In this paper, a sharp version of the Schwarz–Pick Lemma for hyperbolic derivatives is provided for holomorphic selfmappings on
the unit disk with fixed multiplicity for the zero at the origin, hence extending a recent result due to Beardon. A property of preserving hyperbolic distances also studied by Beardon is here completely characterized.Dirección General de Enseñanza Superior (DGES). EspañaJunta de Andalucí

Bernal González, Luis

Calderón Moreno, María del Carmen

English

idUS. Depósito de Investigación Universidad de Sevilla

Two hyperbolic Schwarz lemmasbyL. BERNAL-GONZA´LEZ and M.C. CALDERO´N–MORENO∗AbstractIn this paper, a sharp version of the Schwarz–Pick Lemma forhyperbolic derivatives is provided for holomorphic selfmappings onthe unit disk with fixed multiplicity for the zero at the origin, henceextending a recent result due to Beardon. A property of preser-ving hyperbolic distances also studied by Beardon is here completelycharacterized.11 Introduction and notationThe Schwarz Lemma and its hyperbolic version (= the Schwarz–PickLemma) continues attracting the attention of many mathematicians. Ouraim in this paper is to prove two sharp versions of the former results as-suming that the multiplicity for the zero at the origin of the holomorphicfunction under consideration is fixed. Our results will extend a recent onedue to Beardon, see below.First of all, we need to fix some notation. The symbols N, C, R, D,D(c, r) will denote, as usual, the set of positive integers, the complex plane,the real line, the open unit disk and the euclidean closed disk {z ∈ C :|z− c| ≤ r} (c ∈ C, r > 0), respectively. As for function spaces, H(D) is theclass of all holomorphic functions on D and Aut(D) will stand for the group∗The authors have been partially supported by DGES Grant PB96-1348 and the Juntade Andaluc´ıa.12000 Mathematics Subject Classification: Primary 30F45. Secondary 30C80.Key words and phrases: Schwarz–Pick Lemma, higher order hyperbolic derivative, hy-perbolic distance, multiplicity at a point, m–automorphism of the unit disk.1of conformal automorphisms of D. If f ∈ H(D) and a ∈ D then µ(f, a) willrepresent the multiplicity for the zero at a of the function f(z)− f(a). Form ∈ N we introduce the notationsFm = {f ∈ H(D) : |f | < 1, f(0) = f ′(0) = . . . = f (m−1)(0) = 0}= {f ∈ H(D) : |f | < 1, f(0) = 0, µ(f, 0) ≥ m}andzmAut(D) = {zmf(z) : f ∈ Aut(D)}.For the sake of convenience, we agree that z0Aut(D) = Aut(D). We say thata function f is an m-rotation whenever there exists a constant c with |c| = 1such that f(z) = czm. The set of all m-rotations will be represented by Rm.It is clear that zm−1Aut(D) ∩ Fm = Rm. For a ∈ D we denote by ϕa thespecial automorphism ϕa(z) =a−z1−az . In fact, Aut(D) = {kϕa : |a| < 1 =|k|}. Note that ϕ−1a = ϕa. We define an m-automorphism of D as a functionf of the form f = ψ ◦R ◦ϕ with ϕ, ψ ∈ Aut(D) and R ∈ Rm We denote byAutm(D) the set of m-automorphisms of D. Obviously, Aut1(D) = Aut(D).It is straightforward to see thatAutm(D) = {ϕb ◦R ◦ ϕa : a, b ∈ D, R ∈ Rm}and{f ∈ Autm(D) : f(a) = b} = {ϕb ◦R ◦ ϕa : R ∈ Rm}.The symbol ρ will stand for the hyperbolic (or Poincare’s) distance on D,that is,ρ(z, w) = tanh−1 |ϕz(w)| = 12log1 + | z−w1−zw |1− | z−w1−zw |.If f : D→ D is holomorphic then the hyperbolic derivative of order m of fat z as introduced by Peschl is defined asf [m](z) = (ϕf(z) ◦ f ◦ ϕz)(m)(0).By using the fact ϕ′a(t) =|a|2 − 1(1− at)2 together with Faa di Bruno’s formula(see, for instance, [3]) for the mth derivative of a composite function it isnot difficult to check thatf [1](z) =(1− |z|2)f ′(z)1− |f(z)|22and that, for m ≥ 2,f [m](z) = (−1)m+1 (1− |z|2)f (m)(z)1− |f(z)|2 + α(z),where α(z) is a finite sum of terms each of them containing at least onefactor among f ′(z), . . . , f (m−1)(z). Hence if µ(f, z) ≥ m then α(z) = 0, sowe havef [m](z) = (−1)m+1 (1− |z|2)f (m)(z)1− |f(z)|2 .Hyperbolic derivatives are invariant in the sense that |(S ◦ f ◦ T )[m]| =|f [m]| ◦ T whenever S and T are conformal automorphisms of D.The Schwarz-Pick Lemma is a non-Euclidean version of the classicalSchwarz Lemma. It asserts thatρ(f(z), f(w)) ≤ ρ(z, w) and |f [1](z)| ≤ 1,for all z, w ∈ D and every f ∈ H(D). Furthermore, the equality ρ(f(z), f(w)) =ρ(z, w) holds for every pair z, w ∈ D if and only if this equality holds forsome pair z, w ∈ D (z 6= w) if and only if |f [1](z)| = 1 for every z ∈ D if andonly if |f [1](z)| = 1 for some z ∈ D if and only if f ∈ Aut(D).Recently, Beardon [1] has given an interesting new version of the Schwarz(or the Schwarz-Pick) Lemma. Specifically, he proved Theorem 1.1 be-low (see [1, Theorem] and notes following [1, Lemma 1]), which is a non-Euclidean version of a result due to Dieudonne´ [2] that establishes the fol-lowing Schwarz Lemma for derivatives: If f ∈ F1 then|f ′(z)| ≤{1 if |z| ≤ √2− 1(1+|z|2)24|z|(1−|z|2) if |z| >√2− 1.This inequality is the best possible in terms of |z|. We now transcribe thestatements of Beardon in our terminology (he denoted f [1] = f ∗). Recallthat f [1](z) ∈ D if f 6∈ Aut(D). Beardon realized that this allowed tomeasure the hyperbolic distance between two hyperbolic derivatives.Theorem 1.1. Assume that f ∈ F1 \ Aut(D). We have:(a) The inequalityρ(f [1](0), f [1](z)) ≤ 2ρ(0, z) (1)is satisfied for all z ∈ D.3(b) If equality holds in (1) for some z ∈ D \ {0} then f ∈ zAut(D).(c) If f(z) = z2 then equality holds in (1) for all z ∈ D.2 A preliminary resultBefore stating our theorems, we need an elementary lemma which is an“m-order” generalization of the Schwarz-Pick Lemma.Lemma 2.1. Assume that m ∈ N, a ∈ D, f ∈ H(D), |f | < 1 on D andµ(f, a) ≥ m. Then we have∣∣∣∣∣ f(z)− f(a)1− f(a)f(z)∣∣∣∣∣ ≤∣∣∣∣ z − a1− az∣∣∣∣m (z ∈ D) (2)and |f [m](a)| ≤ m!.Further, equality holds in (2) for all z ∈ D if and only if it holds forsome z 6= a if and only if |f [m](a)| = m! if and only if f ∈ Autm(D).Proof. If a ∈ D satisfies µ(f, a) ≥ m then the functionF (t) =(ϕb ◦ f ◦ ϕa)(t)tm(t ∈ D \ {0})has a holomorphic extension to D because µ(ϕb ◦ f ◦ ϕa, 0) = µ(f, a) ≥ m,where we have denoted b = f(a). Fix r ∈ (0, 1). Then |F (t)| ≤ 1rmon|t| = r, hence an application of the Maximum Modulus Principle yieldssup{|F (t)| : |t| ≤ r} ≤ 1rm. Letting r → 1 we get sup{|F (t)| : |t| < 1} ≤ 1,that is, |F (t)| ≤ 1 on D or, equivalently, |ϕb◦f ◦ϕa(t)| ≤ |t|m (t ∈ D), whichbecomes (2) after the change of variable z = ϕa(t). Note that the value of(the extension of) F at the origin isF (0) = limt→01tm· b− f(ϕa(t))1− bf(ϕa(t))= limz→a(1− az)m1− f(a)f(z) ·f(z)− f(a)(z − a)m · (−1)m+1 =(−1)m+1 · (1− |a|2)m1− |f(a)|2 · limz→af(z)− f(a)(z − a)m =f [m](a)m!,after using the L’Hopital rule together with the definition of the hyperbolicderivative of order m and the fact that f ′(a) = . . . = f (m−1)(a) = 0. Anewby the Maximum Modulus Principle, |F (0)| ≤ 1, whence |f [m](a)| ≤ m!.4Assume now that f is an m-automorphism. Since f(a) = b we havef = ϕb ◦ R ◦ ϕa where R ∈ Rm, i.e., R(t) = ctm for some c with |c| = 1.Therefore F (t) = c on D, so |F | = 1 on D and the equality holds in (2) for allz ∈ D (hence for some z 6= a). Moreover, |f [m](a)| = m! · |F (0)| = m! · |c| =m!. Conversely, suppose that |f [m](a)| = m!. Then |F (0)| = 1 and theMaximum Modulus Principle tells us that F (t) = c for some unimodularconstant c, but this yields (ϕb ◦ f ◦ ϕa)(t) = ctm for all t ∈ D, which inturn implies that f = ϕb ◦ R ◦ ϕa with R as before. Consequently, f is anm-automorphism of D.Finally, assume that equality holds in (2) for some z 6= a. Then thechange z = ϕa(t) shows that |ϕb ◦ f ◦ ϕa(t)| = |t|m for some t 6= 0, whence|F (t)| = 1 for some t ∈ D. Another application of the Maximum ModulusPrinciple drives us to F (t) = c on D for some unimodular constant c, andthis implies as above that f ∈ Autm(D). This concludes the proof.3 Main resultsWe are now ready to state our theorems. Like in [1], we can estimate thehyperbolic distance between two normalized hyperbolic derivatives of higherorder under obvious conditions.Theorem 3.1. Assume that f ∈ Fm \ Autm(D). We have:(a) If a ∈ D and µ(f, a) ≥ m thenρ((−1)m+1f[m](0)m!,f [m](a)m!) ≤ 2ρ(0, a). (3)(b) If there exists a ∈ D \ {0} for which µ(f, a) ≥ m such that equalityholds in (3) then f ∈ zmAut(D).Proof. Observe that if µ(f, a) ≥ m and f ∈ Fm \ Autm(D) then the values(−1)m+1f[m](0)m!,f [m](a)m!are in D by Lemma 2.1, so the hyperbolic distancebetween them makes sense. As for (a), since µ(f, a) ≥ m ≤ µ(f, 0) thefunctionsg(z) =f(z)zm(z ∈ D \ {0})5andh(z) =(1− aza− z)m· f(a)− f(z)1− f(a)f(z) (z ∈ D \ {a})have holomorphic extensions on the whole D if we setg(0) =f (m)(0)m!= (−1)m+1f[m](0)m!andh(a) = limz→ah(z) =(1− |a|2)m1− |f(a)|2 · (−1)m+1 · f(m)(a)m!=f [m](a)m!.We may start with a 6= 0, since the case a = 0 is trivial.Note that by Lemma 2.1 (as applied on points 0, a) we get |g| ≤ 1,|h| ≤ 1 on D, and in fact |g| < 1, |h| < 1 on D since f is not an m-automorphism. On the other hand, g(a) = f(a)amand h(0) = f(a)am. If we applythe Schwarz-Pick Lemma to g and h then one obtains ρ(g(0), g(a)) ≤ ρ(0, a)and ρ(h(0), h(a)) ≤ ρ(0, a). Consequently, observing that g(a) = h(0), thetriangle inequality yieldsρ((−1)m+1f[m](0)m!,f [m](a)m!) = ρ(g(0), h(a)) ≤ ρ(0, a) + ρ(0, a) = 2ρ(0, a),which proves (a). In order to prove (b), assume that equality in (3) holdsfor some a ∈ D \ {0} with µ(f, a) ≥ m. Then2ρ(0, a) = ρ(g(0), h(a)) ≤ ρ(g(0), g(a)) + ρ(h(0), h(a)) ≤ 2ρ(0, a),whence ρ(g(0), g(a)) = ρ(0, a) because both terms in the last sum are notgreater than ρ(0, a). But the Schwarz-Pick Lemma tells that g ∈ Aut(D),hence f ∈ zmAut(D). The proof is finished.Corollary 3.2. If m is even and f ∈ Fm \Autm(D) then f (m)(0) = 0. Inother words, f ∈ Fm+1.Proof. From (3) and the fact that µ(f, 0) ≥ m we obtainρ(f [m](0)m!,−f[m](0)m!) ≤ 2ρ(0, 0) = 0,therefore f [m](0) = −f [m](0). Consequently, f (m)(0) = (−1)m+1f [m](0) =0.6It should be noted that parts (a)-(b) of Theorem 1.1 are covered with thecase m = 1 in Theorem 3.1 (observe that always µ(f, a) ≥ 1). An extensionof part (c) of Theorem 1.1 makes no sense because if m ≥ 2 then the set{z ∈ D : µ(f, z) ≥ m} is discrete in D except for the trivial case f ≡ 0. Inview of parts (b)-(c) of Theorem 1.1 one can wonder whether f ∈ zAut(D)implies equality in (1) for some (or even for all) z ∈ D \ {0}. In fact, wehave been able to discover the exact conditions under which equality holds in(1). This will be accomplished in the following theorem, which strengthensBeardon’s result.Theorem 3.3. Suppose that f ∈ F1 \ Aut(D). We have:(a) The inequalityρ(f [1](0), f [1](z)) ≤ 2ρ(0, z)is satisfied for all z ∈ D.(b) The equalityρ(f [1](0), f [1](z)) = 2ρ(0, z) (4)holds for some z ∈ D \ {0} if and only if it holds for all points of adiameter of D if and only if f ∈ zAut(D).(c) The above equality holds for all z ∈ D \ {0} if and only if it holds fortwo nonzero points lying in two distinct diameters of D if and only iff is a 2-rotation.Proof. Part (a) is as in Theorem 1.1. It has been transcribed for the sakeof completeness. As for (b)-(c), if equality (4) holds for some z ∈ D \ {0}then we already know that f ∈ zAut(D) by Theorem 1.1(b). Assume nowthat f ∈ zAut(D). Then f is either a rotation kz2 (|k| = 1) or a functionof the form kzϕa(z) with 0 < |a| < 1 = |k|. Without loss of generalitywe can suppose k = 1 because ρ(kz, kw) = ρ(z, w) for all z, w ∈ D if|k| = 1. If f(z) = z2 then (4) holds on the unit disk by Theorem 1.1(c). Iff(z) = zϕa(z) with a 6= 0 then a direct computation givesf [1](z) =1− |z|21− ∣∣z a−z1−az∣∣2 · a− 2z + az2(1− az)2 . (5)7On the other hand, (4) means that12log1 +∣∣∣ f [1](0)−f [1](z)1−f [1](0)f [1](z)∣∣∣1−∣∣∣ f [1](0)−f [1](z)1−f [1](0)f [1](z)∣∣∣ = log 1 + |z|1− |z| ,which is equivalent to∣∣∣∣∣∣∣a− 1−|z|21−|z a−z1−az |2 ·a−2z+az2(1−az)21− a 1−|z|21−|z a−z1−az |2 ·a−2z+az2(1−az)2∣∣∣∣∣∣∣ =2|z|1 + |z|2 (6)due to (5) and to the fact that ψ(|z|)2) = ψ(2|z|1 + |z|2)where ψ is thefunction ψ(t) = 1+t1−t , which is one-to-one on (0, 1). The left-hand side of (6)can be written (after some minutes of heavy and careful calculations) as∣∣∣∣a(|1− az|2 − |z(a− z)|2)(1− az)− (1− |z|2)(1− az)(a− 2z + az2)(|1− az|2 − |z(a− z)|2)(1− az)− a(1− |z|2)(1− az)(a− 2z + az2)∣∣∣∣ =∣∣∣∣ (1− |a|2)(1− |z|2)((2z − a|z|2 − az2)(1− |a|2)(1− |z|2)(−az − az|z|2 + 1 + |z|2)∣∣∣∣ .Therefore, after squaring, (6) is equivalent to(2z − a|z|2 − az2)(2z − a|z|2 − az2)(1 + |z|2)2 =(−az − az|z|2 + 1 + |z|2)(−az − az|z|2 + 1 + |z|2)4|z|2.New heavy simplifications lead us to the equivalence of (6) to−2|a|2|z|4(|z|4 − 2|z|2 + 1) + (a2z2 + a2z2)(|z|6 − 2|z|4 + |z|2) = 0,or, what is the same,|z|2(1− |z|2)2[−2|a|2|z|2 + a2z2 + a2z2] = 0.If z 6= 0 (otherwise, (6) is trivial) the last equality is the same as−2aazz + (az)2 + (az)2 = 0,8that is, (az − az)2 = 0, or, equivalently, az = az. In other words, (6) holdsif and only if az ∈ R, which in turn means that z ∈ aR, that is, (6) holds ifand only if z belongs to the diameter D ∩ aR passing through a.With this we have proved (b) and the fact that f is a 2-rotation if andonly if (4) holds on all of D. The remaining of (c) is easy, for if (4) holds fortwo nonzero points lying in two distinct diameters of D then f must be inzAut(D) but it cannot be of the form kzϕa(z) with a 6= 0. Consequently, fis a 2-rotation and the theorem is proved.References[1] A.F. Beardon, The Schwarz–Pick Lemma for derivatives, Proc. Amer.Math. Soc. 125 (1997), 3255–3256.[2] J. Dieudonne´, Recherches sur quelques proble`mes relatifs aux polynoˆmeset aux fonctions borne´es d’une variable complexe, Ann. Sci. E´coleNorm. Sup. 48 (1931), 247–358.[3] W.F. Donoghe, Jr., Distributions and Fourier Transforms, AcademicPress, New York, 1966.LUIS BERNAL GONZA´LEZ MARI´A DEL CARMEN CALDERO´N MORENODEPARTAMENTO DE ANA´LISIS MATEMA´TICO DEPARTAMENTO DE ANA´LISIS MATEMA´TICOFACULTAD DE MATEMA´TICAS, APDO. 1160 FACULTAD DE MATEMA´TICAS, APDO. 1160AVENIDA REINA MERCEDES AVENIDA REINA MERCEDES41080 SEVILLA, SPAIN 41080 SEVILLA, SPAINE–mail: lbernal@us.es E–mail: mccm@us.es9

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Two hyperbolic Schwarz lemmas

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Two hyperbolic Schwarz lemmas

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