Hyponormal dual Toeplitz operators on the orthogonal complement of the Harmonic Bergman space
aa r X i v : . [ m a t h . F A ] J a n HYPONORMAL DUAL TOEPLITZ OPERATORS ON THE ORTHOGONALCOMPLEMENT OF THE HARMONIC BERGMAN SPACE
CHONGCHAO WANG AND XIANFENG ZHAO Abstract.
In this paper, we mainly study the hyponormality of dual Toeplitz operators on theorthogonal complement of the harmonic Bergman space. First we show that the dual Toeplitzoperator with bounded symbol is hyponormal if and only if it is normal. Then we obtain a necessaryand sufficient condition for the dual Toeplitz operator S ϕ with the symbol ϕ ( z ) = az n z m + bz n z m ( n , n , m , m ∈ N and a, b ∈ C ) to be hyponormal. Finally, we show that the rank of thecommutator of two dual Toeplitz operators must be an even number if the commutator has a finiterank. Introduction
A bounded linear operator T on some Hilbert space H is called hyponormal if the commutator T ∗ T − T T ∗ of T ∗ (the adjoint of T ∗ ) and T is positive, i.e., h ( T ∗ T − T T ∗ ) f, f i > f ∈ H . This definition was introduced by Halmos in 1950 and generalizes the concept of anormal operator (where T ∗ T = T T ∗ ). Normal operators are completely understood by people. Infact, it is possible to give a model for an arbitrary normal operator, see [1, Chapter 2] if necessary.The class of hyponormal operators is an important class of non-normal operators. The problem“when a hyponormal operator is normal” has been investigated by many authors. The theoryof normal and hyponormal operators is an extensive and highly developed area, which has madeimportant contributions to a number of problems in functional analysis, operator theory, analyticfunction theory and mathematical physics.Normality and hyponormality of Toeplitz operators on analytic or harmonic function spaces havecaptured people’s attention for a long time. Here, for the basic knowledge about Toeplitz operatorson the Hardy space and on the Bergman space we refer to the books [12] and [29]. The normalToeplitz operator on the Hardy space was characterized by Brown and Halmos [2] in 1964. For theclassical results about hyponormal Toeplitz operators on the Hardy space, one can consult [4] forthe case of bounded symbols and [5], [15], [22], [27, 28] for the case of (trigonometric) polynomialsymbols. The hyponormality of Toeplitz operators with some harmonic polynomial symbols onthe Bergman space was studied in [10], [13, 14] and [17]. Recently, a sufficient condition for theToeplitz operator with the symbol | z | n + c | z | s ( n, s ∈ N , c ∈ C ) to be hyponormal on certainweighted Bergman space was established in [16]. Furthermore, hyponormality of block Toeplitzoperators on the vector-valued Hardy space was investigated in [6]. Concerning Toeplitz operatorson the harmonic function space, Choe and Lee obtained a complete characterization for Toeplitzoperators with bounded harmonic symbols to be normal on the harmonic Bergman space [7].The purpose of this paper is to describe the normality and hyponormality of dual Toeplitzoperators on the orthogonal complement of the harmonic Bergman space. We will elaborate onthis class of operators in the following. Date : January 28, 2021.2010
Mathematics Subject Classification.
Key words and phrases. dual Toeplitz operator; harmonic Bergman space; normal; hyponormal.
Let D be the unit disk in the complex plane C and dA = π dxdy be the normalized area measureon D . L ( D , dA ) denotes the space of the Lebesgue measurable functions on D with the norm k f k = (cid:18)Z D | f ( z ) | dA ( z ) (cid:19) < + ∞ . Clearly, L ( D ) = L ( D , dA ) is a Hilbert space with the inner product h f, g i = Z D f ( z ) g ( z ) dA ( z ) . The Bergman space L a is the closed subspace of L ( D ) consisting of all analytic functions on D . Similarly, the harmonic Bergman space L h is the closed subspace of L ( D ) consisting of allharmonic functions on D . Let P and Q be the orthogonal projections from L ( D ) onto L a and L h ,respectively. Then P and Q have the following relationship: Q ( f ) = P ( f ) + P ( f ) − P ( f )(0) , f ∈ L ( D ) . (1.1)According to the decomposition L ( D , dA ) = L h ⊕ ( L h ) ⊥ , the multiplication operator M ϕ withsymbol ϕ ∈ L ∞ ( D ) can be represented as M ϕ = (cid:18) T ϕ H ∗ ϕ H ϕ S ϕ (cid:19) , where the operator S ϕ u = ( I − Q )( ϕu )is bounded and linear on the Hilbert space ( L h ) ⊥ . We call S ϕ the dual Toeplitz operator withsymbol ϕ on ( L h ) ⊥ . From the definition of S ϕ , the following elementary properties about dualToeplitz operators are easily verified:(a) S ∗ ϕ = S ϕ ;(b) k S ϕ k k ϕ k ∞ ;(c) S αϕ + βψ = αS ϕ + βS ψ for all ϕ, ψ ∈ L ∞ ( D ) and all constants α, β ∈ C .Unlike the Bergman space and the harmonic Bergman space, the orthonormal basis for theHilbert space ( L h ) ⊥ can not be written explicitly. But we are able to study dual Toeplitz operatorson this function space via the functions of the form z n z m with m, n ∈ N . Observe that M = span n ( I − Q )( z n z m ) : m, n = 1 , , · · · o is a dense subspace of ( L h ) ⊥ . Using (1.1), then elementary calculations give us that( I − Q )( z n z m ) = z n z m − m − n + 1 m + 1 z m − n , m > n,z n z m − n − m + 1 n + 1 z n − m , m n. (1.2)Moreover, it is easy to check that(1.3) h z n z m , z k z l i = m + n + k + l + 2 , n − m = k − l, , otherwise , where ( n, m ) and ( k, l ) are two pairs of non-negative integers.Indeed, the concept of “ dual Toeplitz operator ” was first introduced by Stroethoff and Zheng [24].The algebraic and spectral properties of dual Toeplitz operator on the orthogonal complement of theBergman space space ( L a ) ⊥ were characterized in [25]. Ding, Wu and the second author studied theinvertibility and spectral properties of dual Toeplitz operators on ( L a ) ⊥ . Moreover, they obtained a YPONORMAL DUAL TOEPLITZ OPERATORS 3 necessary and sufficient condition for dual Toeplitz operators with bounded harmonic symbols to behyponormal on ( L a ) ⊥ , see [11] if necessary. In addition, dual Toeplitz operators on other functionspaces also have been investigated by many authors in recent years, see [3] and [18, 19, 20, 21] formore details.However, little is known about dual Toeplitz operators on the complement of the harmonicBergman space ( L h ) ⊥ . In 2015, Yang and Lu completely solved the commuting problem of two dualToeplitz operators on ( L h ) ⊥ with bounded harmonic symbols [26]. Therefore, they characterizedthe normal dual Toeplitz operator on ( L h ) ⊥ with a bounded harmonic symbol. Recently, Pengand the second author [23] characterized the boundedness, compactness, spectral structure andalgebraic properties of dual Toeplitz operators on ( L h ) ⊥ . As far as we know, the researches on thenormality and hyponormality of dual Toeplitz operators with non-harmonic symbols on ( L h ) ⊥ havenot been reported.In this paper, we first show that the dual Toeplitz operator on ( L h ) ⊥ with bounded symbolis hyponormal if and only if it is normal, see Theorem 2.2. Moreover, we obtain a necessaryand sufficient condition for the dual Toeplitz operator S ϕ with non-harmonic symbol of the form ϕ ( z ) = az n z m + bz n z m to be hyponormal on ( L h ) ⊥ , where n , n , m , m are nonnegativeintegers and a, b are complex constants, see Theorem 3.1 for the details. Moreover, we consider therank of the commutator of two dual Toeplitz operators with general bounded symbols in Section 4.We show that the rank of the commutator of two dual Toeplitz operators must be an even numberif the commutator has a finite rank, see Theorem 4.1 in the last section.2. Preliminaries
In this section, we will show that the normality and hyponormality for dual Toeplitz operator on( L h ) ⊥ are equivalent. To do so, we need the following well-known result about self-adjoint operators(see [9] if needed). Lemma 2.1.
Suppose that T is a self-adjoint operator on some Hilbert space H and h T h, h i = 0 for all h ∈ H . Then h T f, g i = 0 for any f, g ∈ H , i.e., T = 0 . The first main result of this paper is contained in the following theorem.
Theorem 2.2.
Suppose that ϕ ∈ L ∞ ( D ) . Then the dual Toeplitz operator S ϕ is hyponormal on ( L h ) ⊥ if and only if it is normal.Proof. Clearly, S ϕ is hyponormal if it is normal. Now we assume that S ϕ is hyponormal, i.e., S ∗ ϕ S ϕ − S ϕ S ∗ ϕ > . From the definition of the projection Q , we have Q ( ϕf ) = P ( ϕf ) + P ( ϕf ) − P ( ϕf )(0)= P ( ϕf ) + P ( ϕf ) − P ( ϕf )(0)= Q ( ϕf )and Q ( ϕf ) = Q ( ϕf )for any f ∈ ( L h ) ⊥ . CHONGCHAO WANG AND XIANFENG ZHAO
This gives that k S ϕ f k = k ( I − Q )( ϕf ) k = (cid:13)(cid:13)(cid:13) ( I − Q )( ϕf ) (cid:13)(cid:13)(cid:13) = (cid:13)(cid:13) S ∗ ϕ f (cid:13)(cid:13) and (cid:13)(cid:13) S ∗ ϕ f (cid:13)(cid:13) = k ( I − Q )( ϕf ) k = (cid:13)(cid:13)(cid:13) ( I − Q )( ϕf ) (cid:13)(cid:13)(cid:13) = (cid:13)(cid:13) S ϕ f (cid:13)(cid:13) for each f ∈ ( L h ) ⊥ . Thus we have 0 (cid:10) ( S ∗ ϕ S ϕ − S ϕ S ∗ ϕ ) f, f (cid:11) = k S ϕ f k − (cid:13)(cid:13) S ∗ ϕ f (cid:13)(cid:13) = (cid:13)(cid:13) S ∗ ϕ f (cid:13)(cid:13) − (cid:13)(cid:13) S ϕ f (cid:13)(cid:13) = −h ( S ∗ ϕ S ϕ − S ϕ S ∗ ϕ ) f , f i , where the last inequality follows from that f is also in ( L h ) ⊥ . This implies that h ( S ∗ ϕ S ϕ − S ϕ S ∗ ϕ ) f, f i = 0for all f ∈ ( L h ) ⊥ . By Lemma 2.1, we have S ∗ ϕ S ϕ − S ϕ S ∗ ϕ = 0, which means that S ϕ is normal. Thiscompletes the proof of Theorem 2.2. (cid:3) Combining the above theorem with [26, Corollary 2.1], we obtain the following characterizationfor the hyponormal dual Toeplitz operators with bounded harmonic symbols on ( L h ) ⊥ . Corollary 2.3.
Suppose that ϕ ∈ L ∞ ( D ) is a harmonic function. Then S ϕ is hyponormal if andonly if aϕ + bϕ is constant on D , where a and b are complex constants, not both zero. We end this section by discussing the hyponormality of the special dual Toeplitz operator S z n z m (where m, n ∈ N ) in the following proposition, which is useful for us to prove our main theorem inthe next section. Proposition 2.4.
Let n, m be two nonnegative integers, then S z n z m is hyponormal (normal) on ( L h ) ⊥ if and only if n = m .Proof. If n = m , then S z n z m is self-adjoint, and hence is hyponormal. Next, we will show m = n provided that S z n z m is hyponormal. In view of Theorem 2.2, we suppose that S z n z m is normal.Let k be an arbitrary positive integer such that k > max { m, n } . By (1.2) in Section 1, we have f k ( z ) := z k z − Q ( z k z ) = z k z − kk + 1 z k − and f k ∈ ( L h ) ⊥ . Thus we obtain(2.1) S z n z m ( f k )= S z n z m (cid:16) z k z − kk + 1 z k − (cid:17) = ( I − Q )( z n + k z m +1 ) − kk + 1 ( I − Q )( z n + k − z m )= z n + k z m +1 − n + k − mn + k + 1 z n + k − m − − kk + 1 z n + k − z m + k ( n + k − m )( k + 1)( n + k ) z n + k − m − = z n + k z m +1 − kk + 1 z n + k − z m − n ( n + k − m )( k + 1)( n + k )( n + k + 1) z n + k − m − . YPONORMAL DUAL TOEPLITZ OPERATORS 5
Similarly, we have(2.2) S ∗ z n z m ( f k ) = S z m z n ( f k )= z m + k z n +1 − kk + 1 z m + k − z n − m ( m + k − n )( k + 1)( m + k )( m + k + 1) z m + k − n − . Before computing k S z n z m ( f k ) k and k S ∗ z n z m ( f k ) k , some elementary calculations are required. Directcomputations give us the following equalities: k z n + k z m +1 k = k z m + k z n +1 k = 1 n + k + m + 2 , k z n + k − z m k = k z m + k − z n k = 1 n + k + m , k z n + k − m − k = 1 n + k − m , k z m + k − n − k = 1 m + k − n , D z n + k z m +1 , z n + k − z m E = D z m + k z n +1 , z m + k − z n E = 1 n + k + m + 1 , D z n + k − z m , z n + k − m − E = 1 n + k , D z m + k − z n , z m + k − n − E = 1 m + k , D z m + k z n +1 , z m + k − n − E = 1 m + k + 1 , D z n + k z m +1 , z n + k − m − E = 1 n + k + 1 . Combining the above equalities, we obtain that k S z n z m ( f k ) k = h S z n z m ( f k ) , S z n z m ( f k ) i = 1 n + k + m + 2 − k ( k + 1)( n + k + m + 1) − n ( n + k − m )( k + 1)( n + k )( n + k + 1) + k ( k + 1) ( n + k + m )+ 2 kn ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) and k S ∗ z n z m ( f k ) k = 1 m + k + n + 2 − k ( k + 1)( m + k + n + 1) − m ( m + k − n )( k + 1)( m + k )( m + k + 1) + k ( k + 1) ( m + k + n )+ 2 km ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) + m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) . Since S z n z m is normal, we have0 = h ( S ∗ z n z m S z n z m − S z n z m S ∗ z n z m ) f k , f k i = k S z n z m ( f k ) k − k S ∗ z n z m ( f k ) k = 2 kn ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) − n ( n + k − m )( k + 1)( n + k )( n + k + 1) − km ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) − m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) + 2 m ( m + k − n )( k + 1)( m + k )( m + k + 1) = − n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) . CHONGCHAO WANG AND XIANFENG ZHAO
It follows that(2.3) n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) = m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) for each k > max { m, n } . Multiplying both sides of (2.3) by( k + 1) ( n + k ) ( n + k + 1) ( m + k ) ( m + k + 1) gives us(2.4) n ( n + k − m )( m + k ) ( m + k + 1) = m ( m + k − n )( n + k ) ( n + k + 1) for every k > max { m, n } .Let p ( z ) = n ( n + z − m )( m + z ) ( m + z + 1) − m ( m + z − n )( n + z ) ( n + z + 1) . Then p is a polynomial and p has finitely many zeros in the complex plane. But (2.4) implies that p must be a zero polynomial. This yields that the coefficients of z is zero, that is, n − m = 0 . This implies that n = m , hence the proof of Proposition 2.4 is finished. (cid:3) dual Toeplitz operator with the symbol ϕ ( z ) = az n z m + bz n z m This section is devoted to establishing a necessary and sufficient condition for dual Toeplitzoperator with the symbol ϕ ( z ) = az n z m + bz n z m to be hyponormal (normal) on the orthogonalcomplement of the harmonic Bergman space. In view of Proposition 2.4, we need only to considerthe case of ab = 0 for the symbol ϕ . The main result of this section is the following theorem. Theorem 3.1.
Let ϕ ( z ) be the nonzero function az n z m + bz n z m , where a, b are nonzero con-stants and n , n , m , m are positive integers. Then the dual Toeplitz operator S ϕ is hyponormal(normal) on ( L h ) ⊥ if and only if one of the following conditions holds: (1) n = m = n = m ; (2) n = m , n = m , n = n and arg( a ) = arg( b ) ; (3) n = m , m = n , n = m and | a | = | b | . As the proof of Theorem 3.1 is complicated, we shall divide its proof into several lemmas.
Lemma 3.2.
Let ϕ ( z ) be the nonzero function z n z m + αz n z m , where α is a nonzero complexconstant and n , n , m , m are positive integers. Suppose that ( n − m )( n − m ) > . Then n − m = n − m if S ϕ is normal on ( L h ) ⊥ .Proof. Suppose, to the contrary, that n − m = n − m . Using the same argument as in theproof of Proposition 2.4, we first compute k S ϕ f k k and k S ∗ ϕ f k k , where k is an integer such that k > max { n , n , m , m } and f k is defined by f k ( z ) = z k z − Q ( z k z ) = (cid:16) z k z − kk + 1 z k − (cid:17) ∈ ( L h ) ⊥ . By (2.1) and (2.2) in the proof of Proposition 2.4, we have S z n z m ( f k ) = z n + k z m +1 − kk + 1 z n + k − z m − n ( n + k − m )( k + 1)( n + k )( n + k + 1) z n + k − m − ,S z n z m ( f k ) = z n + k z m +1 − kk + 1 z n + k − z m − n ( n + k − m )( k + 1)( n + k )( n + k + 1) z n + k − m − , YPONORMAL DUAL TOEPLITZ OPERATORS 7 S ∗ z n z m ( f k ) = z m + k z n +1 − kk + 1 z m + k − z n − m ( m + k − n )( k + 1)( m + k )( m + k + 1) z m + k − n − and S ∗ z n z m ( f k ) = z m + k z n +1 − kk + 1 z m + k − z n − m ( m + k − n )( k + 1)( m + k )( m + k + 1) z m + k − n − . Since n − m = n − m , we observe by (1.3) that S z n z m ( f k ) ⊥ S z n z m ( f k ) and S ∗ z n z m ( f k ) ⊥ S ∗ z n z m ( f k ) . This leads to k S ϕ f k k = k S z n z m ( f k ) k + | α | k S z n z m ( f k ) k = 1 n + k + m + 2 − k ( k + 1)( n + k + m + 1) − n ( n + k − m )( k + 1)( n + k )( n + k + 1) + k ( k + 1) ( n + k + m )+ 2 kn ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + | α | n + k + m + 2 − | α | k ( k + 1)( n + k + m + 1) − | α | n ( n + k − m )( k + 1)( n + k )( n + k + 1) + | α | k ( k + 1) ( n + k + m )+ 2 | α | kn ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + | α | n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) . Similarly, we also have k S ∗ ϕ f k k = k S ∗ z n z m ( f k ) k + | α | k S ∗ z n z m ( f k ) k = 1 m + k + n + 2 − k ( k + 1)( m + k + n + 1) − m ( m + k − n )( k + 1)( m + k )( m + k + 1) + k ( k + 1) ( m + k + n )+ 2 km ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) + m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) + | α | m + k + n + 2 − | α | k ( k + 1)( m + k + n + 1) − | α | m ( m + k − n )( k + 1)( m + k )( m + k + 1) + | α | k ( k + 1) ( m + k + n )+ 2 | α | km ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) + | α | m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) . CHONGCHAO WANG AND XIANFENG ZHAO
Combining the above two equalities gives(3.1) 0 = (cid:10) ( S ∗ ϕ S ϕ − S ϕ S ∗ ϕ )( f k ) , f k (cid:11) = k S ϕ f k k − k S ∗ ϕ f k k = k S z n z m ( f k ) k + | α | k S z n z m ( f k ) k − k S ∗ z n z m ( f k ) k − | α | k S ∗ z n z m ( f k ) k = − n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) + −| α | n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + | α | m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) . Therefore, we conclude that(3.2) n ( n + k − m )( n + k ) ( n + k + 1) + | α | n ( n + k − m )( n + k ) ( n + k + 1) = m ( m + k − n )( m + k ) ( m + k + 1) + | α | m ( m + k − n )( m + k ) ( m + k + 1) for any integer k > max { n , n , m , m } .In order to deduce the desired conclusion, we will use the same method as the one in the proofof Proposition 2.4. Multiplying both sides of (3.2) by[( n + k )( n + k + 1)( n + k )( n + k + 1)( m + k )( m + k + 1)( m + k )( m + k + 1)] , now we obtain a polynomial p ( z ) with infinitely many zeros. Thus p must be a zero polynomialand (3.2) holds for all k ∈ N . By substituting k = 0 into (3.2), we get n − m ( n + 1) + | α | ( n − m )( n + 1) = − n − m ( m + 1) − | α | ( n − m )( m + 1) . (3.3)If n = m or n = m , then we have n − m = n − m = 0 , since α = 0. This is a contradiction. Otherwise, we have by our assumption that( n − m )( n − m ) > . In this case, the left-hand side and the right-hand side of (3.3) have opposite signs. This is alsoimpossible. Thus we finish the proof of this lemma. (cid:3)
Under the assumption in Lemma 3.2, we can further show that n = m and n = m . Lemma 3.3.
Let n , n , m , m be positive integers such that ( n − m )( n − m ) > . Suppose that ϕ ( z ) = z n z m + αz n z m is a nonzero function, where α is a nonzero constant. Then n = m and n = m if S ϕ is normal on ( L h ) ⊥ .Proof. Recall that we have shown in the above lemma that n − m = n − m . Using the same notation as the one in Lemma 3.2, we will calculate the norms k S ϕ f k k and k S ∗ ϕ f k k .To do this, the following equalities are needed (which can be calculated by using (1.3)): h z n + k z m +1 , z n + k z m +1 i = h z m + k z n +1 , z m + k z n +1 i = 2 n + m + n + m + 2 k + 4 , h z n + k z m +1 , z n + k − z m i = h z m + k z n +1 , z m + k − z n i YPONORMAL DUAL TOEPLITZ OPERATORS 9 = h z n + k − z m , z n + k z m +1 i = h z m + k − z n , z m + k z n +1 i = 2 n + m + n + m + 2 k + 2 , h z n + k − z m , z n + k − z m i = h z m + k − z n , z m + k − z n i = 2 n + m + n + m + 2 k , h z n + k z m +1 , z n + k − m − i = 1 n + k + 1 , h z n + k − z m , z n + k − m − i = 1 n + k , h z n + k − m − , z n + k z m +1 i = 1 n + k + 1 , h z n + k − m − , z n + k − z m i = 1 n + k , h z m + k z n +1 , z m + k − n − i = 1 m + k + 1 , h z m + k − z n , z m + k − n − i = 1 m + k , h z m + k − n − , z m + k z n +1 i = 1 m + k + 1 , h z m + k − n − , z m + k − z n i = 1 m + k . Since n − m = n − m , we also have h z n + k − m − , z n + k − m − i = 1 n − m + k = 1 n − m + k and h z m + k − n − , z m + k − n − i = 1 m − n + k = 1 m − n + k . From the expressions of S z n z m ( f k ), S z n z m ( f k ), S ∗ z n z m ( f k ) and S ∗ z n z m ( f k ) obtained inLemma 3.2, we observe that h S z n z m ( f k ) , S z n z m ( f k ) i and h S ∗ z n z m ( f k ) , S ∗ z n z m ( f k ) i are bothreal. Thus we have k S ϕ f k k = (cid:10) S z n z m ( f k ) + αS z n z m ( f k ) , S z n z m ( f k ) + αS z n z m ( f k ) (cid:11) = k S z n z m ( f k ) k + ( α + α ) h S z n z m ( f k ) , S z n z m ( f k ) i + | α | k S z n z m ( f k ) k and k S ∗ ϕ f k k = (cid:10) S ∗ z n z m ( f k ) + αS ∗ z n z m ( f k ) , S ∗ z n z m ( f k ) + αS ∗ z n z m ( f k ) (cid:11) = k S ∗ z n z m ( f k ) k + ( α + α ) h S ∗ z n z m ( f k ) , S ∗ z n z m ( f k ) i + | α | k S ∗ z n z m ( f k ) k . It follows that(3.4) 0 = k S ϕ f k k − k S ∗ ϕ f k k = k S z n z m ( f k ) k + | α | k S z n z m ( f k ) k − k S ∗ z n z m ( f k ) k − | α | k S ∗ z n z m ( f k ) k + ( α + α ) (cid:2) h S z n z m ( f k ) , S z n z m ( f k ) i − h S ∗ z n z m ( f k ) , S ∗ z n z m ( f k ) i (cid:3) = − n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) + −| α | n ( n + k − m )( k + 1) ( n + k ) ( n + k + 1) + | α | m ( m + k − n )( k + 1) ( m + k ) ( m + k + 1) + − ( α + α ) n n ( n + k − m )( k + 1) ( n + k )( n + k + 1)( n + k )( n + k + 1)+ ( α + α ) m m ( m + k − n )( k + 1) ( m + k )( m + k + 1)( m + k )( m + k + 1) , where the third equality comes from (3.1). Note that the above equality holds for all k > max { n , n , m , m } . Using the same techniques as in the proof of Lemma 3.2, we see that (3.4)is, in fact, an identity. Thus (3.4) can be reduced to( m − n ) "(cid:12)(cid:12)(cid:12)(cid:12) n + 1 + αn + 1 (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) m + 1 + αm + 1 (cid:12)(cid:12)(cid:12)(cid:12) = 0if we take k = 0. This implies that n = m or (cid:12)(cid:12)(cid:12)(cid:12) n + 1 + αn + 1 (cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) m + 1 + αm + 1 (cid:12)(cid:12)(cid:12)(cid:12) = 0 . Suppose that n = m , then we would have − α = n + 1 n + 1 = m + 1 m + 1 , to obtain n + m + n m = n + m + n m . Since n − m = n − m , we have n m = n m . Thus( n + m ) = n + m + 2 n m = ( n + m ) = n + m + 2 n m , which gives that n − m = n − m . As n − m = n − m = 0, we have n + m = n + m . Butthis implies that n = n and m = m . In this case, ϕ can be rewritten as ϕ ( z ) = (1 + α ) z n z m . By our assumption that ϕ is a nonzero function, we now conclude by Proposition 2.4 that n = m .This is a contradiction. Thus we obtain that n = m and n = m , to complete the proof ofLemma 3.3. (cid:3) For functions of the form ϕ ( z ) = z n z m + αz n z m , we will discuss in the next lemma for thecase that ( n − m )( n − m ) < Lemma 3.4.
Let ϕ ( z ) = z n z m + αz n z m be a nonzero function, where α is a nonzero complexconstant and n , n , m , m are positive integers. Suppose that ( n − m )( n − m ) < . Then we have | α | = 1 , n = m and n = m if S ϕ is normal on ( L h ) ⊥ .Proof. Without loss of generality, we may assume that n > m and n < m . Since n − m = n − m , we observe that (3.2) in the proof of Lemma 3.2 is still valid in this case,i.e., n ( n + k − m )( n + k ) ( n + k + 1) + | α | n ( n + k − m )( n + k ) ( n + k + 1) = m ( m + k − n )( m + k ) ( m + k + 1) + | α | m ( m + k − n )( m + k ) ( m + k + 1) for any integer k > max { n , n , m , m } . For convenience, we write a ( k ) = n ( n + k − m )( n + k ) ( n + k + 1) , b ( k ) = n ( n + k − m )( n + k ) ( n + k + 1) ,c ( k ) = m ( m + k − n )( m + k ) ( m + k + 1) , d ( k ) = m ( m + k − n )( m + k ) ( m + k + 1) . YPONORMAL DUAL TOEPLITZ OPERATORS 11
Let F ( k ) = a ( k ) + | α | b ( k ) − c ( k ) − | α | d ( k )and G ( k ) = (cid:2) ( n + k )( n + k + 1)( n + k )( n + k + 1)( m + k )( m + k + 1)( m + k )( m + k + 1) (cid:3) . Then F ( k ) = 0 for all k > max { n , n , m , m } . Let H ( x ) = F ( x ) G ( x ) . Note that H ( x ) is a polynomial of x and H ( x ) has infinitely many zeros. This yields that H ( x ) isa zero polynomial. Considering the coefficient of x in H ( x ), we have that(3.5) n + | α | n = m + | α | m . Now we are going to show that n = m . If not, we first assume that n > m . Recalling that n > m and n < m , we obtian n > max { n , m , m } . This gives that b ( − n − G ( − n −
1) = c ( − n − G ( − n − d ( − n − G ( − n − . It follows that F ( − n − G ( − n −
1) = a ( − n − G ( − n −
1) = 0 . On the other hand, a ( − n − G ( − n −
1) = − n (1 + m ) (cid:18) g ( x )( n + x ) ( n + 1 + x ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) x = − n − = 0 . The contradiction gives n m . Similarly, we can show that n > m . So we have n = m .To finish the proof, it remains to show that m = n . If m > n , then n < min { n , m , m } . This implies that a ( − n ) G ( − n ) = c ( − n ) G ( − n ) = d ( − n ) G ( − n ) = 0 , to obtain F ( − n ) G ( − n ) = | α | b ( − n ) G ( − n ) = 0 . However, b ( − n ) G ( − n ) = − m n (cid:18) G ( x )( n + x ) ( n + 1 + x ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) x = − n = 0 , which is a contradiction, so m n . Using the same argument as above, we can also show that m > n . Thus we get n = m . From (3.5) and ( n − m )( n − m ) <
0, we conclude that | α | = 1.This completes the proof of Lemma 3.4. (cid:3) Before presenting the proof of Theorem 3.1, one more lemma is needed.
Lemma 3.5.
Suppose that ϕ = | z | n + α | z | m , where α is a constant, m and n are distinct integers.Then S ϕ is normal if and only if α is real. Proof. If α ∈ R , then S ϕ is self-adjoint, and so is normal. Now we assume that S ϕ is normal. Itfollows that0 = S ∗ ϕ S ϕ − S ϕ S ∗ ϕ = ( S | z | n + αS | z | m )( S | z | n + αS | z | m ) − ( S | z | n + αS | z | m )( S | z | n + αS | z | m )= ( α − α )( S | z | n S | z | m − S | z | m S | z | n ) . For f ( z ) = ( I − Q )( | z | ) = (cid:16) | z | − (cid:17) ∈ ( L h ) ⊥ , we have 0 = ( α − α )( S | z | n S | z | m − S | z | m S | z | n )( f )= ( α − α ) n ( I − Q ) (cid:2) | z | n ( I − Q )( | z | m f ) (cid:3) − ( I − Q ) (cid:2) | z | m ( I − Q )( | z | n f ) (cid:3) o = ( α − α )( I − Q ) (cid:2) | z | m Q ( | z | n f ) − | z | n Q ( | z | m f ) (cid:3) . On the one hand, | z | m Q ( | z | n f ) − | z | n Q ( | z | m f ) = | z | m Q ( | z | n +1) − | z | n ) − | z | n Q ( | z | m +1) − | z | m )= | z | m (cid:18) n + 2 − n + 1) (cid:19) − | z | n (cid:18) m + 2 − m + 1) (cid:19) = n | z | m n + 1)( n + 2) − m | z | n m + 1)( m + 2) . On the other hand, we have Q (cid:2) | z | m Q ( | z | n f ) − | z | n Q ( | z | m f ) (cid:3) = Q (cid:20) n | z | m n + 1)( n + 2) − m | z | n m + 1)( m + 2) (cid:21) = n n + 1)( n + 2)( m + 1) − m m + 1)( m + 2)( n + 1)= n − m ( n + 1)( n + 2)( m + 1)( m + 2) . Since m = n , we derive that( I − Q ) (cid:2) | z | m Q ( | z | n f ) − | z | n Q ( | z | m f ) (cid:3) = 0 , to obtain α = α . This completes the proof of Lemma 3.5. (cid:3) We are now in position to prove Theorem 3.1. Let ϕ ( z ) = az n z m + bz n z m . Recall that weneed to show that S ϕ is is normal on ( L h ) ⊥ if and only if one of the following holds:(1) n = m = n = m ;(2) n = m , n = m , n = n and arg( a ) = arg( b );(3) n = m , n = m , n = m and | a | = | b | . Proof of Theorem 3.1.
Suppose that (1) or (2) holds. We conclude that S ϕ is normal by Propo-sition 2.4 and Lemma 3.5, respectively. If (3) holds, then there exists θ ∈ [0 , π ] such that b = ae iθ .In this case, we have S ∗ ϕ S ϕ − S ϕ S ∗ ϕ = | a | ( S z n z m + e − iθ S z n z m )( S z n z m + e iθ S z n z m ) − | a | ( S z n z m + e iθ S z n z m )( S z n z m + e − iθ S z n z m )= | a | e − iθ ( e iθ S z n z m + S z n z m )( e − iθ S z n z m + S z n z m ) e iθ − | a | ( S z n z m + e iθ S z n z m )( S z n z m + e − iθ S z n z m ) YPONORMAL DUAL TOEPLITZ OPERATORS 13 = 0 , which implies that S ϕ is a normal operator.The necessary part of the theorem follows from Lemmas 3.3, 3.4 and 3.5 immediately. Thiscompletes the proof of Theorem 3.1. (cid:3) Finite rank commutator
In the last section, we will study the rank of the commutator of two dual Toeplitz operators on( L h ) ⊥ . For f, g ∈ ( L h ) ⊥ , recall that the rank-one operator f ⊗ g on ( L h ) ⊥ is defined by( f ⊗ g )( h ) = h h, g i f, h ∈ ( L h ) ⊥ . Our main result of this section is the following theorem, which is analogous to the characterizationfor finite rank commutator of Toeplitz operators on the harmonic Bergman space [8, Theorem 2.2].
Theorem 4.1.
Let ϕ and ψ be in L ∞ ( D ) . If the commutator [ S ϕ , S ψ ] = S ϕ S ψ − S ψ S ϕ has a finite rank, then the rank of [ S ϕ , S ψ ] is an even number.Proof. Suppose that the rank of [ S ϕ , S ψ ] is n . Then we can find two linearly independent sequences { s i } ni =1 and { t i } ni =1 in (cid:0) L h (cid:1) ⊥ , which satisfy that[ S ϕ , S ψ ] = n X i =1 s i ⊗ t i . Define an anti-unitary operator U on (cid:0) L h (cid:1) ⊥ as the following: U f = f , f ∈ (cid:0) L h (cid:1) ⊥ . Noting that ( I − Q )( f ) = ( I − Q )( f ) for any f ∈ L ( D ), we have( S ϕ S ψ − S ψ S ϕ ) ∗ ( h ) = ( S ψ S ϕ − S ϕ S ψ )( h )= ( S ψ S ϕ − S ϕ S ψ )( h )= U ( S ψ S ϕ − S ϕ S ψ ) U ( h )= − U [ S ϕ , S ψ ] U ( h )for each h ∈ (cid:0) L h (cid:1) ⊥ . This gives that n X i =1 s i ⊗ t i ! ∗ = − U n X i =1 s i ⊗ t i ! U, which is equivalent to n X i =1 t i ⊗ s i = − n X i =1 s i ⊗ t i . Thus we can find { a ij } nj =1 ⊂ C such that t i = n X j =1 a ij s j for each t i . Then we have n X i =1 t i ⊗ s i = n X i =1 ( n X j =1 a ij s j ) ⊗ s i = n X i =1 n X j =1 a ij s j ⊗ s i = X j =1 s j ⊗ n X i =1 a ij s i = − n X i =1 s i ⊗ t i . It follows that t i = − n X i =1 a ji s j , to obtain t i = − n X i =1 a ji s j = n X j =1 a ij s j . Since { s , s , · · · , s n } is linearly independent, we conclude that a ij = − a ji . Denote the matrix ( a ij ) n × n by A , then det( A ) = 0. Hence A = − A T anddet( A ) = ( − n · det( A T ) = ( − n · det( A ) , where A T is the transpose of the matrix A . This implies that ( − n = 1 as det( A ) = 0, so n mustbe even. This completes the proof of Theorem 4.1. (cid:3) Acknowledgment.
This work was partially supported by NSFC (grant number: 11701052). Thesecond author was partially supported by the Fundamental Research Funds for the Central Uni-versities (grant numbers: 2020CDJQY-A039, 2020CDJ-LHSS-003).
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