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

Transcription

1 A New Necessary Condition for the Hyponormality of Toeplitz Operators on the Bergman Space Raúl Curto IWOTA 2016 Special Session on Multivariable Operator Theory St. Louis, MO, USA; July 19, 2016 (with Željko Čučković, Univ. of Toledo, USA) ~rcurto

2 Abstract It is a well known result of C. Cowen 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 1. In this talk we will consider possible versions of this result in the Bergman space case.

3 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. As a consequence, we obtain a rather sharp inequality involving the above mentioned data: α 2 n 2 + β 2 m 2 γ 2 p 2 δ 2 q 2 2 ᾱβmn γδpq. This result is intended to be a precursor of basic necessary conditions for joint hyponormality of commuting tuples of Toeplitz operators acting on Bergman spaces in several complex variables.

4 Notation and Preliminaries T L(H): algebra of bounded operators on a Hilbert space H normal if T T = TT quasinormal if T commutes with T T subnormal if T = N H, where N is normal and NH H hyponormal if T T TT 2-hyponormal if (T, T 2 ) is (jointly) hyponormal (k 1) ( ) [T, T] [T 2, T] [T, T 2 ] [T 2, T 2 ] 0 normal quasinormal subnormal 2-hypon. hypon.

5 L L (T); H H (T); L 2 L 2 (T); H 2 H 2 (T), P : L 2 H 2 orthogonal projection For ϕ L, the Toeplitz operator with symbol ϕ is T ϕ : H 2 H 2, given by T ϕ f := P(ϕf) (f H 2 ) T ϕ is said to be analytic if ϕ H Halmos s Problem 5 (1970): Is every subnormal Toeplitz operator either normal or analytic? C. Cowen and J. Long (1984): No

6 C. Cowen (1988) ϕ L, ϕ = f + g (f, g H 2 ) T ϕ is hyponormal f = c + T hg, for some c C, h H, h 1. Nakazi-Takahashi (1993) For ϕ L, let E(ϕ) := {k H : k 1 and ϕ k ϕ H }. Then T ϕ is hyponormal E(ϕ).

7 Natural Questions: 1) When is T ϕ subnormal? At present, there s no known characterization of subnormality in terms of the symbol ϕ. 2) Characterize 2-hyponormality for Toeplitz operators Sample Result: Theorem (RC and WY Lee, 2001) Every 2-hyponormal trigonometric Toeplitz operator is subnormal.

8 Hyponormal Toeplitz Pairs In 2001, RC and W.Y. Lee completely characterized the hyponormality of Toeplitz pairs T (T ϕ, T ψ ), when both symbols ϕ and ψ are trigonometric polynomials. Key building block: For ϕ and ψ trigonometric polynomials, the hyponormality of T (T ϕ, T ψ ) forces the co-analytic parts of ϕ and ψ to necessarily coincide up to a constant multiple; equivalently, [ECAP] ϕ βψ H 2 f or some β C. ECAP: Equality (up to a constant multiple) of Co-Analytic Parts

9 Hyponormal Toeplitz Operators on the Bergman Space L L (D); H H (D); L 2 L 2 (D); A 2 A 2 (D), P : L 2 A 2 orthogonal projection 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.

10 Trigonometric Symbols in the Hardy Space Case Proposition Let ϕ be a trigonometric polynomial of the form N ϕ(z) = a n z n. n= m (i) (ii) (D. Farenick & W.Y. Lee, 1996) If T ϕ is a hyponormal operator then m N and a m a N. (D. Farenick & W.Y. Lee, 1996) If T ϕ is a hyponormal operator then N m rank[tϕ, T ϕ ] N.

11 Trigonometric Symbols in the Hardy Space Case Proposition (cont.) (iii) (RC & W.Y. Lee, 2001) The hyponormality of T ϕ is independent of the particular values of the Fourier coefficients a 0, a 1,...,a N m of ϕ. Moreover, for T ϕ hyponormal, the rank of the self-commutator of T ϕ is independent of those coefficients.

12 Trigonometric Symbols in the Hardy Space Case Proposition (cont.) (iv) (D. Farenick & W.Y. Lee, 1996) 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 1 a 2.. = a m ā N m+1 ā N m+2... (hypon.) a m ā N In this case, the rank of [T ϕ, T ϕ ] is N m.

13 Trigonometric Symbols in the Hardy Space Case Proposition (cont.) (v) (D. Farenick & W.Y. Lee, 1996) T ϕ is normal if and only if m = N, a N = a N, and (hypon.) holds with m = N.

14 A Revealing Example Let ϕ z 2 + 2z. On the Hardy space H 2 (T), T ϕ is not hyponormal, because m = 2, N = 1, and m>n. 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,. 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

15 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 for all f A 2 (D). A calculation, using the result on the next page, shows that this happens precisely when α 2. As a result, T z2 +2z is hyponormal.

16 Key Difference Between Hardy and Bergman Cases Lemma For u, v 0, we have { P( z u z v ) = 0 v < u (v u+1) v+1 z v u v u. Proof. P( z u z v ) = = = z u z v z j z j, z j z j j=0 z u z v, z j z j j=0 z j 2 = (j + 1) z v, z u+j z j j=0 { 0 v < u v u+1 v+1 zv u v u.

17 Corollary For v u and t w, we have P( z u z v ), P( z w z t ) v u + 1 = v + 1 zv u, t w + 1 z t w t + 1 (t w + 1) = (v + 1)(t + 1) δ u+t,v+w.

18 Some Known Results (H. Sadraoui, PhD Thesis, Purdue Univ., 1992) If ϕ ḡ + f, the following are equivalent: (i) T ϕ is hyponormal on A 2 (D); (ii) H ḡ Hḡ H f H f ; (iii) Hḡ = CH f, where C is a contraction on A 2 (D). (I.S. Hwang, J. Korean Math. Soc., 2005) Let ϕ a m z m + a N z N + a m z m + a N z N a m ā N = a m a N, then T ϕ is hyponormal if and only if 1 N + 1 ( a N 2 a N 2 ) (0 < m < N) satisfying 1 m+1 ( 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.

19 (I.S. Hwang, J. Korean Math. Soc., 2008) Let ϕ 4 z z 2 + z + z + 2z 2 +βz 3 ( β = 4). Then T ϕ is hyponormal if and only if β = 4. (I.S. Hwang, J. Korean Math. Soc., 2008) Let ϕ 8 z 3 + z 2 +β z +γz + 7z 2 + 2z 3 ( β = γ ). Then T ϕ is not hyponormal. (P. Ahern & Ž Čučković, Pacific J. Math., 1996) Let ϕ ḡ + f L (D), and assume that T ϕ is hyponormal. Then T ϕ u u, where u := f 2 g 2.

20 (P. Ahern & Ž Čučković, Pacific J. Math., 1996) Let ϕ ḡ + f L infty (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 then f (ζ) g (ζ). (I.S. Hwang & J. Lee, Math. Ineq. Appl., 2012) studied hyponormality on weighted Bergman spaces. (Y. Lu & C. Liu, J. Korean Math. Soc., 2009) considered radial symbols. (Y. Lu & Y. Shi, IEOT, 2009) studied the weighted Bergman space case.

21 Hyponormality of Toeplitz Operators on the Bergman Space The self-commutator of T ϕ is C := [Tϕ, T ϕ ] We seek necessary and sufficient conditions on the symbol ϕ to ensure that C 0. The following result gives a flavor of the type of calculations we face when trying 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.

22 Proposition Assume k,l max{a, b}. Then [T z a, T z b](z k + cz l ), z k + cz l [ ] = a 2 1 (k + 1) 2 (k + 1+a) + 1 c2 (l+1) 2 δ a,b (l+1+a) k l+a +ac[ (a+k + 1)(k + 1)(l+1) δ a+k,b+l l k + a + (a+l+1)(k + 1)(l+1) δ a+l,b+k]

23 Corollary 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 1 (k + 1) 2 (k + 1+a) + 1 c2 (l+1) 2 (l+1+a)

24 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. 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 [(Tαzn +βz m +γz p +δz q), T αzn +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,

25 Cf, f A Re(A 10 c)+2 Re(A 01 d)+a 20 cc+2 Re(A 11 cd)+a 02 dd. Alternatively, the matricial form of (1) is A 00 A 10 A A 10 A 20 A 11 c, c. (2) A 01 A 11 A 02 d d (1)

26 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 αzn +βz m +γz p +δz q), T αzn +βz m +γz p +δz q]zk, z k. Similarly, A 10 = Cz l, z k, A 01 = Cz r, z k, A 20 = Cz l, z l, A 11 = Cz r, z l, A 02 = Cz r, z r.

27 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 [T z n, T z n]zk, z k, [T z m, T z m]z k, z k, [T z p, T z p]zk, z k and [T z q, T z q]zk, z k. Applying Corollary 7 we see that ( ) 1 α 2 n 2 A 00 = (k + 1) 2 k + n+1 + β 2 m 2 k + m+1 γ 2 p 2 k + p + 1 δ 2 q 2. k + q + 1

28 Similarly, A 10 1 = αβ( l+m+1 k m+1 (k + 1)(l+1) )δ n+k,m+l 1 +αβ( l+n+1 k n+1 (k + 1)(l+1) )δ m+k,n+l 1 γδ( l+q + 1 k q + 1 (k + 1)(l+1) )δ q+k,p+l 1 γδ( l+p + 1 k p + 1 (k + 1)(l+1) )δ p+k,q+l. (3) 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 10 1 = αβ( l+m+1 k m+1 (k + 1)(l+1) )δ n+k,m+l 1 γδ( l+q + 1 k q + 1 (k + 1)(l+1) )δ q+k,p+l. (4)

29 In a completely analogous way, we obtain A 01 1 = αβ( r + m+1 k m+1 (k + 1)(r + 1) )δ n+k,m+r 1 γδ( r + q + 1 k q + 1 (k + 1)(r + 1) )δ q+k,p+r, (5) A 11 1 = αβ( l+n+1 r n+1 (r + 1)(l+1) )δ m+r,n+l 1 γδ( l+p+ 1 r p + 1 (r + 1)(l+1) )δ p+r,q+l, (6) A 20 = and A 02 = ( ) 1 α 2 n 2 (l+1) 2 l+n+1 + β 2 m 2 l+m+1 γ 2 p 2 l+p+ 1 δ 2 q 2, l+q + 1 ( ) 1 α 2 n 2 (r + 1) 2 r + n+1 + β 2 m 2 r + m+1 γ 2 p 2 r + p+ 1 δ 2 q 2. r + q + 1

30 Recall again that k < l < r and assume, for the sake of simplicity, that l := k + n 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 01 are zero, and thus A 01 = 0. Moreover, A 10 1 = αβ( l+m+1 k m+1 (k + 1)(l+1) ) 1 γδ( l+q + 1 k q + 1 (k + 1)(l+1) )δ q+k,p+l. Similarly, A 11 1 = αβ( l+n+1 r n+1 (r + 1)(l+1) )δ m+r,n+l 1 γδ( l+p + 1 r p+ 1 (r + 1)(l+1) ).

31 The 3 3 matrix associated with C becomes A 00 A 10 0 A 10 A 20 A A 11 A 02 Surprisingly, A 00, A 02 and A 20 all have the same limit as k. Also, A 10 and A 11 have the same limit. It follows that the asymptotic behavior is associated with a tri-diagonal matrix of the form

32 a ρ 0 ρ a ρ. 0 ρ a Here and a := α 2 m 2 + β 2 n 2 γ 2 p 2 δ 2 q 2 ρ := ᾱβmn γδpq

33 Now, if instead of using a vector of the form f := z k + cz l + dz r (k < l < r), we use a longer vector, f := z k 1 + c 2 z k 2 + +c m z km (k 1 < k 2 < < k m ), then the associated asymptotic matrix would still be tridiagonal, with a in the diagonal and ρ in the superdiagonal.

34 Lemma The spectrum of the infinite tridiagonal matrix a ρ 0 ρ a ρ M :=. 0 ρ a is [a 2 ρ,a+2 ρ ].

35 As a consequence, if M is positive (as an operator on l 2 (Z)), then a 2 ρ. Theorem Assume that T ϕ is hyponormal on A 2 (D), with ϕ := αz n +βz m +γz p +δz q (n > m; p < q). Then α 2 n 2 + β 2 m 2 γ 2 p 2 δ 2 q 2 2 ᾱβmn γδpq.

36 Specific Case When p = m and q = n in ϕ := αz n +βz m +γz p +δz q (n > m; p < q). the inequality α 2 n 2 + β 2 m 2 γ 2 p 2 δ 2 q 2 2 ᾱβmn γδpq. reduces to 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.