Some Applications of Higher-Order Hyperbolic Derivatives

Complex Anal. Oper. TheoryDOI 10.1007/s11785-011-0172-z

Complex Analysisand Operator Theory

Some Applications of Higher-Order HyperbolicDerivatives

Patrice Rivard

Received: 17 November 2010 / Accepted: 11 July 2011© Springer Basel AG 2011

Abstract The notion of higher-order hyperbolic derivatives was recently intro-duced. As the given definition does not lend to simple calculations, we first give arecursive formula to calculate them, which may be useful for a possible numerical use.Next, as a first application of hyperbolic derivatives of higher order, we develop sometools for solving a particular interpolation problem involving them and which allowus to consider the constrained Nevanlinna–Pick problem. Finally, we generalize andinterpret some classical results of complex analysis in this new setting: Dieudonné’sLemma and Rogosinski’s Lemma.

Keywords Higher-order hyperbolic derivatives · Dieudonné’s lemma ·Rogosinski’s lemma · Interpolation problem

Mathematics Subject Classification (2010) Primary 30F45; Secondary 30C80

1 Introduction

Our first aim is to present the definitions and notations necessary for a convenientunderstanding of the present work. We will refer the reader to [2] and [11] for moredetails. Throughout the text, we will denote by D the unit disk and by T the unit circle.We also use the notation

Communicated by H. Turgay Kaptanoglu.

The author was supported by the NSERC Postgraduate Scholarships program.

P. Rivard (B)Département de mathématiques et de statistique, Université Laval, Québec, Canadae-mail: [email protected]

P. Rivard

D := D ∪ T, D := D ∪ {∞}.

In this paper we shall be interested primarily in the class H of analytic functionsof the unit disk D into itself,

H := { f : f holomorphic in D, | f (z)| < 1, z ∈ D},

and the Schur class S, defined by

S := { f : f holomorphic in D, | f (z)| ≤ 1, z ∈ D}.

A Blaschke product of degree n ≥ 1 is a function of the form

B(z) := eiθn

∏

j=1

z − z j

1 − z j z, |z j | < 1, θ ∈ R.

The unimodular constants are considered to be Blaschke products of degree 0. Forn ≥ 0, we will denote the class of Blaschke products of degree n by Bn .

Let z1, z2 be points of D. The hyperbolic distance ρ between the points z1 and z2is given by

ρ(z1, z2) = log

(

1 + p(z1, z2)

1 − p(z1, z2)

)

, where p(z1, z2) :=∣

∣

∣

∣

z1 − z2

1 − z2z1

∣

∣

∣

∣

.

The quantity p(z1, z2) is called the pseudo-hyperbolic distance of the points z1, z2.In the case where at least one of the two points belongs to T, we set

ρ(z1, z2) :=

⎧

⎪

⎪

⎨

⎪

⎪

⎩

∞ if z1 ∈ D, z2 ∈ T;∞ if z2 ∈ D, z1 ∈ T;∞ if z1, z2 ∈ T and z1 �= z2;0 if z1, z2 ∈ T and z1 = z2.

(1.1)

Definition 1.1 Let z1, z2 ∈ D. We define

[z1, z2] :=

⎧

⎪

⎨

⎪

⎩

z2 − z1

1 − z2z1if z1, z2 ∈ D, z2z1 �= 1;

z2 if z1 = z2 ∈ T;∞ otherwise.

By the Schwarz–Pick lemma, and the formula for the hyperbolic metric, one seesthat for f ∈ H and z1 ∈ D,

|[ f (z), f (z1)]| ≤ |[z, z1]|

Some Applications of Higher-Order Hyperbolic Derivatives

and, therefore

z �→ [ f (z), f (z1)][z, z1] (1.2)

is in S. In particular, the function given in (1.2) belongs to H, if f ∈ H\B1, with aremovable singularity at the point z1, and belongs to B0, if f ∈ B1. This motivatesthe following.

Definition 1.2 Let f ∈ S and fix k pairwise distinct points z1, . . . , zk ∈ D. Set

�0 f (z) := f (z), z ∈ D.

We define recursively for j = 1, . . . , k the hyperbolic divided difference of order jof the function f with parameters z1, . . . , z j , denoted� j f (z; z1, . . . , z j ), as follows

� j f (z; z1, . . . , z j ) := [� j−1 f (z; z1, . . . , z j−1),�j−1 f (z j ; z1, . . . , z j−1)]

[z, z j ] ,

(1.3)

provided � j−1 f (z; z1, . . . , z j−1) �∈ B0. The quotient in the right-hand side of (1.3)has to be interpreted as a limit if z = z j . If � j−1 f (z; z1, . . . , z j−1) ∈ B0, we define

� j f (z; z1, . . . , z j ) := � j−1 f (z j ; z1, . . . , z j−1).

We also introduce the operators � jz1,...,z j , which act on f in S as follows

�jz1,...,z j f (z) := � j f (z; z1, . . . , z j ), z ∈ D, j = 1, . . . , k.

The above recursive definition of � j f (z; z1, . . . , z j ) can be rewritten as

� j f (z; z1, . . . , z j ) = �1z j

(

�j−1z1,...,z j−1 f

)

(z), j = 1, . . . , k.

It was shown in [2] that � jz1,...,z j f is well-defined and belongs to S for all f ∈ S.

The hyperbolic divided differences of a function f ∈ S, where we suppose thatthey are defined by (1.3), can be conveniently evaluated within a table. We give anexample in the case of order two in Table 1.

We observe that, for a fixed column in Table 1, the computation of each entryinvolves the lower left element in the previous column and a “leading entry”, which isthe first element of that column. Furthermore, the method to evaluate the hyperbolicdivided differences is clearly similar to the one of the classical divided differencesused in the Newton’s interpolatory divided-difference formula.

Hyperbolic divided differences are natural tools for studying maps of the class S.Indeed, we showed in [2] that they can be used to characterize finite Blaschke prod-ucts: a function of the Schur class is a Blaschke product of degree n if and only if the

P. Rivard

Table 1 Table of hyperbolic divided differences of order two of a function f

zi f (zi ) Order one Order two

z1 f (z1)

�1z1

f (z2) = [ f (z2), f (z1)][z2, z1]

z2 f (z2) �2z1,z2

f (z3) = [�1z1

f (z3),�1z1

f (z2)][z3, z1]

�1z1

f (z3) = [ f (z3), f (z1)][z3, z1]

z3 f (z3) �2z1,z2

f (z)[�1

z1f (z),�1

z1f (z2)]

[z, z2]�1

z1f (z) = [ f (z), f (z1)]

[z, z1]z f (z)

hyperbolic divided differences of order k is a Blaschke product of degree n − k, ifn > k and of degree zero if n ≤ k [2, Theorem 3.1]. As a consequence, they play thesame rôle on Blaschke products as the derivatives on polynomials. We also showedhow one can read out from a table of hyperbolic divided differences whether a givenNevanlinna–Pick problem is solvable, and construct solutions, in perfect analogy withNewton interpolation for polynomials.

We now state another definition fundamental for the theory presented here.

Definition 1.3 Let f ∈ S, z ∈ D and n ≥ 1. For ζ ∈ D, we define the hyperbolicdivided difference of order n of f with parameter z, denoted �n

z f (ζ ), by

�nz f (ζ ) ≡ �n f (ζ ; z, . . . , z) := lim

zn→zlim

zn−1→z. . . lim

z1→z�n f (ζ ; z1, . . . , zn) (ζ �= z)

and

�nz f (z) = �n f (z; z, . . . , z) := lim

ζ→z�n f (ζ ; z, . . . , z).

Remark 1.4 Formally, we have

�nz ( f )(ζ ) = lim

zn→zlim

zn−1→z. . . lim

z1→z�1

zn(�n−1

z1,...,zn−1f )(ζ ). (1.4)

For brevity, we simply write

�nz f (ζ ) = �1

z (�n−1z f )(ζ ),

rather than (1.4).

We remark that the previous quantity is also well-defined and �nz f ∈ S, for f ∈

S, n ≥ 1 and z ∈ D [11]. Moreover, the hyperbolic divided differences with oneparameter have the same effect on Blaschke product as the hyperbolic divided differ-ences with many parameters, as we have mentioned earlier [11, Theorem 2.4].


Table 2 Table of hyperbolic divided differences of a function f with one parameter

Order one Order two Order three

z f (z)

H1 f (z)

z f (z) H2 f (z)

H1 f (z) �3z f (ζ ) = [�2

z f (ζ ), H2 f (z)][ζ, z]

z f (z) �2z f (ζ ) = [�1

z f (ζ ), H1 f (z)][ζ, z]

�1z f (ζ )

ζ f (ζ )

2 A Recursive Formula for Hyperbolic Derivatives

The notion of hyperbolic derivative of a function f ∈ H\B1 at a point z ∈ D, denotedf h(z), is already known [4,11] and is given by

f h(z) = f ′(z)(1 − |z|2)1 − | f (z)|2 . (2.1)

Using the definition of hyperbolic divided differences with one parameter we canintroduce the notion of hyperbolic derivatives of higher order.

Definition 2.1 Let f ∈ S and n ≥ 1. The hyperbolic derivative of order n of f atthe point z ∈ D, denoted Hn f (z), is defined by

Hn f (z) := �nz f (z).

For f ∈ H\B1 and z ∈ D, clearly we have

H1 f (z) = f h(z).

By Remark 1.4 and Definition 2.1, the hyperbolic divided differences of a functionζ �→ f (ζ ) belonging to S with parameter z ∈ D can be tabulated as in Table 2, in thecase where all the quantities involved are different from a unimodular constant.

For f ∈ S, we have the following inequality

|Hn f (z)| ≤ 1, z ∈ D. (2.2)

Thus, the higher-order hyperbolic derivatives of a function in the Schur class arebounded by one. This upper bound is obtained essentially from the fact that for thesefunctions, the hyperbolic divided difference with one parameter also belongs to S. Thebound (2.2) is sharp and is attained by Blaschke products of degree less than or equalto n. The case of classical derivatives is more complicated. Indeed, it is of interest

P. Rivard

to find a bound for classical derivatives of higher order of function belonging to theSchur class [12,1]. We have the following estimate.

Theorem 2.2 [12, Theorem 2] Let f ∈ S. Then, for all z ∈ D

| f (n)(z)| ≤ n!(1 − | f (z)|2)(1 − |z|2)n (1 + |z|)n−1. (2.3)

Moreover, for n > 1 and z �= 0, equality holds in (2.3) only for f ≡ λ, |λ| = 1.

For n = 1, the bounds (2.2) and (2.3) are equivalent and equality holds only forBlaschke products of degree less than or equal to one. This result is known in theliterature as Pick’s Lemma. An interesting question arises for n > 1 : can we obtain(2.3) using (2.2)?

We also mention that all the tools of hyperbolic geometry and what the authorshave developped in [2] can be used for interpolation problem with derivatives, sincethe hyperbolic derivatives of higher order are bounded by one. This will be done in thenext section. Before that, we now develop an algorithm for computing the hyperbolicderivatives.

Let f ∈ H. For z ∈ D, we define Dn f (z) by the Taylor series expansion of thefunction

ζ �→ g(ζ ) := −[ f ([−ζ, z]), f (z)] =f

(

z + ζ

1 + zζ

)

− f (z)

1 − f (z) f

(

z + ζ

1 + zζ

) =∞∑

n=1

Dn f (z)

n! ζ n,

(2.4)

where

D1 f (z) := g′(0), D2 f (z) := g′′(0), . . . .

For example,

D1 f (z) = (1 − |z|2) f ′(z)1 − | f (z)|2 ;

D2 f (z) = f ′′(z)(1 − |z|2)21 − | f (z)|2 − 2z f ′(z)(1 − |z|2)

1 − | f (z)|2 + 2 f (z) f ′(z)2(1 − |z|2)2(1 − | f (z)|2)2 .

These derivatives Dn f were first introduced by Peschl [10] and are usually referredto as differential invariants and this terminology is explained by the fact that giventwo automorphisms of the unit disk, φ and ψ , we have

|Dn(ψ ◦ f ◦ φ)(z)| = |Dn( f )(φ(z))|, z ∈ D.


It is possible to derive expressions for Dn f in terms of f (n), the lower-order deriv-atives of f and the derivatives D1 f, . . . , Dn−1 f . These relations appeared in [8,Corollary 7.5].

The following definition will be useful for establishing an iterative formula for thehyperbolic derivatives.

Definition 2.3 Let k > 1, z ∈ D and f ∈ S\ ∪k−1j=0 B j . Set

1(ζ, z) := −g(−ζ ) and 1(ζ, z) := ζ,

where g is the function given in (2.4). Define k(ζ, z) and k(ζ, z) recursively by

k(ζ, z) := Hk−1 f (z)k−1(ζ, z)−k−1(ζ, z);k(ζ, z) := ζ

(

k−1(ζ, z)− Hk−1 f (z)k−1(ζ, z))

.

Remark 2.4 By construction, it is clear thatk(ζ, z),k(ζ, z) are holomorphic func-tions of the variable ζ ∈ D, for each z ∈ D.

Remark 2.5 We impose the condition f ∈ S\ ∪k−1j=0 B j to avoid the case where the

hyperbolic derivative of the function is a unimodular constant.

Lemma 2.6 Let k ≥ 1 and f ∈ S\ ∪k−1j=0 B j . Let z, ζ ∈ D. Then

�kz f ([ζ, z]) := �k f ([ζ, z]; z, . . . , z) = k(ζ, z)

k(ζ, z). (2.5)

Proof We will prove (2.5) by induction on k. For k = 1, we have

�1z f ([ζ, z]) = [ f ([ζ, z]), f (z)]

[[ζ, z], z] = −g(−ζ )ζ

= 1(ζ, z)

1(ζ, z).

Suppose now that the result is true for k. For k + 1, it follows that

�k+1 f ([ζ, z]; z, . . . , z) = [�k f ([ζ, z]; z, . . . , z),�k f (z; z, . . . , z)][[ζ, z], z]

= [�k f ([ζ, z]; z, . . . , z), Hk f (z)]ζ

= Hk f (z)k(ζ, z)−k(ζ, z)

ζ(

k(ζ, z)− Hk f (z)k(ζ, z))

= k+1(ζ, z)

k+1(ζ, z),

and this proves the result. �

P. Rivard

Lemma 2.7 Let k ≥ 1, f ∈ S\ ∪k−1j=0 B j and z, ζ ∈ D. Then

Hk f (z) = limζ→0

k(ζ, z)

k(ζ, z).

Proof By definition Hk f (z) = limζ→z �k f (ζ ; z, . . . , z). This is equivalent to

Hk f (z) = limζ→0�k f ([ζ, z]; z, . . . , z). Now apply Lemma 2.6. �

The next result presents the recursive formula which can be useful for evaluatingthe hyperbolic derivatives of higher order.

Theorem 2.8 Let n > 1, f ∈ S\ ∪n−1j=0 B j and z ∈ D. Then

Hn f (z) = Hn−1 f (z)(n)n−1(0, z)−(n)n−1(0, z)

n(

(n−1)n−1 (0, z)− Hn−1 f (z)(n−1)

n−1 (0, z)) , (2.6)

where (n)n−1 and (n)n−1 denote the n-th derivative with respect to ζ .

Proof By Lemma 2.7, we have

Hn f (z) = limζ→0

n(ζ, z)

n(ζ, z). (2.7)

To evaluate the limit in (2.7), we will apply L’Hôpital’s rule n times. But, to do this,we have to prove

limζ→0

( j)n (ζ, z) = 0, j = 0, . . . , n − 1; (2.8)

limζ→0

( j)n (ζ, z) = 0, j = 0, . . . , n − 1; (2.9)

and

limζ→0

(n)n (ζ, z) �= 0. (2.10)

We will use induction on n to prove (2.6), (2.8), (2.9) and (2.10).For n = 2, it follows from (2.4) and (2.1) that

limζ→0

2(ζ, z) = limζ→0

(1)2 (ζ, z) = 0;

limζ→0

2(ζ, z) = limζ→0

(1)2 (ζ, z) = 0;

limζ→0

(2)2 (ζ, z) = 2 − 2|H1 f (z)|2 �= 0,

because f �∈ B0 ∪ B1.


Suppose that (2.6), (2.8), (2.9) and (2.10) are true for n. We will show that this is alsothe case for n + 1. We will first prove that limζ→0

( j)n+1(ζ, z) = 0 for j = 0, . . . , n.

For j = 0, . . . , n − 1, we have, by the definition and the induction hypothesis

limζ→0

( j)n+1(ζ, z) = lim

ζ→0

[

Hn f (z)( j)n (ζ, z)−

( j)n (ζ, z)

]

= 0.

Let us show now that limζ→0(n)n+1(ζ, z) = 0. We have

limζ→0

(n)n+1(ζ, z) = lim

ζ→0

[

Hn f (z)(n)n (ζ, z)−(n)n (ζ, z)]

− limζ→0

[

Hn f (z)(

ζn−1(ζ, z)− ζHn−1 f (z)n−1(ζ, z))(n)

−(

Hn−1 f (z)n−1(ζ, z)−n−1(ζ, z))(n)

]

= Hn f (z)[

n(n−1)n−1 (0, z)− nHn−1 f (z)(n−1)

n−1 (0, z)]

−Hn−1 f (z)(n)n−1(0, z)

+(n)n−1(0, z)

= 0,

where the last equality is a consequence of the induction hypothesis and (2.6).We now show that limζ→0

( j)n+1(ζ, z) = 0, for j = 0, . . . , n. By definition, we get

limζ→0

( j)n+1 = lim

ζ→0

[

(ζn(ζ, z))( j) − Hn f (z)(ζn(ζ, z))( j)]

.

If j = 0, then it is clear that limζ→0n+1(ζ, z) = 0. For j ≥ 1, we have

limζ→0

( j)n+1 = lim

ζ→0

[

ζ( j)n (ζ, z)+ j( j−1)

n (ζ, z)

− Hn f (z)(

ζ( j)n (ζ, z)+ j( j−1)

n (ζ, z))]

= 0,

by the induction hypothesis. It remains to show that limζ→0(n+1)n+1 (ζ, z) �= 0.

limζ→0

(n+1)n+1 (ζ, z) = lim

ζ→0

[

(ζn(ζ, z))(n+1) − Hn f (z)(ζn(ζ, z))(n+1)]

= n(n)n (0, z)− nHn f (z)(n)n (0, z).

Suppose that limζ→0(n+1)n+1 (ζ, z) = 0. This means that

(n)n (0, z) = Hn f (z)(n)n (0, z).

P. Rivard

By the induction hypothesis, (n)n (0, z) �= 0. Therefore

Hn f (z)(n)n (0, z)

(n)n (0, z)

= 1. (2.11)

Also, by Lemma 2.7, we have


n(ζ, z)

n(ζ, z).

Using the induction hypothesis and L’Hôpital’s rule, applied n times, we obtain


(n)n (ζ, z)

(n)n (ζ, z)

= (n)n (0, z)

(n)n (0, z)

. (2.12)

It follows from (2.11) and (2.12) that |Hn f (z)| = 1, which is in contradiction withthe hypothesis, that f �∈ ∪n

j=0B j . Therefore, limζ→0(n+1)n+1 (ζ, z) �= 0.

Finally, we have to prove

Hn+1 f (z) = Hn f (z)(n+1)n (0, z)−

(n+1)n (0, z)

(n + 1)((n)n (0, z)− Hn f (z)(n)n (0, z)). (2.13)

By Lemma 2.7, we have

Hn+1 f (z) = limζ→0

n+1(ζ, z)

n+1(ζ, z).

By what precedes, we can apply L’Hôpital’s rule (n + 1) times, and we get

Hn+1 f (z) = limζ→0

(n+1)n+1 (ζ, z)

(n+1)n+1 (ζ, z)

= limζ→0

Hn f (z)(n+1)n (ζ, z)−

(n+1)n (ζ, z)

(ζn(ζ, z))(n+1) − Hn f (z)(ζn(ζ, z))(n+1)

= limζ→0

Hn f (z)(n+1)n (ζ, z)−(n+1)

n (ζ, z)(

ζ(n+1)n (ζ, z)+(n+1)(n)n (ζ, z)

)

−Hn f (z)(

(n+1)ζ(n+1)n (ζ, z)+(n+1)ζ(n)n (ζ, z)

) .

Evaluating the previous limit gives (2.13). �

Using the formulae given in (2.6), it is easy to exhibit an explicit form of hyperbolicderivatives. The next corollary presents the first cases.


Corollary 2.9 Let z ∈ D.

1. If f ∈ H, then

H1 f (z) = D1 f (z);

2. If f ∈ H\B1, then

H2 f (z) = D2 f (z)

2(1 − |H1 f (z)|2) ;

3. If f ∈ H\(B1 ∪ B2), then

H3 f (z) = −3H1 f (z)H2 f (z)D2 f (z)+ D3 f (z)

6 − 6|H1 f (z)|2 + 3H2 f (z)D2 f (z);

4. If f ∈ H\(B1 ∪ B2 ∪ B3), then

H4 f (z) =12H1 f (z)H3 f (z)D2 f (z)+ 4H2 f (z)H3 f (z)D3 f (z)+ 4H1 f (z)H2 f (z)D3 f (z)+ D4 f (z)

24 − 24|H1 f (z)|2 − 12H2 f (z)D2 f (z)−12H1 f (z) H3 f (z)H2 f (z)D2 f (z)−4H3 f (z)D3 f (z).

Proof We have already proved (1) and (2) in [11]. It remains to show (3) and (4).This essentially follows from Theorem 2.8, by straightforward, though tedious, cal-culations. �

3 Interpolation with Hyperbolic Derivatives

In this section, we shall present the essential tools which lead to a generalization ofDieudonné’s Lemma [6,7,9] and which can be used to consider in a new mannerRogosinski’s Lemma [7,9]. To do that, we need to look at a particular interpolationproblem. Indeed, with the results obtained in [2] and [11], we can determine a solu-tion to an interpolation problem involving some points together with the hyperbolicderivatives up to a certain order at one of the points. But before, we will recall thefollowing definition.

Given pairwise distinct points z1, . . . , zn ∈ D and w1, . . . , wn ∈ D, we define thehyperbolic divided differences

�kj ≡ �k

j (w1, . . . , wn; z1, . . . , zn), k = 0, . . . , n − 1, j = k + 1, . . . , n,

by the following recursive procedure. Let �0j := w j for j = 1, . . . , n. Assume that,

for some k with 1 ≤ k ≤ n − 1 and j = k + 1, . . . , n, the divided differences �k−1j

and�k−1k are given. Then, with the convention that |∞| := ∞ is greater than any real

number,

P. Rivard

Table 3 Table of hyperbolic divided differences

Points 1 2 3 … n − 2 n − 1

z1 w1 = �01

�12

z2 w2 = �02 �2

3

�13 �3

4

z3 w3 = �03 �2

4

. . .

�14 �3

5 �n−2n−1

z4 w4 = �04 �2

5

.

.

. �n−1n

�15

.

.

. �n−2n

z5 w5 = �05

.

.

. . ..

.

.

....

.

.

. �3n

zn−1 wn−1 = �0n−1 �2

n

�1n

zn wn = �0n

�kj :=

[

�k−1j ,�k−1

k

]

[

z j , zk] if

∣

∣

∣[�k−1j ,�k−1

k ] |≤| [z j , zk]∣

∣

∣ , (3.1)

�kj := �k−1

k if �k−1j = �k−1

k ∈ T, (3.2)

�kj := ∞ in all other cases. (3.3)

The hyperbolic divided differences can be conveniently arranged in a triangulartable as in Table 3.

In [2], it was proved that for a function f ∈ S such that f (z j ) = w j for j =1, . . . , n, then Definition 1.2 corresponds to the definition above of hyperbolic divideddifferences. This link is crucial when we consider interpolation problem of the unitdisk into itself, such as the Nevanlinna–Pick problem treated in [2], which consistsin finding functions f analytic in D with | f (z)| ≤ 1 satisfying the interpolationconditions

f (z j ) = w j , j = 1, . . . , n, (3.4)

where z1, . . . , zn ∈ D are distinct points and w1, . . . , wn ∈ D.A natural generalization of the interpolation problem (3.4) is to require that

the interpolating function matches with given hyperbolic derivatives at the points.Thus, we will be interested by the following problem: for k,m, two integers suchthat k ≥ 1,m ≥ 1, given pairwise distinct points z, u1 . . . , um ∈ D and points


w, v1, . . . , vm, μ1, . . . , μk ∈ D, determine all functions f ∈ S which satisfy

f (ui ) = vi , f (z) = w, H j f (z) = μ j , i = 1, . . . ,m and j = 1, . . . , k. (3.5)

One can see that the problem (3.5) is similar to the Hermite interpolation problemfor interpolating polynomials. In that case, the construction of the solution is basedessentially on divided differences, adapted to involve derivatives, and the Newtonmethod for interpolation.

In the following, we will see that the method used to solve the problem (3.5) isgreatly related to the one for Hermite interpolation problem. For our purpose, wewill now give a new interpretation of the hyperbolic divided differences for datapoints in order to include hyperbolic derivatives. For k ≥ 1 and m ≥ 1, we con-sider the specific case where n = k + 1 + m and z1 := z2 := · · · := zk+1 :=z, zk+2 := u1, zk+3 := u2, . . . , zn := um, w1 := w2 := · · · := wk+1 := w,wk+2 :=v1, wk+3 := v2, . . . , wn := vm . Given the numbers μ1, μ2, . . . , μk ∈ D, we define

�ji := μ j , 1 ≤ j ≤ k, j + 1 ≤ i ≤ k + 1. (3.6)

For other hyperbolic divided differences, the definition is still the same, and is givenby (3.1), (3.2) or (3.3). Therefore, we obtain Table 4.

The next lemma characterizes the table of hyperbolic divided differences in thisparticular situation.

Lemma 3.1 Let k ≥ 1 and m ≥ 1 be integers. Let z, u1, . . . , um ∈ D be distinctpoints and w, v1, . . . , vm, μ1, μ2, . . . , μk ∈ D.

1. If there exists an integer l ≥ 1 such that �li = ∞ for an integer i, k + 2 ≤ i ≤

k + 1 + m, then for all l + 1 ≤ j ≤ k + m, there exists an integer i such thatk + 2 ≤ i ≤ k + 1 + m and � j

i = ∞.2. If there exists an integer l ≥ 1 such that �l

i = λ for an integer i, k + 2 ≤ i ≤k + 1 + m, with |λ| = 1, then either � j

i = λ for all l + 1 ≤ j ≤ k + m and allk + 2 ≤ i ≤ k + 1 + m, or for all l + 1 ≤ j ≤ k + m, there exists an integer isuch that k + 2 ≤ i ≤ k + 1 + m and � j

i = ∞.

Proof 1. If the l-th column of the table contains an infinite entry, then by construc-tion of the hyperbolic divided differences � j

i , with l + 1 ≤ j ≤ k + m andk + 2 ≤ i ≤ k + 1 + m, it follows that all the subsequent columns contain also aninfinite entry in them.

2. Let�li = λ, with |λ| = 1 for some l and i , where 1 ≤ l ≤ k + m and k + 2 ≤ i ≤

k + 1 + m. Suppose that μl �= λ. Then either |[�li , μl ]| = 1 or [�l

i , μl ] = ∞.This implies that neither the condition in (3.1) nor in (3.2) is satisfied. Hence�l+1

i = ∞. By applying (1) the result follows. Suppose now that μl = λ. The

same argument shows that either � ji = λ for all l + 1 ≤ j ≤ k + m and all

k + 2 ≤ i ≤ k + 1 + m, or there exists such an i with �l+1i = ∞. �

P. Rivard

Table 4 Table of hyperbolic divided differences in the case of hyperbolic derivatives

Points 1 2 … k k + 1 . . . k + m − 1 k + m

z w = �01

μ1

z w = �02 μ2

μ1

z w = �03 μ2

. . .

μ1 μk

z w = �04 μ2

.

.

. �k+1k+2

μ1

.

.

.. . . �k+m−1

k+m

�kk+2

.

.

. �k+mk+1+m

z w = �05

.

.

. . .. ... . ..

�k+m−1k+1+m

.

.

....

.

.

. �k+1k+1+m

z w = �0k+1 �2

k+2 �kk+1+m

�1k+2

u1 v1 = �0k+2

. ..

.

.

.

.

.

....

.

.

. �2k+1+m

um−1 vm−1 = �0k+m

�1k+1+m

um vm = �0k+1+m

Given an analytic function f ∈ S which satisfies (3.5), we will show that the tableof hyperbolic divided differences is compatible with the definition of� j

z f (ui ) for all1 ≤ i ≤ m.

Lemma 3.2 Let k ≥ 1 and m ≥ 1 be integers. Let z, u1, . . . , um ∈ D be distinctpoints and f ∈ S such that f (z) = w, f (ui ) = vi for i = 1, . . . ,m. Suppose thatH j f (z) = μ j , j = 1, . . . , k. Then

�jz f (ui ) = �

jk+1+i , 1 ≤ j ≤ k + 1 and 1 ≤ i ≤ m. (3.7)

Proof We will prove the result in the case j = 1. For an arbitrary integer i such that1 ≤ i ≤ m, if | f (ζ )| < 1 for all ζ ∈ D, then we have

�1z f (ui ) = [ f (ui ), f (z)]

[ui , z] = [vi , w][ui , z] = �1

k+1+i .


If f (ζ ) ≡ λ, ζ ∈ D, where λ is an unimodular constant, then

�1z f (ui ) = λ = �1

k+1+i .

Suppose now that the result is true for j . If f is a Blaschke product of degree at mostj , then � j

z f ≡ λ and μ j ≡ λ, where |λ| = 1. Consequently � j+1z f ≡ λ. It fol-

lows, by the induction hypothesis, that � j+1z f (ui ) = λ. By Definition 1.2, we have

�j+1k+1+i = λ. Thus � j+1

z f (ui ) = �j+1k+1+i in that case. Suppose now that f is not a

Blaschke product of degree at most j . Then f ∈ H. By Theorem 2.5 of [11], it followsthat

|[� jk+1+i , μ j ]| = |[� j

z f (ui ),�jz f (z)]|

≤ |[ui , z]|.

Thus � j+1k+1+i is defined by the expression given in (3.1), which coincides with the

definition of � j+1z f (ui ). As we have chosen i arbitrarily, the result is true for all

1 ≤ i ≤ m. �We are now able to give a solution to the interpolation problem (3.5). This is the

next theorem.

Theorem 3.3 Let k ≥ 1 and m ≥ 1 be integers. Let z, u1, . . . , um ∈ D be distinctpoints and w, v1, . . . , vm, μ1, μ2, . . . , μk ∈ D. The problem (3.5) has

1. infinitely many solutions if and only if |�k+mk+1+m | < 1;

2. a unique solution, which is a Blaschke product of degree strictly less than k+1+m,if and only if |�k+m

k+1+m | = 1;

3. no solution if and only if �k+mk+1+m = ∞.

Proof To construct the table of hyperbolic divided differences in that case, let zi := zfor 1 ≤ i ≤ k +1, zi := ui−k−1 for k +2 ≤ i ≤ k +1+m, wi := w for 1 ≤ i ≤ k +1and wi := vi−k−1 for k + 2 ≤ i ≤ k + 1 + m. The hyperbolic divided differences aredefined using (3.6), (3.1), (3.2) and (3.3). Thus we obtain Table 4.

1. Suppose that |�k+mk+1+m | < 1. By Lemma 3.1, it follows that

|� ji | < 1, 1 ≤ j ≤ k + m, k + 2 ≤ i ≤ k + 1 + m. (3.8)

We will show that there exists infinitely many solutions to the problem. We will do soby constructing explicitly the solutions satisfying (3.5).

Let f0 := g, for some g ∈ S. Define the functions f1, . . . , fk+1+m by

f j+1(ζ ) := [[ζ, zk+1+m− j ] · fm(ζ ),�k+m− jk+1+m− j ], 0 ≤ j ≤ k + m.

The inequalities (3.8) imply that the functions f j belong to S, 1 ≤ j ≤ k + 1 + m.We begin to prove by induction on j that

f j (zi ) = �k+1+m− ji , k + 1 + m − j + 1 ≤ i ≤ k + 1 + m.

P. Rivard

For j = 1, we get

f1(zk+1+m) = [[zk+1+m, zk+1+m] · g(ζ ),�k+mk+1+m]

= �k+mk+1+m .

Suppose now the result is true for j and i = k + m + 1 − j + 1, . . . , k + 1 + m. Forj + 1, we obtain for zi such that zi �= zk+m+1− j

f j+1(zi ) =[

[zi , zk+m+1− j ] · fm(zi ),�k+m− jk+1+m− j

]

,

=[

[zi , zk+1+m− j ] ·�k+1+m− ji ,�

k+m− jk+1+m− j

]

=⎡

⎣[zi , zk+1+m− j ] ·[

�k+m− ji ,�

k+m− jk+1+m− j

]

[zi , zk+1+m− j ] ,�k+m− jk+1+m− j

⎤

⎦

= �k+m− jk+1+m− j ,

where the last equality follows from the identity

[[a, w], w] = a, a ∈ D, w ∈ D.

For zi such that zi = zk+m+1− j , we have

f j+1(zi ) =[

[zi , zi ] · fm(zi ),�k+m− jk+1+m− j

]

,

=[

0 ·�k+1+m− ji ,�

k+m− jk+1+m− j

]

= �k+m− jk+1+m− j ,

which completes the induction. In particular, for j = k + m, it follows thatfk+1+m(zi ) = �0

i for i = 1, . . . , k + m + 1. Thus,

fk+1+m(z) = w;fk+1+m(ui ) = vi , i = 1, . . . ,m.

It remains to prove that H j fk+1+m(z) = μ j , for 1 ≤ j ≤ k. To do this, we will

first show that � jz fk+1+m(ζ ) = fk+1+m− j (ζ ), for ζ ∈ D. We will also use induction

in that case. For j = 1, we have

�1z fk+1+m(ζ ) = [ fk+1+m(ζ ), fk+1+m(z)]

[ζ, z]= [[[ζ, z] · fk+m(ζ ), w], w]

[ζ, z]= fk+m(ζ ).


In particular, we have H1 fk+1+m(z) = �1z fk+1+m(z) = fk+m(z). Suppose now

that the result is true for j , i.e., � jz fk+1+m(ζ ) = fk+1+m− j (ζ ) and in particular

H j fk+1+m(z) = �jz fk+1+m(z) = fk+1+m− j (z). Hence

�j+1z fk+1+m(ζ ) = [� j

z fk+1+m(ζ ),�jz fk+1+m(z)]

[ζ, z]= [ fk+1+m− j (ζ ), fk+1+m− j (z)]

[ζ, z]= [[[ζ, z] · fk+m− j (ζ ), μ j ], [[z, z] · fk+m− j (z), μ j ]]

[ζ, z]= fk+m− j (ζ ).

Finally, for 1 ≤ j ≤ k, we have

fk+1+m− j (z) = [[z, z] · fk+m− j (z), μ j ]= μ j .

Consequently, H j fk+1+m(z) = μ j , for 1 ≤ j ≤ k and therefore, the function fk+1+m

solves our interpolation problem.

2. Suppose now that |�k+mk+1+m | = 1. It follows from Lemma 3.1 that there exists a

unimodular constant λ, |λ| = 1 and an integer l, 1 ≤ l ≤ k + m, such that

|� ji | < 1, 1 ≤ j ≤ l − 1, k + 2 ≤ i ≤ k + 1 + m;�

ji = λ, l ≤ j ≤ k + 1, k + 2 ≤ i ≤ k + 1 + m.

As previously, we will construct the interpolating function which is a unique solutionto the problem using the fact that condition (3.9) is satisfied. If there exists a solutionf ∈ S, then we have that |�l

z f (z)| = |μl | = |λ| = 1. By the Maximum Principle,for some minimal l,�l

z f is a unimodular constant. By Theorem 2.4 of [11], it fol-lows that f is a Blaschke product of degree at most l. If f is a Blaschke product ofdegree strictly less than l, then |�l−1

z f (z)| = |�l−1l | = 1, a contradiction. Thus, if

there exists a solution, the solution is necessarily a Blaschke product of degree l. LetBk+1+m−l(ζ ) = λ, a Blaschke product of degree zero. Define

B j+1(ζ ) :=[

[ζ, zk+1+m− j ] · B j (ζ ),�k+m− jk+1+m− j

]

, k + 1 + m − l ≤ j ≤ k + m.

By construction, the function Bm is a Blaschke product of degree j + l −k −m −1. Inparticular, Bk+1+m , a Blaschke product of degree l, solves the interpolation problem.

3. If �k+mk+1+m = ∞, then there exists no solution to the problem. Indeed, if there

exists a function f ∈ S solution of (3.5), then the modulus of all the entries ofTable 4 would be less than or equal to one.

P. Rivard

Conversely, if there exists infinitely many solutions, the only possibility is that|�k+m

k+1+m | < 1. If there exists a unique solution, then necessarily |�k+mk+1+m | = 1

and finally, if there exists no solution then �k+mk+1+m = ∞. �

Remark 3.4 There are only three possible cases regarding the solution of the interpo-lation problem (3.5). Indeed, the problem cannot have a finite number of solutions,since we can construct infinitely many solutions, as we showed in the proof.

The problem (3.5) is said to be regular if there are infinitely many solutions, sin-gular if there exists a unique solution which is a Blaschke product of degree strictlyless than k + 1 + m and void if there is no solution.

Remark 3.5 If we consider the interpolation problem (3.5) and if we suppose that theproblem is regular, then it is possible to determine a solution which is a Blaschkeproduct of degree k + m + 1. Indeed, it is sufficient to choose f0(ζ ) = eiθ , θ ∈ R.

Theorem 3.6 Let k ≥ 1 and m ≥ 1 be integers. Let z, u1, . . . , um ∈ D be distinctpoints and w, v1, . . . , vm, μ1, μ2, . . . , μk ∈ D. Set μ0 := w. Then, the interpolationproblem (3.5) is

1. regular if and only if one of the following conditions is satisfied

|�k+mk+1+m | < 1; (3.9)

⎧

⎪

⎨

⎪

⎩

|� jk+1+i | < 1, 1 ≤ j ≤ k, 1 ≤ i ≤ m;

|�k+ jk+i | < 1, 1 ≤ j < i ≤ m + 1;

(3.10)

⎧

⎪

⎨

⎪

⎩

ρ(μ j ,�jk+1+i ) < ρ(z, ui ), 0 ≤ j ≤ k, 1 ≤ i ≤ m;

ρ(�k+ jk+ j+1,�

k+ jk+i+1) < ρ(u j , ui ) 1 ≤ j < i ≤ m;

(3.11)

2. singular and has a solution which is a Blaschke product of degree l, with l ≥ 1,(a) if and only if, in the case that l ≤ k, one of the following conditions is satisfied,

|�k+mk+1+m | = 1; (3.12)

⎧

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎩

|� jk+1+i | < 1, 1 ≤ j ≤ l − 1, 1 ≤ i ≤ m;

|� jk+1+i | = 1, l ≤ j ≤ k, 1 ≤ i ≤ m;

|�k+ jk+i | = 1, 1 ≤ j < i ≤ m;

(3.13)


⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

ρ(μ j ,�jk+1+i ) < ρ(z, ui ), 0 ≤ j ≤ l − 2, 1 ≤ i ≤ m;

ρ(μl−1,�l−1k+1+i ) = ρ(z, ui ), 1 ≤ i ≤ m;

ρ(μ j ,�jk+1+i ) < ρ(z, ui ), l ≤ j ≤ k, 1 ≤ i ≤ m;

ρ(�k+ jk+1+ j ,�

k+ jk+1+i ) < ρ(u j , ui ), 1 ≤ j < i ≤ m;

(3.14)

(b) if and only if, in the case that l > k, one of the following conditions is satisfied,

|�k+mk+1+m | = 1; (3.15)

⎧

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎩

|� jk+1+i | < 1, 1 ≤ j ≤ k, 1 ≤ i ≤ m;

|�k+ jk+i | < 1, 1 ≤ j ≤ l − 1 − k, j < i ≤ m;

|�k+ jk+i | = 1, l − k ≤ j < i ≤ m;

(3.16)

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

ρ(μ j ,�jk+1+i ) < ρ(z, ui ), 0≤ j ≤k, 1≤ i ≤m;

ρ(�k+ jk+1+ j ,�

k+ jk+1+i ) < ρ(u j , ui ), 1≤ j ≤ l − 2 − k, j < i ≤m;

ρ(�l−1k+1+ j ,�

l−1k+1+i ) = ρ(u j , ui ), l − 1 − k ≤ j < i ≤m;

ρ(�k+ jk+1+ j ,�

k+ jk+1+i ) < ρ(u j , ui ), l − k ≤ j < i ≤m.

(3.17)

Proof 1. The condition (3.9) was demonstrated in Theorem 3.3. The equivalence of(3.9) and (3.10) follows from Lemma 3.1. Suppose that the condition (3.10) istrue, then we have for 1 ≤ j ≤ k + 1 and 1 ≤ i ≤ m

1 > |� jk+1+i | = |[� j−1

k+1+i , μ j−1]||[ui , z]| = p(� j−1

k+1+i , μ j−1)

p(ui , z). (3.18)

For 1 ≤ j < i ≤ m, we obtain

1 > |�k+ jk+i | = |[�k+ j−1

k+i ,�k+ j−1k+ j ]|

|[ui , u j ]| = p(�k+ j−1k+i ,�

k+ j−1k+ j )

p(ui , u j ). (3.19)

As the hyperbolic distance depends of the pseudo-hyperbolic distance monotoni-cally, then it is clear by (3.18) and (3.19) that the condition (3.10) is equivalent to(3.11).

P. Rivard

2. (a) The condition (3.12) was also proved in Theorem 3.3. The fact that the solu-tion is a Blaschke product of degree exactly l has been justified in the proof ofthe same theorem. The equivalence of conditions (3.12) with (3.13) followsfrom Lemma 3.1. Suppose that the condition (3.13) is true. By Lemma 3.1,the modulus of all the entries of columns 1 to (l − 1) is strictly less than one.This implies that for 1 ≤ j ≤ l − 1 and 1 ≤ i ≤ m

1 > |� jk+1+i | = |[� j−1

k+1+i , μ j−1]||[ui , z]| = p(� j−1

k+1+m, μ j−1)

p(ui , z).

From the l-th column and for all the subsequent columns, the entries of thesecolumns are the same and are equal to a unimodular constant. In particular,for j = l and 1 ≤ i ≤ m, we have

1 = |�lk+1+i | = |[�l−1

k+1+i , μl−1]||[ui , z]| = p(�l−1

k+1+m, μl−1)

p(ui , z).

For the remaining, it is sufficient to use (1.1). This shows the equivalence of(3.13) and (3.14).

(b) Similarly, the equivalence of (3.15), (3.16) and (3.17) can be proved in thesame manner. �

It is now possible, with the theory developed precedently, to give a necessary con-dition to the existence of a solution of the constrained Nevanlinna–Pick problem [5].This is a generalized interpolation problem where the values of the derivatives are pre-scribed up to a certain order at each point and not only the values of the function itself,as in the case of the classical Nevanlinna–Pick problem. We will study the problem interms of the hyperbolic derivatives instead of the classical derivatives.

Let k ≥ 0, n ≥ 1 be integers, z1, z2, . . . , zn ∈ D be distinct points andμ0

1, μ02, . . . , μ

0n, μ

11, . . . , μ

1n, . . . , μ

k1, . . . , μ

kn ∈ D. We are interested in the existence

of a function f ∈ S satisfying

f (zi ) = μ0i , H j f (zi ) = μ

ji , for j = 1, . . . , k and i = 1, . . . , n. (3.20)

Choose a point zl , 1 ≤ l ≤ n. Considering Table 4, we set the following:

z := zl , w := μ0l , μ1 := μ1

l , . . . , μk := μkl , ui := zi , vi := μ0

i , 1 ≤ i ≤ n, i �= l,

and we use the notation l�ji , 1 ≤ j ≤ k +n −1, k +2 ≤ i ≤ k +n, for the hyperbolic

divided differences constructed with these points. Once again, these differences areconveniently obtained by forming Table 5.

Theorem 3.7 Let f be a function of S satisfying (3.20). Then,

|l�k+n−1k+n | ≤ 1, 1 ≤ l ≤ n. (3.21)


Table 5 Table of hyperbolic divided differences for the constrained Nevanlinna–Pick problem

Points 1 2 3 … k k + 1 . . . k + n − 2 k + n − 1

zl μ0l

μ1l

zl μ0l μ2

l

μ1l μ3

l

zl μ0l μ2

l

. . .

μ1l μ3

l μkl

zl μ0l μ2

l

.

.

. l�k+1k+2

μ1l

.

.

....

. . .

l�k+n−2k+n−1

zl μ0l

.

.

. . .. l�kk+2

.

.

.... l�k+n−1

k+n

.

.

....

.

.

. l�3k+2

l�k+n−2k+n

zl μ0l

l�2k+2

.

.

. . ..

l�1k+2

.

.

. l�k+1k+n

z1 w1

.

.

. l�kk+n

.

.

. . ..

.

.

....

.

.

. l�3k+n

zn−1 wn−1l�2

k+nl�1

k+n

zn wn

Proof This is a direct consequence of Theorem 3.3, applied to a point zl and all theother points. �

Considering Theorem 3.7, we can give a necessary condition to the existenceof a solution to the interpolation problem (3.20). One has to check whether eachpoint satisfies the condition (3.21). Otherwise, a solution cannot exists. Thus, withthe iterative formula given in Sect. 2 for the calculation of hyperbolic derivatives,it is now possible to consider interpolation with hyperbolic derivatives numeri-cally.

Remark 3.8 Theorem 3.7 begs the following question: is the condition (3.21) also suf-ficient for the existence of a function which is a solution to the interpolation problem?

P. Rivard

4 Dieudonné’s Lemma for Hyperbolic Derivatives of Order n

Classically, the Dieudonné Lemma is stated under the following form.

Dieudonné’s Lemma Let u and v be given points in D, with u �= 0 and such that|v| < |u|. Then, for all functions f analytic and satisfying | f (z)| < 1 in D, withf (0) = 0 and f (u) = v, we have

∣

∣

∣ f ′(u)− v

u

∣

∣

∣ ≤ |u|2 − |v|2|u|(1 − |v|2) . (4.1)

This result first appeared in 1931 and is essentially due to Dieudonné [6]. Moreover,all values in the closed disk (4.1) correspond to f ′(u) for some f . In other words, ifwe denote by F , the set of analytic functions f : D → D such that f (0) = 0 andf (u) = v, then

{

f ′(u) : f ∈ F} ={

ζ ∈ C :∣

∣

∣ζ − u

v

∣

∣

∣ ≤ |u|2 − |v|2|u|(1 − |u|2)

}

. (4.2)

More than seventy years later, Beardon and Minda became interested again on thislemma [4,3]. Using a Schwarz–Pick Lemma involving three points, they have beenable to give a simple and elegant proof of Dieudonné’s Lemma which has the advan-tage of being more geometric, in the hyperbolic sense. Furthermore, the authors havefurnished a different interpretation of Dieudonné’s Lemma in terms of the hyperbolicderivative of first order. For this, we introduce the following notation: we denote byDh(c, r) the open hyperbolic disk of center c and radius r :

Dh(c, r) := {z ∈ D : ρ(z, c) < r},

and by Dh(c, r) the closed hyperbolic disk:

Dh(c, r) := {z ∈ D : ρ(z, c) ≤ r}.

Hence Beardon and Minda have obtained

{H1 f (u) : f ∈ F} = Dh(v /u , ρ(0, u)). (4.3)

Equality (4.3) can be expressed in Euclidean terms as follows.

{H1 f (u) : f ∈ F} ={

ζ ∈ C :∣

∣

∣

∣

ζ − (1 − |u|2)v /u

1 − |v|2∣

∣

∣

∣

≤ |u| − |v|2|u|(1 − |v|2)

}

.

Recently Baribeau, Rivard and Wegert [2] have obtained a multi-point Schwarz–Pick Lemma based on the hyperbolic divided differences. As these lead naturally tothe notion of higher-order hyperbolic derivatives [11], we now present a generalizedversion of the Dieudonné Lemma which can be applied to the hyperbolic derivatives


Table 6 Table of hyperbolic divided differences for the generalized Dieudonné Lemma

Points of D 1 2 3 … n − 1 n n + 1

u v

μ1

u v μ2

μ1 μ3

u v μ2. . .

μ1 μ3 μn−1

u v μ2 Hn f (u)

μ1

.

.

. μn−1 �n+1n+2

u v... �n

n+2

.

.

. �n−1n+2

.

.

....

.

.

. . ..

�3n+2

u v �2n+2

�1n+2

0 0

of order n ≥ 1. In this case, we consider the family of holomorphic functions wherethe first (n − 1)-th hyperbolic derivatives are prescribed, but under certain conditions.We can then locate the n-th hyperbolic derivative in a specfic area, which does notdepend of the function but only on the prescribed derivatives and the fixed points.

Theorem 4.1 (Generalized Dieudonné’s Lemma) Let u, v ∈ D be such that |v| < |u|and n > 1 be an integer. In Table 6, choose n − 1 points μ1, . . . , μn−1, whereμ j ∈ Dh(�

jn+2, ρ(0, u)) for j = 1, . . . , n −1. We define Fn−1 the set of holomorphic

functions by

Fn−1 := { f : D → D : f (0)=0, f (u)=v, H1 f (u)=μ1, . . . , Hn−1 f (u)=μn−1}.

Then,

{Hn f (u) : f ∈ Fn−1} ={

ζ ∈ C :∣

∣

∣

∣

∣

ζ − (1 − |u|2)�nn+2

1 − |u|2|�nn+2|2

∣

∣

∣

∣

∣

≤ |u|(1 − |�nn+2|2)

1 − |u|2|�nn+2|2

}

.

(4.4)

It is important to mention that the set of functions Fn−1 is not empty by the assump-tions of Theorem 4.1 and by Theorem 3.6. Also, the proof of Theorem 4.1 is omit-

P. Rivard

ted, since we shall prove below a more general result from which the generalizedDieudonné Lemma can be derived as a particular case.

The next corollary is concerned with the particular case n = 2 of Theorem 4.1, butinterpreted in terms of the classical derivative of order two instead of the hyperbolicderivative of order two.

Corollary 4.2 Let u, v ∈ D be such that |v| < |u|. Fix μ ∈ Dh(v/u, ρ(0, u)). Sup-pose that f ∈ F1, where

F1 = { f : D → D : f (0) = 0, f (u) = v, H1 f (u) = μ}.

Then,

∣

∣ f ′′(u)− c(u, v, μ)∣

∣ ≤ R(u, v, μ), (4.5)

where

c(u, v, μ) := 2uμ(1 − |v|2)(1 − |u|2)2 − 2vμ2(1 − |v|2)

(1 − |u|2)2 + 2(1 − |μ|2)(1 − |v|2)�24

(1 − |u|2)(1 − |u|2|�24|2)

;

R(u, v, μ) := 2|u|(1 − |�24|2)(1 − |μ|2)(1 − |v|2)

(1 − |u|2|�24|2)(1 − |u|2)2 .

Proof Inequality (4.5) follows from (4.4) in the case n = 2 with μ1 = μ, using thealgebraic expression of H2 f (u), given by Corollary 2.9. �

In the foregoing, we have presented results concerning essentially the origin andanother point of the unit disk. Dieudonné’s Lemma has its equivalent for the case ofany points of D. This result is called the Dieudonné–Pick Lemma ([7, Corollary 5.6]).

Dieudonné–Pick Lemma The region of values of f ′(A), for all analytic functionsf : D → D satisfying f (a) = b and f (A) = B, is the closed disk of center c andradius r , where

c = 1 − ∣

∣(A − a)/

(1 − Aa)∣

∣

2

1 − ∣

∣(B − b)/

(1 − Bb)∣

∣

2

(B − b)/

(1 − Bb)

(A − a)/

(1 − Aa)

1 − |B|21 − |A|2 ;

r =∣

∣(A − a)/

(1 − Aa)∣

∣

2 − ∣

∣(B − b)/

(1 − Bb)∣

∣

2

∣

∣(A − a)/

(1 − Aa)∣

∣ (1 − ∣

∣(B − b)/

(1 − Bb)∣

∣

2)

1 − |B|21 − |A|2 .

We can now state another important result of this section which concerns the gen-eralization of the previous theorem involving the hyperbolic derivatives of all order.

Theorem 4.3 (Generalized Dieudonné–Pick Lemma) Let u, z, v, w ∈ D be suchthat ρ(w, v) < ρ(z, u) and n > 1 be an integer. In Table 7, choose n − 1 points


Table 7 Table of hyperbolic divided differences for the generalized Dieudonné–Pick Lemma

Points of D 1 2 3 … n − 1 n n + 1

u v

μ1

u v μ2

μ1 μ3

u v μ2. . .

μ1 μ3 μn−1

u v μ2 Hn f (u)

μ1

.

.

. μn−1 �n+1n+2

u v... �n

n+2

.

.

. �n−1n+2

.

.

....

.

.

. . ..

�3n+2

u v �2n+2

�1n+2

z w

μ1, . . . , μn−1, where μ j ∈ Dh(�jn+2, ρ(u, z)) for j = 1, . . . , n − 1 and consider

Gn−1, the set of analytic functions

Gn−1 :={ f : D → D : f (z)=w, f (u)=v, H1 f (u) = μ1, . . . , Hn−1 f (u) = μn−1}.

Then,

{Hn f (u) : f ∈ Gn−1}

={

ζ ∈ C :∣

∣

∣

∣

∣

ζ − (1 − |[z, u]|2)�nn+2

1 − |[z, u]|2|�nn+2|2

∣

∣

∣

∣

∣

≤ (1 − |�nn+2|2)|[z, u]|

1 − |[z, u]|2|�nn+2|2

}

. (4.6)

Remark 4.4 The expressions given on the right-hand side of the equality in (4.6) donot depend on the choice of the function f in the set Gn−1, but only on the prescribedpoints u, v, z, w,μ1, . . . , μn−1.

Proof By assumption and the choice of μ j for 1 ≤ j ≤ n − 1, we have

ρ(w, v) < ρ(z, u); (4.7)

ρ(μ j ,�jn+2) < ρ(z, u), j = 1, . . . , n − 1. (4.8)

Consequently, by Theorem 3.6, the set Gn−1 is not empty. Let f ∈ Gn−1. It followsfrom (4.7) and (4.8) that f cannot be a Blaschke product of degree less than or equal

P. Rivard

to n. Therefore |Hn f (u)| < 1 and |�nn+2| < 1. Theorem 3.6 implies that

ρ(Hn f (u),�nn+2) ≤ ρ(z, u),

which is equivalent to Hn f (u) ∈ Dh(�nn+2, ρ(u, z)). Hence

∣

∣

∣

∣

∣

Hn f (u)−�nn+2

1 −�nn+2 Hn f (u)

∣

∣

∣

∣

∣

≤∣

∣

∣

∣

z − u

1 − uz

∣

∣

∣

∣

. (4.9)

Inequality (4.9) represents an Euclidean disk of center and radius respectively

1 − |[z, u]|21 − |[z, u]|2|�n

n+2|2�n

n+2 and1 − |�n

n+2|21 − |[z, u]|2|�n

n+2|2|[z, u]|.

It follows that

Hn f (u)∈{

ζ ∈ C :∣

∣

∣

∣

∣

ζ− 1 − |[z, u]|21 − |[z, u]|2|�n

n+2|2�n

n+2

∣

∣

∣

∣

∣

≤ 1 − |�nn+2|2

1 − |[z, u]|2|�nn+2|2

|[z, u]|}

.

As f ∈ Gn−1 has been chosen arbitrarily, we have

{Hn f (u) : f ∈ Gn−1}

⊂{

ζ ∈ C :∣

∣

∣

∣

∣

ζ − (1 − |[z, u]|2)�nn+2

1 − |[z, u]|2|�nn+2|2

∣

∣

∣

∣

∣

≤ (1 − |�nn+2|2)|[z, u]|

1 − |[z, u]|2|�nn+2|2

}

. (4.10)

We will show the reverse inclusion, i.e. each point of the disk given in (4.10) can bethe hyperbolic derivative of order n of a function belonging to the class Gn−1. This isequivalent to an interpolation problem. Choose α, an arbitrary point of the disk givenin (4.10). Clearly, α ∈ Dh(�

nn+2, ρ(u, z)). By the assumptions, and the choice of α,

we have

ρ(v,w) < ρ(u, z); (4.11)

ρ(μ j ,�jn+2) < ρ(u, z), j = 1, . . . , n − 1; (4.12)

ρ(α,�nn+2) ≤ ρ(u, z). (4.13)

The inequalities (4.11), (4.12) and (4.13) are precisely the hypothesis of Theorem 3.6.Therefore, there exists an analytic function f ∈ Gn−1 such that, in particular,Hn f (u) = α. Finally, we have proved equality (4.6). �

The proof of Theorem 4.3 shows that the Dieudonné Lemma is a result fundamen-tally hyperbolic. Indeed, it is possible, with the hyperbolic derivatives of higher order,to give an analogous version of the classical one. Also, it seems to suggest that thenotion of hyperbolic derivative is more natural to studying analytic functions of theunit disk into itself. We believe that to obtain Corollary 4.2 without using the preceding


tools would be less convenient than the methods used. The equivalent for the classicalderivatives of higher order follows easily from the hyperbolic case.

It is of interest to compare the hyperbolic disk given in the case of the hyperbolicderivative and the Euclidean disk given in the case of the classical derivative. Indeed,in the case of the derivative of order one, we have by (4.3) and (4.2), that the center ofthe Euclidean disk of the classical derivative is the same as the center of the hyperbolicdisk of the hyperbolic derivative. The authors had suggested in [4] that there couldpossibly exist a deep reason explaining this phenomenon. In the case of order two, weobtain

{H2 f (u) : f ∈ F1} = Dh(�24, ρ(z, 0)),

where

�24 = [v /u , μ]

u, (4.14)

by the generalized Dieudonné Lemma and the proof of Theorem 4.3, in the case n = 2.We can easily check that the center of the Euclidean disk of the derivative of secondorder given by (4.5) is different from the center of the hyperbolic disk of the hyper-bolic derivative of second order given by (4.14). Thus, the phenomenon seems to beparticular to the case of derivative of order one.

5 Rogosinski’s Lemma

We begin by recalling a second result of classical complex analysis. This is the cele-brated Rogosinski Lemma [7,9].

Rogosinski’s Lemma Let f : D → D, an analytic function such that f (0) = 0.Then for z ∈ D and z �= 0, we have

∣

∣

∣

∣

f (z)− z f ′(0)(1 − |z|2)1 − |z|2| f ′(0)|2

∣

∣

∣

∣

≤ |z|2 1 − | f ′(0)|21 − |z|2| f ′(0)|2 . (5.1)

This result has its own interest because it is an improved version of the well-knownSchwarz Lemma. The restriction f (0) = 0 can be avoided and an analogous versionexists for any points of the unit disk. This is called the Rogosinski–Pick Lemma. Wewill show that we can treat this lemma in a more hyperbolic manner, which is nearerto the essence of the present work. Indeed, the theory presented in Sect. 2 has inspireda new demonstration using hyperbolic divided differences and hyperbolic derivatives.In fact, the Rogosinski–Pick Lemma asserts that for a holomorphic function f of theunit disk into itself and for a fixed point u ∈ D, we can determine a region, dependingin particular on the hyperbolic derivative of first order at the point u, to locate theimage under the function f of any point of the disk different from u.

Rogosinski–Pick Lemma Let f : D → D be an analytic function and u ∈ D.Suppose that f (u) = v. Let z ∈ D, z �= u.

P. Rivard

Table 8 Table of hyperbolicdivided differences for theRogosinski–Pick Lemma

Points of D 1 2

u v

H1 f (u)

u v �23

�13

z f (z)

1. If v = 0, then

| f (z)+ c| ≤ r; (5.2)

2. if v �= 0, then

∣

∣

∣

∣

f (z)− (1 − |v|2)(vc − 1)+ |vc − 1|2 − |v|2r2

v(|vc − 1|2 − |v|2r2)

∣

∣

∣

∣

≤ (1 − |v|2)r||vc − 1|2 − |v|2r2)| ,

(5.3)

where

c= 1 − |[z, u]|21 − |[z, u]|2|H1 f (u)|2 H1 f (u)[z, u] and r = 1 − |H1 f (u)|2

1 − |[z, u]|2|H1 f (u)|2 |[z, u]|2.

Proof In the following, we will consider Table 8, which represents the table of hyper-bolic divided differences in that case. First we prove the lemma for f ∈ B1. By thedefinitions and Theorem 2.4 of [11], we have

�13 = H1 f (u) = �2

3 = λ, |λ| = 1.

In particular, c = λ[z, u], r = 0 and

�13 = [ f (z), v]

[z, u] = λ,

whence

f (z) = v − λ[z, u]1 − vλ[z, u] . (5.4)

If v = 0, then f (z) = −λ[z, u], and this gives (5.2). If v �= 0, then the set in (5.3) isalso reduced to a single point which is exactly (5.4).

Suppose now that f �∈ B1 and consider Table 8, which represents the table of hyper-bolic divided differences in this particular case. By Theorem 3.3, we have |�2

3| ≤ 1.


Thus,

∣

∣

∣

∣

∣

�13 − H1 f (u)

1 − H1 f (u)�13

∣

∣

∣

∣

∣

≤∣

∣

∣

∣

u − z

1 − uz

∣

∣

∣

∣

. (5.5)

Inequality (5.5) implies that �13 ∈ Dh(H1 f (u), ρ(z, u)). This is equivalent to the

fact that �13 belongs to an Euclidean disk of center and radius respectively

1 − |[z, u]|21 − |[z, u]|2|H1 f (u)|2 H1 f (u) and

1 − |H1 f (u)|21 − |[z, u]|2|H1 f (u)|2 |[z, u]|.

Using the definition of hyperbolic divided differences, it follows that

∣

∣

∣

∣

[ f (z), v][z, u] − 1 − |[z, u]|2

1 − |[z, u]|2|H1 f (u)|2 H1 f (u)

∣

∣

∣

∣

≤ 1 − |H1 f (u)|21 − |[z, u]|2|H1 f (u)|2 |[z, u]|,

which in turn is equivalent to

∣

∣

∣

∣

v − f (z)

1 − v f (z)− 1 − |[z, u]|2

1 − |[z, u]|2|H1 f (u)|2 H1 f (u)[z, u]∣

∣

∣

∣

≤ 1 − |H1 f (u)|21 − |[z, u]|2|H1 f (u)|2 |[z, u]|2. (5.6)

Let ϕv(z) := (v−z)/

(1 − vz) . We can deduce from inequality (5.6) that ϕv( f (z))belongs to an Euclidean disk, denoted DE , of center c and radius r defined by

c := 1 − |[z, u]|21−|[z, u]|2|H1 f (u)|2 H1 f (u)[z, u] and r := 1 − |H1 f (u)|2

1−|[z, u]|2|H1 f (u)|2 |[z, u]|2.

Therefore f (z) ∈ ϕ−1v (DE (c, r)) = ϕv(DE (c, r)). If v = 0, then we obtain (5.2) and

this proves (1). If v �= 0, then the disk ϕv(DE (c, r)) is of center and radius respectively

(1 − |v|2)(vc − 1)+ |vc − 1|2 − |v|2r2

v(|vc − 1|2 − |v|2r2)and

(1 − |v|2)r||vc − 1|2 − |v|2r2| .

Finally, we get (5.3), and this proves (2), as desired. �Acknowledgements The author wishes to thank sincerely his doctoral thesis advisor, Professor LineBaribeau, for her many valuable suggestions and comments that have greatly improved the present paper.

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Some Applications of Higher-Order Hyperbolic Derivatives