PROPERTIES OF EXPOSED POINTS IN THE UNIT BALL OF H^{1}.

Alexei G. Poltoratski

Abstract. We investigate properties of exposed points in the unit ball of the Hardy space H^{1} in connection with the well known conjecture by D. Sarason. We construct a counterexample for the remaining part of the conjecture.

Introduction

This note is devoted to the discussion of the properties of the set of exposed points in the unit ball of the Hardy space H^{1}. A point of a convex set in a vector space is called exposed if there exists a linear functional attaining its maximum over the set at this point and nowhere else. The Hardy space H^{p}, 1 ≤ p < ∞ is the space of analytic functions f in the unit disk 𝔻 = {|z| < 1} such that the norm

||f||_{p} = sup_{0<r<1} (∫_{𝕋} |f(rξ)|^{p}dm(ξ))^{1/p} ,

where 𝕋 = {|z| = 1} and m is the normalized Lebesgue measure on 𝕋, is finite. Every function from H^{p} can be defined a. e. on 𝕋 by its nontangential boundary values.

A nonzero function f ∈ H^{1} is called rigid if the set of all functions g ∈ H^{1} such that arg g = arg f a. e. on 𝕋 consists only of positive scalar multiples of f (here and throughout the paper arg stands for the principal branch of argument taking values in [0; 2π)). As was observed in [BJH] and [S1], a function f from the unit ball of H^{1} is exposed iff it is rigid and ||f||_{1} = 1. Indeed, since all linear functionals on H^{1} are induced by L^{∞} functions on 𝕋, the only functional of unit norm that is equal to 1 on f is the one induced by function |f|/f. It is left to notice that such a functional will attain the same value on all functions of unit norm with the same argument as f.

Rigid (exposed) functions can also be characterized in terms of Toeplitz operators, namely if f is an outer function in H^{2} then f^{2} is rigid iff the Toeplitz operator T_{f̄/f} has a trivial kernel (see [BJH]). Kernels of such Toeplitz operators corresponding to nonrigid functions f play an important role in the study of so-called almost-unitary operators [M]. The connections between rigidity and properties of de Branges-Rovnyak spaces are investigated in [S1]. Rigid functions also appear in the problem of characterizing spectral densities of completely nondeterministic Gaussian processes in prediction theory (see

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 30E20.

Key words and phrases. analytic functions, Hardy classes, exposed points.

[BJH]). Namely, a Gaussian process is completely nondeterministic if and only if its spectral measure is equal to |h|m, where h is an exposed point in the unit ball of H^{1}. For other references on rigid functions and related topics see [S1].

It is not difficult to notice that all exposed points in the unit ball of H^{1} are outer (see [S1]). Indeed, if f = θg and θ is inner, then (1 − θ)^{2}g has the same argument, and therefore f is not rigid. The same trick produces examples of outer but not rigid functions: if f = (1 − θ)^{2} then f is outer but not rigid because θ has the same argument. Since all exposed points are extreme, the fact that exposed functions are outer also follows from a Theorem by W. Rudin and K. de Leeuw, which says that all extreme functions in the unit ball of H^{1} are outer.

In [S2] D. Sarason formulated a conjecture on the structure of the set of the exposed points in the unit ball of H^{1}. Before we present the conjecture we need the following definition.

Let μ be a positive Borel measure on the unit circle 𝕋, ||μ|| = 1. Denote by Kμ the Cauchy integral of μ in the unit disk 𝔻:

Kμ = ∫_{𝕋} \frac{1 + ξ̄z}{1 − ξ̄z} dμ(ξ).

Since μ is positive, Kμ takes its values in the right half-plane {ℜz > 0}. Consider the family {Ω_{α}}_{α∈𝕋} of Möbius transforms of the right half-plane into itself:

Ω_{α}(z) = \frac{(α − 1) + (α + 1)z}{(α + 1) + (α − 1)z} .

Then for any α ∈ 𝕋 function Ω_{α}(Kμ) has positive real part in the unit disk 𝔻 and is equal to 1 at 0. Thus there exists a positive Borel measure μ_{α} on 𝕋 such that

Kμ_{α} = Ω_{α}(Kμ).

We will denote by M_{μ} the family {μ_{α}}_{α∈𝕋} of all such measures corresponding to μ (note that then μ_{1} = μ).

Families M_{μ} appear in problems of perturbation theory and mathematical physics. We will discuss one of such problems in Section 2. In [S1] such families of measures are studied in connection with the de Branges–Rovnyak spaces of analytic functions in the unit disk, which will be defined later in the introduction.

The following conjecture was stated in [S2]. Let f be an outer function from H^{1}. Then the following 3 conditions are equivalent:

1) f is rigid.

2) There is no nonconstant inner function θ such that f/(1 − θ)^{2} is in H^{1}.

3) The family M_{μ}, where μ = |f|m, consists of absolutely continuous measures.

As we discussed before, the implication (1) ⇒ (2) is simple. The implications (1) ⇒ (3) and (2) ⇒ (3) are established in [S2] and [S3].

The implications (2) ⇒ (1) and (3) ⇒ (1) are shown to hold in certain cases in [S4] and [S3]. Also, V. Kapustin [K] has recently shown that f is rigid iff f is outer and \sqrt{f} is not divisible in H^{2} by any Cauchy integral of a real measure (the part (2) is equivalent to the statement that \sqrt{f} is not divisible by any Cauchy integral of a positive measure).

On the negative side, an example showing that the implication (2) ⇒ (1) fails in general was constructed by J. Inoue in [I].

The problem of verification of the remaining implication (3) ⇒ (2) is contained in [S]. We will solve this problem in Section 2.

As we mentioned before, the inverse implication (1) ⇒ (3) follows from the results on the de Branges–Rovnyak spaces. For ϕ ∈ H^{∞}(𝔻), ||ϕ||_{∞} ≤ 1, ϕ(0) = 0 consider the Toeplitz operator T_{ϕ} on H^{2}(𝔻). Let ψ be an outer function such that |ψ| = (1 − |ϕ|^{2})^{1/2} on 𝕋, ψ(0) > 0 and let T_{ψ} be the corresponding Toeplitz operator. Define spaces H(ϕ) and M(ψ) as

H(ϕ) = range(I − T_{ϕ}^{∗}T_{ϕ})^{1/2}

with the norm which makes (I − T_{ϕ}^{∗}T_{ϕ})^{1/2} an isometry from H^{2} to H(ϕ) and

M(ψ) = rangeT_{ψ}

with the norm which makes T_{ψ} an isometry from H^{2} to M(ψ). Space H(ϕ) has the following integral representation. Consider the family of positive measures {μ_{α}}_{α∈𝕋} such that

Kμ_{α} = \frac{α + ϕ}{α − ϕ}.

(Note, that then {μ_{α}}_{α∈𝕋} = M_{μ_{1}}.) Then for each α ∈ 𝕋 the integral operator

V_{μ_{α}} : f ↦ \frac{Kfμ_{α}}{Kμ_{α}}

is unitary from L^{2}(μ_{α}) to H(ϕ). As was shown in [S1], if μ_{1} = |f|m for some rigid f then M(ψ) is dense in H(ϕ), in which case all μ_{α} must be absolutely continuous. This proves the implication (1) ⇒ (3) in Sarason’s conjecture.

However, as shown in Section 2, the implication (3) ⇒ (2), and therefore (3) ⇒ (1), fails. There exists a function f divisible by (1−θ)^{2} in H^{1} for which all associated measures are absolutely continuous.

As we will see in Section 1, for every nonrigid outer function f there exists n, 1 ≤ n ≤ ∞ such that for some nonconstant inner functions θ_{k}, 1 ≤ k ≤ n the product

(1)  f × ∏_{1}^{n} (i\frac{1 + θ_{k}}{1 − θ_{k}} )^{2}

is an outer function from H^{1}. Define the order of a nonrigid function f, ||f||_{1} = 1 as the minimum of such n. Then, since (1 + θ)(1 − θ)^{−1} + 1 = 2(1 − θ)^{−1}, the part (1) ⇔ (2) of the conjecture is equivalent to the statement that the order of a nonrigid function is always 1.

The simplest example of a nonrigid function of order 1 is f(z) = (1 − z)^{2}. It is not difficult to check that the nonrigid function constructed in the counterexample in [I] has order ∞. In fact, all known examples of non-exposed points give nonrigid functions of orders 1 or ∞. Moreover, as we will see in Section 1 (Theorem 1) any nonrigid function having only isolated zeros on 𝕋 (in Inoue’s example it has only one zero) must have order 1 or ∞.

However, in more complicated situations nonrigid functions of other finite orders emerge. In Section 1 we will prove that for any positive integer n there exists a nonrigid function of order n.

The difference between the case of functions with isolated zeros (in which only orders 1 or ∞ are possible) and the general case, reflects a certain difference between the case of onepoint spectrum and the general case in the study of almost-unitary operators mentioned above.

In Section 2 we start with a discussion of relations between the part (2) ⇔ (3) of Sarason’s conjecture and Perturbation Theory of self-adjoint operators. Let A be a cyclic self-adjoint operator with absolutely continuous spectrum on (0; 1), ϕ be a cyclic vector and μ be the spectral measure of A corresponding to ϕ. Denote by A_{λ}, λ ∈ ℝ the rank one perturbations of A:

A_{λ} = A + λ(·, ϕ)ϕ.

Then the part (2) ⇔ (3) of Sarason’s conjecture is equivalent to the following statement:

2′) For any inner function θ in the upper half-plane (1 − θ)^{−2} is not summable on (0; 1) with respect to μ.

⇕

3′) All A_{λ} have absolutely continuous spectrum on (0; 1).

If we replace arbitrary θ with c exp(ix) in (2’) and ”absolutely continuous” with ”continuous” in (3’), we will obtain the result proved by B. Simon and T. Wolff in [S-W] (see Theorem 10 in Section 2). The implication (2′) ⇒ (3′) follows from the implication (2) ⇒ (3) established in [S3]. However, despite all the existing evidence, the implication (3′) ⇒ (2′) ((3) ⇒ (2)) fails in general, as follows from the example presented in Section 2.

1. The order of a nonrigid function

As we mentioned in the introduction, a function from the unit sphere of H^{1} is an exposed point in the unit ball iff it is rigid. We also saw that every rigid function is outer. We will now discuss what makes an outer function rigid.

Recall that by the definition an outer function f ∈ H^{1} is nonrigid iff there is a function g ∈ H^{1} such that g/f is positive a. e. on 𝕋. Function g can also be chosen outer: if g = θG, where θ is inner, then (1 − θ)^{2}G has the same argument. If f and g are outer then

(2)  \frac{g}{f} = exp(2πKv)

where v is a real summable function on 𝕋. Since g/f is positive a. e. on 𝕋, the conjugate function u = ṽ, defined by the conjugate Poisson integral

u = Qv = ∫_{𝕋} \frac{2ℑξ̄z}{|ξ − z|^{2}} v(ξ)dm(ξ),

satisfies u(𝕋) ⊂ ℤ (here by u(𝕋) we denote the set of all essential values of u on 𝕋).

Denote by S the set of analytic in 𝔻 nonconstant functions representable in the form of the right hand side in (2), i. e.

S = {exp(2πKv)|v ∈ L^{1}, v ≠ const, ṽ(𝕋) ⊂ ℤ}.

Note, that when u = ṽ is also summable, we have v = −ũ and the corresponding function h ∈ S admits two representations:

h = exp(2πKv) = exp(2πiKu).

As one can see from (2), an outer function f ∈ H^{1} is nonrigid iff there exists h ∈ S such that fh ∈ H^{1}.

Let h ∈ S, h = exp(2πKv), u = ṽ. Denote γ(h) = ess sup u − ess inf u, i. e. γ(h)/2 is the norm of u in the factor space L^{∞}(𝕋)/ℂ.

Definition of the order of a nonrigid outer function. If f is a nonrigid outer function from H^{1}, we define its order δ(f) as

δ(f) = min{γ(h)|h ∈ S, fh ∈ H^{1}}.

As one can see, δ(f) is an integer number between 1 and ∞. In this section we will discuss which values δ(f) can attain in different situations.

We will now supply an equivalent definition in terms of inner functions. Note that if h ∈ S and γ(h) = n < ∞ then there exist inner functions θ_{1}, ..., θ_{n} such that

(3)  h = ∏_{1}^{n} (i\frac{1 + θ_{k}}{1 − θ_{k}} )^{2} .

Indeed, if h = exp(i2πKu) then u = l + ∑_{1}^{n} u_{k} where l is an integer and u_{k} are functions on 𝕋, each taking only values 0 and 1. Then for k = 1, ..., n the function U_{k} = exp(iπKu_{k}) has positive imaginary part in 𝔻 and is real a. e. on 𝕋. Therefore the functions θ_{k} = (iU_{k} + 1)/(iU_{k} − 1) are inner. Since exp(iπKu_{k}) = i\frac{1+θ_{k}}{1−θ_{k}}, we obtain (3).

A similar argument can be carried out in the case γ(h) = ∞, although in this case one needs to care about the convergence of the product in (3).

Presentation (3) implies that the following definition is equivalent to the last one.

Another definition of the order of a nonrigid outer function. If f is a nonrigid outer function in H^{1}, then its order δ(f) is the minimal positive integer n for which there exist inner functions θ_{1}, ..., θ_{n} such that the product

(4)  ∏_{1}^{n} (i\frac{1 + θ_{k}}{1 − θ_{k}} )^{2}

is nonconstant and

f × ∏_{1}^{n} (i\frac{1 + θ_{k}}{1 − θ_{k}} )^{2}

is from H^{1}. If the set of such integers n is empty, we say that δ(f) = ∞.

This definition makes it possible to establish connections with Sarason’s conjecture. Namely, as we mentioned in the introduction, the part (1) ⇔ (2) of the conjecture is equivalent to the statement that all nonrigid functions have order 1. This is true for all known explicit examples of nonrigid functions except for the Inoue’s example, where the order is infinity. The following Theorem says that orders 1 and ∞ are indeed quite common.

As usual, we assume that our analytic functions are defined a. e. on 𝕋. In particular, a function f has a zero at ξ ∈ 𝕋 if f(z) tends to 0 as z tends to ξ nontangentially. We will say that a function f ∈ H^{1} has an isolated zero at ξ ∈ 𝕋 if f has a zero at ξ and there exists a neighborhood V ⊂ 𝕋 of ξ such that for any other neighborhood W ⊂ 𝕋 of ξ such that m(V \ W) > 0

ess inf_{V \W} |f| > 0.

Theorem 1. Let f be a nonrigid outer function. Suppose that all zeros of f on 𝕋 are isolated. Then the order of f is either 1 or ∞.

In general, however, δ(f) can attain other integer values.

Theorem 2. For any n ∈ ℤ̂ = ℤ ∪ {∞}, n ≥ 1 there exists a nonrigid outer function f of order n.

Proofs of Theorems 1 and 2 require some preparation.

When we say that a function is increasing (decreasing) on an arc of 𝕋 we mean that it is essentially increasing (decreasing) in the counterclockwise direction. I. e. u is increasing on I ⊂ 𝕋 if there is a set J ⊂ I of measure 0 such that the values of u are increasing when we move along I \ J counterclockwise.

The following Lemma can be easily deduced from the results of [NN].

Lemma 3. Let f ∈ S, f = exp(2πKv), u = ṽ. Let I be an arc on 𝕋. If

∫_{I} |f(ξ)|dm(ξ) < ∞

then u is nondecreasing on I.

Lemma 4. Let u ∈ L^{∞}(𝕋), u(𝕋) ⊂ ℤ. Then there exists a singular measure μ on 𝕋 such that for μ-a. e. point ξ ∈ 𝕋

\frac{Kμ(z)}{exp(iπKu)(z)} = O(1)

as z tends nontangentially to ξ.

Proof. WLOG u ≥ 0. Suppose ||u||_{∞} = n. Put v_{1} = χ_{{u=n}} and v_{2} = (u − nv_{1})/(n − 1). Then there exist a positive singular measure μ and a positive measure ν on 𝕋 such that i exp(iπKv_{1}) = Kμ and i exp(−iπKv_{2}) = −Kν. Then

\frac{Kμ}{exp(iπKu)} = \frac{Kμ}{exp(iπnKv_{1}) exp(−iπ(n − 1)Kv_{2})} = (−1)^{n−1} (\frac{Kν}{Kμ})^{n−1} .

But the ratio

\frac{Kν}{Kμ}

has finite nontangential boundary values μ-a. e. as was proved in [P1].

■

In a similar way one can prove the following statement.

Lemma 5. Let I be an open arc on 𝕋, u ∈ L^{∞}, u(𝕋) ⊂ ℤ. Then u is nondecreasing on I iff

lim sup_{r→1−} Ku(rξ)

is finite for any ξ ∈ I.

We will also need the following Lemma.

Lemma 6 [P1]. Let f ∈ H^{1} and μ be a finite complex singular measure on 𝕋. Then

lim_{z⟶_{≯}ξ} \frac{f(z)}{Kμ(z)} = 0

for μ-a. e. ξ.

Proof of Theorem 1. Suppose that the order of f is less than ∞. Then there exist h ∈ S, h = exp(2iπKu), u ∈ L^{∞} such that g = fh ∈ H^{1}. By Lemma 4 there exists a singular measure μ on 𝕋 such that

(5)  \frac{Kμ(z)}{h(z)} = O(1)

as z tends nontangentially to ξ, for μ-a. e. point ξ ∈ 𝕋. For such a measure μ we must have f = 0 μ-a. e. on 𝕋. Indeed,

lim_{z⟶_{≯}ξ} \frac{g(z)}{Kμ(z)} = 0

for μ-a. e. ξ by Lemma 6. But

\frac{g(z)}{Kμ(z)} = f \frac{h(z)}{Kμ(z)}.

Therefore by (5) f = 0 μ-a.e. Since f is outer and has only isolated zeros, the set {f = 0} on 𝕋 is countable. Hence, μ is a discrete measure and there exists an isolated zero ξ of f such that μ({ξ}) > 0. Let I be a small open arc centered at ξ such that for any neighborhood W of ξ, if m(I \ W) > 0 then

(6)  ess inf_{I\W} |f| > 0.

Denote by I_{+} and I_{−} the halves of I, i. e. disjoint open subarcs of I such that I_{+} ∪I_{−} = I \ {ξ}. Since |Kμ| tends to ∞ nontangentially at ξ, h must also tend to ∞ nontangentially at ξ by (5). Therefore by Lemma 5, u can not be nondecreasing on I. But it is nondecreasing on each half I_{+} and I_{−} by Lemma 3, because hf ∈ H^{1} and by (6). Hence u has a jump discontinuity at ξ.

Suppose ξ = exp(iϕ). Define v to be equal to 1 on {exp(iψ)|ϕ < ψ < ϕ + π} and to be equal to 0 elsewhere on 𝕋. Then v = nu a. e. in some neighborhood of ξ, where n is the height of the jump of u at ξ. Hence f exp(2πiKv) must belong to H^{1}. Therefore the order of f is 1.

■

Let ξ = exp(iϕ) ∈ 𝕋, ϵ > 0. We will denote by V_{ϵ}(ξ) the ϵ-neighborhood of ξ:

V_{ϵ}(ξ) = {exp(iψ)|ϕ − ϵ < ψ < ϕ + ϵ}.

We will denote by C the standard Cantor set on the unit circle constructed as follows. Put C_{0} = J_{0,1} = 𝕋; C_{1} = J_{1,1} = 𝕋 \ I_{0,1}, where I_{0,1} is the \frac{π}{3} -neighborhood of 1; C_{k} = J_{k,1} ∪ J_{k,2}... ∪ J_{k,2^{k−1}} for k ≥ 2, with J_{k,2n−1} ∪ J_{k,2n} = J_{k−1,n} \ I_{k−1,n} (n ≤ 2^{k−2}), where I_{k−1,n} is the open arc of the size |J_{k−1,n}|/3 centered at the middle of J_{k−1,n}. The Cantor set C is defined as C = ∩_{k}C_{k}.

We also define the standard Cantor measure γ on C as the ∗-weak limit of the sequence of measures

\frac{m|_{C_{k}}}{m(C_{k})}.

Denote by P γ the Poisson integral of γ in 𝔻:

P γ(z) = ∫_{𝕋} \frac{1 − |z|^{2}}{|ξ − z|^{2}} dγ(ξ).

The following property of the Cantor measure γ is not difficult to check.

Lemma 7. For any ξ ∈ C

|P γ((1 − \frac{π}{3^{n}})ξ)| > A (\frac{3}{2})^{n}

for n = 1, 2, ... for some absolute constant A.

We will also need the following (probably, well known) property of outer functions with bounded arguments.

Lemma 8. Let f = exp(iKu) where u is a real function from L^{∞}(𝕋). Suppose that for some sequence {r_{n}} of positive numbers monotonically increasing to 1 we have

|f(r_{n})| → ∞

as n → ∞. Then for any ϵ > 0 there exist c > 0 and N ∈ ℕ such that for n > N

\frac{|V_{(1−r_{n})}(1) ∩ \{ξ ∈ 𝕋| |f(ξ)| < c|f(r_{n})|\}|}{1 − r_{n}} < ϵ.

Proof. If

g = exp(−\frac{π}{4||u||_{∞}} iKu)

then the statement translates into the following: for any ϵ > 0 there exists c_{1} > 0 and N such that for any n > N

\frac{|V_{(1−r_{n})}(1) ∩ \{ξ ∈ 𝕋| |g(ξ)| > c_{1}|g(r_{n})|\}|}{1 − r_{n}} < ϵ.

But 0 ≤ ℜg ≤ |g| ≤ \sqrt{2}ℜg and therefore we need to show that for any ϵ > 0 there exists c_{2} and N such that for any n > N

\frac{|V_{(1−r_{n})}(1) ∩ \{ξ ∈ 𝕋| ℜg(ξ) > c_{2}ℜg(r_{n})\}|}{1 − r_{n}} < ϵ.

Since ℜg is the Poisson integral of a positive function, the last inequality follows from the standard properties of Poisson kernel.

■

We will also need the following technical Lemma.

Lemma 9. Let u ∈ L^{∞}(𝕋), u(𝕋) ⊂ ℤ, ||u||_{∞} = n and f = exp(iπKu). Suppose that for some ξ ∈ 𝕋 and ϵ > 0, u is constant on the ϵ/2-neighborhood V_{ϵ/2}(ξ) and nondecreasing on the ϵ-neighborhood V_{ϵ}(ξ) of ξ. Then

(7)  max_{V_{ϵ/2}(ξ)} f ≤ cf(ξ)

for some constant c depending only on n.

Note that since u is constant on V_{ϵ/2}(ξ), f has a nontangential limit at ξ and therefore the value f(ξ) is correctly defined.

Proof. To prove the Lemma let us estimate the difference Qu(ζ) − Qu(ξ) for ζ ∈ V_{ϵ/2}(ξ). Denote ξ = exp(iϕ). From the properties of conjugate Poisson kernel, it is not difficult to deduce that this difference achieves its maximum, over all functions u satisfying the conditions of the Lemma, when u = u_{0}, where u_{0} is defined as u_{0} = n on the arc {exp(iψ)|ϕ − ϵ < ψ < ϕ + ϵ/2} and u_{0} = 0 elsewhere on 𝕋. Once again, from the definition of conjugate Poisson kernel one can easily see that

Qu(ζ) − Qu(ξ) ≤ Qu_{0}(exp(i(ϕ + ϵ/2)) − Qu_{0}(ξ) ≤ c_{1} + 2n ln 2

which implies the statement.

■ Definition. We will say that a function f on 𝕋 is locally unsummable at ξ ∈ 𝕋 if for any ϵ > 0

∫_{V_{ϵ}(ξ)} |f(ζ)|dm(ζ) = ∞.

Proof of Theorem 2.

Case n = 1. It is not difficult to check that (1 − z)^{2} is a nonrigid function of order 1. Case n = ∞. The example we present here is essentially the same as the one in [I],

although in our construction we use different techniques. We include this construction to make the paper more self-contained. Those readers familiar with the example from [I] may notice that by Theorem 1 the function constructed there has order ∞, and skip the present part of the proof.

We start by constructing a function g from S.

Consider a real function u on 𝕋 defined as u = 0 on {exp(iϕ)|π ≤ ϕ ≤ 2π}, and u = −n on {exp(iϕ)|π/2^{n−1} ≥ ϕ ≥ π/2^{n}} for n = 1, 2, .... Then Qu ∈ L^{1} and therefore g = exp(2πiKu) belongs to S.

Function g/(z − exp(iπ/2^{n}))^{2} is bounded in some neighborhood of exp(iπ/2^{n}) for any n ∈ ℕ. Indeed, one can easily check that

(z − exp(iπ/2^{n}))^{2} = exp(2πiKχ_{{ξ|π/2^{n}<arg ξ<π/2^{n}+π}})

and therefore

g/(z − exp(iπ/2^{n}))^{2} = exp(2πiK(u − χ_{{ξ|π/2^{n}<arg ξ<π/2^{n}+π}})).

It is left to notice that

u − χ_{{ξ|π/2^{n}<arg ξ<π/2^{n}+π}}

is constant in some vicinity of exp(iπ/2^{n}).

Since g/(z−exp(iπ/2^{n}))^{2} is bounded in some neighborhood of exp(iπ/2^{n}), 1/g is locally unsummable at exp(iπ/2^{n}) for any n ∈ ℕ. Also, |1/g| is bounded away from 0 outside of any neighborhood of 1. Therefore there exists a positive function ϕ on 𝕋 such that ϕ < |1/g|, ϕ, ln ϕ ∈ L^{1}, ϕ/(1 − z)^{2} ∉ L^{1} and ϕ is bounded away from zero outside of any neighborhood of 1.

Let f be an outer function such that |f| = ϕ. Then f ∈ H^{1}. Also fg ∈ H^{1}, and therefore f is nonrigid. Also, f can not be divided in L^{1} by (1 − θ)^{2} for any inner θ. Indeed, since f has only one zero at 1, we only have to show that f can not be divided (1 − z)^{2}. But f/(1 − z)^{2} ∉ L^{1}, by our construction, and therefore the order of f is not 1. Hence, by Theorem 1, f has order infinity.

Case 1 < n < ∞. Once again, we will start by constructing a function g ∈ S.

Let C = 𝕋 \ ⋃ I_{k,l} be the Cantor set constructed before Lemma 7. On each complimentary arc I_{k,l} consider points a_{k,l}^{1}, ..., a_{k,l}^{n} dividing I_{k,l} into n + 1 equal subarcs, i. e. if I_{k,l} = {0 < ϕ_{k,l} < arg z < ψ_{k,l} < 2π} then arg a_{k,l}^{m} = ϕ_{k,l} + m(ψ_{k,l} − ϕ_{k,l})/(n + 1). Let us define function u on 𝕋 as follows: on each I_{k,l} = {0 < ϕ_{k,l} < arg z < ψ_{k,l} < 2π}

u is equal to 0 on {ϕ_{k,l} < arg z < arg a_{k,l}^{1}}, u is equal to m on {arg a_{k,l}^{m} < arg z < arg a_{k,l}^{m+1}} for 1 ≤ m ≤ n − 1 and u is equal to n on {arg a_{k,l}^{n} < arg z < ψ_{k,l}}. Put g = exp(2πiKu). Then g has simple zero at each a_{k,l}^{m}. Denote

V_{k,l}^{m} = V_{\frac{π}{3^{k}2(n+1)}} (a_{k,l}^{m}).

Then V_{k,l}^{m} ⊂ I_{k,l}, |V_{k,l}^{m}| = |I_{k,l}|/4(n + 1) and all V_{k,l}^{m} are disjoint. Define a positive function f such that

1) ln f, f ∈ L^{1}(𝕋),

2) |f| < |1/g|,

3) f is bounded away from zero on any closed subset of 𝕋 \ C,

4) for any k, l, m ∫_{V_{k,l}^{m}} |f| = α_{k}, where a_{k} > 0 is a sequence of numbers such that

∑ 2^{k}α_{k} < ∞ but ∑(\frac{9}{4})^{k}α_{k} = ∞.

Note that it is possible to satisfy (2) and (3) simultaneously because g is bounded on any closed subset of 𝕋 \ C. It is also possible to choose f satisfying (4) because 1/g is locally unsummable at every point a_{k,l}^{m} (which can be proved in the same way as in the previous part of the proof). The condition ∑ 2^{k}α_{k} < ∞ is necessary for f to belong to L^{1} because for any fixed k there are n2^{k} arcs V_{k,l}^{m}.

Let F be an outer function such that |F | = f. We claim that F is a nonrigid function of order n. Indeed, δ(F) ≤ n because F g ∈ H^{1} by condition (2) above. Suppose that δ(F) < n. Then there exists a function v ∈ L^{∞}(𝕋), 0 ≤ v ≤ n − 1, v(𝕋) ∈ ℤ such that F h ∈ H^{1} for h = exp(2πiKv).

By Lemma 4 there exists a singular measure μ on 𝕋 such that |h| grows no slower (up to a constant factor) than |Kμ|^{2} on μ-a. e. radius. By Lemma 3 and condition 3) imposed on f, v must be nondecreasing on every complementary arc I_{k,l}. Therefore, by Lemma 5, μ is supported on C. As was proved in [P1], the ratio

\frac{Kγ}{Kμ}

has finite nontangential limits μ-a. e. This means that there exists ξ ∈ C such that |Kμ(rξ)|^{2} grows no slower (up to a constant factor) than |Kγ(rξ)|^{2} as r → 1−. Therefore by Lemma 7

(8)  |h((1 − \frac{π}{3^{k}})ξ)| ≥ c_{1}(\frac{9}{4})^{k}

for some constant c_{1} > 0.

For any k there exists l_{k} such that the arc I_{k,l_{k}} lies in \frac{π}{3^{k−1}} -neighborhood of ξ. Now we will show, that (8) implies that for any k there exists V_{k,l_{k}}^{m_{k}} ⊂ I_{k,l_{k}} such that

(9)  min_{V_{k,l_{k}}^{m_{k}}} |h| ≥ c_{2}(\frac{9}{4})^{k}

for some c_{2} > 0. Since ∑(\frac{9}{4})^{k}α_{k} = ∞ this will imply that F h ∉ H^{1}.

To prove that there exists V_{k,l_{k}}^{m_{k}} satisfying (9), for each m ≤ n consider W_{k,l_{k}}^{m}, the \frac{π}{3^{k}(n+1)}-neighborhood of a_{k,l_{k}}^{m}. Then W_{k,l_{k}}^{m} ⊃ V_{k,l_{k}}^{m}, |W_{k,l_{k}}^{m}| = 2|V_{k,l_{k}}^{m}| and W_{k,l_{k}}^{m} are still pairwise disjoint. Since v has at most n − 1 jumps on each I_{k,l_{k}}, there exists at least one arc W_{k,l_{k}}^{m_{k}} on which v is constant. Since v is nondecreasing on each I_{k,l_{k}}, by Lemma 9 for each ξ ∈ V_{k,l}^{m_{k}}

max_{V_{k,l_{k}}^{m_{k}}} |h| ≤ max_{V_{|V{m_{k} \atop k,l_{k}} |}(ξ)} |h| < c_{4}|h(ξ)|

and therefore

(10)  max_{V_{k,l_{k}}^{m_{k}}} |h| ≤ c_{4} min_{V_{k,l_{k}}^{m_{k}}} |h|

for some c_{4} > 0. Since each I_{k,l_{k}} is in \frac{π}{3^{k−1}} -neighborhood of point ξ satisfying (6) and |V_{k,l_{k}}^{m_{k}}|/|I_{k,l_{k}}| > c_{5} > 0, by Lemma 8 there exists c_{6} > 0 such that

max_{V_{k,l_{k}}^{m_{k}}} |h| > c_{6}(\frac{9}{4})^{k}.

By (10) this implies (9). Therefore F h ∉ H^{1} and we have a contradiction. Thus F has order n.

■

Remark. Let f be a nonrigid outer function. Define spectrum of f as

σ(f) = {n ∈ ℕ| there exists h ∈ S such that γ(h) = n and fh ∈ H^{1}}.

We just proved that δ(f) = min σ(f) can attain any value from ℕ̂ = ℕ ∪ {∞}. In a similar manner one can show that, more generally, σ(f) can be any subset of ℕ̂. This fact may prove to be useful in the study of almost-unitary operators acting on the spaces generated by nonrigid functions [M].

2. A counterexample to the part (2) ⇔ (3) of the main conjecture

We start this section with a discussion of connections of our topic with Perturbation Theory. Let A be a cyclic self-adjoint operator acting in a separable Hilbert space, ϕ its cyclic vector, μ the spectral measure of A corresponding to ϕ:

μ(B) = (E_{B}(A)ϕ, ϕ)

for any Borel set B ∈ ℝ, where E_{B}(A) is the spectral projection of A. Consider self-adjoint rank one perturbations of A:

A_{λ} = A + λ(·, ϕ)ϕ, λ ∈ ℝ.

Let μ_{λ} be the spectral measure of A_{λ} corresponding to ϕ. Then the following statement, which easily follows from the results of [SW], holds true.

Theorem 10. All A_{λ} are continuous (all μ_{λ} have no point masses) on (0; 1) iff

∫_{0}^{1} \frac{dμ(x)}{(x − y)^{2}} = ∞

for any y from (0; 1).

If λ ∈ ℝ we will denote by Γ_{λ} the following Möbius transform of the upper half-plane ℂ_{+} onto itself:

Γ_{λ}(z) = \frac{z}{1 + λz} .

Denote by Kμ the Cauchy integral of μ in ℂ_{+}:

Kμ(z) = ∫_{ℝ} \frac{dμ(x)}{x − z}

Then the spectral measures μ_{λ} satisfy the relation

Kμ_{λ} = Γ_{λ}(Kμ)

(see [A]). Therefore the family of measures {μ_{λ}}_{λ∈ℝ} is an upper half-plane analogy of family M_{μ} defined in the introduction. In particular, it is not difficult to see that Theorem 10 above is equivalent to the following statement (see [S1]): all μ_{α} ∈ M_{μ} are continuous iff

(11)  ∫_{𝕋} \frac{dμ(ξ)}{|z − ξ|^{2}} = ∞

for any z ∈ 𝕋.

If we want to replace ”continuous” with ”absolutely continuous” in the last statement, it seems natural to try to replace z with arbitrary inner θ in (11). Indeed, if θ(0) = 0 then θ − ξ has the same distribution on 𝕋 as z − ξ, but instead of having 0 at a single point on 𝕋, function θ − ξ can have more complicated zero sets. After such a replacement, the statement becomes the part (2) ⇔ (3) of Sarason’s conjecture.

I. e., the part (2) ⇔ (3) of the conjecture, formulated in the case of the unit disk, can be ”mapped” to the upper half-plane to obtain the following equivalent statement:

(2′)For any inner function θ in the upper half-plane (1 − θ)^{−2} is not summable on (0; 1) with respect to μ.

⇕

(3′)All A_{λ} are absolutely continuous on (0; 1).

For more details on the relations between families of spectral measures μ_{λ} and families M_{μ} in the unit disk see [P2].

As we mentioned before, the implication (2) ⇒ (3), and therefore the implication (2′) ⇒ (3′), is established in [S3]. The implication (3) ⇒ (1), and therefore (3) ⇒ (2), was shown to hold in certain cases in [S4].

Here we show that the implication (3) ⇒ (2) fails in general by presenting the example of a positive function f ∈ L^{1}(𝕋) and an inner function θ such that f/(1 − θ)^{2} ∈ L^{1} but M_{fm} consists of absolutely continuous measures.

Example 11. We will need the following two Lemmas. If μ is a measure we will denote by μ_{s} the singular component of μ. We will denote by \frac{dμ}{dm} the usual Radon-Nikodim derivative of μ with respect to the normalized Lebesgue measure m on 𝕋. We will write \frac{dμ}{dm}(ξ) ≠ 0 if the derivative at point ξ ∈ 𝕋 is nonzero or does not exist.

Lemma 12. Let μ be a positive finite Borel measure on 𝕋, μ_{α}, α ≠ 1 be one of the measures from the family M_{μ} (defined in the introduction). Then the derivative \frac{dμ}{dm} is equal to 0 (μ_{α})_{s}-almost everywhere.

Proof. Recall that

Kμ_{α} = \frac{(α − 1) + (α + 1)Kμ}{(α + 1) + (α − 1)Kμ}

by the definition of M_{μ}. Since |Kμ_{α}(rξ)| tends to ∞ as r → 1− for (μ_{α})_{s}-a. e. ξ, |Kμ(rξ)| must tend to (1 + α)/(1 − α) for (μ_{α})_{s}-a. e. ξ. Then P μ(rξ) must tend to ℜ(1 + α)/(1 − α) = 0 as r → 1− for (μ_{α})_{s}-a. e. ξ. The statement now follows from the properties of Poisson kernels.

■

It is also not difficult to prove the following statement.

Lemma 13. Let E ⊂ 𝕋 is a set such that for any arc I ⊂ 𝕋 of nonzero Lebesgue measure m(I ∩ E) ≠ 0. Let Σ ⊂ 𝕋 be a set of Lebesgue measure 0. Then there exists an absolutely continuous positive finite measure μ supported on E such that \frac{dμ}{dm} ≠ 0 on Σ.

Let now θ be an inner function whose spectrum contains the whole circle 𝕋. Put f_{1} = |1 − θ|^{2}. Denote Σ = {dμ_{1}/dm = 0}, where μ_{1} = f_{1}m. Then m(Σ) = 0 (in fact one can show that Σ = {θ = 1}). Denote E = {ξ ∈ 𝕋|ℜθ(ξ) < 0}. Then for every arc I ⊂ 𝕋 of nonzero Lebesgue measure m(I ∩ E) ≠ 0. Therefore by Lemma 13 there exists a summable positive function f_{2} supported on E such that for the measure μ_{2} = f_{2}m we have dμ_{2}/dm ≠ 0 on Σ. Put f = f_{1} + f_{2}. Note that f_{2}/(1 − θ)^{2} ∈ L^{1} because 1/(1 − θ)^{2} is bounded on E and f_{2} is supported on E. Therefore f/(1 − θ)^{2} ∈ L^{1}. But for μ = μ_{1} + μ_{2} = (f_{1} + f_{2})m we have {dμ/dm = 0} = ∅ and hence M_{μ} consists of absolutely continuous measures by Lemma 12.

References

[A] N. Aronszajn, On a problem of Weyl in the theory of singular Sturm- Liouville equations, Amer. J. Math. 79 (1957), 597-610.

[BHJ] P. Bloomfield, N. P. Jewell and E. Hayashi, Characterizations of completely nondetermunistic stochastic processes, Pacific J. Math. 107 (1983), 307-317.

[I] J. Inoue, An example of a non-exposed extremal function in the unit ball of H^{1}, Proc. Edinburg Math Soc. 37 (1993), 47-51.

[K] V. Kapustin, Real functions in weighted Hardy spaces I, Preprint.

[M] N. Makarov, Private communications.

[NN] J. Neuwirth and D. J. Newman, Positive H^{1/2} functions are constants, Proc. Amer. Math. Soc. 18 (1987), 958.

[P1] A. Poltoratski, On the boundary behavior of pseudocontinuable functions, St. Petersburg Math. J. 5 (1994), 389-406.

[P2] A. Poltoratski, The Krein spectral shift and rank one perturbations of spectra, MSRI Preprint No. 02-1996.

[S1] D. Sarason, Sub-Hardy Hilbert Spaces in the unit disk, The University of Arkansas lecture notes in the mathematical sciences; v. 10, J. Wiley and Sons, New York, 1994.

[S2] D. Sarason, Exposed points in H^{1}, I, Operator Theory: Advances and Applications 41 (1989), 485-496.

[S3] D. Sarason, Exposed points in H^{1}, II, Operator Theory: Advances and Applications 48 (1990), 333-347.

[S4] D. Sarason, Making an outer function from two inner functions, Contemporary Mathematics 137 (1992), 407-414.

[SW] B. Simon, T. Wolff, Singular continuous spectrum under rank one perturbations and localization for random Hamiltonians, Comm. Pure Appl. Math. 39 (1986), 75-90.

Dept. of Math. 2-171, Massachusetts Institute of Technology, 77 Massachusetts Ave., Cambridge, MA 02139

E-mail address: agp@math.mit.edu