Normalized univalent function ($\mathcal{S}=\{f:\mathbb{D}\to \mathbb{C}\text{ univalent},f(0)=0,f^{\prime }(0)=1\}$)

A normalized univalent function is a holomorphic function on the unit disc which is injective and has the displayed normalization. Its Taylor series begins $z+a_2{z}^2+\cdots$. Normalization makes coefficient bounds and image-size estimates independent of translation and dilation; the Koebe quarter theorem is the sharp universal image bound.

For $f(z)=z+{\sum }_{n\ge 2}{a}_n{z}^n$ in the normalized univalent function class, the integral mean bound for an odd square-root transform gives $(2\pi {)}^{-1}\int |f(r{e}^{i\theta })|d\theta \le r/(1-r)$. The Cauchy integral formula for Taylor coefficients therefore yields $|a_n|\le r^{1-n}/(1-r)$. Choosing $r=(n-1)/n$ and applying the logarithm tangent bound gives the displayed bound for every $n\ge 2$.

The odd square-root transform of a normalized univalent function has reciprocal exterior expansion $1/h(1/\zeta )=\zeta -(a_2/2){\zeta }^{-1}+\cdots$. Its first negative Laurent coefficient has modulus at most one by the area theorem for univalent functions, proving the bound. If $w$ is omitted by $f$, the normalized map $wf/(w-f)$ has second coefficient $a_2+1/w$. Applying the bound twice gives $|1/w|\le 4$, the Koebe quarter theorem.

Because $f(z)/z$ is holomorphic and nowhere zero on the unit disc, its square root with value one at zero exists. Defining $h(z)=z\sqrt{f(z^2)/z^2}$ gives an odd normalized univalent function. If $h(z)=h(w)$, squaring gives $z^2=w^2$; the possibility $z=-w$ and oddness force both to be zero. Applying the area theorem for univalent functions to $1/h(1/\zeta )$ yields $|a_2|\le 2$.

Suppose $h(z{)}^2=f(z^2)$ is the odd square-root transform of a normalized univalent function. The Koebe growth theorem gives $|h(z)|\le |z|/(1-|z{|}^2)$, so its radius-$r$ image has area at most $\pi r^2/(1-r^2{)}^2$. Write $h=\sum c_n{z}^n$ and $M(r)=\sum |c_n{|}^2{r}^{2n}$, the mean-square boundary integral by the Parseval identity. The area formula for a univalent power series gives $M^{\prime }(r)=2\mathrm{Area}(h(r\mathbb{D}))/(\pi r)\le 2r/(1-r^2{)}^2$. Integrating from zero proves the bound. The relation $|h(\sqrt{r}{e}^{i\theta }){|}^2=|f(r{e}^{2i\theta })|$ then gives the corresponding first absolute mean bound $r/(1-r)$ for $f$.