Two hyperbolic Schwarz lemmas

Two hyperbolic Schwarz lemmas by L. BERNAL-GONZ´ ALEZ and M.C. CALDER´ ON–MORENO∗ Abstract 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.1 1 Introduction and notation The Schwarz Lemma and its hyperbolic version (= the Schwarz–Pick Lemma) continues attracting the attention of many mathematicians. Our aim in this paper is to prove two sharp versions of the former results assuming that the multiplicity for the zero at the origin of the holomorphic function under consideration is fixed. Our results will extend a recent one due 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 the class of all holomorphic functions on Dand Aut(D) will stand for the group ∗The authors have been partially supported by DGES Grant PB96-1348 and the Junta de Andaluc´ıa. 12000 Mathematics Subject Classification: Primary 30F45. Secondary 30C80. Key words and phrases: Schwarz–Pick Lemma, higher order hyperbolic derivative, hyperbolic distance, multiplicity at a point, m–automorphism of the unit disk. 1 of conformal automorphisms of D. If f∈H(D) and a∈Dthen µ(f, a) will represent the multiplicity for the zero at aof the function f(z)−f(a). For m∈Nwe introduce the notations Fm={f∈H(D) : |f|<1, f(0) = f0(0) = . . . =f(m−1)(0) = 0} ={f∈H(D) : |f|<1, f(0) = 0, µ(f, 0) ≥m} and zmAut(D) = {zmf(z) : f∈Aut(D)}. For the sake of convenience, we agree that z0Aut(D) = Aut(D). We say that a function fis an m-rotation whenever there exists a constant cwith |c|= 1 such 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∈Dwe denote by ϕathe special automorphism ϕa(z) = a−z 1−az . In fact, Aut(D) = {kϕa:|a|<1 = |k|}. Note that ϕ−1 a=ϕa. We define an m-automorphism of Das a function fof the form f=ψ◦R◦ϕwith ϕ, ψ ∈Aut(D) and R∈ RmWe denote by Autm(D) the set of m-automorphisms of D. Obviously, Aut1(D) = Aut(D). It is straightforward to see that Autm(D) = {ϕb◦R◦ϕa:a, b ∈D, R ∈ Rm} and {f∈Autm(D) : f(a) = b}={ϕb◦R◦ϕa:R∈ Rm}. The symbol ρwill stand for the hyperbolic (or Poincare’s) distance on D, that is, ρ(z, w) = tanh−1|ϕz(w)|=1 2log 1 + |z−w 1−zw | 1−| z−w 1−zw |. If f:D→Dis holomorphic then the hyperbolic derivative of order mof f at zas introduced by Peschl is defined as f[m](z) = (ϕf(z)◦f◦ϕz)(m)(0). By using the fact ϕ0 a(t) = |a|2−1 (1 −at)2together with Faa di Bruno’s formula (see, for instance, [3]) for the mth derivative of a composite function it is not difficult to check that f[1](z) = (1 −|z|2)f0(z) 1−|f(z)|2 2 and 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 one factor among f0(z), . . . , f(m−1)(z). Hence if µ(f, z)≥mthen α(z) = 0, so we have f[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]|◦Twhenever Sand Tare conformal automorphisms of D. The Schwarz-Pick Lemma is a non-Euclidean version of the classical Schwarz Lemma. It asserts that ρ(f(z), f(w)) ≤ρ(z, w) and |f[1](z)| ≤ 1, for all z, w ∈Dand every f∈H(D). Furthermore, the equality ρ(f(z), f(w)) = ρ(z, w) holds for every pair z, w ∈Dif and only if this equality holds for some pair z, w ∈D(z6=w) if and only if |f[1](z)|= 1 for every z∈Dif and only if |f[1](z)|= 1 for some z∈Dif 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 below (see [1, Theorem] and notes following [1, Lemma 1]), which is a nonEuclidean version of a result due to Dieudonn´e [2] that establishes the following Schwarz Lemma for derivatives: If f∈ F1then |f0(z)| ≤ (1 if |z| ≤ √2−1 (1+|z|2)2 4|z|(1−|z|2)if |z|>√2−1. This inequality is the best possible in terms of |z|. We now transcribe the statements of Beardon in our terminology (he denoted f[1] =f∗). Recall that f[1](z)∈Dif f6∈ Aut(D). Beardon realized that this allowed to measure the hyperbolic distance between two hyperbolic derivatives. Theorem 1.1. Assume that f∈ F1\Aut(D). We have: (a) The inequality ρ(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) = z2then equality holds in (1) for all z∈D. 2 A preliminary result Before 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|<1on Dand µ(f, a)≥m. Then we have  f(z)−f(a) 1−f(a)f(z)≤ z−a 1−az  m (z∈D) (2) and |f[m](a)| ≤ m!. Further, equality holds in (2) for all z∈Dif and only if it holds for some z6=aif and only if |f[m](a)|=m!if and only if f∈Autm(D). Proof. If a∈Dsatisfies µ(f, a)≥mthen the function F(t) = (ϕb◦f◦ϕa)(t) tm(t∈D\{0}) has a holomorphic extension to Dbecause µ(ϕb◦f◦ϕa,0) = µ(f, a)≥m, where we have denoted b=f(a). Fix r∈(0,1). Then |F(t)| ≤ 1 rmon |t|=r, hence an application of the Maximum Modulus Principle yields sup{|F(t)|:|t| ≤ r} ≤ 1 rm. Letting r→1 we get sup{|F(t)|:|t|<1} ≤ 1, that is, |F(t)| ≤ 1 on Dor, equivalently, |ϕb◦f◦ϕa(t)|≤|t|m(t∈D), which becomes (2) after the change of variable z=ϕa(t). Note that the value of (the extension of) Fat the origin is F(0) = lim t→0 1 tm·b−f(ϕa(t)) 1−bf(ϕa(t)) = lim z→a (1 −az)m 1−f(a)f(z)·f(z)−f(a) (z−a)m·(−1)m+1 = (−1)m+1 ·(1 −|a|2)m 1−|f(a)|2·lim z→a f(z)−f(a) (z−a)m=f[m](a) m!, after using the L’Hopital rule together with the definition of the hyperbolic derivative of order mand the fact that f0(a) = . . . =f(m−1)(a) = 0. Anew by the Maximum Modulus Principle, |F(0)| ≤ 1, whence |f[m](a)| ≤ m!. 4 Assume now that fis an m-automorphism. Since f(a) = bwe have f=ϕb◦R◦ϕawhere R∈ Rm, i.e., R(t) = ctmfor some cwith |c|= 1. Therefore F(t) = con D, so |F|= 1 on Dand the equality holds in (2) for all z∈D(hence for some z6=a). Moreover, |f[m](a)|=m!·|F(0)|=m!·|c|= m!. Conversely, suppose that |f[m](a)|=m!. Then |F(0)|= 1 and the Maximum Modulus Principle tells us that F(t) = cfor some unimodular constant c, but this yields (ϕb◦f◦ϕa)(t) = ctmfor all t∈D, which in turn implies that f=ϕb◦R◦ϕawith Ras before. Consequently, fis an m-automorphism of D. Finally, assume that equality holds in (2) for some z6=a. Then the change z=ϕa(t) shows that |ϕb◦f◦ϕa(t)|=|t|mfor some t6= 0, whence |F(t)|= 1 for some t∈D. Another application of the Maximum Modulus Principle drives us to F(t) = con Dfor some unimodular constant c, and this implies as above that f∈Autm(D). This concludes the proof. 3 Main results We are now ready to state our theorems. Like in [1], we can estimate the hyperbolic distance between two normalized hyperbolic derivatives of higher order under obvious conditions. Theorem 3.1. Assume that f∈ Fm\Autm(D). We have: (a) If a∈Dand µ(f, a)≥mthen ρ((−1)m+1 f[m](0) m!,f[m](a) m!)≤2ρ(0, a).(3) (b) If there exists a∈D\ {0}for which µ(f, a)≥msuch that equality holds in (3) then f∈zmAut(D). Proof. Observe that if µ(f, a)≥mand f∈ Fm\Autm(D) then the values (−1)m+1 f[m](0) m!,f[m](a) m!are in Dby Lemma 2.1, so the hyperbolic distance between them makes sense. As for (a), since µ(f, a)≥m≤µ(f, 0) the functions g(z) = f(z) zm(z∈D\{0}) 5 and h(z) = 1−az a−zm ·f(a)−f(z) 1−f(a)f(z)(z∈D\{a}) have holomorphic extensions on the whole Dif we set g(0) = f(m)(0) m!= (−1)m+1 f[m](0) m! and h(a) = lim z→ah(z) = (1 −|a|2)m 1−|f(a)|2·(−1)m+1 ·f(m)(a) m!=f[m](a) m!. We may start with a6= 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 Dsince fis not an mautomorphism. On the other hand, g(a) = f(a) amand h(0) = f(a) am. If we apply the Schwarz-Pick Lemma to gand hthen 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), the triangle inequality yields ρ((−1)m+1 f[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) holds for some a∈D\{0}with µ(f, a)≥m. Then 2ρ(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 not greater 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 mis even and f∈ Fm\Autm(D)then f(m)(0) = 0. In other words, f∈ Fm+1. Proof. From (3) and the fact that µ(f, 0) ≥mwe obtain ρ(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. 6 It should be noted that parts (a)-(b) of Theorem 1.1 are covered with the case m= 1 in Theorem 3.1 (observe that always µ(f, a)≥1). An extension of 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 Dexcept for the trivial case f≡0. In view 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, we have been able to discover the exact conditions under which equality holds in (1). This will be accomplished in the following theorem, which strengthens Beardon’s result. Theorem 3.3. Suppose that f∈ F1\Aut(D). We have: (a) The inequality ρ(f[1](0), f[1](z)) ≤2ρ(0, z) is satisfied for all z∈D. (b) The equality ρ(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 a diameter of Dif and only if f∈zAut(D). (c) The above equality holds for all z∈D\{0}if and only if it holds for two nonzero points lying in two distinct diameters of Dif and only if fis a 2-rotation. Proof. Part (a) is as in Theorem 1.1. It has been transcribed for the sake of 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 now that f∈zAut(D). Then fis either a rotation kz2(|k|= 1) or a function of the form kzϕa(z) with 0 <|a|<1 = |k|. Without loss of generality we can suppose k= 1 because ρ(kz, kw) = ρ(z, w) for all z,w∈Dif |k|= 1. If f(z) = z2then (4) holds on the unit disk by Theorem 1.1(c). If f(z) = zϕa(z) with a6= 0 then a direct computation gives f[1](z) = 1−|z|2 1−za−z 1−az  2·a−2z+az2 (1 −az)2.(5) 7 On the other hand, (4) means that 1 2log 1 +  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|2 1−|za−z 1−az |2·a−2z+az2 (1−az)2 1−a1−|z|2 1−|za−z 1−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 ψis the function ψ(t) = 1+t 1−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 z6= 0 (otherwise, (6) is trivial) the last equality is the same as −2aazz + (az)2+ (az)2= 0, 8 that is, (az −az)2= 0, or, equivalently, az =az. In other words, (6) holds if and only if az ∈R, which in turn means that z∈aR, that is, (6) holds if and only if zbelongs to the diameter D∩aRpassing through a. With this we have proved (b) and the fact that fis a 2-rotation if and only if (4) holds on all of D. The remaining of (c) is easy, for if (4) holds for two nonzero points lying in two distinct diameters of Dthen fmust be in zAut(D) but it cannot be of the form kzϕa(z) with a6= 0. Consequently, f is 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. Dieudonn´e, Recherches sur quelques probl`emes relatifs aux polynˆomes et aux fonctions born´ees d’une variable complexe, Ann. Sci. ´ Ecole Norm. Sup. 48 (1931), 247–358. [3] W.F. Donoghe, Jr., Distributions and Fourier Transforms, Academic Press, New York, 1966. LUIS BERNAL GONZ´ ALEZ MAR´ IA DEL CARMEN CALDER´ ON MORENO DEPARTAMENTO DE AN´ ALISIS MATEM´ ATICO DEPARTAMENTO DE AN´ ALISIS MATEM´ ATICO FACULTAD DE MATEM´ ATICAS, APDO. 1160 FACULTAD DE MATEM´ ATICAS, APDO. 1160 AVENIDA REINA MERCEDES AVENIDA REINA MERCEDES 41080 SEVILLA, SPAIN 41080 SEVILLA, SPAIN E–mail: lb[email protected] E–mail: [email protected] 9