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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

# Recursive interpolating sequences

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

## Abstract

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) = $\begin{array}{}{w}_{1}^{\prime }\end{array}$ and f′(zn+1) = $\begin{array}{}{a}_{n}^{\prime }\end{array}$ [(1 − |zn|2)/(1 − |zn+1|2)] f′(zn) + $\begin{array}{}{w}_{n+1}^{\prime },\end{array}$ where ($\begin{array}{}{a}_{n}^{\prime }\end{array}$) is bounded and ($\begin{array}{}{w}_{n}^{\prime }\end{array}$) 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.

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

$ψ(z,w)=z−w1−z¯w,$

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

$B(z)=∏n|zn|znψ(zn,z).$

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

$|Bm(zm)|≥c∀m∈N.$(1)

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

$|f(z)−f(w)|≤c‖f‖∞ρ(z,w)$(2)

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

([5]).

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 ($\begin{array}{}{w}_{n}^{\prime }\end{array}$) satisfying ($\begin{array}{}{w}_{n}^{\prime }\end{array}$(1 − |zn|2)) ∈ l, there is fH such that f(zn) = wn and f′(zn) = $\begin{array}{}{w}_{n}^{\prime }\end{array}$. 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:

$Γ(tm)=ρ(zm,zm+1)+|1−tm|Π(tm)=‖T‖∞ρ(zm,zm+1)+ρ(zm+1,zm+2)+|tm−tm+1|Λ(tm)=ρ(zm,zm+1){ρ(zm,zm+1)+|1+tm|}.$

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) ≤ $\begin{array}{}{c}_{T,Z}^{\prime }\end{array}$ Π(tm) for some constants cT, Z, $\begin{array}{}{c}_{T,Z}^{\prime }\end{array}$ > 0 and for all m ∈ ℕ.

From now on A = (an) and A′ = ($\begin{array}{}{a}_{n}^{\prime }\end{array}$) 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

$|(∑i=1naiai+1…anwi)+wn+1|≤c$(3)

and

$|wn+1|≤cΓ(an),$(4)

there exists fH such that

${f(z1)=w1f(zn+1)=anf(zn)+wn+1.$(5)

Clearly if ‖A < 1, then the sum in (3) is bounded by $\begin{array}{}\frac{‖\left({w}_{n}\right){‖}_{\mathrm{\infty }}}{1-‖A{‖}_{\mathrm{\infty }}}.\end{array}$

#### Definition 1.4

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

$|wn+1−wn+2|≤cΠ(an),$(6)

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 ($\begin{array}{}{w}_{n}^{\prime }\end{array}$) verifying

$|∑i=1nai′ai+1′…an′(1−|zi|2)wi′+(1−|zn+1|2)wn+1′|≤c$(7)

and

$|wn+1′|(1−|zn+1|2)≤cΓ(an′),$(8)

there exists fH satisfying recursion (5) and

${f′(z1)=w1′f′(zn+1)=an′1−|zn|21−|zn+1|2f′(zn)+wn+1′.$(9)

#### Definition 1.6

We say that Z is zero and A′-interpolating if given ($\begin{array}{}{w}_{n}^{\prime }\end{array}$) verifying (7),

$|w1′|(1−|z1|2)≤cΛ(a1′)$(10)

and

$|wn+1′|(1−|zn+1|2)≤cΛ(an′),$(11)

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

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

${μ1=w1μn+1=(∑i=1naiai+1…anwi)+wn+1,$(12)

and recursion (9) is equivalent to f′(zn) = $\begin{array}{}{\mu }_{n}^{{}^{\prime }}\end{array}$, with

${μ1′=w1′μn+1′=∑i=1nai′ai+1′…an′(1−|zi|2)wi′1−|zn+1|2+wn+1′.$(13)

Thus, we must have (3) and (7) to state that sequences (μn) and ($\begin{array}{}{\mu }_{n}^{{}^{\prime }}\end{array}$(1 − |zn|2)) are bounded. We impose that data sequences (wn) and ($\begin{array}{}{w}_{n}^{\prime }\end{array}$) 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

$|wn+1|≤|f(zn)−f(zn+1)|+|f(zn)||1−an|$

and

$|wn+1−wn+2|≤|an||f(zn)−f(zn+1)|+|f(zn+1)−f(zn+2)|+|f(zn+1)||an−an+1|,$

respectively. On the other hand,

$|f′(z)(1−|z|2)−f′(w)(1−|w|2)|≤c‖f‖∞ρ(z,w),$(14)

$|f(z)−f(w)+f′(w)(1−|w|2)ψ(z,w)|≤c‖f‖∞ρ(z,w)2$(15)

and

$|f(z)−f(w)−[f′(z)(1−|z|2)+f′(w)(1−|w|2)]ψ(z,w)2|≤c‖f‖∞ρ(z,w)3.$(16)

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

$|wn+1′|(1−|zn+1|2)≤|f′(zn+1)(1−|zn+1|2)−f′(zn)(1−|zn|2)|+|f′(zn)|(1−|zn|2)|1−an′|.$

If on the right of this last inequality, we put

$|f′(zn+1)(1−|zn+1|2)+f′(zn)(1−|zn|2)|+|f′(zn)|(1−|zn|2)|1+an′|,$

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

$|Bm+1(zm+1)|≥cΓ(am)∀m∈N,$

and condition (D) if

$|Bm+1(zm+1)|≥cΠ(am)∀m∈N.$

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

$|Bm+1(zm+1)|2≥cΛ(am′)∀m∈N.$

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 $\begin{array}{}\mathit{\Gamma }\left({a}_{n}^{\prime }\right)=O\left(\mathit{\Gamma }\left({a}_{n+1}^{\prime }\right)\right),\end{array}$ then Z is u.s.

4. If Z is zero and A′-interpolating and $\begin{array}{}\mathit{\Lambda }\left({a}_{n}^{\prime }\right)=O\left(\mathit{\Lambda }\left({a}_{n+1}^{\prime }\right)\right),\end{array}$ 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

$|an+1|≤r|an|$(17)

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), $\begin{array}{}\mathit{\Lambda }\left({a}_{n}^{\prime }\right)=O\left(\mathit{\Lambda }\left({a}_{n+1}^{\prime }\right)\right)\end{array}$ and

$|a1′a2′⋯an′|Λ(a1′)+∑i=2n|ai′ai+1′⋯an′|Λ(ai−1′)≤cΛ(an−1′)∀n≥2,$(18)

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

$|wm+2|≤‖A‖∞Γ(am)≤cA,Z‖A‖∞Γ(am+1)$

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,

$Γ(am)=|fm(zm+1)|≤c|Bm+1(zm+1)|$

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),

$|wm+1−wm+2|=|1+am+1|wm+1≤(1+‖A‖∞)Π(am),|wm−wm+1|=wm+1=Π(am)≤cA,Z′Π(am−1) if m≥2,|wm+2−wm+3|=wm+2≤‖A‖∞Π(am)≤cA,Z‖A‖∞Π(am+1).$

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

3. Let (wn) be defined as in (i) and let ($\begin{array}{}{w}_{n}^{\prime }\end{array}$) be defined by $\begin{array}{}{w}_{m+1}^{\prime }=\frac{\mathit{\Gamma }\left({a}_{m}^{\prime }\right)}{1-|{z}_{m+1}{|}^{2}},{w}_{m+2}^{\prime }=-{a}_{m+1}^{\prime }\phantom{\rule{thinmathspace}{0ex}}\frac{\mathit{\Gamma }\left({a}_{m}^{\prime }\right)}{1-|{z}_{m+2}{|}^{2}}\end{array}$ and $\begin{array}{}{w}_{n}^{\prime }\end{array}$ = 0 if nm + 1, m + 2. We have that $\begin{array}{}{w}_{m+2}^{\prime }\end{array}$ also satisfies (8), because

$|wm+2′|(1−|zm+2|2)≤‖A′‖∞Γ(am′)≤cA′,Z‖A′‖∞Γ(am+1′).$

Proceeding as in (i), there exists gmH such that $\begin{array}{}{f}_{m}={g}_{m}{B}_{m+1}^{2}.\end{array}$ Since ρ(zm, zm+1) ≤ Γ(am) (see definition of Γ) and |Bm+1(zm+1)| < ρ(zm, zm+1), we have

$ρ(zm,zm+1)≤Γ(am)=|fm(zm+1)|≤c|Bm+1(zm+1)|2

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

4. Let ($\begin{array}{}{w}_{n}^{\prime }\end{array}$) be defined as in (iii) replacing Γ by Λ. Since $\begin{array}{}\mathit{\Lambda }\left({a}_{n}^{\prime }\right)=O\left(\mathit{\Lambda }\left({a}_{n+1}^{\prime }\right)\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{then}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{w}_{m+2}^{\prime }\end{array}$ verifies (11). Proceeding as in (i), there is gmH such that fm = gm BBm+1. A simple calculation gives

$|B′(zm+1)|=|Bm+1(zm+1)|1−|zm+1|2$(19)

and then,

$Λ(am′)1−|zm+1|2=|fm′(zm+1)|=|(gmB′Bm+1)(zm+1)|≤c|Bm+1(zm+1)|21−|zm+1|2.$

Thus, (M) holds.

□

#### Proof of proposition 2.2

Let (μn) be as in (12) and ($\begin{array}{}{\mu }_{n}^{{}^{\prime }}\end{array}$) 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

$λm=μ2m+1−g(z2m+1)BE(z2m+1),$

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,

$|λm|≤c|μ2m+1−μ2m|+|g(z2m)−g(z2m+1)|Γ(a2m).$

Taking into account that

$μ2m+1=a2mμ2m+w2m+1$(20)

and using (3) and (4),

$|μ2m+1−μ2m|≤|1−a2m||μ2m|+|w2m+1|≤cΓ(a2m).$

By (2)

$|g(z2m)−g(z2m+1)|≤cρ(z2m,z2m+1)≤cΓ(a2m).$

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

$f(z)=f1(z)+(μ1−f1(z1))B1(z)B1(z1),$

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

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

$|λm|≤c|μ2m+1−μ2m+2|+|g(z2m+2)−g(z2m+1)|Π(a2m).$

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

$|μ2m+1−μ2m+2|≤|w2m+1−w2m+2|+|μ2m+1||a2m−a2m+1|+|a2m|(|μ2m+1|+|μ2m|)≤c{Π(a2m)+|a2m|}$

and by (17), it turns out that $\begin{array}{}|{a}_{2m}|\le \frac{|{a}_{2m}-{a}_{2m+1}|}{1-r}\le c\phantom{\rule{thinmathspace}{0ex}}\mathit{\Pi }\left({a}_{2m}\right).\end{array}$ On the other hand, by (2)

$|g(z2m+2)−g(z2m+1)|≤cρ(z2m+1,z2m+2)≤cΠ(a2m).$

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) = $\begin{array}{}{\mu }_{n}^{{}^{\prime }}\end{array}$.

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) = $\begin{array}{}{\mu }_{2m}^{\prime }.\end{array}$ We look for a function h1H such that h1(z2m+1) = λm, where

$λm=−g(z2m+1)BE2(z2m+1),$

because then $\begin{array}{}{h}_{2}=g+{B}_{E}^{2}{h}_{1}\end{array}$ is in H, vanishes on Z ∖ {z1} and $\begin{array}{}{h}_{2}^{\prime }\left({z}_{2m}\right)={\mu }_{2m}^{\prime }.\end{array}$ We have

$|λm|≤c|g(z2m+1)|Λ(a2m′).$

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

$(1−|z2|2)|μ2′|≤|a1′|(1−|z1|2)|w1′|+(1−|z2|2)|w2′|≤cΛ(a1′).$(21)

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

$(1−|zn+1|2)|μn+1′|≤c{|a1′a2′…an′|Λ(a1′)+∑i=2n|ai′ai+1′…an′|Λ(ai−1′)}+(1−|zn+1|2)|wn+1′|≤c{Λ(an−1′)+Λ(an′)}≤cΛ(an′).$(22)

Then, using (15), (21), (22) and $\begin{array}{}\mathit{\Lambda }\left({a}_{n}^{\prime }\right)=O\left(\mathit{\Lambda }\left({a}_{n+1}^{\prime }\right)\right),\end{array}$

$|g(z2m+1)|=|g(z2m)−g(z2m+1)−g′(z2m)(1−|z2m|2)ψ(z2m,z2m+1)|+(1−|z2m|2)|μ2m′|ρ(z2m,z2m+1)≤c{ρ(z2m,z2m+1)2+Λ(a2m′)ρ(z2m,z2m+1)}.$

Thus,

$|λm|≤c(ρ(z2m,z2m+1)2Λ(a2m′)+ρ(z2m,z2m+1))≤c$

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

$h2c(z)=h2(z)−h2(z1)B12(z)B12(z1),$

because then h2c also vanishes on z1. Finally, we look for h3H such that h3(z2m+1) = $\begin{array}{}{\lambda }_{m}^{\prime },\end{array}$ where

$λm′=μ2m+1′−h2c′(z2m+1)(B′BE)(z2m+1),$

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

$|λm′|≤c{|μ2m+1′+μ2m′|+|h2c′(z2m)+h2c′(z2m+1)|}(1−|z2m+1|2)Λ(a2m′).$

An easy computation shows that

$|z−w|1−|z|2<4ρ(z,w),$(23)

if ρ(z, w) < min$\begin{array}{}\left(|z|,\frac{1}{2}\right);\end{array}$ then, using (21), (22), (23) and $\begin{array}{}\mathit{\Lambda }\left({a}_{n}^{\prime }\right)=O\left(\mathit{\Lambda }\left({a}_{n+1}^{\prime }\right)\right),\end{array}$

$|μ2m′|||z2m|2−|z2m+1|2|≤cΛ(a2m−1′)|z2m−z2m+1|1−|z2m|2≤cΛ(a2m′)ρ(z2m,z2m+1)$(24)

and taking into account (21), (22), (24) and $\begin{array}{}\mathit{\Lambda }\left({a}_{n}^{\prime }\right)=O\left(\mathit{\Lambda }\left({a}_{n+1}^{\prime }\right)\right),\end{array}$

$|μ2m+1′+μ2m′|(1−|z2m+1|2)≤|μ2m+1′|(1−|z2m+1|2)+|μ2m′|(1−|z2m|2)+|μ2m′|||z2m|2−|z2m+1|2|≤c{Λ(a2m′)+Λ(a2m′)ρ(z2m,z2m+1)}.$

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

$|h2c′(z2m)+h2c′(z2m+1)|(1−|z2m+1|2)≤|h2c′(z2m)(1−|z2m|2)+h2c′(z2m+1)(1−|z2m+1|2)|+|μ2m′|||z2m|2−|z2m+1|2|≤c{ρ(z2m,z2m+1)2+Λ(a2m′)ρ(z2m,z2m+1)}.$

Then,

$|λm′|≤c(1+ρ(z2m,z2m+1)2Λ(a2m′)+ρ(z2m,z2m+1))≤c$

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

$fc(z)=f(z)−(f′(z1)−μ1′)(BB1)(z)(B′B1)(z1),$

because then $\begin{array}{}{f}_{c}^{\prime }\left({z}_{1}\right)={\mu }_{1}^{\prime }\end{array}$ and so fc performs the desired interpolation on Z.

□

## References

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Hayman, W.K., Interpolation by bounded functions, Ann. Inst. Fourier, 1958, 8, 277-290

• [2]

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## About the article

Accepted: 2018-03-28

Published Online: 2018-04-30

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 461–468, ISSN (Online) 2391-5455,

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