Here's a version of the theorem: If $1 < p, r < \infty$ and $ 0 < \alpha < n $ be such that $ \frac{1}{p} + \frac{ \alpha }{ n} = \frac{1}{r} + 1 $. Then for any $ f \in L^p ( \mathbb R ^n )$, the function $$I_\alpha f (x): = \int_{ \mathbb R^n } \frac{f (y) }{ |x - y| ^\alpha } d y $$ is well defined almost everywhere and lies in $L^r ( \mathbb R^n )$. Moreover, $$|| I_ \alpha f ||_{ L^r ( \mathbb R ^n )} \le C_{p ,\alpha , n } ||f||_{ L^ p(\mathbb R^n )} $$ for some constant $ C_{ p ,\alpha , n } > 0$.
I want to show the theorem fails for $ p =1 ,r = \infty$, or $\alpha = n$. I tried a long time but don't make any progress. Any help is appreciated.
The problem with $\alpha=n$ is that the kernel is not locally integrable. See Existence of a function in $L^{p}$ with a certain property: the characteristic function of unit ball is a counterexample.
The problem with $p=1$ is that then $r= n/\alpha$, and the potential does not decay at infinity quickly enough to be in $L^{n/\alpha}$. The same example $f=\chi_B$ works, because $I_\alpha f \sim |x|^{-\alpha}$ at infinity.
For $r=\infty$, let $f(x) = |x|^{-n/p}|\log|x||^{1/\sqrt{p}}$ near the origin $x=0$ (kill it away from $0$ to keep things locally integrable). Note that $f\in L^p$, but the convolution with $|x|^{-\alpha}$, where $\frac{1}{p}+\frac{\alpha}{n} = 1$, blows up at $0$: $$ I_\alpha f(0) = \int |x|^{-\alpha - n/p}|\log|x||^{1/\sqrt{p}} = \int |x|^{-n}|\log|x||^{1/\sqrt{p}} =\infty $$ Since $I_\alpha$ is lower semicontinuous, it follows that $ I_\alpha f(x)\to\infty$ as $x\to 0$. Thus, $I_\alpha$ is not in $L^\infty$.