Odd square-root transform of a normalized univalent function ($h(z)=\sqrt{f(z^2)}=z+\frac{1}{2}{a}_2{z}^3+\cdots$)

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 $f$.