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Worked solution: De Branges's proof of the Bieberbach conjecture via Milin's conjecture (1984)

Step 1 of 8: State the conjecture and the extremal Koebe function
In plain words

A function ff is univalent (one-to-one) on the unit disk if it never sends two different points to the same place, like a rubber sheet stretched and bent but never folded onto itself. Bieberbach's 1916 conjecture said that for such functions, normalized to start as f(z)=z+a2z2+a3z3+⋯f(z) = z + a_2 z^2 + a_3 z^3 + \cdots, the size of each coefficient ana_n can never exceed nn.

The bound is tight: the Koebe function k(z)=z(1−z)2k(z) = \frac{z}{(1-z)^2}, which stretches the disk onto the whole plane minus a slit along the negative real axis, achieves an=na_n = n exactly for every nn, so no bound better than nn could ever hold.

∣an∣≤n for univalent f(z)=z+∑n=2∞anzn on ∣z∣<1,equality only for rotations of k(z)=z(1−z)2|a_n| \le n \ \text{for univalent} \ f(z) = z + \sum_{n=2}^{\infty} a_n z^n \ \text{on} \ |z|<1, \quad \text{equality only for rotations of} \ k(z) = \frac{z}{(1-z)^2}
Detailed analysis

In 1916, Ludwig Bieberbach conjectured that every univalent (injective) holomorphic function f(z)=z+∑n=2∞anznf(z) = z + \sum_{n=2}^{\infty} a_n z^n on the open unit disk {z∈C:∣z∣<1}\{z \in \mathbb{C} : |z| < 1\} satisfies ∣an∣≤n|a_n| \le n for every n≥2n \ge 2, with equality only for rotations of the Koebe function k(z)=z(1−z)2k(z) = \frac{z}{(1-z)^2}, which maps the disk onto the entire plane minus a slit along the negative real axis from −∞-\infty to −14-\tfrac{1}{4}.

Direct attacks on the nn-th coefficient succeeded only case by case: Bieberbach himself for n=2n=2 (1916), Karl Loewner for n=3n=3 (1923, inventing his differential equation for the purpose), Garabedian and Schiffer for n=4n=4 (1955), and by 1972 the cases up to n=6n=6; no method in sight could handle every nn simultaneously, and the conjecture stood open for 68 years.

The eventual proof, found by Louis de Branges in 1984, does not attack ∣an∣≤n|a_n| \le n directly for each nn; instead it proves a stronger, uniform conjecture of Isaak Milin about a different set of coefficients, which forces all the Bieberbach bounds at once — the subject of the next step.

Terms in this step
univalent function
A holomorphic function ff on a domain (here the unit disk) that is injective: f(z1)=f(z2)f(z_1) = f(z_2) only when z1=z2z_1 = z_2, so it never maps two different points to the same image.
Koebe function
The function k(z)=z(1−z)2=∑n=1∞nznk(z) = \frac{z}{(1-z)^2} = \sum_{n=1}^{\infty} n z^n, the unique (up to rotation) extremal example achieving equality ∣an∣=n|a_n| = n in the Bieberbach conjecture for every nn.
Knowledge used in this step