Denote $B(0,1)=\{x\in{\bf R}^n\mid |x|<1\}$ and let $\{r_k\}$ be a countable dense subset of $U:=B(0,1)$. Consider the following almost everywhere defined function on $U$: $$ u(x):=\sum_{k=1}^\infty\frac{1}{2^k}|x-r_k|^{-\alpha},\quad (x\in U) $$ where $\alpha>0$ is some fixed real number.
Here is my question:
For what $\alpha$ do we have $L^p(U)$?
[Some thoughts] It is not difficult to show that for $\alpha p<n$, the map $x\mapsto |x-r_k|^{-\alpha}$ is in $L^p(U)$ for each $k$. Eventually, one wants $$ \int_U\big(\sum_{k=1}^\infty\frac{1}{2^k}|x-r_k|^{-\alpha}\big)^p\,dx <\infty\tag{*} $$ I find that things boil down to the following:
- Bound up to a constant the integral in ($*$) by $\sum_{k=1}^\infty \frac{1}{2^k}\int_U|x-r_k|^{-\alpha p}\,dx$;
- Show that $\{a_k\}$ is a bounded set where $$ a_k=\int_U|x-r_k|^{-\alpha p}\,dx. $$
As in my comment, a necessary condition is that $\alpha p <n.$ Is this sufficient? Hint: By Jensen,
$$ \left (\sum_{k=1}^\infty\frac{1}{2^k}|x-r_k|^{-\alpha} \right )^p \le \sum_{k=1}^\infty\frac{1}{2^k}(|x-r_k|^{-\alpha})^p.$$