A New Necessary Condition for the Hyponormality of Toeplitz Operators on the Bergman Space

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1 A New Necessary Condition for the Hyponormality of Toeplitz Operators on the Bergman Space Željko Čučković Department of Mathematics, University of Toledo, Toledo, Ohio Raúl E Curto Department of Mathematics, The University of Iowa, Iowa City, Iowa Abstract A well known result of C Cowen states that, for a symbol ϕ L, ϕ f + g (f,g H 2 ), the Toeplitz operator T ϕ acting on the Hardy space of the unit circle is hyponormal if and only if f = c + T hg, for some c C, h H, h In this note we consider possible versions of this result in the Bergman space case Concretely, we consider Toeplitz operators on the Bergman space of the unit disk, with symbols of the form ϕ αz n +βz m +γz p +δz q, where α,β,γ,δ C and m,n,p,q Z +, m < n and p < q By letting T ϕ act on vectors of the form z k +cz l +dz r (k < l < r), we study the asymptotic behavior of a suitable matrix of inner products, as k Asaresult, weobtainasharpinequalityinvolvingtheabovementioned data: α 2 n 2 + β 2 m 2 γ 2 p 2 δ 2 q 2 2 ᾱβmn γδpq This inequality improves a number of existing results, and it is intended to be a precursor of basic necessary conditions for joint hyponormality of tuples of Toeplitz operators acting on Bergman spaces in one or several complex variables addresses: zeljkocuckovic@utoledoedu (Željko Čučković), rcurto@mathuiowaedu (Raúl E Curto ) URL: (Željko Čučković), (Raúl E Curto ) The second named author was partially supported by NSF Grants DMS and DMS Preprint submitted to Elsevier October 7, 206

2 Keywords: hyponormality, Toeplitz operators, Bergman space, commutators 200 Mathematics Subject Classification Primary: 47B35, 47B20, 32A36; Secondary: 47B36, 47B5, 47B47 Notation and Preliminaries A bounded operator acting on a complex, separable, infinite dimensional Hilbert space H is said to be normal if T T = TT ; quasinormal if T commutes with T T; subnormal if T = N H, where N is normal on a Hilbert space K which contains H and NH H; hyponormal if T T TT ; and 2-hyponormal if (T,T 2 ) is (jointly) hyponormal, that is ( ) [T,T] [T 2,T] [T,T 2 ] [T 2,T 2 0 ] Clearly, normal quasinormal subnormal 2-hyponormal hyponormal In this paper we focus primarily on the cases H 2 (T) and A 2 (D), the Hardy space on the unit circle T and the Bergman space on the unit disk D, respectively For these Hilbert spaces, we look at Toeplitz operators, that is, the operators obtained by compressing multiplication operators on the respective L 2 spaces to the above mentioned Hilbert spaces We consider possible versions, in the Bergman space context, of C Cowen s characterization of hyponormality for Toeplitz operators on Hardy space of the unit circle Concretely, we consider Toeplitz operators on the Bergman space of the unit disk, with symbols of the form ϕ αz n +βz m +γz p +δz q, where α,β,γ,δ C and m,n,p,q Z +, m < n and p < q By letting T ϕ act on vectors of the form z k +cz l +dz r (k < l < r), we study the asymptotic behavior of a suitable matrix of inner products, as k Asaresult, weobtainasharpinequalityinvolvingtheabovementioned data We begin with a brief survey of the known results in the Hardy space context 2 The Hardy Space Case Let L 2 (T) denote the space of square integrable functions with respect to the Lebesgue measure on the unit circle, and let H 2 (T) denote the subspace consisting of functions with vanishing negative Fourier coefficients; equivalently, H 2 (T) is the L 2 (T)-closure of the space of analytic polynomials We also 2

3 let L (T) and H (T) denote the corresponding Banach spaces of essentially bounded functions on T The orthogonal projection from L 2 (T) onto H 2 (T) will be denoted by P Given ϕ L (T), the Toeplitz operator with symbol ϕ is T ϕ : H 2 (T) H 2 (T), given by T ϕ f := P(ϕf) (f H 2 (T)) T ϕ is said to be analytic if ϕ H (T) PR Halmos s Problem 5 ([9]) asks whether every subnormal Toeplitz operator is either normal or analytic In 984, C Cowen and J Long answered this question in the negative [4] Along the way, C Cowen obtained a characterization of hyponormality for Toeplitz operators, as follows [3]: if ϕ L, ϕ = f +g (f,g H 2 ), then T ϕ is hyponormal f = c+t hg, for some c C, h H, and h T Nakazi and K Takahashi [7] later found an alternative description: For ϕ L, let E(ϕ) := {k H : k and ϕ k ϕ H }; then T ϕ is hyponormal E(ϕ) (For a generalization of Cowen s result, see [8]) In this note we take a first step toward finding suitable generalizations of these results to the case of Toeplitz operators on the Bergman space over the unit disk We also wish to pursue appropriate generalizations of the results on joint hyponormality of pairs of Toeplitz operators on the Hardy space, obtained in [6] and [5] At present, there is no known characterization of subnormality of Toeplitz operators in the unit circle in terms of the symbol However, we do know that every 2-hyponormal Toeplitz operator with a trigonometric symbol is subnormal [6] Thus, a suitable intermediate goal is to find a characterization of 2-hyponormality in terms of the symbol, perhaps using as a starting point either Cowen s or Nakazi-Takahashi s characterizations of hyponormality For Toeplitz operators with trigonometric symbols, the following results describe hyponormality Proposition 2 [20] Suppose c n ϕ(z) n a k z k + k=0 n b k z k, k=0 with a n 0 Let c 0 a a 2 a n a n b c = a 2 a 3 a n 0 b 2 a n Then T ϕ is hyponormal if and only if Φ k (c 0,,c k ) (0 k n ), where Φ k denotes the Schur function introduced in [9] Proposition 22 Let ϕ be a trigonometric polynomial of the form ϕ(z) = N n= m a n z n b n 3

4 (i) (D Farenick and WY Lee [7]) If T ϕ is a hyponormal operator then m N and a m a N (ii) (D Farenick and WY Lee [7]) If T ϕ is a hyponormal operator then N m rank[t ϕ,t ϕ ] N (iii) (R Curto and WY Lee [6]) The hyponormality of T ϕ is independent of the particular values of the Fourier coefficients a 0,a,,a N m of ϕ Moreover, for T ϕ hyponormal, the rank of the self-commutator of T ϕ is independent of those coefficients (iv) (D Farenick and WY Lee [7]) If m N and a m = a N 0, then T ϕ is hyponormal if and only if the following equation holds: ā N a a 2 a m = a m ā N m+ ā N m+2 ā N In this case, the rank of [T ϕ,t ϕ ] is N m (hyponormal) (v) (D Farenick and WY Lee [7]) T ϕ is normal if and only if m = N, a N = a N, and (hyponormal) holds with m = N 3 The Bergman Space Case By analogy with the case of the unit circle, let L L (D), H H (D), L 2 L 2 (D) and A 2 A 2 (D) denote the relevant spaces in the case of the unit disk D Similarly, let P : L 2 A 2 denote the orthogonal projection onto the Bergman space For ϕ L, the Toeplitz operator on the Bergman space with symbol ϕ is T ϕ : A 2 (D) A 2 (D), given by T ϕ f := P(ϕf) (f A 2 ) T ϕ is said to be analytic if ϕ H 3 A Revealing Example Let ϕ z 2 +2z On the Hardy space H 2 (T), T ϕ is not hyponormal, because m = 2, N =, and m>n (see Proposition 22) However, on the Bergman space A 2 (D) T ϕ is hyponormal, as we now prove Consider a slight variation of the symbol, that is, 4

5 ϕ z 2 +αz (α C) Observe that [T ϕ,t ϕ ]f,f = α 2 [T z,t z ]+[T z 2,T z 2]f,f so that T ϕ is hyponormal if and only if α 2 zf 2 + P( z 2 f), z 2 f α 2 P( zf), zf + z 2 f 2 (3) for all f A 2 (D) Acalculationnowshowsthatthishappenspreciselywhen α 2, asfollows For, given f A 2 (D),f 0 b nz n, one can apply Lemma 3 and obtain and Thus, (3) becomes 0 zf 2 = P( zf) 2 = z 2 f 2 = P( z 2 f) 2 = 0 2 b n 2 n+2, b n 2 n (n+) 2, α 2 b n 2 n+2 + b n 2 n (n+) 2 α b n 2 n+3, b n 2 n (n+) 2 b n 2 n (n+) 2 + b n 2 n+3 (32) Consider now a sequence (b n ) with b 0 := and b n := 0 for all n By (3), we have α If we now take b 0 := 0, b := and b n := 0 for all n 2, then (3) yields α 2 3, respectively Finally, if we use a sequence (b n ) with b 0 := 0, b := 0, b 2 := and b n := 0 for all n For n 3, then (32) becomes α 2 n+2 + n (n+) 2 n α 2 (n+) 2 + n+3, which immediately leads to the condition α 2 4 n+2 n+3 (all n 2), that is, α 2 4 As a result, T ϕ is hyponormal if and only if α 2 It follows that T z2 +2z is hyponormal 0 5

6 32 A Key Difference Between the Hardy and Bergman Cases Lemma 3 For u,v 0, we have Proof P( z u z v ) = { P( z u z v ) = = = j=0 j=0 { z u z v, 0 v < u (v u+) v+ z v u v u z j z j z j z j z u z v,z j z j z j 2 = (j +) z v,z u+j z j j=0 0 v < u v u+ v+ zv u v u Corollary 33 For v u and t w, we have P( z u z v ),P( z w z t ) v u+ = z v u, t w+ z t w v + t+ (t w +) = (v +)(t+) δ u+t,v+w 33 Some Known Results In this subsection, we briefly summarize a number of partial results relating to the Bergman space case (H Sadraoui [8]) If ϕ ḡ +f, the following are equivalent: (i) T ϕ is hyponormal on A 2 (D); (ii) H ḡhḡ H fh f; (iii) Hḡ = CH f, where C is a contraction on A 2 (D) (IS Hwang [0]) Let ϕ a m z m +a N z N +a m z m +a N z N (0 < m < N) satisfying a m ā N = a m a N, then T ϕ is hyponormal if and only if N + ( a N 2 a N 2 ) m+ ( a m 2 a m 2 ) (if a N a N ) N 2 ( a N 2 a N 2 ) m 2 ( a m 2 a m 2 ) (if a N a N ) The last condition is not sufficient (IS Hwang []) Let ϕ 4 z 3 +2 z 2 + z+z+2z 2 +βz 3 ( β = 4) Then T ϕ is hyponormal if and only if β = 4 6

7 (IS Hwang []) Let ϕ 8 z 3 + z 2 +β z +γz +7z 2 +2z 3 ( β = γ ) Then T ϕ is not hyponormal (P Ahern and Ž Čučković []) Let ϕ ḡ+f L (D), and assume that T ϕ is hyponormal Then Bu u, where B denotes the Berezin transform and u := f 2 g 2 (P Ahern and Ž Čučković []) Let ϕ ḡ+f L (D), and assume that T ϕ is hyponormal Then lim z ζ ( f (z) 2 g (z) 2 ) 0 for all ζ T In particular, if f and g are continuous at ζ T then f (ζ) g (ζ) (IS Hwang and J Lee [4]) The authors obtain some basic results on hyponormality of Toeplitz operators on weighted Bergman spaces (Y Lu and C Liu [5]) The authors obtain necessary and sufficient conditions for the hyponormality of T ϕ in the case when ϕ is a radial symbol (Y Lu and Y Shi [6]) The authors study the weighted Bergman space case, and prove the following result Theorem 34 (cf [6, Theorem 24(ii)]) Let ϕ := αz n +βz m +γ z m + δ z n, with n > m Then m 2 ( β 2 γ 2 )+n 2 ( α 2 δ 2 ) mn ᾱβ γδ 34 Hyponormality of Toeplitz Operators on the Bergman Space The self-commutator of T ϕ is C := [T ϕ,t ϕ ] We seek necessary and sufficient conditions on the symbol ϕ to ensure that C 0 Thefollowingresultgivesaflavorofthetypeofcalculationswefacewhentrying to decipher the hyponormality of a Toeplitz operator acting on the Bergman space Although the calculation therein will be superseded by the calculations in the following section, it serves both as a preliminary example and as motivation for the organization of our work 7

8 Proposition 35 Assume k, l max{a, b} Then [T z a,t z b](z k +cz l ),z k +cz l [ ] = a 2 (k +) 2 (k ++a) +c2 (l+) 2 δ a,b (l++a) k l+a +ac[ (a+k +)(k +)(l+) δ a+k,b+l l k +a + (a+l+)(k +)(l+) δ a+l,b+k] Proof Keeping in mind that k,l max{a,b}, we calculate the action of the commutator on the binomial z k +cz l [T z a,t z b](z k +cz l ),z k +cz l = T z at z b(z k +cz l ),z k +cz l T z bt z a(z k +cz l ),z k +cz l = z b+k +cz b+l,z a+k +cz a+l P( z a z k +c z a z l ),P( z b z k +c z b z l ) = = = = δ a,b δ a,b a+k + +c δ a+k,b+l a+k + +c δ a+l,b+k a+l+ +c2 a+l+ (k b+) (k +) 2 δ (k a+) a,b c (k +)(l+) δ a+l,b+k (k b+) c (k +)(l+) δ a+k,b+l c 2(l b+) (l+) 2 δ a,b [ (k b+) a+k + +c2 a+l+ (k +) 2 c 2 (l b+) ] (l+) 2 δ a,b [ ] (k b+) +c δ a+k,b+l a+k + (k +)(l+) [ ] (k a+) +c δ a+l,b+k a+l+ (k +)(l+) [ (k +)(b a)+ab (k +) 2 (k ++a) +c2 (l+)(b a)+ab ] (l+) 2 δ a,b (l++a) [ +c a+k + (l a+) ] δ a+k,b+l (k +)(l+) [ ] (k a+) +c δ a+l,b+k a+l+ (k +)(l+) [ a 2 a 2 ] (k +) 2 (k ++a) +c2 (l+) 2 δ a,b (l++a) (if a b then the whole expression is 0) k l+a +ac (a+k +)(k +)(l+) δ a+k,b+l l k +a +ac (a+l+)(k +)(l+) δ a+l,b+k 8

9 [ ] = a 2 (k +) 2 (k ++a) +c2 (l+) 2 δ a,b (l++a) k l+a +ac[ (a+k +)(k +)(l+) δ a+k,b+l l k +a + (a+l+)(k +)(l+) δ a+l,b+k], as desired Corollary 37 Assume a = b, k,l a and k l Then [T z a,t z a](z k +cz l ),z k +cz l [ ] = a 2 (k +) 2 (k ++a) +c2 (l+) 2 (l++a) 35 Self-Commutators We focus on the action of the self-commutator C of certain Toeplitz operators T ϕ on suitable vectors f in the space A 2 (D) The symbol ϕ and the vector f are of the form and ϕ := αz n +βz m +γz p +δz q (n > m; p < q) f := z k +cz l +dz r (k < l < r), respectively, with l and r to be determined later We also assume that n m = q p Our ultimate goal is to study the asymptotic behavior of this action as k goes to infinity Thus, we consider the expression Cf,f, given by [(Tαz n +βz m +γz p +δz q),t αz n +βz m +γz p +δz q](zk +cz l +dz r ),z k +cz l +dz r, for large values of k (and consequently large values of l and r) It is straightforward to see that Cf,f is a quadratic form in c and d, that is, Cf,f A Re(A 0 c)+2 Re(A 0 d)+a 20 cc+2 Re(A cd)+a 02 dd (33) Alternatively, the matricial form of (33) is A 00 A 0 A 0 A 0 A 20 A c, c A 0 A A 02 d d (34) We now observe that the coefficient A 00 arises from the action of C on the monomial z k, that is, A 00 = Cz k,z k [(T αz n +βz m +γz p +δz q),t αz n +βz m +γz p +δz q]zk,z k 9

10 Similarly, A 0 = Cz l,z k, A 0 = Cz r,z k, (35) A 20 = Cz l,z l, A = Cz r,z l, (36) A 02 = Cz r,z r To calculate A 00 explicitly, we first recall that the algebra of Toeplitz operators with analytic symbols is commutative, and therefore T z n commutes with T z m, T z p and T z q We also recall that two monomials z u and z v are orthogonal whenever u v As a result, the only nonzero contributions to A 00 must come from the inner products [Tz n,t z n]zk,z k, [Tz m,t z m]zk,z k, [Tz p,t z p]zk,z k and [T z q,t z q]z k,z k Applying Corollary 37 we see that ( ) α 2 n 2 A 00 = (k +) 2 k +n+ + β 2 m 2 k +m+ γ 2 p 2 k +p+ δ 2 q 2 k +q + Similarly, A 0 = αβ( l+m+ k m+ (k +)(l+) )δ n+k,m+l +αβ( l+n+ k n+ (k +)(l+) )δ m+k,n+l γδ( l+p+ k p+ (k +)(l+) )δ q+k,p+l k q + γδ( l+q + (k +)(l+) )δ p+k,q+l (37) Now recall that m < n and k < l, so that m + k < n + l, and therefore δ m+k,n+l =0 Also, p < q implies p + k < q + l, so that δ p+k,q+l = 0 As a consequence, A 0 = αβ( l+m+ k m+ (k +)(l+) )δ n+k,m+l γδ( l+p+ k p+ (k +)(l+) )δ q+k,p+l (38) Consider now A 0, as described in (35) We wish to imitate the calculation for A 0 Observe that k < r, so that the vanishing of the relevant δ s in (37) still holds Thus, we obtain A 0 = αβ( r +m+ k m+ (k +)(r +) )δ n+k,m+r γδ( r +p+ k p+ (k +)(r +) )δ q+k,p+r (39) 0

11 In short, A 0 can be obtained from A 0 by replacing l by r In a completely analogous way, we can calculate A, by replacing l by r and k by l in (37): A = αβ( r +m+ l m+ (l+)(r +) )δ n+l,m+r γδ( r +p+ l p+ (r +)(l+) )δ q+l,p+r (30) Also, A 20 and A 02 follow the pattern of A 00 : ( ) α 2 n 2 A 20 = (l+) 2 l+n+ + β 2 m 2 l+m+ γ 2 p 2 l+p+ δ 2 q 2, l+q + and A 02 = ( ) α 2 n 2 (r +) 2 r +n+ + β 2 m 2 r +m+ γ 2 p 2 r +p+ δ 2 q 2 r +q + Recall again that k < l < r We now make a judicious choice to simplify the forms of A 0, A and A 0 That is, we let l := n+k m and r := l+q p It follows that n+k = m+l < m+r and q +k < q +l = p+r Therefore, both Kronecker deltas appearing in A 0 are zero, and thus A 0 = 0 Moreover, and A 0 A = αβ( l+m+ k m+ (k +)(l+) ) (3) γδ( l+p+ k p+ (k +)(l+) )δ q+k,p+l (32) = αβ( r +m+ l m+ (r +)(l+) )δ n+l,m+r γδ( r +p+ l p+ (r +)(l+) ) The 3 3 matrix associated with C becomes A 00 A 0 0 M := A 0 A 20 A 0 A A 02 We now wish to study the asymptotic behavior of k 3 M as k Surprisingly, k 3 A 00, k 3 A 02 and k 3 A 20 all have the same limit as k Also, k 3 A 0 and k 3 A have the same limit To see this, observe that k 3 A 00 = k 2 (k +) 2 ( ) k α 2 n 2 k +n+ + k β 2 m 2 k +m+ k γ 2 p 2 k +p+ k δ 2 q 2, k +q +

12 so that a := lim k k3 A 00 = α 2 n 2 + β 2 m 2 γ 2 p 2 δ 2 q 2 Then lim k k 3 A 20 = lim k k 3 A 02 = a In terms of the remaining entries of k 3 M, recall the assumption n m = q p, and let g := n m = q p It follows that l = k+g and r = l+g = k+2g By using these values in (3), we obtain k 3 k 3 mn A 0 = ᾱβ (k +)(k +g +)(k +g +m+) γδ k 3 pq (k +)(k +g +)(k +g +p+), so that ρ := lim k k3 A 0 = ᾱβmn γδpq The calculation for A is entirely similar, and one gets lim k k 3 A = ρ It follows that the asymptotic behavior of k 3 M is given by the tridiagonal matrix a ρ 0 ρ a ρ 0 ρ a Now, if instead of using a vector of the form f := z k +cz l +dz r (k < l < r), with l = k +g and r = l+g = k +2g (that is, a vector of the form f := z k +cz k+g +dz k+2g we were to use a longer vector with similar power structure, f N := z k +c z k+g +c 2 z k+2g + +c N z k+ng, it is not hard to see that the asymptotic behavior of the associated matrix would still be given by the tridiagonal matrix with a in the diagonal and ρ in the super-diagonal To see this, one only need to observe that the entries of the matrix P associated with Cf,f will follow the same pattern as the entries in M For example, when N = 3 the (3,4)-entry of P will follow the pattern of A 0 above, with l and k replaced by k +3g and k +2g, resp Similarly, the (2,4)-entry of P will follow the pattern of A 0 above, with r and k replaced by k +3g and k +g, resp As a result, it is straightforward to see that, like A 0, the entry P 24 will be zero As for P 34, one gets P 34 = αβ( k +3g +m+ k +2g m+ (k +2g +)(k +3g +) )δ n+k+2g,m+k+3g γδ( k +3g +p+ k +2g p+ (k +3g +)(k +2g +) )δ q+k+2g,p+k+3g (33) 2

13 As before, k 3 P 34 k 3 mn = ᾱβ (k +2g +)(k +3g +)(k +3g +m+) k 3 pq γδ (k +2g +)(k +3g +)(k +3g +p+), and once again, lim k k3 P 34 = ᾱβmn γδpq = ρ In summary, the hyponormality of T ϕ, detected by the positivity of the self-commutator C, leads to the positive semi-definiteness of the tridiagonal (N +) (N +) matrix P Since this must be true for all N, it follows that a necessary condition for the hyponormality of T ϕ is the positive semidefiniteness of the infinite tridiagonal matrix Q := a ρ 0 ρ a ρ 0 ρ a We now consider the spectral behavior of Q as an operator on l 2 (Z + ) Lemma 38 For a R and ρ C, the spectrum of the infinite tridiagonal matrix Q is [a 2 ρ,a+2 ρ ] Proof This result is well known; we present a proof for the sake of completeness Observe that Q is the canonical matrix representation of the Toeplitz operator on H 2 (T) with symbol ϕ(z) := a + 2Re( ρz) Since the symbol is harmonic, it follows that the spectrum of T ϕ ai + T ρz+ρ z is the set a+2re({ ρz : z D} ) = a+2[ ρ, ρ ], as desired As a consequence, if Q is positive (as an operator on l 2 (Z + )), then 36 Main Result a 2 ρ Theorem 30 Assume that T ϕ is hyponormal on A 2 (D), with ϕ := αz n +βz m +γz p +δz q (n > m; p < q) Assume also that n m = q p Then α 2 n 2 + β 2 m 2 γ 2 p 2 δ 2 q 2 2 ᾱβmn γδpq 3

14 37 A Specific Case When p = m and q = n in the inequality reduces to ϕ := αz n +βz m +γz p +δz q (n > m; p < q) α 2 n 2 + β 2 m 2 γ 2 p 2 δ 2 q 2 2 ᾱβmn γδpq n 2 ( α 2 δ 2 )+m 2 ( β 2 γ 2 ) 2mn ᾱβ γδ This not only generalizes previous estimates, but also sharpens them, since previous results did not include the factor 2 in the right-hand side 4 When is T ϕ normal? We conclude this paper with a description of those symbols ϕ in Theorem 30 which produce a normal operator T ϕ We first recall a result of S Axler and Ž Čučković Lemma 4 ([2]) Let ϕ be harmonic and bounded Then T ϕ is normal if and only if there exist a pair of complex numbers a and b such that (a,b) (0,0) and F := aϕ+b ϕ is constant on D In what follows, we write a harmonic symbol as ϕ f + ḡ, with f and g analytic A straightforward calculation using ( z ) and ( z ) shows that ( a 2 b 2 ) f z = 0 If f is constant, a similar calculation shows that g is also constant, and a fortiori ϕ is constant, and T ϕ is normal Thus, without loss of generality, we can assume that f is not constant, and therefore a = b > 0 If we write a = a e iθ and b = a e iη, it is not hard to see that ϕ + e i(η θ) ϕ is constant, and then e i 2 (η θ )[e i 2 (η θ )ϕ+e i 2 (η θ )ϕ is constant on D We conclude that, up to a constant multiple of modulus, we can always assume that ϕ + ϕ is constant on D Theorem 42 Let ϕ αz n +βz m +γ z p +δ z q and assume that F := ϕ+ ϕ is constant on D Then T ϕ is normal if and only if ϕ is of one of exactly three types: (i) ϕ = αz n ᾱ z n (when n = p); (ii) ϕ = αz n +βz m β z m ᾱ z n (when m = p); or (iii) ϕ = βz m β z m (when m = q) Proof Since F is a constant trigonometric polynomial, with analytic monomials z m, z n, z p and z q, and since ϕ is a nonconstant harmonic function, in the above mentioned list of four monomials we must necessarily have at least two identical monomials Since m < n, p < q and n m = q p, we are led to consider the following three cases: (i) n = p (and therefore m < n = p < q); here F = βz m +(α+ γ)z n + δz q +βz m +(α+ γ)z n + δz q 4

15 fromwhichiteasilyfollowsthatβ = 0, γ = αandδ = 0 Thenϕ = αz n αz n, as desired (ii) m = p (and therefore m = p < q = n); here F = (α+ δ)z n +(β + γ)z m +(α+ δ)z n +(β + γ)z m, so that α+ δ = 0 and β + γ = 0 We then get ϕ = αz n +βz m αz n +βz m, as desired (iii) m = q, which leads to ϕ = βz m βz m The proof is complete References [] P Ahern and Ž Čučković, A mean value inequality with applications to Bergman space operators, Pacific J Math 73(996), [2] S Axler and Ž Čučković, Commuting Toeplitz operators with harmonic symbols, Integral Equations Operator Theory 4(99), 2 [3] C Cowen, Hyponormality of Toeplitz operators, Proc Amer Math Soc 03(988), [4] C Cowen and J Long, Some subnormal Toeplitz operators, J Reine Angew Math 35(984), [5] RE Curto, IS Hwang and WY Lee, Matrix Functions of Bounded Type: An Interplay Between Function Theory and Operator Theory, preprint 206; submitted to Memoirs Amer Math Soc [6] RE Curto and WY Lee, Joint hyponormality of Toeplitz pairs, Memoirs Amer Math Soc 72, Amer Math Soc, Providence, 200 [7] D Farenick and WY Lee, Hyponormality and spectra of Toeplitz operators, Trans Amer Math Soc 348(996), [8] C Gu, A generalization of Cowen s characterization of hyponormal Toeplitz operators, J Funct Anal 24(994), [9] PR Halmos, Ten problems in Hilbert space, Bull Amer Math Soc 76(970), [0] IS Hwang, Hyponormal Toeplitz operators on the Bergman space, J Korean Math Soc 42(2005), [] IS Hwang, Hyponormality of Toeplitz operators on the Bergman space, J Korean Math Soc 45(2008), [2] IS Hwang and WY Lee, Hyponormality of trigonometric Toeplitz operators, Trans Amer Math Soc 354(2002),

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