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

Step 4 of 8: De Branges's weight functions

\sigma_k(t)

and the target monotonicity

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In plain words

The auxiliary quantity is one carefully designed functional $\Phi_n(t)$ for each target index $n$ , not an independent claim about every summand in Milin's sum. It starts at $0$ and its endpoint controls the whole weighted inequality, so the proof must establish monotonicity of that functional rather than term-by-term decrease.

\Phi_n(t) \ \text{along the Loewner flow}, \qquad \Phi_n(0)=0, \quad \Phi_n(1)\le 0 \ \Longrightarrow\ \text{Milin's inequality}

Detailed analysis

De Branges associates to each Milin index $n$ a single functional $\Phi_n(t)$ along the Loewner chain, normalized by $\Phi_n(0)=0$ , whose endpoint inequality is exactly the weighted Milin sum. Thus the required statement is $\Phi_n(1)\le0\Longrightarrow\sum_{k=1}^{n}(n+1-k)k(|\gamma_k|^2-1/k^2)\le0$ . It is important that this is a statement about the full weighted sum, not separate inequalities for its individual summands.

Terms in this step

weight function $\sigma_k(t)$: An auxiliary quantity de Branges attaches to each $k$ and each time $t$ of the Loewner flow, engineered to start at $0$ and to encode, at the end of the flow, exactly the $k$ -th term of Milin's inequality.

Knowledge used in this step

Functions of a complex variable Special functions