Checking that a function is in Hardy-2 space of the unit disc around the origin

Question

I want to check that the function $f(s) = \frac{1}{(s-1)^{0.3}}$ is in $H^2(\mathcal{D(0,1)})$ which is a vector space of holomorphic functions that satisfy $$ \sup _{0 \leqslant r<1}\left(\frac{1}{2 \pi} \int_0^{2 \pi}\left|f\left(r e^{i \theta}\right)\right|^2 \mathrm{~d} \theta\right)^{\frac{1}{2}}<\infty $$ So, first I want to evaluate the integral inside the supremum as a function of r: $\int_0^{2 \pi}\left|f\left(r e^{i \theta}\right)\right|^2 \mathrm{~d} \theta$: $$ \int_0^{2 \pi}\left|f\left(r e^{i \theta}\right)\right|^2 \mathrm{~d} \theta = \int_0^{2 \pi}\frac{1}{(r e^{i \theta} - 1)^{0.3}}\overline{\frac{1}{(r e^{i \theta} - 1)^{0.3}}} \mathrm{~d} \theta = \int_0^{2 \pi}\frac{1}{(r^2 -2r\cos(\theta) + 1)^{0.3}} \mathrm{~d} \theta $$ So it boils down to evaluating the last integral. This is where I'm stuck. Any suggestions are welcome

Answer 1

I thought about it more. Consider the equivalent recast of the integral

$$I = \oint\limits_{\partial D(0,r)}\frac{|dz|}{|z-1|^{0.6}} = \oint\limits_{\partial D(-1,r)} \frac{|dz|}{|z|^{0.6}}$$

By using

$$|dz| = \frac{1+\bar{z}}{ir}dz$$

and the formula

$$\oint_{\partial\Omega}f\:dz = 2i \iint\limits_{\Omega}\frac{\partial f}{\partial \bar{z}}\:dx\wedge dy$$

we get that

$$I = \frac{2}{r}\iint\limits_{D(-1,r)}\frac{dx\wedge dy}{|z|^{0.6}}\left(0.7 - \frac{0.3}{\bar{z}}\right)$$

Converting to polar coordinates (and subtracting $\pi$ from the angular bounds), the integral is

$$I = \frac{2}{r}\int_{-\sin^{-1}r}^{\sin^{-1}r}\int_{\cos\phi-\sqrt{r^2-\sin^2\phi}}^{\cos\phi+\sqrt{r^2-\sin^2\phi}}s^{0.4}\:ds\:d\phi\left(0.7 + \frac{0.3e^{-i\phi}}{s}\right)$$

$$= \frac{1}{r}\int_{0}^{\sin^{-1}r}d\phi \:\Bigr[s^{0.4}(2s+3\cos\phi)\Bigr]_{\cos\phi-\sqrt{r^2-\sin^2\phi}}^{\cos\phi+\sqrt{r^2-\sin^2\phi}}$$

where we utilized even and odd symmetry to simplify the integral as well. The expression from here is not unbounded anywhere on $r\in(0,1)$ as can be verified by using the substitution $\phi = r\theta$:

$$\lim_{r\to 0}I(r) = \lim_{r\to 0}\int_0^{\frac{\sin^{-1}r}{r}}d\theta\:\Bigr[s^{0.4}(2s+3\cos r\theta)\Bigr]_{\cos r\theta-\sqrt{r^2-\sin^2r \theta}}^{\cos r\theta+\sqrt{r^2-\sin^2r\theta}} = \int_0^1 d\theta \cdot [0] = 0$$

I will leave finding the true sup to you, though I suspect it occurs at $r = 1$:

$$I(1) = 7\cdot 2^{0.4}\cdot\int_0^{\frac{\pi}{2}}d\theta\:(\cos\theta)^{1.4} \leq \frac{7\pi}{2^{0.6}}$$