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Research ArticleDistortion Type Theorems for Functions inthe Logarithmic Bloch Space
Armando J. GarcĆa-OrtĆz,1 Milton del Castillo Lesmes Acosta,2
and Julio C. Ramos-FernƔndez3
1 AĢrea de MatemaĢtica, Universidad Nacional Experimental de Guayana, Pto. Ordaz, BolĢıvar, Venezuela2Proyecto Curricular de MatemaĢticas, Facultad de Ciencias y EducacioĢn, Universidad Distrital Francisco JoseĢ de Caldas,Carrera 3 No. 26 A-40, BogotaĢ, Colombia3Departamento de MatemaĢtica, Universidad de Oriente, 6101 CumanaĢ, Sucre, Venezuela
Correspondence should be addressed to Julio C. Ramos-FernaĢndez; [email protected]
Received 23 January 2017; Revised 22 March 2017; Accepted 2 April 2017; Published 16 April 2017
Academic Editor: John R. Akeroyd
Copyright Ā© 2017 Armando J. GarcĢıa-OrtĢız 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 celebrateddistortionās theorem established by Koebe and Bieberbach[1, 2] at the beginning of the twentieth century. Koebe andBieberbach showed that the range of any function š in theclass S of all conformal functions on D, the open unit diskof the complex plane C, normalized such that š(0) = 0 =š(0) ā 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 Blochās 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 š ā S and š§ ā D the followingestimations hold.
(1) Growth theorem:|š§|
(1 + |š§|)2 ā¤ š (š§) ā¤|š§|
(1 ā |š§|)2 (1)(2) Distortion theorem:
1 ā |š§|(1 + |š§|)3 ā¤
š (š§) ā¤ 1 + |š§|(1 ā |š§|)3 (2)
with equality if and only if š is a rotation of the Koebefunction defined by
š¾ (š§) = š§(1 ā š§)2 , (š§ ā 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 š-Bloch spaceBš for all š ā„ 3 (see Section 3 for the definition ofBš). 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 š(0) = 0 = š(0) ā 1 (recall that a holomorphicfunction š is locally schlicht on D if š(š§) Ģø= 0 for all š§ āD). 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 š in the classical Bloch spaceB satisfyingthe conditions š(0) = 0, š(0) = 1, and āšāB = 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
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2 Journal of Function Spaces
give sharp lower bounds for |š(š§)| and for Reš(š§) and asconsequence of their results they obtain a lower bound forBlochās constant.Determination of the (locally schlicht) Blochconstant is still an open problem. By Landauās 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 š-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 conditionš(0) = 0 = š(0)ā1.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: Juliaās 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; šD denotes the boundary of D. Thespace of all complex and holomorphic functions on D, as isusual, is denoted byš»(D). A function š ā š»(D) is said to benormalized if š(0) = 0 and š(0) = 1 and š is locally schlichtor locally univalent if š(š§) Ģø= 0 for all š§ ā D. A point š§0 is abranch point for š if š(š§0) = 0. For š > 0, we define
Ī (1, š) = {š§ ā D : |1 ā š§|21 ā |š§|2 < š} . (4)Ī(1, š) is known as a horodisk D; that is, it is an Euclideandisk contained inDwhich is tangent to šD at 1. Furthermore,Ī(1, š) has center at 1/(1+š) and radius š/(1+š). The closureof Ī(1, š) relative toD is denoted by Ī(1, š). Observe that 1 āĪ(1, š) but (1 ā š)/(1 + š) ā Ī(1, š). With these notations, wecan enunciate the well known Juliaās Lemma; the reader canconsult the excellent book of Ahlfors [13] for its proof.
Lemma 1 (Juliaās Lemma). Suppose that š¤ is a complex andholomorphic function onDāŖ{1} such thatš¤mapsD intoH+ ={š§ ā C : Re(š§) > 0}, the right half-plane, and š¤(1) = 0. Then,for any š > 0, the functionš¤maps the horodisk Ī(1, š) into theEuclidean disk {š§ ā C : |š§ ā šš| < šš}, where š = āš¤(1) > 0.Furthermore, a boundary point of the first disk is mapped onthe boundary of the second disk if and only if š¤ is a conformalfunction mapping D onto H+ and satisfying š¤(1) = 0.
In 1992, Liu andMinda [6] established distortion theoremfor functions in the Bloch space; they showed the followingresults which are consequences of Juliaās Lemma. We includethe proof of the first one to illustrate the application of JuliaāsLemma.
Lemma2 ([6, corollary in Section 1]). Letš¤ be a holomorphicfunction onDāŖ{1}. Suppose thatš¤mapsD into the right half-plane H+ and that š¤(1) = 0. Then š = āš¤(1) > 0 and
Reš¤ (š„) ā¤ 2š1 ā š„1 + š„ , (5)for all š„ ā (ā1, 1), with equality for some š„ ā (ā1, 1) if andonly if
š¤ (š§) = 2š1 ā š§1 + š§ , (6)for all š§ ā D.Proof. Indeed, let us fix š„ ā (ā1, 1); then š = (1āš„)/(1+š„) >0 and by Juliaās Lemma, š¤ maps Ī(1, š) into the Euclideandisk š·(šš, šš). In particular, since š„ ā Ī(1, š), then š¤(š„) āš·(šš, šš); this fact implies that Reš¤(š„) ā¤ 2šš. Furthermore,š„ ā šĪ(1, š); hence if Reš¤(š„) = 2šš, then we conclude thatš¤(š„) = 2šš ā šš·(šš, šš) and, by Juliaās Lemma, this last factoccurs if š¤ is the conformal map from D onto H+ such thatš¤(1) = 0; that is,š¤(š§) = 2š((1āš§)/(1+š§)) for all š§ ā D. Thisshows the lemma.
Lemma 3 ([6, corollary to Theorem 3]). Let š be a holomor-phic function onDāŖ{1}. Suppose thatš(D) ā D,š(1) = 1 andthat all the zeros of š have multiplicity at least š. If š(1) = š,then
(1) |š(š„)| ā„ š„š for all š„ ā [0, 1), with equality for someš„ ā [0, 1) if and only if š(š§) = š§š for all š§ ā D;(2) Re(š(š„)) ā„ š„š for all (š ā 1)/(š + 1) ā¤ š„ < 1, with
equality for some š„ ā [(š ā 1)/(š + 1), 1) if and only ifš(š§) = š§š for all š§ ā 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 š„ ā [0, 1) be fixed and š·š„ the Euclidean diskwith center at (1 ā š„)/(1 + š„) and radius (1 ā š„)/(1 + š„); then
min {Re (exp (āš§)) : š§ ā š·š„} = exp (ā21 ā š„1 + š„) . (7)Proof. Let š = (1 ā š„)/(1 + š„) for simplicity and let š(š§) =šš(š§ā1). Since 1 + š§š(š§)/š(š§) = 1 ā šš§ has positive real parton |š§| < 1, the functionš is convex. In particular, Re(š(š§)) >š(ā1) = šā2š, 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
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that a function weight š on D is a bounded, positive, andcontinuous function defined on D. Given a weight š on D,š-Bloch space, denoted by Bš, consists of all holomorphicfunctions š on D such that
šš fl supš§āD
š (š§) š (š§) < ā. (8)It is known that if the weight š is radial, that is, š(š§) = š(|š§|)for all š§ ā D, then Bš is a Banach space with the normāšāBš = |š(0)| + āšāš. When š(š§) = 1ā |š§|2, with š§ ā D,Bšbecomes the Bloch space which is denoted byB, while whenš(š§) = (1 ā |š§|2)š, with š§ ā D and š > 0 fixed, we obtainš-Bloch space which is denoted byBš.
Clearly, the function šlog, defined byšlog (š§) = [log( š1 ā |š§|2)]
ā1 , (9)defines a weight on D. Hence, the space Blog = Bšlog is aBanach space with the norm
šBlog = š (0) + šlogfl š (0) + sup
š§āD
š (š§)log (š/ (1 ā |š§|2)) .
(10)
We callBlog as the logarithmic Bloch space. In the next resultwe are going to show that Blog is a subspace of Bš for allš ā„ 1.Proposition 5. The spaceBlog is contained inBš, for all š ā„1. Furthermore, šBš ā¤ šBlog , (11)for all function š ā Blog.Proof. It is enough to show that for š ā„ 1 fixed
(1 ā |š§|2)š log( š1 ā |š§|2) ā¤ 1, (12)for all š§ ā D. But, this last inequality is true since the function
ā (š”) = š”š log(šš” ) ā 1, (13)with š” ā (0, 1], is increasing and ā(1) = 0.
Also, we have the following very useful identity (seeLemma 3.3 in [12]).
Lemma 6. If š ā Blog, š(0) = 0, š(0) = 1, and āšālog ā¤ 1,then š(0) = 0.Proof. Suppose that š ā Blog, š(0) = 0, š(0) = 1, andāšālog ā¤ 1. Then, for each š§ ā D, we have
š (š§) ā¤ log( š1 ā |š§|2) . (14)
Taylorās theorem implies that
1 + š (0) š§ + š (|š§|)2 ā¤ log2 ( š1 ā |š§|2)= (1 + |š§|2 + š (|š§|))2 ,
(15)
as š§ ā 0. But since1 + š (0) š§ + š (|š§|)2
= 1 + 2Re (š (0) š§) + š (|š§|) , (16)as š§ ā 0, and
(1 + |š§|2 + š (|š§|))2 = 1 + š (|š§|) , (17)as š§ ā 0, we obtain from (15) that
2Re (š (0) š§) ā¤ š (|š§|) , (18)as š§ ā 0. Now, if we consider š§ = šš(0)/|š(0)| with š > 0small in (18), we conclude |š(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(š¤) to denote the principal logarithmic of the complexnumber š¤ Ģø= 0. Observe that the principal logarithmic isa holomorphic function on š·(1, 1), the Euclidean disk withcenter at 1 and radius 1:
(1) For each š ā N, we setš¹š (š§) = ā«š§
0(1 ā š /šš1 ā ššš )
š (1 ā 2 log (1 ā ššš )) šš , (19)where šš = āš/(š + 2) and š§ ā D. Clearly, š¹š ā š»(D) for allš ā N, š¹š(0) = 0, and š¹š(0) = 1.
(2) For š§ ā D, we defineš¹ (š§) = ā«š§
0exp (ā 2š 1 ā š ) (1 ā 2 log (1 ā š )) šš . (20)
We can see that š¹ ā š»(D), š¹(0) = 0, and š¹(0) = 1.Also we have that the function š¹ satisfies the following
properties.
Proposition 7. The function š¹ belongs toBlog. Furthermore,supš§āDšlog(š§)|š¹(|š§|)| = 1 but āš¹ālog > 1.Proof. We see that āš¹ālog > 1. Indeed, we have
āš¹ālog ā„š¹ (š/2)1 ā log (1 ā |š/2|2)
= exp (2/5)1 ā log (3/4)ā(1 ā log(54))2 + 4 arctan2 (12)
ā 1.4014837 > 1.
(21)
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Now, we are going to show that š¹ ā Blog. Since thefunction exp(ā2š§/(1 ā š§)) is holomorphic on D, then themodulus maximum principle tells us that its maximum valueis attained in the boundary šD. But if |š§| = 1 then |1 ā š§|2 =2(1 ā Re(š§)) and hence
supš§āD
exp(ā 2š§1 ā š§) = sup|š§|=1 exp(2 ā
2 (1 ā Re (š§))|1 ā š§|2 )
= š.(22)
On the other hand, for each š§ ā D, we have |arg(1āš§)| < š/2and
supš§āD
arg (1 ā š§)1 ā log (1 ā |š§|2) ā¤š2 . (23)
Furthermore, using elementary calculus, we can see that thereal functionš»(š”) = š2 ā (1 ā š”)(1 + š”)3 is nonnegative for allš” ā [0, 1] (its minimum value isš»(1/2) = š2 ā 27/16 ā 5.70).Hence for any š” ā [0, 1) we obtain
1 ā 2 log (1 ā š”) ā 3 (1 ā log (1 ā š”2))= log((1 ā š”) (1 + š”)3š2 ) ā¤ 0.
(24)
This last implies that
1 ā 2 log (1 ā š”)1 ā log (1 ā š”2) ā¤ 3, (25)for all š” ā [0, 1). We conclude that for any š§ ā D such that|1 ā š§|2 ā¤ š
log (š/ |1 ā š§|2)1 ā log (1 ā |š§|2) =1 ā 2 log (|1 ā š§|)1 ā log (1 ā |š§|2)
ā¤ 1 ā 2 log (1 ā |š§|)1 ā log (1 ā |š§|2) ā¤ 3,(26)
while for š§ ā D such that |1 ā š§|2 > š we havelog (š/ |1 ā š§|2)1 ā log (1 ā |š§|2) =
log (|1 ā š§|2 /š)1 ā log (1 ā |š§|2) ā¤ log (4) ā 1. (27)
These last inequalities, (22) and (23), imply that
āš¹ālog ā¤ š (3 + log (4) ā 1 + 2 (š2 ))= š (2 + log (4) + š)
(28)
which shows that š¹ ā Blog.Now, we are going to show that supš§āDšlog(š§)|š¹(|š§|)| = 1.
Observe that šlog(0)|š¹(0)| = 1. Also the real functionš»(š”) =exp(ā2š”/(1āš”))(1ā2log(1āš”))ā1+ log(1āš”2)with š” ā [0, 1)
satisfies š»(0) = 0, š»(š”) ā āā as š” ā 1ā and it is strictlydecreasing since
š» (š”)= ā 2š”1 ā š”2
ā 21 ā š” ( 21 ā š” log( š(1 ā š”)2) ā 1) exp (ā2š”1 ā š”)
< 0,
(29)
for all š” ā [0, 1). Hence we conclude that š»(š”) ā¤ 0 for allš” ā [0, 1) which shows the affirmation.For the sequence {š¹š}, we have the following properties.
Proposition 8. Functions š¹š with š ā N belong to Blog andsatisfy
limšāā
š¹š (š§) = š¹ (š§) , (30)for each š§ ā D. Furthermore, for each š ā N āš¹šālog > 1, infact, supš§āDšlog(š§)|š¹š(|š§|)| > 1.Proof. Clearly, for any š ā N, the function š¹š belongsto Blog since š¹š ā š»(D). We are going to show thatsupš§āDšlog(š§)|š¹š(|š§|)| > 1. It is enough to show that thereexists a š”0 ā (šš, 1) such thatš»(š”0) > 0, where
š»(š”) = (š”/šš ā 11 ā ššš” )š
log( š(1 ā ššš”)2)ā log( š1 ā š”2 ) ,
(31)
with š” ā (šš, 1). Observe that, for š”š = ššš with š = 2/(1 +š2š) > 1, we have š”š ā (šš, 1), š ā 1 = 1 ā šš2š , and 1 ā š2š2š =(1 ā šš2š)2 which implies thatš»(š”š) = 0. Also,š» (š”)
= š (š”/šš ā 11 ā ššš” )šā1 1/šš ā šš(1 ā ššš”)2 log(
š(1 ā ššš”)2)
+ (š”/šš ā 11 ā ššš” )š 2šš1 ā ššš” ā
2š”1 ā š”2 ;(32)
hence
š» (š”š) = š 1/šš ā šš(1 ā šš2š)2 log(š
(1 ā šš2š)2) +2šš1 ā šš2š
ā 2ššš1 ā š2š2š =š
(1 ā šš2š)2 (1šš ā šš)
ā (log( š(1 ā šš2š)2) ā1š
2š2š1 ā š2š)
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= š(1 ā šš2š)2 (1šš ā šš)
ā (log( š(1 ā šš2š)2) ā 1) ,(33)
since š ā 1 = 1 ā šš2š , 1 ā š2š2š = (1 ā šš2š)2, and š(š2š + 1) = 2and we have used that šš = āš/(š + 2) in the last equality.Thus, we conclude that š»(š”š) > 0 and since š»(š”š) = 0, thenthere exists š”0 ā (š”š, 1) such that š»(š”0) > 0. This shows theaffirmation. The other properties of š¹šās 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 š½(ā)log the class ofall holomorphic functions š ā Blog such that š is locallyschlicht, š(0) = 0, š(0) = 1, and āšālog ā¤ 1. With thesenotations, we have the following result.
Theorem 9. If š ā š½(ā)log then we have the following:(1) |š(š§)| ā„ š¹(|š§|) = exp(ā2|š§|/(1 ā |š§|))(1 ā 2 log(1 ā|š§|)) for all š§ ā D. There is not a function š0 ā š½(ā)log
such that |š0(š§0)| = š¹(|š§0|) for some š§0 ā D \ {0}.(2) Reš(š§) ā„ š¹(|š§|) = exp(ā2|š§|/(1 ā |š§|))(1 ā 2log(1 ā|š§|)) for all |š§| ā¤ 1/2. There is not a function š0 ā š½(ā)log
such thatReš0(š§0) = š¹(|š§0|) for some š§0 ā š·(0, 1/2)\{0}.Proof. (1) Suppose that š ā š½(ā)log . Let us fix |š| = 1 and we setthe function
š (š¢) = (1 ā 2 log(1 + š¢2 ))ā1 š (1 ā š¢2 š) , (34)
with š¢ ā D, where log(š¤) denotes the principal logarithmicof š¤ ā š·(1, 1). Clearly š is holomorphic on D \ {ā1} andš(1) = 1 because š(0) = 1. Since š is locally schlicht on D,we have that š(š§) Ģø= 0 for all š§ ā D. Furthermore,
š (š¢)= 21 + š¢ (1 ā 2 log(1 + š¢2 ))
ā2 š (1 ā š¢2 š)+ (1 ā 2 log(1 + š¢2 ))
ā1 š (1 ā š¢2 š)(āš2) .(35)
In particular, š(1) = 1 since š(0) = 1 and š(0) = 0 (byLemma 6). Also, for any š¢ ā D, we have
š (š¢) = 1 ā 2 log(1 + š¢2 )ā1 š (1 ā š¢2 š)
ā¤ 1Re (1 ā 2 log ((1 + š¢) /2))ā log( š1 ā |(1 ā š¢) /2|2) =
1log (š/ |(1 + š¢) /2|2)
ā log (š/1 ā |(1 ā š¢) /2|2) < 1
(36)
and šmapsD intoD \ {0}. Hence, there exists a holomorphicfunction š¤mapping the unit disk D into the right half-planeH+ = {š§ : Re(š§) > 0} and such that š(š¢) = exp{āš¤(š¢)}for all š¢ ā D. Observe that š¤(1) = 0 since š(1) = 1 andš = āš¤(1) = 1. Invoking Lemma 2, it follows that
Re (š¤ (š„)) ā¤ 21 ā š„1 + š„ , (37)for all š„ ā (ā1, 1). Henceš (š„) = exp {āš¤ (š„)} = exp (āRe {š¤ (š„)})
ā„ exp(ā21 ā š„1 + š„) ,(38)
for all š„ ā (ā1, 1). That is,(1 ā 2 log(1 + š„2 ))
ā1 š (1 ā š„2 š)
ā„ exp(ā21 ā š„1 + š„) ,(39)
for all š„ ā (ā1, 1).Making the change š = (1āš„)/2, we obtain that š ā (0, 1)
and š (šš) ā„ exp(ā2 š1 ā š) (1 ā 2 log (1 ā š)) . (40)Therefore, if we consider š = |š§| with š§ ā D Ģø= 0 and we takeš = š§/|š§|, we conclude that
š (š§) ā„ exp(ā2 |š§|1 ā |š§|) (1 ā 2 log (1 ā |š§|))= š¹ (|š§|) .
(41)
This shows inequality (1).Now, if there existsš0 ā š½(ā)log such that |š0(š§0)| = š¹(|š§0|)
for some š§0 ā D\{0}, then arguing as in the proof of inequality(1), for š0 = š§0/|š§0|, the function,
š0 (š§) = (1 ā 2 log(1 + š§2 ))ā1 š0 (1 ā š§2 š0) , (42)
maps D into D \ {0}. Hence, there exists a holomorphicfunction š¤ mapping D into H+ such that š¤(1) = 0, āš¤(1) =1 > 0, and
š0 (š§) = exp (āš¤ (š§)) , (43)
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for all š§ ā D. In particular, for š„ = 1ā2|š§0| ā (ā1, 1), we haveexp (āReš¤ (š„)) = š0 (š„) =
š0 (((1 ā š„) /2) š0)1 ā 2 log ((1 + š„) /2)=
š0 (š§0)1 ā 2 log (1 ā š§0)= š¹ (š§0)1 ā 2 log (1 ā š§0)= exp(ā 2 š§01 ā š§0)= exp(ā21 ā š„1 + š„) .
(44)
Thus,
Reš¤ (š„) = 21 ā š„1 + š„ , (45)for some š„ ā (ā1, 1) and, by Lemma 2, we conclude that
š¤ (š§) = 21 ā š§1 + š§ (46)for all š§ ā D. Therefore,
š0 (1 ā š§2 š0) = (1 ā 2 log(1 + š§2 ))š0 (š§)= (1 ā 2 log(1 + š§2 )) exp(ā21 ā š§1 + š§) ,
(47)
for all š§ ā D. Hence, changing (1āš§)/2 by š§, which belongs toš·(1/2, 1/2), and using the identity principle for holomorphicfunctions, we obtain that
š0 (š§š0) = š¹ (š§) , (48)for all š§ ā D. This last relation implies that āš¹ālog ā¤ 1 whichis a contradiction to Proposition 7. This complete the proofof item (1).
(2) Arguing as in the proof of part (1), for |š| = 1 fixed, weset the function
š (š¢) = (1 ā 2 log(1 + š¢2 ))ā1 š (1 ā š¢2 š) , (49)
with š¢ ā D. We have shown that there exists a holomorphicfunction š¤ such that š(š¢) = exp{āš¤(š¢)} for all š¢ ā D.Furthermore, š¤ satisfies the hypothesis of Juliaās Lemma(Lemma 1); that is, š¤ is a holomorphic function in D āŖ {1},whichmapsD intoH+,š¤(1) = 0, and āš¤(1) = š = 1. Hence,for š = (1 ā š„)/(1 + š„) with š„ ā [0, 1) fixed, š¤ maps thehorodisk Ī(1, š) into the open Euclidean disk with center at(1 ā š„)/(1 + š„) and radius (1 ā š„)/(1 + š„). In particular, sinceš„ ā Ī(1, š), we have thatš¤(š„) ā š·š„ = š·((1 ā š„)/(1 + š„), (1 āš„)/(1 + š„)). Thus, Lemma 4 allows us to write
Re (š (š„)) = Re (exp (āš¤ (š„)))ā„ min {Re (exp (āš§)) : š§ ā š·š„}= exp(ā21 ā š„1 + š„) ,
(50)
for all š„ ā [0, 1). This last inequality is equivalent to writingRe((1 ā 2 log(1 + š„2 ))
ā1 š (1 ā š„2 š))ā„ exp (ā21 ā š„1 + š„) ,
(51)
for all š„ ā [0, 1) and from here we haveRe(š (1 ā š„2 š))
ā„ exp(ā21 ā š„1 + š„)(1 ā 2 log(1 + š„2 )) .(52)
Making the change š = (1 ā š„)/2, we obtain š ā (0, 1/2] sinceš„ ā [0, 1) and alsoRe (š (šš)) ā„ exp (ā2 š1 ā š) (1 ā 2 log (1 ā š)) . (53)
We conclude, as before, that
Re (š (š§)) ā„ exp(ā2 |š§|1 ā |š§|) (1 ā 2 log (1 ā |š§|)) , (54)for all |š§| ā¤ 1/2. This shows the inequality in the second partof the theorem.
Now, if there exists a function š0 ā š½(ā)log such thatReš(š§0) = š¹(|š§0|) for some š§0 ā š·(0, 1/2), then we candefine š0 = š§0/|š§0| and the function
š0 (š§) = (1 ā 2 log(1 + š§2 ))ā1 š0 (1 ā š§2 š0) , (55)
which maps D into D \ {0}. Hence, as before, there exists aholomorphic functionš¤mappingD intoH+ such thatš¤(1) =0, āš¤(1) = 1 > 0, and š0(š§) = exp(āš¤(š§)) for all š§ ā D. Inparticular, for š„ = 1 ā 2|š§0| ā [0, 1), we haveRe exp (āš¤ (š„)) = Reš0 (š„) = Reš0 (((1 ā š„) /2) š0)1 ā 2 log ((1 + š„) /2)
= Reš0 (š§0)1 ā 2 log (1 ā š§0)= š¹ (š§0)1 ā 2 log (1 ā š§0)= exp(ā 2 š§01 ā š§0)= exp (ā21 ā š„1 + š„) ;
(56)
that is,š¤(š„) is the value inš·š„ where Re exp(āš¤(š„)) attain itsminimum value, but by the proof of Lemma 4, we know thatthis happens if Imš¤(š„) = 0 andReš¤(š„) = ā2((1āš„)/(1+š„)).Now, by Lemma 2, we conclude thatš¤(š§) = 2((1āš§)/(1+š§))for all š§ ā D. As before, this last fact implies that š0(š§š0) =š¹(š§) for all š§ ā D and therefore āš¹ālog ā¤ 1 which is acontradiction to Proposition 7. This completes the proof ofitem (2).
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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 š ā N, we denote by š½(š)log the class of all holomorphicfunctions š ā Blog such that š(0) = 0, š(0) = 1, āšālog ā¤ 1and if š(š) = 0 for some š ā D then š(š)(š) = 0 for allš = 1, 2, . . . , š. Clearly we have
š½(ā)log =āāš=1
š½(š)log. (57)With these notations, we have the following result.
Theorem 10. For š ā N fixed, we set šš = āš/(š + 2). Thenfor every š ā š½(š)log we have the following:
(1) |š(š§)| ā„ š¹š(|š§|) = ((ššā|š§|)/(ššāš2š|š§|))š(1ā2 log(1āšš|š§|)), for all |š§| ā¤ šš.There is not a functionš0 ā š½(š)logsuch that |š0(š§0)| = š¹š(|š§0|) for some š§0 ā š·(0, šš) \{0}.
(2) Reš(š§) ā„ š¹š(|š§|) = ((šš ā |š§|)/(šš ā š2š|š§|))š(1 ā2 log(1 ā šš|š§|)), for all |š§| ā¤ āš(š + 2)/(2š + 1).Furthermore, there is not a function š0 ā š½(š)log suchthat Reš0(š§0) = š¹š(|š§0|) for some š§0 ā š·(0,āš(š + 2)/(2š + 1)) \ {0}.
Proof. (1) Let us fix |š| = 1, š ā N and šš = āš/(š + 2) < 1.We set the function
š (š¢) = (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš¢))ā1
ā š (šš ā ššš¢1 ā š2šš¢ š) ,(58)
with š¢ ā D, where log(š¤) denotes the principal logarithmicof the complex number š¤ ā š·(1, 1). Clearly the function šis holomorphic on D āŖ {1} and š(1) = 1 because š(0) = 1.Also, we have
š (š¢) = ā (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš¢))ā2
ā ( ā2š2š1 ā š2šš¢)š (šš ā ššš¢1 ā š2šš¢ š)
+ (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš¢))ā1
ā š (šš ā ššš¢1 ā š2šš¢ š)[šš (š2š ā 1)(1 ā š2šš¢)2 ] š.
(59)
And hence š(1) = 2š2š/(1 ā š2š) = š since š(0) = 1 andš(0) = 0 (by Lemma 6). Furthermore, since š ā š½(š)log andš(š¢0) = 0 if and only if š(((šš ā ššš¢0)/(1 ā š2šš¢0))š) = 0, we
conclude that all the zeros of the function š have multiplicityat less š.
On the other hand, since āšālog ā¤ 1 we haveš (š¢) ā¤ 1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš¢)ā1
ā log( š1 ā (šš ā ššš¢) / (1 ā š2šš¢)2)ā¤ 1Re (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš¢))ā log( š1 ā (šš ā ššš¢) / (1 ā š2šš¢)2)
= [log( š(1 ā š2š) / (1 ā š2šš¢)2)]ā1
ā log( š1 ā (šš ā ššš¢) / (1 ā š2šš¢)2) < 1,
(60)
for all š¢ ā D, since1 ā š2š1 ā š2šš¢
2 < 1 ā
šš ā ššš¢1 ā š2šš¢2 , (61)
for all š¢ ā D. Hence, we have shown that š(D) ā D. InvokingLemma 3, we conclude that |š(š„)| ā„ š„š for all š„ ā [0, 1).Therefore, for each š„ ā [0, 1), the following estimation holds:
1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš„)ā1
ā š (šš ā ššš„1 ā š2šš„ š)
ā„ š„š. (62)
That is,
š (šš ā ššš„1 ā š2šš„ š)
ā„ š„š (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš„)) ,
(63)
for all š„ ā [0, 1) since šš ā (0, 1).Next, we make the change š = (šš ā ššš„)/(1 ā š2šš„). Thenš ā (0, šš] since š„ ā [0, 1), š„ = (1/šš)((šš ā š)/(1 ā ššš)) and
we can write
š (šš) ā„ ( 1šššš ā š1 ā ššš)
š
ā (1 ā 2 log (1 ā š2š) + 2 log(1 ā š2š 1šššš ā š1 ā ššš))
= ( 1šššš ā š1 ā ššš)
š (1 ā 2 log (1 ā ššš)) .(64)
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8 Journal of Function Spaces
Finally, if we set š = š§/|š§| and š = |š§|, we conclude thatš (š§) ā„ ( 1šš
šš ā |š§|1 ā šš |š§|)š (1 ā 2 log (1 ā šš |š§|))
= (1 ā |š§| /šš1 ā šš |š§| )š (1 ā 2 log (1 ā šš |š§|))
= š¹š (|š§|) ,(65)
for all |š§| ā (0, šš]. 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 š0 ā š½(š)log such that|š0(š§0)| = š¹š(|š§0|) for some š§0 ā š·(0, šš) \ {0}, then we setš = š§0/|š§0| and the functionš0 (š¢) = (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš¢))ā1
ā š0 (šš ā ššš¢1 ā š2šš¢ š)(66)
withš¢ ā D.Wehave showed thatš0 satisfies all the hypothesisof Lemma 3. Furthermore, choosing š„ ā [0, 1) such that
šš ā ššš„1 ā š2šš„ =š§0 (67)
we obtainš0 (šš ā š§0šš ā š2š š§0)
=š0 (š„)
= (1 ā 2 log (1 ā š2š) + 2 log( 1 ā š2š1 ā šš š§0))
ā1
ā š0 (š§0) = š¹š (š§0)1 ā 2 log (1 ā šš š§0)
= ( šš ā š§0šš ā š2š š§0)š .
(68)
By Lemma 3, we conclude that š0(š§) = š§š for all š§ ā D.Hence,
š0 (šš ā ššš§1 ā š2šš§ š)= (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš§)) š§š,
(69)
for all š§ ā D. Changing (šš ā ššš§)/(1 ā š2šš§) by š§, we obtainthat
š0 (š§š) = (1 ā 2 log (1 ā ššš§)) ( šš ā š§šš ā š2šš§)š
= š¹š (š§) ,(70)
for all š§ ā š·(0, šš) and consequently for all š§ ā D. Thislast equality implies that āš¹šālog = āš0ālog ā¤ 1 which is acontradiction to Proposition 8.
(2) As before, for š ā N, we set šš = āš/(š + 2), we fix|š| = 1, and we define the functionš (š¢) = (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš¢))ā1
ā š (šš ā ššš¢1 ā š2šš¢ š) ,(71)
with š¢ ā D. In the first part we have shown that this functionsatisfies the hypothesis of Lemma 3. Hence Reš(š„) ā„ š„š forall (š ā 1)/(š + 1) ā¤ š„ < 1. Therefore
Re((1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš„))ā1
ā š (šš ā ššš„1 ā š2šš„ š)) ā„ š„š,
(72)
and thus we have
(1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš„))ā1
ā Re(š (šš ā ššš„1 ā š2šš„ š)) ā„ š„š, (73)
which is the same as
Re(š (šš ā ššš„1 ā š2šš„ š))
ā„ š„š log( š(1 ā š2š) / (1 ā š2šš„)2) .(74)
As before, we make the change š = (šš ā ššš„)/(1 ā š2šš„); then0 < š ā¤ āš(š + 2)/(2š + 1), š„ = (1/šš)((šš ā š)/(1 ā ššš)),and
Reš (šš) ā„ ( 1šššš ā š1 ā ššš)
š
ā log( š[(1 ā š2š) / (1 ā š2š (1/šš) ((šš ā š) / (1 ā ššš)))]2)
= ( 1šššš ā š1 ā ššš)
š (1 ā 2 log (1 ā ššš)) .(75)
Setting š = š§/|š§| and š = |š§|, we conclude thatReš (š§) ā„ ( 1šš
šš ā |š§|1 ā šš |š§|)š (1 ā 2 log (1 ā šš |š§|)) , (76)
for all |š§| ā¤ āš(š + 2)/(2š + 1). This shows the inequality inpart (2).
If there exists a function š0 ā š½(š)log such that Reš0(š§0) =š¹š(|š§0|) for some š§0 ā š·(0,āš(š + 2)/(2š + 1)) \ {0}, then weset š = š§0/|š§0| and the function
š0 (š¢) = (1 ā 2 log (1 ā š2š) + 2 log (1 ā š2šš¢))ā1
ā š0 (šš ā ššš¢1 ā š2šš¢ š) ,(77)
-
Journal of Function Spaces 9
withš¢ ā D.Wehave showed thatš0 satisfies all the hypothesisof Lemma 3. Furthermore, choosing š„ ā [0, 1) such that
šš ā ššš„1 ā š2šš„ =š§0 , (78)
we obtain
š„ = 1šššš ā š§01 ā šš š§0 , (79)
and since 0 < |š§0| ā¤ šš((š + 2)/(2š + 1)) we can see that(š ā 1)/(š + 1) ā¤ š„ < 1. Furthermore,Reš0 ( šš ā
š§0šš ā š2š š§0) = Reš0 (š„)
= (1 ā 2 log (1 ā š2š) + 2 log( 1 ā š2š1 ā šš š§0))
ā1
ā Reš0 (š§0) = š¹š (š§0)1 ā 2 log (1 ā šš š§0)
= ( šš ā š§0šš ā š2š š§0)š .
(80)
By Lemma 3, we conclude that š0(š§) = š§š for all š§ ā D.Hence, as before, this last fact implies that āš¹šālog = āš0ālog ā¤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 š is aholomorphic function on D and š§0 ā D, šš (š§0, š) denotethe radius of the largest schlicht disk on the Riemann surfaceš(D) centered at š(š§0) (a schlicht disk on š(D) centered atš(š§0) means that š maps an open subset of D containing š§0conformally onto this disk). With this notation, we have thefollowing results.
Corollary 11. If š ā š½(ā)log , thenšš (0, š) ā„ ā«1
0š¹ (|š§|) š |š§|
= ā«10exp(ā 2š”1 ā š”) (1 ā 2 log (1 ā š”)) šš”.
(81)
Proof. From the definition of šš (0, š), it follows the fact thatthere exists a simply connected domain šø ā D containing thezero such that š maps šø conformally onto an Euclidean diskwith center at š(0) and radius šš (0, š). This Euclidean diskmust meet the boundary of š(D) because, in other cases, theboundary of the set šø is a Jordan curve in the interior of Dand we can find an open set š ā D where š is univalent;hence š(š) contain an Euclidean disk with center at š(0)and radius greater than šš (0, š), which contradict with thedefinition of šš (0, š). We conclude then that there is a radial
segment Ī jointing š(0) to the boundary of š(D). Let š¾ beinverse image of Ī under š; then š¾ joint the point 0 to theboundary of D. Thus, fromTheorem 9, it follows that
šš (0, š) = ā«Ī|šš¤| = ā«
š¾
š (š§) |šš§|ā„ ā«10š¹ (š¾ (š”)) š¾ (š”) šš”
= ā«10š¹ (š¾ (š”))
š¾ (š”) š¾ (š”)š¾ (š”) šš”ā„ ā«10š¹ (š¾ (š”)) š¾ (š”) ā š¾
(š”)š¾ (š”) šš” = ā«1
0š¹ (š) šš
= ā«10exp (ā 2š1 ā š) (1 ā 2 log (1 ā š)) šš ā 0.4104136111,
(82)
where we have used Cauchy-Schwarzās inequality in thefourth line, š¾(š”) ā š¾(š”) is the scalar product of š¾(š”) and š¾(š”),and we have made the change š = |š¾(š”)| = āš¾(š”) ā š¾(š”), whereš ā 1ā as š” ā 1ā. This shows the result.
While for functions in the classš½(š)log wehave the following.Corollary 12. Suppose that š ā N is fixed. If š ā š½(š)log, then
šš (0, š) ā„ ā«āš/(š+2)0
š¹š (š”) šš”= ā«āš/(š+2)0
(1 ā š”/šš1 ā ššš” )š (1 ā 2 log (1 ā ššš”)) šš”.
(83)
Proof. Indeed, arguing as in the proof of Corollary 11, fromTheorem 10, it follows that
šš (0, š) = ā«Ī|šš¤| = ā«
š¾
š (š§) |šš§|
ā„ ā«āš/(š+2)0
š¹š (š) šš= ā«āš/(š+2)0
( šš ā ššš ā š2šš)š (1 ā 2 log (1 ā ššš)) šš”,
(84)
where we have used that š = |š¾(š”)| ā šš as š” ā 1ā. 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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