Assume \(f(z)\) is analytic in \({\mathbb D}\) and \(f(0)=0\) and is not a rotation (i.e. \(f(z) \neq e^{i \theta} z\)). Show that \(\displaystyle \sum_{n=1}^\infty f^{n}(z)\) converges uniformly to an analytic function on compact subsets of \({\mathbb D}\), where \(f^{n+1}(z) = f(f^{n}(z))\).
Solution
Goal: Let \(f: \mathbb{D} \to \mathbb{D}\) be a holomorphic self-map of the unit disk with \(f(0) = 0\) that is not a rotation (\(f(z) \neq e^{i\theta} z\)). Prove that the series of iterates \(\sum_{n=1}^\infty f^n(z)\) converges locally uniformly on \(\mathbb{D}\) to a holomorphic function, where \(f^1(z) = f(z)\) and \(f^{n+1}(z) = f(f^n(z))\).
Step 1: Strict Contraction Bound from the Schwarz Lemma
1. Apply the Schwarz Lemma to \(f\).
1.1. By the Schwarz Lemma, for any holomorphic \(f: \mathbb{D} \to \mathbb{D}\) with \(f(0) = 0\), \(|f(z)| \leq |z|\) for all \(z \in \mathbb{D}\).
The Schwarz Lemma states that a holomorphic self-map \(f \colon \mathbb{D} \to \mathbb{D}\) with \(f(0) = 0\) satisfies \(|f(z)| \le |z|\) for all \(z \in \mathbb{D}\) and \(|f'(0)| \le 1\).
1.2. The auxiliary function \(h(z) = \frac{f(z)}{z}\) for \(z \neq 0\) and \(h(0) = f'(0)\) is holomorphic on \(\mathbb{D}\) and satisfies \(|h(z)| \leq 1\) for all \(z \in \mathbb{D}\).
Since \(f(0) = 0\), the quotient \(\frac{f(z)}{z}\) has a removable singularity at \(0\) with value \(f'(0)\), so \(h\) is holomorphic on \(\mathbb{D}\); the Schwarz Lemma bound \(|f(z)| \le |z|\) gives \(|h(z)| = \frac{|f(z)|}{|z|} \le 1\) for \(z \neq 0\), and \(|h(0)| = |f'(0)| \le 1\).
1.3. If \(|h(z_0)| = 1\) for some \(z_0 \in \mathbb{D}\), the Maximum Modulus Principle implies \(h(z) \equiv e^{i\theta}\) is constant, so \(f(z) = e^{i\theta} z\).
A holomorphic function attaining its maximum modulus at an interior point is constant; since \(|h| \le 1\) and \(|h(z_0)| = 1\), \(h\) is constant of modulus \(1\), so \(h \equiv e^{i\theta}\) and \(f(z) = z h(z) = e^{i\theta} z\).
1.4. Since \(f\) is given to not be a rotation, \(h(z)\) cannot attain the value \(1\) anywhere in \(\mathbb{D}\).Therefore, \(|h(z)| < 1\) for all \(z \in \mathbb{D}\).
If \(|h(z_0)| = 1\) for some \(z_0\), then by 3 \(f\) would be a rotation \(z \mapsto e^{i\theta} z\), contradicting the hypothesis; hence \(|h(z)| < 1\) everywhere.
1.5. Q.E.D.
2. Uniform geometric bound on compact subsets \(K \subset \mathbb{D}\).
2.1. Let \(K \subset \mathbb{D}\) be an arbitrary compact subset.
We fix an arbitrary compact subset \(K\) of \(\mathbb{D}\) on which to establish the uniform bound.
2.2. Since \(K\) is compact, \(r_0 \coloneqq \sup_{z \in K} |z| < 1\).
The function \(z \mapsto |z|\) is continuous, so it attains its maximum on the compact set \(K\); since \(K \subset \mathbb{D}\), this maximum \(r_0\) is strictly less than \(1\).
2.3. Choose \(R\) such that \(r_0 \leq R < 1\).The closed disk \(\overline{D}(0, R) = \{|z| \leq R\}\) is compact and contains \(K\).
Any \(R\) with \(r_0 \le R < 1\) works; then \(|z| \le r_0 \le R\) for every \(z \in K\), so \(K \subseteq \overline{D}(0, R) \subset \mathbb{D}\).
2.4. Since \(|h(z)| < 1\) is continuous on the compact set \(\overline{D}(0, R)\), it achieves its maximum: \[c \coloneqq \max_{|z| \leq R} |h(z)| = \max_{|z| \leq R} \left| \frac{f(z)}{z} \right| < 1.\]
The continuous function \(|h|\) attains its maximum \(c\) on the compact disk \(\overline{D}(0, R)\) by the Extreme Value Theorem; since \(|h(z)| < 1\) for every \(z \in \mathbb{D}\) by <1>1.<2>4>, this maximum satisfies \(c < 1\).
2.5. Thus for all \(z \in \overline{D}(0, R)\), \(|f(z)| \leq c |z|\).
For \(z \neq 0\), \(|f(z)| = |z| \cdot |h(z)| \le c |z|\); at \(z = 0\) both sides vanish, so the inequality holds for all \(z \in \overline{D}(0, R)\).
2.6. Q.E.D.
Step 2: Exponential Decay of Iterates \(f^n(z)\)
3. Induction bound on \(|f^n(z)|\) on \(K\).
3.1. Base case \(n = 1\): For all \(z \in K \subseteq \overline{D}(0, R)\), \(|f(z)| \leq c |z| \leq c R \leq R\) (since \(c < 1\)).
By <1>2.<2>5>, \(|f(z)| \le c|z|\); since \(|z| \le R\) and \(c < 1\), this gives \(|f(z)| \le cR \le R\).
3.2. In particular, \(f(z) \in \overline{D}(0, R)\) whenever \(z \in \overline{D}(0, R)\).
The bound \(|f(z)| \le c|z| \le |z| \le R\) shows \(f\) maps \(\overline{D}(0, R)\) into itself.
3.3. Inductive step: Assume \(|f^k(z)| \leq c^k |z|\) for all \(z \in K\).
This is the induction hypothesis.
3.4. Then \(f^k(z) \in \overline{D}(0, R)\), so by <1>2.<2>5>: \[|f^{k+1}(z)| = |f(f^k(z))| \leq c |f^k(z)| \leq c \cdot (c^k |z|) = c^{k+1} |z| \leq c^{k+1} R.\]
Since \(|f^k(z)| \le c^k |z| \le c^k R \le R\), the point \(f^k(z)\) lies in \(\overline{D}(0, R)\); applying the contraction bound <1>2.<2>5> to this point gives \(|f(f^k(z))| \le c |f^k(z)|\), and the induction hypothesis gives \(|f^k(z)| \le c^k |z|\).
3.5. Therefore, for all \(n \geq 1\) and all \(z \in K\): \[|f^n(z)| \leq R \cdot c^n.\]
Induction on \(n\) using the base case 1 and the step 4 yields \(|f^n(z)| \le c^n |z| \le c^n R\).
3.6. Q.E.D.
Step 3: Uniform Convergence via the Weierstrass M-Test
4. Apply the Weierstrass M-test to \(\sum_{n=1}^\infty f^n(z)\).
4.1. Each iterate \(f^n(z)\) is a composition of holomorphic functions, hence holomorphic on \(\mathbb{D}\).
\(f^n\) is the \(n\)-fold composition of the holomorphic map \(f\) with itself, and a composition of holomorphic functions is holomorphic.
4.2. On the compact set \(K\), \(|f^n(z)| \leq M_n \coloneqq R c^n\).
This is exactly the bound established in <1>3.<2>5>.
4.3. Since \(0 \leq c < 1\), the geometric series \(\sum_{n=1}^\infty M_n = R \sum_{n=1}^\infty c^n = \frac{R c}{1 - c} < \infty\) converges.
The geometric series \(\sum_{n=1}^\infty c^n\) converges to \(\frac{c}{1-c}\) because its ratio \(c\) satisfies \(|c| < 1\).
4.4. By the Weierstrass M-Test, the series \(\sum_{n=1}^\infty f^n(z)\) converges uniformly and absolutely on \(K\).
The Weierstrass M-test states that if \(|f^n(z)| \le M_n\) on \(K\) and \(\sum M_n\) converges, then \(\sum f^n(z)\) converges uniformly and absolutely on \(K\); 2 and 3 supply exactly these hypotheses.
4.5. Since \(K \subset \mathbb{D}\) was arbitrary, the series converges locally uniformly on \(\mathbb{D}\).
Local uniform convergence means uniform convergence on every compact subset; since \(K\) was arbitrary, the series converges uniformly on every compact subset of \(\mathbb{D}\).
4.6. By Weierstrass’s theorem on analytic series, the locally uniform limit of holomorphic functions is holomorphic.
A locally uniform limit of holomorphic functions is holomorphic: uniform convergence on compact sets permits interchanging the limit with contour integrals, so Morera’s theorem applies.
4.7. Q.E.D.