Question

Let \(\Omega\) be an open connected subset of \(\mathbb{C}\) containing \(U = \{z \in \mathbb{C} : |z| \le \frac{1}{2}\}\).
Let \(\mathcal{F} = \{ f : \Omega \to \mathbb{C} : f \text{ is analytic and } \sup_{z,w \in U} |f(z) - f(w)| = 1 \}\).
Consider the following statements:
P: There exists \(f \in \mathcal{F}\) such that \(|f'(0)| \ge 2\).
Q: \(|f^{(3)}(0)| \le 48\) for all \(f \in \mathcal{F}\), where \(f^{(3)}\) denotes the third derivative of \(f\).
Then

• A. P is TRUE
• B. Q is FALSE
• C. P is FALSE
• D. Q is TRUE

Hint:

For problems involving bounds on analytic functions or their derivatives on a disk, always consider the Schwarz Lemma (if the function maps the disk to itself and fixes the origin) and Cauchy's Inequality/Integral Formula. Often, a simple change of variables can transform the given disk into the unit disk where these theorems are most easily applied.

Answer 1

The Correct Option is C, D

Step 1: Understanding the Concept:
This problem tests properties of analytic functions defined on a disk, specifically bounds on their derivatives. The condition \(\sup_{z,w \in U} |f(z) - f(w)| = 1\) states that the diameter of the image of the disk \(U\) under \(f\) is 1. We can use this information with scaled versions of the Schwarz Lemma and Cauchy's Integral Formula for derivatives to evaluate the statements.
Step 3: Detailed Explanation:
Analysis of Statement P:
Let \(f \in \mathcal{F}\). Define a new function \(g(z) = f(z/2)\) for \(|z| \le 1\). This function g is analytic on the closed unit disk. The image of the unit disk under g, \(g(\{|z|\le 1\})\), is the same as the image of the disk U under f, \(f(U)\).
The diameter of the image of g is given as 1: \[ \text{diam}(g(\{|z|\le 1\})) = \sup_{|z_1|\le 1, |z_2|\le 1} |g(z_1) - g(z_2)| = \sup_{|w_1|\le 1/2, |w_2|\le 1/2} |f(w_1) - f(w_2)| = 1 \] Now, define another function \(h(z) = g(z) - g(0)\). This function is also analytic on the unit disk, and \(h(0) = 0\). For any \(z\) in the unit disk, \(|h(z)| = |g(z) - g(0)| \le \text{diam}(g(\{|z|\le 1\})) = 1\).
So, \(h(z)\) maps the unit disk into itself and fixes the origin. By the Schwarz Lemma, we must have \(|h'(0)| \le 1\).
Let's relate this back to \(f'(0)\).
\(h'(z) = g'(z)\), so \(|g'(0)| \le 1\).
From the definition \(g(z) = f(z/2)\), we use the chain rule to get \(g'(z) = f'(z/2) . \frac{1}{2}\).
At \(z=0\), this gives \(g'(0) = f'(0) . \frac{1}{2}\).
Substituting this into the inequality from the Schwarz Lemma: \[ |f'(0) . \frac{1}{2}| \le 1 \implies |f'(0)| \le 2 \] This inequality holds for every function \(f \in \mathcal{F}\). This means there cannot exist any function in \(\mathcal{F}\) for which \(|f'(0)| \ge 2\) unless equality holds. Equality in the Schwarz Lemma holds only for rotations, i.e., \(h(z)=e^{i\theta}z\). This would imply \(f(w) = 2e^{i\theta}w + c\). For such a function, \(\sup_{w_1,w_2 \in U} |f(w_1)-f(w_2)| = \sup |2e^{i\theta}(w_1-w_2)| = 2\sup|w_1-w_2| = 2(1)=2\), which is not 1.
So the function for which equality holds is not in \(\mathcal{F}\).
Therefore, we must have \(|f'(0)|<2\) for all \(f \in \mathcal{F}\).
Statement P, which claims there exists an \(f\) with \(|f'(0)| \ge 2\), is therefore FALSE. This makes option (C) correct.
Analysis of Statement Q:
Let \(f \in \mathcal{F}\). Let \(c = f(0)\) and define \(g(z) = f(z) - c\). Then \(g(0) = 0\) and \(f^{(3)}(0) = g^{(3)}(0)\).
For any \(z \in U\), we have \(|g(z)| = |f(z) - f(0)| \le \sup_{w \in U} |f(z) - f(w)| \le 1\).
So \(g(z)\) is an analytic function on \(\Omega\) such that \(|g(z)| \le 1\) for all \(z\) on the boundary of U, i.e., for \(|z|=1/2\).
By Cauchy's Integral Formula for the third derivative at the origin: \[ g^{(3)}(0) = \frac{3!}{2\pi i} \oint_{|z|=1/2} \frac{g(z)}{z^4} dz \] We can bound its magnitude: \[ |g^{(3)}(0)| \le \frac{3!}{2\pi} \oint_{|z|=1/2} \frac{|g(z)|}{|z|^4} |dz| \] On the path of integration, \(|z|=1/2\) and we have the bound \(|g(z)| \le 1\). \[ |f^{(3)}(0)| = |g^{(3)}(0)| \le \frac{6}{2\pi} . \frac{\sup_{|z|=1/2} |g(z)|}{(1/2)^4} . (\text{Length of path}) \] \[ |f^{(3)}(0)| \le \frac{3}{\pi} . \frac{1}{1/16} . (2\pi . 1/2) = \frac{3}{\pi} . 16 . \pi = 48 \] This inequality holds for all \(f \in \mathcal{F}\). Therefore, statement Q is TRUE. This makes option (D) correct.
Step 4: Final Answer:
P is FALSE and Q is TRUE. The correct statements from the list are (C) and (D).