Let \(f\) be a non-constant analytic function on \(\mathbb D\) with \(f(\mathbb D) \subseteq \mathbb D\). Use \(\psi_{a} (f(z))\) (where \(a=f(0)\), \(\displaystyle \psi_a(z) = \frac{a - z}{1 - \bar{a}z}\)) to prove that \(\displaystyle \frac{|f(0)| - |z|}{1 + |f(0)||z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)||z|}\).
Solution
Goal: Let \(f: \mathbb{D} \to \mathbb{D}\) be a holomorphic self-map of the unit disk with \(a = f(0) \in \mathbb{D}\). Using the Blaschke automorphism \(\psi_a(w) = \frac{a - w}{1 - \bar{a}w}\), prove that for all \(z \in \mathbb{D}\): \[\frac{|a| - |z|}{1 + |a||z|} \leq |f(z)| \leq \frac{|a| + |z|}{1 - |a||z|}.\]
Step 1: Application of the Schwarz Lemma to the Composite Map
1. Define \(g(z) = \psi_a(f(z))\). Then \(g\) satisfies the hypotheses of the Schwarz Lemma.
1.1. The Möbius transformation \(\psi_a(w) = \frac{a - w}{1 - \bar{a}w}\) is an automorphism of \(\mathbb{D}\) (i.e. \(\psi_a \in \text{Aut}(\mathbb{D})\)) with \(\psi_a(a) = 0\) and \(\psi_a^{-1} = \psi_a\).
For \(|a| < 1\), the map \(\psi_a\) is a Blaschke factor: it maps \(\mathbb{D}\) to itself, maps the unit circle to itself, satisfies \(\psi_a(a) = 0\), and is its own inverse since \(\psi_a(\psi_a(w)) = w\).
1.2. Since \(f(\mathbb{D}) \subseteq \mathbb{D}\), the composite map \(g(z) = \psi_a(f(z))\) is holomorphic from \(\mathbb{D}\) to \(\mathbb{D}\).
\(g\) is the composition of the holomorphic map \(f\) with the holomorphic automorphism \(\psi_a\), and its image lies in \(\mathbb{D}\) because both factors map into \(\mathbb{D}\).
1.3. \(g(0) = \psi_a(f(0)) = \psi_a(a) = 0\).
By hypothesis \(f(0) = a\), and \(\psi_a(a) = 0\) by 1.
1.4. By the Schwarz Lemma, \(|g(z)| \leq |z|\) for all \(z \in \mathbb{D}\).
\(g \colon \mathbb{D} \to \mathbb{D}\) is holomorphic with \(g(0) = 0\), so the Schwarz Lemma applies and gives \(|g(z)| \le |z|\).
1.5. Q.E.D.
2. Express \(f(z)\) in terms of \(g(z)\).
2.1. Since \(\psi_a\) is an involution, \(g(z) = \psi_a(f(z)) \implies f(z) = \psi_a(g(z)) = \frac{a - g(z)}{1 - \bar{a} g(z)}\).
Applying \(\psi_a\) to both sides of \(g(z) = \psi_a(f(z))\) and using \(\psi_a \circ \psi_a = \operatorname{id}\) gives \(f(z) = \psi_a(g(z))\).
2.2. Define \(w = g(z)\), which satisfies \(|w| \leq |z| < 1\).
By <1>1.<2>4>, \(|g(z)| \le |z|\), and \(|z| < 1\) since \(z \in \mathbb{D}\).
2.3. Q.E.D.
Step 2: Upper Bound for \(|f(z)|\)
3. Prove \(|f(z)| \leq \frac{|a| + |z|}{1 - |a||z|}\).
3.1. Using the triangle inequality on the numerator of \(f(z) = \frac{a - w}{1 - \bar{a}w}\): \[|a - w| \leq |a| + |w| \leq |a| + |z|.\]
The triangle inequality gives \(|a - w| \le |a| + |w|\), and \(|w| \le |z|\) by <1>2.<2>2>.
3.2. Using the reverse triangle inequality on the denominator: \[|1 - \bar{a}w| \geq 1 - |\bar{a}w| = 1 - |a||w| \geq 1 - |a||z| > 0.\]
The reverse triangle inequality gives \(|1 - \bar{a}w| \ge 1 - |\bar{a}w| = 1 - |a||w|\); since \(|w| \le |z| < 1\) and \(|a| < 1\), this is at least \(1 - |a||z| > 0\).
3.3. Combining the bounds for numerator and denominator: \[|f(z)| = \frac{|a - w|}{|1 - \bar{a}w|} \leq \frac{|a| + |z|}{1 - |a||z|}.\]
The numerator is at most \(|a| + |z|\) and the positive denominator is at least \(1 - |a||z|\), so the quotient is at most \(\frac{|a| + |z|}{1 - |a||z|}\).
3.4. Q.E.D.
Step 3: Lower Bound for \(|f(z)|\)
4. Prove \(|f(z)| \geq \frac{|a| - |z|}{1 + |a||z|}\).
4.1. If \(|z| \geq |a|\), then \(\frac{|a| - |z|}{1 + |a||z|} \leq 0\), so the inequality \(|f(z)| \geq 0 \geq \frac{|a|-|z|}{1+|a||z|}\) holds.
The modulus of any complex number is non-negative, and \(\frac{|a| - |z|}{1 + |a||z|} \le 0\) when \(|z| \ge |a|\), so \(|f(z)| \ge 0\) dominates it.
4.2. Now assume \(|z| < |a|\).By the reverse triangle inequality on the numerator: \[|a - w| \geq |a| - |w| \geq |a| - |z| > 0.\]
The reverse triangle inequality gives \(|a - w| \ge |a| - |w|\); since \(|w| \le |z| < |a|\), this is at least \(|a| - |z| > 0\).
4.3. By the triangle inequality on the denominator: \[|1 - \bar{a}w| \leq 1 + |\bar{a}w| = 1 + |a||w| \leq 1 + |a||z|.\]
The triangle inequality gives \(|1 - \bar{a}w| \le 1 + |\bar{a}w| = 1 + |a||w|\), and \(|w| \le |z|\).
4.4. Combining the bounds: \[|f(z)| = \frac{|a - w|}{|1 - \bar{a}w|} \geq \frac{|a| - |z|}{1 + |a||z|}.\]
The numerator is at least the positive quantity \(|a| - |z|\) and the denominator is at most \(1 + |a||z|\), so the quotient is at least \(\frac{|a| - |z|}{1 + |a||z|}\).
4.5. Q.E.D.
Step 4: Conclusion
5. The double inequality holds for all \(z \in \mathbb{D}\).
5.1. By 3 and 4, for all \(z \in \mathbb{D}\): \[\frac{|f(0)| - |z|}{1 + |f(0)||z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)||z|}.\]
Combining the upper bound <1>3.<2>3> and the lower bound <1>4.<2>4>, and substituting \(a = f(0)\), gives the double inequality.
5.2. Q.E.D.