How to determine $p$ for which $\frac{1}{1-z} \in H^p(\mathbb{D})$

Question

I want to figure out for which $p \in (0,\infty)$ it is true that $\frac{1}{1-z} \in H^p(\mathbb{D})$, where $H^p(\mathbb{D})$ is the space of analytic functions $f$ on $\mathbb{D}$ such that $\sup_{0 < r < 1} (\frac{1}{2 \pi} \int_{0}^{2\pi}|f(re^{i \theta})|^p d \theta)^{\frac{1}{p}} < \infty$. I believe I have shown that $\frac{1}{1-z} \notin H^1(\mathbb{D})$. My argument is that by the reverse triangle inequality we have that $\frac{1}{|1-re^{i \theta}|} \geq \frac{1}{1 - |re^{i \theta}|} = \frac{1}{1 - |r|}$ for all $re^{\theta t} \in \mathbb{D}$. Then it follows that $ \int_{0}^{2\pi}\frac{1}{|1-re^{i \theta}|} d \theta \geq \int_{0}^{2\pi}\frac{1}{1 - |r|} d \theta$. But as $r \rightarrow 1^-$ then $\frac{1}{1 - |r|} \rightarrow \infty$, so the integral in question is unbounded, hence $\frac{1}{1-z} \notin H^1(\mathbb{D})$. Does this hold up? And furthermore, how would do I deal with other $p$?

Answer 1

Begin by assuming the function is in $H^p$. One of the properties of $H^p$ functions is that they have radial limits $$f^*(\theta) = \lim_{r\uparrow1} f(re^{i\theta})$$ almost everywhere, and the $H^p$ norm is obtained by integrating them: $$ \|f\|_{H^p}^p = \frac{1}{2\pi}\int_0^{2\pi} |f^*(\theta)|^p\,d\theta $$ Since $1/(1-e^{i\theta})$ is comparable to $|\theta|^{-1}$, the integral converges if and only if $p<1$.