The following prelim question bothers me:
Question. Let $f\in L^{\frac{3}{2}+h}(\mathbb{R}^{3})\cap L^{\frac{3}{2}-h}(\mathbb{R}^{3})$, for some small $h>0$. Show that $u=f\ast\frac{1}{|x|}\in L^{\infty}(\mathbb{R}^{3})$ and give a bound on $\|u\|_{L^{\infty}}$ in terms of only $\|f\|_{L^{\frac{3}{2}+h}}$.
It seems that controlling $\|u\|_{L^{\infty}}$ by only $\|f\|_{L^{3/2+h}}$ is impossible by scaling. Indeed, suppose we could. Set $f_{\lambda}:=f(\lambda\cdot)$. Then $f_{\lambda}\ast|x|^{-1}=\lambda^{-3+1}u_{\lambda}$. So by dilation invariance, we obtain $$\lambda^{-2}\|u\|_{L^{\infty}}=\lambda^{-2}\|u_{\lambda}\|_{L^{\infty}}=\|f_{\lambda}\ast|x|^{-1}\|_{L^{\infty}}\lesssim\|f_{\lambda}\|_{L^{3/2+h}}=\lambda^{-\frac{3}{3/2+h}}\|f\|_{L^{3/2+h}}$$ Since $h>0$, the exponents don't cancel and we can let $\lambda\rightarrow 0$ to obtain a contradiction.
By breakinging up $|x|^{-1}$ into a local piece supported around the origin and a global piece supported outside a neighborhood of the origin, we can control $\|u\|_{L^{\infty}}$ in terms of $\|f\|_{L^{3/2+h}}$ and $\|f\|_{L^{3/2-h}}$, but this is not what the question asks.
Well, the last time we thought one of these problems was wrong we were wrong. But this time it seems there's a simple counterexample.
Let $p=3/2+h$. For $1<R<\infty$ define $$f_R(x)=|x|^{-2}\chi_{1<|x|<R}.$$If I got the $p$'s and $q$'s right then $||f_R||_p$ is bounded but $||u_R||_\infty\to\infty$ as $R\to\infty$.
You may want to check my work...