Research article · 2020 · Full text
International Journal of Statistics and Applied MathematicsSchwarz lemma applications
Volume 5(3) · Pages 42–46
Research areasAnalytic and geometric function theoryHolomorphic and Operator Theory
Abstract
Author-writtenIn this Paper, we have studied the Schwarz’s lemma and arbitrary analytic functions on the open unit disk D and we have proved that range of all the analytic functions on the open unit disk always lies in the closed annulus with centre origin and inner and outer radius depends on the modulus of the function at origin and also discuss its one more application.
Summary
Source-checkedPlain-language reading of the full paper, prepared with AI and checked against the text. Each point links to the authors' own words.
Why it matters
- The study focuses on Schwarz's lemma and the behavior of arbitrary analytic functions within the open unit disk D. View evidence ↗
- An analytic function is defined as one that is differentiable at a point and has a continuous derivative. View evidence ↗
- In complex analysis, a region is defined as an open connected subset of the complex plane C. View evidence ↗
The question
- The paper aims to prove that the range of all analytic functions on the open unit disk is contained within a closed annulus centered at the origin, with radii determined by the function's modulus at the origin. View evidence ↗
What they did
- The author utilizes Schwarz's lemma, which applies to analytic functions mapping the open unit disk D into itself such that the origin is mapped to the origin. View evidence ↗
- The Maximum Modulus Theorem is used as a foundational tool to determine if a function must be constant based on its maximum value in a region. View evidence ↗
- The proof for the range of analytic functions is conducted by analyzing three distinct cases based on the modulus of the function at the origin: |f(0)| = 1, |f(0)| = 0, and |f(0)| ≠ 0,1. View evidence ↗
What they found
- In the case where |f(0)| = 0, the modulus of the function |f(z)| lies within an annulus defined by the inner radius r(z) = -|z| and outer radius R(z) = |z|. View evidence ↗
- For a non-constant analytic function on the disk whose real part is always non-zero, it is proven that the real part Ref(z) must be strictly greater than 0 for all z in D. View evidence ↗
What it means
- The author concludes that the modulus of any analytic function on the open unit disk is always contained within an annulus formed by two circles. View evidence ↗
- The author states that these findings can be used to help prove other standard results in complex analysis, such as Bloch's theorem and the Little Picard Theorem. View evidence ↗
Visual summary
1 visualsIllustrations explain the paper; they do not show study data.
Analytic Function Range as an Annulus
AI-generated conceptual illustration (Geometric representation of a complex plane mapping). It explains the idea and does not show study data.
View source evidence ↗Introduction
1.1 Analytic Function Let G be an open set in the complex plane C then a function f defined on G is said to be analytic at a ∈ G if f is differentiable at z = a and f0(a) is continuous. Note that: a function f defined on an open subset G of C is said to be analytic on G if f is analytic at each point of G. 1.2 Region An subset G of C is said connected if any two points in G are joined by a continuous function whose image lies completely in G. An open connected subset G of C is said to be a region in C. [1] Notation 1.3 Maximum Modulus Theorem Let f be analytic function on some region G ⊆ C if there exists some a ∈ G such that |f (z)| ≤ |f (a)| ∀ z ∈ G. Then f must be a constant function. 2.
Schwarz’s lemma Let f be analytic function from D into D (i.e |f(z)| ≤ 1 ∀z ∈ D) mapping 0 to 0 then Moreover [2], equality occurs in one of the inequality above for some 0 6= z ∈ D then ∃ a constant c ∈ C with |c| = 1 with F (w) = cw ∀ w ∈ D Maths 2020; 5(3): 42-46 Praveen Sharma P.G. Student, Department of mathematics, University of Delhi, Delhi, India Praveen Sharma P.G. Student, Department of mathematics, University of Delhi, Delhi, India http://www.mathsjournal.com Theorem 2.1. Let f: D → D be analytic function then |f(z)| ∈ ann{0,r(z),R(z)} ∀ z ∈ D where r and R are function on the open unit disk D given by ∀ z ∈ D Proof. • Case1: when |f(0)| = 1 Since f(z) ∈ D ⇒ |f(z)| ≤ 1 ∀z ∈ D ∴ f attains its maximum modulus in the open connected set D then by Maximum modulus theorem (1.3) f must be a constant function with modulus 1. |f(z)| = 1 ∀z ∈ D Then Case 2: when |f(0)| = 0 Since f is analytic function ⇒ f(0) = 0 Then f is analytic function mapping D into D mapping 0 to 0 Then By Schwarz’s Lemma we have |f(z)| ≤ |z| ∀z ∈ D ⇒ −|z| ≤ |f(z)| ≤ |z| ∀z ∈ D Where r(z) = −|z| and R(z) = |z| ⇒ |f(z)| ∈ ann{0,−|z|,|z|} ∀ z ∈ D Case3: When |f(0)| 6= 0,1 Define a function h on D given by Clearly h(0) = 0 To show |h(z)| ≤ 1 z ∈ D http://www.mathsjournal.com Since Now ∴ h is analytic function from the open unit disk D into D mapping 0 to 0.
Then By Schwarz’s Lemma we have By Triangle Inequality we have ||f(z)| − |f(0)|| ≤ |f(z) − f(0)| ≤ |z|(|1 − |f(0)||f(z)|) ∀z ∈ D Taking first two parts of inequality (1) them for each z ∈ D we have Taking last two inequalities of (1) for each z ∈ D we have ∴ from (2) and (3) we have |f(z)| ∈ ann{0,r(z),R(z)} ∀ z ∈ D http://www.mathsjournal.com Theorem 2.2. Let f is a non constant analytic function on the disk whose real part is always non zero Then Show that Ref (z) > 0 ∀z ∈ D Moreover if f (0) = 1 Then Proof. Suppose if possible Ref (z0) = 0 for some z0 ∈ D Define a function h on D by h(z) = exp(−f(z)) ∀z ∈ D Then |h(z)| = |exp(−f(z)| = |exp(−Ref(z))| ≤ 1 ∀z ∈ D (∵ Ref(z) ≥ 0) Also |h(z0)| = |exp(−f(z0))| = 1 ∴ h attains its maximum modulus on the region D By maximum modulus Theorem (1.3) h must be a constant function on the disk D ⇒ f must be a constant function on the disk D Which is a contradiction since f is a non-constant analytic function ∴ Ref (z) > 0 ∀z ∈ D Let f(0) = 1 Then Define a function g on D by Clearly To show |g(z)| ≤ 1 z ∈ D Let z = x + ιy ∈ D where x > 0 (∵ Ref(z) > 0) Then ∴ g is analytic function from D into D mapping 0 to 0 Then By Schwarz’s Lemma we have http://www.mathsjournal.com Then by Triangle inequality we have ||f(z)| − 1| ≤ |f(z) − 1| ≤ |z|(|f(z) + 1|) ≤ |z|(|f(z)| + 1) ∀z ∈ D ⇒ ||f(z)| − 1| ≤ |z|(|f(z)| + 1) ∀z ∈ D ⇒ −|z|(1 + |f(z)|) ≤ |f(z)| − 1 ≤ |z|(1 + |f(z)|) ∀z ∈ D (1) By Taking first two inequalities of (1) for each z ∈ D we have (2) By taking last two inequalities of (1) for each z ∈ D we have (3) Frpm (2) and (3) we have
Conclusion
In this paper, we conclude that modulus of any analytic function on the open unit disk always lies in the annulus of two circles which has various applications to study the range of various type of analytic functions on the disks of large radius. This result also helps to prove the some typical results in the complex analysis like Little Picard Theorem and Bloch’s theorem.
References
3 in the original paper- James Ward Brown, Ruel Vance Churchill et al. Complex variables and applications. Boston: McGraw-Hill Higher Education, 2009.
- John B Conway. Functions of one complex variable II, volume
- Springer Science & Business Media, 2012.