THE SCHWARZ LEMMA AND ITS APPLICATION AT A BOUNDARY POINT
THE SCHWARZ LEMMA AND ITS APPLICATION AT A BOUNDARY POINT
The Pure and Applied Mathematics. 2014. Aug, 21(3): 219-227
• Received : May 31, 2014
• Accepted : July 15, 2014
• Published : August 31, 2014 PDF e-PUB PubReader PPT Export by style
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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.
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1. Introduction
One of the very basic theorems in complex analysis is the following Schwarz lemma  .
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  ).
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  .
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
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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  . In  , 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  .
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
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and hence PPT Slide
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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
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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  showed that the following inequality PPT Slide
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holds where F(z) = PPT Slide
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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  ).
Recently, Örnek  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
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Moreover, the equality in (1.5) holds if and only if PPT Slide
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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  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
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for | z | ＜ 1. Moreover , PPT Slide
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The equality in (2.1) for some nonzero z U or in (2.2) holds if and only if PPT Slide
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for some real θ.
Proof . Let g(z) = f(z) − 1 and let PPT Slide
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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
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It implies that PPT Slide
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Therefore, we have the inequality (2.1).
On the other hand, the facts that PPT Slide
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and | w ′(0)| ≤ 1 by the Schwarz lemma induce that PPT Slide
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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
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that is, PPT Slide
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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
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The equality in (2.6) holds if and only if PPT Slide
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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
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The inequality | w ′( z 0 )| ≥ 1 in (1.2) implies that PPT Slide
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If | f ′( z 0 )| = PPT Slide
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, then | w ′( z 0 )| = 1 and so by Theorem 1.4, w(z) = ze for some real θ . It means that PPT Slide
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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
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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  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
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The equality in (2.8) holds if and only if PPT Slide
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where θ satisfies e = 1/ PPT Slide
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.
Proof . By using the formula of w in (2.4), PPT Slide
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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
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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
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The equality in (2.10) holds for the function PPT Slide
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with z 0 = 1 where b = PPT Slide
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and a = 1, i.e ., PPT Slide
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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
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Since, PPT Slide
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we have PPT Slide
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Hence, PPT Slide
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For the equality in (2.10), choose arbitrary b satisfying 0 ≤ b ≤ 1 and let PPT Slide
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Then, PPT Slide
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and it implies that PPT Slide
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Therefore the equality in (2.10) holds at z 0 = 1 where b = PPT Slide
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.
On the other hand, PPT Slide
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In order to satisfy that f (1) = 2, the condition a = 1 should be satisfied. So, PPT Slide
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where b = PPT Slide
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= | 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
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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