# Properties of Commutativity of Dual Toeplitz Operators on the Orthogonal Complement of Pluriharmonic Dirichlet Space over the Ball

Authors: Yinyin Hu; Yufeng Lu; Tao Yu
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We completely characterize the pluriharmonic symbols for (semi)commuting dual Toeplitz operators on the orthogonal complement of the pluriharmonic Dirichlet space in Sobolev space of the unit ball. We show that, forf and g pluriharmonic functions, SfSg=SgSf on (Dh)⊥ if and only if f and g satisfy one of the following conditions: (1) both f and g are holomorphic; (2) both f¯ and g¯ are holomorphic; (3) there are constants α and β, both not being zero, such that αf+βg is constant.


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## Body

## 1. Introduction

For any integern>1, let Bn denote the open unit ball in Cn. The boundary of Bn is the sphere Sn and the closure of Bn with the Euclidean metric on Cn is denoted by Bn¯. Let dν denote the Lebesgue volume measure on the unit ball Bn of Cn, normalized so that the measure of Bn equals 1. The Sobolev space W1,2=W1,2(Bn,dν) is the completion of the collection of all smooth functions f on Bn for which(1)f=∫Bnfzdνz2+∑i=1n∫Bn∂f∂zi2+∂f∂zi¯2dνz1/2<∞,where ∂/∂zi, ∂/∂zi¯ is the weak partial derivative. The W1,2(Bn,dν) is a Hilbert space with the inner product(2)f,g=∫Bnfzdνz∫Bngz¯dνz+∑i=1n∂f∂zi,∂g∂zi2+∂f∂zi¯,∂g∂zi¯2,where 〈·,·〉2 denotes the inner product in the Lebesgue space L2(Bn,dν). The Dirichlet space D=D(Bn,dν) is the closed subspace of W1,2(Bn,dν) consisting of all holomorphic functions, and let P denote the orthogonal projection from W1,2(Bn,dν) onto D(Bn,dν). Then P is an integral operator represented by(3)Pfw=f,Kw=∫Bnfdν+∑i=1n∫Bn∂f∂zi∂Kw∂zi¯dν,where Kw(z)=K(z,w) is the reproducing kernel of D. By computation, we know that(4)Kz,w=1+∑α∈Nn-0α+n-1!n!α!αzαw¯α,where {0}=(0,…,0),α=(α1,…,αn)∈Nn,α!=α1!⋯αn!,zα=z1α1⋯znαn,|α|=∑i=1n‍αi and N is the set of nonnegative integers. The pluriharmonic Dirichlet space Dh is the closed subspace of W1,2(Bn,dν) consisting of all pluriharmonic functions. Let Q denote the orthogonal projection from W1,2 onto Dh; then (Qf)(z)=〈f,Rz〉, where Rz=Kz+Kz¯-1. In fact,(5)Qfz=Pfz+Pf¯z¯-Pf0.Let(6)W1,∞Bn=φ∈W1,2Bn,dν:φ,∂φ∂zi,∂φ∂zi¯∈L∞Bn,dν,i=1,2,…,n.Given a function f∈W1,∞(Bn), the multiplication operator Mf, the Toeplitz operator Tf, the Hankel operator Hf, the dual Toeplitz operator Sf, and the dual Hankel operator Rf with symbol f are defined, respectively, by(7)Mf:W1,2⟶W1,2,Mfh=fh,h∈W1,2;Tf:Dh⟶Dh,Tfh=Qfh,h∈Dh;Hf:Dh⟶Dh⊥,Hfh=I-Qfh,h∈Dh;Sf:Dh⊥⟶Dh⊥,Sfh=I-Qfh,h∈Dh⊥;Rf:Dh⊥⟶Dh,Rfh=Qfh,h∈Dh⊥.They are all bounded linear operators. Under the decomposition W1,2=Dh⊕(Dh)⊥, the multiplication operator Mf is represented as(8)TfRfHfSf.This shows close relationships among the above four types of operators. Many studies for dual Toeplitz operators offer some insights into the study for Toeplitz operators. So it is reasonable to focus on the dual Toeplitz operators. Although dual Toeplitz operators differ in many ways from Toeplitz operators, they do have some of the same properties. The general problem that we are interested in is the following: what is the relationship between their symbols when two dual Toeplitz operators commute?For Toeplitz operators, this problem has been studied for a long time. In the case of the classical Hardy space, Brown and Halmos [1] showed that two Toeplitz operators with general bounded symbols commute if and only if either both symbols are analytic, both symbols are conjugate analytic, or a nontrivial linear combination of the symbols is constant.Initiated by Brown and Halmos’s pioneering work, the problem of characterizing when two Toeplitz operators commute has been one of the topics of constant interest in the study of Toeplitz operators on classical function spaces over various domains. On the Bergman space of the unit disk, Axler and Čučković [2] studied commuting Toeplitz operators with harmonic symbols and obtained a similar result to that of Brown and Halmos. Stroethoff [3] later extended that result to essentially commuting Toeplitz operators. Axler et al. [4] showed that if two Toeplitz operators commute and the symbol of one of them is nonconstant analytic, then the other one must be analytic. Čučković and Rao [5] studied Toeplitz operators that commute with Toeplitz operators with monomial symbols. On the Bergman space of several complex variables, by making use of M-harmonic function theory, Zheng [6] characterized commuting Toeplitz operators with pluriharmonic symbols on the Bergman space of the unit ball. Choe and Lee [7–9] studied commuting and essentially commuting Toeplitz operators with pluriharmonic symbols on the unit ball. Lu [10] characterized commuting Toeplitz operators on the bidisk with pluriharmonic symbols. Choe et al. [11] obtained characterizations of (essentially) commuting Toeplitz operators with pluriharmonic symbols on the Bergman space of the polydisk.The fact that the product of two harmonic functions is no longer harmonic adds some mystery to the study of operators on harmonic Bergman space. Many methods which work for the operators on analytic Bergman space lose their effectiveness on harmonic Bergman space. On the harmonic Bergman space of the unit disk, Ohno [12] first characterized the commutativity of Tf and Tz, where f is an analytic function. Choe and Lee [13] studied commuting Toeplitz operator with harmonic symbols and one of the symbols is a polynomial. In [14], Choe and Lee proved that if f,g∈H∞ and supposedly one of them is noncyclic, then TfTg¯=TgTf¯ if and only if either f or g is constant. On the pluriharmonic Bergman space of the unit ball, commuting Toeplitz operators were studied in [15, 16].However, the study on the problem for dual Toeplitz operators started recently. Stroethoff and Zheng [17] characterized the commutativity of dual Toeplitz operators with bounded symbols on the orthogonal complement of the Bergman space of the unit disk and studied algebraic and spectral properties of dual Toeplitz operators. On the Bergman space of the unit ball and the polydisk, commuting dual Toeplitz operators were studied in [18–20]. Yang and Lu [21] gave complete characterization for the (semi)commuting dual Toeplitz operators with harmonic symbols on harmonic Bergman space.In recent years the Dirichlet space has received a lot of attention from mathematicians in the areas of modern analysis, probability, and statistical analysis. Many mathematicians are interested in function theory and operator theory on the Dirichlet space. Yu and Wu [22, 23] investigated commuting dual Toeplitz operators with harmonic symbols on the Dirichlet space. Yu [24] obtained the commutativity of dual Toeplitz operators with general symbols on Dirichlet space.In this paper, we want to characterize commuting dual Toeplitz operators with pluriharmonic symbols on the orthogonal complement of the pluriharmonic Dirichlet space in Sobolev space of the unit ball.We state our main result now. We postpone the proofs of these theorems until Section3.Theorem 1.
Suppose thatf,g∈W1,∞(Bn) are pluriharmonic functions; then Sfg=SfSg if and only if one of the following statements holds:(1)
Bothf and g are holomorphic.(2)
Bothf¯ and g¯ are holomorphic.(3)
Eitherf or g is constant.Theorem 2.
Suppose thatf,g∈W1,∞(Bn) are pluriharmonic functions, then SgSf=SfSg if and only if one of the following statement holds:(1)
Bothf and g are holomorphic.(2)
Bothf¯ and g¯ are holomorphic.(3)
There are constantsα and β, both not being zero, such that αf+βg is constant.Forn=1, two dual Toeplitz operators with harmonic symbols always commute on the orthogonal complement of harmonic Dirichlet space; that is, SfSg=SgSf holds for all harmonic functions f and g.A pluriharmonic function in the unit ball is the sum of a holomorphic function and the conjugate of a holomorphic function. It is clear that all pluriharmonic functions onBn are M-harmonic. A good reference for the function theory of the unit ball is Rudin’s book [25].The difficult part of the proof of Theorem2 is to answer the following question about pluriharmonic functions.Question. If f1,…,fN and g1,…,gN are holomorphic functions in Bn, when is f1g1¯+⋯+fNgN¯ pluriharmonic?This question is very subtle. IfN=2, this question is a special case of Theorem 5.6 in [6]. In [26], Choe et al. gave a necessary and sufficient condition for this question in Lemma 4.7, which is useless to the proof of Theorem 2. In this paper, we give another characterization to the question and induce the proof of Theorem 2.


## 2. Some Lemmas

The following Lemma has been known to be true forn=1 in [24]. For n>1, the following lemma may be known, but we cannot find its proof; for completeness, we give its proof.Lemma 3.
The set of all polynomials inz and z¯ is dense in W1,2(Bn).Proof.
We will discuss it in the case of real variables. Foru∈W1,2(Bn) and zj=xj+yj, since ∂u/∂zj=1/2∂u/∂xj-i∂u/∂yj and ∂u/∂zj¯=1/2∂u/∂xj+i∂u/∂yj, one can see that the norm of u is equivalent to the following norm:(9)ur=1M∫Bnudx1dy1⋯dxndyn2+1M∫Bn∂u∂x12+∂u∂y12+⋯+∂u∂xn2+∂u∂yn2dx1dy1⋯dxndyn1/2,where M=∫Bn‍dx1dy1⋯dxndyn. For any f∈W1,2(Bn) and ɛ>0, by Theorem 3.18 in [27], there exists a smooth function u∈C0∞(R2n) such that f-u<ɛ. Choose a constant R≥1 such that the support set of u is contained in(10)K=x1,y1,…,xn,yn:xj≤R,yj≤Rforj=1,…,n.It follows that the support set of ∂2nu/∂x1∂y1⋯∂xn∂yn is also in K. Let p be a polynomial such that(11)px1,y1,…,xn,yn-∂2nux1,y1,…,xn,yn∂x1∂y1⋯∂xn∂yn<ϵ4nR2nfor all (x1,y1,…,xn,yn) in K, and let(12)p1x1,y1,…,xn,yn=∫-Ry1∫-Rx2⋯∫-Rxn∫-Rynpx1,t2,…,t2ndt2⋯dt2n.Similarly, we also can define pj for j=1,…,2n. It is obtained that(13)p1x1,y1,…,xn,yn-∂u∂x1≤∫-Ry1∫-Rx2⋯∫-Rxn∫-Rynp-∂2nu∂x1∂y1⋯∂xn∂ynx1,t2,…,t2ndt2⋯dt2n≤ɛ2R.Similarly, we have |p2j-1-∂u/∂xj|≤ɛ/2R and |p2j-∂u/∂yj|≤ɛ/2R for any j=1,…,n.
Letq denote the polynomial ∫-Rx1‍p1(t,y1,…,xn,yn)dt. Also we have(14)∫-Rx1p1t,y1,…,xn,yndt=⋯=∫-Rynp2nx1,y1,…,xn,tdt.Similar to the above one can see that(15)qx1,y1,…,xn,yn-ux1,y1,…,xn,yn<ɛfor all (x1,y1,…,xn,yn) in K. Thus we have q-ur<2n+1ɛ. This completes the proof.For two multi-indexesα=(α1,…,αn) and β=(β1,…,βn), the notation α>β means that(16)α≠β,αi≥βi,i=1,…,n.The standard orhonormal basis for Cn consists of the vectors d1,d2,…,dn, where dk is the ordered n-tuple that has 1 in the kth spot and 0 everywhere else. A direct computation gives that(17)Qzαz¯β=α!α-β!n+α-β-1!n+α-1!zα-β,α>β;n!α!n+α!,α=β;β!β-α!n+β-α-1!n+β-1!z¯β-α,α<β;0,else.LetN=span{zαz¯β-Q(zαz¯β):α,β≥0} and we have the following Lemma.Lemma 4.
SetN is dense in Dh⊥.Proof.
Since polynomials are dense inW1,2 by Lemma 3 and I-Q is a bounded operator, we get that N is dense in Dh⊥.The following lemma will be useful for the proof of the main theorem.Lemma 5.
Suppose thatf∈W1,∞(Bn) is holomorphic; then we have Rf(Dh⊥)⊂D, Rf¯(Dh⊥)⊂D¯.Proof.
SinceN is dense in (Dh)⊥, it suffices to prove Rf[zαz¯β-Q(zαz¯β)]∈D for α,β∈Nn-{0}. Since f∈W1,∞ is holomorphic, we have f=∑m≥0‍amzm. For α=β, it follows that(18)Rfzαz¯α-Qzαz¯α=Rfzαz¯α-n!α!n+α!=Q∑m≥0amzm+αz¯α-n!α!n+α!∑m≥0amzm=∑m≥0amm+α!m!n+m-1!n+m+α-1!-n!α!n+α!zmin the Dirichlet space. For α>β, a direct computation gives that(19)Rfzαz¯β-α!α-β!n+α-β-1!n+α-1!zα-β=∑m≥0amα+m!n+α+m-β-1!α+m-β!n+α+m-1!-α!n+α-β-1!α-β!n+α-1!zα+m-β,which is also in D. For α<β, it is obtained that(20)Rfzαz¯β-β!β-α!n+β-α-1!n+β-1!z¯β-α=Q∑m≥0amzα+mz¯β-β!β-α!n+β-α-1!n+β-1!∑m≥0amzmz¯β-α=-aβ-αn!β!n+β!αn+β-α+∑m>β-αcm,β,αamzm+α-β,where cm,β,α = (m+α)!(n+|m|+|α|-|β|-1)!/(m+α-β)!(n+|α|+|m|-1)! − β!/(β-α)!(n+|β|-|α|-1)!/(n+|β|-1)! m!(n+|m|+|α|-|β|-1)!/(m+α-β)!(n+|m|-1)!.
The last case is similar; we omit the proof. Hence we get that iff∈W1,∞ and f is holomorphic, we have Rf((Dh)⊥)⊂D. As the same discussion, we can deduce that Rf¯((Dh)⊥)⊂D¯.In the following proposition, we give an answer to the question that when a dual Toeplitz operator equals zero.Proposition 6.
Suppose thatf∈W1,∞ is a pluriharmonic function. Then Sf=0 if and only if f≡0.Proof.
Assume thatSf=0. Let(21)h1=zd1z¯d1+⋯+zdnz¯dn-nn+1∈Dh⊥.A direct computation gives that (22)Sfh1z=I-Qfh1z=fzz2-nn+1-1n+1fz=fzz2-1=0.Since |z|<1, it follows that f≡0. The converse part is easy to see.Iff,g,h, and k are holomorphic functions in Bn, when is fg¯-hk¯  M-harmonic? In [6], Zheng gives a necessary and sufficient condition for this question. In the following lemma, we give a generalization. For z,w∈Cn, the inner product of z and w is defined by 〈z,w〉Cn=∑j=1n‍zjwj¯.Lemma 7.
Suppose thatf1,…,fN and g1,…,gN are holomorphic functions. Then f1g1¯+⋯+fNgN¯ is pluriharmonic if and only if there is N×N unitary matrix:(23)U=u11¯⋯u1N¯⋮⋱⋮uN1¯⋯uNN¯=u1¯⋮uN¯and some 1≤k≤N+1 such that 〈(f1,…,fN),uj〉CN are constants for 1≤j≤k-1, and 〈(g1,…,gN),uj〉CN are constants for k≤j≤N.Proof.
To prove the sufficient part, suppose thatU is the above unitary matrix U=(u1¯,…,uN¯)⊤ such that for some 1≤k≤N+1, 〈(f1,…,fN),uj〉CN=cj for 1≤j≤k-1 and 〈(g1,…,gN),uj〉CN=cj for k≤j≤N, where ci are constants. Let f=(f1,…,fN) and g=(g1,…,gN). It follows that(24)f1g1¯+⋯+fNgN¯=f,gCN=Uf,UgCN=c1,…,ck-1,f,ukCN,…,f,uNCN,g,u1CN,…,g,uk-1CN,ck,…,cNCN=∑j=1k-1cjg,ujCN¯+∑j=kNcj¯f,ujCN.It is obtained that f1g1¯+⋯+fNgN¯ is pluriharmonic.
Conversely, assume thatf1g1¯+⋯+fNgN¯ is pluriharmonic. There exist two holomorphic functions h1 and h2 on Bn such that (25)f1g1¯+⋯+fNgN¯+h1+h2¯=0.By complexifying (25) (see Lemma 2 in [28]), for all z and w in Bn, we get (26)f1zg1w¯+⋯+fNzgNw¯+h1z+h2w¯=0.It follows that for z and w in Bn, we have(27)f1z,…,fNz,h1z,1⊥g1w,…,gNw,1,h2w.Then there is an orthonormal basis e1,…,eN+2 of CN+2 such that for some 0≤k≤N+1,(28)f1z,…,fNz,h1z,1,eiCN+2=0,1≤i≤k;g1z,…,gNz,1,h2z,ejCN+2=0,k+1≤j≤N+2for all z∈Bn. By Gauss elimination, we eliminate h1 and h2. Then we get the following equations:(29)f1z,…,fNz,ui′CN=ci′,1≤i≤k-1;g1z,…,gNz,uj′CN=cj′,k≤j≤N.Since (ui′,0,ci′) are orthogonal to (uj′,cj′,0) for 1≤i≤k-1, k≤j≤N, it follows that ui′ and uj′ are orthogonal for 1≤i≤k-1 and k≤j≤N. Also the rank of the matrix u1′,c1′⋮uk-1′,ck-1′ is k-1. Since the equations(30)f,ui′CN=ci′,1≤i≤k-1have solutions, it follows that the rank of {u1′,…,uk-1′} equals k-1. After orthonormalization, we get orthonormal bases u1,…,uk-1 such that 〈f,ui〉=ci for 1≤i≤k-1. The case of {uk,…,uN} is similar. Then we get N×N unitary matrix U=u1¯⋮uN¯ satisfying the lemma.


## 3. Proofs of Main Theorems

In this section, we will present the proofs of the main results.Proof of Theorem1.
If(1) holds, we have the fact that Rg(Dh⊥) is contained in H(Bn). It follows that HfRg=0. The desired result follows from the equation Sfg=HfRg+SfSg. Case (2) is similar. Case (3) is easy to get the desired result.
To prove the necessity, suppose thatSfg=SfSg. Then we have HfRg=0. Since f and g are pluriharmonic functions, there exist holomorphic functions f1,f2,g1,g2 such that f=f1+f2¯, g=g1+g2¯. Without loss of generality, we assume that f(0)=g(0)=0. And g1=∑α>0‍aαzα, g2=∑β>0‍bβzβ. Let(31)h1=zd1z¯d1+⋯+zdnz¯dn-nn+1∈Dh⊥.By a direct calculation, we have(32)Qg1h1=Q∑α>0aαzα+d1z¯d1+⋯+zα+dnz¯dn-nn+1zα=∑α>0aαzαα+d1!n+α-1!α!n+α!+⋯+α+dn!n+α-1!α!n+α!-nn+1=1n+1∑α>0aαzα=1n+1g1.Similarly, we have Q(g2¯h1)=1/n+1g2¯. Since HfRgh1=0, it follows(33)1n+1I-Qf1+f2¯g1+g2¯=0.Hence f1g2¯+g1f2¯∈Dh is obtained. By Theorem 5.6 in [6], we have f1g2¯+g1f2¯∈Dh implying that one of the following statements holds:(1)
Bothf and g are holomorphic.(2)
Bothf¯ and g¯ are holomorphic.(3)
Eitherf or g is constant.(4)
There is a nonzero constantt1 such that f1-t1g1 and f2+t1¯g2 are constants.Then it suffices to prove that t1=0 in condition (4) when both g1 and g2 are not constants. For fixed 1≤j≤n, let(34)h2=zdjz¯dj-1n+1,h3=z2djz¯2dj-2n+1n+2∈Dh⊥.In the same way, we get the following:(35)f1∑β>0bββj+1n+βz¯β+f2¯∑α>0aααj+1n+αzα∈Dh,f1∑β>0bββj+1βj+2n+βn+β+1z¯β+f2¯∑α>0aααj+1αj+2n+αn+α+1zα∈Dh.Applying Theorem 5.6 in [6] again, there exist two constants t2, t3 such that(36)f1=t2∑α>0aααj+1n+αzα,f1=t3∑α>0aααj+1αj+2n+αn+α+1zα.Therefore we have(37)t1aα=t2aααj+1n+α=t3aααj+1αj+2n+αn+α+1for all α∈Nn-{0} and 1≤j≤n.
Case  1. If there exist two multi-indexes α≠β such that aα≠0, aβ≠0. Then from (37) we have(38)t1=t2αj+1n+α=t3αj+1αj+2n+αn+α+1,t1=t2βj+1n+β=t3βj+1βj+2n+βn+β+1.If t1≠0, then(39)αj+1n+α=βj+1n+β,αj+1αj+2n+αn+α+1=βj+1βj+2n+βn+β+1,which induces αj=βj for all 1≤j≤n. That is a contradiction.
Case 2. If g1 is a monomial amzm for some m∈Nn-{0} with am≠0. Suppose that α≠β such that(40)m1+α1+1n+m+α-α1+1n+α≠0,m1+β1+1n+m+β-β1+1n+β≠0.Let(41)h4=zα+d1z¯d1-α1+1n+αzα,h5=zβ+d1z¯d1-β1+1n+βzβ;a direct calculation in the same way above gives (42)f1=t4amm1+α1+1n+m+α-α1+1n+αzα+m,f1=t5amm1+β1+1n+m+β-β1+1n+βzm+β.It follows that t4=t5=0. Then the fact that f1-t1g1=-t1g1 is a constant implies that t1=0 which is a contradiction. Hence we get the desired result.
Suppose thatf,g are pluriharmonic functions and f=f1+f2¯, g=g1+g2¯ where f1,f2,g1,g2 are holomorphic functions. We are ready to prove Theorem 2.Proof of Theorem2.
From the equationSfg=HfRg+SfSg, it follows that(43)SfSg-SgSf=HgRf-HfRg.Then SfSg=SgSf if and only if HgRf=HfRg.
Assume thatSfSg=SgSf. Then for any v∈Dh⊥, we have HgRfv=HfRgv. It is obtained that (44)I-Qf1+f2¯Qg1v+g2¯v=I-Qg1+g2¯Qf1v+f2¯v.By Lemma 5, we have the fact that Q(g1v),Q(f1v) are holomorphic and Q(g2¯v)¯,Q(f2¯v)¯ are holomorphic. Then we get(45)I-Qf1Qg1v+f2¯Qg2¯v=0,I-Qg1Qf1v+g2¯Qf2¯v=0.It follows that(46)I-Qf1Qg2¯v+f2¯Qg1v-g1Qf2¯v-g2¯Qf1v=0.If one of f1,g1,f2¯,g2¯ is a constant function, without loss of generality, assume that f1 is a constant function; this follows for any v∈(Dh)⊥; we get(47)I-Qf2¯Qg1v-g1Qf2¯v=0.We have the fact that f2¯Q(g1v)-g1Q(f2¯v) is pluriharmonic for all v∈(Dh)⊥. By Theorem 5.6 in [6], one of the following holds:(1)
Bothg1 and f2¯ are constants.(2)
Bothg1 and Q(g1v) are constants.(3)
BothQ(f2¯v) and f2¯ are constants.(4)
BothQ(f2¯v) and Q(g1v) are constants.(5)
There is a nonzero constantt such that g1-tQ(g1v) and f2¯-tQ(f2¯v) are constants.If g1 is a constant function, we have the fact that both f¯ and g¯ are holomorphic. If f2¯ is a constant function, then f is a constant function. Assume that neither f2¯ nor g1 is constant. Then for all v∈(Dh)⊥, f2¯Q(g1v)-g1Q(f2¯v) is pluriharmonic if and only if one of the following holds:(1)
BothQ(f2¯v) and Q(g1v) are constants.(2)
There is a nonzero constantt such that g1-tQ(g1v) and f2¯-tQ(f2¯v) are constants.Since g1 is holomorphic, g1=∑m≥0‍amzm. And g1 is not a constant; there exists a multi-index β>0 such that aβ≠0. For any multi-index α>β, let vα=zα+d1z¯d1-α1+1/n+|α|zα∈(Dh)⊥. A direct computation gives(48)Qg1vα=zα∑m≥0amm1+α1+1n+m+α-α1+1n+αzm.We choose a α′>β such that β1+α1′+1/n+|β|+|α′|-α1′+1/n+|α′|≠0. Since aβ≠0, it follows that Q(g1vα′) is not a constant. Then we get that there is a nonzero constant t such that g1-tQ(g1vα′) is constant. Since α′>β, from the fact that g1-tQ(g1vα′) is constant, we get aβ=0, which is a contradiction. Hence if f1 is a constant function, we have either both f¯ and g¯ are holomorphic or f is a constant function.
In the following proof, assume that none off1,g1,f2¯,andg2¯ is a constant function. It follows that f1Q(g2¯v)+f2¯Q(g1v)-g1Q(f2¯v)-g2¯Q(f1v)∈Dh. By Lemma 7, we get that there is a 4×4 unitary matrix Uv such that for some 1≤k≤5, 〈(f1,Q(g1v),g1,-Q(f1v)),uj〉C4 are constants for 1≤j≤k-1 and 〈(Q(g2¯v),f2¯,-Q(f2¯v),g2¯),uj〉C4 are constants for k≤j≤4.
Case  1. If there exists v∈(Dh)⊥ such that k=1 or k=5, it follows that f1, g1 are constants or f2¯, g2¯ are constants since U is a unitary matrix. Hence we get that both f and g are holomorphic or both f¯ and g¯ are holomorphic.
Case  2. If there exists v∈(Dh)⊥ such that k=2 or k=4. We just prove the case of k=4; the case of k=2 is similar. Since 〈(f1,Q(g1v),g1,-Q(f1v)),uj〉C4 are constants for 1≤j≤3, it follows that there exist a nonzero constant t1 and a constant c1 such that(49)f1z=t1g1z+c1.Then by (46), we get(50)I-Qt1g1Qg2¯v+f2¯Qg1v-g1Qf2¯v-t1g2¯Qg1v=0,which implies that(51)g1t1Qg2¯v-Qf2¯v+Qg1vf2¯-t1g2¯is pluriharmonic. By Theorem 5.6 in [6], one of the following holds:(1)
Botht1Q(g2¯v)-Q(f2¯v) and Q(g1v) are constants.(2)
Botht1Q(g2¯v)-Q(f2¯v) and f2¯-t1g2¯ are constants.(3)
There is a nonzero constantt2 such that Q(g1v)-t2g1 and [t1Q(g2¯v)-Q(f2¯v)]+t2[f2¯-t1g2¯] are constants.If f2¯-t1g2¯ is a constant, it follows easily that f=t1g+c. Assume that f2¯-t1g2¯ is not a constant. Then for all v∈(Dh)⊥, g1[t1Q(g2¯v)-Q(f2¯v)]+Q(g1v)[f2¯-t1g2¯] is pluriharmonic if and only if one of the following holds:(1)
Botht1Q(g2¯v)-Q(f2¯v) and Q(g1v) are constants.(2)
There is a nonzero constantt2 such that Q(g1v)-t2g1 and [t1Q(g2¯v)-Q(f2¯v)]+t2[f2¯-t1g2¯] are constants.Since g1 is not a constant, similar to the previous proof, we can find vα′∈(Dh)⊥ such that neither Q(g1v) nor Q(g1v)-t2g1 is constant, which is a contradiction. Hence we get that f=t1g+c.
Case  3. For all v∈(Dh)⊥, we have k=3. For each v, there exist constants t1, t2 and c1 such that(52)Qf1v=t1f1+t2g1+c1.Suppose that f1=∑amzm and g1=∑‍bmzm. For multi-index α, let v=zα+d1z¯d1-α1+1/n+|α|zα; there exists a holomorphic function h such that Q(f1v)=zαh. Then for all multi-index m<α, we get t1am+t2bm=0.
Note thatf1 and g1 are not constants. If for every multi-index m, ambm=0. Suppose that am1≠0 and bm2≠0, where m1≠m2. For any multi-index α satisfying α>m1 and α>m2, let vα=zα+d1z¯d1-α1+1/n+|α|zα. Then there exist constants t1,α,t2,α and cα such that(53)Qf1vα=t1,αf1+t2,αg1+cα.From the above computation, for all nonzero m<α, we have t1,αam+t2,αbm=0. It follows that t1,α=t2,α=0. Then Q(f1vα)=cα for all vα with α>m1 and α>m2. A direct computation gives(54)Qf1vα=zα∑m≥0amm1+α1+1n+m+α-α1+1n+αzm=zα∑m≥0amα1n+α-mα1+1n+m+αn+αzm.It follows that, for all multi-index m, we have amα1(n+|α|)-|m|(α1+1)/(n+|m|+|α|)(n+|α|)=0. For each m, we can find a vα such that α1(n+|α|)-|m|(α1+1)≠0. Hence am=0 for all nonzero multi-index m. Then f1 is a constant and this leads a contradiction.
Then there exists a multi-indexm0 such that am0≠0 and bm0≠0. For any α>m0, let vα=zα+d1z¯d1-α1+1/n+|α|zα. Similarly, we still have t1,αam0+t2,αbm0=0. If t1,α=0, we get t2,α=0. Fix a multi-index β>m0, suppose that, for all vα with α>β, we have t1,α=t2,α=0. It follows that Q(f1vα)=cα for all α>β, which implies that f1 is a constant. That is a contradiction. Suppose that there exists a multi-index β1>m0 such that t1,β1≠0. It follows that t2,β1≠0 and, for all m<β1, t1,β1am+t2,β1bm=0. If am=0, it follows that bm=0. If am≠0, we get bm=-t1,β1/t2,β1am. We also can find a multi-index β2 with β2>β1 such that t1,β2≠0. Similarly, for all m<β2, if am=0, we have bm=0. If am≠0, we get bm=-t1,β2/t2,β2am. Clearly, -t1,β1/t2,β1=-t1,β2/t2,β2. For each nonzero multi-index m, we can find a vβ such that bm=tam, where t is a nonzero constant. Then we have f1=tg1+c, and similar to the proof of Case  2, we get that f=tg+c.
By Lemma5, the converse is easy to see. The proof is complete.



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*Source: 1054768-2016-01-20.xml*