= Integral mean bound for an odd square-root transform
{title2=$\frac1{2\pi}\int_0^{2\pi}|h(re^{i\theta})|^2d\theta\leq\frac{r^2}{1-r^2}$}
Suppose $h(z)^2=f(z^2)$ is the <odd square-root transform of a normalized univalent function>. The <Koebe growth theorem> gives $|h(z)|\leq|z|/(1-|z|^2)$, so its radius-$r$ image has <area> at most $\pi r^2/(1-r^2)^2$. Write $h=\sum c_nz^n$ and $M(r)=\sum|c_n|^2r^{2n}$, the mean-square boundary <integral> by the <Parseval identity>. The <area formula for a univalent power series> gives $M'(r)=2\operatorname{Area}(h(r\mathbb D))/(\pi r)\leq2r/(1-r^2)^2$. Integrating from zero proves the bound. The relation $|h(\sqrt r e^{i\theta})|^2=|f(re^{2i\theta})|$ then gives the corresponding first absolute mean bound $r/(1-r)$ for $f$.
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