If $u\in L^{\frac{3}{2}+\delta}(\mathbb{R}^3)\cap L^{\frac{3}{2}-\delta}(\mathbb{R}^3)$ show that $v=u\ast \frac{1}{|x|}\in L^\infty(\mathbb{R}^3)$

Question

This is an old pre lim problem. Suppose $u\in L^{\frac{3}{2}+\delta}(\mathbb{R}^3)\cap L^{\frac{3}{2}-\delta}(\mathbb{R}^3)$ for $\delta>0$ sufficiently small. Show that $v=u\ast \frac{1}{|x|}\in L^\infty(\mathbb{R}^3)$ and provide a bound for $||v||_{L^\infty}$ which depends only on $||u||_{L^{\frac{3}{2}+\delta}(\mathbb{R}^3)}$.

Now I've thought about using Young's inequality, but this clearly doesn't work since the second term isn't in $L^{p}(\mathbb{R}^3)$ since either integrability fails at zero or at infinity. Plus I still haven't been able to incorporate the fact that $f\in L^{\frac{3}{2}-\delta}(\mathbb{R}^3)$. I've tried just manipulating the integral itself, but that also doesn't seem to get me anywhere.

EDIT: It seems the original problem might have a typo. In particular at the end. The following revised version is one that is solvable.

Suppose $u\in L^{\frac{3}{2}+\delta}(\mathbb{R}^3)\cap L^{\frac{3}{2}-\delta}(\mathbb{R}^3)$ for $\delta>0$ sufficiently small. Show that $v=u\ast \frac{1}{|x|}\in L^\infty(\mathbb{R}^3)$ and provide a bound for $||v||_{L^\infty}$ which depends only on $||u||_{L^{\frac{3}{2}+\delta}(\mathbb{R}^3)}$ and $||u||_{L^{\frac{3}{2}-\delta}(\mathbb{R}^3)}$.

Answer 1

Hint. Write

$$\frac{1}{|y|} = \frac{1}{|y|}\mathbf{1}_{\{|y|<1\}} + \frac{1}{|y|}\mathbf{1}_{\{|y|\geq1\}} =: \varphi_0(y) + \varphi_{\infty}(y). $$

Then it follows that

• If $1\leq q < 3$, we have $\int_{\mathbb{R}^3} \varphi_0 (y)^q \, dy = \frac{4\pi}{3-q} < \infty $ and hence $\varphi_0 \in L^q(\mathbb{R}^3)$.

• If $q > 3$, we have $\int_{\mathbb{R}^3} \varphi_{\infty} (y)^q \, dy = \frac{4\pi}{q-3} < \infty $ and hence $\varphi_{\infty} \in L^q(\mathbb{R}^3)$.

Now can you show that both $u * \varphi_0$ and $u * \varphi_{\infty}$ are in $L^{\infty}$ using Young's inequality?