A generalization of the Brown-Halmos theorems for the unit ball
aa r X i v : . [ m a t h . F A ] J a n A GENERALIZATION OF THE BROWN–HALMOS THEOREMSFOR THE UNIT BALL
TRIEU LE AND AKAKI TIKARADZE
Abstract.
In this paper we generalize the classical theorems of Brown andHalmos about algebraic properties of Toeplitz operators to the Bergman spaceover the unit ball in several complex variables. A key result, which is ofindependent interest, is the characterization of summable functions u on theunit ball whose Berezin transform B ( u ) can be written as a finite sum P j f j ¯ g j with all f j , g j being holomorphic. In particular, we show that such a functionmust be pluriharmonic if it is sufficiently smooth and bounded. We also settlean open question about M -harmonic functions. Our proofs employ techniquesand results from function and operator theory as well as partial differentialequations. Introduction and main results
In their seminal work [BH64], Brown and Halmos classified all pairs of commutingToeplitz operators on the Hardy space over the unit disc, as well as characterizedall triples of Toeplitz operators ( T f , T g , T h ) such that T f T g = T h . They showedthat the product of two Toeplitz operators is zero if and only if one of them iszero. These theorems are commonly referred to as the Brown–Halmos theorems.Extending these results to the Bergman space setting and to Hilbert spaces ofholomorphic functions on more general domains in several complex variables hasbeen one of the central themes of research in the theory of Toeplitz operators inthe last few decades.On the Bergman space over the unit disc, the first results in the spirit of theBrown–Halmos theorems were obtained by Axler and ˇCuˇckovi´c [A ˇC91] and Ahernand ˇCuˇckovi´c [A ˇC01]. It was shown in these papers that Brown–Halmos theoremshold true on the Bergman space for Toeplitz operators with bounded harmonicsymbols. Subsequently, using his study of the range of the Berezin transform, Ahern[Ahe04] improved the main result in [A ˇC01]. Guo, Sun and Zheng [GSZ07] laterstudied finite rank semi-commutators and commutators of Toeplitz operators withharmonic symbols. It was showed that semi-commutators and commutators havefinite rank if and only if they are actually zero. As a consequence, characterizationsof the symbols were given. ˇCuˇckovi´c [ ˇCuˇc07] obtained criteria for T f T g − T h n tohave finite rank, where f, g and h are bounded harmonic. More general resultsin this direction were investigated in [CKL08]. In a recent paper, Ding, Qin andZheng [DQZ17] provided a more complete answer to the possible rank of T f T g − T h under the assumption that f, g are bounded harmonic and h is a C -functions and(1 − | z | ) ∆ h is integrable. A complete characterization of these functions was thenobtained. Date : January 12, 2021.
Researchers have also investigated Brown–Halmos theorems in the setting of sev-eral complex variables. A classification of pairs of commuting Toeplitz operatorswith pluriharmonic symbols on the unit ball was given by Zheng in [Zhe98]. Sub-sequently, Choe and Koo [CK06] studied the zero product problem for Toeplitzoperators on the unit ball with harmonic symbols having continuous extensions topart of the boundary. Finite sums of products of Toeplitz operators with pluri-harmonic or n -harmonic symbols on the Bergman space over the polydisks wereinvestigated in the papers by Choe et al. [CLNZ07, CKL09]. The same prob-lem on the Hardy space over the unit sphere was considered in [CKL11]. On theother hand, there has not much progress in proving Brown–Halmos type results forToeplitz operators with plurharmonic symbols on the ball. It is our main goal tooffer such results.While it is not the focus of the current paper, we would like to mention thatthere is a vast literature on the study of Toeplitz operators with non-harmonic,non-pluriharmonic symbols. Researchers have investigated algebraic properties ofToeplitz operators whose symbols are radial, quasihomogeneous, or finite sumsof quasihomogeneous functions in one and several variables. See, for example,[ ˇCL08, GKV03, GKV04, LR08, LRZ15, LSZ06] and the references therein.Throughout the paper, N denotes a positive integer. We write B N for the openunit ball in C N . We use H ( B N ) to the denote the algebra of all functions holo-morphic on B N . For p >
0, the Bergman space A p ( B N ) consists of all functions in H ( B N ) that are p -integrable with respect to the normalized Lebesgue volume mea-sure dV . The reader is referred to [Zhu05, Chapter 2] for an excellent introductionto these Bergman spaces. We use L p ( B N ) to denote the usual L p -space with respectto dV . Clearly, A p ( B N ) = H ( B N ) ∩ L p ( B N ). Given f ∈ L ∞ ( B N ), one defines thecorresponding Toeplitz operator T f : A ( B N ) −→ A ( B N ) by T f ( h ) = P ( f h ) forall h ∈ A ( B N ). Here P is the orthogonal projection from L ( B N ) onto A ( B N ). Itis immediate that the operator T f is bounded and k T f k ≤ k f k ∞ . It is well knownthat T f can be expressed as an integral operator. In fact, we have T f ( h )( z ) = Z B N f ( w ) h ( w )(1 − h z, w i ) N +1 dV ( w ) , z ∈ B N . For f ∈ L ( B N ), the above integral is also well defined for all bounded holomorphicfunctions h on the ball. As a consequence, one may define Toeplitz operators for L -symbols. It is well known that if f ∈ L ( B N ) is bounded on a set { z : r < | z | < } for some 0 < r <
1, then T f extends to a bounded operator on A ( B N ). See thediscussion on [A ˇC01, p. 204] for the one dimensional case. The general setting ofseveral variables is similar.We recall here some basic properties of Toeplitz operators. For bounded func-tions φ, ψ and complex numbers a and b , we have T aφ + bψ = aT φ + bT ψ , T ∗ φ = T ¯ φ . It is also well known that if ψ or ¯ φ is holomorphic, then T φ T ψ = T φψ . However,this property fails for general symbols, which is one of the reasons why the studyof Toeplitz operators has attracted a great deal of attention.The following theorem is the main result of the paper. It is a vast generalizationof the aforementioned results and in a sense represents the best possible result onecan hope for in the spirit of Brown–Halmos theorems for Toeplitz operators withpluriharmonic symbols. Recall that for two functions x, y ∈ A ( B N ), we use x ⊗ y ROWN–HALMOS TYPE THEOREMS ON THE BALL 3 to denote the operator ( x ⊗ y )( h ) = h h, y i x, h ∈ A ( B N ) . Theorem 1.1.
Let φ j , ψ j be bounded pluriharmonic functions for ≤ j ≤ n and h be a C N +2 bounded function on B N . Let x ℓ , y ℓ ∈ A ( B N ) for ≤ ℓ ≤ r . Write φ j = f j + ¯ g j , ψ j = u j + ¯ v j where f j , g j , u j and v j are holomorphic. Then n X j =1 T φ j T ψ j = T h + r X ℓ =1 x ℓ ⊗ y ℓ if and only if h − P nj =1 ¯ g j u j is pluriharmonic and n X j =1 φ j ψ j = h + (1 − | z | ) N +1 r X ℓ =1 x ℓ ¯ y ℓ . As an immediate corollary, we have the following direct generalization of theBrown–Halmos theorems, which in particular settles the zero product problem forToeplitz operators with pluriharmonic functions. The zero product problem forgeneral symbols is a long standing open problem in the area of Toeplitz operators,which has resisted researchers’ attempts even for the unit disc. Our result here inthe single variable setting reduces to [GSZ07, Theorem 7].
Corollary 1.2.
Let φ, ψ be bounded pluriharmonic functions on B N .(a) If T φ T ψ = T h for some h ∈ C N +2 ( B N ) ∩ L ∞ ( B N ) , then ¯ φ or ψ is holomor-phic and φψ = h .(b) If T φ T ψ has a finite rank, then φ or ψ must be zero. Another direct consequence of our main result is a strengthening of the aforemen-tioned Zheng’s theorem about commuting Toeplitz operators with pluriharmonicsymbols. In the case of a single variable, we recover [GSZ07, Theorem 6].
Corollary 1.3.
Let φ, ψ be bounded pluriharmonic functions on B N . The commu-tator [ T φ , T ψ ] has a finite rank if and only if both φ, ψ are holomorphic, or both areanti-holomorphic, or there are constants c , c , not both zero, such that c φ + c ψ is constant on B N . The main tool for showing our results is the Berezin transform. Recall that theBergman space A ( B N ) is a reproducing kernel Hilbert space with kernel K z ( w ) = K ( w, z ) = 1(1 − h w, z i ) N +1 , z, w ∈ B N . Given a function u ∈ L ( B N ), one defines the Berezin transform of u as follows B ( u )( z ) = h uk z , k z i L ( B N ) = (1 − | z | ) N +1 Z B N u ( ξ ) | − h z, ξ i| N +1) dV ( ξ ) , where k z ( w ) = K ( w, z ) p K ( z, z )is the normalized reproducing kernel. More generally, given a bounded operator S : A ( B N ) → A ( B N ), one defines similarly its Berezin transform B ( S )( z ) = h S ( k z ) , k z i L ( B N ) . TRIEU LE AND AKAKI TIKARADZE
It is well known that the Berezin transform is an injective map. That is, if B ( S )( z ) = B ( S )( z ) for all z ∈ B N , then S = S . The Berezin transform playsan important role in the theory of Toeplitz operators. In fact, it has been used asthe main tool in the study of Brown–Halmos theorems for Toeplitz operators withpluriharmonic symbols in most of the references we have mentioned so far.It is clear that for u ∈ L ( B N ), the Berezin transform B ( u ) is real analytic on B N . As a result, we may expand B ( u ) as a series B ( u )( z ) = X α,β c α,β z α ¯ z β . We say that B ( u ) has a finite rank if the infinite matrix of coefficients [ c α,β ] α,β has a finite rank. This happens if and only if there exist holomorphic functions f , . . . , f n and g , . . . , g n such that B ( u ) = n X j =1 f j ¯ g j . Besides being interesting on its own right, the following natural question is im-portant in regards to algebraic properties of Toeplitz operators with pluriharmonicsymbols.
Question.
For which u ∈ L ( B N ) does B ( u ) have a finite rank?For the unit disc on the complex plane, N. V. Rao [Rao18] provided a full reso-lution of the above question. Rao’s result asserts that for an L -function u on theunit disc, B ( u ) has finite rank if and only if u is harmonic except at a finite set ofpoints. In particular, if u is also assumed to be locally bounded, then it must beharmonic. In higher dimensions, the situation turns out to be more complicatedand high dimensional phenomena do occur. In the theorem below, we completelydescribe B ( u ) whenever it is of finite rank. The proof of Theorem 1.1 relies heav-ily on this result. In addition, we answer an open question about M -harmonicfunctions raised in [CKL11]. We recall here that M -harmonic functions are thoseannihilated by the invariant Laplacian (see Section 2). It is well known that suchfunctions are fixed points of the Berezin transform. Theorem 1.4.
Suppose u ∈ L ( B N ) such that B ( u ) has a finite rank. Then thereexists a finite set Λ ⊂ B N , a collection { P w : w ∈ Λ } of polynomials in z and ¯ z oftotal degree at most N + 1 , and a pluriharmonic function h such that for z ∈ B N , B ( u )( z ) = h ( z ) + X w ∈ Λ P w (cid:16) z − h z, w i (cid:17) . Furthermore,(a) If u also belongs to C N +2 ( B N ) , then Λ ⊂ ∂ B N , the unit sphere.(b) If u belongs to L N +2 ( B N ) , then Λ ⊂ B N .(c) If B ( u ) = f ¯ g + · · · + f d ¯ g d , where f ℓ ∈ A N +2 ( B N ) and g ℓ ∈ H ( B N ) for all ℓ ,then Λ ⊂ B N .As a consequence, if both (a) and (b), or both (a) and (c) hold, then u is plurihar-monic. ROWN–HALMOS TYPE THEOREMS ON THE BALL 5
Remark 1.5.
As shown by Ahern and Rudin in [AR91] and can be verified directly,for N ≥
3, the function u ( z ) = z ¯ z | − z | , which belongs to L ( B N ), is M -harmonic. For such a function, we have B ( u ) = u ,so B ( u ) has a finite rank but u is not pluriharmonic. In the case N = 2, it can alsobe verified that u ( z ) = z ¯ z (1 − z )(1 − ¯ z ) −
12 ¯ z z (1 − z )(1 − ¯ z ) is an M -harmonic L -function which is not pluriharmonic. It should be notedthat Ahern and Rudin already showed that in the case of two complex variables, if u = f ¯ g (and f, g are holomorphic) is M -harmonic, then u is actually pluriharmonic.As a result, some type of regularity near the boundary is required to conclude that u is pluriharmonic, as in Theorem 1.1. We would like to alert the reader that theexistence of a smooth integrable function u such that B ( u ) has a finite rank and u is not pluriharmonic is a high dimensional phenomenon. Indeed, it follows fromthe aforementioned result of Rao that if u ∈ L ( B ) is locally bounded (without anyother assumption on regularity) so that B ( u ) has a finite rank, then u is harmonic.Our proof is influenced by Rao’s idea to reformulate the finite rank property ofthe Berezin transform of u in terms of a certain distribution associated to u havinga finite rank, which allows the usage of a result due to Alexandrov and Rozenblum[AR09]. In extending this approach to the case of the unit ball in C N , significantcomplications do arise. We overcome these difficulties by establishing various iden-tities for differential operators related to the invariant Laplacian and making useof a regularity result on integrable solutions of partial differential equations (seeSection 2).Brown and Halmos proved that the zero operator is the only compact Toeplitzoperator on the Hardy space over the unit disc. On Bergman spaces, there are manynontrivial compact Toeplitz operators. Indeed, whenever f is a bounded functionwith a compact support contained in the unit ball, the operator T f is compact. Onthe other hand, the problem of determining nonzero finite rank Toeplitz operatorswas open for quite some time. In [Lue08], Luecking settled this question in thenegative by showing that whenever ν is a compactly supported finite measure on C for which the matrix of moments [ R C z ℓ ¯ z k dν ( z )] ℓ,k has finite rank, then ν is alinear combination of finitely many point masses. Luecking’s theorem has beengeneralized to several complex variables [Cho09, RS10] as well as to distributionalsymbols. We end this section by recalling the following result, which is crucial toour approach. Theorem 1.6 (Alexandrov-Rosenblum) . Let F be a compactly supported distribu-tion on C N . If the matrix (cid:2) F ( z ℓ ¯ z k ) (cid:3) ℓ,k has a finite rank, then the support of F consists of finitely many points. Some results on invariant Laplacian and radial derivative
In this section we establish some results associated with certain differential op-erators on the unit ball. Besides playing a crucial role in our study of the Berezintransform, these identities are also interesting in their own right.
TRIEU LE AND AKAKI TIKARADZE
Ahern and ˇCuˇckovi´c [A ˇC01] and subsequently Ahern [Ahe04], Rao [Rao18] cru-cially used the following property of the kernel function of the Berezin transform(referred as a “marvelous identity” by Ahern)∆ z (cid:16) (1 − | z | ) | − z ¯ ξ | (cid:17) = ∆ ξ (cid:16) (1 − | ξ | ) | − z ¯ ξ | (cid:17) . The setting of several variables gets more complicated. We offer here several al-ternative identities which are important in our proofs. We shall make use of thefollowing notation: E z = N X j =1 z j ∂∂z j , ¯ E z = N X j =1 ¯ z j ∂∂ ¯ z j , ∆ z = N X j =1 ∂ ∂z j ∂ ¯ z j . For any real number s , we write | E z + s | = ( E z + s )( ¯ E z + s ). We use E ξ , ¯ E ξ , and∆ ξ to denote the corresponding operators acting on the variable ξ . Lemma 2.1.
For any integer m ≥ and z, ξ ∈ B N , we have (cid:0) | E ξ + m | − ∆ ξ (cid:1) · · · ( | E ξ | − ∆ ξ ) n | − h z, ξ i| o = | E z | n ( m !) (1 − | z | ) m +1 | − h z, ξ i| m +1) o , (2.1) and (cid:0) | E ξ + m | − ∆ ξ (cid:1) · · · ( | E ξ | − ∆ ξ ) n − | ξ | | − h z, ξ i| o = (cid:0) | E z | − ∆ z (cid:1)n ( m !) (1 − | z | ) m +1 | − h z, ξ i| m +1) o . (2.2) Proof.
A direct calculation shows that( | E ξ | − ∆ ξ ) n | − h z, ξ i| o = |h z, ξ i| − | z | | − h z, ξ i| = | E z | n − | z | | − h z, ξ i| o , (2.3) (cid:16) | E ξ | − ∆ ξ (cid:17)n − | ξ | | − h z, ξ i| o = ( N − | − h z, ξ i| + (1 − | z | )(1 − | ξ | ) | − h z, ξ i| = (cid:16) | E z | − ∆ z (cid:17)n − | z | | − h z, ξ i| o , (2.4)and for any real number s , (cid:16) | E ξ + s | − ∆ ξ (cid:17)n | − h z, ξ i| s o = s (1 − | z | ) | − h z, ξ i| s +1) . (2.5)Applying (cid:0) | E ξ + m | − ∆ ξ (cid:1) · · · ( | E ξ + 1 | − ∆ ξ ) to (2.3) and using (2.5) repeatedlyfor s = 1 , . . . , m give (2.1). Finally, applying the same operator to (2.4) and using(2.5) give (2.2). (cid:3) We will use e ∆, as usual, to denote the invariant Laplacian on C ( B N ) whichsatisfies e ∆ = (1 − | z | )(∆ z − | E z | ) . Recall that e ∆ can also be defined using the ordinary Laplacian and automorphismsof the unit ball. For more information on e ∆ and its properties, see [Rud80, Chap-ter 4]. However, the reader should be aware that the Laplacian defined there isactually four times our Laplacian. ROWN–HALMOS TYPE THEOREMS ON THE BALL 7
Lemma 2.2.
For any integer m ≥ and z, ξ ∈ B N , we have (1 − | ξ | ) − m − p m ( e ∆ ξ ) (cid:26) − | ξ | | − h z, ξ i| (cid:27) = ( | E z | − ∆ z ) (cid:26) (1 − | z | ) m +1 | − h z, ξ i| m +1) (cid:27) , (2.6) where p m ( t ) = 1( m !) m Y j =0 (cid:0) j ( j − N ) − t (cid:1) . As a consequence, ( m !) (1 − | ξ | ) − m − p m ( e ∆ ξ ) = (cid:0) | E ξ + m | − ∆ ξ (cid:1) · · · ( | E ξ | − ∆ ξ ) . (2.7) Proof.
Put h ( ξ ) = 1 − | ξ | . A direct but tedious calculation shows that for anypositive integer j ≥ j h j +1 = (cid:0) j ( j − N ) − e ∆ (cid:1) ( h j ) , which implies h m +1 = 1( m !) m Y j =1 (cid:16) j ( j − N ) − e ∆ (cid:17) ( h ) . Therefore, − e ∆ h m +1 = 1( m !) ( − e ∆) m Y j =1 (cid:16) j ( j − N ) − e ∆ (cid:17) ( h ) = p m ( e ∆) h. Since both sides are radial functions that depend only on the modulus of the vari-able, for any z, ξ ∈ B N , we have − ( e ∆ h m +1 ) ◦ ϕ ξ ( z ) = ( p m ( e ∆) h ) ◦ ϕ z ( ξ ) . On the other hand, the invariance of e ∆ under the automorphisms of B N gives( e ∆ h m +1 ) ◦ ϕ ξ ( z ) = e ∆ z ( h m +1 ◦ ϕ ξ ( z )) = e ∆ z (cid:0) (1 − | ϕ ξ ( z ) | ) m +1 (cid:1) , and ( p m ( e ∆) h ) ◦ ϕ z ( ξ ) = p m ( e ∆ ξ )( h ◦ ϕ z ( ξ )) = p m ( e ∆ ξ ) (cid:0) − | ϕ z ( ξ ) | (cid:1) . Consequently, p m ( e ∆ ξ ) (cid:16) − | ϕ z ( ξ ) | (cid:17) = − e ∆ z (cid:16) (1 − | ϕ ξ ( z ) | ) m +1 (cid:17) . (2.8)Since 1 − | ϕ z ( ξ ) | = 1 − | ϕ ξ ( z ) | = (1 − | z | )(1 − | ξ | ) | − h z, ξ i| and − e ∆ z = (1 − | z | )( | E z | − ∆ z ), the identity (2.6) now follows from (2.8).From Lemma 2.1 and (2.6), we conclude that the differential operators on bothsides of (2.7) agree on functions of the form −| ξ | | −h z,ξ i| for all z ∈ B N . Takingpartial derivatives in z, ¯ z and setting z = 0, we see that the two operators agreeon all polynomials of the form (1 − | ξ | ) q ( ξ, ¯ ξ ), where q is a polynomial. Since anysmooth functions on B N can be approximated by such polynomials (in the topologyof uniform convergence on compact sets of all derivatives up to order 2 m + 2), weobtain the required identity. (cid:3) TRIEU LE AND AKAKI TIKARADZE
We end the section with a result about singularities of L -solutions to PDEs.While we think this result may be known in the literature, due to the lack of anappropriate reference, we provide here a proof. Let us first recall some notation.For any distribution F and any function φ in the domain of F , we shall use hF , φ i to denote F ( φ ), the action of F on φ . For any differential operator L with smoothcoefficients, we use L ∗ to denote its formal adjoint which satisfies h L ∗ ( F ) , φ i = hF , L ( φ ) i for any distribution F and any function φ in the domain of F . Proposition 2.3.
Let L be a differential operator of order µ with smooth coeffi-cients on R n . Suppose u ∈ L ( R n ) having a compact support such that the distri-bution L ( u ) is supported at finitely many points. Then the order of L ( u ) is at most µ − . Remark 2.4.
For a general function u , the distribution L ( u ) may have order µ .The point here is that if the support of L ( u ) has only finitely many elements, thenits order must be strictly smaller than µ . Proof.
Let D j = − i ∂ j and D = ( D , . . . , D n ). Write the adjoint operator L ∗ inthe form L ∗ = X | α |≤ µ c α ( x ) D α , (2.9)where each c α is smooth. Since u has a compact support, the Fourier transform of L ( u ) can be computed by F{ L ( u ) } ( ζ ) = Z R n u ( x ) L ∗ x ( e − i h ζ,x i ) dx = X | α |≤ µ ζ α Z R n u ( x ) c α ( x ) e − i h ζ,x i dx = X | α |≤ µ ζ α F{ uc α } ( ζ ) . Because c α is locally bounded, the function uc α belongs to L ( R n ). As a conse-quence, F{ uc α } ( ζ ) → | ζ | → ∞ . It follows that for any ζ ∈ R n \{ } ,lim t →∞ F{ L ( u ) } ( tζ ) t µ = 0 . (2.10)Now let { a , . . . , a s } be the support of L ( u ). Then there are differential operators L , . . . , L s with constant coefficients such that L ( u ) = s X j =1 L j ( δ a j ) , where δ a denotes the Dirac distribution at a . Since L ( u ) has order at most µ , each L j has order at most µ as well. As a result, there are homogeneous polynomials p j ROWN–HALMOS TYPE THEOREMS ON THE BALL 9 of degree µ such that L j = p j ( D ) + lower order derivatives. We then have F{ L ( u ) } ( ζ ) = s X j =1 F{ L j ( δ a j ) } ( ζ )= s X j =1 e − i h ζ,a j i · (cid:0) p j ( ζ ) + lower order terms in ζ (cid:1) . It now follows from (2.10) thatlim t →∞ s X j =1 e − it h ζ,a j i p j ( ζ ) = 0 (2.11)for all ζ ∈ R n \{ } .Claim: for all ζ ∈ R n such that h ζ, a j i 6 = h ζ, a k i for all j = k , (2.11) forces p j ( ζ ) = 0.Since the set of all ζ in the claim is dense in R n , we conclude that p j = 0 andhence L j is of order at most µ − j . Consequently, L ( u ) has order at most µ − Proof of the claim.
We believe that the claim should be well known but we sketchhere a proof. To simplify the notation, put λ j = −h ζ, a j i and b j = p j ( ζ ). Note thatthe values λ , . . . , λ s are pairwise distinct so there exists a real number c such that e iλ c , . . . , e iλ s c are pairwise distinct. Define f ( t ) = P sj =1 b j e iλ j t for t ∈ R . Then(2.11) gives lim t →∞ f ( t ) = 0 and hence, lim t →∞ f ( t + ℓc ) = 0 for all 0 ≤ ℓ ≤ s .Note that f ( t + ℓc ) = s X j =1 ( e iλ j c ) ℓ b j e iλ j t so each b j e iλ j t can be expressed as a linear combination of f ( t ) , f ( t + c ) , . . . , f ( t +( s − c ) via Vandermonde determinant. It then follows that for each 1 ≤ j ≤ s ,we have lim t →∞ b j e iλ j t = 0, which implies b j = 0. (cid:3) Finite rank Berezin transform
The goal of this section is to study finite rank Berezin transform B ( u ), whichcan be written in the form B ( u ) = P nj =1 f j ¯ g j for holomorphic functions f j and g j .To simplify the notation, we define the following differential operator which playsan important role in our proof D = (1 − | ξ | ) − ( N +1) N Y j =0 (cid:16) j ( j − N ) − e ∆ (cid:17) . Despite the rational factor (1 − | ξ | ) − ( N +1) , the operator D is in fact a differentialoperator with polynomial coefficients. Indeed, by Lemma 2.2, we have D = (1 − | · | ) − ( N +1) p N ( e ∆) = ( | E + N | − ∆) · · · ( | E | − ∆) . Since the adjoint operator of E + s is − E + s − N , the differential operator D isself-adjoint in the sense that for any distribution φ on C N with a compact supportand any ψ ∈ C ∞ ( C N ), hD ( φ ) , ψ i = h φ, D ( ψ ) i . For a distribution L on C N , we say that L is finitely supported (or L has a finitesupport ) if its support is a finite set. It is well known that there then exist finitely many points q , . . . , q m ∈ C N and differential operators with constant coefficients L j , such that L = P mj =1 L j ( δ q j ), where δ q denotes the point mass distribution at q ∈ C N .For a function u ∈ L ( B N ), we use uχ B N to denote the corresponding tempereddistribution on C N defined as φ Z B N u ( ξ ) φ ( ξ ) dV ( ξ ) , φ ∈ C ∞ ( C N ) . The proof of Theorem 1.4 is divided into several steps. We are now ready toprove the first part and statement (a) in the theorem.
Proposition 3.1.
Let u ∈ L ( B N ) such that B ( u ) has a finite rank. Then thereexists a finite set Λ ⊂ B N , a collection { P w : w ∈ Λ } of polynomials in z and ¯ z oftotal degree at most N + 1 , and a pluriharmonic function h such that for z ∈ B N , B ( u )( z ) = h ( z ) + X w ∈ Λ P w (cid:16) z − h z, w i (cid:17) . If, furthermore, u also belongs to C N +2 ( B N ) , then Λ ⊂ ∂ B N , the unit sphere.Proof. Recall the formula for the Berezin transform B ( u )( z ) = Z B N u ( ξ ) (1 − | z | ) N +1 | − h z, ξ i| N +1) dV ( ξ ) . Applying ( N !) | E z | to both sides and using Lemma 2.1, we conclude that( N !) | E z | ( B ( u )( z )) = Z B N u ( ξ ) D ξ n | − h z, ξ i| o dV ( ξ ) (3.1)= Z B N u ( ξ ) X k,l (cid:18) | k | k (cid:19)(cid:18) | l | l (cid:19) z k ¯ z l D ξ ( ¯ ξ k ξ l ) dV ( ξ )= X k,l n(cid:18) | k | k (cid:19)(cid:18) | l | l (cid:19) Z B N u ( ξ ) D ξ ( ¯ ξ k ξ l ) dV ( ξ ) o z k ¯ z l . (3.2)Since B ( u ) is real analytic, we may write B ( u )( z ) = X k,l a k,l z k ¯ z l , which implies | E z | ( B ( u )( z )) = X k,l | k || l | a k,l z k ¯ z l . It follows that B ( u ) has a finite rank if and only if | E z | ( B ( u )) has a finite rank.Using (3.2), we conclude that B ( u ) has finite rank ⇐⇒ h(cid:18) | k | k (cid:19)(cid:18) | l | l (cid:19) Z B N u ( ξ ) D ξ ( ¯ ξ k ξ l ) dV ( ξ ) i k,l has finite rank ⇐⇒ h Z B N u ( ξ ) D ξ ( ¯ ξ k ξ l ) dV ( ξ ) i k,l has finite rank . Applying Theorem 1.6 with the distribution F = D ( uχ B N ) given as F ( φ ) = hD ( uχ B N ) , φ i = Z B N u ( ξ ) D ( φ ) dV ( ξ ) ROWN–HALMOS TYPE THEOREMS ON THE BALL 11 for φ ∈ C ∞ ( C N ), we see that B ( u ) has a finite rank if and only if D ( uχ B N ) is afinitely supported distribution. In particular, D ( u ) = 0 for all except finitely manypoints on B N . Since the differential operator | E |− ∆ is elliptic on B N , the operator D is also elliptic. It follows that u is real analytic except at finitely many points on B N .Let { w , . . . , w s } ⊂ B N be the support of D ( uχ B N ). By Proposition 2.3, D ( uχ B N )has order at most 2 N + 1 since D is a differential operator of order 2 N + 2. Fromformula (3.1), we see that there are complex constants a j,α,β for 1 ≤ j ≤ s and | α | + | β | ≤ N + 1 such that | E z | ( B ( u )( z )) = D ( uχ B N ) n | − h z, ·i| o = X ≤ j ≤ s | α | + | β |≤ N +1 a j,α,β z α ¯ z β (1 − h z, w j i ) | α | (1 − h w j , z i ) | β | . (3.3)A direct calculation shows that for any w ∈ B N and | α | , | β | ≥ − h z, w i − h z, w i − h z, w i = E z n log 11 − h z, w i o ,z α (1 − h z, w i ) | α | = E z n | α | z α (1 − h z, w i ) | α | o . Thus, for such α, β , the functions z α ¯ z β (1 − h z, w i ) | α | (1 − h w, z i ) | β | , (cid:16) − h z, w i − (cid:17) ¯ z β (1 − h w, z i ) | β | , z α (1 − h z, w i ) | α | (cid:16) − h w, z i − (cid:17) belong to the range of | E z | . Hence, the identity (3.3) implies that the plurihar-monic function X ≤ j ≤ s | α |≥ a j,α, z α (1 − h z, w j i ) | α | + X ≤ j ≤ s | β |≥ a j, ,β ¯ z β (1 − h w j , z i ) | β | + X ≤ j ≤ s a j, , (cid:16) − h z, w j i + 11 − h w j , z i − (cid:17) is the image, under | E z | , of a real analytic function. Using power series, we seethat zero is the only pluriharmonic function belonging to the range of | E z | . It thenfollows that for all j , we have a j,α,β = 0 whenever | α | = 0 or | β | = 0. Consequently, | E z | ( B ( u )) = X ≤ j ≤ s | α |≥ , | β |≥ a j,α,β z α ¯ z β (1 − h z, w j i ) | α | (1 − h w j , z i ) | β | = | E z | n X ≤ j ≤ s | α |≥ , | β |≥ a j,α,β | α | | β | z α ¯ z β (1 − h z, w j i ) | α | (1 − h w j , z i ) | β | o , which gives B ( u )( z ) = h ( z ) + X ≤ j ≤ s | α |≥ , | β |≥ a j,α,β | α | | β | z α ¯ z β (1 − h z, w j i ) | α | (1 − h w j , z i ) | β | , for some pluriharmonic function h on B N . Defining P j ( z ) = X | α |≥ , | β |≥ | α | + | β |≤ N +1 a j,α,β | α | | β | z α ¯ z β , we obtain the required representation for B ( u ).If u is 2 N + 2 times differentiable on B N , then the support of D ( uχ B N ), being afinite set of points, must be contained on the unit sphere. As a result, | w j | = 1 forall 1 ≤ j ≤ s . (cid:3) From the proof of Proposition 3.1 we have the following result which might beof independent interest.
Corollary 3.2.
Let u ∈ L ( B N ) be of the form u = P nj =1 f j ¯ g j with holomorphic f j , g j . If u is an eigenfunction of the invariant Laplacian with eigenvalue λ , then λ = j ( j − N ) for some j ∈ { , , . . . , N } . Proof.
It is well known [Rud80, Theorem 4.2.4] that eigenfunctions of e ∆ are alsoeigenfunctions of B . Therefore, B ( u ) has a finite rank. From the proof of Proposi-tion 3.1 as above, we have that D ( u ) = 0 on B N . The desired result is immediatefrom the definition of D , which we recall here as D = (1 − | ξ | ) − ( N +1) N Y j =0 (cid:16) j ( j − N ) − e ∆ (cid:17) . (cid:3) Applying Proposition 3.1 to the case where u belongs to L N +2 ( B N ), we provestatement (b) in Theorem 1.4. Proposition 3.3.
Suppose u ∈ L N +2 ( B N ) and B ( u ) has a finite rank. Then thereexist finitely many points w , . . . , w s ∈ B N , polynomials Q , . . . , Q s in C [ z, ¯ z ] withtotal degrees at most N + 1 , and a pluriharmonic function h such that B ( u )( z ) = h ( z ) + s X j =1 Q j ◦ ϕ w j ( z ) . Proof.
We know that there exist holomorphic functions h , h on B N and finitelymany points w , . . . , w s ∈ B N such that B ( u )( z ) = h ( z ) + ¯ h ( z ) + X ≤ j ≤ s ≤| β |≤ N Q j,β (cid:16) z (1 − h z, w j i ) (cid:17) · ¯ z β (1 − h w j , z i ) | β | , (3.4)where each Q j,β is a holomorphic polynomial of degree at most 2 N + 1 − | β | with Q j,β (0) = 0. We prove first that Q j,β = 0 whenever | w j | = 1.Complexifying (3.4) gives B ( u )( z, ζ ) − ¯ h ( ζ ) = h ( z ) + X ≤ j ≤ s ≤| β |≤ N Q j,β (cid:16) z (1 − h z, w j i ) (cid:17) · ¯ ζ β (1 − h w j , ζ i ) | β | ROWN–HALMOS TYPE THEOREMS ON THE BALL 13 for all z, ζ ∈ B N , where we define B ( u )( z, ζ ) = (1 − h z, ζ i ) N +1 Z B N u ( ξ )(1 − h z, ξ i ) N +1 (1 − h ξ, ζ i ) N +1 dV ( ξ )Since the set { } ∪ n ¯ ζ β (1 − h w j , ζ i ) | β | : 1 ≤ | β | ≤ N, ≤ j ≤ s o is linearly independent, it follows that each Q j,β ( z −h z,w j i ) can be written as a linearcombination of finitely many functions in the set n B ( u )( · , ζ ) − ¯ h ( ζ ) : ζ ∈ B N o . Note that for each ζ ∈ B N , the function B ( u )( z, ζ ) is the product of (1 − h z, ζ i ) N +1 with the Bergman projection of u ( ξ )(1 − h ξ, ζ i ) − N − , which belongs to L N +2 ( B N )by the assumption about u . It is well known that the Bergman projection maps L p ( B N ) into itself for 1 < p < ∞ . Therefore, the function B ( u )( · , ζ ) − ¯ h ( ζ ) belongsto L N +2 ( B N ). This implies that each Q j,β ( z −h z,w j i ) belongs to L N +2 ( B N ). ByLemma 3.5, for any j with w j on the unit sphere, Q j,β must be constant, hence,identically zero since Q j,β vanishes at the origin. As a result, we may assume that | w j | < ≤ j ≤ s .To complete the proof, we show that for ω ∈ B N and 1 ≤ j ≤ N , the rationalfunction z j −h z,ω i is a linear combination of 1 and the components of ϕ w ( z ). Therequired representation then follows from (3.4).For z, ζ ∈ B N , [Rud80, Theorem 2.2.2] provides the identity1 − h ϕ ω ( z ) , ϕ ω ( ζ ) i = (1 − | ω | )(1 − h z, ζ i )(1 − h z, ω i )(1 − h ω, ζ i ) , which is equivalent to1 − h z, ζ i − h z, ω i = 1 − h ω, ζ i − | ω | (cid:0) − h ϕ ω ( z ) , ϕ ω ( ζ ) i (cid:1) . Setting ζ = 0 then ζ = e j and subtracting the two quantities, we have z j − h z, ω i = ω j − | ω | + 11 − | ω | h ϕ ω ( z ) , − ω + (1 − ω j ) ϕ ω ( e j ) i . Note that the right hand-side is an affine function in ϕ ω ( z ). As a consequence, forany multi-indexes α and β , the rational function z α ¯ z β (1 − h z, ω i ) | α | (1 − h ω, z i ) | β | is a polynomial in ϕ ω ( z ) and ϕ ω ( z ) of total degree | α | + | β | . (cid:3) We now obtain a proof of statement (c) in Theorem 1.4.
Proposition 3.4.
Suppose u ∈ L ( B N ) and B ( u ) = f ¯ g + · · · + f d ¯ g d , where f ℓ , g ℓ ∈ H ( B N ) and f ℓ ∈ L N +2 ( B N ) for each ℓ . Then there exist finitely manypoints w , . . . , w s ∈ B N , polynomials Q , . . . , Q s in C [ z, ¯ z ] with total degrees atmost N + 1 , and a pluriharmonic function h such that B ( u )( z ) = h ( z ) + s X j =1 Q j ◦ ϕ w j ( z ) . Proof.
We know that there exist holomorphic functions h , h on B N and finitelymany points w , . . . , w s ∈ B N and holomorphic polynomials Q j,β of degree at most2 N + 1 − | β | with Q j,β (0) = 0 such that h ( z ) + ¯ h ( z ) + X ≤ j ≤ s ≤| β |≤ N Q j,β (cid:16) z (1 − h z, w j i ) (cid:17) · ¯ z β (1 − h w j , z i ) | β | = B ( u )( z )= f ( z ) g ( z ) + · · · + f d ( z ) g d ( z ) . Complexifying as in the proof of Theorem 3.3 shows that each Q j,β (cid:0) z −h z,w j i (cid:1) be-longs to the linear span of { } ∪ n g ( ζ ) f + · · · + g s ( ζ ) f s : ζ ∈ B N o , which is contained in L N +2 ( B N ) by the hypothesis. Now the same argument as inthe proof of Theorem 3.3 may be used to finish the proof. (cid:3) Lemma 3.5.
Let Q be a polynomial in C [ z , . . . , z N ] . If Q (cid:0) z −h z,ω i (cid:1) belongs to L N +2 ( B N ) for some ω on the unit sphere, then Q is a constant.Proof. Since the case of a single complex variable may be regarded as a specialcase of two or more variables, we consider N ≥ ω = (0 , . . . , , z [ N − to denote( z , . . . , z N − ) ∈ C N − . Then Q can be written as Q ( z ) = X | α |≥ Q α ( z N ) z α [ N − , where the sum is finite over α ∈ Z N − and each Q α is a holomorphic polynomialin z N . We have F ( z ) = Q (cid:16) z − h z, ω i (cid:17) = X | α |≥ Q α (cid:16) z N − z N (cid:17) z α [ N − (1 − z N ) | α | . Since F belongs to L N +2 ( B N ), for each α , the function F α ( z ) = Q α (cid:16) z N − z N (cid:17) z α [ N − (1 − z N ) | α | = Z [0 , π ] N − F ( e iθ z , . . . , e iθ N − z N − , z N ) e − i ( α θ + ··· + α N − θ N − ) dθ π · · · dθ N − π must belong to L N +2 ( B N ). However, for | α | ≥
1, if Q α is not identically zero, thenfor z near (0 , . . . , , | F α ( z ) | dominates a nonzero constant multipleof (cid:12)(cid:12) z α [ N − (1 − z N ) | α | (cid:12)(cid:12) . It can be showed, using the techniques in [Rud80, Section 1.4],that such functions do not belong to L N +2 ( B N ). On the other hand, if Q isnot a constant, then for z near (0 , . . . , , | F ( z ) | = | Q ( z ) | dominates a nonzeroconstant multiple of | − z N | , which again does not belong to L N +2 ( B N ). As aconsequence, Q α = 0 for all | α | ≥ Q is a constant. Therefore, Q is aconstant, as desired. (cid:3) ROWN–HALMOS TYPE THEOREMS ON THE BALL 15
Besides its important applications in the theory of Toeplitz operators as we shallsee in the next section, Theorem 1.4 also helps answer open questions about M -harmonic functions. In the early nineties, Ahern and Rudin [AR91] completelycharacterized holomorphic functions f, g on the ball for which f ¯ g is M -harmonic.Nearly a decade later, Zheng [Zhe98] showed that for f, g, h and k belonging to theHardy space H N ( ∂ B N ), the function f ¯ g − h ¯ k is M -harmonic if and only if it ispluriharmonic. About ten years ago, making use of Ahern–Rudin’s characterization,Choe et al. [CKL11, Lemma 4.5] proved a single-product version of Zheng’s resultunder a slightly weaker hypothesis. They only assumed that one of the factorbelongs to H N ( ∂ B N ). The problem of generalizing this and Zheng’s result tofinite sums of more than two products has been opened since then, see [CKL11,Question 6.1]. Our Theorem 1.4 offers a far-reaching answer. Theorem 3.6.
Suppose for each ≤ j ≤ s , the functions f j , g j are holomorphicon B N and f j belongs to A N +2 ( B N ) . If u = P sj =1 f j ¯ g j is an eigenfunction of e ∆ ,then u must be pluriharmonic.Proof. By [Rud80, Theorem 4.2.4], u is an eigenfunction of the Berezin transform,that is, there exists λ ∈ C such that B ( u ) = λu = s X j =1 λf j ¯ g j . Since u is clearly a C N +2 -function and f j ∈ A N +2 ( B N ) for all j , Theorem 1.4parts (a) and (c) hold, which implies that u is pluriharmonic. A careful examinationof the proof of Theorem 1.4(c) (see Proposition 3.4 and Lemma 3.5) shows that theconclusion also holds if A N +2 ( B N ) is replaced by H N ( ∂ B N ). (cid:3) Brown–Halmos type results
We first recall the following standard lemma characterizing when a function ofthe form P j ¯ g j u j (with holomorphic g j , u j ) is pluriharmonic. The one-dimensionalversion was already proved in [CKL08, Theorem 3.3] but our proof here is muchsimpler. Lemma 4.1.
Let u , . . . , u s and g , . . . , g s be holomorphic functions on B N . Then P sj =1 ¯ g j u j is pluriharmonic on B N if and only if s X j =1 (cid:0) g j − g j (0) (cid:1)(cid:0) u j − u j (0) (cid:1) = 0 , which is equivalent to s X j =1 ¯ g j u j = s X j =1 (cid:16) ¯ g j u j (0) + g j (0) u j − g j (0) u j (0) (cid:17) . Proof.
Without loss of generality, we may assume that u j (0) = g j (0) = 0 for all j .Using power expansions, we have s X j =1 u j ( z )¯ g j ( z ) = X | α |≥ , | β |≥ c α,β z α ¯ z β , which is pluriharmonic if and only if it is identically zero. (cid:3) We are now ready to prove Theorem 1.1, which is restated below for the reader’sconvenience.
Theorem 4.2.
Let φ j , ψ j for ≤ j ≤ n be bounded pluriharmonic functions and h be a C N +2 bounded function on B N . Let x ℓ , y ℓ ∈ A ( B N ) for ≤ ℓ ≤ r . Write φ j = f j + ¯ g j , ψ j = u j + ¯ v j where f j , g j , u j , v j are holomorphic. Then n X j =1 T φ j T ψ j = T h + r X ℓ =1 x ℓ ⊗ y ℓ (4.1) if and only if h − P nj =1 ¯ g j u j is pluriharmonic and n X j =1 φ j ψ j = h + (1 − | z | ) N +1 r X ℓ =1 x ℓ ¯ y ℓ . (4.2) Proof.
For any functions x, y ∈ A ( B N ), we compute the Berezin transform B ( x ⊗ y )( z ) = (1 − | z | ) N +1 x ( z ) y ( z ) , z ∈ B N . Also, if φ = f + ¯ g and ψ = u + ¯ v are bounded pluriharmonic, where f, g, u, v areholomorphic functions (which might not be bounded but they all belong to L p ( B N )for all p ), then it is well known that B ( T φ T ψ ) = φψ − ¯ gu + B (¯ gu ) . Therefore, B (cid:16) n X j =1 T φ j T ψ j − T h (cid:17) = n X j =1 ( φ j ψ j − ¯ g j u j ) + B (cid:0) n X j =1 ¯ g j u j − h (cid:1) . Using the linearity and injectivity of the Berezin transform, we conclude that (4.1)holds if and only if B (cid:16) n X j =1 T φ j T ψ j − T h (cid:17) = r X ℓ =1 B (cid:0) x ℓ ⊗ y ℓ (cid:1) , which is equivalent to B (cid:16) n X j =1 ¯ g j u j − h (cid:17) = n X j =1 ( − φ j ψ j + ¯ g j u j ) + (1 − | z | ) N +1 r X ℓ =1 x ℓ ¯ y ℓ . (4.3)We now show that this equation is equivalent to the two conditions stated in thetheorem. Put u = P nj =1 ¯ g j u j − h . Suppose first that (4.3) holds. Then the Berezintransform B ( u ) has finite rank. Since u belongs to C ∞ ( B N ) ∩ L N +2 ( B N ), Theorem1.4 implies that it is pluriharmonic on B N and B ( u ) = u . As a consequence, h − P nj =1 ¯ g j u j is pluriharmonic and n X j =1 ¯ g j u j − h = n X j =1 ( − φ j ψ j + ¯ g j u j ) + (1 − | z | ) N +1 r X ℓ =1 x ℓ ¯ y ℓ , which means h = n X j =1 φ j ψ j − (1 − | z | ) N +1 r X ℓ =1 x ℓ y ℓ . (4.4) ROWN–HALMOS TYPE THEOREMS ON THE BALL 17
Conversely, if u is pluriharmonic and (4.4) holds, then using the fact that the Berezintransform fixes pluriharmonic functions, we conclude that (4.3) holds, which implies(4.1) as desired. (cid:3) Remark 4.3.
In this remark, we discuss a construction of functions that satisfy thetwo conditions in Theorem 4.2. As before, write φ j = f j + ¯ g j , ψ j = u j + ¯ v j , where f j , g j , u j , v j are holomorphic and f j (0) = v j (0) = 0. Assume that (4.2) holds, then h − n X j =1 ¯ g j u j = n X j =1 ( f j ¯ v j + f j u j + ¯ g j ¯ v j ) − (1 − | z | ) N +1 r X ℓ =1 x ℓ ¯ y ℓ , which, by Lemma 4.1, is pluriharmonic if and only if n X j =1 f j ¯ v j − (1 − | z | ) N +1 r X ℓ =1 x ℓ ¯ y ℓ = r X ℓ =1 − x ℓ (0)¯ y ℓ − x ℓ y ℓ (0) + x ℓ (0) y ℓ (0) . The above identity is equivalent to n X j =1 f j ( z )¯ v j ( z ) = r X ℓ =1 (cid:26)(cid:0) x ℓ ( z ) − x ℓ (0) (cid:1)(cid:0) ¯ y ℓ ( z ) − y ℓ (0) (cid:1) (4.5)+ X ≤| α |≤ N +1 ( − | α | (cid:18) | α | α (cid:19)(cid:0) z α x ℓ ( z ) (cid:1)(cid:0) z α y ℓ ( z ) (cid:1)(cid:27) . Let x ℓ , y ℓ (1 ≤ ℓ ≤ r ) be any finite collection of bounded holomorphic functions.We can easily choose bounded holomorphic functions f j , v j (1 ≤ j ≤ n ) for some n such that f j (0) = v j (0) = 0 and (4.5) holds. For each j , choose arbitrary boundedholomorphic functions g j and u j and set φ j = f + ¯ g j and ψ j = u j + ¯ v j . Put h = n X j =1 φ j ψ j − (1 − | z | ) N +1 r X ℓ =1 x ℓ ¯ y ℓ . We then have n X j =1 T φ j T ψ j = T h + r X ℓ =1 x ℓ ⊗ y ℓ . The problem becomes more delicate if one imposes a restriction on n . The recentpaper [DQZ17] considered the case n = 1 in the setting of a single variable. Itwas shown that for bounded harmonic functions φ, ψ , and smooth h , if T φ T ψ − T h has rank one, then it must be zero. On the other hand, for any r ≥
2, exampleswere constructed so that T φ T ψ − T h has rank exactly r . It would be interesting togeneralize the results in [DQZ17] to the setting of several variables. Proof of Corollary 1.2.
Write φ = f + ¯ g, ψ = u + ¯ v with holomorphic f, g, u, v and f (0) = v (0) = 0.(a) By Theorem 1.1, if T φ T ψ = T h , then h = φψ and h − ¯ gu is pluriharmonic.It follows that f ¯ v = ( h − ¯ gu ) − f u − ¯ g ¯ v is pluriharmonic. Lemma 4.1 implies that f ¯ v = 0 which forces either f = 0 or v = 0. Therefore, either ¯ φ or ψ must beholomorphic.(b) Now suppose that T φ T ψ has a finite rank. Then there exist functions x ℓ , y ℓ ∈ A ( B N ) , ≤ ℓ ≤ r so that T φ T ψ = P ℓ x ℓ ⊗ y ℓ . Using Theorem 1.1 with h = 0,we obtain that φψ = (1 − | z | ) N +1 P ℓ x ℓ ¯ y ℓ , and ¯ gu is pluriharmonic, which implieseither g or u is constant. Therefore, either φ or ¯ ψ is holomorphic. Taking operator adjoints if necessary, we may assume that φ is holomorphic. Assume further that φ is not identically zero. Then T φ is injective. Since T φ T ψ has a finite rank, it followsthat T ψ must have a finite rank, hence ψ = 0, by the multivariable Luecking’sTheorem. (cid:3) We now apply Theorem 1.1 to characterize when a sum of products of Hankeloperators with pluriharmonic symbols has a finite rank. Recall that for a boundedsymbol φ , the Hankel operator H φ : A ( B N ) → L ( B N ) ⊖ A ( B N ) is defined as H φ =( I − P ) M φ | A ( B N ) , where M φ is the multiplication by φ and P is the orthogonalprojection from L ( B N ) onto A ( B N ). The crucial identity relating properties ofToeplitz and Hankel operators is given by H ∗ ¯ φ H ψ = T φψ − T φ T ψ . Proposition 4.4.
Let φ j , ψ for ≤ j ≤ n be bounded pluriharmonic functions on B N . Then the followings are equivalent:(1) P nj =1 H ∗ ¯ φ j H ψ j = 0 .(2) P nj =1 H ∗ ¯ φ j H ψ j = T F for some F ∈ C N +2 ( B N ) ∩ L ∞ ( B N ) .(3) P nj =1 H ∗ ¯ φ j H ψ j has a finite rank.(4) P nj =1 P ( φ j ) · ( ψ j − P ( ψ j )) is pluriharmonic.Proof. It is clear that (1) implies (2). Now assume that (2) holds. Then n X j =1 T φ j T ψ j = n X j =1 ( T φ j ψ j − H ∗ ¯ φ j H ψ j ) = T h − T F = T h − F , where h = P nj =1 φ j ψ j . By Theorem 1.1, we have n X j =1 φ j ψ j = h − F, which implies F = 0. Therefore, (3) (and (1) as well) follow.Now assume that (3) holds, that is, the operator T = P nj =1 H ∗ ¯ φ j H ψ j has finiterank. The same argument as above gives P nj =1 T φ j T ψ j = T h − T . By Theorem 1.1again, the function h − n X j =1 ( φ j − P ( φ j )) P ( ψ j )is pluriharmonic, which then implies (4).Finally, assume that (4) holds. Setting h = P nj =1 φ j ψ j and x ℓ = y ℓ = 0, we seethat both conditions in Theorem 1.1 are satisfied and so P nj =1 T φ j T ψ j = T P nj =1 φ j ψ j ,which gives (1). This completes the proof of the proposition. (cid:3) Proof of Corollary 1.3.
The sufficient direction is well known and not difficult toprove. To show the necessary direction, replacing φ by φ − φ (0) and ψ by ψ − ψ (0)if necessary, we may assume that φ (0) = ψ (0) = 0. As before, write φ = f + ¯ g and ψ = u + ¯ v with holomorphic f, g, u, v satisfying f (0) = g (0) = u (0) = v (0) = 0. Wehave H ∗ ¯ φ H ψ − H ∗ ¯ ψ H φ = ( T φψ − T φ T ψ ) − ( T ψφ − T ψ T φ ) = − [ T φ , T ψ ] . Therefore, if [ T φ , T ψ ] is of finite rank, then by Proposition 4.4, P ( φ )( ψ − P ( ψ )) − P ( ψ )( φ − P ( φ )) = f ¯ v − u ¯ g ROWN–HALMOS TYPE THEOREMS ON THE BALL 19 is pluriharmonic. We may apply [Zhe98, Theorem 5.6 and Lemma 6.8] to completethe proof. Here, we provide a direct argument. Indeed, Lemma 4.1 implies f ¯ v = u ¯ g which, by complexifying, gives f ( z )¯ v ( w ) = u ( z )¯ g ( w ) for all z, w ∈ B N . If u = v = 0, then ψ = cφ with c = 0. If u = 0 and v is not identically zero,then f = 0 so both φ and ψ are anti-holomorphic. Similarly, if v = 0 and u isnot identically zero, then g = 0 so both φ and ψ are holomorphic. On the otherhand, if neither of u nor v is identically zero, then there exists z ∈ B N such that u ( z ) v ( z ) = 0 and it follows that f = cu and ¯ g = c ¯ v , where c = ¯ g ( z )¯ v ( z ) = f ( z ) u ( z ) . Hence, φ − cψ = 0. This completes the proof of the corollary. (cid:3) We end this section with another important application of Theorem 1.1.
Corollary 4.5.
Let φ j , ψ j ∈ L ∞ ( B N ) be pluriharmonic functions and let h ∈ L ( B N ) be locally bounded if N = 1 , and C N +2 -smooth and bounded if N ≥ . Write φ j = f j + ¯ g j , ψ j = u j + ¯ v j where f j , g j , u j , v j are holomorphic. Then P nj =1 T φ j T ψ j = T h if and only if h = P nj =1 φ j ψ j and X j (cid:0) f j − f j (0) (cid:1)(cid:0) ¯ v j − v j (0) (cid:1) = 0 . For
N >
1, Corollary 4.5 follows immediately from Theorem 1.1 and Lemma 4.1.The proof in the case N = 1 goes along the same lines except that we use Rao’stheorem which shows that for a locally bounded function u , the Berezin transform B ( u ) has finite rank if and only if u is harmonic.5. Polynomials in the range of Berezin transform and applications
In this section we first describe all polynomials in the range of the Berezintransform. We then construct examples which show that the conclusion of Theorem1.1 may fail for N ≥ h is dropped. Lastly, weshow that the product of two Toeplitz operators with polynomial symbols, undera certain additional condition on the degrees, is always equal to another Toeplitzoperator with an integrable symbol.In the setting of a single variable, Ahern [Ahe04] showed that if p and q areholomorphic polynomials such that the degree of pq is at most 3, then p ¯ q is theBerezin transform of an L -function. The following theorem generalizes this resultto several variables. Since calculations cannot be performed explicitly as in thesingle variable case, the proof here is considerably more complicated. Theorem 5.1.
Let f be a polynomials in z and ¯ z . Then f = B ( u ) for some u ∈ L ( B N ) if and only if for any ≤ j, ℓ ≤ N , the derivative ∂ z j ¯ ∂ z ℓ f has totaldegree at most N − .As a consequence, if w , . . . , w s belongs to B N and Q , . . . , Q s are polynomials in C [ z, ¯ z ] with total degrees at most N + 1 , then there exists a function u ∈ L ( B N ) such that B ( u ) = P sj =1 Q j ◦ ϕ w j . Proof.
Throughout the proof, we write ran( B ) to denote the image of L ( B N ) underthe Berezin transform. Suppose that f = B ( u ) for some u ∈ L ( B N ). By Theorem3.4, there exists a pluriharmonic function h and a polynomial Q ∈ C [ z, ¯ z ] of degreeat most 2 N + 1 such that f = B ( u ) = h + Q . It follows that for any 1 ≤ j, ℓ ≤ N , ∂ z j ¯ ∂ z ℓ f = ∂ z j ¯ ∂ z ℓ h + ∂ z j ¯ ∂ z ℓ Q = ∂ z j ¯ ∂ z ℓ Q, which has total degree at most 2 N − ≤ j, ℓ ≤ N , the derivative ∂ z j ¯ ∂ z ℓ f has totaldegree at most 2 N −
1. Since f is a polynomial, there exists a pluriharmonicpolynomial h and complex coefficients c α,β for | α | ≥ , | β | ≥ f ( z ) = h ( z ) + X | α |≥ , | β |≥ c α,β z α ¯ z β . The assumption implies that c α,β = 0 whenever | α | + | β | > N + 2. Consequently,we may write f = h + Q , where h is pluriharmonic and Q has total degree at most2 N + 1. Thus, it remains to show that Q belongs to ran( B ).Let α, β be two multiindexes and ℓ be a non-negative integer such that | α | + | β | + 2 ℓ ≤ N + 1. We shall show that the polynomial ¯ z α z β (1 − | z | ) ℓ belongs toran( B ). Taking complex conjugates if necessary, we may assume that | β | ≤ | α | .Using [Rud80, Proposition 1.4.9] and the rotation invariant of the surface mea-sure on ∂ B N , we see that for any integer s ≥ z ∈ B N , Z ∂ B N |h z, ζ i| s dσ ( ζ ) = Γ( N ) Γ( s + 1)Γ( N + s ) | z | s . Replacing s by s + | α | and applying ∂ αz ( s + | α | ) ··· ( s +1) to both sides of the above identitygives Z ∂ B N ¯ ζ α h z, ζ i s h ζ, z i s + | α | dσ ( ζ ) = Γ( N ) Γ( s + | α | + 1)Γ( N + s + | α | ) ¯ z α | z | s . Applying Γ( s + | α |−| β | +1)Γ( s + | α | +1) ¯ ∂ βz , we have Z ∂ B N ¯ ζ α ζ β h z, ζ i s h ζ, z i s + | α |−| β | dσ ( ζ ) = Γ( N ) Γ( s + | α | − | β | + 1)Γ( N + s + | α | ) ¯ ∂ βz (cid:16) ¯ z α | z | s (cid:17) . Now let u ∈ L ( B N ) be of the form u ( z ) = ¯ z α z β ϕ ( | z | ), where ϕ is a function on[0 ,
1) to be defined later. Integration in polar coordinates (using ξ = rζ ) togetherwith the above identity gives Z B N ¯ ξ α ξ β ϕ ( | ξ | ) h z, ξ i s h ξ, z i s + | α |−| β | dV ( ξ )= 2 N Z r N +2 s +2 | α |− ϕ ( r ) dr Z ∂ B N ¯ ζ α ζ β h z, ζ i s h ζ, z i s + | α |−| β | dσ ( ζ )= Γ( N + 1) Γ( s + | α | − | β | + 1)Γ( N + s + | α | ) (cid:16) Z r N + s + | α |− ϕ ( r ) dr (cid:17) ¯ ∂ βz (cid:0) ¯ z α | z | s (cid:1) = Γ( N + 1) Γ( s + | α | − | β | + 1)Γ( N + s + | α | ) b ϕ ( N + s + | α | ) ¯ ∂ βz (cid:0) ¯ z α | z | s (cid:1) , where b ϕ denotes the Mellin transform of ϕ given by b ϕ ( ζ ) = Z r ζ − ϕ ( r ) dr. ROWN–HALMOS TYPE THEOREMS ON THE BALL 21
It follows that 1Γ( N + 1) Γ( s + | α | − | β | + 1) Z B N u ( ξ ) h z, ξ i s h ξ, z i s + | α |−| β | dV ( ξ )= 1Γ( N + s + | α | ) b ϕ ( N + s + | α | ) ¯ ∂ βz (cid:0) ¯ z α | z | s (cid:1) . (5.1)We now compute, for z ∈ B N , Z B N u ( ξ ) | − h z, ξ i| N +1) dV ( ξ )= ∞ X s,t =0 Γ( N + 1 + s )Γ( N + 1) Γ( s + 1) · Γ( N + 1 + t )Γ( N + 1) Γ( t + 1) Z B N u ( ξ ) h z, ξ i s h ξ, z i t dV ( ξ ) . Since the integral vanishes unless t = s + | α | − | β | , we may rewrite the abovesummation as Z B N u ( ξ ) | − h z, ξ i| N +1) dV ( ξ )= ∞ X s =0 Γ( N + 1 + s )Γ( N + 1) Γ( s + 1) · Γ( N + 1 + s + | α | − | β | )Γ( N + 1) Γ( s + | α | − | β | + 1) ×× Z B N u ( ξ ) h z, ξ i s h ξ, ξ i s + | α |−| β | dV ( ξ ) (5.2)= ¯ ∂ βz ( ¯ z α · ∞ X s =0 Γ( N + 1 + s ) Γ( N + 1 + s + | α | − | β | )Γ( N + 1) Γ( s + 1) Γ( N + s + | α | ) b ϕ ( N + s + | α | ) | z | s ) . The last identity follows from formula (5.1). To simplify the notation we now set M = N + 1 − | β | − ℓ . Since | α | + | β | + 2 ℓ ≤ N + 1 and | β | ≤ | α | , we have1 ≤ M ≤ N + 1 − | β | . Let us choose ϕ such that b ϕ ( ζ ) = Γ( N + 1)Γ( N + 1 − | ℓ | ) · Γ( ζ ) Γ( ζ + 1 − | α | − | β | − ℓ )Γ( ζ + 1 − | α | ) Γ( ζ + 1 − | β | ) (5.3)= Γ( N + 1)Γ( M + | β | ) · Γ( ζ ) Γ( ζ + M − N − | α | )Γ( ζ − | α | + 1) Γ( ζ − | β | + 1) . The existence of such a function ϕ will be established below. Since for all integers s ≥ b ϕ ( N + s + | α | ) = Γ( N + 1)Γ( M + | β | ) · Γ( N + s + | α | ) Γ( M + s )Γ( N + s + 1) Γ( N + s + | α | − | β | + 1) , formula (5.2) simplifies to Z B N u ( ξ ) | − h z, ξ i| N +1) dV ( ξ ) = Γ( M )Γ( M + | β | ) ¯ ∂ βz ( ¯ z α · ∞ X s =0 Γ( M + s )Γ( M ) Γ( s + 1) | z | s ) = Γ( M )Γ( M + | β | ) ¯ ∂ βz (cid:8) ¯ z α (1 − | z | ) − M (cid:9) . It follows that B ( u )( z ) = Γ( M )Γ( M + | β | ) (1 − | z | ) N +1 · ¯ ∂ βz (cid:8) ¯ z α (1 − | z | ) − M (cid:9) . (5.4) We now explain the existence of ϕ and show that the corresponding function u belongs to L ( B N ). First note that if | α | = 0, then | β | = 0 as well since we assumedthat | β | ≤ | α | and so in this case, formula (5.3) becomes b ϕ ( ζ ) = Γ( N + 1)Γ( N + 1 − ℓ ) · Γ( ζ ) Γ( ζ + 1 − ℓ )Γ( ζ + 1) Γ( ζ + 1)= ζ if ℓ = 0 , Γ( N +1)Γ( N +1 − ℓ ) · ζ · ζ +1 − ℓ ) ··· ζ if ℓ ≥ . In the first case, ϕ = 1. In the second case, ϕ is a linear combination of log( r ) and r − , . . . , r − ℓ .Now assume | α | ≥
1. Then second factor on the right hand-side of (5.3) reducesto a proper rational function of the form ζ −| β |− ℓ ) ··· ( ζ −| β | ) if | α | = 1 , ( ζ +1 −| α | ) ··· ( ζ − ζ +1 −| α |−| β |− ℓ ) ··· ( ζ −| β | ) if | α | ≥ , whose numerator has degree | α |− | α | + ℓ > | α |−
1. Therefore, ϕ ( r ) exists and it is a linear combination of r −| α |−| β |− ℓ , . . . , r −| β | .In all cases, we have ϕ ( r ) = O ( r −| α |−| β |− ℓ ) as r → + , which implies that for any ζ ∈ ∂ B N , u ( rζ ) = r | α | + | β | ϕ ( r )¯ ζ α ζ β = O ( r −| α |−| β |− ℓ ) . Since (2 N −
1) + 2 − | α | − | β | − ℓ = 2 N + 1 − | α | − | β | − ℓ ≥
0, using integrationby polar coordinates, we conclude that u ∈ L ( B N ).Choosing | β | = 0 in (5.4) shows that ¯ z α (1 − | z | ) ℓ belongs to ran( B ) whenever | α | + 2 ℓ ≤ N + 1. It then follows that ¯ z α | z | s (and hence z α | z | s , after takingcomplex conjugates) belongs to ran( B ) whenever | α | + 2 s ≤ N + 1.Generally, whenever | α | + | β | + 2 ℓ ≤ N + 1, we may use (5.4) to conclude thatran( B ) contains the functionΓ( M )Γ( M + | β | ) (1 − | z | ) N +1 · ¯ ∂ βz (cid:8) ¯ z α (1 − | z | ) − M (cid:9) = ¯ z α z β (1 − | z | ) ℓ + X µ + ν = β | µ |≥ c µ,ν ¯ z α − µ z ν (1 − | z | ) N +1 − M −| ν | = ¯ z α z β (1 − | z | ) ℓ + X µ + ν = β | µ |≥ c µ,ν ¯ z α − µ z ν (1 − | z | ) | β | + ℓ −| ν | , where c µ,ν ’s are constants. Note that each term in the summation has total degreeat most | α | + | β | +2 ℓ ≤ N +1 and the degree in z is | ν | < | β | . As a consequence, aninduction in | β | shows that ¯ z α z β (1 −| z | ) ℓ belongs to ran( B ) whenever | α | + | β | +2 ℓ ≤ N +1. Letting ℓ = 0, we conclude that ¯ z α z β ∈ ran( B ) whenever | α | + | β | ≤ N +1.For each 1 ≤ j ≤ s , we showed above the existence of a function u j ∈ L ( B N )such that B ( u j ) = Q j . Using the commutativity of the Berezin transform and ROWN–HALMOS TYPE THEOREMS ON THE BALL 23 automorphisms of the unit ball (see [AFR93, Proposition 2.3], for example), wehave B ( u j ◦ ϕ w j ) = B ( u j ) ◦ ϕ w j = Q j ◦ ϕ w j . It then follows that B ( P sj =1 u j ◦ ϕ w j ) = P sj =1 Q j ◦ ϕ w j as required. (cid:3) Remark 5.2.
Using (5.4) in the case ℓ = 1, | β | = 0 and | α | ≥ M = N )shows that for u ( z ) = ¯ z α ϕ ( | z | ) with b ϕ given by (5.3), we obtain B (cid:16) ¯ z α | z | − | α | − ¯ z α (cid:17) = | α | N ¯ z α (1 − | z | ) , which implies B (cid:16) | α | + NN ¯ z α − ¯ z α | z | − | α | (cid:17) = | α | N ¯ z α | z | . This identity is valid for all 1 ≤ | α | < N . For N = 1 and α = 1, this is the formulagiven in [Ahe04, Lemma 1].As is well known in the literature, properties of the Berezin transform haveconsequences in the theory of Toeplitz operators. We discuss here a few examples. Remark 5.3.
Setting α = β = (1 , , . . . ,
0) and ℓ = 0 (hence, M = N ) in (5.4)gives B ( u )( z ) = 1 N (1 − | z | ) N +1 · ¯ ∂ z { ¯ z (1 − | z | ) − N } = 1 N (1 − | z | ) N +1 · (cid:8) (1 − | z | ) − N + N | z | (1 − | z | ) − N − (cid:9) = 1 N (1 − | z | ) + | z | . (5.5)Here, u ( z ) = | z | ϕ ( | z | ) with b ϕ ( ζ ) = Γ( ζ ) Γ( ζ − ζ ) Γ( ζ ) = 1 ζ − d r − ( ζ ) . It follows that u ( z ) = | z | | z | for z ∈ B N \{ } , which is bounded and is not plurihar-monic if N ≥ N − | z | − N X j =2 | z j | = B ( h )with h ( z ) = − N | z | | z | . It follows that( N − T z T ¯ z − N X j =2 T z j T ¯ z j = T h , where h is a bounded function and h ( z ) = ( N − | z | − P Nj =2 | z j | . It is importantto note that this phenomenon cannot occur for N = 1 due to Corollary 4.5.We end the paper by showing that the product of two Toeplitz operators withpolynomial symbols, under a certain condition on the degrees, is again a Toeplitzoperator. However, we note that the symbol of the resulting Toeplitz operator isnot always a polynomial. Proposition 5.4.
Let α and β be two multiindexes such that | α | ≥ and | β | ≥ .Then T z β T ¯ z α = T u for some u ∈ L ( B N ) if and only if | α | + | β | ≤ N + 1 .As a consequence, if f and g are polynomials in z and ¯ z such that the sum ofthe degree of f in z and the degree of g in ¯ z is at most N + 1 , then there exists h ∈ L ( B N ) such that T f T g = T h .Proof. Suppose T z β T ¯ z α = T u for some u ∈ L ( B N ). Taking Berezin transformsgives B ( u ) = B ( T u ) = B ( T z β T ¯ z α ) = ¯ z α z β . Write α = ( α , . . . , α N ) and β = ( β , . . . , β N ). Since | α | ≥ | β | ≥
1, there exist j, ℓ such that β j = 0 and α ℓ = 0. It follows that the total degree of ∂ z j ¯ ∂ z ℓ (¯ z α z β )is exactly | α | + | β | −
2. Proposition 5.1 implies that | α | + | β | − ≤ N −
1, whichgives | α | + | β | ≤ N + 1.Conversely, if | α | + | β | ≤ N + 1, then by Theorem 5.1, there exists a function u ∈ L ( B N ) such that ¯ z α z β = B ( u ). This implies that B ( T z β T ¯ z α ) = B ( u ), whichgives T z β T ¯ z α = T u . For any holomorphic polynomials p and q , using the well-knownproperties of Toeplitz operators, we have T ¯ p ( z ) z β T q ( z )¯ z α = T ¯ p (cid:0) T z β T ¯ z α (cid:1) T q = T ¯ p T u T q = T ¯ puq . The last statement of the proposition now follows. (cid:3)
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Department of Mathematics and Statistics, University of Toledo, Toledo, OH 43606
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Department of Mathematics and Statistics, University of Toledo, Toledo, OH 43606
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