Advanced
THE SCHWARZ LEMMA AND ITS APPLICATION AT A BOUNDARY POINT
THE SCHWARZ LEMMA AND ITS APPLICATION AT A BOUNDARY POINT
Journal of the Korean Society of Mathematical Education Series B The Pure and Applied Mathematics. 2014. Jul, 21(3): 219-227
Copyright © 2014, Korean Society of Mathematical Education
  • Received : May 31, 2014
  • Accepted : July 15, 2014
  • Published : July 28, 2014
Download
PDF
e-PUB
PubReader
PPT
Export by style
Share
Article
Author
Metrics
Cited by
TagCloud
About the Authors
MOONJA JEONG

Abstract
In this note we study the Schwarz lemma and inequalities for some holomorphic functions on the unit disc. Also, we obtain the inequality of the derivative of holomorphic maps at a boundary point of the unit disc and find a holomorphic map to satisfy the equality.
Keywords
1. Introduction
One of the very basic theorems in complex analysis is the following Schwarz lemma [4] .
Theorem 1.1 (Schwarz Lemma) . Let f be a holomorphic function of the open unit disc U = { z ∈ ℂ : | z | < 1} into U with f (0) = 0. Then | f ′(0)| ≤ 1 and | f(z) | ≤ | z | for all z U with the equality only for f(z) = ez with real θ .
If, in addition, f has multiple zeroes at z = 0, then the Schwarz lemma results in the following (see [7] ).
Corollary 1.2. Let f be a holomorphic function of U into U with f (0) = f ′(0) = … = f (n−1) (0) = 0. Then | f(z) | ≤ | z | n for all | z | < 1 with the equality only for f(z) = ezn with real θ .
More generally, the Schwarz lemma can be applied to a function with the information f ( α ) = β for some α, β with | α | < 1, | β | < 1 instead of f (0) = 0 and it is called the Schwarz-Pick lemma [2] .
Corollary 1.3 (Schwarz-Pick Lemma) . Let f be a holomorphic function of U into U with f(α) = β for some α, β with | α | < 1, | β | < 1. Then
PPT Slide
Lager Image
for | α | < 1.
If f in Corollary 1.3 fixes α , then | f ′( α )| ≤ 1. Note that the equality in the above corollary holds only for Möbius transformation mapping the open unit disc into itself.
The Schwarz lemma looks a simple result, but it is highly influential in the function theory of the complex analysis. It is used to get properties of holomorphic functions of the unit disc into itself at a boundary point of the unit disc. For historical background about the Schwarz lemma and its applications on the boundary of the unit disc, we refer to Boas [1] . In [3] , we find a holomorphic self map defined on the closed unit disc with fixed points only on the boundary of the unit disc.
Now, our concern is for holomorphic functions mapping the unit disc into itself at a boundary point of the unit disc. From the Schwarz lemma, it is known that if a holomorphic function f of the unit disc into itself with f (0) = 0 extends continuously to a boundary point z 0 with | z 0 | = 1, | f ( z 0 )| = 1, and f ′( z 0 ) exists, then | f ′( z 0 )| ≥ 1. The following theorem called the boundary Schwarz lemma can be found in [6] .
Theorem 1.4 (The boundary Schwarz Lemma) . Let f be a holomorphic function of U into U with f (0) = 0. Assume that for some point z 0 with | z 0 | = 1, f extends continuously to z 0 , | f ( z 0 )| = 1, and f ′( z 0 ) exists. Then
PPT Slide
Lager Image
and hence
PPT Slide
Lager Image
The equality in (1.2) holds if and only if f(z) = ze for some real θ .
If, in addition, the function f has the property f (0) = f ′(0) = … = f (n−1) (0) = 0, n ∈ ℕ, then
PPT Slide
Lager Image
The equality in (1.3) holds if and only if f(z) = zne for some real θ .
Remark 1.5. Under the same hypothesis as in Theorem 1.4 except f (0) = 0, Osserman [6] showed that the following inequality
PPT Slide
Lager Image
holds where F(z) =
PPT Slide
Lager Image
instead of the inequality (1.1).
The assumption in Theorem 1.4 that f extends continuously to z 0 with | z 0 | = 1, | f ( z 0 )| = 1, and f ′( z 0 ) exists can be changed to the assumption that f has a radial limit w 0 at z 0 with | z 0 | = 1, | w 0 | = 1, f has a radial derivative at z 0 . (see [6] ).
Recently, Örnek [5] proved the following inequality at a boundary point of the unit disc.
Theorem 1.6. Let f be a holomorphic function in U with f (0) = 1 and | f ( z )− | < for | z | < 1 and 1/2 < ≤ 1. If for some z 0 with | z 0 | = 1, f has an angular limit f ( z 0 ) at z 0 , f ( z 0 ) = 2 , then
PPT Slide
Lager Image
Moreover, the equality in (1.5) holds if and only if
PPT Slide
Lager Image
for a real θ.
In this paper, we show some inequalities at a boundary point for different form of holomorphic functions and find the condition for equality.
2. The Schwarz Lemma and its Application at a Boundary Point
The Schwarz lemma means that any holomorphic function of the unit disc into itself with zero fixed maps each disc centered at zero into a smaller one. Moreover it maps each disc centered at zero into a strictly smaller one if it is not a rotation. From now on, more generally we consider a holomorphic function with zero not fixed. Örnek [5] considered a holomorphic function f on U with f (0) = 1, | f(z) − 𝜖| < 𝜖 where 𝜖 > 1/2. We consider a different form of holomorphic functions and get the following proposition by the similar method.
Proposition 2.1. Let f be a holomorphic function on U satisfying | f(z) − 1| < 1 with f (0) = a where 0 < a < 2. Then, f satisfies the inequality
PPT Slide
Lager Image
for | z | < 1. Moreover ,
PPT Slide
Lager Image
The equality in (2.1) for some nonzero z U or in (2.2) holds if and only if
PPT Slide
Lager Image
for some real θ.
Proof . Let g(z) = f(z) − 1 and let
PPT Slide
Lager Image
for z U .
Then g and w are holomorphic functions on U with | g(z) | < 1 and | w(z) | < 1 for | z | < 1 and w (0) = 0. Hence w satisfies the condition for the Schwarz lemma.
By the Schwarz lemma, | w(z) | ≤ | z | for | z | < 1. Hence,
PPT Slide
Lager Image
It implies that
PPT Slide
Lager Image
Therefore, we have the inequality (2.1).
On the other hand, the facts that
PPT Slide
Lager Image
and | w ′(0)| ≤ 1 by the Schwarz lemma induce that
PPT Slide
Lager Image
Hence, | f ′(0)| ≤ 1 − ( a − 1) 2 = a (2 − a ).
The equality in (2.1) for some nonzero z U or in (2.2) holds if and only if
PPT Slide
Lager Image
that is,
PPT Slide
Lager Image
for some real θ .                                    ⧠
Remark 2.2. From the formula (2.5),
  • |f(z)−a| = |f(z)| −a
  • and |(1 −a)f(z)+a| = |(1 −a)f(z)| +a
should be satisfied at some nonzero point z U that the equality in (2.1) holds, i.e., 0 < a < 1 and f(z) is real with a < f(z) < 2 at the above nonzero point z U .
Therefore, if f is the function in (2.3) and 0 < a < 1, then at some nonzero point z U satisfying a < f(z) < 2 the equality in (2.1) holds.
Theorem 2.3. Let f be a holomorphic function on U satisfying | f(z) −1| < 1 with f (0) = a where 0 < a < 2. Assume that for some z 0 with | z 0 | = 1, f extends continuously to z 0 , f ( z 0 ) = 2, and f ′( z 0 ) exists. Then
PPT Slide
Lager Image
The equality in (2.6) holds if and only if
PPT Slide
Lager Image
where some real θ satisfies that e = 1/ z 0 .
Proof . Let w be the function in (2.4). Then w′(z) satisfies that
PPT Slide
Lager Image
The inequality | w ′( z 0 )| ≥ 1 in (1.2) implies that
PPT Slide
Lager Image
If | f ′( z 0 )| =
PPT Slide
Lager Image
, then | w ′( z 0 )| = 1 and so by Theorem 1.4, w(z) = ze for some real θ . It means that
PPT Slide
Lager Image
for some real θ . By the condition f ( z 0 ) = 2,
  • a(1 +z0eiθ) = 2{1 − (1 −a)z0eiθ}.
Hence, z 0 e = 1 and so θ satisfies that e = 1/ z 0 . Conversely, for the given function
PPT Slide
Lager Image
where θ satisfies that e = 1/ z 0 , the equality in (2.6) holds.                                ⧠
Remark 2.4. The inequalities in Proposition 2.1 and Theorem 2.3 with a = 1 coincide with the inequalities in Theorem 1.6 and a theorem in Örnek [5] with 𝜖 = 1.
Now, we consider a holomorphic function f with f (0) − a = f ′(0) = … = f (n−1) (0) = 0. A function given by
  • f(z)=a+cnzn+cn+1zn+1+ … ,n∈ ℕ
with cn ≠ 0, is such a holomorphic function. If we change the role | w ′( z 0 )| ≥ 1 by | w ′( z 0 )| ≥ n in the proof of Theorem 2.3, then the following corollary holds.
Corollary 2.5. Let f be a holomorphic function defined on U by
  • f(z)=a+cnzn+cn+1zn+1+ … ,n≥ 1
satisfying | f(z) − 1| < 1 on U where 0 < a < 2 and cn ≠ 0. Assume that for some z 0 with | z 0 | = 1, f extends continuously to z 0 , f ( z 0 ) = 2, and f ′( z 0 ) exists. Then
PPT Slide
Lager Image
The equality in (2.8) holds if and only if
PPT Slide
Lager Image
where θ satisfies e = 1/
PPT Slide
Lager Image
.
Proof . By using the formula of w in (2.4),
PPT Slide
Lager Image
where bn , b n+1 , … ∈ ℂ and bn = cn /{1 − ( a − 1) 2 } ≠ 0.
Hence, w (0) = w ′(0) = … = w (n−1) (0) = 0. By Theorem 1.4 and (2.7),
PPT Slide
Lager Image
holds and we get (2.8).                                             ⧠
The following theorem provides a refined inequality at a boundary point z 0 than Theorem 2.3. If we apply the inequality (2.2), we find that the following inequality (2.10) implies the inequality (2:6).
Theorem 2.6. Let f be a holomorphic function on U satisfying | f ( z )−1| < 1 with f (0) = a where 0 < a < 2. Assume that for some z 0 with | z 0 | = 1, f extends continuously to z 0 , f ( z 0 ) = 2, and f ′( z 0 ) exists. Then
PPT Slide
Lager Image
The equality in (2.10) holds for the function
PPT Slide
Lager Image
with z 0 = 1 where b =
PPT Slide
Lager Image
and a = 1, i.e .,
PPT Slide
Lager Image
with z 0 = 1 where 0 ≤ b = | f ′(0)| ≤ 1.
Proof . Let w be the function in (2.4). By applying the inequality (1.1) to w′(z) and the equation (2.7),
PPT Slide
Lager Image
Since,
PPT Slide
Lager Image
we have
PPT Slide
Lager Image
Hence,
PPT Slide
Lager Image
For the equality in (2.10), choose arbitrary b satisfying 0 ≤ b ≤ 1 and let
PPT Slide
Lager Image
Then,
PPT Slide
Lager Image
and it implies that
PPT Slide
Lager Image
Therefore the equality in (2.10) holds at z 0 = 1 where b =
PPT Slide
Lager Image
.
On the other hand,
PPT Slide
Lager Image
In order to satisfy that f (1) = 2, the condition a = 1 should be satisfied. So,
PPT Slide
Lager Image
where b =
PPT Slide
Lager Image
= | f ′(0)| ≤ 1.                                             ⧠
Remark 2.7. The function f in (2.11) can be represented by the following interesting Mclaurin series at zero.
PPT Slide
Lager Image
PPT Slide
Lager Image
where 0 ≤ b = | f ′(0)| = f ′(0) ≤ 1 for | z | < 1.
  • For example, ifb= 0, thenf(z)= 1 +z2.
  • Ifb= 1, thenf(z)= 1 +z.
  • Ifb=, thenf(z)== 1 +z+z2+ … .
References
Boas H.P. 2010 Julius and Julia: Mastering the Art of the Schwarz Lemma Amer. Math. Monthly 11 (7) 770 - 785    DOI : 10.4169/000298910X521643
Greene R. , Krantz S. 2002 Function theory of one complex variable, Graduate studies on Mathematics Amer. Math. Soc. Providence
Jeong M. 2011 The Schwarz lemma and boundary fixed points J. Korean. Soc. Math. Educ. Ser. B: Pure Appl. Math. 18 (3) 275 - 284
Nehari Z. 1952 Conformal Mapping Dover publications, Inc. New York
Örnek B. 2013 Scharpened forms of the Schwarz lemma on the boundary Bull. Korean Math. Soc. 50 2053 - 2059    DOI : 10.4134/BKMS.2013.50.6.2053
Osserman R. 2000 A Sharp Schwarz Inequality on the boundary Proc. Amer. Math. Soc. 128 3513 - 3517    DOI : 10.1090/S0002-9939-00-05463-0
Silverman H. 1975 Complex Variables Houghton Mifflin