Reduced Cowen Sets

New York J. Math. 7(2001)217–222.

Reduced Cowen Sets

Ra´ul E. Curto and Woo Young Lee

Abstract. Forf∈H², let

G_f:={g∈zH²:f+g∈L^∞andTf+gis hyponormal}.

In 1988, C. Cowen posed the following question: Ifg∈G_fis such thatλ g /∈G_f (allλ∈C,|λ|>1), isgan extreme point ofG_f? In this note we answer this question in the negative. At the same time, we obtain a general sufficient condition for the answer to be affirmative; that is, whenf∈H^∞is such that rankH_f <∞.

Contents

1. Introduction 217

2. Main results 219

References 222

1. Introduction

A bounded linear operatorAon a Hilbert space is said to be hyponormal if its self-commutator [A^∗, A] : =A^∗A−AA^∗is positive (semidefinite). Givenϕ∈L^∞(T), the Toeplitz operator with symbolϕis the operatorT_ϕon the Hardy spaceH²(T) of the unit circle T ≡ ∂D defined by Tϕf := P(ϕ·f), where f ∈ H²(T) and P denotes the orthogonal projection that maps L²(T) onto H²(T). Let H^∞(T) : = L^∞∩H², that is,H^∞is the set of bounded analytic functions onD. The problem of determining which symbols induce hyponormal Toeplitz operators was solved by C. Cowen [Co2] in 1988. Cowen’s method is to recast the operator-theoretic problem of hyponormality for Toeplitz operators as a functional equation involving the operator’s symbol.

Received February 15, 2001.

Mathematics Subject Classification. Primary 47B35; Secondary 47B20, 30D50.

Key words and phrases. Toeplitz operators, Hankel operators, hyponormal operators, reduced Cowen sets, Hermite-Fej´er interpolation problem.

The work of the first author was partially supported by NSF research grant DMS-9800931.

The work of the second author was partially supported by grant No. 2000-1-10100-002-3 from the Basic Research Program of the KOSEF.

ISSN 1076-9803/01

217

Suppose that ϕ∈ L^∞(T) is arbitrary and consider the following subset of the closed unit ball ofH^∞(T),

E(ϕ) : ={k∈H^∞(T) :||k||_∞≤1 andϕ−kϕ∈H^∞(T)}.

Cowen’s Theorem states that T_ϕ is hyponormal if and only if E(ϕ) is nonempty [Co2], [NT]. We also recall the connection between Hankel and Toeplitz operators.

ForϕinL^∞, theHankel operatorHϕ:H²→H²is defined byHϕf :=J(I−P)(ϕf), whereJ : (H²)^⊥ →H²is given by Jz⁻ⁿ =zⁿ⁻¹ forn≥1. The following are two basic identities:

T_ϕψ−T_ϕT_ψ=H_ϕ^∗H_ψ (ϕ, ψ∈L^∞) and H_ϕh=T_h^∗H_ϕ (h∈H^∞), (1)

where forζ ∈L^∞, we defineζ(z) : = ζ(z). From this we can see that if k∈ E(ϕ) then

[T_ϕ^∗, T_ϕ] =H_ϕ^∗H_ϕ−H_ϕ^∗H_ϕ=H_ϕ^∗H_ϕ−H_{k ϕ}^∗ H_{k ϕ}=H_ϕ^∗(1−T_kT_k^∗)H_ϕ, which implies that kerH_ϕ⊆ker [T_ϕ^∗, T_ϕ].

To describe the set of gsuch that Tf+g is hyponormal for a given f, C. Cowen [Co1] defined the setG_f as follows. IfH :={h∈zH^∞:||h||2≤1}, let

G_f :=

g∈zH²:sup

h0∈H|hh0, f| ≥ sup

h0∈H|hh0, g| for everyh∈H²

. To see how this definition is relevant to hyponormality of Toeplitz operators, we assume thatf+g∈L^∞. Note that iff ∈H²thenH_f makes sense whenf has an L^∞-conjugateg∈H², that is,f+g∈L^∞. For, givenh∈H²we haveH_f+g(h) = J(I−P)(fh+gh) =J(I−P)(fh) =:H_fh. Iff+g∈L^∞(f ∈H², g∈zH²) and h∈H²then

hsup0∈H|hh0, f|= sup

h0∈H

Thh0f dµ = sup

h0∈H

T(I−P)(fh+gh)h0dµ

= sup

h0∈H|(I−P)fh, h0|= sup

h0∈H|J(I−P)fh, h0|

=||H_fh||

and similarly,

hsup0∈H|hh₀, g|=||H_gh||.

Recall ([Ab, Lemma 1]) that if ϕ = f +g ∈ L^∞ (f ∈ H², g ∈ zH²) then the following are equivalent:

(a) T_ϕis hyponormal;

(b) ||H_fh|| ≥ ||Hgh||for everyh∈H². Therefore we can see that forf ∈H²,

G_f =

g∈zH²:f+g∈L^∞ andTf+g is hyponormal . (2)

We callG_f thereduced Cowen set forf. To avoid some technical difficulties using the original definition of G_f when dealing with hyponormality of T_f+g, hereafter we assume thatf+g∈L^∞and adopt (2) as our definition ofG_f; this appears to

be natural when studying the setG_f. We can easily see that G_f is balanced and convex. Write

∇G_f :=

g∈G_f : λ g /∈G_f (allλ∈C, |λ|>1)

and extG_f for the set of all extreme points ofG_f. In [Co1] the following question was posed:

Question. Is∇G_f ⊆extG_f?

In [CCL] an affirmative answer to the above question was given in case f is an analytic polynomial. In this note we answer the above question in the negative, and give a general sufficient condition for the answer to be affirmative:If rankH_f <∞ then∇G_f ⊆extG_f. In [CCL], our ploy was to use the Carath´eodory-Schur Inter- polation Problem to deal with the case of an analytic polynomialf. By comparison, we here resort to the classical Hermite-Fej´er Interpolation Problem.

2. Main results

If ϕ ∈ L^∞, write ϕ₊ = P(ϕ) ∈ H² and ϕ₋ = (I−P)(ϕ) ∈ zH². Thus ϕ = ϕ₊+ϕ₋ is the decomposition of ϕ into its analytic and co-analytic parts.

We first reformulate Cowen’s Theorem. Suppose that ϕ ∈ L^∞ is of the form ϕ(z) =_∞

n=−∞a_nzⁿ and thatk(z) =_∞

n=0c_nzⁿ is inH². Thenϕ−k ϕ∈H^∞if and only if













a₁ a₂ a₃ . . . a_n . . . a₂ a₃ . . . a_n . . . a₃ . . . .

... a_n . . . a_n . . .

...























 c₀ c₁ c₂ ...

...













=











 a₋₁ a₋₂ a₋₃ ...

...











 , (3)

that is,H_ϕ+k=zϕ₋. Thus by Cowen’s Theorem we have:

Lemma 1([CuL]). If ϕ ≡ ϕ+ +ϕ− ∈ L^∞, then E(ϕ) = ∅ if and only if the equation H_ϕ+k=zϕ₋ admits a solutionk satisfying||k||_∞≤1.

Recall that a functionϕ∈L^∞ is of bounded type (or in the Nevanlinna class) if it can be written as the quotient of two functions inH^∞(D), that is, there are functionsψ₁, ψ₂ inH^∞(D) such that

ϕ(z) =ψ₁(z)

ψ2(z) for almost allz∈T.

For example, rational functions in L^∞ are of bounded type. By an argument of M. Abrahamse [Ab, Lemma 3], the function ϕ is of bounded type if and only if kerHϕ = {0}. Thus if ϕ ≡ ϕ++ϕ− ∈ L^∞ and ϕis not of bounded type then kerH_ϕ+= kerH_ϕ ={0}, so that the equationH_ϕ+k=zϕ₋ has a unique solution whenever it is solvable; in other words, if ϕ is not of bounded type, and T_ϕ is hyponormal, thenE(ϕ) has exactly one element.

We now have:

Theorem 2. Suppose that ψ∈H^∞ is such thatψis not of bounded type, and let f :=z³ψ. Then∇G_f extG_f.

Proof. By assumption, f ∈H^∞ and f is not of bounded type; indeed, iff were of bounded type thenf = ^g_h (g, h∈H^∞(D)), and soψ=^z_h^3g would be of bounded type. Observe now that by definition and Lemma1,

G_f={g∈zH²:f+g∈L^∞ andH_fk=zg for somek∈H^∞ with||k||∞≤1}.

Sincef ∈z³H^∞, we have thatzf,z²f, ¹₂(z+z²)fall are inzH^∞. A straightforward calculation shows that

H_f(q) =zqf forq=z, z², 1

2(z+z²).

Since||q||_∞ ≤1 and qf=qf∈zH^∞ we have that{z f, z²f, ¹₂(z+z²)f} ⊆G_f. We will now show that ¹₂(z+z²)f ∈ ∇G_f, which proves∇G_f extG_f. Sincef is not of bounded type (so kerH_f ={0}), we know that for|λ|>1 andq:=¹₂(z+z²), the unique solution of the equation H_fk=λzqfisk =λ q. But ||λ q||∞ >1, so λ q f /∈G_f and therefore ¹₂(z+z²)f ≡qf∈ ∇G_f. For a concrete example satisfying the hypotheses of Theorem 2, let ψ be a Riemann mapping of the unit disk onto the interior of the ellipse with vertices

±i(1−α)⁻¹ and passing through±(1 +α)⁻¹, where 0< α <1. Thenψis inH^∞, andψis not of bounded type ([CoL, Corollary 2]).

In [CCL], an affirmative answer to Cowen’s Question was given in casef is an analytic polynomial. We now establish that the answer is also affirmative in the more general instances of rankH_f <∞.

To see this we need the following auxiliary lemma.

Lemma 3. Let qbe a finite Blaschke product, letk∈H^∞, and let G≡G(q, k) : ={b∈k+qH^∞: ||b||∞≤1}.

If Gcontains at least two functions then it contains a functionb with||b||_∞<1.

Proof. Write

q≡e^iθn

i=1

bⁿ_iⁱ, where b_i≡ z−α_i

1−α_iz, θ∈[0,2π), andα1,· · ·, αn are distinct points inD. If we define

x_i,j:= z^j

(1−α_iz)^j+1 for 1≤i≤n and 0≤j < n_i,

then the functionsx_i,j form a basis forH²qH² (cf. [FF, Lemma X.1.1]). Write k=k₁+k₂, wherek₁∈H²qH²andk₂∈qH². Note thatk₁is entirely determined by the values ofk^(j)₁ (α_i) (1≤i≤n, 0≤j < n_i), and also that

k^(j)(αi) =k₁^(j)(αi) for 1≤i≤n and 0≤j < ni.

Therefore the problem of finding a functionbink+qH^∞with||b||_∞≤1 is equiv- alent to the problem of finding a functionb∈H^∞ satisfying

(a) b^(j)(αi) =k^(j)₁ (αi) for 1≤i≤nand 0≤j < ni;

(b) ||b||∞≤1.

This is exactly the classical Hermite-Fej´er Interpolation Problem (HFIP) (Ifn= 1, this is the Carath´eodory–Schur Interpolation Problem and ifni= 1 for alli, this is the Nevanlinna-Pick Interpolation Problem; cf. [FF]). Then by [FF, Theorem X.5.6 and Corollary X.5.7], there exists a solution to HFIP if and only if the Hermite- Fej´er matrixMk1 associated withk1 is a contraction, and furthermore the solution is unique if and only if||Mk1||= 1. (Mk1is thed×dlower triangular matrix whose entries involve the values ofk₁^(j)(α_i), whered=_n

i=1n_i.) Suppose thatGcontains two functions. Then the Hermite-Fej´er matrixM_k1 has norm less than 1. We can then choose a positive number λ > 1 for which ||M_λk1|| < 1. This implies that

||λk₁+qh||_∞ ≤1 for some h∈ H^∞. Letb := k₁+_λ¹qh; then b ∈k+qH^∞ and

||b||_∞≤_λ¹ <1. This proves Lemma3.

In Section1we noticed that ifϕ≡ϕ₊+ϕ₋ ∈L^∞is such thatT_ϕis a hyponormal operator then kerH_ϕ+= kerH_ϕ⊆ker [T_ϕ^∗, T_ϕ]. Thus we can see that ifϕ=f+g, where f ∈ H^∞ and g ∈G_f and if rankH_f <∞ then rank [T_ϕ^∗, Tϕ]≤rankH_f^∗ = rankH_f.

We now have:

Theorem 4. If f ∈H^∞ is such that rankH_f <∞then ∇G_f ⊆extG_f.

Proof. Suppose that rankH_f = N. By the above considerations, if g ∈ G_f and ϕ := f +g then rank [T_ϕ^∗, Tϕ] ≤ N. We observe that if g ∈ ∇G_f then every solutionk of the equation H_fk=zg has exactly norm 1; for, ifk is a solution of the equation H_fk =zg with ||k||∞ <1 then _||k||^k_∞ ∈ E(ψ) for ψ:=f +g/||k||∞, and hence _||k||¹_∞ ·g= _||k||^g_∞ ∈G_f, a contradiction. We now claim that ifg∈ ∇G_f then E(f+g) consists of exactly one finite Blaschke product. To see this observe that by Beurling’s Theorem, kerH_f = q H² for some inner function q. (Recall that the second identity in (1) implies that z(kerH_ϕ) ⊆kerH_ϕ for allϕ ∈L^∞.) Since rankH_f <∞, q must be a finite Blaschke product. Furthermore ifk is in E(f +g), that is, k is a solution of the equation H_fk = zg and ||k||∞ ≤1, then E(f+g) =G(q, k) ={b∈k+q H^∞: ||b||_∞≤1}. By the above considerations and Lemma 3, E(f +g) then contains exactly one element. Since [T_ϕ^∗, Tϕ] is of finite rank it follows from an argument of T. Nakazi and K. Takahashi [NT, Theorem 10] that E(f+g) contains a finite Blaschke product, and consequently, E(f +g) consists of one finite Blaschke product.

To prove ∇G_f ⊆ extG_f, we now assume, without loss of generality, that g₁, g2, ¹₂(g1+g2)∈ ∇G_f; it will suffice to show that g1 =g2. By what we have just discussed, there exist finite Blaschke products b1 and b2 corresponding to g1 and g2, respectively. SinceH_fbi=zgifori= 1,2, it follows that ¹₂(b1+b2) is a solution of the equationH_fk= ¹₂z(g₁+g₂). Further since||¹₂(b₁+b₂)||_∞≤1, we have that

12(b₁+b₂)∈ E(f+¹₂(g₁+g₂)). But since ¹₂(g₁+g₂)∈ ∇G_f, it follows that ¹₂(b₁+b₂) is a finite Blaschke product. However since Blaschke products are extreme points of the unit ball of H^∞ (cf. [Ga, p. 179]), we can conclude that b₁ = b₂, which impliesg₁ =g₂. (In fact, by an argument of K. deLeeuw and W. Rudin [dLR], if f ∈H^∞, ||f||_∞= 1, thenf is an extreme point of the unit ball ofH^∞ if and only if

log(1− |f(e^iθ)|)dθ=−∞.) This completes the proof of Theorem4.

References

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Department of Mathematics, University of Iowa, Iowa City, IA 52242 [email protected] http://www.math.uiowa.edu/˜curto/

Department of Mathematics, Sungkyunkwan University, Suwon 440-746, Korea [email protected]

This paper is available via http://nyjm.albany.edu:8000/j/2001/7-13.html.