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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 $f$.

The quotient $q(w)=\varphi (w)/w$ extends to a holomorphic function at zero with value one. Since $\varphi$ is injective and vanishes at zero, $q$ has no zero in the unit disc. A holomorphic logarithm $\ell$ exists there, with $\ell (0)=0$. Define the odd square-root transform of a normalized univalent function byIt satisfies $h(z{)}^2=\varphi (z^2)$, $h(0)=0$, $h^{\prime }(0)=1$ and $h(-z)=-h(z)$. To check injectivity, $h(z)=h(w)$ gives $\varphi (z^2)=\varphi (w^2)$, hence $z^2=w^2$. If $w=-z$, oddness gives $h(z)=0$, so $z=w=0$. Thus the two arguments always coincide, and $h\in \mathcal{S}$. Choosing the square root by its derivative at zero removes its sign ambiguity.

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.