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I have been trying to follow the lines in a proof I saw online. You can read the snippet on page 6. The part of the proof I am stuck on can be stated as follows

Let $r$ ble close to $0$ and $p>1$ then for every $z \in D$ $$ \int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}\right|^q} \leq \left( 1 - \frac{|z|}{r} \right)^{-1-q} $$

To give some background, the inequality is part of giving an estimate in how the functions in the hardy space are bounded.

Lemma 1.4 Let $f \in H^p(D)$, Then, for every $z \in D$, $$ |f(z)| \leq C_p \frac{\|f\|_{H^p}}{\bigl(1-|z|\bigr)^{1/p}}. $$

The proof starts by defining the Cauchy integral formula, rewriting using $w = re^{it}$ and $z = |z|e^{i\theta}$: $$ f(z) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{f(re^{it})}{1 - \frac{|z|}{r}e^{i(\theta-t)}}\,\mathrm{d}t $$ The Author then applies Hölders inequality and the change of variables $u \mapsto \theta - t$: $$ |f(z)| \leq \|f\|_{H^p} \left( \int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left|1 - \frac{|z|}{r}e^{iu}\right|^q} \right)^{1/q} $$ My problem is when the author starts estimating the integral inside the last parenthesis. What it essentially boils down to is showing

$$ \int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}\right|^{q}} \leq \left( 1 - \frac{|z|}{r} \right)^{-1-q} $$

Does anyone have any idea on how to prove this inequality? In the paper mentioned above, the author proves a similar statement to the one above. However, I was unable to follow that proof as It to me was written somewhat convoluted. I have tried a few things, but alas I have made no progress on the inequality above.

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  • $\begingroup$ is |z|/r<1? ???? $\endgroup$ – tired Jan 10 '17 at 14:16
  • $\begingroup$ This should hold for small r, basically we are letting $r \to 1$ later, and as $|z|<1$ I think it is safe to assume that $|z|/r < 1$. $\endgroup$ – N3buchadnezzar Jan 10 '17 at 19:08
  • $\begingroup$ ok...that means the modolus in the denominator can be dropped $\endgroup$ – tired Jan 10 '17 at 19:10
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It's regretted, however, that the inequality does not hold. Take $q=2,|z|/r=1/2$ for instance. Then LHS is $$ \int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}\right|^{q}}= \int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\frac{5}{4}-\cos u}=\frac{8\pi}{3},$$ and RHS is $$ \left( 1 - \frac{|z|}{r} \right)^{-q+1}=2<\frac{8\pi}{3}. $$ (RHS of your inequality should be $\left( 1 - \frac{|z|}{r} \right)^{-q+1}.$ A typo, I think.)
However we can prove $$ \int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \frac{|z|}{r}e^{iu}\right|^{q}}\le A\left( 1 - \frac{|z|}{r} \right)^{-q+1},\quad \left(1>\frac{|z|}{r}\ge \frac{1}{4}\right)\tag{1} $$ where $A$ is a constant independent of $z$, and $(1)$ is enough for us to prove Lemma 1.4.

Let $\rho=|z|/r>1/4.$ Then \begin{align} |1-\rho e^{iu}|^2&=1+\rho ^2-2\rho \cos u=(1-\rho )^2+4\rho \sin^2 \frac{u}{2}\\ &\ge (1-\rho )^2+\sin^2 \frac{u}{2}\\ &\ge \frac{1}{2}\left(1-\rho +\left|\sin\frac{u}{2}\right|\right)^2,\\ |1-\rho e^{iu}|&=\frac{1}{\sqrt{2}}\left(1-\rho +\left|\sin\frac{u}{2}\right|\right)\\ &\ge \frac{1}{\sqrt{2}}\left(1-\rho +\frac{1}{\pi}|u|\right)\ge \frac{1}{2\pi}\left(1-\rho +|u|\right). \end{align}

Therefore we have \begin{align} \int_{-\pi}^{\pi}\frac{\mathrm{d}u}{\left| 1 - \rho e^{iu}\right|^{q}}&\le \int_{-\pi}^{\pi}\left(\frac{2\pi}{1-\rho+|u| }\right)^q\mathrm{d}u\\ &=2(2\pi)^q\int_0^{\pi}\frac{1}{(1-\rho+u)^q}\mathrm{d}u\\ &=\frac{2(2\pi)^q}{q-1}\left(\frac{1}{(1-\rho )^{q-1}}-\frac{1}{(\pi+1-\rho )^{q-1}}\right)\\ &\le \frac{A}{(1-\rho )^{q-1}}. \end{align}

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