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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo

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Volume 16, Issue 1


Volume 13 (2015)

Recursive interpolating sequences

Francesc Tugores
Published Online: 2018-04-30 | DOI: https://doi.org/10.1515/math-2018-0044


This paper is devoted to pose several interpolation problems on the open unit disk 𝔻 of the complex plane in a recursive and linear way. We look for interpolating sequences (zn) in 𝔻 so that given a bounded sequence (an) and a suitable sequence (wn), there is a bounded analytic function f on 𝔻 such that f(z1) = w1 and f(zn+1) = an f(zn) + wn+1. We add a recursion for the derivative of the type: f′(z1) = w1 and f′(zn+1) = an [(1 − |zn|2)/(1 − |zn+1|2)] f′(zn) + wn+1, where (an) is bounded and (wn) is an appropriate sequence, and we also look for zero-sequences verifying the recursion for f′. The conditions on these interpolating sequences involve the Blaschke product with zeros at their points, one of them being the uniform separation condition.

Keywords: Interpolating sequence; Uniformly separated sequence; Bounded analytic function

MSC 2010: 30E05; 30H05; 30J10

1 Introduction

Interpolation problems on the unit disk 𝔻 are a classical branch of complex analysis. Several types of interpolating sequences for different classes of analytic functions have been addressed since the middle of the last century, beginning with the celebrated works of W.K. Hayman [1], D.J. Newman [2] and L. Carleson [3] about the so-called “universal” interpolation problem, which consists in characterizing the sequences (zn) in 𝔻 verifying that for any bounded sequence (wn), there is a bounded analytic function f on 𝔻 such that f(zn) = wn.

On the other hand, recursion appears in many areas of mathematics: formulas, algorithms, optimization…, providing alternative definition procedures. Since there exists a specific theory for recursive numerical sequences and interpolating sequences have not been studied from a recursive perspective, we think it is interesting to pose recursive-type interpolation problems.

We want to emphasize that our approach converts the universal interpolation problem and other problems related to it into trivial cases of those that we introduce. Furthermore, most conditions involved are new and depend not only on the separation of the points of the sequence in 𝔻, but also on the sequences that we employ to define recursion.

We begin with the necessary notation. Let H be the space of all analytic functions f on 𝔻 such that ‖f = supz∈𝔻|f(z)| < ∞ and let l be the Banach space of all sequences of complex numbers (wn) such that ‖(wn)‖ = supn |wn| < ∞. We put Z = (zn) for any sequence of different points in 𝔻 verifying the Blaschke condition ∑n(1 − |zn|) < ∞, which characterizes the zero-sequences of functions in H. For two points z and w in 𝔻, we write


so that ρ = |ψ| is their pseudo-hyperbolic distance. Let B be the Blaschke product with zeros at Z, that is,


If E is a subsequence of Z, we put BE for the Blaschke product with zeros at E and for a fixed m ∈ ℕ, we denote BZ∖{zm} by Bm. We write c for strictly positive constants that may change from one occurrence to the next one.

First, we recall that Z is interpolating if given any (wn) ∈ l, there exists fH such that f(zn) = wn. Interpolating sequences are characterized by the well-known Carleson’s theorem:

Theorem 1.1

([3]). Z is interpolating if and only if


Sequences satisfying (1) are called uniformly separated (u.s.). From the Schwarz lemma, it follows that


and thus, it is said that Z is interpolating in differences if given (wn) verifying |wiwj| ≤ c ρ(zi, zj), there is fH such that f(zn) = wn. These sequences are the union of two u.s. [4], characterized as follows.

Lemma 1.2


For a sequence Z, the following are equivalent

  1. Z is the union of two u.s. sequences.

  2. For each zi, there exists zj such that |Bi(zi)| ≥ c ρ(zi, zj).

  3. Z is either u.s. or it can be rearranged Z = (αn) ∪ (βn), where (αn) and (βn) are u.s. sequences, ρ(αn,βn) = ρ(αn,Z ∖ {αn}) → 0, and ρ(αn, zi) ≥ c, if ziβn.

Finally, since |f′(z)|(1 − |z|2) ≤ cf, it is said that Z is double interpolating if given (wn) ∈ l and (wn) satisfying (wn(1 − |zn|2)) ∈ l, there is fH such that f(zn) = wn and f′(zn) = wn. It is proved in [6] that these sequences are also the u.s. ones.

Next we consider the following three quantities for the terms of a sequence T = (tn) ∈ l:


We need them to take suitable target spaces in our interpolation problems and they also appear in the results we get (Section 2).

We write Γ(tm) = O(Γ(tm+1)) (resp. Λ(tm) = O(Λ(tm+1))) if there exists a constant cT, Z > 0 such that Γ(tm) ≤ cT, Z Γ(tm+1) (resp. Λ(tm) ≤ cT, Z Λ(tm+1)) for all m ∈ ℕ. We write Π(tm) ∼ Π(tm+1) if Π(tm) ≤ cT, Z Π(tm+1) and Π(tm+1) ≤ cT,Z Π(tm) for some constants cT, Z, cT,Z > 0 and for all m ∈ ℕ.

From now on A = (an) and A′ = (an) will denote sequences in l. Our purpose is to examine the following distinguished sequences of 𝔻:

Definition 1.3

We say that Z is A-interpolating if given (wn) verifying




there exists fH such that


Clearly if ‖A < 1, then the sum in (3) is bounded by (wn)1A.

Definition 1.4

We say that Z is A-interpolating in differences if given (wn) satisfying (3) and


there is fH verifying recursion (5).

Definition 1.5

We say that Z is (A,A′)-interpolating if given (wn) satisfying (3) and (4) and (wn) verifying




there exists fH satisfying recursion (5) and


Definition 1.6

We say that Z is zero and A′-interpolating if given (wn) verifying (7),




there is fH vanishing on Z and satisfying recursion (9).

Recursion (5) is equivalent to f(zn) = μn, where


and recursion (9) is equivalent to f′(zn) = μn, with


Thus, we must have (3) and (7) to state that sequences (μn) and (μn(1 − |zn|2)) are bounded. We impose that data sequences (wn) and (wn) verify (4), (6), (8) and (11), because they are intrinsic to recursions (a technical reason justifies (10)). In effect, (4) and (6) are obtained taking into account (2) in the inequalities




respectively. On the other hand,





See [7] for (14), [8] for (15) and [9] for (16). Thus, (8) is obtained using (14) in the inequality


If on the right of this last inequality, we put


then (11) is obtained using (16) for the first summand and (15) for the second one.

We introduce these interpolating sequences because they provide a generalization of the usual interpolation problems, in the sense that (0)-interpolating sequences, (0)-interpolating in differences and ((0), (0))-interpolating are interpolating, interpolating in differences and double interpolating, respectively.

Extending recursion to an arbitrary order or increasing the degree of derivability are projects certainly cumbersome, so that we confine ourselves to order one and the first derivative. Nevertheless, we think it would be interesting to consider these types of sequences for other spaces of analytic functions, such as the Lipschitz class and the Bloch space, for which the pseudo-hyperbolic distance in (2) is replaced by the Euclidean and hyperbolic distance, respectively (interpolating sequences for these spaces are characterized in [10] and [11]).

While the proofs of results turn out to be rather standard (Carleson’s theorem is used repeatedly), we appreciate the following separation conditions, which are consistent with the problems posed and appear in a natural way.

Definition 1.7

We say that (Z, A) satisfies condition (S) if


and condition (D) if


We say that (Z, A′) satisfies condition (M) if


We name (S), (D) and (M) to the above conditions because these are the initials of simple, differences and mixed, respectively, and we will see in the next section that condition (S) is related to interpolating sequences in a simple sense; (D), in a differences sense, and (M), in a mixed sense (zero and interpolating).

Since ρ(zm, zm+1) > |Bm+1(zm+1)|, it follows that if (Z, A′) verifies (M), then (Z, −A′) satisfies (S). All conditions imply (b) in Lemma 1.2 so that Z is the union of two u.s. sequences. Note that if ‖A < 1 (resp. ‖ A′‖ < 1), then (S) (resp. (M)) ⇒ u.s.

2 Statement of results

Our results are the following ones.

Proposition 2.1

  1. If Z is A-interpolating and Γ(an) = O(Γ(an+1)), then (Z, A) verifies (S).

  2. If Z is A-interpolating in differences and Π(an) ∼ Π(an+1), then (Z, A) verifies (D).

  3. If Z is (A, A′)-interpolating, Γ(an) = O(Γ(an+1)) and Γ(an)=O(Γ(an+1)), then Z is u.s.

  4. If Z is zero and A′-interpolating and Λ(an)=O(Λ(an+1)), then (Z, A′) verifies (M).

Proposition 2.2

  1. If (Z, A) verifies (S), then Z is A-interpolating.

  2. If (Z, A) verifies (D) and A is such that


    for some r ∈ (0, 1), then Z is A-interpolating in differences.

  3. If Z is u.s., then Z is (A, A′)-interpolating for any sequences A and A′.

  4. If (Z, A′) verifies (M), Λ(an)=O(Λ(an+1)) and


    then Z is zero and A′-interpolating.

3 Proof of results

Proof of proposition 2.1

  1. For a fixed m ∈ ℕ, let (wn) be defined by wm+1 = Γ(am), wm+2 = −am+1 wm+1 and wn = 0 if nm + 1, m + 2. Since Γ(an) = O(Γ(an+1)), it follows that


    and wm+2 also verifies (4). Since the operator

    R:H{(wn)/(wn) verifies (3) and (4)}

    defined by 𝓡(f) = (un), where u1 = f(z1) and un+1 = f(zn+1) −an f(zn), is linear and onto, a standard argument using the closed graph theorem gives the existence of fmH such that 𝓡(fm) = (wn) and ‖ fmc ‖ (wn)‖ = c. Since fm(zm+1) = wm+1 and fm(zn) = 0 if nm + 1, there is gmH such that fm = gm Bm+1 and ‖ gm = ‖ fm. Thus,


    and (S) holds.

  2. For a fixed m ∈ ℕ, let (wn) be defined by wm+1 = Π(am) and the other terms as in (i). This sequence satisfies (6), because taking into account that Π(an) ∼ Π(an+1),

    |wm+1wm+2|=|1+am+1|wm+1(1+A)Π(am),|wmwm+1|=wm+1=Π(am)cA,ZΠ(am1) if m2,|wm+2wm+3|=wm+2AΠ(am)cA,ZAΠ(am+1).

    Condition (D) is obtained proceeding exactly as in the proof of (i).

  3. Let (wn) be defined as in (i) and let (wn) be defined by wm+1=Γ(am)1|zm+1|2,wm+2=am+1Γ(am)1|zm+2|2 and wn = 0 if nm + 1, m + 2. We have that wm+2 also satisfies (8), because


    Proceeding as in (i), there exists gmH such that fm=gmBm+12. Since ρ(zm, zm+1) ≤ Γ(am) (see definition of Γ) and |Bm+1(zm+1)| < ρ(zm, zm+1), we have


    Thus, it follows that Z ∖ {z1} is u.s. and so is Z.

  4. Let (wn) be defined as in (iii) replacing Γ by Λ. Since Λ(an)=O(Λ(an+1)),thenwm+2 verifies (11). Proceeding as in (i), there is gmH such that fm = gm BBm+1. A simple calculation gives


    and then,


    Thus, (M) holds.


Proof of proposition 2.2

Let (μn) be as in (12) and (μn) as in (13).

  1. We situate ourselves in (c) of Lemma 1.2. If Z is u.s., then Carleson’s theorem provides f performing the interpolation f(zn) = μn. Otherwise, Z is the union of E = (z2m) and F = (z2m+1), both u.s. Let gH such that g(z2m) = μ2m. We look for a function hH such that h(z2m+1) = λm, where


    because then f1 = g + BEh is in H and performs the above interpolation on Z ∖ {z1}. Since |BE(z2m+1)| > |B2m+1(z2m+1)| and |B2m+1(z2m+1)| ≥ c Γ(a2m) (condition (S)), we have |BE(z2m+1)| ≥ c Γ(a2m). Then,


    Taking into account that


    and using (3) and (4),


    By (2)


    Thus, (λn) ∈ l and Carleson’s theorem provides h. Putting


    it follows that f is in H and performs the above interpolation on Z.

  2. The proof is as in (i). Now we have


    By (20) and using (3) and (6),


    and by (17), it turns out that |a2m||a2ma2m+1|1rcΠ(a2m). On the other hand, by (2)


    Thus, (λn) ∈ l and the proof continues as in (i).

  3. Since Z is u.s., then Z is double interpolating and there is f in H performing f(zn) = μn and f′(zn) = μn.

  4. In case that Z = EF, where E = (z2m) and F = (z2m+1) are u.s., we know that there is gH vanishing on E and satisfying g′(z2m) = μ2m. We look for a function h1H such that h1(z2m+1) = λm, where


    because then h2=g+BE2h1 is in H, vanishes on Z ∖ {z1} and h2(z2m)=μ2m. We have


    It follows from (13), taking into account (10) and (11), that


    For n ≥ 2, by (13), (10), (11) and (18)


    Then, using (15), (21), (22) and Λ(an)=O(Λ(an+1)),




    and Carleson’s theorem provides h1 performing the above interpolation. We replace h2 by h2c, where h2c is given by


    because then h2c also vanishes on z1. Finally, we look for h3H such that h3(z2m+1) = λm, where


    because then f = h2c + BBEh3 is in H and performs the desired interpolation on Z ∖ {z1}. Using (19),


    An easy computation shows that


    if ρ(z, w) < min(|z|,12); then, using (21), (22), (23) and Λ(an)=O(Λ(an+1)),


    and taking into account (21), (22), (24) and Λ(an)=O(Λ(an+1)),


    On the other hand, by (16) and (24)




    and Carleson’s theorem provides h3 performing the above interpolation. We replace f by fc defined by


    because then fc(z1)=μ1 and so fc performs the desired interpolation on Z.



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About the article

Received: 2017-02-28

Accepted: 2018-03-28

Published Online: 2018-04-30

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 461–468, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0044.

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© 2018 Tugores, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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