Algebraic Properties of Dual Toeplitz Operators on Harmonic Hardy Space over Polydisc

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1 Journal of Mathematical Research with Applications Nov., 2016, Vol. 36, No. 6, pp DOI: /j.issn: Algebraic Properties of Dual Toeplitz Operators on Harmonic Hardy Space over Polydisc Yufeng LU, Fatemeh AZARI KEY, Hewei YANG, Liu LIU School of Mathematical Sciences, Dalian University of Technology, Liaoning , P. R. China Abstract In this paper, we introduce the harmonic Hardy space on T n and study some algebraic properties of dual Toeplitz operator on the harmonic Hardy space on T n. Keywords Hardy space; Toeplitz operator; spectrum; semi-commutative MR(2010) Subject Classification 47B35 1. Introduction Let T be the unit circle on the complex plane C. For a fixed positive integer n 1, the C n and T n are the cartesian products of n copies of C and T, respectively. Denote by z = (z 1,..., z n ) the coordinates on C n. Let dσ be the normalized Haar measure on T n and L 2 (T n ) be the square integral functions with respect to dσ. The Hardy space H 2 (T n ) is the closure of analytic polynomials in L 2 (T n ), that is H 2 (T n ) = clos{p(z 1,..., z n ) : p is anlytic polynomials}. In the setting of classical Hardy space on T, it is well known that H 2 (T) + H 2 (T) = L 2 (T), where ( ) is the complex conjugate. However, in the higher dimension (n 2), the situation is completely different, indeed, H 2 (T n ) + H 2 (T n ) is much smaller than L 2 (T n ). Denote h 2 (T n ) = H 2 (T n ) + H 2 (T n ) and call it the harmonic Hardy space on T n. The reader may not confuse that h 2 (T n ) does not contain all harmonic functions. In the whole paper, P denotes the orthogonal projection from L 2 (T n ) onto h 2 (T n ) and Q = 1 P. Let L (T n ) be the set of essentially bounded measurable functions on T n. For φ L (T n ), the Toeplitz operator T φ on h 2 (T n ) is defined by T φ f = P (φf), f h 2 (T n ). The Toeplitz operators on analytic and harmonic function spaces have been widely studied [1]. Received September 19, 2016; Accepted October 12, 2016 Supported by the National Natural Science Foundation of China (Grant Nos ; ). * Corresponding author address: lyfdlut@dlut.edu.cn (Yufeng LU); fa.azari@mail.dlut.edu.cn (Fatemeh AZARI KEY); beth.liu@ dlut.edu.cn (Liu LIU)

2 704 Yufeng LU, Fatemeh AZARI KEY, Hewei YANG and et al. The Hankel operator and dual Toeplitz operator can also be defined as follows: H φ : h 2 (T n ) (h 2 (T n )) H φ f = Q(φf), f h 2 (T n ). S φ : (h 2 (T n )) (h 2 (T n )) S φ f = Q(φf ), f h 2 (T n ). One can check that Hφ f = P (φf ), f h2 (T n ), and S φ (f) 2 = Q(φ f) 2 φ f 2 φ f 2, where is the essential sup norm and 2 is the norm of L 2 (T n ). The following algebraic properties of dual Toeplitz operators are also easy to check. For φ, ψ L (T n ), α, β C, we have S φ = S φ, S αφ+βψ = αs φ + β S ψ. For dual Toeplitz operator, Stroethoff and Zheng [2] studied algebraic and spectral properties of dual Toeplitz operators with general symbols on the Bergman space of the unit disk. Lu [3] characterized commuting dual Toeplitz operators on the Bergman space of the unit ball with pluriharmonic symbols. Lu and Shang [4] studied commutativity, essential commutativity and essential semi-commutativity of dual Toeplitz operators with general symbols on the polydisk. Lu and Yang [5] studied the properties of dual Toeplitz operators with general symbols on the weighted Bergman space of the unit ball. In the several-variable situation, the study of dual Toeplitz is much more complicated [6 8]. The Toeplitz operator, Hankel operator and dual Toeplitz operator have close relationships through the multiplication operators on L 2 (T n ). Under the decomposition L 2 (T n ) = h 2 (T n ) (h 2 (T n )), the multiplication operator M φ, φ L (T n ) can be represented as follows ( ) M φ = For φ, ψ L (T n ), the identity M φ M ψ = M φψ = M ψ M φ implies that Equation (1) will be used frequently. 2. Properties of spectrum T φ H φ H φ S φ. S φψ = S φ S ψ + H φ H ψ = S ψs φ + H ψ H φ. (1) Characterizations of spectrum is one of important properties for bounded linear operators. For α = (α 1, α 2,..., α n) Z n +, we denote z α = z α 1 1 zα 2 2 z α n n. Lemma 2.1 Let ψ C(T n ), where C(T n ) is the set of continuous functions on T n. Then for

3 Algebraic properties of Dual Toeplitz operators on harmonic Hardy space over polydisc 705 any z T n, we have 1 ψ(ζ) 1 + z 2m dσ(ζ) ψ(z), as m, λ m T n where λ m = T n z α ζ,z n z α ζ,z n+1 2m dσ(ζ), α = (2, 0, 0,..., 0). Proof Firstly, let us show that z 2m dσ(ζ) 0, as m, (2) λ m T n \V z(ε) for any neighborhood of z with following form, On T n \V z (ε), whence V z (ε) = { ζ T n : 1 z < ε }. n + 1 ( z 1 + z ) sup 1 + z = ρ < 2, (3) ζ T n \V z(ε) n + 1 T n \V z(ε) 1 + z 2m dσ(ζ) ρ 2m σ(t n \V z ). (4) If 0 < δ < 2 ρ < 1, considering another neighborhood V z (δ) of z, we can get λ m 1 + z 2m dσ(ζ) (1 δ) 2 (2 δ) 2m σ(v δ ). (5) V z(δ) From inequalities (3) (5), we see that, as m, 1 z z 2m dσ(ζ) c ( ρ ) 2m 0. 2 δ Therefore, λ m 1 λ m T n \V z (ε) V z(ε) 1 + z 2m dσ(ζ) 1, as m. (6) Since λ m is independent of z in T n, we have 1 ψ(ζ) 1 + z 2m dσ(ζ) ψ(z) λ m T n sup ψ(ζ) ψ(z) z 2m dσ(ζ)+ ζ T n \V z (ε) λ m T n \V z (ε) sup ψ(ζ) ψ(z) z 2m dσ(ζ). ζ V z (ε) λ m V z(ε) The continuity of ψ with conclusion (2) and (6) yields the desired result. Lemma 2.2 Let φ L (T n ). If S φ is invertible in (h 2 (T n )), then φ is invertible in L (T n ). Proof The assumption that S φ is invertible implies that there is a constant k > 0 satisfying S φ f k f, f (h 2 (T n )). Considering the projection has norm 1, we can see φ f k f, f (h 2 (T n )). (7)

4 706 Yufeng LU, Fatemeh AZARI KEY, Hewei YANG and et al. Particularly, for ( f (z) = z 1 + z ) m, ζ T n, m 1, α = (2, 0, 0,..., 0), which is clearly an element of (h 2 (T n )), the inequality (7) yields φ(z) z 2m dσ(z) k 2 λ m. T n It follows that for any nonnegative ψ C(T n ), one has 1 φ(z) 2 ψ(ζ) 1 + z 2m dσ(z)dσ(ζ) λ m T n T n k 2 ψ(ζ)dσ(ζ) = k 2 ψ(z)dσ(z). T n T n Hence, invoking Lemma 2.1, we obtain T n ( φ(z) 2 k 2 )ψ(z)dσ(z) 0, which implies that φ(z) k > 0 a.e., in T n, hence φ(z) is invertible in L (T n ). An immediate consequence is the following spectral inclusion theorem. But firstly, let us denote by R( f ) the essential range of the essentially bounded function f, and by σ(t ), r(t ) respectively the spectrum and the spectral radius of an operator T. Theorem 2.3 If φ is in L (T n ), then R(φ) σ(s φ ). 3. Semi-commuting dual Toeplitz operator When will the product of two dual Toeplitz operators be semi-commutative? This question has been solved well in Hardy space and harmonic Bergman space [2 5]. But the crucial question is what are the conditions we need if the product of two dual Toeplitz operators with the special form we have discussed is a dual Toeplitz operator in harmonic Hardy space. several theorems have been observed. H ψ The following Equation (1) suggests that S φ and S ψ commute if φ or ψ is constant, that is H φ = 0 or = 0. If a non-trivial linear combination of φ and ψ is constant, they do commute as well. Lemma 3.1 If φ H (D n ), then H φ ((h2 ) ) H 2, H φ((h 2 ) ) H 2. Proof Let f = k=1 α, β =k a α,βz α z β, the parameters α, β Z n +. To satisfy the condition f (h 2 ), for any positive integer k, there exist i and j such that α i > β i, α j < β j. Let Hφ ( f ) = g + h, g be analytic, and h be co-analytic, then we can get h is constant (If not, then α i + c β i, c 0, this is a contradiction), which yields the first desired result. Similarly, the second result can be proved. Combining this lemma with Eq. (1), we get the following proposition. Proposition 3.2 If the symbols φ and ψ are both analytic or co-analytic, then the dual Toeplitz operators S φ and S ψ are commutative, i.e., S φ S ψ = S ψ S φ.

5 Algebraic properties of Dual Toeplitz operators on harmonic Hardy space over polydisc 707 Proof If φ and ψ are both analytic, by Lemma 3.1, for all f (h 2 (T n )), there exists g H 2 such that H φ H ( f ) = H ψ φ(g). Since the product of two analytic functions is still analytic, H φ (g) = Q(φg) 0, namely, H ψ Hφ 0, which is equivalent to the fact that S φψ = S φ S ψ = S ψ S φ. The same result also can be proved when φ and ψ are both co-analytic through a similar method. Theorem 3.3 Suppose that φ = f + z m1 w n1, ψ = g + z m2 w n2, where f, g H (D 2 ), and the parameters m 1, m 2, n 1, n 2 are non-zero positive integers. Then S φ S ψ = S φψ if and only if both f and g are constants. Proof By the definition of dual Toeplitz operator, we can get that S φ S ψ = S f S g + S f S z m 2 w n 2 + S z m 1 w n 1 S g + S z m 1 w n 1 S z m 2 w n 2 = S φψ = S f g+f z m 2 w n 2 +z m 1 w n 1 g+z m 1 w n 1 z m 2 w n 2. (8) Equation (1) and Proposition 3.2 can reduce Eq. (8) to the following formula: which is equivalent to S f z m 2 w n 2 H f H z m 2 w n 2 + S z m 1 w n 1 g H z m 1 w n 1 H g = S f z m 2 w n 2 + S z m 1 w n 1 g (9) H f H z m 2 w n 2 + H z m 1 w n 1 H g = 0. (10) For the reason of the Eqs. (8) (10), S φ and S ψ are semi-commutative if and only if Eq. (10) is set up on (h 2 (T 2 )). Let f = i 0,j 0 a ij z i w j, g = k 0,l 0 b kl z k w l, (z, w) T 2. Firstly, assume that f and g are both constants. It is clear that H f = H g = Hg = 0, so the Eq. (10) is set up, whence S φ S ψ =S φψ. Now we assume that S φ S ψ =S φψ. A little more computation gives that H f Hz m 2 w n 2 (z α w β ) = H z m 1 w n 1 Hg (z α w β ) = H f Hz m 2 w n 2 (z α w β ) = 0, α > m 2, a ij z α m2+i w β+n2 j + i>m 2 α 0 j<β+n 2 b kl z α m1+k w β+n1 l, α > m 1, k 0 β l<β+n 1 b kl z α m1+k w β+n1 l + k>m 1 α β l<β+n 1 0, β > n 2, a ij z i α m 2 w n2 β j + i>m 2 +α 0 j<n 2 β 0 i<m 2 α j>β+n 2 a ij z m2 α i w j β n2, α m 2, (11) 0 k<m 1 α l>β+n 1 b kl z m1 α k w l β n1, α m 1, 0 i<m 2 +α j>n 2 β (12) a ij z m 2+α i w β n 2+j, β n 2, (13)

6 708 Yufeng LU, Fatemeh AZARI KEY, Hewei YANG and et al. b kl z α+m1 k w β n1+l, β > n 1, H z m 1 w n 1 Hg (z α w β α k<α+m 1 l 0 ) = b kl z k α m1 w n1 β l + b kl z α+m1 k w β n1+l, β n 1. We distinguish several cases. k>m 1 +α 0 l<n 1 β α k<α+m 1 l>n 1 β Case 1 m 1 = m 2 = m 1, n 1 = n 2 = n 1. For β > n, since (13) + (14) = 0, we can get 0 + b kl z α+m k w β n+l = 0. α k<α+m l 0 By the linear independence and the arbitrariness of α, it follows (14) b kl = 0, k > 0, l 0, (15) which means that g is only about w. For β = n, (13) + (14) = 0 implies a ij z m+α i w j = 0, hence which means that f is only about z. hence 0 i<m+α j>0 a ij = 0, i 0, j > 0, (16) If α = m in the case (11) + (12) = 0, together with the conclusion (16), a i0 z i w β+n = 0, i>0 j=0 a i0 = 0, i 1. (17) If α > m in the case (11) + (12) = 0, together with the conclusion (16), then by the same way, we can get so 0 + k=0 β l<β+n b 0l z α m w β+n l = 0, Considering the conclusions (15) (18), both f and g are constants. b 0l = 0, l > 0. (18) Case 2 m 1 = m 2 = m 1, n 2 n 1. Without loss of generality, we can assume n 2 > n 1 1. For β > n 2 > n 1, it follows from (13) + (14) = 0 that 0 + b kl z α+m k w β n1+l = 0, so α k<α+m l 0 b kl = 0, k > 0, l 0. (19)

7 Algebraic properties of Dual Toeplitz operators on harmonic Hardy space over polydisc 709 For β = n 2 > n 1, a ij z m+α i w j + 0 i<m+α j>0 which means that α k<α+m l 0 b kl z α+m k w β n 1+l = We also need to consider the condition (11) + (12) = 0. hence 0 i<m+α j>0 a ij z m+α i w j = 0, a ij = 0, i 0, j > 0. (20) For α = m, together with (19) and (20), the equation can be reduced to a i0 z i w β+n = 0, i>0 j=0 a i0 = 0, i > 0. (21) For α > m 1, b 0l = 0, l > 0. (22) By the conclusions (19) (22), the result that f and g are both constants can be observed immediately. Case 3 m 1 m 2, n 1 = n 2 = n 1. It is easy to complete the proof by the similar way used in Case 2. Case 4 m 1 m 2, n 1 n 2. Without loss of generality, we can assume that m 2 > m 1 1, n 1 > n 2 1. For α > m 2 > m 1, which means that 0 + α m b klz 1+k w β+n1 l = 0, k 0 β l<β+n 1 b kl = 0, k 0, l > 0. (23) For α = m 2 > m 1, together with (23), the following equation is checked easily: which implies that 0 + a ijz i w β+n2 j = 0, i>0 0 j<β+n 2 a ij = 0, i > 0, j 0. (24) Through a similar discussion of β, we can get a 0j = 0, j > 0; b k0 = 0, k > 0. (25) Therefore, the proof can be completed together with the conclusions (23) (25). In Theorem 3.3, each parameter should be non-zero positive integer. Next theorem considers the case of m 1 = m 2 = 0.

8 710 Yufeng LU, Fatemeh AZARI KEY, Hewei YANG and et al. Theorem 3.4 Suppose that φ = f+w n 1, ψ = g+w n 2, where f, g H (D 2 ), and the parameters n 1, n 2 are non-zero positive integers. Then the following conclusions can be established: (i) If n 1 = n 2, S φ S ψ = S φψ if and only if f and g are both only about z, and f +g = a+bz; (ii) If n 1 n 2, S φ S ψ = S φψ if and only if f = a 0 + a 1 z, g = b 0 + b 1 z. Proof Let m 1 = m 2 = 0 in the equation that (11) + (12) = (13) + (14) = 0. Then the theorem can be proved by a discussion of α and β as we have done before in this paper. References [1] A. BROWN, P. R. HALMOS. Algebraic properties of toeplitz operators. J. Reine Angew. Math., 1964, 213(1): [2] K. STROETHOFF, Dechao ZHENG. Algebraic and spectral properties of dual Toeplitz operators. Trans. Amer. Math. Soc., 2002, 354(6): [3] Yufeng LU. Commuting dual Toeplitz operators with pluriharmonic symbols. J. Math. Anal. Appl., 2005, 302(1): [4] Yufeng LU, Shuxia SHANG. Commuting dual Toeplitz operators on the polydisk. Acta Math. Sin. (Engl. Ser.), 2007, 23(5): [5] Jingyu YANG, Yufeng LU. Commuting dual Toeplitz operators on the harmonic Bergman space. Sci. China Math., 2015, 58(7): [6] Tao YU, Shiyue WU. Algebraic properties of dual Toeplitz operators on the orthogonal complement of the Dirichlet space. Acta Math. Sin. (Engl. Ser.), 2008, 24(11): [7] Y. J. LEE. Commuting dual Toeplitz operators on the polydisk. J. Math. Anal. Appl., 2008, 341(1): [8] G. HOCINE. Dual Toeplitz operators on the sphere. Acta Math. Sin. (Engl. Ser.), 2013, 29(9):