∑k=1m-1fkhm-k)zm+∑l=2∞hl(∑k=1l-1g¯kzl-k)=∑m=2∞(∑k=1m-1fkhm-k)zm+∑m=1∞∑k=1∞g¯khk+mzm.
Furthermore,(10)∂(Tf+g¯h)∂z=∑m=1∞(m+1)(∑k=1mfkhm-k+1)zm+∑m=0∞(m+1)(∑k=1∞g¯khk+m+1)zm=∑k=1∞g¯khk+1+∑m=1∞(m+1)×(∑k=1mfkhm-k+1+∑k=1∞g¯khk+m+1)zm.
Therefore(11)∥Tf+g¯h∥2=|∑k=1∞g¯khk+1|2+∑m=1∞|∑k=1mfkhm-k+1+∑k=1∞g¯khk+m+1|2.
Similarly, we have(12)Tf+g¯*h(z)=〈Tf+g¯*h,Kz〉=〈h,Tf+g¯Kz〉=∑m=2∞1m(∑l=1m-1lgm-lhl)zm+∑m=1∞(∑k=1∞m+kmf¯khk+m)zm,∂(Tf+g¯*h)∂z=∑k=1∞(k+1)f¯khk+1+∑m=1∞[∑k=1∞(m+k+1)f¯khm+k+1+∑l=1mlgm-l+1hl]zm,∥Tf+g¯*h∥2=|∑k=1∞(k+1)f¯khk+1|2+∑m=1∞1m+1|∑k=1∞(m+k+1)fk¯hm+k+1+∑l=1mlgm-l+1hl|2.
Denote ei(z)=(1/i)zi for i≥1. Since Tu is hyponormal, we have
(13)∥Tuei∥2-∥Tu*ei∥2≥0fori≥1.
For i≥2, ∥Tuei∥2-∥Tu*ei∥2≥0 implies that
(14)1i2(∑k=1i-1|gk|2+∑l=1∞|fl|2)≥∑k=1i-11i-k|fk|2+∑l=1∞1i+l|gl|2≥1i|f1|2.
Hence
(15)(∑i=2N1i2)(∑l=1∞|gl|2+∑l=1∞|fl|2)≥(∑i=2N‍1i)|f1|2forN≥2.
Letting N→∞, since ∑i=2N‍(1/i2) and (∑l=1∞‍|gl|2+∑l=1∞‍|fl|2) are convergent and ∑i=2N‍(1/i) is disconvergent, we get f1=0. Similarly, by choosing i, we get fl=0 for l≥1. Note that ∥Tue1∥2-∥Tu*e1∥2≥0 implies that ∑l=1∞‍|fl|2≥∑l=1∞‍(1/(l+1))|gl|2. Thus gl=0 for l≥1 and the proof is finished.The following corollary generalizes Theorems 1 and 2 in [10]. Denote
(16)Ω={u:u=f+g¯,f,g∈H∞(𝔻),|f|2dAisa𝒟-Carleson measure},
where H∞(𝔻) is the space of the bounded analytic functions on 𝔻.Corollary 3.
Letu∈Ω. Then Tu is hyponormal on 𝒟 if and only if u is a constant function.


## 3. Case on the Harmonic Dirichlet Space

In this section, we will characterize the hyponormality ofTu with u∈W1,∞(𝔻) on 𝒟h.The harmonic Dirichlet space𝒟h consists of all harmonic functions in W1,2(𝔻). It is a closed subspace of W1,2(𝔻), and hence it is a Hilbert space with the following reproducing kernel:
(17)Rz(w)=Kz(w)¯+Kz(w)+1=ln11-zw¯+ln11-z¯w+1.LetQ be the orthogonal projection of W1,2(𝔻) onto 𝒟h. Q is an integral operator represented by
(18)Q(f)(z)=〈f,Rz〉=∫𝔻∂f∂w(∂Kz∂w)¯dA(w)+∫𝔻∂f∂w¯(∂Kz¯∂w¯)¯dA(w)+∫𝔻fdA.
For u∈W1,∞(𝔻), the Toeplitz operator Tu~ with symbol u is defined by
(19)Tu~h=Q(uh)h∈𝒟h.Tu~ is a bounded operator for u∈W1,∞(𝔻) on 𝒟h (see [11]).Theorem 4.
Letu=u0+c+f+g¯∈W1,∞(𝔻) with u0+c∈𝒜, f,g∈H(𝔻), and f(0)=g(0)=0. Then Tu~ is hyponormal on 𝒟h if and only if u∈𝒜.Proof.
By Proposition1, we only need to prove the necessity with u(z)=f(z)+g(z)¯=∑k=1∞‍fkzk+∑k=1∞‍g¯kz¯k.
Leth(z)=a0+∑k=1∞‍akzk+∑k=1∞‍bkz¯k=∑k=0∞‍akzk+∑k=1∞‍bkz¯k∈𝒟h. Since Tu~ is hyponormal on 𝒟h, we have ∥T~f+g¯h∥2-∥T~f+g¯*h∥2≥0. Note that
(20)T~f+g¯h(z)=Q[(f+g¯)h](z)=〈(f+g¯)h,Rz〉=∫𝔻∂[(f+g¯)h]∂wz1-zw¯dA+∫𝔻∂[(f+g¯)h]∂w¯z¯1-z¯wdA+∫𝔻(f+g¯)hdA=∑k=1∞1k+1(bkfk+akg¯k)+[a0f1+∑k=1∞(bkfk+1+ak+1g¯k)]z+∑m=2∞[∑k=1m-1am-kfk+a0fm+∑k=1∞(bkfk+m+ak+mg¯k)]zm+[a0g¯1+∑k=1∞(akg¯k+1+bk+1fk)]z¯+∑m=2∞[∑k=1m-1bm-kg¯k+a0g¯m+∑k=1∞(akg¯k+m+bk+mfk)]z¯m.
Thus(21)∥T~f+g¯h∥2=|∫𝔻T~f+g¯hdA|2+∫𝔻|∂T~f+g¯h∂z|2dA+∫𝔻|∂T~f+g¯h∂z¯|2dA=|∑k=1∞1k+1(bkfk+akg¯k)|2+|a0f1+∑k=1∞(bkfk+1+ak+1g¯k)|2+∑m=2∞|∑k=1m-1am-kfk+a0fm+∑k=1∞(bkfk+m+ak+mg¯k)|2+|a0g¯1+∑k=1∞(akg¯k+1+bk+1fk)|2+∑m=2∞|∑k=1m-1bm-kg¯k+a0g¯m+∑k=1∞(akg¯k+m+bk+mfk)|2.
Similarly, we have(22)T~f+g¯*h(z)=〈T~f+g¯*h,Rz〉=〈h,T~f+g¯Rz〉=〈h,(f+g¯)Rz〉=∑k=1∞k(akf¯k+bkgk)+{a02g1+∑k=1∞[kbkgk+1+(k+1)ak+1fk¯]}z+∑m=2∞1m{∑k=1m-1kakgm-k+a0m+1gm+∑k=1∞[(m+k)ak+mf¯k+kbkgk+m]}zm+{a02f¯1+∑k=1∞[kakf¯k+1+(k+1)bk+1gk]}z¯+∑m=2∞1m{∑k=1m-1kbkf¯m-k+a0m+1f¯m+∑k=1∞‍[(m+k)bk+mgk+kakf¯k+m]}z¯m∥T~f+g¯*h∥2=|∫𝔻T~f+g¯*hdA|2+∫𝔻|∂(T~f+g¯*h)∂zdA|2+∫𝔻|∂(T~f+g¯*h)∂z¯|2dA=|∑k=1∞k(akf¯k+bkgk)|2+|a02g1+∑k=1∞[kbkgk+1+(k+1)ak+1fk¯]|2+∑m=2∞1m|∑k=1m-1kakgm-k+a0m+1gm+∑k=1∞[(m+k)ak+mf¯k+kbkgk+m]|2+|a02f¯1+∑k=1∞[kakf¯k+1+(k+1)bk+1gk]|2+∑m=2∞1m|∑k=1m-1kbkf¯m-k+a0m+1f¯m+∑k=1∞[(m+k)bk+mgk+kakf¯k+m]|2.
For i≥2, let ai=1/i and aj=0 for j≠i. It follows that
(23)1i2(∑k=1∞|fk|2+∑k=1i-1|gk|2+|1i+1gi|2+∑k=i+1∞|gk|2)≥∑k=1∞1i+k|gk|2+|fi|2+∑k=1i-11k|fi-k|2+∑k=1∞1k|fi+k|2.
Therefore,
(24)1i2(∑k=1∞|fk|2+∑k=1∞|gk|2)≥∑k=1∞1i+k(|fk|2+|gk|2).
For every k≥1, we have
(25)(∑i=2N1i2)(∑l=1∞|fl|2+∑l=1∞|gl|2)≥(∑i=2N1i+k)(|fk|2+|gk|2),
where N≥2. Letting N→∞, Since ∑i=2N‍(1/i2) and (∑l=1∞‍|fl|2+∑l=1∞‍|gl|2) are convergent and ∑i=2N‍(1/(i+k)) (k≥1 is fixed) is disconvergent, we get |fk|=|gk|=0 for k≥1. The proof is finished.The following corollary generalizes Theorem 3 in [10].Corollary 5.
Suppose thatu=f+g¯∈W1,∞(𝔻) with f,g∈H(𝔻). Then Tu is hyponormal on 𝒟h if and only if u is a constant function.



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*Source: 186326-2013-11-17.xml*