Radial limits of interpolating Blaschke products

transcript

DOI: 10.1007/s00208-004-0588-0

Math. Ann. 331, 417–444 (2005) Mathematische Annalen

Pamela Gorkin · Raymond Mortini

Received: 6 November 2003 / Revised version: 1 June 2004 /Published online: 15 October 2004 – © Springer-Verlag 2004

Abstract. It is shown that if (λn) is a sequence of distinct points on the unit circle, then, for asequence (an) of points in the closed unit disk, there exists an interpolating Blaschke productB with B∗(λn) = an for all n if and only if (an) is bounded away from zero. This complementsresults of Cargo, Carroll, Colwell, Belna and Piranian on prescribing radial limits for Blaschkeproducts.

Mathematics Subject Classification (2000): 30D50

1. Introduction

A Blaschke product with zero sequence (zn) in the open unit disk D is a functionof the form

B(z) = eiθ zN

∞∏

|zn|zn − z

1 − znz,

where (zn) satisfy∑

n(1 − |zn|) < ∞. The function B is said to be normalized,if B(0) > 0. We denote the zero set of B by Z(B). A sequence (zn) in D is calledan interpolating sequence if for every bounded sequence of complex numbers,(wn), there exists a bounded analytic function f in D such that f (zn) = wn forall n ∈ N. In 1958, Carleson [C] proved that a sequence of points (zn) in D is aninterpolating sequence if and only if

infk∈N

j :j �=k

∣∣∣∣zj − zk

1 − zj zk

∣∣∣∣ ≥ δ > 0.

A Blaschke product for which the zero sequence, (zn), is an interpolating sequenceis called an interpolating Blaschke product with uniform separation constant δ(B)

defined by

P. GorkinDepartment of Mathematics, Bucknell University, PA 17837 Lewisburg, USA(e-mail: pgorkin@bucknell.edu)

R. MortiniDepartement de Mathematiques, Universite de Metz, Ile du Saulcy, F-57045 Metz, France(e-mail: mortini@math.univ-metz.fr)

418 P. Gorkin, R. Mortini

δ(B) := infk∈N

j :j �=k

∣∣∣∣zj − zk

1 − zj zk

∣∣∣∣.

We note that

δ(B) = infn

(1 − |zn|2)|B ′(zn)|.

By convention, a finite Blaschke product with simple zeros will be consideredan interpolating Blaschke product with associated separation constant as definedabove.

Interpolating Blaschke products play a crucial role in function theory; forexample, in the theory of Douglas algebras and in Hoffman’s theory on the struc-ture of the maximal ideal space of H∞. (See [Ga] for an exposition of theseresults.)

Interpolation problems for finite sets of points were among the first to bestudied. Thus, given a finite set of distinct points in D, {z1, . . . , zn}, and dataw1, . . . , wn, in D, we may ask the following question:When can we find a boundedanalytic function f : D → D such that f (zj ) = wj for all j? For z1, . . . , zn inD, the answer is the Nevanlinna-Pick theorem (see [Ga], p. 7). It tells us that asolution exists if and only if the matrix (

1−wj wk

1−zj zk)nj,k=1 is nonnegative. Moreover,

in that case there is a Blaschke product of degree at most n solving the problem.In this paper, we consider interpolation problems on ∂D, the boundary of the

unit disk. It is known that if z1, . . . , zn and w1, . . . , wn lie on the unit circle,then a finite Blaschke product will always solve the interpolation problem (see,for example, [CP], [JR], [Yo]). Of course, if we consider data with z1, . . . , zn in∂D and w1, . . . , wn in D, we must look beyond finite Blaschke products for asolution.

In 1979, G. T. Cargo [Ca] showed that given n distinct points z1, . . . , zn on ∂D

and w1, . . . , wn in D, there always exists a Blaschke product with radial limit atzj equal to wj (for j = 1, . . . , n). Belna, Carroll, and Piranian [BCaP] extendedCargo’s result by showing that if {λj : j ∈ N} is a countable set of distinctpoints on ∂D and (wj ) a sequence in D, then there exists a Blaschke product withradial limit wj at each λj . The following natural question now arises. Given a set{λj : j ∈ N} of distinct points in ∂D and a sequence (wj ) in D, can we find aninterpolating Blaschke product with radial limit wj at λj ? To fix notation, recallthat if f ∈ H∞, then f ∗(λ) denotes the radial limit function evaluated at the pointλ ∈ ∂D. Thus, if we write f ∗(λ) = w, we mean that the radial limit limr→1 f (rλ)

exists and equals w.It is interesting to note that results for interpolation by Blaschke products do

not always extend to interpolating Blaschke products. For example, Nicolau hasshown ([Ni2], p. 200) that for every Blaschke sequence (zn) there is an interpola-tion problem with data in D of the form f (zn) = wn (for all n) that has a Blaschkeproduct as solution, but no interpolating Blaschke product will solve the problem.

Radial limits of interpolating Blaschke products 419

On the other hand, he also presents a positive result ([Ni2], p. 202): Recall that asequence (zn) and the associated Blaschke product B is called thin, if

(1 − |zn|2)|B ′(zn)| = 1.

Nicolau shows that if (zn) is a thin sequence and (an) is a sequence of points inD for which supn |an| < 1, then there is a (thin) interpolating Blaschke productthat solves the interpolation problem f (zn) = an with f ∈ H∞ and ||f ||∞ ≤ 1,provided the problem has more than one solution.

Another interesting interpolation result is due to J. P. Earl. Letting b(�∞) denotethe unit ball of �∞, and letting (zn) be an interpolating sequence in the unit disk D

with uniform separation constant δ, Earl [E] extended Carleson’s theorem aboveby showing that any interpolation problem in the disk

f (zn) = an with (an) ∈ b(�∞), (IP)

can be solved by a multiple of an interpolating Blaschke product. More precisely,

he proved that if M ≥(

1+√1−δ2

)2, then there exists a (not necessarily normalized)

interpolating Blaschke product B such that MB(zn) = an for all n.Complementing results of Belna, Carroll and Piranian [BCaP], we shall now

prove that boundary interpolation problems

f ∗(λn) = an with (an) ∈ b(�∞) (BIP)

are always solvable by interpolating Blaschke products provided that (an) isbounded away from zero. It will also be shown that in order to find interpolatingBlaschke products solving the boundary interpolation problem (BIP) above, it isnecessary that (an) is bounded away from zero. So it is this condition on the (an)

that tells us precisely when interpolation can be done by an interpolating Blaschkeproduct, as opposed to a general Blaschke product.

We have already singled out thin interpolating sequences in our discussionof Nicolau’s work on interpolation. The associated thin Blaschke products areimportant for a variety of reasons. For example, T. Wolff [Wo] showed that thinsequences are the universal interpolating sequences for the algebra of boundedanalytic functions of vanishing mean oscillation. Nicolau’s result indicates whythin sequences are desirable in interpolation problems. For this reason, it is worththe extra effort to obtain thin products when possible. In fact, for unimodular data(an) we shall construct a thin Blaschke product B solving the boundary interpo-lation problem B∗(λn) = an.

Finally, we refer the reader to papers of A. Nicolau [Ni1] and R. Berman,T. Nishiura [BN1], [BN2] for further results on boundary interpolation by innerfunctions. They considered interpolation problems on other classes of sets of theunit circle.

2. Interpolation by Mobius transforms and finite Blaschke products

For the reader’s convenience, we will begin with some elementary backgroundmaterial on interpolation by Mobius transforms; that is, functions of the form

f (z) = eiθ z0 − z

1 − z0z, where |z0| < 1. We shall denote the set of Mobius transfor-

mations by M.The first three lemmas are standard results from an introductory course on

function theory, but the fourth one, which will be our main tool for the construc-tion of interpolating Blaschke products solving infinite boundary interpolationproblems, does not seem to appear explicitly in the literature. The fifth result,dealing with finite boundary interpolation, has been proven by Jones and Rusche-weyh [JR], with preliminary versions due to Cantor and Phelps [CP] and Younis[Yo]. We also refer to Sarason’s paper [Sa] on Nevanlinna-Pick interpolation onthe boundary of the disk or half-plane.

The following estimation will be used frequently to change estimations involv-ing products to estimations involving sums. It also appears in our papers [GM1]and [GM2].

Lemma 2.1. Let n ∈ N and let aj and bj be complex numbers with |aj | ≤ 1,

|bj | ≤ 1. Then ∣∣∣∣∣∣

aj −n∏

∣∣∣∣∣∣≤

|aj − bj |.

Proof. The proof will be by induction: The result is clearly true if n = 1. Soassume the result holds for a product of n factors. Let Pn(a) = ∏n

j=1 aj , and usethe following inequalities:

∣∣∣∣∣∣

n+1∏

aj −n+1∏

∣∣∣∣∣∣≤ |an+1

(Pn(a) − Pn(b)

)| + |an+1 − bn+1||Pn(b)|

≤ |Pn(a) − Pn(b)| + |an+1 − bn+1|,from which the result now follows. �

For a ∈ D, we let La denote the normalized Blaschke factor La(z) = |a|a

a − z

1 − az.

If a = 0, we take |a|/a = −1.

Lemma 2.2. Let a, b ∈ D and α, β ∈ ∂D be distinct numbers. Then there is a

unique Mobius transformation f (z) = eiθ z0 − z

1 − z0zwith f (a) = b and f (α) = β.

Proof. Let �1(z) = La(α)La(z) and �2(z) = Lb(β)Lb(z). Then f := �−12 ◦�1

maps a to b and α to β. To check uniqueness, suppose that there is a second such

transformation g. Then f ◦g−1 fixes b and β. Since a Mobius transformation witha fixed point in D and a fixed point on the unit circle must be the identity, we haveestablished uniqueness. �Lemma 2.3. Let (α1, α2, α3) and (β1, β2, β3) be triples of distinct points on theunit circle. Then there exists f ∈ M with f (αj ) = βj for j = 1, 2, 3 if and onlyif the triples determine the same orientation on T.

Proof. There is a unique Mobius transformation T with T (αj ) = βj for j =1, 2, 3. A circle is uniquely determined by three points on it. The orientationpreserving property of Mobius transformations now yields the assertion. �Lemma 2.4. Let (α1, α2) and (β1, β2) be pairs of distinct points on the unit circle.Then there exist infinitely many Mobius transformations f such that f (αj ) = βj

for j = 1, 2. The zero of f can be chosen arbitrarily on an arc joining α1 with α2

that cuts ∂D at an angle of π/2 − |θ |, where

θ = 1

[| arg β2 − arg β1| − π

], (mod 2π), |θ | ≤ π/2. (1)

Proof. Let β = β2β1. Then β �= 1. Choose an argument of β satisfying 0 <

arg β < 2π . We need to determine all functions f ∈ M such that f (1) = 1 andf (−1) = β, and then we will modify f to get the function we want. To this end,we begin with the following.

Let b ∈ D and f (z) =(

−1 − b

1 − b

)b − z

1 − bz. Then f (1) = 1 and f (−1) = β

if and only if

(1 + b

1 − b

)/(1 + b

1 − b

)= −β. If we write (1 + b)/(1 − b) = reiθ

with r > 0 and θ ∈] − π/2, π/2[, we see that we must have e2iθ = −β, but r

can be an arbitrary positive number. To obtain the desired argument, we chooseθ = (arg β −π)/2. Letting T denote the map defined by T (z) = (1 + z)/(1 − z),we see that our b can be chosen from the set of solutions to T (b) = reiθ , wherer > 0. In other words, b can be chosen from the set T −1({reiθ : r > 0}). Now thisis just the image of the ray {reiθ : r > 0} under the map T −1(z) = (z−1)/(z+1).This image is an arc in D, which we denote here by C, joining −1 to 1 and makingan angle of |θ | degrees with respect to the line segment [−1, 1]. Thus any pointon this arc will work as a choice for b and

f (z) = �b,β(z) :=(

−1 − b

1 − b

)b − z

1 − bz

will solve the problem

�b,β(1) = 1 and �b,β(−1) = β.

For γ ∈ ∂D, we let Rγ denote the rotation given by Rγ (z) = γ z. Let α = α1α2.Now consider the function g defined by g = Rβ1 ◦ �b,β ◦ �−1

a,α ◦ Rα1 , where

a ∈ D is chosen so that

(1 + a

1 − a

)/(1 + a

1 − a

)= −α. Then g(αj ) = βj for

j = 1, 2, and the zero of g lies on the curve (Rα1 ◦ �a,α)(C), which is part ofa circle passing through α1 and α2 and making an angle π/2 − |θ | with the unitcircle. Conversely, any point on this curve can be chosen to be a zero of a functiong satisfying g(αj ) = βj , (j = 1, 2). �The following interpolation result is due to Jones and Ruscheweyh [JR].

Proposition 2.5. [JR] Let α1, . . . , αN be distinct points of ∂D and let β1, . . . ,

βN ∈ ∂D. Then there exists a Blaschke product of degree less than or equal toN − 1 such that B(αj ) = βj for j = 1, . . . , N .

The following corollary is a consequence of the preceding proposition. Thoughits proof can be found in [GLMR], we present a short proof here for convenience.

Corollary 2.6. Let α1, . . . , αN be distinct points of the unit circle and let β1, . . . ,

βN ∈ ∂D. Then there exists a Blaschke product B of degree less than or equal toN such that B(0) = 0 and B(αj ) = βj for j = 1, . . . , N . If βj = 1 for all j ,then the degree of this B is necessarily N .

Proof. By Proposition 2.5 there is a Blaschke product B1 of degree n , with n lessthan N − 1, such that B1(αj ) = αjβj for j = 1, 2, . . . , N . Then the Blaschkeproduct B defined by B(z) = zB1(z) is the solution we were looking for. Forj = 1, . . . , n + 1, let zj denote the zeros of B. Because eitB ′(eit )/B(eit ) =∑n+1

j=11−|zj |2|eit−zj |2 > 0, we see that each point on ∂D has exactly n + 1 distinct pre-

images under B on ∂D. Thus we see that the degree of B is N in the case whereall the βj = 1. �

The next two lemmas and their proofs can be found in [GM1]. We remind the

reader that the function La is the function given by La(z) = |a|a

a − z

1 − az.

Lemma 2.7. Let a ∈ D. Then the following assertions hold.

(i) La converges locally uniformly in D to 1 as |a| → 1.(ii) Let λ ∈ ∂D and let (an) denote a sequence in D converging to λ. Then for

every neighborhood, U(λ), of λ the sequence (Lan) converges uniformly to 1

on D \ U(λ).

Lemma 2.8. Let θ be chosen with 0 < θ < π/4 and let β = eiθ . Suppose thatσ ∈ ∂D and that a ∈ D satisfies 1 − |a| <

|1−β|2 and La(σ ) = β. Then the image

of {rσ : r ∈ [0, 1]} under La is contained in the disk |z − 1| ≤ |1 − β|.

The next lemma is one that we will use to obtain information on the distancebetween the zeros of a Blaschke product. We will measure the distance using thepseudohyperbolic distance defined on points a and b of D by

ρ(a, b) =∣∣∣∣

a − b

1 − ab

∣∣∣∣ .

Lemma 2.9. Let 0 < s < 1. Suppose that for j = 1, 2, . . . , n

Qj := Qj(s, eitj ) := {reit : 0 < 1 − r < s, |t − tj | < s}

are pairwise disjoint Carleson boxes. Let d be the minimal arc distance betweenthese squares divided by π and let B be a Blaschke product of degree n withexactly one zero in each of these boxes. If s < d

1+d, then the uniform separation

constant, δ(B), of B can be estimated as follows:

δ(B) > δn(s, d) :=(

(1 − s)d − s

4s + d

)n−1

Proof. For j �= k, let a ∈ Qj and b ∈ Qk Suppose that α = a/|a| and β = b/|b|.By Lemma 4.1 in [MN] we have |1 − ab| ≤ 3(1 −|a|)+ 1 −|b|+ |α −β|. Usingthis and the fact that δn(s, d) is increasing in d, we obtain

ρ(a, b) =∣∣∣∣

|a|α − |b|β1 − |a| |b|αβ

∣∣∣∣ ≥∣∣∣∣|a|(α − β) + β(|a| − |b|)

4s + |α − β|∣∣∣∣ ≥

≥ (1 − s)|α − β| − s

4s + |α − β| ≥ (1 − s)d − s

4s + d.

Since 0 < s < d/(1+d), the numerator is positive. So, if B(z) = eiθ∏n

j=1aj −z

1−aj z,

where aj ∈ Qj , then δ(B) = infk

∏nj=1,j �=k ρ(aj , ak) ≥ δn(s, d), completing the

proof. �Later we will use the fact that lims→0 δn(s, d) = 1 uniformly in d ∈ [ε, 1]. Theconvergence is monotonic.

Recall that a subset E of D is said to be ρ-separated with respect to the pseudo-hyperbolic metric, if ρ(x, y) ≥ ρ > 0 for all x, y ∈ E, with x �= y.

We need two more results before we turn to our result (Theorem 3.3) on infiniteboundary interpolation of unimodular data by interpolating Blaschke products.The first of these two results is the following technical lemma.

Lemma 2.10. Let s be a real number satisfying 0 < s < 1. Suppose thatα0, α1, α2, . . . , αn are distinct points on the unit circle and let β ∈ ∂D \ {1}.Then for every δ with 0 < δ < 1 and m ∈ N there exists a Blaschke product Bn

of degree n such that

Bn(αj ) = 1, for j = 0, 1, 2, . . . , n − 1, (3)

Bn(αn) = β, (4)

|1 − Bn(z)| < 1/2m+2 for |z| ≤ s, (5)

|1 − Bn(rαj )| < 1/2m for j = 0, 1, 2, . . . , n − 1 and 0 < r ≤ 1, (6)

δ(Bn) ≥ δ. (7)

In addition, the following also hold:

(i) If E is a ρ-separated set in D, then the zero of Bn closest to αn can be chosento be at a pseudohyperbolic distance of at least ρ/3 to the points of E.

(ii) It is possible to choose the zeros a1, . . . , an of Bn so that

1−|aj ||aj −αk | ≤ 2−m for j = 1, . . . , n − 1 and k = 0, 1, 2, . . . , n (8)

and such that an is very close to αn.

Note that (5) and (7) tell us that the zeros of Bn may be chosen as close to the unitcircle and as far apart from each other as we wish.

Proof. By Corollary 2.6 there exists a Blaschke product Rn of degree n such thatRn(0) = 0 and Rn(αj ) = 1 for j = 0, 1, . . . , n − 1. Let α = Rn(αn). Thenα �= 1, because 1 has only n preimages under the (degree n) Blaschke productRn.

Using Lemma 2.4, we obtain a set of Mobius transformations, Ma , satisfyingMa(a) = 0, Ma(1) = 1, Ma(α) = β, and the zero a lying on the arc, A, joining1 with α as described in the lemma. Let Bn = Ma ◦ Rn. Then Bn satisfies (3) and(4). Note that though Bn depends on the parameter a, at this point a ∈ A is freeand thus we will be able to choose a so that Bn will also satisfy (5)–(7).

For (5), we note that Rn(0) = 0 and therefore, by Schwarz’s Lemma, |Rn(z)| ≤|z|. Thus, for fixed a we have

{(Ma ◦ Rn)(z) : |z| ≤ s} ⊆ {Ma(w) : |w| ≤ s}.

Recalling that for some λa ∈ ∂D we have Ma(z) = λa

(a − z

1 − az

)and that

Ma(1) = 1, for |w| ≤ s we obtain

|1 − Ma(w)| =∣∣∣∣λa

(a − 1

1 − a

)− λa

(a − w

1 − aw

)∣∣∣∣ =∣∣∣∣

(a − 1

1 − a

(a − w

1 − aw

)∣∣∣∣

= (1 − |a|2)|w − 1||(1 − a)(1 − aw)| ≤ 2(1 − |a|2)

(1 − s)|1 − a| → 0

as a → α �= 1. Therefore, choosing a close enough to α, we get |Bn(z) − 1| <

1/2m+2 whenever |z| ≤ s. This gives (5).

Now we show how to obtain (7). Let us label the n distinct points where Rn

assumes the value α as σ1, σ2, . . . , σn. We know that αn is one of these points, sayσn = αn. Note that the σj are distinct from the αk, for k = 0, 1, . . . , n − 1, sincethe values of Rn taken at these points are α and 1, respectively, and α �= 1. Letd∗ be one third of the minimum of the angular distances between the σj dividedby π . Using the notation from Lemma 2.9, we choose s∗ ∈ ]0, d∗

1+d∗ [ so close to0 that δn(s, d) > δ whenever d > d∗ and s < s∗.

Now Rn maps the closure, Qj , of a small Carleson box Qj about each of thesepoints σj onto a neighborhood (in D) of α. Since there are only finitely many suchpoints, we may choose a so close to α that each of the n points where Rn = a,denoted a1, . . . , an, are distinct and each one will be close to some σj . Usingthis, we may assume that aj ∈ Qj . Without loss of generality we may chooseour boxes so small that the arc length s is smaller than s∗, the minimal angulardistance πd between distinct boxes is bigger than πd∗ and

Qj ∩n−1⋃

{rαk : 0 < r < 1} = ∅.

(See figure).

Fig. 1. Carleson boxes

By Lemma 2.7, if b → σj , then |b|b

(b−z

1−bz

)→ 1 uniformly on D \ Qj . Thus,

requiring a to be even closer to α if necessary, we may assume that

1 −n∏

|aj | < 1/2m+2 and (9)

∣∣∣1 − ∏nj=1

|aj |aj

(aj −z

1−aj z

) ∣∣∣ < 1/2m+1 for all z ∈ D \ ⋃nj=1 Qj. (10)

Note that for any a the zeros of Bn = Ma ◦ Rn will occur at the points whereRn = a, which we have denoted a1, . . . , an. Thus, there exists λ ∈ ∂D such that

Bn(z) = λ

|aj |aj

(aj − z

1 − ajz

Since aj ∈ Qj , Lemma 2.9 shows that δ(Bn) ≥ δn(s, d) > δ; that is, (7) issatisfied.

We must still check (6). Obviously, (Ma ◦Rn)(0) = Bn(0) = λ∏n

j=1 |aj |. But

|1 − λ| ≤∣∣∣∣∣∣1 − λ

|aj |∣∣∣∣∣∣+

∣∣∣∣∣∣λ

|aj | − λ

∣∣∣∣∣∣

≤ |1 − Bn(0)| + (1 −n∏

|aj |)(9)≤(5)

2−m−2 + 2−m−2 = 2−m−1.

For k = 0, 1, . . . , n − 1 we have {rαk : 0 < r < 1} ⊆ D \ ⋃nj=1 Qj , so using

|1 − Bn(rαk)| =∣∣∣∣∣∣1 − λ

|aj |aj

(aj − rαk

1 − aj rαk

)∣∣∣∣∣∣

=∣∣∣∣∣∣(1 − λ) +

λ − λ

|aj |aj

(aj − rαk

1 − aj rαk

∣∣∣∣∣∣

≤ 1/2m+1 +∣∣∣∣∣∣1 −

|aj |aj

(aj − rαk

1 − aj rαk

)∣∣∣∣∣∣< 1/2m,

completing the proof of (6).We now turn to the proof of (i) and (ii). For the proof of (i), recall that the

zeros a1, . . . , an of Bn were chosen to be contained in Carleson boxes centeredat σ1, . . . , σn and that the points σj (j = 1, . . . , n − 1), are distinct from theinterpolation nodes αk for k = 0, 1, 2, . . . , n − 1, except for σn, which coincideswith αn. Now the zero an = an(a) of Bn had the representation R−1

n (a) for somebranch of R−1

n . Since this branch is a continuous function of a, we see that if theimage of R−1

n hits two disjoint pseudohyperbolic disks of radius ρ/3 centered atpoints of E, then as a runs through a connected subset of D, this image must meetthe complement of the union of all pseudohyperbolic disks of radius ρ/3 centeredat points in E. It suffices to take such an a to get the zero of Bn we were lookingfor.

Assertion (ii) follows from the first part of the proof presented here. �

Lemma 2.10 dealt with solutions to interpolation problems of the form f (αj ) =βj , (j = 1, . . . , n), where all but one βj = 1. Now we must consider the gen-eral case, where the βj are chosen randomly from the unit circle. Proposition 2.5guarantees the existence of a Blaschke solution of degree less than n, but it doesnot tell us where the zeros will be. This will be dealt with in Theorem 2.11 andallow us to control the location of the zeros. In general, if we choose a solution ofdegree n to the interpolation problem, we cannot expect the zeros to be close tothe boundary. For example, as was shown in ([GLMR], p. 267), if |ξ | = 1, thereexists a Blaschke product Bn of degree n with B(λj ) = ξ, (j = 1, . . . , n) andB(0) = B(a) = 0, if and only if a lies in the convex hull of the λj (j = 1, . . . , n).By taking the n preimages µ1, . . . , µn of a second point η ∈ ∂D, we see that a

lies in the intersection of these two closed convex hulls; hence is far away fromthe boundary of D (see also [DGM] for related material).

Thus, in order to obtain solutions of our boundary interpolation problemB(αj ) = βj , j = 1, . . . , n, with zeros close to ∂D, we have to consider degreeshigher than n.

Theorem 2.11. Let α1, . . . , αn be distinct points on the unit circle and let |βj | = 1for j = 1, . . . , n. Suppose that E is a ρ-separated subset in D whose cluster setis contained in the set of interpolation nodes, {αj : j = 1, 2, . . . , n}. Then forevery δ ∈ ]0, 1[ and m ∈ N there exists a finite Blaschke product Bn, of degreeless than or equal to n2, such that

Bn(αj ) = βj for j = 1, 2, . . . , n; (11)

δ(Bn) ≥ δ; (12)

ρ(Z(Bn), E) ≥ ρ/3. (13)

Moreover, if p ≥ 2 and rn ∈ [0, 1[, then Bn can be chosen so that for k ∈{1, 2, . . . , n} and r ∈ [0, 1]

|Bn(rαk) − 1| ≤ |1 − βk| + 2−m, whenever | arg βk| < π/4, (14)

and |Bn(z) − 1| ≤ p−m−2 for |z| ≤ rn. (15)

Proof. The proof will be by induction. Choose δj ∈ ]0, 1[ and q ∈ N, q ≥ 2,such that

k=1k �=j

(1 − q−k) > δ, (16)

q−m−4 < p−m−2, (17)

|1 − βk| ≥ 2q−m−2 for every k with βk �= 1, (18)

β ∈ ∂D and min{k:| arg βk |<π/4}

|βk − β| < q−m implies | arg β| < π/4. (19)

Let s1 ∈ ]rn, 1[. Use Lemma 2.10 to choose a Blaschke product b1 of degree n

such thatb1(α1) = β1;

b1(αi) = 1 for i = 2, . . . , n;

|b1(z) − 1| < q−m−5 for |z| ≤ s1;|b1(rαi) − 1| ≤ q−m−4 for r ∈ [0, 1] and i = 2, 3, . . . , n and

δ(b1) > δ1.

Assume now that for k = 1, . . . , j −1, where j ≤ n, numbers s1 < s2 < . . . < sk

and Blaschke products bk of degree n have been chosen to satisfy

bk(αk) = βk; (20)

bk(αi) = 1 for 1 ≤ i ≤ n, i �= k; (21)

|bk(z) − 1| < q−m−4−k for |z| < sk; (22)

|bk(rαi) − 1| < q−m−3−k for r ∈ [0, 1] and 1 ≤ i ≤ n, i �= k ; (23)

δ(bk) > δk (24)

whenever βk �= 1 and bk ≡ 1 if βk = 1.We shall now proceed with the induction step (which is done on j ). Since

the bk are finite Blaschke products, we may choose sj ∈ ]sj−1, 1[ so that for allk ∈ {1, 2, . . . , j − 1} we get |bk| ≥ 1 − q−k on {z : |z| ≥ sj }. By Lemma 2.10,there exists bj satisfying (20)–(24) with k replaced by j above. This completes theinduction. By our choice of sj , it follows that these Blaschke products bk have theproperty that |bk| ≥ 1−q−k on the zeros of the remaining bj , j = 1, . . . , n, j �= k.

Let Bn = b1b2 . . . bn. Then Bn(αj ) = βj for j = 1, . . . , n, so (11) is satis-fied. Further, the degree of Bn is less than or equal to n2. The uniform separationconstant of Bn can be estimated as follows: Let bj (z) = 0. Then, by the choice ofq given in (16),

(1 − |z|2)|B ′n(z)| =

k:k �=j

|bk(z)|

(1 − |z|2)|b′j (z)|

k:k �=j

(1 − q−k)

δj ≥ δ,

where the first inequality follows from the fact that δ(bj ) > δj .We are almost done, but our Bn must still be modified before we can conclude

that Bn satisfies (13). To do this we will choose bj more carefully, using (i) and(ii) of Lemma 2.10. Fix j . For ν = 1, . . . , n let aν denote the zeros of bj . These

zeros lie in certain Carleson boxes, say aν ∈ Qν , centered at points σν , whereσj = αj , and σk /∈ {α1, . . . , αn} for k �= j . Now, (i) and (ii) of Lemma 2.10 tellus that the zero closest to αj , which we denote by aj , can be chosen to satisfyρ(E, aj ) ≥ ρ/3. Recall that the Carleson boxes Qν are pairwise disjoint. Usingthe hypothesis that E clusters only at the interpolation nodes αl, l = 1, . . . , n, wesee that for k �= j whenever Qk shrinks to σk we have ρ(Qk, E) → 1. Hence,for k �= j , the pseudohyerbolic distance to E of ak can be made as close to 1 as wewant. Thus ρ(Z(bj ), E) ≥ ρ/3. Since this can be done for every j ∈ {1, . . . , n},we see that ρ(Z(Bn), E) ≥ ρ/3, which is (13).

Assertion (15) is seen as follows. Let z be chosen with |z| ≤ rn. Becausern < s1 < sj , (22) and (17) yield

|Bn(z) − 1| ≤n∑

|bj (z) − 1| ≤n∑

q−m−4−j ≤ q−m−4 ≤ p−m−2.

To prove (14), we have to choose the zeros of bk yet more carefully. Recall thatBn = ∏n

j=1 bj , bk(αk) = βk and that bk = λk

∏nν=1 Laν

, where the zeros aν

depend on k. We may assume that βk �= 1. By Lemma 2.7 and the method ofproof in Lemma 2.10, the aν could have been chosen so that 1 − |aν | ≤ q−m−2

2and |Laν

− 1| < (1/n)q−m−3 outside a small neighborhood of σν , where σν �∈{αi : i �= k} and |σν | = 1. In particular, this holds on the rays {rαi : 0 ≤ r ≤1}, i �= k. Moreover, we may suppose that |λk − 1| ≤ q−m−3 (see the proof ofLemma 2.10). Without loss of generality, we may let ak be the zero of bk closestto αk. So,

|Lak(αk) − βk| = |Lak

(αk) − bk(αk)| = |Lak(αk)|

∣∣∣∣∣∣∣1 − λk

ν=1ν �=k

Laν(αk)

∣∣∣∣∣∣∣

≤|1−λk|+n∑

ν=1ν �=k

|Laν(αk)−1| ≤ q−m−3+

ν=1ν �=k

(1/n)q−m−3 ≤q−m−2. (25)

Let βk = Lak(αk). Since by hypothesis | arg βk| < π/4, from (25) and the choice

of q in (19) we see that | arg βk| < π/4. To apply Lemma 2.8, we have to verifythat 1 − |ak| ≤ 1

2 |1 − βk|. But this follows from the following estimates: sinceβk �= 1, we have

|1 − βk| ≥ |1 − βk| − |βk − βk| ≥(25)

|1 − βk| − q−m−2

≥(18)

q−m−2 ≥ 2(1 − |ak|).

Thus, by Lemma 2.8, for any r ∈ [0, 1]:

|Lak(rαk) − 1| ≤ |1 − βk| ≤ |1 − βk| + |βk − βk| ≤ |1 − βk| + q−m−2.

Now we know that bk(αk) = βk, |λk − 1| ≤ q−m−3, and q ≥ 2, so we may applyLemma 2.1 to obtain with (23)

|Bn(rαk) − 1| ≤n∑

j=1j �=k

|bj (rαk) − 1| + |bk(rαk) − 1|

≤n∑

j=1j �=k

q−m−3−j + |λk − 1| +n∑

ν=1ν �=k

|Laν(rαk) − 1| + |Lak

(rαk) − 1|

≤ q−m−3∞∑

q−j + q−m−3 + q−m−3 + q−m−2 + |1 − βk|

≤ 2−m + |1 − βk|. �

3. Boundary interpolation: data on ∂D

Proposition 3.1. Let (Bn) be a sequence of Blaschke products. Suppose that∑n |Bn − 1| converges locally uniformly on D. Then the infinite product B =∏n Bn converges locally uniformly and unconditionally to a Blaschke product.

Proof. It is well known that our assumptions imply that∏

Bn converges locallyuniformly in D to a function, denoted here by B, in the unit ball of H∞. Thus, weneed only show that B is a Blaschke product. Let B be the normalized Blaschkeproduct associated with the zeros of B. We have to show that B = eiθ B. First wenote that by our assumptions, at most finitely many Bn can vanish at 0, and that ifthere is a zero of order m of B at zero, then there is a zero of order m of B at zero.So we may assume that neither vanish at 0, and we consider the function B/B.Due to the unconditional convergence, we can rearrange and regroup the factors

for B in such a way that B = ∏n

|Bn(0)|Bn. In particular, |B(0)| = ∏n |Bn(0)|.

Now B/B is also a function in the unit ball and since |B(0)| = |B(0)|, it must beconstant, which is what we were aiming to prove. �Lemma 3.2. Let (Bn) be a sequence of normalized interpolating Blaschke prod-ucts with δ(Bn) ≥ δ > 0. Let εn ∈ ]0, 1[ be such that η := ∏

n(1−εn) converges.Suppose that

⋃n Z(Bn) is a Blaschke sequence and that, for each n, |Bn| ≥ 1−εn

j �=n Z(Bj ). Then∏

Bn is an interpolating Blaschke product with separationconstant bigger than infn δ(Bn)η.

Proof. Let B = ∏a∈∪Z(Bn) La be the normalized Blaschke product associated

with the set⋃

Z(Bn). Then the convergence of this infinite product is locallyuniform and unconditional in D. Hence its factors can be rearranged in an arbi-trary manner. Since Bn has only simple zeros, we see that B = ∏

Bn. Nowsuppose that z ∈ D is a zero of B. Then there exists n such that Bn(z) = 0. So

(1 − |z|2)|B ′(z)| = (1 − |z|2)|B ′n(z)| |

j �=n

Bj (z)| ≥ infn

δ(Bn)∏

j �=n

(1 − εj )

η infn

δ(Bn) ≥ ηδ > 0.

Hence B is an interpolating Blaschke product. �Theorem 3.3. Let (αj ) be a sequence of distinct points on ∂D and let (βj ) be asequence of points on ∂D. Then there exists a thin interpolating Blaschke productB such that B∗(αj ) = βj for all j .

Proof. We will use induction on n to construct Blaschke products Bn of degreen such that the uniform separation constant of Bn is “big” in a certain sense, theBlaschke product Bn has the right radial behavior, and such that |Bn| is close to1 on the zeros of the remaining Blaschke products Bj for j �= n. The Blaschkeproduct we are looking for will be the product of these Bn. The proof will becompleted in three steps.

Step 1. To begin our construction, we let (δn) be a sequence in ]0, 1[ convergingto one.

Let 0 < r1 < 1. To simplify our proof, we add α0, a unimodular constantdistinct from the other αj , to our list. If β1 �= 1, use Lemma 2.10 to choose aBlaschke product of degree 1, called B1, such that B1(α0) = 1, B1(α1) = β1,|1 − B1(z)| ≤ 1/24 for |z| ≤ r1 and |1 − B1(rα0)| < 1/22 for 0 < r < 1. Ifβ1 = 1, let B1 ≡ 1.

Suppose that for k = 1, . . . , n − 1 the k-th degree Blaschke product, Bk, andthe positive numbers r1, . . . , rk have been chosen so that r1 < r2 < · · · < rk and

Z(Bk) ⊆k−1⋂

{|Bj | > 1 − 2−j }; (26)

|Bk| ≥ 1 − 2−k on the zeros of B1, . . . , Bk−1; (27)

Bk(αj ) = 1, for j = 1, 2, . . . , k − 1; (28)

Bk(αk) =

k−1∏

Bj(αk)

βk; (29)

|1 − Bk(z)| < 1/2k+1 for |z| ≤ rk; (30)

|1 − Bk(rαj )| < 1/2k for j = 1, 2, . . . , k − 1 and 0 < r ≤ 1; (31)

δ(Bk) ≥ δk → 1, (32)

if βk

∏k−1j=1 Bj(αk) �= 1; otherwise, let Bk ≡ 1.

We now proceed with the induction step. Since the Bj are finite Blaschkeproducts, we may choose rn ∈ ]0, 1[ with rn > max{rn−1, 1 − 2−n} so that

|Bj(z)| > 1 − 2−j for rn ≤ |z| ≤ 1 and j = 1, . . . , n − 1. By Lemma 2.10 thereis a Blaschke product Bn of degree n satisfying (28)–(32) for k = n. Due to (30)and the choice of rn, this Bn automatically satisfies (26) and (27). That concludesour inductive construction.

Step 2. Now we form the infinite product B = ∏n Bn. Since rn → 1, (30) implies

the local uniform convergence of∑ |1 − Bn| in D. By Proposition 3.1, B is a

Blaschke product (though not necessarily normalized). We claim that thisBlaschke product does the boundary interpolation.

To see this, fix n ∈ N, choose ε > 0, use Lemma 2.1 and consider the follow-ing. There exists an integer m such that m > n and 1/2m < ε/3. By Lemma 2.1,for all r ∈ [0, 1[ we have

|1 −∞∏

Bj(rαn)| ≤∞∑

|1 − Bj(rαn)| (31)<

∞∑

2−j = 1/2m < ε/3.

Now, as seen by (29),∏n

j=1 Bj has radial limit βn at αn. Thus there exists ρ0 =ρ0(ε, n) such that for r > ρ0 we have

Bj(rαn) − βn| < ε/3.

Also, for j = n + 1, n + 2, . . . , m the Blaschke product Bj has radial limit 1 atαn , so there exists ρ1 = ρ1(ε, n) such that for r > ρ1 we have

Bj(rαn) − 1| < ε/3.

So choose r with r > max{ρ0, ρ1}. Then, by Lemma 2.1 and the three precedingestimates, for all such r we have

|B(rαn) − βn| ≤∣∣∣

Bj(rαn) − βn

∣∣∣

+∣∣∣∣∣∣

Bj(rαn) − 1

∣∣∣∣∣∣+

∞∑

|Bj(rαn) − 1|

≤ ε/3 + ε/3 + ε/3 = ε.

Hence B∗(αn) = βn for all n.

Step 3. We are now going to show that B is a thin interpolating Blaschke product.We have to prove that B has no multiple zeros and that if (zn) is the zero sequenceof B, then lim

(1 −|zn|2

)|B ′(zn)| = 1 for B(zn) = 0 and n → ∞. To this end, let

ε > 0. Since δ(Bn)(32)→ 1, there exists n0 so that

∏j≥n0

(1 − 2−j ) ≥ 3√

1 − ε and

δ(Bn) ≥ 3√

1 − ε for n ≥ n0. Take r0 = r0(ε) so close to 1 that |∏j≤n0Bj(z)| ≥

1 − ε whenever r0 ≤ |z| < 1. Then, by (26) and (27), for those z with B(z) = 0for which Bn(z) = 0, |z| > r0 and n > n0 we obtain

(1 − |z|2)|B ′(z)| =∣∣∣∣∣∣

j≤n0

∣∣∣∣∣∣

n−1∏

j=n0+1

∣∣∣∣∣∣

Bj (z)

∣∣∣∣∣∣(1 − |z|2)|B ′

≥∣∣∣∣∣∣

j≤n0

∣∣∣∣∣∣

∞∏

j=n0+1j �=n

(1 − 2−j )

(1 − |z|2)|B ′n(z)|

≥ (3√

1 − ε)3 = 1 − ε.

Thus B is a thin Blaschke product. Since the zeros of the finite Blaschke product∏n0j=1 Bj are all simple, and distinct from those of

∏j>n0

Bj (see (27)), we getthat B is an interpolating Blaschke product. (This last assertion also follows fromthe proof of Lemma 3.2.) Moreover, δ(B) ≥ infn δ(Bn)

∏∞j=1(1 − 2−j ). �

We remark that it is well known that no thin Blaschke product can have a nonuni-modular radial limit. This follows for example from Lemma 4.3. Thus, in orderto obtain a thin Blaschke product doing the interpolation, it is necessary that thevalues we interpolate are unimodular.

We also remark that by choosing the δj appropriately and by replacing the2−j by q−j for some suitable q ≥ 2 the uniform separation constant of the thinBlaschke product B above, can be taken as close to one as desired.

4. Boundary interpolation by finite products of interpolating Blaschkeproducts

Let u be an inner function. By Sing u we denote the set of all those points on theunit circle, for which u has no analytic continuation. If u is a Blaschke product,then Sing u is simply the set of cluster points of the zeros of u. Recall that theN -th tail of a Blaschke product B(z) = eiθ zp

∏∞n=1

|an|an−z

1−anzis the function T

defined by

T (z) =∞∏

|an|an − z

1 − anz.

Proposition 4.1. Let (bn) be a sequence of interpolating Blaschke products withSing bn ∩ Sing bk = ∅ for n �= k. Suppose that δ := infn δ(bn) > 0. Then thereexist tails, Tn, of bn such that

∏Tn is an interpolating Blaschke product.

Proof. Take any sequence of numbers εn ∈ ]0, 1[ for which η := ∏n(1−εn) > 0.

We shall construct the tails, Tn, inductively so that

|Tn| > 1 − εn on⋃

{k:k �=n}Z(Tk), and (33)

{z:z∈Z(Tn)}(1 − |z|) ≤ 2−n. (34)

So, let T1 be a tail of b1 satisfying∑

{z:z∈Z(T1)}(1 − |z|) ≤ 2−1. Assume that fork = 1, 2, . . . , n − 1 the tails Tk have been chosen to satisfy

|Tk| > 1 − εk on Z(Tj ) for j = 1, . . . , k − 1, and (35)

|Tj | > 1 − εj on Z(Tk) for j = 1, . . . , k − 1. (36)

Now for j < n, we assume that Sing bn ∩ Sing bj = ∅, so bn is analytic andtherefore |bn| = 1 on Sing Tj . Hence |bn| > 1 − εn in a neighborhood Uj,n ofSing Tj (within D). Since Blaschke products converge locally uniformly in D, wemay drop off finitely many zeros of bn to obtain a tail Tn of bn that is close to 1on each of the compact sets Z(Tj ) \ Uj,n, for j = 1, . . . , n − 1, and that satisfies(34). Thus we obtain a tail satisfying (34) and (35) for k = n; that is, |Tn| > 1−εn

on Z(Tj ) for j < n.On the other hand, we note that for j < n the function bj (and hence Tj ) is

analytic and unimodular on Sing Tn. By considering yet another tail, Tn, of Tn, forj < n we may assume that Tj has modulus bigger than 1 − εj on Z(Tn). HenceTn satisfies (36) with k replaced by n.

Thus, this set of tails, {Tn : n ∈ N}, satisfies (33) and (34). This ends ourconstruction of tails.

By (34), it is clear that ∪Z(Tn) is a Blaschke sequence. Hence, due to uncon-ditional convergence, the associated Blaschke product B can be written as B =∏

Tn, where all the Tn are normalized. The only thing left to do is to show that B

is an interpolating Blaschke product. But (33) and δ(Tn) ≥ δ(bn) ≥ δ > 0 assureus that the hypotheses of Lemma 3.2 are fulfilled and we see that B is indeed aninterpolating Blaschke product. Moreover, we get that δ(B) ≥ δη. �Observe that if b = ∏

bn is an interpolating Blaschke product, then triviallyδ(bn) ≥ δ(b) for each n. Our proof above shows that the uniform separationconstant of B can be chosen to be as close to the inf δ(bn) as we wish.

The following version of Hoffman’s Lemma will be used to estimate the uni-form separation constant of the solution to our interpolation problem.

Lemma 4.2. [Ho], [Ga] Let B be an interpolating Blaschke product with uniformseparation constant δ(B). Let δ and η be positive numbers satisfying 0 < δ <

δ(B), 0 < η <1 − √

1 − δ2

δ. Then for z ∈ D the inequality ρ(z, Z(B)) ≥ η

implies that |B(z)| > η2.

The next lemma also appears in our paper [GM1]. For the reader’s conveniencewe represent its short proof.

Lemma 4.3. Let B be an interpolating Blaschke product with δ(B) > δ > 0.Then for every eiθ ,

lim supr→1

|B(reiθ )| ≥(

1 − √1 − δ2

In other words, if K denotes the radial cluster set of B at eiθ , then

max{|z| : z ∈ K} ≥(

1−√1−δ2

Proof. Let τ = 1−√1−δ2

δand let (zn) denote the zeros of B. By Hoffman’s Lemma,

the disks Dρ(zn, τ ) are pairwise disjoint. Moreover, |B(z)| > τ 2 whenever z /∈⋃n Dρ(zn, τ ). For r0 ∈]0, 1[ this union does not cover the entire ray {reiθ : r0 ≤

r < 1}, so we see that lim supr→1 |B(reiθ )| ≥ τ 2. �Corollary 4.4. Let B be an interpolating Blaschke product. Suppose that fora sequence (λn) on the unit circle the radial limits an := B∗(λn) exist. Theninfn |an| > 0.

We note that Corollary 4.4 also holds for finite products of interpolating Blaschkeproducts. Indeed, if B is a finite product of interpolating Blaschke products, thenby ([MN], Theorem 2.2), there exists ε > 0 such that for all a with |a| < ε theinner function a−B

1−aBis again a finite product of interpolating Blaschke products.

But, by that same theorem, finite products of interpolating Blaschke products can-not have radial limits 0 and so 0 cannot be a radial limit of a−B

1−aBwhenever |a| < ε.

Therefore, if |a| < ε, we see that a cannot be a radial limit of B, and consequentlyinfn |an| ≥ ε.

We now turn to some information about the atomic singular inner functionS defined by S(z) = exp

(− 1+z1−z

). It is well known that for any a ∈ D \ {0} the

Frostman transform of S given by ba = a − S

1 − aSis an interpolating Blaschke

product with δ(ba) = 2|a| log(1/|a|)1 − |a|2 . The radial limit of ba at the point z = 1

is a. It is this property that will guide us in constructing interpolating Blaschkeproducts with prescribed radial limits. We will also use the following facts: Thezeros of ba cluster only at the point 1. So, for any neighborhood U of 1, thisBlaschke product will converge uniformly on D \ U and, in particular, the tails ofba converge uniformly to the function 1 on this set.

We now turn to the construction of a Blaschke product that solves our inter-polation problem for the case in which the interpolated values are bounded awayfrom 0 (and lie in D.)We will be able to show that the function that does the interpo-lation is a product of two interpolating Blaschke products. The result also follows

from our main theorem (Theorem 5.1), but by proving this weaker assertion, wehope to provide some insight into the proof of the general problem.

This proof will use a function, called F for the remainder of this paper, defined

on ]0, 1[ by F(x) := 2x log 1/x

1 − x2. The function F is strictly increasing on ]0, 1[,

limx→0 F(x) = 0 and limx→1 F(x) = 1. Further, x < F(x) ≤ 2√

1+x, where the

upper estimate follows from the fact that

F(x)/(2√

1 + x) =

√x log(1/x)

1 − x= F(

√x) ≤ 1.

Proposition 4.5. Let (λn) be a sequence of distinct points on the unit circle andlet (an) be a sequence in D that is bounded away from zero. Then

(1) there exists an interpolating Blaschke product B such that for every n theradial limit of B at λn exists and has modulus |an|. If A = inf{|an| : n ∈ N},then the uniform separation constant of B can be chosen to be bigger than A,and

(2) there exists a Blaschke product C that is a product of two interpolatingBlaschke products and satisfies C∗(λn) = an for every n.

Proof. For the proof of (1), let Sn(z) = S(λnz) and consider the Frostman shifts

bn = an − Sn

1 − anSn

. Since composing an interpolating Blaschke product with a rota-

tion on the right does not change the uniform separation constant, we see from theremarks above that δ(bn) = F(|an|) ≥ F(A) > A > 0. Moreover Sing(bn) ={λn}, and therefore the Sing(bj ) are pairwise disjoint sets. Hence, by Proposition4.1 there are tails Tn of bn such that B = ∏

Tn is an interpolating Blaschke product.By comments following the proof of the Proposition 4.1, we may suppose thatδ(B) > A.

To obtain the existence of a radial limit with correct modulus at λn, we haveto choose the tails Tn of bn satisfying

|Tn − 1| ≤ 2−n on D \ Un, (37)

where Un is a small neighborhood of λn in C that does not contain the radialsegments {rλj : 0 ≤ r ≤ 1} for j = 1, . . . , n− 1. Let fn = B/Tn. We claim that

limr→1 |fn(rλn)| = 1 for every n ∈ N. (38)

To see this, fix n. Let ε > 0 and choose an integer N > n so large that

∏j≥N(1 − 2−j ) ≥ 1 − ε. Since

j=1j �=n

Tj is analytic at λn, it is unimodular at

λn, and so there exists r0 ∈ ]0, 1[ such that |N∏

j=1j �=n

Tj (rλn)| ≥ 1 − ε for r0 ≤ r < 1.

Hence, by (37), for those r satisfying r0 ≤ r < 1 we have

|fn(rλn)| =

∣∣∣∣∣∣∣

j=1j �=n

Tj (rλn)

∣∣∣∣∣∣∣

∣∣∣∣∣∣

Tj (rλn)

∣∣∣∣∣∣≥ (1 − ε)2.

This proves (38).Next we show that the radial limit, limr→1 fn(rλn), exists. Indeed, by (37), for

every ε > 0 there exists an integer N > n such that for r ∈ ]0, 1[∑

|Tj (rλn) − 1| ≤ 2−N+1 < ε.

Again, due to the analyticity of∏N

j=1j �=n

Tj at λn and Lemma 2.1, there exists r1 ∈]0, 1[ such that for all r, s ∈ [r1, 1[ we have

|fn(rλn) − fn(sλn)| ≤

∣∣∣∣∣∣∣

j=1j �=n

Tj (rλn) −N∏

j=1j �=n

Tj (sλn)

∣∣∣∣∣∣∣

+∞∑

|Tj (rλn) − 1| +∞∑

|Tj (sλn) − 1| ≤ ε + ε + ε = 3ε.

Thus the limit, limr→1 fn(rλn), exists, and by (38), this limit is unimodular. Notic-ing that the radial limit at λn of Tn exists and has modulus |an|, we conclude thatlimr→1 B(rλn) exists and has modulus |an|. This gives assertion (1).

Turning to the proof of (2), we let βn = an

|an| and γn = B∗(λn)

|an| . Then |γn| = 1.

Using Theorem 3.3, there exists an interpolating Blaschke product B such thatB∗(λn) = γnβn for every n. Hence the Blaschke product BB has radial limit an

at λn, and is, of course, a product of two interpolating Blaschke products. �If the zeros of B = ∏

Bn could be chosen to be ρ-separated from those of B, itwould follow (as in [Ga], p. 314) that BB is an interpolating Blaschke product.Unfortunately, Lemma 2.10 tells us, that only one zero of each of the n-th degreeBlaschke products Bn can be chosen to be at a fixed distance of a ρ-separated set.Thus, in order to prove our main result, we must combine the methods in sections2 and 3. Our solution to the general interpolation problem B∗(λn) = an for all n

(which will be an interpolating Blaschke product) will have the form

B =∞∏

b1,nb2,n . . . bn,nTn,

where the Tn control the moduli of the radial limits (as above), the n-th degreeBlaschke products bj,n solve interpolation problems with right side of the form1, 1, . . . , 1, ωj , 1, . . . , 1 and where we are able to control the location of thezeros as discussed in Lemma 2.10 and so that b1,nb2,n . . . bn,nTn = 1 at the pointsλj for j = 1, . . . , n − 1 and

b1,nb2,n . . . bn,nT∗n = an at λn.

The new feature of the proof will be that the Tn are chosen to have a certaindependence on the Blaschke products Bn := b1,nb2,n . . . bn,n (and conversely, ofcourse).

5. Boundary interpolation: data in D

We are now ready to prove the main result of this paper.

Theorem 5.1. Let (λn) be a sequence of distinct points on the unit circle and let(an) be a sequence of points in D. Then there exists an interpolating Blaschkeproduct B whose radial limit at λn is an if and only if (an) is bounded away fromzero.

Proof. Suppose that B is an interpolating Blaschke product with the prescribedradial limits an. By Corollary 4.4 we know that (an) is bounded away from zero.

So suppose now that A := inf |an| > 0. Let (δn) be a sequence in ]0, 1[converging to one, and satisfying δn > F(A) (where F(x) = 2x log(1/x)

1−x2 ). Weshall construct pairs of functions, Bn and Tn, where Bn is a finite Blaschke prod-uct of degree n2 with uniform separation constant bigger than δn that is bigon the zeros of the remaining Bj and where Tn is a tail of the Frostman shiftan − Sn

1 − anSn

of Sn(z) = S(λnz) that is big on the zeros of the other Tj and such

that ρ

(Z(Bn), Z(

∏j≤n Tj )

)> A/3. Moreover, BnTn will be an interpolating

Blaschke product with uniform separation constant bigger than A3/9 such thatBnTn has the right radial behavior at λ1, . . . , λn. Our final solution will be theproduct

∏n BnTn.

Step 1. Since F(A) > A, we can choose an integer q > 2 such thatF(A)

∏j (1 − q−j ) > A. Let r1 be chosen with 0 < r1 < 1. We would like

to apply Lemma 2.10, and in order to do so, we add a unimodular point, λ0,

to our list with λ0 �= λj . If |a1| < 1, let T1 = a1 − S1

1 − a1S1. If |a1| = 1, but

a1 �= 1, use Lemma 2.10 to choose a Blaschke product of degree 1, called B1,such that B1(λ0) = 1, B1(λ1) = a1, |1 − B1(z)| ≤ 1/q3 for |z| ≤ r1 and|1 − B1(rλ0)| < 1/q for 0 < r < 1. If a1 = 1, let B1 ≡ 1.

Suppose that for k = 1, . . . , n − 1 the Blaschke products, Bk, of degree k2,

the tails Tk ofak − Sk

1 − akSk

, and the numbers 0 < r1 < r2 < · · · < rk have been

chosen to satisfy

Z(Bk) ⊆k−1⋂

{|Bj | > 1 − q−j }; (39)

Z(Tk) ⊆k−1⋂

{|Tj | > 1 − q−j } ∩k−1⋂

{|Bj | > 1 − q−j }; (40)

|Bk| ≥ 1 − q−k on the zeros of B1, . . . , Bk−1; (41)

|Tk| ≥ 1 − q−k on the zeros of T1B1, . . . , Tk−1Bk−1; (42)

BkTk(λj ) = 1, for j = 1, 2, . . . , k − 1; (43)

BjT∗j )(λk) = ak; (44)

|1 − Bk(z)| < 1/2k+1 and |1 − Tk(z)| < 1/2k+1 for |z| ≤ rk; (45)

|1 − BkTk(rλj )| < 1/2k−1 for j = 1, 2, . . . , k − 1 and 0 < r ≤ 1; (46)

δ(Bk) ≥ δk → 1; (47)

ρ(Z(Bk), Z(Tj )) ≥ A/3 for j ≤ k; (48)

δ(BkTk) ≥ A3/9. (49)

We now proceed with the induction step. For j = 1, . . . , n − 1 the Bj are finiteBlaschke products, so we may choose rn ∈ ]0, 1[ with rn > max{rn−1, 1 − 2−n}so that |Bj | > 1 − q−j on {rn ≤ |z| ≤ 1}. Let Un be a neighborhood of λn in D

that does not contain the rays {rλj : 0 ≤ r ≤ 1} for j = 1, . . . , n − 1, that isdisjoint from the disk {z : |z| ≤ rn} and such that for j = 1, . . . , n − 1 we have|Tj | > 1 − q−j on Un.

If |an| = 1, let Tn ≡ 1. For n �= j , we know that λn /∈ Sing Tj , so if |an| < 1

we may choose a tail Tn ofan − Sn

1 − anSn

such that

|Tn − 1| ≤ q−n−2 on(D \ Un

) ∪ ⋃n−1j=1 Z(BjTj ). (50)

This choice of Tn forces Z(Tn) ⊆ Un ⊆ {|z| > rn}. We will also need an estimateon δ(

∏j≤n Tj ) (to obtain (56) below), so we show here that δ(

∏j≤n Tj ) > A,

and, in particular Z(∏

j≤n Tj ) is A-separated. To see this, suppose Tj (z) = 0.Then

(1 − |z|2)∣∣∣∣∣(

1≤k≤n

Tk)′(z)

∣∣∣∣∣ = (1 − |z|2)|T ′j (z)|

1≤k≤nk �=j

|Tk(z)|

≥ δ(Tj )∏

k �=j

(1 − q−k) ≥ F(A)∏

k �=j

(1 − q−k) > A.

Noticing that |Tn − 1| (50)< q−n−2 at λj (for 1 ≤ j ≤ n − 1) implies that

| arg Tn(λj )| < π/4 , we may useTheorem 2.11 to conclude that there is a Blaschkeproduct Bn of degree less than or equal n2 satisfying

Bn(λj ) = Tn(λj ) for j = 1, . . . , n − 1; (51)

Bn(λn) = βn :=[(∏n−1

j=1 BjTj

)(λn)

] (T ∗

n (λn)/|T ∗n (λn)|

)(an/|an|); (52)

|Bn(z) − 1| ≤ q−n−2 for |z| ≤ rn; (53)

|Bn(rλj )−1|≤2−n+|1−Tn(λj )| for r ∈ [0, 1[ and j =1, . . . , n−1; (54)

δ(Bn) ≥ δn; (55)

ρ(Z(Bn), Z(

Tj )) ≥ A/3. (56)

Since the radial limit T ∗n (λn) of Tn at λn exists and has modulus |an|, we

BjT∗j )(λn) = (

n−1∏

BjTj )(λn) (BnT∗n )(λn)

n−1∏

BjTj )(λn) (

n−1∏

BjTj )(λn) · T ∗n (λn)

|T ∗n (λn)|

|an| T ∗n (λn)

= |T ∗n (λn)|2

|T ∗n (λn)|

|an| = |an| an

|an| = an.

Putting all this together we see that Bn and Tn satisfy (39)–(45) and (47)–(48)(with the k appearing there replaced by n).

We now show that (46) is satisfied, too. So let 1 ≤ j ≤ n − 1. Then for0 ≤ r < 1 we may use (54) and (50) to see that

|Bn(rλj ) − 1| ≤ 2−n + |1 − Tn(λj )| ≤ 2−n + 2−n−2,

and hence by Lemma 2.1

It remains to show that (49) holds; that is, that BnTn is an interpolating Blaschkeproduct with δ(BnTn) ≥ A3/9. For this we note that

A/3 <1 − √

1 − A2

A< A ≤ F(A) ≤ F(|an|) ≤ δ(Tn).

Hence, by Hoffmann’s Lemma 4.2, ρ(Z(Bn), Z(Tn)) ≥ A/3 implies that |Tn| ≥A2/9 on Z(Bn). Similarly |Bn| ≥ A2/9, since we know that δ(Bn) ≥ δn > A.Thus, if Bn(z) = 0 we obtain

(1 − |z|2)|(BnTn)′(z)| = (1 − |z|2)|B ′

n(z)| |Tn(z)| ≥ δ(Bn)A2/9 ≥ A3/9

and if Tn(z) = 0 we also obtain that (1 − |z|2)|(BnTn)′(z)| ≥ A3/9.

That concludes our inductive construction. Let us note (for later purposes) that(40) and (42) imply that

|Tj | ≥ 1 − q−j on⋃

k �=j Z(Tk); (57)

and that (39) and (41) imply that

|Bj | ≥ 1 − q−j on⋃

k �=j Z(Bk). (58)

Step 2. Let Pn = BnTn. Now we form the infinite product B = ∏n Pn. By Lemma

2.1 and (45), we have |1 − BnTn| ≤ |1 − Bn| + |1 − Tn| ≤ 2−n for |z| ≤ rn.Since rn → 1, this implies the local uniform convergence of

∑ |1 − Pn| in D. ByProposition 3.1, we know that B is a Blaschke product, though not necessarilynormalized. We claim that this Blaschke product does the boundary interpolation.To see this, fix n ∈ N, choose ε > 0, and consider the following. There exists aninteger m = m(ε), m > n, such that 1/2m−1 < ε/3 and for all r ∈ [0, 1[ wehave (using Lemma 2.1)

|1 −∞∏

Pj(rλn)| ≤∞∑

|1 − Pj(rλn)|

∞∑

2−j+1 = 1/2m−1 < ε/3.

Now∏n

j=1 Pj was chosen to have radial limit an at λn, so there exists ρ0 = ρ0(ε, n)

such that for r > ρ0 we have

Pj(rλn) − an| < ε/3.

Also, Pj has radial limit 1 at λn for j = n + 1, n + 2, . . . , m, so there existsρ1 = ρ1(ε, n) such that for r > ρ1 we have

Pj(rλn) − 1| < ε/3,

So choose r with r > max{ρ0, ρ1}. Then, by Lemma 2.1, for all those r we have

|B(rλn) − an| ≤∣∣∣

Pj(rλn) − an

∣∣∣

+∣∣∣∣∣∣

Pj(rλn) − 1

∣∣∣∣∣∣+

∞∑

|Pj(rλn) − 1| ≤ ε/3 + ε/3 + ε/3 = ε.

Hence B∗(λn) = an for all n.

Step 3. To conclude the proof, we are now going to show that B is an interpolatingBlaschke product. Since δ(Tn) ≥ F(A), we obtain by Lemma 3.2, (57) and thechoice of q that T = ∏

n Tn is an interpolating Blaschke product with

δ(T ) > infn

δ(Tn)∏

j �=n

(1 − q−j ) ≥ F(A)∏

j �=n

(1 − q−j ) > A.

Similarly, since δ(Bn) ≥ δn > F(A), we see by (58) and the proof of Lemma 3.2that

∏Bn is an interpolating Blaschke product with δ(

∏n Bn) > A.

By (48) ρ(Z(Bn), Z(

∏j≤n Tj )

) ≥ A/3 and so ρ(Z(

∏j≥n Bj ), Z(Tn)

) ≥A/3. Since

A/3 <1 − √

1 − A2

A< A < min

δ(∏

Tj ), δ(∏

by Hoffman’s Lemma 4.2 we obtain

|∏j≤n Tj | ≥ (A/3)2 on Z(Bn) and |∏j≥n Bj | ≥ (A/3)2 on Z(Tn). (59)

Now we estimate the uniform separation constant of B = ∏BnTn. So let z ∈ D

with B(z) = 0.

Case 1. Tn(z) = 0. Then by (40) |Bj(z)| ≥ 1 − q−j for j < n and by (57)

(1 − |z|2)|B ′(z)| = (1 − |z|2)|T ′n(z)|

j �=n

|Tj (z)|∏

|Bj(z)|∣∣∣∣∣∣

Bj (z)

∣∣∣∣∣∣

≥ δ(Tn)∏

j �=n

(1 − q−j )∏

(1 − q−j ) (A/3)2 ≥ C > 0.

Case 2. Bn(z) = 0. Then by (42) |Tj (z)| > 1 − q−j for j > n and hence, due to(58)

(1 − |z|2)|B ′(z)| = (1 − |z|2)|B ′n(z)|

j �=n

|Bj(z)|∣∣∣∣∣∣

Tj (z)

∣∣∣∣∣∣

|Tj (z)|

≥ δ(Bn)∏

j �=n

(1 − q−j ) (A/3)2∏

(1 − q−j ) ≥ C > 0.

�We remark that by taking δ(Bn) close to 1 and q very large, then a lower boundfor δ(B) can be taken to be as close to A3/9 as we wish.

Acknowledgements. The authors thank the Mathematisches Forschungsinstitut Oberwolfachfor the support and for the kind hospitality they always receive. The work presented here ispart of their 2002–2004 project “Inner functions”. The first author also thanks Universitat Bernwhere she spent her sabbatical, as well as Universite de Metz, where she was “professeur invite”for one month.

The second author presented this work in May 2004 at the fourth congress of the Europeannetwork “Analysis and Operators” in Dalfsen, the Netherlands. He gratefully acknowledgesthe support he received from the European community’s human potential program, contractHPRN-CT-2000-00116.

References

[BCaP] Belna, C., Carroll, F., Piranian, G.: Strong Fatou-1-points of Blaschke products.Trans. Amer. Math. Soc. 280, 695–702 (1983)

[BCoP] Belna, C., Colwell, P., Piranian, G.: The radial behavior of Blaschke products. Proc.Amer. Math. Soc. 93, 267–271 (1985)

[BN1] Berman, R., Nishiura, T.: Interpolation for inner functions on dispersed subsets ofthe unit circle. J. London Math. Soc. 38, 463–484 (1988)

[BN2] Berman, R., Nishiura, T.: Some mapping properties of the radial limit function of aninner function. J. London Math. Soc. 52, 375–390 (1995)

[CP] Cantor, D., Phelps, R.: An elementary interpolation theorem. Proc. Amer. Math. Soc.16, 523–525 (1965)

[Ca] Cargo, G.: Blaschke products and singular inner functions with prescribed boundaryvalues. J. Math. Analysis Appl. 71, 287–296 (1979)

[C] Carleson, L.: An interpolation problem for bounded analytic functions. Amer. J.Math. 80, 921–930 (1958)

[DGM] Daepp, U., Gorkin, P., Mortini, R.: Ellipses and finite Blaschke products.Amer. Math.Monthly 109, 785–795 (2002)

[E] Earl, J.P.: On the interpolation of bounded sequences by bounded functions. J. LondonMath. Soc. 2, 544–548 (1970)

[Ga] Garnett, J.: Bounded Analytic Functions. Academic Press, New York 1981[GL] Glader, C., Lindstrom, M.: Finite Blaschke product interpolation on the closed unit

disc. J. Math. Anal. Appl. 273, 417–427 (2002)[GLMR] Gorkin, P., Laroco, L., Mortini, R., Rupp, R.: Composition of inner functions. Results

in Math. 25, 252–269 (1994)[GM1] Gorkin, P., Mortini, R.: Cluster sets of interpolating Blaschke products. Preprint[GM2] Gorkin, P., Mortini, R.:Value distribution of interpolating Blaschke products. Preprint

[Ho] Hoffman, K.: Bounded analytic functions and Gleason parts. Ann. Math. 86, 74–111(1967)

[JR] Jones, W., Ruscheweyh, S.: Blaschke product interpolation and its application to thedesign of digital filters. Constr. Approx. 3, 405–409 (1987)

[KL] Kerr-Lawson, A.: Some lemmas on interpolating Blaschke products and a correction.Canad. J. Math. 21, 531–534 (1969)

[MN] Mortini, R., Nicolau, A.: Frostman shifts of inner functions, to appear in J. d’AnalyseMath.

[Ni1] Nicolau, A.: Blaschke products with prescribed radial limits. Bull. London Math.Soc. 23, 249–255 (1991)

[Ni2] Nicolau, A.: Interpolating Blaschke products solving Pick-Nevanlinna problems. J.d’Analyse Math. 62, 199–224 (1994)

[Sa] Sarason, D.: Nevanlinna-Pick interpolation with boundary data. Integral EquationsOperator Theory 30, 231–250 (1998)

[Yo] Younis, R.: Interpolation by a finite Blaschke product. Proc. Amer. Math. Soc. 78,451–452 (1980)

[Wo] Wolff, T.: Some theorems on vanishing mean oscillation. PH. D.Thesis, Univ. ofCalifornia, Berkeley, 1979

Radial limits of interpolating Blaschke products

Documents