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Boundary Behaviour ofFunctions of Nevanlinna Class

A. BOURHIM, O. EL–FALLAH & K. KELLAY

ABSTRACT. We consider a function f of Nevanlinna class suchthat its radial limit f∗ is square summable on the unit circle T.We give a growth condition on the Taylor coefficients of f andthe rate of decrease of the negatively indexed Fourier coefficientsof f∗ in order that f be in the Hardy space H2(D).

1. INTRODUCTION

Let H∞(D) be the algebra of bounded analytic functions on the open unit disc Dand let H2(D) denote the Hardy space of analytic functions f on D for which

‖f‖H2(D) :=(

sup0≤r<1

∫ 2π

0|f(reit)|2dt

)1/2< +∞.

Let N denote the Nevanlinna class of analytic functions f on D for which

supr<1

∫ 2π

0ln+ |f(reit)|dt < +∞.

If f ∈ N, then by Fatou’s [8, Th. 1.3], the radial limit f∗(eit) = limr→1 f(reit)exists almost everywhere on the unit circle T. We also have ln |f∗| ∈ L1(T) andf = g1/g2, where g1, g2 ∈ H∞(D) (see [8]).

Let N+ be the Smirnov class of analytic functions f ∈ N such that

supr<1

∫ 2π

0ln+ |f(reit)|dt =

∫ 2π

0ln+ |f∗(eit)|dt.

If f ∈ N+, then f = g1/g2, where g1, g2 ∈ H∞(D) and g2 is an outer function,i.e., has the form

g2(z) = exp1

2π

∫ 2π

0

eit + zeit − z log |g2∗(eit)|dt , z ∈ D,

347Indiana University Mathematics Journal c©, Vol. 53, No. 2 (2004)

348 A. BOURHIM, O. EL–FALLAH & K. KELLAY

(see [8]). It is well known that if f ∈ N+ is such that f∗ ∈ L2(T), then f ∈H2(D) and therefore f∗(n) = 0 for every n < 0, where f∗(n) is the nth Fouriercoefficient of f∗ (see [8, Theorem 2.11]). However, this result does not remainvalid if the assumption f ∈ N+ is replaced by f ∈ N, as a counter-example,we can consider f(z) = e−(z+1)/(z−1), z ∈ D. So, one may ask, given a vectorsubspace X of

N∗ := f ∈ N : f∗ ∈ L2(T),what additional conditions can be imposed on the radial limit of a function f ∈ Xso that f ∈ H2(D)? The corresponding problem for X = N∗ was studiedfirst by Shapiro [24]. To be more precise, let (α(n))n≥0 be a positive log–concave sequence, that is α(n)2 ≥ α(n + 1)α(n − 1) for all n ≥ 1, such thatlimn→+∞α(n) = +∞ and let

Nα :=f ∈ N∗ :

∑n≥1

|f∗(−n)|2α(n)2 < +∞.

Shapiro proved by the method of the polynomial approximation that if α(n) =ec√n, n ≥ 0, for some positive constant c, then Nα = H2(D). In [22], Shamoyan

extended this result and proved that Nα = H2(D) if and only if

(1.1)∑n≥1

lnα(n)n3/2 = +∞.

In fact, the result of Shamoyan can also be obtained by using Shapiro’s techniquestogether with Nikolskii’s polynomial approximation theorem (see [20, §2.6] and[4]).

In the present work, we will study this kind of problem for large class ofsubspaces of N∗. Let

Nα,β =f(z) =

∑n≥0

anzn ∈ N : f∗ ∈ L2(T),∑n≥1

|an|2β(n)2

< +∞

and∑n≥1

|f∗(−n)|2α(n)2 < +∞,

where(α(n)

)n≥0 and

(β(n)

)n≥0 are two positive log–concave sequences which

tend to infinity. We establish a condition between the sequences(α(n)

)n≥0 and(

β(n))n≥0 for which Nα,β = H2(D). In fact, as we shall see, the results obtained

in the present paper depend on the growth of the sequence (α(n))n≥0.Obviously, we have H2(D) ⊂ Nα,β ⊂ Nα. As for every function f(z) =∑

n≥0 anzn, (z ∈ D), in N there is a positive constant c such that |an| =O(ec√n), n→ +∞, we note that Nα,β = Nα whenever

limn→+∞

lnβn√n= +∞ .

Boundary Behaviour of Functions of Nevanlinna Class 349

This shows that we deal with a more general problem than the one considered byH. S. Shapiro and F. A. Shamoyan.

Let (α(n))n≥0 and (β(n))n≥0 be two positive sequences which tend to infin-ity. We also would like to point out that Kahane–Katznelson proved in [14] thatfor every g ∈ L2(T) which satisfies

∑n≥1 |g(−n)|2α(n)2 < +∞, there exists an

analytic function on D, f(z) =∑n≥0 anzn, z ∈ D, satisfying

∑n≥0

|an|2β(n)2

< +∞

and f∗ = g. This shows that the assumption f ∈ N in the above definition ofNα,β is necessary.

In order to state our results we need the following definition.

Definition 1.1. A positive sequence (σ(n))n≥0 is said to satisfy the condition(R) if(1) σ(0) = 1, the sequence (σ(n))n≥0 is log-concave, and σ(n) → +∞ as n →

+∞.

(2) The sequence(

lnσ(n)na

)n≥1

is non-increasing for some 0 < a < 12 .

(3) supn≥0

σ(n)σ(n+ 1)

= 1.

Theorem A. Suppose that(α(n)

)n≥0 and

(β(n)

)n≥0 are two positive sequences

satisfying the condition (R) such that

lim infn→+∞

lnα(n)lnn

> 0 and supn≥1

lnα(n2)lnα(n)

< +∞ .

Then Nα,β = H2(D) if and only if

(1.2)∑n≥1

lnα(n+ 1)− lnα(n)lnβ(n)

= +∞.

As a consequence of Theorem A, we obtain the following corollary.

Corollary 1.2. Suppose that for n ≥ 0, α(n) = ec(ln(n+1))q , where c is a posi-tive constant and q ≥ 1. If

(β(n)

)n≥0 is a positive sequence satisfying the condition

(R), then Nα,β = H2(D) if and only if

(1.3)∑n≥1

(lnn)q−1

n lnβ(n)= +∞.

The case when 0 < q < 1 will be treated in Theorem C. In the case wheresupn lnα(n2)/ lnα(n) is not bounded, we have the following theorem.


Theorem B. Let 0 < τ < 12 and α(n) = ecnτ , where c is a positive constant,

and let(β(n)

)n≥0 be a positive sequence satisfying the condition (R). If

(1.4)∑n≥1

1n(1−2τ)/(1−τ)(lnβ(n))1/(1−τ)

= +∞,

then H2(D) =Nα,β.On the other hand, if there is ε > 0 such that

(1.5)∑n≥1

(lnn)τ/(1−τ)+ε

n(1−2τ)/(1−τ)(lnβ(n))1/(1−τ)< +∞,

then H2(D)$Nα,β.

As a consequence, we obtain the following.

Corollary 1.3. Let 0 < a < 1/2, α(n) = ec1na and β(n) = ec2nb , where b,c1 and c2 are positive constants. Then Nα,β = H2(D) if and only if b ≤ a.

Note that if α(n) = ecnτ with τ ≥ 12 , then by Shapiro’s theorem [24],Nα,β =

H2(D).In order to state the case where α(n) increases more slowly than any positive

power of n, we need the following definition.

Definition 1.4. A positive sequence (σ(n))n≥0 is said to satisfy the condition(S1) if(1) σ(0) = 1, the sequence

(σ(n)2

)n≥0 is concave, and σ(n) → +∞ as n →

+∞.

(2) The sequence(lnσ(n)/ ln(n+ 1)

)n≥0 is decreasing to 0.

(3) supn≥0 σ(n)/σ(n+ 1) = 1.The sequence (σ(n))n≥0 is said to satisfy the condition (S2) if(1) σ(0) = 1, the sequence (σ(n)2)n≥0 is concave, and σ(n) → +∞ as n →

+∞ ,(2) limn→+∞ σ(n)2/n = 0 .

Note that if (σ(n))n≥0 is a positive sequence satisfying the condition (S1),then it satisfies both conditions (R) and (S2), such that

supn≥1

lnσ(n2)lnσ(n)

< +∞ .

As examples of positive sequences satisfying the condition (S1), one can considerσ(n) = e(ln(n+1))p with 0 < p < 1 and σ(n) = ln ln(n+ ee).


Theorem C. Let(α(n)

)n and

(β(n)

)n be two positive sequences satisfying (S1)

and (S2) respectively. If β(n) = O(α(n)p) for some p > 0, then Nα,β = H2(D).On the other hand, if

(1.6)∑n≥1


< +∞,

then H2(D)$Nα,β.

In order to give the main ideas used in the proofs of the above results, weintroduce some important notations and definitions that will be used throughoutthis paper.1. We associate with two given positive sequences (α(n))n≥0 and (β(n))n≥1 the

following functions:

1.1. Λα(t) := supn≥1

(n ln(1− t)+ lnα(n)

), t ∈ (0,1).

1.2. hα(t) := tΛα(t), t ∈ (0,1).1.3. Kα(n) := lnα(n)

n, n ≥ 1 and linear on every interval

[n, n+1

[

1.4. ωα,β(n) :=1/α(n) if n ≥ 0

β(−n) if n < 0.

2. Let A(D) denote the disc algebra and A∞(D) := f ∈ A(D) : f (n) ∈ A(D).We put

Z(f) := z ∈ D : f(z) = 0

for f ∈ A(D) ,

Z∞(g) :=+∞⋂n=0

Z(g(n)) for g ∈ A∞(D).

3. A closed subset E of T is said to be Λα-Carleson set if∫ 2π

0Λα(dist(eit, E))dt < +∞.

4. The expression f g means that there is a positive constant C such that(1/C)f ≤ g ≤ Cf .

This paper is organized as follows : In Section 2, we give some basic factsabout hyperfunction. In Section 3, we describe the elements of H2(D) that arecyclic for the unilateral shift, Mz : f → zf , on the weighted Bergman spaces

B2α :=

f(z) =

∑n≥0

anzn analytic on D : ‖f‖2B2α=∑n≥0

|an|2α(n)2

< +∞.


A function f ∈ B2α is called cyclic in B2

α if Spanznf : n ≥ 0 is dense in B2α.

For a positive singular measure µ on T, we associate the singular inner function

Uµ(z) = exp

(1

2π

∫ 2π

0

z + eitz − eit dµ(t)

)for z ∈ D.

In Section 3, we prove the following result.

Theorem D. Let α = (α(n))n≥0 be a positive sequence satisfying the condition(R). If µ is a positive singular measure on T such that µ(E) = 0 for every Λα–Carlesonset E, then the singular inner function Uµ is cyclic in B2

α.

This result is a generalization of Korenblum-Roberts’s theorem [18], [21] con-cerning cyclic singular inner functions on the standard Bergman space (the casewhere α(n) = (1 + n)γ , γ > 0 and Λα(t) ln 1/t). In Section 4, we re-duce our problem to the study of bicyclicity of singular inner functions for bi-lateral shift on the weighted Hilbert space. More precisely, let (α(n))n≥0 and(β(n))n≥0 be two positive sequences satisfying certain regularity conditions anddenote by Sωα,β : (un)n∈Z 7 -→ (un−1)n∈Z the usual bilateral shift on the Hilbertspace `2

ωα,β(Z) of all sequences u = (un)n∈Z such that

‖u‖ωα,β =( ∑n∈Z

|un|2ωα,β(n)2)1/2

< +∞.

A vector u ∈ `2ω(Z) is called bicyclic if [u]`2

ωα,β (Z):= SpanSnωα,βu : n ∈ Z =

`2ωα,β(Z). We prove in this section that Nα,β = H2(D) if and only if Uµ :=(Uµ(n))n≥0 is bicyclic in `2

ωα,β(Z) for all positive singular measures µ on T suchthat µ(E) = 0 for every Λα–Carleson set E. Section 5 is devoted to the proofsof Theorem A and Theorem B. We establish a discrete version of Hruscev’s max-imum principal [12]. Afterwards, we apply this discrete version to prove thatNα,β = H2(D) if either the series (1.2) or the series (1.4) diverges. Using Taylor-Williams’s Theorem (see [30]), we also prove that for every Λα–Carleson set Ethere is an outer function F ∈ A∞(D) such that Z(F) = Z∞(F) = E and∑n≥0 |F(n)|2α(n)2 < +∞. We therefore show that if the series (1.2) converges,

then there is a positive singular measure µ on T supported by a Λα–Carlesonset E and an outer function F vanishing on E such that F/Uµ ∈ Nα,β\H2(D);this establishes the converse of Theorem A. By a similar argument, the converseof Theorem B is proved by using a non–uniqueness theorem of Vinogradov (see[13]) instead of Taylor-Williams’s theorem. In Section 6, we present the proof ofTheorem C. For a positive constant C, we consider

Lipe−CΛα (T) :=f ∈ C(T) : |f(ei(t+h) − f(eit)| = O(e−CΛα(h)), h→ 0

.

We first assume that β(n) = O(α(n)p) for some p > 0 and prove thatLipe−CΛα (T) ⊂ `2

ωα,β(Z) for some positive constant C in the sense that if f ∈


Lipe−CΛα (T), then (f (n))n∈Z ∈ `2ωα,β(Z). Then we let µ be a positive singu-

lar measure on T which is supported by a Λα–Carleson set E and deduce fromShirokov’s division theorem [28, Theorem 9, p. 137], that there is an outer func-tion F ∈ A(D) ∩ Lipe−CΛα (T) such that Z(F) = E and FUµ ∈ Lipe−CΛα (T) ⊂`2ωα,β(Z). Since F = UµFUµ and F is an outer function, we get that Uµ :=(Uµ(n))n∈Z is bicyclic in `2

ωα,β(Z) and then Nα,β = H2(D). Conversely, whenthe series (1.6) converges, there is a positive singular measure µ on T supportedby a Λα–Carleson set E such that its modulus of continuity satisfies ωµ(δ) =O(hβ(δ)), δ → 0+. Just as in the first part of Theorem C, we prove thatthere is a positive constant C such that

∑n∈Z |f (n)|2α(n)2 < +∞ for all f ∈

Lipe−CΛα (T). We use this together with Shirokov’s division theorem to prove thatthere is an outer function F such that F/Uµ ∈ Nα,β\H2(D). In Section 7, weprove that if (α(n))n≥0 is a positive sequence satisfying either the condition (R)with supn≥1

(lnα(n2)/ lnα(n)

)< +∞ or the condition (S1), then the converse

of generalized Korenblum-Roberts’s theorem holds. Finally, in the Appendix, wegive some elementary technical lemmas which we use frequently throughout thispaper.

We end this introduction by mentioning that this work was announced in [5].

2. DUALITY AND HYPERFUNCTIONS

A weight is a mapω : Z→ ]0,+∞[ which satisfies the following properties:

(1) ω(0) = 1,(2) 0 < infn∈Z

(ω(n+1)/ω(n)

) ≤ supn∈Z(ω(n+1)/ω(n)

)< +∞,

(3) lim|n|→+∞ ω(n)1/|n| = 1 where ω(n) = supm∈Z(ω(n+m)/ω(m)).

Letω be a weight, and let `2ω(Z) be the Hilbert space of all complex sequences

u = (un)n∈Z such that

‖u‖ω =( ∑n∈Z

|un|2ω(n)2)1/2

< +∞.

We denote by Sω : (un)n∈Z 7 -→ (un−1)n∈Z the standard invertible bilateral shifton `2

ω(Z). For u ∈ `2ω(Z), we denote by [u]`2

ω(Z) the closed vector subspace of`2ω(Z) generated by

Snωu : n ∈ Z. It is the smallest common closed invariant

subspace for Sω and for its inverse which contains u. If [u]`2ω(Z) = `2

ω(Z), weshall say that u is a bicyclic vector in `2

ω(Z).Let ω := (ω(n))n∈Z be a weight, and letω∗(n) :=ω(−n)−1, n ∈ Z. The

dual space of `2ω(Z) can be identified with `2

ω∗(Z) by the following formula

〈u,v〉 =∑n∈Z

unv−n, u = (un)n∈Z ∈ `2ω(Z) , v = (vn)n∈Z ∈ `2

ω∗(Z) .


Observe that for u = (un)n∈Z ∈ `2ω(Z) and v = (vn)n∈Z ∈ `2

ω∗(Z),

〈Snωu,v〉 =∑m∈Z

umvn−m = (u∗ v)n, n ∈ Z .

Therefore, v is orthogonal to [u]`2ω(Z) if and only if u∗v := ((u∗v)n)n∈Z = 0.

Recall that a hyperfunction on T is an analytic function on C\T which vanishesat infinity, and denote byH (T) the set of all hyperfunctions on T. Recall also thatthe support of a hyperfunction ψ ∈ H (T), denoted by Suppψ, is the smallestclosed subset E of T such that ψ has an analytic extension on C\E. For ψ ∈H (T), we set ψ+ := ψ|D, ψ− := ψ|C\D, and ψ := (ψ+,ψ−) and define asequence (ψ(n))n∈Z by

ψ+(z) :=∑n≥1

ψ(n)zn−1, z ∈ D ,

ψ−(z) := −∑n≤0ψ(n)zn−1, z ∈ C\D.

Assume that ω := (ω(n))n∈Z is a weight and set

Hω(T) :=ψ ∈H (T) : ‖ψ‖ω∗ =

( ∑n∈Z

|ψ(n)|2ω(−n)2

)1/2< +∞

.

For u = (un)n∈Z ∈ `2ω(Z) andψ ∈Hω(T), we define the hyperfunction-product,

u ·ψ, as follows

Çu.ψ(n) :=∑k∈Zukψ(n− k), n ∈ Z.

For v = (vn)n∈Z ∈ `2ω∗(Z), we set v = (v+, v−), where

v+(z) :=∑n≥1vnzn−1 z ∈ D ,

v−(z) := −∑n≤0vnzn−1 z ∈ C\D.

The following transformation F : ψ = (ψ+,ψ−) , (ψ(n))n∈Z is an isometryfrom Hω(T) onto `2

ω∗(Z) and its inverse is given by F−1(v) = v for everyv = (vn)n∈Z ∈ `2

ω∗(Z). This shows that one can identify the dual space of`2ω(Z) with Hω(T), and deduces that if u ∈ `2

ω(Z) and ψ ∈ Hω(T), then ψ isorthogonal to [u]`2

ω(Z) if and only if u.ψ = 0.It should be noted that ifω := (ω(n))n∈Z is a weight such that supn≥0ω(n)

<∞, then the following map

H2(D) -→ `2ω(Z)

f 7 -→ f := (f (n))n∈Z


is continuous; so that the Hardy space H2(D) can be regarded as a subspace of`2ω(Z).

Before stating a result of [10, Proposition 3.12], we add some more notations.For f ∈ N, we denote by D(f) the G.C.D of all factors of inner functions g ∈H∞(D) such that gf ∈ H∞(D). For a hyperfunction ψ ∈ H (T) such that theradial limits of ψ+ and ψ− exist a.e. on T, we set ψ∗ = (ψ+)∗ − (ψ−)∗.

Proposition 2.1. Let f ∈ H2(D), f 6= 0, and ψ ∈H (T) such that

∑n<0

|ψ(n)|2 < +∞.

Then f ·ψ = 0 if and only if the following properties hold.(i) ψ+ ∈ N and D(ψ+) is a divisor of the singular inner factor of f .

(ii) ψ∗ = 0 a.e. on T.

3. CYCLIC VECTORS IN WEIGHTED BERGMAN SPACES

3.1. Λ–Carleson sets. A non–decreasing function ρ on [0, a], a > 0, is saidto be a modulus of continuity if ρ(0) = 0, the function t , ρ(t)/t is decreasingand limt→0+ ρ(t)/t = +∞.

Let Λ be a positive function on (0,1] such that the function t , tΛ(t) is amodulus of continuity. A closed subset E of T is said to be a Λ-Carleson set if

∫ 2π

0Λ(dist(eit, E))dt < +∞.

Denote by | · | the normalized Lebesgue measure on T, and assume that E isa closed subset of T such that |E| = 0. If there is some constant 0 < γ < 1such that the function t , tγΛ(t) is non-decreasing on (0,1], then the followingstatements are equivalent:

(i) E is a Λ−Carleson set.

(ii)∑n |In|Λ(|In|) < +∞, where (In)n is the sequence of the component arcs

of T\E.

(iii)∫ 1

0|Et|dΛ(t) > −∞, where Et := ξ ∈ T : dist(ξ, E) < t.

If Λ(t) = ln 1/t, t ∈ (0,1], then the Λ-Carleson sets are the classicalBeurling–Carleson sets which were introduced by Beurling and studied by Car-leson [6]. It is shown in [31] that a closed subset E of T is a Carleson set if andonly if it is a zero set of a non-identically zero outer function F of A∞(D).

Every closed subset of T which has a Lebesgue measure zero is a Λ–Carlesonset for some function Λ (see [3]). Dolzenko [7] proved that for every Λ–Carlesonset E, there is an outer function f ∈ A(D) such that |f(z)| ≤ e−Λ(dist(z,E)) for


every z ∈ D, in particular, f vanishes on E. Other interesting results of this kindwere obtained by Hruscev [13] and Shirokov [28].

3.2. Generalization of Korenblum–Roberts’s Theorem. This section is de-voted to the study of cyclic singular inner functions for the unilateral shift,Mz : f 7 -→ zf , on the Hilbert space

B2α :=

f =

∑n≥0anzn analytic in D : ‖f‖2

B2α=∑n≥0

|an|2α(n)2

< +∞.

Here α = (α(n))n≥0 is a positive sequence such that limn→+∞α(n) = +∞ andsupn≥0α(n)/α(n + 1) < +∞. Recall that a function f ∈ B2

α is said to be cyclicfor B2

α if the smallest closed invariant subspace of Mz containing f , denoted by[f ]B2

α, coincides with B2

α.Motivated by the description of z-invariant subspaces in Banach spaces of an-

alytic functions, several authors studied the cyclicity of singular inner functions inBanach spaces of analytic function which contain H∞(D). For more informationabout cyclic vectors in Banach spaces of analytic functions, we refer for exampleto the monographs [11], and [26]. It is known that if α(n) = (1+n)p, n ≥ 0,with p > 0, then a singular inner function

Uν(z) := exp

(1

2π

∫ 2π

0

z + eitz − eit dν(t)

), z ∈ D,

is cyclic in the Bergman space, B2α, if and only if its associated positive singular

measure ν puts no mass on Beurling–Carleson sets. The necessity is due to H.S. Shapiro [23] and the sufficiency was proved independently by Korenblum [18]and Roberts [21]. In [20, Section 2.6, Theorem 2], N. K. Nikolskii showed thatif (α(n))n≥0 is a positive log–concave sequence such that

∑n≥1

lnα(n)n3/2 = +∞,

then every singular inner function is cyclic in B2α. Thus, we are particularly in-

terested in the case where α is a positive sequence for which α(n) = O(ena) forsome 0 < a < 1

2 and shall extend the result of Korenblum-Roberts.Now, we are able to state the main result of this section which will be useful

later. A converse of this theorem will be given in Section 7.

Theorem 3.1. Assume that α = (α(n))n≥0 is a positive sequence satisfying thecondition (R). If µ is a positive singular measure on T vanishing on any Λα-Carlesonset, then the singular inner function Uµ is cyclic in B2

α.

Until the end of this section, we shall assume that α = (α(n))n≥0 is a positivesequence satisfying the condition (R) and that Λ is a positive function on (0,1)such that the function t , tΛ(t) is a modulus of continuity.


We need some lemmas and use them together with the arguments of J. Robertsto prove Theorem 3.1.

Remark 3.2. Note that, since limn→+∞α(n) = +∞, the Hardy space H2(D)is continuously embedded in B2

α. Note also that, since (α(n))n≥0 satisfies thecondition (R),

limn→+∞

(supk≥0

α(k)α(n+ k)

)1/n

= supn≥0

α(n)α(n+ 1)

= 1.

Therefore, it follows from [27, Theorem 10 (iii) and (vii)] that every functionφ ∈ H∞(D) induces a linear bounded operator

Mφ : B2α -→ B2

αf 7 -→ φf

such that ‖Mφ‖ = ‖φ‖∞. Finally, note that for every uniformly bounded sequence(φn)n of H2(D) which tends to 0 pointwise, we have limn→+∞ ‖φn‖B2

α= 0.

For any sequence of positive integers (ni)i, we set

DΛ[(ni)i] := exp(−2

3Λ( 1n1

))+∑i≥2

exp(

13

(Λ( 1n1

)+ · · · +Λ( 1

ni−1

)))exp

(23Λ( 1ni

)) .

Lemma 3.3. For every ε > 0, there is a sequence of positive integers (ni)i≥1 suchthat

(i) ni divides ni+1, i ≥ 1.

(ii) 2Λ( 1ni

)≤ Λ( 1

ni+1

)≤ 4Λ( 1

ni

), i ≥ 1.

(iii) DΛ[(ni)i] < ε.Proof. Since t , tΛ(t) is a modulus of continuity, we have

(3.1) Λ(t) ≤ Λ( t2

)≤ 2Λ(t) for t near 0.

Let n1 ∈ N be a fixed positive integer and let

k1 = maxs ∈ N : Λ( 1

2sn1

)≤ 4Λ( 1

n1

).


Set n2 = 2k1n1. It is obvious that n1 divides n2. On the other hand, it is easy tosee that

2Λ( 1n1

)≤ Λ( 1

n2

)≤ 4Λ( 1

n1

).

By induction, we construct a positive sequence (ni)i of integers satisfying asser-tions (i) and (ii). Now, let us prove that the series

(3.2)∑i≥2

exp(

13

(Λ( 1n1

)+ · · · +Λ( 1

ni−1

)))exp

(23Λ( 1ni

))

converges. Indeed, by setting

Ai =exp

(13

(Λ( 1n1

)+ · · · +Λ( 1

ni−1

)))exp

(23Λ( 1ni

)) , i ≥ 2 ,

we have

Ai+1

Ai= exp

(Λ( 1ni

)− 2

3Λ( 1ni+1

))≤ exp

(−1

3Λ( 1ni

))≤ exp

(−1

3Λ( 1n1

)).

Thus, the series (3.2) converges. Therefore, for ε > 0, there is an integer i0 suchthat

exp

(−2

3Λ( 1

ni0

))+∑i≥i0

exp

(13

(Λ( 1ni0

)+ · · · +Λ( 1

ni−1

)))exp

(23Λ( 1ni

)) ≤ ε.

Thus, the sequence(ni+i0−1

)i≥1 satisfies (i), (ii), and (iii). This completes the

proof.

Recall that the modulus of continuity of a positive measure µ on T is given by

ωµ(δ) := supµ(I) : |I| ≤ δ,

where the supremum is taken over all closed arcs I on T such that |I| ≤ δ.


Definition 3.4. A positive measure µ on T is said to be Λ–smoothly decom-posable if for every ε > 0, there is a sequence of positive measures (µi)i on T anda sequence of positive integers (ni)i such that

(i) µ = ∑i µi.(ii) ωµi(1/ni) ≤ (1/ni)Λ(1/ni) for every i.

(iii) DΛ[(ni)i] < ε.The next result plays a fundamental role in the proof of Theorem 3.1. The

proof of this result holds by using analogue arguments like those of J. Roberts’sarguments [21]. For the sake of completeness, we put it here.

Proposition 3.5. A positive measure on T is Λ–smoothly decomposable wheneverit puts no mass on any Λ–Carleson set.

For the proof of Proposition 3.5, it is convenient to introduce first the follow-ing definition.

Definition 3.6. Let µ be a positive measure on T, and let P = I1, ..., In bea partition of T into n closed arcs of equal length. We say that

Ii is

Λ–light if µ(Ii) ≤ Λ(1/n)

2n,

Λ–heavy if µ(Ii) >Λ(1/n)

2n.

The positive measure, µ1, defined for each Borel subset E of Ii by

µ1(E) =

µ(E) if Ii is Λ–light,µ(E)

2µ(Ii)Λ(1/n)n

If Ii is Λ–heavy

is called P −Λ–grating of µ.

Remark 3.7. Let µ be a positive measure on T, and let P = I1, ..., In be apartition of T into n closed arcs of equal length. If µ1 is P −Λ–grating of µ, then(a) µ1 ≤ µ and Supp(µ − µ1) is contained in the union of all Λ–heavy arcs.

(b) µ1(Ii) =(Λ(1/n))/(2n) when Ii is a Λ–heavy arc.

(c) ωµ1(1/n) ≤(Λ(1/n))/n.

Proof of Proposition 3.5. Suppose that the measure µ is not Λ–smoothly de-composable, and let ε > 0. By Lemma 3.3, there is a sequence (ni)i≥1 of positiveintegers such that

(i) Each ni divides ni+1.


(ii) 2Λ( 1ni

)≤ Λ( 1

ni+1

)≤ 4Λ( 1

ni

)for all i ≥ 1.

(iii) DΛ[(ni)i] < ε.Let (Pi)i≥1 be a collection of partitions of T such each Pi consists of ni closedarcs of equal length. Moreover, in view of (i), we may assume that each Pi+1refines Pi. Let (µi)i≥1 be the sequence of positive measures on T such that µ1 isthe P1 −Λ–grating of µ and for every i ≥ 2, the measure µi is the Pi −Λ–gratingof µ − (µ1 + ...+ µi−1). Now, set ν :=∑i≥ µi and note that ν ≤ µ, and that

ωµi(

1ni

)≤Λ( 1ni

)ni

for all i ≥ 1

(see Remark 3.7(c)). Since µ is not Λ–smoothly decomposable, we can assumethat ν 6= µ. Let K := ⋂

i≥1Hi, where each Hi is the union of all Λ–heavy arcs inPi with respect to µ − (µ1 + · · · + µi). It is clear that (Hi)i≥1 is a non-increasingsequence of closed subsets of T, and that Supp

(µ − (µ1 + · · · + µi)

) ⊂ Hi for alli ≥ 1 (see Remark 3.7(a)). Hence, K is a closed subset of T, and Supp(µ−ν) ⊂ K.Since ν 6= µ, we have µ(K) > 0. On the other hand, in view of Remark 3.7(b),we have for all i ≥ 1

(3.3) µ(T) ≥ µi(T) ≥ µi(Hi) =∑

I⊂Hi:I heavy in Pi

µi(I) = 12|Hi|Λ( 1

ni

),

so that |K| = limi→+∞ |Hi| = 0. Now, let K0 := T\⋃i≥2 Li such that each Lidenotes the union of the interiors of those Λ–light arcs in Pi which lie in Hi−1.We clearly see that K0 is a closed subset of T containing K; in particular, we haveµ(K0) ≥ µ(K) > 0. As K0\K consists of endpoints of two adjacent Λ–light arcs,we have |K0| = 0. What remains to show here is that K0 is a Λ–Carleson set.Indeed, we have |Li| ≥ 1/ni and Li ⊂ Hi−1, which gives

∑i≥2

|Li|Λ(|Li|) ≤ ∑i≥2

|Li|Λ( 1ni

)

≤∑i≥2

|Hi−1|Λ( 1ni

)

=∑i≥1

|Hi|Λ( 1ni + 1

).


In view of (ii) and (3.3), we have

∑i≥2

|Li|Λ( 1ni

)≤ 4

∑i≥1

|Hi|Λ( 1ni

)

≤ 8∑i≥1

µi(T)

≤ 8µ(T) < +∞,

and therefore, K0 is a Λ–Carleson set. We have a contradiction to the assumptionthat µ puts no mass on Λ–Carleson sets, and this completes the proof.

Recall that the Corona theorem (see [19]) shows that there is a constantA > 0such that if f1, f2 ∈ H∞(D) with ‖fi‖∞ ≤ 1, i = 1, 2, and

|f1| + |f2| ≥ δ on D,

where 0 < δ ≤ 1, then there exist g1, g2 ∈ H∞(D) satisfying

f1g1 + f2g2 = 1 and ‖gi‖∞ ≤ 1δA, i = 1, 2 .

Recall also that Shapiro’s inequality (see [25]) says that there is a constant c1 > 0such that for every positive singular measure µ on T,

(3.4) |Uµ(z)| ≥ exp(− c1

ωµ(1− |z|)1− |z|

), z ∈ D .

Lemma 3.8. Let (α(n))n≥0 be a positive sequence satisfying the condition (R).Then there exist a constant c > 0 and an integer N such that if ν is a positive singularmeasure on T for which

ων(1n) ≤ c

2

nΛα( 1

n) for some n ≥ N,

then there is g ∈ H∞(D) such that

‖g‖∞ ≤ exp(c3Λα ( 1

n

))and ‖1− gUν‖B2

α ≤ exp(−2c

3Λα ( 1

n

)).

Proof. Let A, c1, a, and b be the constants given in the Corona theorem, in(3.3), in Definition 1.1(2), and in Lemma 8.1(vi) respectively. Let

c ≤ min

13Ac1

,

√3

2c1,(b4

(4c1

3

)a)1/(1−2a) ,


and let N be a positive integer such that

−32≤ n ln

(1− 1

n

)and 1 ≤ B

2Λα ( 1

n

)for every n ≥ N,

where B = 23c

2c1 ≤ 1. Now, fix an integer n ≥ N and let ν be a positive singularmeasure on T for which

ων(

1n

)≤ c

2

nΛα ( 1

n

).

It follows from Shapiro’s inequality that,

|Uν(z)| ≥ exp(−c2c1Λα ( 1

n

))for |z| ≤ 1− 1

n.

On the other hand, for k = n[BΛα(1/n)], where [t] denotes the integer part oft, and for |z| > 1− 1/n, we have

|zk| ≥(

1− 1n

)k= exp

(n[BΛα ( 1

n

)]ln(

1− 1n

))≥ exp

(−3

2

[BΛα ( 1

n

)])≥ exp

(−c2c1Λα ( 1

n

)).

Hence,

|Uν(z)| + |zk| ≥ exp(−c2c1Λα ( 1

n

))for z ∈ D.

Keeping in mind that c ≤ 1/(3Ac1), it follows from the Corona theorem thatthere are g, h ∈ H∞(D) such that

‖h‖∞, ‖g‖∞ ≤ exp(c3Λα ( 1

n

))and gUν + zkh = 1.

Therefore,

‖1− gUν‖B2α = ‖zkh‖B2

α ≤ ‖zk‖B2α‖h‖∞ =

‖h‖∞α(k)

.

Thus , the desired estimate will be verified if we prove that

1α(k)

≤ exp(−cΛα ( 1

n

)).


Indeed, we have

1α(k)

= exp(− kKα(k))

= exp(−n

[BΛα ( 1

n

)]Kα

(n[BΛα ( 1

n

)])).

Since the sequence(s1−aKα(s)

)s≥1 is non-increasing (see Definition 1.1(2)), we

get

Kα(n[BΛα ( 1

n

)])≥

[Λα ( 1

n

)][BΛα ( 1

n

)]

1−a

Kα(n[Λα ( 1

n

)]).

As c ≤ √3/(2c1) and Kα is a non-increasing function, we have

Kα(n[BΛα ( 1

n

)])≥ (2B)a−1Kα

(nΛα ( 1

n

)).

By Lemma 8.1(vi), we have

Kα(n[BΛα ( 1

n

)])≥ bn(2B)a−1.

Hence,

1α(k)

≤ exp(−b(2B)a−1

[BΛα ( 1

n

)])≤ exp

(−b(2B)a−1

(BΛα ( 1

n

)− 1

)).

Since 1 ≤ (B/2)Λα(1/n), we have

1α(k)

≤ exp(−b

4(2B)aΛα ( 1

n

)).

As c ≤ ((b/4)(4c1/3)a)1/(1−2a), we have c ≤ (b/4)(2B)a. Thus,

1α(k)

≤ exp(−cΛα ( 1

n

)).

The proof is therefore complete.


Proof of Theorem 3.1. Assume that µ is a positive singular measure on T van-ishing on Λα–Carleson sets. Let c and N be the two constants given in Lemma3.8. By Proposition 3.5, the measure µ/c is cΛα–smoothly decomposable. Letε > 0, there is a sequence (µi)i of positive measures on T and a sequence (ni)i ofpositive integers greater than N such that

(i) µ =∑i µi;(ii) ωµi(1/ni) ≤ (c2/ni)Λα(1/ni) for all i;

(iii) DcΛα[(ni)i] < ε.Therefore, it follows from Lemma 3.8 that there is a function g1 ∈ H∞(D) suchthat

‖1− g1Uµ1‖B2α≤ exp

(−2c

3Λα ( 1

n1

)).

By induction one can show that for every k ≥ 1 there is gk ∈ H∞(D) such that

‖1− gkUµ1+···+µk‖B2α≤ exp

(−2c

3Λα ( 1

n1

))

+k∑i=2

exp(c3

(Λα ( 1n1

)+ · · · +Λα ( 1

ni−1

)))exp

(2c3Λα ( 1

ni

)) .

This implies that for every k ≥ 1, we have

dist(1, [Uµ1+···+µk]B2

α

) ≤ DcΛα[(ni)i] < ε.Thus

dist(1, [Uµ]B2

α

) ≤ ε.This concludes the proof.

4. LINKS BETWEEN Nα,β AND SINGULAR INNER FUNCTIONS

Let ω be a weight such that supn≥0ω(n) < +∞ and limn→−∞ω(n) = +∞. Weset

Nω :=f(z) = ∑

n≥0

anzn ∈ N :

f∗ ∈ L2(T),∑n≥1

|an|2ω2(−n) < +∞ and

∑n≥1

|f∗(−n)|2ω2(n)

< +∞ ,

and recall thatH2(D) can be seen as a subspace of `2ω(Z) by identifying a function

f ∈ H2(D) with the sequence of its Fourier coefficients (f (n))n∈Z.


The following result reduces our problem to the study of the bicyclicity ofsingular inner functions in `2

ω(Z) .

Lemma 4.1. Letω be a weight such that supn≥0ω(n) < +∞ andlimn→−∞ω(n) = +∞. Then Nω = H2(D) if and only if every singular innerfunction is bicyclic in `2

ω(Z).

Proof. Suppose that every singular inner function is bicyclic in `2ω(Z). Let

f ∈ Nω, then there exists a singular inner function U such that g = fU ∈ N+.Consider the following hyperfunction

ϕ(z) =

f(z)−

∑n≥0

f∗(n)zn for |z| < 1,∑n<0f∗(n)zn for |z| > 1.

It is clear that the function ϕ+ ∈ N, D(ϕ+) is a divisor of U and ϕ∗ = 0 a.e. onT. By Proposition 2.1, the hyperfunction–product U ·ϕ = 0. Since U is bicyclicin `2

ω(Z) and ϕ ∈Hω(T), we haveϕ = 0. So, f∗(n) = 0 for every n ≤ −1 andf ∈ H2(D).

Conversely, suppose that there is a non-bicyclic singular inner function U in`2ω(Z), then there is ψ ∈ Hω(T), ψ ≠ 0 such that U ·ψ = 0. From Proposition

2.1 it follows that ψ+ ∈ N and ψ∗ = 0 a.e. on T. Since ψ− ∈ H2(C\D) and(ψ−)∗ = (ψ+)∗ a.e. on T. Therefore (ψ+)∗(−n) = −ψ(n+ 1), n ≤ −1. Thisimplies that ψ+ ∈ Nω\H2(D).

Remark 4.2. Let

N 1ω :=

f(z) = ∑n≥0anzn ∈ N :

f∗ ∈ L1(T),∑n≥1

|an|2ω2(−n) < +∞ and

∑n≥1

|f∗(−n)|2ω2(n)

< +∞ .

Note that N 1ω = H1(D) if and only if Nω = H2(D). Indeed, Nω ⊂ N 1

ω.On the other hand, if Nω = H2(D), then by Lemma 4.1, every singular innerfunction is bicyclic in `2

ω(Z). Let now f ∈N 1ω and considerϕ as in the proof of

Lemma 4.1. Since∑n≥0

f∗(n)zn ∈⋂p<1

Hp(D) ⊂ N+ and∑n<0

|f∗(n)|2 < +∞,

as in the proof of Lemma 4.1, f (n) = 0 for every n ≤ −1 and f ∈ H1(D).


Assume that Λ is a positive function on (0,1] such that the function t ,tΛ(t) is a modulus of continuity. Let Ms(T) be the set of all positive singularmeasures on T and set

Ms(Λ) := µ ∈Ms(T) : Suppµ is a Λ–Carleson set,

and

M0s (Λ) := µ ∈Ms(T) : µ(E) = 0 for every Λ–Carleson set E

.

The following result reduces the triviality ofNω to the bicyclicity of some singularinner functions in `2

ω(Z).Theorem 4.3. Let ω be a weight and let α(n) = 1/ω(n), (n ≥ 0). If

the sequence (α(n))n≥0 satisfies the condition (R), then the following statements areequivalent.

(i) Nω = H2(D).(ii) N 1

ω = H1(D)(iii) Every singular inner function is bicyclic in `2

ω(Z).(iv) For every µ ∈ Ms(Λα), the singular inner function, Uµ, is bicyclic in `2

ω(Z).

Note that, since (α(n))n≥0 satisfies the condition (R), every function φ ∈H∞(D) induces a linear bounded operator, Mφ, on `2

ω(Z) given by

Mφ(x) :=( ∑k≥0

φ(k)xn−k)n∈Z,

(x = (xn)n∈Z ∈ `2

ω(Z)).

Proof. In view of Lemma 4.1 and Remark 4.2, it suffices to prove that (iv)⇒(iii). Assume that Uν is bicyclic in `2

ω(Z) for every ν ∈ Ms(Λα). Let µ ∈ Ms(T);we shall prove that Uµ is bicyclic in `2

ω(Z). Since the union of two Λα–Carlesonsets is a Λα–Carleson set, it is easy to see that there are measures λ ∈ M0

s (Λα)and µi ∈ Ms(Λα) such that µ = λ +∑

iµi (see [26, Lemma 2, page 333]). By

Theorem 3.1, Uλ is cyclic in B2α and each Un := Uλ+µ1+...µn is bicyclic in `2

ω(Z).Therefore, there is a sequence (xn)n of finitely supported elements of `2

ω(Z) suchthat limn→0 ‖Unxn − 1‖ω = 0. If for n ≥ 1, we set Vn = Uµ/Un and yn =Unxn − 1, then

‖Uµxn − 1‖ω = ‖Vn(1+yn)− 1‖ω≤ ‖Vn − 1‖ω + ‖Vnyn‖ω≤ ‖Vn − 1‖ω + ‖Vn‖∞ ‖yn‖ω= ‖Vn − 1‖ω + ‖yn‖ω.

Since the sequence (Vn − 1)n is uniformly bounded in H2(D) and tends to 0pointwise, we have limn→+∞ ‖Vn − 1‖ω = 0 (see Remark 3.2). As

limn→0

‖yn‖ω = limn→0

‖Unxn − 1‖ω = 0,


obviously (Uµxn)n tends to 1 in `2ω(Z). This shows that Uµ is bicyclic in `2

ω(Z),and the proof is complete.

5. PROOF OF THEOREMS A AND B

5.1. Discrete version of Hruscev’s theorem. Let h be a non-decreasing con-tinuous function on [0,1] such that h(0)=0. Such a function is called a determin-ing function. The h-Hausdorff measure of a subset E of T is defined by

Hh(E) := limt→0+

[inf∑ih(|∆i|)],

the infinimum is taken over all finite coverings of E by arcs (∆i)i such that |∆i| < t(see [15]).

Let h be a modulus of continuity which satisfies

(5.1)∫ s

0

h(t)tdt ≤ Const h(s) for s near 0.

Let E be a closed subset of T with a zero Lebesgue measure. We set

Fh(E) :=f ∈H (E) : f|C\D ∈ H2(C\D) and

|f(z)| = O(

exph(1− |z|)

1− |z|)(|z| → 1−)

.

Here, H2(C\D) is the Hardy space on C\D, that is, the space of all analytic func-tions f on C\D satisfying

‖f‖H2(C\D) := sup

0≤r<1

∫ 2π

0

∣∣∣∣∣f(eit

r

)∣∣∣∣∣2

dt

1/2

< +∞.

Theorem ([12, Theorem 9.1]). Let h be a modulus of continuity satisfying (5.1).If E is a closed subset of T with zero Lebesgue measure, then Fh(E) = 0 if and onlyif Hh(E) = 0.

Assume that Λ is a positive function on (0,1] such that the function t ,tΛ(t) is a modulus of continuity, and the function t , tγΛ(t) is non-decreasingon (0,1] for some 0 < γ < 1. Let h be a modulus of continuity satisfying (5.1).It is shown in [1] and [3] that Hh(E) = 0 for every Λ−Carleson set E if and onlyif ∫ 1

0

tdΛ(t)h(t)

= −∞.


Let (β(n))n≥0 be a positive sequence. For every closed subset E of T, we set

HD2β(E) :=

ψ ∈H (E) : sup

n≥0

|ψ(n)|β(n)

< +∞ and ψ|C\D ∈ H2(C\D).

The following result is a discrete version of Hruscev’s theorem for Λ–Carlesonsets.

Theorem 5.1. Let (β(n))n≥0 be a positive sequence satisfying the condition (R)such that limn→+∞(lnβ(n))/ lnn > 0. Assume that Λ is a positive function on(0,1] such that the function t , tΛ(t) is a modulus of continuity, and the functiont , tγΛ(t) is non-decreasing on (0,1] for some 0 < γ < 1. The following statementsare equivalent.

(i) For every Λ–Carleson set E, HD2β(E) = 0.

(ii)∑n≥1

Λ( lnβ(n+ 1)n+ 1

)−Λ( lnβ(n)

n

)lnβ(n)

= +∞.

To deduce Theorem 5.1 from Hruscev’s result, we need the following lemma.

Lemma 5.2. Assume that Λ is a positive function on (0,1] such that the functiont , tΛ(t) is a modulus of continuity. If (β(n))n≥0 is a positive sequence satisfyingcondition (R), then

∑n≥1

Λ( lnβ(n+ 1)n+ 1

)−Λ( lnβ(n)

n

)lnβ(n)

∫ 1

0

−tdΛ(t)hβ(t)

.

Proof. In view of Lemma 8.1(iv) and (v), since (β(n))n≥0 satisfies the con-dition (R), we have Λβ(Kβ(n)) lnβ(n).

Therefore, there is a positive constant C such that for every n ≥ 1, we have

1CΛ(Kβ(n+ 1))−Λ(Kβ(n))

lnβ(n+ 1)≤∫ Kβ(n)Kβ(n+1)

−dΛ(t)Λβ(t)≤ CΛ(Kβ(n+ 1))−Λ(Kβ(n))

lnβ(n).

Since ∫ Kβ(1)0

−tdΛ(t)hβ(t)

=∑n≥1

∫ Kβ(n)Kβ(n+1)

−dΛ(t)Λβ(t) ,


we have ∑n≥1

Λ(Kβ(n+ 1))−Λ(Kβ(n))lnβ(n+ 1)

∫ Kβ(1)

0

−tdΛ(t)hβ(t)

.

This completes the proof.

In order to prove Theorem 5.1, we use the fact that HD2β(E) ⊂ Fhβ(E). By

Theorem 5.1 and Hruscev’s theorem we have (ii) ⇒ (i). On the other hand, if (ii)fails, then by Lemma 5.2, ∫ s

0

tdΛ(t)hβ(t)

> −∞.

It follows from [3] and [1] that there is a positive singular measure µ on T sup-ported by a Λ–Carleson set E such that ωµ(δ) = O(hβ(δ)), δ → 0+. Using theShapiro inequality (3.4), we obtain

1Uµ

− 1Uµ(∞) ∈ HD

2β(E) ≠ 0,

where Uµ(∞) = lim|z|→∞Uµ(z). For further details, see [16, Theorem 2.2].

5.2. Outer functions. The main result of this section is the following.

Theorem 5.3. Assume that Λ is a positive non-increasing function on (0,1)satisfying the following

(i) limt→0

Λ(t)ln 1/t

> 0.

(ii) There is α ∈ (0,1) such that the function t , tαΛ(t) is non-decreasing on(0,1).

For every Λ–Carleson set E, there is a positive function Λ, an outer function F ∈A∞(D) with Z(F) = Z∞(F) = E, and a positive constant B such that

(5.2) limt→0+

Λ(t)Λ(t) = +∞ and |F(n)(z)| ≤ n!BneΛ∗(n),for all n ≥ 0, and z ∈ D,

where Λ∗(n) = supnx − Λ(e− x2 ) : x > 0.We have immediately the following corollary.

Corollary 5.4. Assume that Λ is a positive function on (0,1) satisfying thehypotheses of Theorem 5.3. For every Λ–Carleson set E, there is an outer functionF ∈ A∞(D) with Z(F) = Z∞(F) = E and a positive constant B such that

(5.3) |F(n)(z)| ≤ n!BneΛ∗(n) , for all n ≥ 0, and z ∈ D,where Λ∗(n) = supnx −Λ(e−x/2) : x > 0.


Let E be a closed subset of T and let((eian, eibn)

)n be the collection of com-

plementary arcs of E in T. We set

(5.4) ρ(θ) :=

12π

(1

θ − an +1

bn − θ)−1

if θ ∈ ]an, bn[0 if eiθ ∈ E

Note that

(5.5)1

4πdist(eiθ, E) ≤ ρ(θ) ≤ 1

4dist(eiθ, E).

Theorem 5.3 is a consequence of the next result of Taylor-Williams [30].

Theorem 5.5. [30] Assume that E is a closed subset of T. Let λ be a non-negativeconvex infinitely differentiable function such that ϕ(eiθ) :=λ(−2 ln ρ(θ)) satisfies

(i)∫ 2π

0|ϕ(eiθ)|dθ < +∞.

(ii) There is a constant K > 0 such that for every n ≥ 0, we have

∣∣∣∣dnϕ(eiθ)dθn

∣∣∣∣ ≤ n!Kn+1

dist(eiθ, E)n+1 for all eiθ ∈ T\E.

(iii) For every C > 0, ϕ(eiθ)+ C ln dist(eiθ, E)→ +∞ as dist(eiθ, E)→ 0.Then there exists an outer function F ∈ A∞(D) with Z(F) = Z∞(F) = E such that

|F(n)(z)| ≤ n!Bneλ∗(n) for all n ≥ 0 and z ∈ D,

where B is some positive constant and λ∗(n) = supnx − λ(x) : x > 0.In the proof of Theorem 5.3, we shall need the following elementary lemmas.

Lemma 5.6. Let k be a non-increasing continuous real function on (0,1] suchthat limt→0+ k(t) = +∞. Then for every γ > 0, there is a non-increasing continuousreal function h such that limt→0+ h(t) = +∞, h(t) ≤ k(t), and the function hγ :t , tγh(t) is non-decreasing on (0,1] with hγ(0) = 0.

Proof. It is sufficient to consider the function h defined on (0,1] by

h(t) := 1tγ/2

min√sk(s) : tγ ≤ s ≤ 1

.

Lemma 5.7. Under the hypotheses of Theorem 5.3, for every Λ–Carleson set E,there exists a non-negative infinitely differentiable function, Λ, having the followingproperties.


(i) The function σ , Λ(e−σ/2) is convex.

(ii) limt→0+

Λ(t)Λ(t) = +∞ and E is a Λ–Carleson set.

(iii) There is a constant K > 0 such that |Λ(n)(t)| ≤ n!Kn

tn+1 for every t ∈ (0,1).Proof. There exists a positive decreasing function u on (0,1] satisfying the

following conditions

(a) limt→0+

u(t)Λ(t) = +∞.

(b) There is 0 < β < 1 such that the function uβ : t , tβu(t) is non-decreasingon (0,1) with uβ(0) = 0.

(c)∫ 2π

0u(dist(eit, E))dt < +∞.

Indeed, let In = (eian, eibn) be the component arcs of T\E enumerated such thatthe sequence (|In|)n is non-increasing. We have,

∫ 2π

0Λ(dist(eit, E))dt = 4π

∑n

∫ (bn−an)/20

Λ(t)dt < +∞.Then there is a non-decreasing positive sequence (dn)n which tends to infinitysuch that

(5.6)∑ndn

∫ (bn−an)/20

Λ(t)dt < +∞.It follows from Lemma 5.6 that there is a non-increasing continuous real functionh such that(1) limt→0+ h(t) = +∞.(2) h

((bn − an)/2

) ≤ dn for every n ≥ 0.(3) The function hγ : t 7 -→ tγh(t) is non-decreasing on (0,1], hγ(0) = 0, where

γ is a positive constant for which β := γ +α < 1.Now, we set u(t) := h(t)Λ(t), t ∈ (0,1]. It is clear that the function u satisfiesthe conditions (a) and (b). We claim that the functionu satisfies also the condition(c). For every t ∈ (0,1], we have

∫ t0u(s)ds =

∫ t0sγh(s)s−γΛ(s)ds ≤ h(t)tγ ∫ t

0s−γΛ(s)ds = h(t)f (t),

where f(t) = tγ∫ t

0s−γΛ(s)ds. Since f

′(t) = γtγ−1

∫ t0s−γΛ(s)ds + Λ(t) and

t , tαΛ(t) is a non-decreasing function, it follows that f ′(t) ≤ CΛ(t) for every


t ∈ (0,1]. Thus, for every t ∈ ]0,1], we have

∫ t0u(s)ds ≤ Ch(t)

∫ t0Λ(s)ds.

Therefore, the third condition follows from (5.6).Now, we are able to construct the function Λ. For x > 0, set

Λ(x) :=∫ 1

0

xu(t)x2 + (x − t)2 dt.

It is clear that the function Λ is infinitely differentiable. Using elementary com-putation, we can show that the function σ , Λ(e−σ/2) is convex for large σ ,and

Λ(x) ≥ ∫ xx/2

xu(t)x2 + (x − t)2 dt

≥ 45u(x)x

∫ xx/2dt = 2

5u(x).

On the other hand,

Λ(x) = ∫ x0

xu(t)x2 + (x − t)2

tβ

tβdt +

∫ 1

x

xu(t)x2 + (x − t)2 dt

≤ xβ−1u(x)∫ x

0

dttβ+u(x)

∫ 1

0

xx2 + (x − t)2 dt

≤(

11− β +π

)u(x).

Hence, Λ u. This shows that limt→0+ Λ(t)/Λ(t) = +∞, that E is a Λ–Carlesonset and limt→0+

Λ(t)+ B ln t = +∞ for all B > 0.

For x > 0, we set D(x, 1

8x) = z ∈ C : |z − x| ≤ 1

8x. We have

z2 + (z − t)2 = 2(z − 1+ i

2t)(z − 1− i

2t)

andx2≤∣∣∣∣x − 1± i

2t∣∣∣∣ .

Then for z ∈ D(x, 18x), we have∣∣∣∣z + 1± i

2t∣∣∣∣ = ∣∣∣∣z − x + x − 1± i

2t∣∣∣∣ ≥ ∣∣∣∣x − 1± i

2t∣∣∣∣− 1

8x

≥ 34

∣∣∣∣x − 1± i2t∣∣∣∣ .


Therefore, ∣∣∣∣ zz2 + (z − t)2

∣∣∣∣ ≤ 98

42

32x

x2 + (x − t)2 .

Using the Cauchy inequalities, we see that there is a positive constant B such that∣∣∣∣ dndxn(

xx2 + (x − t)2

)∣∣∣∣ ≤ n!Bn1xn

xx2 + (x − t)2 .

Since u(x) = o(1/xβ), x → 0 and β < 1, we have

|Λ(n)(x)| = ∣∣∣∣∣∫ 1

0

dn

dxn( xx2 + (x − t)2

)u(t)dt

∣∣∣∣∣≤ n!Kn

xn+1 .

Proof of Theorem 5.3. Let Λ be the function given in Lemma 5.7 and let ϕbe the function defined by

ϕ(eiθ) =ϕ(θ) := Λ(ρ(θ)).As limt→+∞Λ(t)/| ln t| > 0, one can easily verify that ϕ satisfies the conditions(i) and (iii) of Theorem 5.5. Since ρ(k)(eiθ) = 0 for every eiθ ∈ T\E and k ≥ 3,it follows from Faa di Bruno’s composition formula that

ϕ(n)(θ) = (Λ ρ)(n)(θ)=

∑k1+2k2=n

n!k1!k2!

Λ(k1+k2)(ρ(θ)

)(ρ′(θ)

)k1(ρ′′(θ)

2

)k2

, n ≥ 1.

In view of Lemma 5.7(iv), there is a constant C ≥ 1 such that for every n ≥ 1, wehave

|ϕ(n)(θ)| ≤∑

k1+2k2=n

n!k1!k2!

Ck1+k2(k1 + k2)!ρ(θ)k1+k2+1

∣∣ρ′(θ)∣∣k1

∣∣∣∣∣ ρ′′(θ)2

∣∣∣∣∣k2

= n!ρ(θ)n+1

∑k1+2k2=n

Ck1+k2(k1 + k2)!k1!k2!

∣∣ρ′(θ)∣∣k1

∣∣∣∣∣ρ(θ) ρ′′(θ)2

∣∣∣∣∣k2

.

Let Im = (eiam, eibm) be the component arcs of T\E, we have∣∣ρ′(θ)∣∣ ≤ 1 and∣∣ρ(θ)ρ′′(θ)∣∣ ≤ 2 for all eiθ ∈ Im. Then for every n ≥ 1

|ϕ(n)(θ)| ≤ n!Cn

ρ(θ)n+1

∑k1+2k2=n

(k1 + k2)!k1!k2!

≤ n!(eC)n

ρ(θ)n+1 .


Hence, in view of (5.5), the functionϕ satisfies the condition (ii) of Theorem 5.5.Therefore, there is an outer function F ∈ A∞(D) with Z(F) = Z∞(F) = E and apositive constant B such that for every n ≥ 0, we have

|F(n)(z)| ≤ n!Bn exp

supt>0(nt − Λ(e− t2 )), z ∈ D.

This finishes the proof.

5.3. Bicyclicity of inner functions. Throughout this section, we need thefollowing elementary lemma.

Lemma 5.8. Let (α(n))n≥0 and (β(n))n≥0 be two positive sequences satisfyingthe condition (R) such that limn→+∞

(lnα(n)

)/ lnn > 0. If the series

(5.7)∑n≥1

Λα ( lnβ(n+ 1)n+ 1

)−Λα ( lnβ(n)

n

)lnβ(n)

converges, then np = O(β(n)) for all p > 0.

Proof. Since limn→+∞(lnα(n)

)/ lnn > 0, there is a positive constant γ such

thatnγ ≤ α(n) for all n ≥ 1.

This implies that there is a positive constant C such that

(5.8) Λα(t) ≥ C| ln t| for all t ∈ (0,1).

We also note that, since (β(n))n≥0 satisfies the condition (R), there is an integerN1 such that

(5.9) lnβ(n) ≤ √n, for all n ≥ N1,

(see Definition 1.1(2)). Now, assume that the series (5.7) converges and let p > 0.We have

limn→+∞

Λα ( lnβ(n)n

)lnβ(n)

= 0.

Indeed, let Kβ(n) =(lnβ(n)

)/n, we have

n∑j=1

Λα(Kβ(j + 1))−Λα(Kβ(j))lnβ(j)

=n∑j=2

(1

lnβ(j − 1)− 1

lnβ(j)

)Λα(Kβ(j))+(Λα(Kβ(n+ 1))

lnβ(n)− Λα(Kβ(1))

lnβ(1)

).


This shows that

∑j≥2

(1

lnβ(j − 1)− 1

lnβ(j)

)Λα(Kβ(j)) < +∞.Therefore,

+∞∑j=n

(1

lnβ(j − 1)− 1

lnβ(j)

)Λα(Kβ(j))≥ Λα(Kβ(n)) +∞∑

j=n

(1

lnβ(j − 1)− 1

lnβ(j)

)

≥ Λα(Kβ(n))lnβ(n− 1)

≥ Λα(Kβ(n))lnβ(n)

→ 0.

Thus, it follows from from (5.8) and (5.9) that

Λα ( lnβ(n)n

)lnβ(n)

≥ lnn− ln lnβ(n)lnβ(n)

≥ lnn2 lnβ(n)

→ 0,

and np = O(β(n))(n → +∞).

Theorem 5.9. Let (α(n))n≥0 and (β(n))n≥0 be two positive sequences satisfy-ing the condition (R) such that limn→+∞

(lnα(n)

)/ lnn > 0. If

(5.10)∑n≥1

Λα ( lnβ(n+ 1)n+ 1

)−Λα ( lnβ(n)

n

)lnβ(n)

= +∞,

then Nα,β = H2(D).

Proof. We first set β(n) = (n+ 1)β(n), n ≥ 0, and note that the sequence(β(n))n≥0 satisfies the condition (R). Moreover, we have limn→+∞

(ln β(n)

)/ lnn >

0 and

(5.11)∑n≥1

Λα(

ln β(n+ 1)n+ 1

)−Λα

(ln β(n)n

)ln β(n)

= +∞.

It follows from Theorem 5.1 that HD2β(E) = 0 for every Λα–Carleson set E.

Indeed, it suffices to prove (5.11). Assume by the way of contradiction that this


series converges. And so,

limn→+∞

Λα(

ln β(n)n

)ln β(n)

= 0.

By Lemma 5.8, we see that ln β(n) lnβ(n) and that

Λα(

ln β(n)n

) Λα ( lnβ(n)

n

).

Therefore,

∑n≥1

Λα(

ln β(n+ 1)n+ 1

)−Λα

(ln β(n)n

)ln β(n)

∑n≥1

Λα(

ln β(n+ 1)n+ 1

)−Λα

(ln β(n)n

)lnβ(n)

∑n≥1

Λα(

ln β(n)n

)(1

lnβ(n)− 1

lnβ(n+ 1)

)

∑n≥1

Λα ( lnβ(n)n

)(1

lnβ(n)− 1

lnβ(n+ 1)

)

∑n≥1

Λα ( lnβ(n+ 1)n+ 1

)−Λα ( lnβ(n)

n

)lnβ(n)

.

We have a contradiction to (5.10), and therefore (5.11) is established.Next, we note that H2(D) ⊂Nα,β ⊂Nα,β. Therefore, to prove that Nα,β =

H2(D), it suffices to show that Nα,β = H2(D). And so, in view of Theorem4.3, one has to prove that Uµ is bicyclic in `2

ωα,β(Z) for all µ ∈ Ms(Λα). Assume

that there is a measure µ ∈ Ms(Λα) such that Uµ is not bicyclic in `2ωα,β

(Z).Then there exists a non-zero hyperfunction ψ ∈Hωα,β(T) such that Uµ ·ψ = 0.And so we have Suppψ ⊂ E := Suppµ (see [9]), ψ ∈ HD2

β(E) and E is a Λα

Carleson-set, which yields a contradiction.


Theorem 5.10. Let (α(n))n≥0 and (β(n))n≥0 be two positive sequences satisfy-ing the condition (R) such that limn→+∞

(lnα(n)

)/ lnn > 0 and supn≥1

(lnα(n2)

)/

lnα(n) < +∞. If

(5.12)∑n≥1

Λα ( lnβ(n+ 1)n+ 1

)−Λα ( lnβ(n)

n

)lnβ(n)

< +∞,

then H2(D)$Nα,β.

Proof. Set β(n) = √β(n), n ≥ 0, and note that the sequence

(β(n)

)n≥0

satisfies the condition (R) and that

(5.13)∑n≥1

Λα(

ln β(n+ 1)n+ 1

)−Λα

(ln β(n)n

)ln β(n)

< +∞.

By Lemma 5.2, we have ∫0

tdΛα(t)hβ(t)

> −∞.

Therefore, it follows from [3] and [1] that there is a positive singular measure µon T supported by a Λα-Carleson set, E, such thatωµ(δ) = O(hβ(δ)), δ→ 0+.So, by Shapiro’s inequality, (3.4), there is a positive constant c1 such that

(5.14)

∣∣∣∣∣ 1Uµ(z)

∣∣∣∣∣ ≤ exp

(c1hβ(1− |z|)

1− |z|

), z ∈ D.

Since (α(n))n≥0 is a positive sequence satisfying the condition (R) andlimn→+∞

(lnα(n)

)/ lnn > 0, the function Λα satisfies both conditions (i) and

(ii) of Theorem 5.3 (see Lemma 8.1(ii)). Therefore, there is an outer functionf ∈ A∞(D) with Z(f) = Z∞(f ) = E and a positive constant B such that forevery integer n ≥ 0, we have

(5.15) |f (n)(z)| ≤ n!BneΛα∗(n), z ∈ D,

where Λα∗(n) = supnx − Λα(e−x/2) : x > 0 and Λα is the function given byLemma 5.7.

Now, let

ψ(z) := f(z)Uµ(z)

=∑n≥0

anzn, z ∈ D.


We claim thatψ ∈Nα,β\H2(D). It is clear thatψ ∈ N\H2(D) andψ∗ ∈ L2(T).So, it remains to prove that

(5.16)∑n≥0

|an|2β(n)2

< +∞,

and

(5.17)∑n≥0

|ψ∗(−n)|2α(n)2 < +∞.

Indeed, since f is a bounded function on D, by Cauchy’s formula, we have

|an| =∣∣∣∣∣ 1

2πi

∫|z|=r

f (ξ)Uµ(ξ)ξn+1 dξ

∣∣∣∣∣ , n ≥ 0 and r < 1 ,

≤ Const

sup|z|=r

∣∣∣∣∣ 1Uµ(z)

∣∣∣∣∣rn

≤ Const β(n).

where the latter inequality follows from (5.14) and Lemma 8.1(i). As n2 =O(β(n)) (see Lemma 5.8), we have

∑n≥0

|an|2β(n)2

≤ Const∑n≥0

1β(n)

< +∞,

and (5.16) is established.Now, let us prove (5.17). By successive derivatives, we show that there is a

positive constant C such that for every n ≥ 0, we have

|U(n)µ (z)| ≤ Cnn!dist(z, E)n+1 , z ∈ D.

Using Taylor’s formula, we get that for every n, k ≥ 0, we have

|f (n)(z)| ≤ 1k!

dist(z, E)kmaxz∈D

|f (n+k)(z)|, z ∈ D.

Applying the Leibniz’s formula, we see that |(fUµ)(n)(eit)| has the same estimatesas in (5.15). On the other hand, for every n ≥ 1 and k ≥ 0, we have

|ψ∗(−n)| =∣∣(fUµ)(−n)∣∣

=∣∣∣∣∣ 1

2π

∫ 2π

0

(fUµ)(k)(eit)(in)k

e−int dt

∣∣∣∣∣


Therefore, for every n ≥ 1, we have

|ψ∗(−n)| ≤ infk

1nk

max|(fUµ)(k)(eit)| : t ∈ [0,2π]

≤ infk

(Bn

)kk! exp

supkx − Λα(e−x/2) : x > 0

= infk

(Bn

)kk! sup

e−Λα(√t)tk

: t < 1

.

In view of Lemma 8.2(i) and the fact that limt→0+ Λα(t)/Λα(t) = +∞ (seeLemma 5.7(ii)), we see that for every C > 0, there is a positive constant K > 0such that

e−Λα(√t) ≤ Ke−Λα(Ct), t ∈ (0,1).If we take C < min(1,1/B), then for every n ≥ 1, we have

|ψ∗(−n)| ≤ K infk

(Bn

)kk! supt<1

e−Λα(Ct)tk

= K infk

(BCn

)kk! supt<C

e−Λα(t)tk

≤ K infk

(BCn

)kk! supt<1

e−Λα(t)tk

.

Therefore, by Lemma 8.3, we have

|ψ∗(−n)| = O(

1n2α(n)

),∑

n≥0|ψ∗(−n)|2α(n)2 < +∞ .

Thus, (5.17) holds and therefore, ψ ∈Nα,β\H2(D). This finishes the proof.

Combining Theorem 5.9, Theorem 5.10 and Lemma 8.2(ii), we obtain thefollowing result.

Corollary 5.11. Assume that (α(n))n≥0 and (β(n))n≥0 are two sequences sat-isfying the condition (R). If limn→+∞

(lnα(n)

)/ lnn > 0 and supn≥1

(lnα(n2)

)/

lnα(n) < +∞, then the following assertions are equivalent.(i) For every measure µ ∈Ms(T), Uµ is bicyclic in `2

ωα,β(Z).

(ii) Nα,β = H2(D).


(iii)∑n≥1

(lnα(n+ 1)− lnα(n)

)/ lnβ(n) = +∞.

Theorem A follows from Corollary 5.11. Now, we are going to prove Theo-rem B.

Theorem 5.12. Assume that α(n) = ecnτ , n ≥ 0, where 0 < c and 0 < τ <12 . Let

(β(n)

)n≥0 be a positive sequence satisfying the condition (R). If

(5.18)∑n≥1

1n(1−2τ)/(1−τ)(lnβ(n))1/(1−τ)

= +∞,

then H2(D) =Nα,β.On the other hand, if there is ε > 0 such that

(5.19)∑n≥1

(lnn)τ/(1−τ)+ε

n(1−2τ)/(1−τ)(lnβ(n))1/(1−τ< +∞,

then H2(D)$Nα,β.

Proof. Assume that α(n) = ecnτ , n ≥ 0, where 0 < c and 0 < τ < 12 . We

have Λα(t) 1tτ/(1−τ)

.

Therefore, (5.10) and (5.18) are equivalent. By Theorem 5.9, we have H2(D) =Nα,β.

Conversely, if (5.19) is satisfied then it follows from Lemma 5.2 that∫ s0

tdΛ(t)hβ(t)

> −∞,

where Λ(t) = t−τ/(1−τ)(ln 1/t)τ/(1−τ)+ε, (t ∈ (0,1)). Therefore, there is a Λ–Carleson set E such that the hβ–Hausdorff measure of E satisfies Hhβ(E) > 0.Since E is Λ–Carleson set, |E| = 0, and

∑n≥0

|In|(1−2τ)/(1−τ)(

ln1|In|

)τ/(1−τ)+ε< +∞,

where (In)n is the sequence of all finite complementary arcs of E. Using thenon–uniqueness theorem of Vinogradov (see [13]), we get an outer function Fvanishing on E such that

|F(n)(eint)| ≤ Bn(n!)(2−τ)/(1−τ) , n ≥ 0 ,

for some constant B > 0. We conclude the rest of the proof as in Theorem5.10.


6. PROOF OF THEOREM C

We begin this section by introducing some notations and preliminary lemmaswhich we will use to prove Theorem C. For a function u in L1(T), we denote byP[u] the Poisson transform of u given by

P[u](z) := 12π

∫ 2π

0Pz(eiθ)u(eiθ)dθ , z ∈ D,

where Pz(ζ) = (1− |z|2)/(|1− ζz|2), ζ ∈ T), is the Poisson kernel.Let (υ(n))n≥0 be a positive sequence satisfying the condition (S2), and let Ωυ

be the step function defined on each interval[1−1/n, 1−1/(n+1)

)by

(6.1) Ωυ(t) := n(n+ 1)(2υ(n+ 1)2 − υ(n+ 2)2 − υ(n)2), n ≥ 1.

Note that Ωυ is a non-negative summable function on [0,1) and that

(6.2) υ(n+ 1)2 − υ(n)2 =∫ 1

1−1/nΩυ(r)dr , n ≥ 1,

(see [17]). Let m denote the normalized area measure on D, and let L2Ωυ(T) beHilbert space of all functions f ∈ L2(T) such that

‖f‖2Ωυ = ∣∣P[f](0)∣∣2 +∫D

P[|f |2](z)− |P[f](z)|21− |z|2 Ωυ(|z|)dm(z) < +∞.

The inner product is given by

〈f , g〉 = P[f](0)P[g](0)

+∫D

P[f g](z)− P[f](z)P[g](z)1− |z|2 Ωυ(|z|)dm(z), f , g ∈ L2Ωυ(T).

The following lemma is due to Aleman [2]; here we give a simple proof.

Lemma 6.1. Let (υ(n))n≥0 be a non-negative sequence satisfying the condition(S2). Then

‖f‖2Ωυ ∑n∈Z

|f (n)|2υ(|n|)2.

Proof. For each integer n ∈ Z, let υ(n) := υ(|n|). Since(eint

)n∈Z is an

orthogonal basis of L2Ωυ(T) and the weight υ is symmetrical, it suffices to provethe lemma for fp(eiθ) = eipθ with p ≥ 1. Indeed, let us fix a positive integer p.Since P[fp](z) = zp, z ∈ D, we have

‖fp‖2Ωυ = 2∫ 1

0

1− r 2p

1− r 2 rΩυ(r)dr .


On the other hand, we have

υ(p)2 − υ(1)2 =p−1∑k=1

υ(k+ 1)2 − υ(k)2

=p−1∑k=1

∫ 1

1−1/kΩυ(r)dr

=∫ 1

0

( p−1∑k=1

χ[1−1/k,1](r))Ωυ(r)dr .

An easy computation shows that

12

1− r 2p

1− r 2 ≤p−1∑k=1

χ[1− 1k ,1](r) ≤ e4 1− r 2p

1− r 2 , r ∈ (0,1),

and therefore the proof is complete.

Let C(T) denote the algebra of all complex continuous functions on T and letρ be a modulus of continuity. We set

Lipρ(T) := f ∈ C(T) : |f(ei(t+h))− f(eit)| = O(ρ(|h|))(h → 0).

Lemma 6.2. Assume that (υ(n))n≥0 is a positive sequence satisfying the condi-tion (S2) and Ωυ is the step function defined in (6.1). If ρ is a modulus of continuitysuch that

(6.3)∫ 1

0

Ωυ(r)1− r 2ρ

2((1− r)1/3)dr < +∞,

then Lipρ(T) ⊂ L2Ωυ(T).Proof. Let f ∈ Lipρ(T), and let A(f)(z) := P[|f |2](z)− |P[f](z)|2, z ∈

D. Note that, in view of (6.3), if we prove that A(f)(z) ≤Const ρ2((1 − |z|)1/3), z ∈ D, then f ∈ L2Ωυ(T). Indeed, for all z ∈ D, wehave

A(f)(z) = 1(2π)2

∫ 2π

0

∫ 2π

0|f(eiθ)− f(eit)|2Pz(eiθ)Pz(eit)dθ dt.

Now assume that z = reia, 0 ≤ r < 1, and a ∈ [0,2π], and set

∆ = θ ∈ [0,2π] : |θ−a| ≤ 12(1−r)1/3

×t ∈ [0,2π] : |t−a| ≤ 12(1−r)1/3

.

Since f ∈ Lipρ(T),


∫∆ |f(eiθ)− f(eit)|2Pz(eiθ)Pz(eit)dθ dt ≤ Const ρ2((1− r)1/3).

As ∫|s−a|≥(1/2)(1−r)1/3

Pz(eis)ds ≤ Const(1− r)2/3,

we obtain∫T\∆ |f(eiθ)− f(eit)|2Pz(eiθ)Pz(eit)dθ dt ≤ Const(1− r)2/3.

Since ρ is a modulus of continuity, we have ρ(t) ≥ Const t for all t ∈ (0,1). Andso, ∫

T\∆ |f(eiθ)− f(eit)|2Pz(eiθ)Pz(eit)dθ dt ≤ Const ρ2((1− r)1/3).

Therefore, A(f)(z) ≤ Const ρ2((1− |z|)1/3) as desired.

Lemma 6.3. Let (σ(n))n≥0 and (υ(n))n≥0 be two positive sequences satisfyingthe conditions (S1) and (S2), respectively. Let c > 0 and let

Iσ,υ(c) :=∑n≥1

nσ(n)c

(2υ(n+ 1)2 − υ(n+ 2)2 − υ(n)2).

Then the following assertions hold.(i) There exist two positive constants a and b such that for all q > 0

14Iσ,υ(bq) ≤

∫ 1

0

Ωυ(r)e−qΛσ ((1−r)1/3)1− r 2 dr ≤ 2Iσ,υ(aq).

(ii) If υ(n) = O(σ(n)p) for some p > 0, then Iσ,υ(2p + 1) < +∞.

Proof. (i) Let q > 0; we have

(6.4)∫ 1

0

Ωυ(r)e−qΛσ ((1−r)1/3)(1− r 2)

dr =∑n≥1

∫ 1−1/(n+1)

1−1/n

Ωυ(r)e−qΛσ ((1−r)1/3)(1− r 2)

dr .

So, what we need here is to estimate

e−qΛσ ((1−r)1/3)(1− r 2)


when r ∈ [1−1/n, 1−1/(n+1)

), n ≥ 1. By Lemma 8.4, we have 1

2e−qΛσ (t) ≤

e−qΛσ (t/2) ≤ e−qΛσ (t), t ∈ (0,1). Thus, for a fixed integer n ≥ 1, we have

n4e−qΛσ ((1/n)1/3) ≤ e−qΛσ ((1−r)

1/3)

(1− r 2)

≤ 2ne−qΛσ ((1/n)1/3), r ∈[

1− 1n, 1− 1

n+ 1

).

On the other hand, it follows from Lemma 8.2(i) that Λσ ((1/n)1/3) lnσ(n).This implies that there exist two positive constants a and b such that σ(n)a ≤eΛσ((1/n)1/3) ≤ σ(n)b, (n ≥ 1). Hence, for every integer n ≥ 1, we have

(6.5)n

4σ(n)bq≤ e

−qΛσ ((1−r)1/3)(1− r 2)

≤ 2nσ(n)aq

, r ∈[

1− 1n, 1− 1

n+ 1

).

Therefore, in view of (6.1), (6.4), and (6.5), the desired statement holds.(ii) Let p > 0 and suppose that υ(n) = O(σ(n)p). For every integer k ≥ 1,

we set

I(k)σ ,υ(2p + 1) =k∑n=1

nσ(n)2p+1

(2υ(n+ 1)2 − υ(n+ 2)2 − υ(n)2).

Fixing an integer k ≥ 1, we have

I(k)σ ,υ(2p + 1)

≤ Constk∑n=1

nυ(n)2+1/p

(υ(n+ 1)2 − υ(n)2)− (υ(n+ 2)2 − υ(n+ 1)2

)

= Const k∑n=2

(n

υ(n)2+1/p −n− 1

υ(n− 1)2+1/p

)(υ(n+ 1)2 − υ(n)2)

+ 1υ(1)2+1/p

(υ(2)2 − υ(1)2)− k

υ(k)2+1/p

(υ(k+ 2)2 − υ(k+ 1)2

).

Since (υ(n)2)n≥0 is a concave sequence, we have υ(n+ 1)2 − υ(n)2 ≤ υ(n)2/n,n ≥ 1. This implies that limn→+∞

(n/(υ(n)2+1/p)

)(υ(n+2)2−υ(n+1)2

) = 0.Therefore,


I(k)σ ,υ(2p + 1)

≤ Constk∑n=2

(n

υ(n)2+1/p −n− 1

υ(n− 1)2+1/p

)(υ(n)2 − υ(n− 1)2

)+ C= Const

k∑n=2

nυ(n− 1)2+1/p − (n− 1)υ(n)2+1/p

υ(n)1/pυ(n− 1)1/p

(1

υ(n− 1)2− 1υ(n)2

)+ C

≤ Constk∑n=2

υ(n− 1)2+1/p

υ(n)1/pυ(n− 1)1/p

(1

υ(n− 1)2− 1υ(n)2

)+ C

≤ Constk∑n=2

υ(n− 1)2−1/p(

1υ(n− 1)2

− 1υ(n)2

)+ C

≤ Const∑n

∫ 1/(υ(n−1)2)

1/(υ(n)2)

dtt1−1/(2p) + C < +∞.

This proves the lemma.

Before proving Theorem C, let us remind its statement.

Theorem 6.4. Suppose that(α(n)

)n≥0 and

(β(n)

)n≥0 are two positive se-

quences satisfying the conditions (S1) and (S2), respectively. If β(n) = O(α(n)p)for some p > 0, then Nα,β = H2(D).

On the other hand, if

(6.6)∑n≥1


< +∞,

then H2(D)$Nα,β.

Proof. For a modulus of continuity ρ, we let

Lipρ(D) := f ∈ A(D) : |f(z)− f(w)| = O(ρ(|z −w|))(|z −w| → 0).

(i) Suppose that β(n) = O(α(n)p) for some p > 0, and let us prove thatNα,β = H2(D). Note that, since (α(n))n≥0 satisfies the condition (R), it issufficient to prove that for every µ ∈ Ms(Λα) the singular inner function, Uµ, isbicyclic in `2

ωα,β(Z) (see Theorem 4.3). Let a be the positive constant given inLemma 6.3(i), and let ρ(t) := exp

(−((2p + 1)/a)Λα(t)), t ∈ (0,1]. In view ofLemma 6.3, we have

∫ 1

0

Ωβ(r)ρ((1− r)1/3)1− r 2 dr ≤ 2Iα,β

(a

2p + 1a

)= 2Iα,β(2p + 1) < +∞.


Since ρ is a modulus of continuity (see Lemma 8.4), it then follows from Lemma6.2 that Lipρ(T) ⊂ L2Ωβ(T). Let µ ∈Ms(Λα), and let E := Suppµ. As∫ 2π

0ln

1ρ(

dist(eit, E)) dt < +∞,

it follows from Shirokov’s theorem [28, Theorem 9] that there exists a non-identicallyzero outer function f ∈ Lipρ(D) vanishing on E such that

fUµ ∈ Lipρ(D).

Since ∣∣(fUµ)(eit)− (fUµ)(eiθ)∣∣= |f(eit)Uµ(eiθ)− f(eiθ)Uµ(eit)|≤ |(f (eit)− f(eiθ))Uµ(eiθ)|

+|(fUµ)(eit)− (fUµ)(eiθ)| + |(f (eit)− f(eiθ))Uµ(eit)|,

fUµ ∈ Lipρ(T) ⊂ L2Ωβ(T). By Lemma 6.1, we have∑n∈Z

|ÅfUµ(n)|2β(|n|)2 < +∞;

and so,(ÅfUµ(n))n∈Z ∈ `2

ωα,β(Z). Since f = UµfUµ andH∞(D) ⊂ mult(`2ωα,β(Z)

),

the function f belongs to [Uµ]`2ωα,β (Z)

. This completes the proof since f is anouter function.

(ii) Suppose that

∑n≥1


< +∞.

By Lemma 8.2(ii) and Lemma 5.2, we have∫0

dΛα(t)Λβ(t) =∫

0

t dΛα(t)hβ(t)

> −∞.

It then follows from [1], [3] that there is a positive singular measure µ ∈ Ms(Λα)such that ωµ(δ) = O(hβ(δ)), δ → 0+. Multiplying µ by a positive scalar, ifnecessary, we can suppose that

(6.7) ωµ(δ) ≤ hβ(δ)2c1= δΛβ(δ)

2c1for all δ ∈

(0,β(0)β(1)

).


where c1 is the positive constant given in (3.4). Let a be the positive constantgiven in Lemma 6.3(i), and let ρ(t) := exp

(−(3/(2a))Λα(t)), t ∈ (0,1). Notethat, since (α(n))n≥0 satisfies the condition (S1), ρ is a modulus of continuity(see Lemma 8.4). By Lemma 6.3, we have∫ 1

0

Ωα(r)ρ2((1− r)1/3)1− r 2 dr =

∫ 1

0

Ωα(r)e−(3/a)Λα((1−r)1/3)1− r 2 dr ≤ 2Iα,α(3) < +∞.

Therefore, in view of Lemma 6.2, we have Lipρ(T) ⊂ L2Ωα(T). Let E := Suppµ.Since ∫ 2π

0ln

1ρ(dist(eit, E))

dt < +∞.

By Shirokov’s Theorem [28, Theorem 9], there is a non-identically zero outerfunction f ∈ Lipρ(D) vanishing on E such that fUµ ∈ Lipρ(D). And so,as before, we have fUµ ∈ Lipρ(T) ⊂ L2Ωα(T). Thus, by Lemma 6.1, we have∑n∈Z |ÅfUµ(n)|2α(|n|)2 < +∞. Moreover, we claim that

g := fUµ

=∑n≥0

anzn ∈Nα,β\H2(D).

Indeed, we clearly have g ∈ N\H2(D), g∗ ∈ L2(T), and∑n≥1

|g∗(−n)|2α(n)2 <+∞. It therefore remains to show that

∑n≥1 |an|2/(β(n)2) < +∞. To do this,

we set

γ(r) :=

e−2Λβ(1−r) if r ∈

(1−β(0)β(1)

, 1

)1 otherwise.

By Lemma 8.5, there is a positive constant C such that

1β(n)2

≤ −C∫ 1

0r 2ndγ(t) for all n ≥ 0.

And so, ∑n≥0

|an|2β(n)2

≤ − C2π

∫ 1

0

∫ 2π

0|g(reiθ)|2 dγ(r)dθ.

Using the inequalities (3.4) and (6.7), for 1− β(0)/β(1) < r < 1, we have

|g(reiθ)| =∣∣∣∣∣ f(reiθ)Uµ(reiθ)

∣∣∣∣∣≤ ‖f‖∞ exp c1

ωµ(1− r)1− r

≤ ‖f‖∞e(1/2)Λβ(1−r).


Thus,

∑n≥0

|an|2β(n)2

≤ − C2π

∫ 1

0

∫ 2π

0|g(reiθ)|2dγ(r)dθ

= − C2π

∫ 1

1−β(0)/β(1)

∫ 2π

0|g(reiθ)|2 dγ(r)dθ

≤ − C2π

‖f‖2∞∫ 1

1−β(0)/β(1)eΛβ(1−r) d(e−2Λβ(1−r))

≤ C2π

‖f‖2∞e−Λβ(β(0)/β(1)) < +∞.This completes the proof.

7. CONVERSE OF GENERALIZED KORENBLUM–ROBERTS THEOREM

We now give a converse of Theorem 3.1.

Theorem 7.1. Assume that (α(n))n≥0 is a positive sequence satisfying eitherthe condition (R) with supn≥0

(lnα(n2)/ lnα(n)

)< +∞ or the condition (S1). If

µ ∈Ms(T), then the singular inner function Uµ is cyclic in B2α if and only if µ(E) = 0

for every Λα–Carleson set E.

Proof. Let µ ∈ Ms(T) such that µ(E) > 0 for certain Λα–Carleson set E. Weshall prove that the singular inner function Uµ is not cyclic in B2

α. Let ν be thesingular measure defined by ν(K) = µ(K ∩ E) for each Borel subset K of T. It isclear that ν is supported by E and [Uµ]B2

α⊂ [Uν]B2

α. Therefore, it is sufficient to

show that Uν is not cyclic in B2α.

Case 1. Suppose that (α(n))n≥0 satisfies the condition (S1). Let a be thepositive constant given in Lemma 6.3(i), and let ρ(t) := exp

(−(3/(2a))Λα(t)),t ∈ (0,1). As it is shown in the proof of Theorem 6.4(ii), we see that ρ is a mod-ulus of continuity and Lipρ(T) ⊂ L2Ωα(T). And so, by Shirokov’s theorem thereis an outer function f ∈ Lipρ(D) with Z(f) = E such that fUν ∈ Lipρ(T) ⊂L2Ωα(T). In particular, in view of Lemma 6.1, we have

(7.1)∑n∈Z

|ÅfUν(n)|2α(|n|)2 < +∞.Set g(eit) := eitf (eit)Uν(eit) and note that, in view of (7.1) and the fact thatlimn→+∞

(α(n)/α(n + 1)

) = 1, we have

(7.2)∑n∈Z

|g(n)|2α(|n|)2 < +∞.


Now, define a linear functional on B2α by Lg(f ) := ∑

n≥0 ang(n), f =∑n≥0 anzn ∈ B2

α. In fact, it follows from (7.2) that Lg is a continuous linearfunctional on B2

α. Moreover, we have Lg ≡ 0 on [Uν]B2α . Indeed, for every n ≥ 0,

we have

Lg(znUν) = 12π

∫ 2π

0eintUν(eit)g(eit) dt

= 12π

∫ 2π

0ei(n+1)tf (eit)dt

= 0,

as required. On the other hand, Lg is not identically zero functional. Indeed,assume that Lg ≡ 0 on B2

α. In particular, for every n ≥ 0, we have

0 = Lg(zn)

= 12π

∫ 2π

0eintg(eit) dt

= 12π

∫ 2π

0ei(n+1)t f (eit)

Uν(eit)dt.

This implies that f/Uν ∈ H∞(D) and we have a contradiction since f is an outerfunction and Uν is a non-constant inner function. This completes the proof ofthe first case.

Case 2. Suppose that the sequence (α(n))n≥0 satisfies the condition (R) suchthat supn≥0

(lnα(n2)/ lnα(n)

)< +∞. By Theorem 5.3, there is an outer func-

tion f ∈ A∞(D) such that Z(f) = Z∞(f ) = E and satisfies (5.2). The samecomputation based on Leibniz formula, as in the proof of Theorem 5.10, showsthat the function g(eit) := eitf (eit)Uν(eit) satisfies

∑n≥0 |g(n)|2α(n)2 < +∞.

Now, the rest of the proof of this case goes as before.

8. APPENDIX

In this section, we shall state and prove several technical lemmas that we used inthis paper.

Lemma 8.1. If (α(n))n≥0 is a positive sequence satisfying the condition (R),then the following assertions hold.

(i) For every n ≥ 1 and t ∈[

1− α(n)α(n+ 1)

, 1−α(n− 1)α(n)

],

Λα(t) = n ln(1− t)+ lnα(n).

(ii) For every b > a/(1− a), there exists sb > 0 such that fb : t , tbΛα(t) is anincreasing continuous function on (0, sb) and fb(0) = 0.


(iii)∫ s

0

hα(t)t

dt ≤ Const hα(s) for s near 0.

(iv) Λα(t) ≤ Λα(1− e−t) ≤ 2Λα(t) for t near 0.

(v) Λα(1− e−Kα(n)) lnα(n).

(vi) There is a positive constant b such that Kα(nΛα ( 1

n

))≥ bn

, for every n ≥ 1.

Proof. The first statement (i) results from the log–concavity of the sequence(α(n))n≥0 and the second one follows from [29, Lemma 4.4]. The third one isan immediate consequence of (ii).

Now, let us establish the statement (iv). Indeed, there is a sufficiently smallt0 > 0 such that t/2 ≤ 1 − e−t ≤ t, for all t ∈ (0, t0). Since the function Λα isdecreasing, we have

Λα(t) ≤ Λα(1− e−t) ≤ Λα ( t2), for all t ∈ (0, t0).

As the function hα : t , tΛα(t) is increasing on (0, s1), we have Λα(t/2) ≤2Λα(t) for all t ∈ (0, s1) and (iv) is satisfied.

In order to show (v), we use the same techniques as in [16]. Note that thefunction Kα is decreasing on [1,+∞) and limx→+∞ Kα(x) = 0. It is shown in theproof of Lemma 2.1 of [16] that there is t0 > 0 such that

(8.1)t2K−1α (2t) ≤ Λα(1− e−t) ≤ tK−1

α

(t2

), for all t ∈ (0, t0).

For a sufficiently large integer n, take t = Kα(n)/2 so that 12 lnα(n) ≤Λα(1−e−Kα(n)/2). AsΛα is decreasing, we have 1

2 lnα(n) ≤ Λα((1−e−Kα(n))/2).Hence,

(8.2)12

lnα(n) ≤ 2Λα(1− e−Kα(n)).On the other hand, since

(lnα(k)/

√k)k≥1 is decreasing, we have Kα(n)/2 ≥

Kα(4n). Now, take t = Kα(n) and keep in mind that Kα is decreasing, we getfrom (8.1) that

(8.3) Λα(1− e−Kα(n)) ≤ 4 lnα(n), for n sufficiently large.

From (8.2) and (8.3), we see that Λα(1− e−Kα(n)) lnα(n), as desired.It follows from (iv) and (8.1) that there is t0 such that (Λα(t))/t ≤ K−1

α (t/2),for all t ∈ (0, t0). Since Kα is decreasing, we have

Kα(Λα(t)

t

)≥ t

2, for all t ∈ (0, t0).

This implies the statement (vi).


Lemma 8.2. Let (α(n))n≥0 and (β(n))n≥0 be two sequences satisfying thecondition (R) such that supn≥1

(lnα(n2)/ lnα(n)

)< +∞. The following statements

hold.(i) Λα(1/n) lnα(n) and Λα(t) Λα(√t).

(ii)∑n≥1


∑n

Λα(Kβ(n+ 1))−Λα(Kβ(n))lnβ(n+ 1)

.

Proof. (i) Note that, since f1 : t , tΛ(t) is a modulus of continuity, thesecond identity is a consequence of the first one. As

(lnα(n)/

√n)n≥1 is a non-

increasing sequence, we have Kα(n) ≥ Kα(n2), n ≥ 1. Since the function Λα isnon–increasing and Λα(t) Λα(t/2), it follows from Lemma 8.1(iv) and (v)that there is a positive constant C such that

1C

lnα(n) ≤ Λα ( 1n

)≤ C lnα(n2) , n ≥ 1.

As supn≥1 lnα(n2)/ lnα(n) < +∞, we see that Λα(1/n) lnα(n), as desired.(ii) Since

(lnβ(n)/

√n)n≥1 is decreasing, we have lnβ(1)/n ≤ Kβ(n) ≤

lnβ(1)/√n, ∀n ≥ 1. And so, by (i), we have

(8.4) Λα(Kβ(n)) lnα(n).

On the other hand, as in the proof of Lemma 5.8, limn→+∞Λα(Kβ(n))/ lnβ(n) =0 whenever one of the series∑

n


,

∑nΛα(Kβ(n))

(1

lnβ(n)− 1

lnβ(n+ 1)

)

converges. Thus,

∑n


∑nΛα(Kβ(n))

(1

lnβ(n)− 1

lnβ(n+ 1)

).

In view of (8.4), we have

∑n


∑n

lnα(n)(

1lnβ(n)

− 1lnβ(n+ 1)

)

∑n≥1


.

Therefore, the desired statement holds.


Lemma 8.3. Let (α(n))n≥1 be a positive log–concave sequence such that np =o(α(n)), n→ +∞ for every p ≥ 1. Let

Mk := exp

supt>0(kt −Λα(e−t)), k ≥ 0,

T(x) := supk≥0

xk

k!Mk, x ≥ 1.

Then for every Q > 1,

α(n) = O(T(Qn)n2

), n→ +∞.

For the proof, see [12, Lemma 6.5].

Lemma 8.4. If (σ(n))n≥0 is a positive sequence satisfying the condition (S1),then for every p > 0, the function t , e−pΛσ (t) is a modulus of continuity.

Proof. Since limn→+∞ σ(n)/σ(n + 1) = 1, it is enough to show that thefunction g : t , t−1/pe−Λσ (t) is non–increasing on each interval

In :=[

1− σ(n)σ(n+ 1)

, 1−σ(n− 1)σ(n)

]for all n large enough. Indeed, by Lemma 8.1(i), we have

g(t) = t−1/p((1− t)nσ(n))−1, t ∈ In.

Since the function t , t−1/p(1 − t)−n is decreasing on(0, 1/(1 + pn)), it re-

mains to prove that 1 − σ(n−1)/σ(n) ≤ 1/(1 + pn) for all sufficiently largen. We have lnσ(n)/ ln(n+1) ≤ lnσ(n−1)/ lnn for every n ≥ 2. Thuslnσ(n)/ lnσ(n−1) ≤ 1+1/(n lnn) for every n ≥ 2. As the sequence

(lnσ(n)/

ln(n + 1))n≥0 decreases to 0, we obtain σ(n)/σ(n − 1) = 1 + o(1/n), as

desired.

Lemma 8.5. Let (β(n))n be a positive sequence satisfying the condition (S2),and let γ be the function defined on (0,1) by

γ(r) :=

e−2Λβ(1−r) if r ∈

(1−β(0)β(1)

, 1

),

1 if r ∈(0, 1−β(0)β(1)

).

Then

β(n)−2 −∫ 1

0r 2n dγ(t).


Proof. Since (β(n)2)n is concave, (β(n)2/n)n is decreasing. An easy com-putation shows that

(8.5) β(n)−2 γ(

1− 1n

).

It is clear that for every n ≥ 0,

2n∫ 1

0r 2n−1γ(r)dr = −

∫ 1

0r 2n dγ(r).

So,

∫ 1

0r 2n−1γ(r)dr ≥

∫ 1−1/n

0r 2n−1γ(r)dr

≥ γ(

1− 1n

)∫ 1−1/n

0r 2n−1 dr

≥ Const1nγ(

1− 1n

).

On the other hand, as in the proof of Lemma 8.4, the function γ(t) = γ(1−t)1/4is a modulus of continuity on [0,1]. Therefore, the function γ is decreasing andt → γ(t)/((1− t)4) is increasing.

∫ 1

0r 2n−1γ(r)dr =

∫ 1−1/n

0r 2n−1 γ(r)

(1− r)4 (1− r)4 dr +

∫ 1

1−1/nr 2n−1γ(r)dr

≤ n4γ(

1− 1n

)∫ 1−1/n

0r 2n−1(1− r)4 dr +Const

1nγ(

1− 1n

)≤ Const n4γ

(1− 1

n

)1n5 +Const

1nγ(

1− 1n

).

This, combined with (8.5), concludes the proof of the lemma.

Acknowledgements. The authors would like to thank the referee for readingthe previous version of the paper carefully, for pointing out some inaccuracies andfor his helpful remarks and suggestions.

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A. BOURHIM, O. EL–FALLAH :Departement de MathematiquesUniversite Mohamed VB.P.1014Rabat, Maroc .E-MAIL: [email protected]: [email protected]

K. KELLAY:CMI, LATPUMR 6632Universite de Provence39, rue F.Joliot–Curie13453 Marseille cedex 13, France .E-MAIL: [email protected]

KEY WORDS AND PHRASES: Nevanlinna class; singular inner functions; outer functions; Hyper-functions; shifts.

2000 MATHEMATICS SUBJECT CLASSIFICATION: Primary 30D55; Secondary 30D60, 30H05,47B37.

Received : November 4th, 2002; revised: June 23rd, 2003.

Boundary behaviour of functions of Nevanlinna classkkellay/p9.pdf · Boundary Behaviour of Functions of Nevanlinna Class ... e—ln—n‡1––p with 0

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