Research ArticleDistortion Type Theorems for Functions inthe Logarithmic Bloch Space

Armando J. García-Ortíz,1 Milton del Castillo Lesmes Acosta,2

and Julio C. Ramos-Fernández3

1 Área de Matemática, Universidad Nacional Experimental de Guayana, Pto. Ordaz, Boĺıvar, Venezuela2Proyecto Curricular de Matemáticas, Facultad de Ciencias y Educación, Universidad Distrital Francisco José de Caldas,Carrera 3 No. 26 A-40, Bogotá, Colombia3Departamento de Matemática, Universidad de Oriente, 6101 Cumaná, Sucre, Venezuela

Correspondence should be addressed to Julio C. Ramos-Fernández; jcramos@udo.edu.ve

Received 23 January 2017; Revised 22 March 2017; Accepted 2 April 2017; Published 16 April 2017

Academic Editor: John R. Akeroyd

Copyright © 2017 Armando J. Garćıa-Ort́ız et al.This is an open access article distributed under theCreativeCommonsAttributionLicense, which permits unrestricted use, distribution, and reproduction in anymedium, provided the originalwork is properly cited.

We establish distortion type theorems for locally schlicht functions and for functions having branch points satisfying a normalizedBloch condition in the closed unit ball of the logarithmic Bloch spaceBlog. As a consequence of our results we have estimations ofthe schlicht radius for functions in these classes.

1. Introduction

One of the most important results in the area of geometrictheory of functions of a complex variable is the celebrateddistortion’s theorem established by Koebe and Bieberbach[1, 2] at the beginning of the twentieth century. Koebe andBieberbach showed that the range of any function 𝑓 in theclass S of all conformal functions on D, the open unit diskof the complex plane C, normalized such that 𝑓(0) = 0 =𝑓(0) − 1 contain the Euclidean disk with center at the originand radius 1/4. This last result is today known as Koebe 1/4Theorem and, in particular, shows that Bloch’s constant (see[3]) is greater than or equal to 1/4. Koebe and Bieberbachfound sharp lower and upper bounds for the growth and thedistortion of conformal maps in the class S; more precisely,they showed that for any 𝑓 ∈ S and 𝑧 ∈ D the followingestimations hold.

(1) Growth theorem:|𝑧|

(1 + |𝑧|)2 ≤ 𝑓 (𝑧) ≤|𝑧|

(1 − |𝑧|)2 (1)(2) Distortion theorem:

1 − |𝑧|(1 + |𝑧|)3 ≤

𝑓 (𝑧) ≤ 1 + |𝑧|(1 − |𝑧|)3 (2)

with equality if and only if 𝑓 is a rotation of the Koebefunction defined by

𝐾 (𝑧) = 𝑧(1 − 𝑧)2 , (𝑧 ∈ D) , (3)which also belongs to the classS. In particular, the distortiontheorem implies that the class S is contained in the closedball with center at the origin and radius 8 of 𝑝-Bloch spaceB𝑝 for all 𝑝 ≥ 3 (see Section 3 for the definition ofB𝑝). Formore properties of conformal maps and distortion theorem,we recommend the excellent books [4, 5].

Although the distortion theorem gives sharp bounds forthe modulus of the derivative of functions in the class S,it cannot be applied to the bigger class of locally schlichtfunctions defined on D satisfying the normalized Blochconditions 𝑓(0) = 0 = 𝑓(0) − 1 (recall that a holomorphicfunction 𝑓 is locally schlicht on D if 𝑓(𝑧) ̸= 0 for all 𝑧 ∈D). Many authors have obtained distortion type theorems orlower bounds for the modulus or real part of the derivative oflocally schlicht functions in Bloch-type spaces. The pioneerwork about this subject appears in 1992 and is due to Liu andMinda [6]. They established distortion theorems for locallyschlicht functions 𝑓 in the classical Bloch spaceB satisfyingthe conditions 𝑓(0) = 0, 𝑓(0) = 1, and ‖𝑓‖B = 1 (seeSection 3 for the definition of Bloch space). Liu and Minda

HindawiJournal of Function SpacesVolume 2017, Article ID 8694516, 10 pageshttps://doi.org/10.1155/2017/8694516

https://doi.org/10.1155/2017/8694516

2 Journal of Function Spaces

give sharp lower bounds for |𝑓(𝑧)| and for Re𝑓(𝑧) and asconsequence of their results they obtain a lower bound forBloch’s constant.Determination of the (locally schlicht) Blochconstant is still an open problem. By Landau’s reduction, itis enough to consider those functions with Bloch seminormnot greater than 1. Hence, it is important to consider certainsubclasses of functions in Bloch spaces having seminorm notgreater than 1.

The results of Liu and Minda in [6] have been extendedto other classes of locally schlicht functions or to functionshaving branch points in the Bloch space by Yanagihara [7],Bonk et al. [8, 9], and Graham andMinda [10].The extensionof the above results to 𝑝-Bloch spaces was obtained by Teradaand Yanagihara [11] and by Zheng and Wang [12]. It is anopen problem to obtain distortion type theorems for locallyschlicht functions in other spaces of analytic functions.

In this article we extend the results of Liu and Minda[6] to the logarithmic Bloch space Blog which we define inSection 3; we obtain lower bounds for the modulus and thereal part of the derivative of locally schlicht functions and forfunctions having branch points in the closed unit ball ofBlogsatisfying a normalized Bloch condition𝑓(0) = 0 = 𝑓(0)−1.Our resultswill be showed in Sections 4 and 5, as consequenceof our results, in Section 6, we obtain lower bounds for theschlicht radius of functions in these classes.

2. Some Preliminaries: Julia’s Lemma

In this section we gather some notations, definitions, andresults that we will need through this note. We denote by Dthe open unit disk in the complex plane C, with center atthe origin and radius 1; 𝜕D denotes the boundary of D. Thespace of all complex and holomorphic functions on D, as isusual, is denoted by𝐻(D). A function 𝑓 ∈ 𝐻(D) is said to benormalized if 𝑓(0) = 0 and 𝑓(0) = 1 and 𝑓 is locally schlichtor locally univalent if 𝑓(𝑧) ̸= 0 for all 𝑧 ∈ D. A point 𝑧0 is abranch point for 𝑓 if 𝑓(𝑧0) = 0. For 𝑟 > 0, we define

Δ (1, 𝑟) = {𝑧 ∈ D : |1 − 𝑧|21 − |𝑧|2 < 𝑟} . (4)Δ(1, 𝑟) is known as a horodisk D; that is, it is an Euclideandisk contained inDwhich is tangent to 𝜕D at 1. Furthermore,Δ(1, 𝑟) has center at 1/(1+𝑟) and radius 𝑟/(1+𝑟). The closureof Δ(1, 𝑟) relative toD is denoted by Δ(1, 𝑟). Observe that 1 ∉Δ(1, 𝑟) but (1 − 𝑟)/(1 + 𝑟) ∈ Δ(1, 𝑟). With these notations, wecan enunciate the well known Julia’s Lemma; the reader canconsult the excellent book of Ahlfors [13] for its proof.

Lemma 1 (Julia’s Lemma). Suppose that 𝑤 is a complex andholomorphic function onD∪{1} such that𝑤mapsD intoH+ ={𝑧 ∈ C : Re(𝑧) > 0}, the right half-plane, and 𝑤(1) = 0. Then,for any 𝑟 > 0, the function𝑤maps the horodisk Δ(1, 𝑟) into theEuclidean disk {𝑧 ∈ C : |𝑧 − 𝑑𝑟| < 𝑑𝑟}, where 𝑑 = −𝑤(1) > 0.Furthermore, a boundary point of the first disk is mapped onthe boundary of the second disk if and only if 𝑤 is a conformalfunction mapping D onto H+ and satisfying 𝑤(1) = 0.

In 1992, Liu andMinda [6] established distortion theoremfor functions in the Bloch space; they showed the followingresults which are consequences of Julia’s Lemma. We includethe proof of the first one to illustrate the application of Julia’sLemma.

Lemma2 ([6, corollary in Section 1]). Let𝑤 be a holomorphicfunction onD∪{1}. Suppose that𝑤mapsD into the right half-plane H+ and that 𝑤(1) = 0. Then 𝑑 = −𝑤(1) > 0 and

Re𝑤 (𝑥) ≤ 2𝑑1 − 𝑥1 + 𝑥 , (5)for all 𝑥 ∈ (−1, 1), with equality for some 𝑥 ∈ (−1, 1) if andonly if

𝑤 (𝑧) = 2𝑑1 − 𝑧1 + 𝑧 , (6)for all 𝑧 ∈ D.Proof. Indeed, let us fix 𝑥 ∈ (−1, 1); then 𝑟 = (1−𝑥)/(1+𝑥) >0 and by Julia’s Lemma, 𝑤 maps Δ(1, 𝑟) into the Euclideandisk 𝐷(𝑑𝑟, 𝑑𝑟). In particular, since 𝑥 ∈ Δ(1, 𝑟), then 𝑤(𝑥) ∈𝐷(𝑑𝑟, 𝑑𝑟); this fact implies that Re𝑤(𝑥) ≤ 2𝑑𝑟. Furthermore,𝑥 ∈ 𝜕Δ(1, 𝑟); hence if Re𝑤(𝑥) = 2𝑑𝑟, then we conclude that𝑤(𝑥) = 2𝑑𝑟 ∈ 𝜕𝐷(𝑑𝑟, 𝑑𝑟) and, by Julia’s Lemma, this last factoccurs if 𝑤 is the conformal map from D onto H+ such that𝑤(1) = 0; that is,𝑤(𝑧) = 2𝑑((1−𝑧)/(1+𝑧)) for all 𝑧 ∈ D. Thisshows the lemma.

Lemma 3 ([6, corollary to Theorem 3]). Let 𝑓 be a holomor-phic function onD∪{1}. Suppose that𝑓(D) ⊂ D,𝑓(1) = 1 andthat all the zeros of 𝑓 have multiplicity at least 𝑛. If 𝑓(1) = 𝑛,then

(1) |𝑓(𝑥)| ≥ 𝑥𝑛 for all 𝑥 ∈ [0, 1), with equality for some𝑥 ∈ [0, 1) if and only if 𝑓(𝑧) = 𝑧𝑛 for all 𝑧 ∈ D;(2) Re(𝑓(𝑥)) ≥ 𝑥𝑛 for all (𝑛 − 1)/(𝑛 + 1) ≤ 𝑥 < 1, with

equality for some 𝑥 ∈ [(𝑛 − 1)/(𝑛 + 1), 1) if and only if𝑓(𝑧) = 𝑧𝑛 for all 𝑧 ∈ D.We finish this section by establishing the following ele-

mentary property of the complex exponential. We thank thereviewer for providing us the following simple demonstrationof this fact.

Lemma 4. Let 𝑥 ∈ [0, 1) be fixed and 𝐷𝑥 the Euclidean diskwith center at (1 − 𝑥)/(1 + 𝑥) and radius (1 − 𝑥)/(1 + 𝑥); then

min {Re (exp (−𝑧)) : 𝑧 ∈ 𝐷𝑥} = exp (−21 − 𝑥1 + 𝑥) . (7)Proof. Let 𝑟 = (1 − 𝑥)/(1 + 𝑥) for simplicity and let 𝑓(𝑧) =𝑒𝑟(𝑧−1). Since 1 + 𝑧𝑓(𝑧)/𝑓(𝑧) = 1 − 𝑟𝑧 has positive real parton |𝑧| < 1, the function𝑓 is convex. In particular, Re(𝑓(𝑧)) >𝑓(−1) = 𝑒−2𝑟, which proves the assertion.3. Logarithmic Bloch Space

In this section we gather the definition and some of theproperties of the logarithmic Bloch space Blog. Let us recall

Journal of Function Spaces 3

that a function weight 𝜇 on D is a bounded, positive, andcontinuous function defined on D. Given a weight 𝜇 on D,𝜇-Bloch space, denoted by B𝜇, consists of all holomorphicfunctions 𝑓 on D such that

𝑓𝜇 fl sup𝑧∈D

𝜇 (𝑧) 𝑓 (𝑧) < ∞. (8)It is known that if the weight 𝜇 is radial, that is, 𝜇(𝑧) = 𝜇(|𝑧|)for all 𝑧 ∈ D, then B𝜇 is a Banach space with the norm‖𝑓‖B𝜇 = |𝑓(0)| + ‖𝑓‖𝜇. When 𝜇(𝑧) = 1− |𝑧|2, with 𝑧 ∈ D,B𝜇becomes the Bloch space which is denoted byB, while when𝜇(𝑧) = (1 − |𝑧|2)𝑝, with 𝑧 ∈ D and 𝑝 > 0 fixed, we obtain𝑝-Bloch space which is denoted byB𝑝.

Clearly, the function 𝜇log, defined by𝜇log (𝑧) = [log( 𝑒1 − |𝑧|2)]

−1 , (9)defines a weight on D. Hence, the space Blog = B𝜇log is aBanach space with the norm

𝑓Blog = 𝑓 (0) + 𝑓logfl 𝑓 (0) + sup

𝑧∈D

𝑓 (𝑧)log (𝑒/ (1 − |𝑧|2)) .

(10)

We callBlog as the logarithmic Bloch space. In the next resultwe are going to show that Blog is a subspace of B𝑝 for all𝑝 ≥ 1.Proposition 5. The spaceBlog is contained inB𝑝, for all 𝑝 ≥1. Furthermore, 𝑓B𝑝 ≤ 𝑓Blog , (11)for all function 𝑓 ∈ Blog.Proof. It is enough to show that for 𝑝 ≥ 1 fixed

(1 − |𝑧|2)𝑝 log( 𝑒1 − |𝑧|2) ≤ 1, (12)for all 𝑧 ∈ D. But, this last inequality is true since the function

ℎ (𝑡) = 𝑡𝑝 log(𝑒𝑡 ) − 1, (13)with 𝑡 ∈ (0, 1], is increasing and ℎ(1) = 0.

Also, we have the following very useful identity (seeLemma 3.3 in [12]).

Lemma 6. If 𝑓 ∈ Blog, 𝑓(0) = 0, 𝑓(0) = 1, and ‖𝑓‖log ≤ 1,then 𝑓(0) = 0.Proof. Suppose that 𝑓 ∈ Blog, 𝑓(0) = 0, 𝑓(0) = 1, and‖𝑓‖log ≤ 1. Then, for each 𝑧 ∈ D, we have

𝑓 (𝑧) ≤ log( 𝑒1 − |𝑧|2) . (14)

Taylor’s theorem implies that

1 + 𝑓 (0) 𝑧 + 𝑜 (|𝑧|)2 ≤ log2 ( 𝑒1 − |𝑧|2)= (1 + |𝑧|2 + 𝑜 (|𝑧|))2 ,

(15)

as 𝑧 → 0. But since1 + 𝑓 (0) 𝑧 + 𝑜 (|𝑧|)2

= 1 + 2Re (𝑓 (0) 𝑧) + 𝑜 (|𝑧|) , (16)as 𝑧 → 0, and

(1 + |𝑧|2 + 𝑜 (|𝑧|))2 = 1 + 𝑜 (|𝑧|) , (17)as 𝑧 → 0, we obtain from (15) that

2Re (𝑓 (0) 𝑧) ≤ 𝑜 (|𝑧|) , (18)as 𝑧 → 0. Now, if we consider 𝑧 = 𝑟𝑓(0)/|𝑓(0)| with 𝑟 > 0small in (18), we conclude |𝑓(0)| = 0 and we are done.

The following functions play a very important role in ourwork; theywill be used to get lower bounds for locally schlichtfunctions and for functions having branch points in certainclasses in the logarithmic Bloch space. From now, we uselog(𝑤) to denote the principal logarithmic of the complexnumber 𝑤 ̸= 0. Observe that the principal logarithmic isa holomorphic function on 𝐷(1, 1), the Euclidean disk withcenter at 1 and radius 1:

(1) For each 𝑛 ∈ N, we set𝐹𝑛 (𝑧) = ∫𝑧

0(1 − 𝑠/𝑎𝑛1 − 𝑎𝑛𝑠 )

𝑛 (1 − 2 log (1 − 𝑎𝑛𝑠)) 𝑑𝑠, (19)where 𝑎𝑛 = √𝑛/(𝑛 + 2) and 𝑧 ∈ D. Clearly, 𝐹𝑛 ∈ 𝐻(D) for all𝑛 ∈ N, 𝐹𝑛(0) = 0, and 𝐹𝑛(0) = 1.

(2) For 𝑧 ∈ D, we define𝐹 (𝑧) = ∫𝑧

0exp (− 2𝑠1 − 𝑠) (1 − 2 log (1 − 𝑠)) 𝑑𝑠. (20)

We can see that 𝐹 ∈ 𝐻(D), 𝐹(0) = 0, and 𝐹(0) = 1.Also we have that the function 𝐹 satisfies the following

properties.

Proposition 7. The function 𝐹 belongs toBlog. Furthermore,sup𝑧∈D𝜇log(𝑧)|𝐹(|𝑧|)| = 1 but ‖𝐹‖log > 1.Proof. We see that ‖𝐹‖log > 1. Indeed, we have

‖𝐹‖log ≥𝐹 (𝑖/2)1 − log (1 − |𝑖/2|2)

= exp (2/5)1 − log (3/4)√(1 − log(54))2 + 4 arctan2 (12)

≈ 1.4014837 > 1.

(21)

4 Journal of Function Spaces

Now, we are going to show that 𝐹 ∈ Blog. Since thefunction exp(−2𝑧/(1 − 𝑧)) is holomorphic on D, then themodulus maximum principle tells us that its maximum valueis attained in the boundary 𝜕D. But if |𝑧| = 1 then |1 − 𝑧|2 =2(1 − Re(𝑧)) and hence

sup𝑧∈D

exp(− 2𝑧1 − 𝑧) = sup|𝑧|=1 exp(2 −

2 (1 − Re (𝑧))|1 − 𝑧|2 )

= 𝑒.(22)

On the other hand, for each 𝑧 ∈ D, we have |arg(1−𝑧)| < 𝜋/2and

sup𝑧∈D

arg (1 − 𝑧)1 − log (1 − |𝑧|2) ≤𝜋2 . (23)

Furthermore, using elementary calculus, we can see that thereal function𝐻(𝑡) = 𝑒2 − (1 − 𝑡)(1 + 𝑡)3 is nonnegative for all𝑡 ∈ [0, 1] (its minimum value is𝐻(1/2) = 𝑒2 − 27/16 ≈ 5.70).Hence for any 𝑡 ∈ [0, 1) we obtain

1 − 2 log (1 − 𝑡) − 3 (1 − log (1 − 𝑡2))= log((1 − 𝑡) (1 + 𝑡)3𝑒2 ) ≤ 0.

(24)

This last implies that

1 − 2 log (1 − 𝑡)1 − log (1 − 𝑡2) ≤ 3, (25)for all 𝑡 ∈ [0, 1). We conclude that for any 𝑧 ∈ D such that|1 − 𝑧|2 ≤ 𝑒

log (𝑒/ |1 − 𝑧|2)1 − log (1 − |𝑧|2) =1 − 2 log (|1 − 𝑧|)1 − log (1 − |𝑧|2)

≤ 1 − 2 log (1 − |𝑧|)1 − log (1 − |𝑧|2) ≤ 3,(26)

while for 𝑧 ∈ D such that |1 − 𝑧|2 > 𝑒 we havelog (𝑒/ |1 − 𝑧|2)1 − log (1 − |𝑧|2) =

log (|1 − 𝑧|2 /𝑒)1 − log (1 − |𝑧|2) ≤ log (4) − 1. (27)

These last inequalities, (22) and (23), imply that

‖𝐹‖log ≤ 𝑒 (3 + log (4) − 1 + 2 (𝜋2 ))= 𝑒 (2 + log (4) + 𝜋)

(28)

which shows that 𝐹 ∈ Blog.Now, we are going to show that sup𝑧∈D𝜇log(𝑧)|𝐹(|𝑧|)| = 1.

Observe that 𝜇log(0)|𝐹(0)| = 1. Also the real function𝐻(𝑡) =exp(−2𝑡/(1−𝑡))(1−2log(1−𝑡))−1+ log(1−𝑡2)with 𝑡 ∈ [0, 1)

satisfies 𝐻(0) = 0, 𝐻(𝑡) → −∞ as 𝑡 → 1− and it is strictlydecreasing since

𝐻 (𝑡)= − 2𝑡1 − 𝑡2

− 21 − 𝑡 ( 21 − 𝑡 log( 𝑒(1 − 𝑡)2) − 1) exp (−2𝑡1 − 𝑡)

< 0,

(29)

for all 𝑡 ∈ [0, 1). Hence we conclude that 𝐻(𝑡) ≤ 0 for all𝑡 ∈ [0, 1) which shows the affirmation.For the sequence {𝐹𝑛}, we have the following properties.

Proposition 8. Functions 𝐹𝑛 with 𝑛 ∈ N belong to Blog andsatisfy

lim𝑛→∞

𝐹𝑛 (𝑧) = 𝐹 (𝑧) , (30)for each 𝑧 ∈ D. Furthermore, for each 𝑛 ∈ N ‖𝐹𝑛‖log > 1, infact, sup𝑧∈D𝜇log(𝑧)|𝐹𝑛(|𝑧|)| > 1.Proof. Clearly, for any 𝑛 ∈ N, the function 𝐹𝑛 belongsto Blog since 𝐹𝑛 ∈ 𝐻(D). We are going to show thatsup𝑧∈D𝜇log(𝑧)|𝐹𝑛(|𝑧|)| > 1. It is enough to show that thereexists a 𝑡0 ∈ (𝑎𝑛, 1) such that𝐻(𝑡0) > 0, where

𝐻(𝑡) = (𝑡/𝑎𝑛 − 11 − 𝑎𝑛𝑡 )𝑛

log( 𝑒(1 − 𝑎𝑛𝑡)2)− log( 𝑒1 − 𝑡2 ) ,

(31)

with 𝑡 ∈ (𝑎𝑛, 1). Observe that, for 𝑡𝑛 = 𝑟𝑎𝑛 with 𝑟 = 2/(1 +𝑎2𝑛) > 1, we have 𝑡𝑛 ∈ (𝑎𝑛, 1), 𝑟 − 1 = 1 − 𝑟𝑎2𝑛 , and 1 − 𝑟2𝑎2𝑛 =(1 − 𝑟𝑎2𝑛)2 which implies that𝐻(𝑡𝑛) = 0. Also,𝐻 (𝑡)

= 𝑛 (𝑡/𝑎𝑛 − 11 − 𝑎𝑛𝑡 )𝑛−1 1/𝑎𝑛 − 𝑎𝑛(1 − 𝑎𝑛𝑡)2 log(

𝑒(1 − 𝑎𝑛𝑡)2)

+ (𝑡/𝑎𝑛 − 11 − 𝑎𝑛𝑡 )𝑛 2𝑎𝑛1 − 𝑎𝑛𝑡 −

2𝑡1 − 𝑡2 ;(32)

hence

𝐻 (𝑡𝑛) = 𝑛 1/𝑎𝑛 − 𝑎𝑛(1 − 𝑟𝑎2𝑛)2 log(𝑒

(1 − 𝑟𝑎2𝑛)2) +2𝑎𝑛1 − 𝑟𝑎2𝑛

− 2𝑟𝑎𝑛1 − 𝑟2𝑎2𝑛 =𝑛

(1 − 𝑟𝑎2𝑛)2 (1𝑎𝑛 − 𝑎𝑛)

⋅ (log( 𝑒(1 − 𝑟𝑎2𝑛)2) −1𝑛

2𝑎2𝑛1 − 𝑎2𝑛)

Journal of Function Spaces 5

= 𝑛(1 − 𝑟𝑎2𝑛)2 (1𝑎𝑛 − 𝑎𝑛)

⋅ (log( 𝑒(1 − 𝑟𝑎2𝑛)2) − 1) ,(33)

since 𝑟 − 1 = 1 − 𝑟𝑎2𝑛 , 1 − 𝑟2𝑎2𝑛 = (1 − 𝑟𝑎2𝑛)2, and 𝑟(𝑎2𝑛 + 1) = 2and we have used that 𝑎𝑛 = √𝑛/(𝑛 + 2) in the last equality.Thus, we conclude that 𝐻(𝑡𝑛) > 0 and since 𝐻(𝑡𝑛) = 0, thenthere exists 𝑡0 ∈ (𝑡𝑛, 1) such that 𝐻(𝑡0) > 0. This shows theaffirmation. The other properties of 𝐹𝑛’s are clear.4. A Distortion Theorem for Locally SchlichtFunctions inBlog

In this section we establish a distortion theorem for locallyschlicht functions in the closed unit ball of Blog satisfyingnormalized Bloch conditions. We denote by 𝛽(∞)log the class ofall holomorphic functions 𝑓 ∈ Blog such that 𝑓 is locallyschlicht, 𝑓(0) = 0, 𝑓(0) = 1, and ‖𝑓‖log ≤ 1. With thesenotations, we have the following result.

Theorem 9. If 𝑓 ∈ 𝛽(∞)log then we have the following:(1) |𝑓(𝑧)| ≥ 𝐹(|𝑧|) = exp(−2|𝑧|/(1 − |𝑧|))(1 − 2 log(1 −|𝑧|)) for all 𝑧 ∈ D. There is not a function 𝑓0 ∈ 𝛽(∞)log

such that |𝑓0(𝑧0)| = 𝐹(|𝑧0|) for some 𝑧0 ∈ D \ {0}.(2) Re𝑓(𝑧) ≥ 𝐹(|𝑧|) = exp(−2|𝑧|/(1 − |𝑧|))(1 − 2log(1 −|𝑧|)) for all |𝑧| ≤ 1/2. There is not a function 𝑓0 ∈ 𝛽(∞)log

such thatRe𝑓0(𝑧0) = 𝐹(|𝑧0|) for some 𝑧0 ∈ 𝐷(0, 1/2)\{0}.Proof. (1) Suppose that 𝑓 ∈ 𝛽(∞)log . Let us fix |𝜁| = 1 and we setthe function

𝑔 (𝑢) = (1 − 2 log(1 + 𝑢2 ))−1 𝑓 (1 − 𝑢2 𝜁) , (34)

with 𝑢 ∈ D, where log(𝑤) denotes the principal logarithmicof 𝑤 ∈ 𝐷(1, 1). Clearly 𝑔 is holomorphic on D \ {−1} and𝑔(1) = 1 because 𝑓(0) = 1. Since 𝑓 is locally schlicht on D,we have that 𝑔(𝑧) ̸= 0 for all 𝑧 ∈ D. Furthermore,

𝑔 (𝑢)= 21 + 𝑢 (1 − 2 log(1 + 𝑢2 ))

−2 𝑓 (1 − 𝑢2 𝜁)+ (1 − 2 log(1 + 𝑢2 ))

−1 𝑓 (1 − 𝑢2 𝜁)(−𝜁2) .(35)

In particular, 𝑔(1) = 1 since 𝑓(0) = 1 and 𝑓(0) = 0 (byLemma 6). Also, for any 𝑢 ∈ D, we have

𝑔 (𝑢) = 1 − 2 log(1 + 𝑢2 )−1 𝑓 (1 − 𝑢2 𝜁)

≤ 1Re (1 − 2 log ((1 + 𝑢) /2))⋅ log( 𝑒1 − |(1 − 𝑢) /2|2) =

1log (𝑒/ |(1 + 𝑢) /2|2)

⋅ log (𝑒/1 − |(1 − 𝑢) /2|2) < 1

(36)

and 𝑔mapsD intoD \ {0}. Hence, there exists a holomorphicfunction 𝑤mapping the unit disk D into the right half-planeH+ = {𝑧 : Re(𝑧) > 0} and such that 𝑔(𝑢) = exp{−𝑤(𝑢)}for all 𝑢 ∈ D. Observe that 𝑤(1) = 0 since 𝑔(1) = 1 and𝑑 = −𝑤(1) = 1. Invoking Lemma 2, it follows that

Re (𝑤 (𝑥)) ≤ 21 − 𝑥1 + 𝑥 , (37)for all 𝑥 ∈ (−1, 1). Hence𝑔 (𝑥) = exp {−𝑤 (𝑥)} = exp (−Re {𝑤 (𝑥)})

≥ exp(−21 − 𝑥1 + 𝑥) ,(38)

for all 𝑥 ∈ (−1, 1). That is,(1 − 2 log(1 + 𝑥2 ))

−1 𝑓 (1 − 𝑥2 𝜁)

≥ exp(−21 − 𝑥1 + 𝑥) ,(39)

for all 𝑥 ∈ (−1, 1).Making the change 𝑟 = (1−𝑥)/2, we obtain that 𝑟 ∈ (0, 1)

and 𝑓 (𝑟𝜁) ≥ exp(−2 𝑟1 − 𝑟) (1 − 2 log (1 − 𝑟)) . (40)Therefore, if we consider 𝑟 = |𝑧| with 𝑧 ∈ D ̸= 0 and we take𝜁 = 𝑧/|𝑧|, we conclude that

𝑓 (𝑧) ≥ exp(−2 |𝑧|1 − |𝑧|) (1 − 2 log (1 − |𝑧|))= 𝐹 (|𝑧|) .

(41)

This shows inequality (1).Now, if there exists𝑓0 ∈ 𝛽(∞)log such that |𝑓0(𝑧0)| = 𝐹(|𝑧0|)

for some 𝑧0 ∈ D\{0}, then arguing as in the proof of inequality(1), for 𝜁0 = 𝑧0/|𝑧0|, the function,

𝑔0 (𝑧) = (1 − 2 log(1 + 𝑧2 ))−1 𝑓0 (1 − 𝑧2 𝜁0) , (42)

maps D into D \ {0}. Hence, there exists a holomorphicfunction 𝑤 mapping D into H+ such that 𝑤(1) = 0, −𝑤(1) =1 > 0, and

𝑔0 (𝑧) = exp (−𝑤 (𝑧)) , (43)

6 Journal of Function Spaces

for all 𝑧 ∈ D. In particular, for 𝑥 = 1−2|𝑧0| ∈ (−1, 1), we haveexp (−Re𝑤 (𝑥)) = 𝑔0 (𝑥) =

𝑓0 (((1 − 𝑥) /2) 𝜁0)1 − 2 log ((1 + 𝑥) /2)=

𝑓0 (𝑧0)1 − 2 log (1 − 𝑧0)= 𝐹 (𝑧0)1 − 2 log (1 − 𝑧0)= exp(− 2 𝑧01 − 𝑧0)= exp(−21 − 𝑥1 + 𝑥) .

(44)

Thus,

Re𝑤 (𝑥) = 21 − 𝑥1 + 𝑥 , (45)for some 𝑥 ∈ (−1, 1) and, by Lemma 2, we conclude that

𝑤 (𝑧) = 21 − 𝑧1 + 𝑧 (46)for all 𝑧 ∈ D. Therefore,

𝑓0 (1 − 𝑧2 𝜁0) = (1 − 2 log(1 + 𝑧2 ))𝑔0 (𝑧)= (1 − 2 log(1 + 𝑧2 )) exp(−21 − 𝑧1 + 𝑧) ,

(47)

for all 𝑧 ∈ D. Hence, changing (1−𝑧)/2 by 𝑧, which belongs to𝐷(1/2, 1/2), and using the identity principle for holomorphicfunctions, we obtain that

𝑓0 (𝑧𝜁0) = 𝐹 (𝑧) , (48)for all 𝑧 ∈ D. This last relation implies that ‖𝐹‖log ≤ 1 whichis a contradiction to Proposition 7. This complete the proofof item (1).

(2) Arguing as in the proof of part (1), for |𝜁| = 1 fixed, weset the function

𝑔 (𝑢) = (1 − 2 log(1 + 𝑢2 ))−1 𝑓 (1 − 𝑢2 𝜁) , (49)

with 𝑢 ∈ D. We have shown that there exists a holomorphicfunction 𝑤 such that 𝑔(𝑢) = exp{−𝑤(𝑢)} for all 𝑢 ∈ D.Furthermore, 𝑤 satisfies the hypothesis of Julia’s Lemma(Lemma 1); that is, 𝑤 is a holomorphic function in D ∪ {1},whichmapsD intoH+,𝑤(1) = 0, and −𝑤(1) = 𝑑 = 1. Hence,for 𝑟 = (1 − 𝑥)/(1 + 𝑥) with 𝑥 ∈ [0, 1) fixed, 𝑤 maps thehorodisk Δ(1, 𝑟) into the open Euclidean disk with center at(1 − 𝑥)/(1 + 𝑥) and radius (1 − 𝑥)/(1 + 𝑥). In particular, since𝑥 ∈ Δ(1, 𝑟), we have that𝑤(𝑥) ∈ 𝐷𝑥 = 𝐷((1 − 𝑥)/(1 + 𝑥), (1 −𝑥)/(1 + 𝑥)). Thus, Lemma 4 allows us to write

Re (𝑔 (𝑥)) = Re (exp (−𝑤 (𝑥)))≥ min {Re (exp (−𝑧)) : 𝑧 ∈ 𝐷𝑥}= exp(−21 − 𝑥1 + 𝑥) ,

(50)

for all 𝑥 ∈ [0, 1). This last inequality is equivalent to writingRe((1 − 2 log(1 + 𝑥2 ))

−1 𝑓 (1 − 𝑥2 𝜁))≥ exp (−21 − 𝑥1 + 𝑥) ,

(51)

for all 𝑥 ∈ [0, 1) and from here we haveRe(𝑓 (1 − 𝑥2 𝜁))

≥ exp(−21 − 𝑥1 + 𝑥)(1 − 2 log(1 + 𝑥2 )) .(52)

Making the change 𝑟 = (1 − 𝑥)/2, we obtain 𝑟 ∈ (0, 1/2] since𝑥 ∈ [0, 1) and alsoRe (𝑓 (𝑟𝜁)) ≥ exp (−2 𝑟1 − 𝑟) (1 − 2 log (1 − 𝑟)) . (53)

We conclude, as before, that

Re (𝑓 (𝑧)) ≥ exp(−2 |𝑧|1 − |𝑧|) (1 − 2 log (1 − |𝑧|)) , (54)for all |𝑧| ≤ 1/2. This shows the inequality in the second partof the theorem.

Now, if there exists a function 𝑓0 ∈ 𝛽(∞)log such thatRe𝑓(𝑧0) = 𝐹(|𝑧0|) for some 𝑧0 ∈ 𝐷(0, 1/2), then we candefine 𝜁0 = 𝑧0/|𝑧0| and the function

𝑔0 (𝑧) = (1 − 2 log(1 + 𝑧2 ))−1 𝑓0 (1 − 𝑧2 𝜁0) , (55)

which maps D into D \ {0}. Hence, as before, there exists aholomorphic function𝑤mappingD intoH+ such that𝑤(1) =0, −𝑤(1) = 1 > 0, and 𝑔0(𝑧) = exp(−𝑤(𝑧)) for all 𝑧 ∈ D. Inparticular, for 𝑥 = 1 − 2|𝑧0| ∈ [0, 1), we haveRe exp (−𝑤 (𝑥)) = Re𝑔0 (𝑥) = Re𝑓0 (((1 − 𝑥) /2) 𝜁0)1 − 2 log ((1 + 𝑥) /2)

= Re𝑓0 (𝑧0)1 − 2 log (1 − 𝑧0)= 𝐹 (𝑧0)1 − 2 log (1 − 𝑧0)= exp(− 2 𝑧01 − 𝑧0)= exp (−21 − 𝑥1 + 𝑥) ;

(56)

that is,𝑤(𝑥) is the value in𝐷𝑥 where Re exp(−𝑤(𝑥)) attain itsminimum value, but by the proof of Lemma 4, we know thatthis happens if Im𝑤(𝑥) = 0 andRe𝑤(𝑥) = −2((1−𝑥)/(1+𝑥)).Now, by Lemma 2, we conclude that𝑤(𝑧) = 2((1−𝑧)/(1+𝑧))for all 𝑧 ∈ D. As before, this last fact implies that 𝑓0(𝑧𝜁0) =𝐹(𝑧) for all 𝑧 ∈ D and therefore ‖𝐹‖log ≤ 1 which is acontradiction to Proposition 7. This completes the proof ofitem (2).

Journal of Function Spaces 7

5. Distortion Theorems for Complex FunctionsinBlog Having Branch Points

In this section we establish a distortion theorem for functionsin the closed unit ball of Blog having branch points andsatisfying a normalized Bloch conditions. More precisely, foreach 𝑛 ∈ N, we denote by 𝛽(𝑛)log the class of all holomorphicfunctions 𝑓 ∈ Blog such that 𝑓(0) = 0, 𝑓(0) = 1, ‖𝑓‖log ≤ 1and if 𝑓(𝑏) = 0 for some 𝑏 ∈ D then 𝑓(𝑘)(𝑏) = 0 for all𝑘 = 1, 2, . . . , 𝑛. Clearly we have

𝛽(∞)log =∞⋂𝑛=1

𝛽(𝑛)log. (57)With these notations, we have the following result.

Theorem 10. For 𝑛 ∈ N fixed, we set 𝑎𝑛 = √𝑛/(𝑛 + 2). Thenfor every 𝑓 ∈ 𝛽(𝑛)log we have the following:

(1) |𝑓(𝑧)| ≥ 𝐹𝑛(|𝑧|) = ((𝑎𝑛−|𝑧|)/(𝑎𝑛−𝑎2𝑛|𝑧|))𝑛(1−2 log(1−𝑎𝑛|𝑧|)), for all |𝑧| ≤ 𝑎𝑛.There is not a function𝑓0 ∈ 𝛽(𝑛)logsuch that |𝑓0(𝑧0)| = 𝐹𝑛(|𝑧0|) for some 𝑧0 ∈ 𝐷(0, 𝑎𝑛) \{0}.

(2) Re𝑓(𝑧) ≥ 𝐹𝑛(|𝑧|) = ((𝑎𝑛 − |𝑧|)/(𝑎𝑛 − 𝑎2𝑛|𝑧|))𝑛(1 −2 log(1 − 𝑎𝑛|𝑧|)), for all |𝑧| ≤ √𝑛(𝑛 + 2)/(2𝑛 + 1).Furthermore, there is not a function 𝑓0 ∈ 𝛽(𝑛)log suchthat Re𝑓0(𝑧0) = 𝐹𝑛(|𝑧0|) for some 𝑧0 ∈ 𝐷(0,√𝑛(𝑛 + 2)/(2𝑛 + 1)) \ {0}.

Proof. (1) Let us fix |𝜁| = 1, 𝑛 ∈ N and 𝑎𝑛 = √𝑛/(𝑛 + 2) < 1.We set the function

𝑔 (𝑢) = (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑢))−1

⋅ 𝑓 (𝑎𝑛 − 𝑎𝑛𝑢1 − 𝑎2𝑛𝑢 𝜁) ,(58)

with 𝑢 ∈ D, where log(𝑤) denotes the principal logarithmicof the complex number 𝑤 ∈ 𝐷(1, 1). Clearly the function 𝑔is holomorphic on D ∪ {1} and 𝑔(1) = 1 because 𝑓(0) = 1.Also, we have

𝑔 (𝑢) = − (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑢))−2

⋅ ( −2𝑎2𝑛1 − 𝑎2𝑛𝑢)𝑓 (𝑎𝑛 − 𝑎𝑛𝑢1 − 𝑎2𝑛𝑢 𝜁)

+ (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑢))−1

⋅ 𝑓 (𝑎𝑛 − 𝑎𝑛𝑢1 − 𝑎2𝑛𝑢 𝜁)[𝑎𝑛 (𝑎2𝑛 − 1)(1 − 𝑎2𝑛𝑢)2 ] 𝜁.

(59)

And hence 𝑔(1) = 2𝑎2𝑛/(1 − 𝑎2𝑛) = 𝑛 since 𝑓(0) = 1 and𝑓(0) = 0 (by Lemma 6). Furthermore, since 𝑓 ∈ 𝛽(𝑛)log and𝑔(𝑢0) = 0 if and only if 𝑓(((𝑎𝑛 − 𝑎𝑛𝑢0)/(1 − 𝑎2𝑛𝑢0))𝜁) = 0, we

conclude that all the zeros of the function 𝑔 have multiplicityat less 𝑛.

On the other hand, since ‖𝑓‖log ≤ 1 we have𝑔 (𝑢) ≤ 1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑢)−1

⋅ log( 𝑒1 − (𝑎𝑛 − 𝑎𝑛𝑢) / (1 − 𝑎2𝑛𝑢)2)≤ 1Re (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑢))⋅ log( 𝑒1 − (𝑎𝑛 − 𝑎𝑛𝑢) / (1 − 𝑎2𝑛𝑢)2)

= [log( 𝑒(1 − 𝑎2𝑛) / (1 − 𝑎2𝑛𝑢)2)]−1

⋅ log( 𝑒1 − (𝑎𝑛 − 𝑎𝑛𝑢) / (1 − 𝑎2𝑛𝑢)2) < 1,

(60)

for all 𝑢 ∈ D, since1 − 𝑎2𝑛1 − 𝑎2𝑛𝑢

2 < 1 −

𝑎𝑛 − 𝑎𝑛𝑢1 − 𝑎2𝑛𝑢2 , (61)

for all 𝑢 ∈ D. Hence, we have shown that 𝑔(D) ⊂ D. InvokingLemma 3, we conclude that |𝑔(𝑥)| ≥ 𝑥𝑛 for all 𝑥 ∈ [0, 1).Therefore, for each 𝑥 ∈ [0, 1), the following estimation holds:

1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑥)−1

⋅ 𝑓 (𝑎𝑛 − 𝑎𝑛𝑥1 − 𝑎2𝑛𝑥 𝜁)

≥ 𝑥𝑛. (62)

That is,

𝑓 (𝑎𝑛 − 𝑎𝑛𝑥1 − 𝑎2𝑛𝑥 𝜁)

≥ 𝑥𝑛 (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑥)) ,

(63)

for all 𝑥 ∈ [0, 1) since 𝑎𝑛 ∈ (0, 1).Next, we make the change 𝑟 = (𝑎𝑛 − 𝑎𝑛𝑥)/(1 − 𝑎2𝑛𝑥). Then𝑟 ∈ (0, 𝑎𝑛] since 𝑥 ∈ [0, 1), 𝑥 = (1/𝑎𝑛)((𝑎𝑛 − 𝑟)/(1 − 𝑎𝑛𝑟)) and

we can write

𝑓 (𝑟𝜁) ≥ ( 1𝑎𝑛𝑎𝑛 − 𝑟1 − 𝑎𝑛𝑟)

𝑛

⋅ (1 − 2 log (1 − 𝑎2𝑛) + 2 log(1 − 𝑎2𝑛 1𝑎𝑛𝑎𝑛 − 𝑟1 − 𝑎𝑛𝑟))

= ( 1𝑎𝑛𝑎𝑛 − 𝑟1 − 𝑎𝑛𝑟)

𝑛 (1 − 2 log (1 − 𝑎𝑛𝑟)) .(64)

8 Journal of Function Spaces

Finally, if we set 𝜁 = 𝑧/|𝑧| and 𝑟 = |𝑧|, we conclude that𝑓 (𝑧) ≥ ( 1𝑎𝑛

𝑎𝑛 − |𝑧|1 − 𝑎𝑛 |𝑧|)𝑛 (1 − 2 log (1 − 𝑎𝑛 |𝑧|))

= (1 − |𝑧| /𝑎𝑛1 − 𝑎𝑛 |𝑧| )𝑛 (1 − 2 log (1 − 𝑎𝑛 |𝑧|))

= 𝐹𝑛 (|𝑧|) ,(65)

for all |𝑧| ∈ (0, 𝑎𝑛]. This shows the inequality in part (1).The proof of the second part is similar to part (1) of

Theorem 9. If there exists a function 𝑓0 ∈ 𝛽(𝑛)log such that|𝑓0(𝑧0)| = 𝐹𝑛(|𝑧0|) for some 𝑧0 ∈ 𝐷(0, 𝑎𝑛) \ {0}, then we set𝜁 = 𝑧0/|𝑧0| and the function𝑔0 (𝑢) = (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑢))−1

⋅ 𝑓0 (𝑎𝑛 − 𝑎𝑛𝑢1 − 𝑎2𝑛𝑢 𝜁)(66)

with𝑢 ∈ D.Wehave showed that𝑔0 satisfies all the hypothesisof Lemma 3. Furthermore, choosing 𝑥 ∈ [0, 1) such that

𝑎𝑛 − 𝑎𝑛𝑥1 − 𝑎2𝑛𝑥 =𝑧0 (67)

we obtain𝑔0 (𝑎𝑛 − 𝑧0𝑎𝑛 − 𝑎2𝑛 𝑧0)

=𝑔0 (𝑥)

= (1 − 2 log (1 − 𝑎2𝑛) + 2 log( 1 − 𝑎2𝑛1 − 𝑎𝑛 𝑧0))

−1

⋅ 𝑓0 (𝑧0) = 𝐹𝑛 (𝑧0)1 − 2 log (1 − 𝑎𝑛 𝑧0)

= ( 𝑎𝑛 − 𝑧0𝑎𝑛 − 𝑎2𝑛 𝑧0)𝑛 .

(68)

By Lemma 3, we conclude that 𝑔0(𝑧) = 𝑧𝑛 for all 𝑧 ∈ D.Hence,

𝑓0 (𝑎𝑛 − 𝑎𝑛𝑧1 − 𝑎2𝑛𝑧 𝜁)= (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑧)) 𝑧𝑛,

(69)

for all 𝑧 ∈ D. Changing (𝑎𝑛 − 𝑎𝑛𝑧)/(1 − 𝑎2𝑛𝑧) by 𝑧, we obtainthat

𝑓0 (𝑧𝜁) = (1 − 2 log (1 − 𝑎𝑛𝑧)) ( 𝑎𝑛 − 𝑧𝑎𝑛 − 𝑎2𝑛𝑧)𝑛

= 𝐹𝑛 (𝑧) ,(70)

for all 𝑧 ∈ 𝐷(0, 𝑎𝑛) and consequently for all 𝑧 ∈ D. Thislast equality implies that ‖𝐹𝑛‖log = ‖𝑓0‖log ≤ 1 which is acontradiction to Proposition 8.

(2) As before, for 𝑛 ∈ N, we set 𝑎𝑛 = √𝑛/(𝑛 + 2), we fix|𝜁| = 1, and we define the function𝑔 (𝑢) = (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑢))−1

⋅ 𝑓 (𝑎𝑛 − 𝑎𝑛𝑢1 − 𝑎2𝑛𝑢 𝜁) ,(71)

with 𝑢 ∈ D. In the first part we have shown that this functionsatisfies the hypothesis of Lemma 3. Hence Re𝑔(𝑥) ≥ 𝑥𝑛 forall (𝑛 − 1)/(𝑛 + 1) ≤ 𝑥 < 1. Therefore

Re((1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑥))−1

⋅ 𝑓 (𝑎𝑛 − 𝑎𝑛𝑥1 − 𝑎2𝑛𝑥 𝜁)) ≥ 𝑥𝑛,

(72)

and thus we have

(1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑥))−1

⋅ Re(𝑓 (𝑎𝑛 − 𝑎𝑛𝑥1 − 𝑎2𝑛𝑥 𝜁)) ≥ 𝑥𝑛, (73)

which is the same as

Re(𝑓 (𝑎𝑛 − 𝑎𝑛𝑥1 − 𝑎2𝑛𝑥 𝜁))

≥ 𝑥𝑛 log( 𝑒(1 − 𝑎2𝑛) / (1 − 𝑎2𝑛𝑥)2) .(74)

As before, we make the change 𝑟 = (𝑎𝑛 − 𝑎𝑛𝑥)/(1 − 𝑎2𝑛𝑥); then0 < 𝑟 ≤ √𝑛(𝑛 + 2)/(2𝑛 + 1), 𝑥 = (1/𝑎𝑛)((𝑎𝑛 − 𝑟)/(1 − 𝑎𝑛𝑟)),and

Re𝑓 (𝑟𝜁) ≥ ( 1𝑎𝑛𝑎𝑛 − 𝑟1 − 𝑎𝑛𝑟)

𝑛

⋅ log( 𝑒[(1 − 𝑎2𝑛) / (1 − 𝑎2𝑛 (1/𝑎𝑛) ((𝑎𝑛 − 𝑟) / (1 − 𝑎𝑛𝑟)))]2)

= ( 1𝑎𝑛𝑎𝑛 − 𝑟1 − 𝑎𝑛𝑟)

𝑛 (1 − 2 log (1 − 𝑎𝑛𝑟)) .(75)

Setting 𝜁 = 𝑧/|𝑧| and 𝑟 = |𝑧|, we conclude thatRe𝑓 (𝑧) ≥ ( 1𝑎𝑛

𝑎𝑛 − |𝑧|1 − 𝑎𝑛 |𝑧|)𝑛 (1 − 2 log (1 − 𝑎𝑛 |𝑧|)) , (76)

for all |𝑧| ≤ √𝑛(𝑛 + 2)/(2𝑛 + 1). This shows the inequality inpart (2).

If there exists a function 𝑓0 ∈ 𝛽(𝑛)log such that Re𝑓0(𝑧0) =𝐹𝑛(|𝑧0|) for some 𝑧0 ∈ 𝐷(0,√𝑛(𝑛 + 2)/(2𝑛 + 1)) \ {0}, then weset 𝜁 = 𝑧0/|𝑧0| and the function

𝑔0 (𝑢) = (1 − 2 log (1 − 𝑎2𝑛) + 2 log (1 − 𝑎2𝑛𝑢))−1

⋅ 𝑓0 (𝑎𝑛 − 𝑎𝑛𝑢1 − 𝑎2𝑛𝑢 𝜁) ,(77)

Journal of Function Spaces 9

with𝑢 ∈ D.Wehave showed that𝑔0 satisfies all the hypothesisof Lemma 3. Furthermore, choosing 𝑥 ∈ [0, 1) such that

𝑎𝑛 − 𝑎𝑛𝑥1 − 𝑎2𝑛𝑥 =𝑧0 , (78)

we obtain

𝑥 = 1𝑎𝑛𝑎𝑛 − 𝑧01 − 𝑎𝑛 𝑧0 , (79)

and since 0 < |𝑧0| ≤ 𝑎𝑛((𝑛 + 2)/(2𝑛 + 1)) we can see that(𝑛 − 1)/(𝑛 + 1) ≤ 𝑥 < 1. Furthermore,Re𝑔0 ( 𝑎𝑛 −

𝑧0𝑎𝑛 − 𝑎2𝑛 𝑧0) = Re𝑔0 (𝑥)

= (1 − 2 log (1 − 𝑎2𝑛) + 2 log( 1 − 𝑎2𝑛1 − 𝑎𝑛 𝑧0))

−1

⋅ Re𝑓0 (𝑧0) = 𝐹𝑛 (𝑧0)1 − 2 log (1 − 𝑎𝑛 𝑧0)

= ( 𝑎𝑛 − 𝑧0𝑎𝑛 − 𝑎2𝑛 𝑧0)𝑛 .

(80)

By Lemma 3, we conclude that 𝑔0(𝑧) = 𝑧𝑛 for all 𝑧 ∈ D.Hence, as before, this last fact implies that ‖𝐹𝑛‖log = ‖𝑓0‖log ≤1 which is a contradiction to Proposition 8.6. Some Estimations for the Schlicht Radius

In this section we present some consequences of the resultsobtained in Sections 4 and 5. We recall that if 𝑓 is aholomorphic function on D and 𝑧0 ∈ D, 𝑟𝑠(𝑧0, 𝑓) denotethe radius of the largest schlicht disk on the Riemann surface𝑓(D) centered at 𝑓(𝑧0) (a schlicht disk on 𝑓(D) centered at𝑓(𝑧0) means that 𝑓 maps an open subset of D containing 𝑧0conformally onto this disk). With this notation, we have thefollowing results.

Corollary 11. If 𝑓 ∈ 𝛽(∞)log , then𝑟𝑠 (0, 𝑓) ≥ ∫1

0𝐹 (|𝑧|) 𝑑 |𝑧|

= ∫10exp(− 2𝑡1 − 𝑡) (1 − 2 log (1 − 𝑡)) 𝑑𝑡.

(81)

Proof. From the definition of 𝑟𝑠(0, 𝑓), it follows the fact thatthere exists a simply connected domain 𝐸 ⊂ D containing thezero such that 𝑓 maps 𝐸 conformally onto an Euclidean diskwith center at 𝑓(0) and radius 𝑟𝑠(0, 𝑓). This Euclidean diskmust meet the boundary of 𝑓(D) because, in other cases, theboundary of the set 𝐸 is a Jordan curve in the interior of Dand we can find an open set 𝑊 ⊂ D where 𝑓 is univalent;hence 𝑓(𝑊) contain an Euclidean disk with center at 𝑓(0)and radius greater than 𝑟𝑠(0, 𝑓), which contradict with thedefinition of 𝑟𝑠(0, 𝑓). We conclude then that there is a radial

segment Γ jointing 𝑓(0) to the boundary of 𝑓(D). Let 𝛾 beinverse image of Γ under 𝑓; then 𝛾 joint the point 0 to theboundary of D. Thus, fromTheorem 9, it follows that

𝑟𝑠 (0, 𝑓) = ∫Γ|𝑑𝑤| = ∫

𝛾

𝑓 (𝑧) |𝑑𝑧|≥ ∫10𝐹 (𝛾 (𝑡)) 𝛾 (𝑡) 𝑑𝑡

= ∫10𝐹 (𝛾 (𝑡))

𝛾 (𝑡) 𝛾 (𝑡)𝛾 (𝑡) 𝑑𝑡≥ ∫10𝐹 (𝛾 (𝑡)) 𝛾 (𝑡) ⋅ 𝛾

(𝑡)𝛾 (𝑡) 𝑑𝑡 = ∫1

0𝐹 (𝜏) 𝑑𝜏

= ∫10exp (− 2𝜏1 − 𝜏) (1 − 2 log (1 − 𝜏)) 𝑑𝜏 ≈ 0.4104136111,

(82)

where we have used Cauchy-Schwarz’s inequality in thefourth line, 𝛾(𝑡) ⋅ 𝛾(𝑡) is the scalar product of 𝛾(𝑡) and 𝛾(𝑡),and we have made the change 𝜏 = |𝛾(𝑡)| = √𝛾(𝑡) ⋅ 𝛾(𝑡), where𝜏 → 1− as 𝑡 → 1−. This shows the result.

While for functions in the class𝛽(𝑛)log wehave the following.Corollary 12. Suppose that 𝑛 ∈ N is fixed. If 𝑓 ∈ 𝛽(𝑛)log, then

𝑟𝑠 (0, 𝑓) ≥ ∫√𝑛/(𝑛+2)0

𝐹𝑛 (𝑡) 𝑑𝑡= ∫√𝑛/(𝑛+2)0

(1 − 𝑡/𝑎𝑛1 − 𝑎𝑛𝑡 )𝑛 (1 − 2 log (1 − 𝑎𝑛𝑡)) 𝑑𝑡.

(83)

Proof. Indeed, arguing as in the proof of Corollary 11, fromTheorem 10, it follows that

𝑟𝑠 (0, 𝑓) = ∫Γ|𝑑𝑤| = ∫

𝛾

𝑓 (𝑧) |𝑑𝑧|

≥ ∫√𝑛/(𝑛+2)0

𝐹𝑛 (𝜏) 𝑑𝜏= ∫√𝑛/(𝑛+2)0

( 𝑎𝑛 − 𝜏𝑎𝑛 − 𝑎2𝑛𝜏)𝑛 (1 − 2 log (1 − 𝑎𝑛𝜏)) 𝑑𝑡,

(84)

where we have used that 𝜏 = |𝛾(𝑡)| → 𝑎𝑛 as 𝑡 → 1−. Thisshows the result.

Conflicts of Interest

The authors declare that they have no conflicts of interestregarding the publication of this paper.

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10 Journal of Function Spaces

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[9] M. Bonk, D. Minda, and H. Yanagihara, “Distortion theoremsfor Bloch functions,” Pacific Journal of Mathematics, vol. 179, no.2, pp. 241–262, 1997.

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