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I trying to understand a proof in Pommerenke's Univalent Functions. Given $f$ analytic on the unit disk and $M(r) := \max\limits_{0 \leq \theta \leq 2\pi}|f(re^{i\theta})|$,

Using the fact that $$M(r) \leq \frac{r}{(1-r)^2},$$ $(0 < r < 1)$, I want to show that $$\int_0^1 \frac{M^p(r)}{r}dr < \infty,$$ for $0 < p < \frac{1}{2}$.

I get $$\frac{M^p(r)}{r} ~\leq~ \frac{r^{p-1}}{(1-r)^{2p}}.$$

The term on the RHS doesn't seem easy to integrate on $(0,1)$...

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1 Answer 1

up vote 1 down vote accepted

Since the function $$r\mapsto\frac{r^{p-1}}{(1-r)^{2p}}$$ is continuous on $[0,1)$ the only problem for integrability could be in $1^-$, but there you have a function which behaves like $$\frac{r^{p-1}}{(1-r)^{2p}}\sim_{r\to1^-}\frac{1}{(1-r)^{2p}}$$and from your hypothesis, $0<2p<1$, hence, from the usual results about singularities of integrals, there is no problem to integrate in $1^-$.

(Remember that you can prove that $\lim_{\varepsilon\to0^+}\int_\varepsilon^1\frac{dr}{r^\alpha}<\infty$ for any $\alpha$ in $[0,1)$ and your problem is just the same problem translated.)

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