Convolution of $L^1 \cap L^q$ and $L^2 \cap L^{\infty}$

Question

This is not from assignment, something on my notes.

With $f \in L^1 \cap L^q$ for a $q > \frac{3}{2}$ and we take its convolution $u = \frac{1}{4 \pi|x|} * f$, we can show that $u$ is in $L^p \cap L^\infty$ for all $p > 3$.

My attempt was: We know that $\frac{1}{4\pi|x|} = (\frac{1}{4\pi^2 |\xi|^2})^\vee$ ($\vee$ means inverse fourier transformation). Since $\frac{1}{4\pi^2 |\xi|^2}$ is in $L^1 \cap L^2$, $\frac{1}{4\pi|x|} \in L^2 + L^\infty$. Write $\frac{1}{4\pi|x|} = G_1 + G_2$, where $G_1 \in L^2$ and $G_2 \in L^\infty$. Then we consider $G_1*f$ and $G_2*F$ separately, and use generalized Young's inequality. However I am only able to prove that $u \in L^p + L^\infty$ this way, which is not what is wanted.

Answer 1

The shorter proof would be using Lorentz spaces and since $\frac{1}{|x|}\in L^{3,\infty}$ and since $f\in L^1\cap L^{q} \subset L^{3/2,1}$ by interpolation, you deduce that $\frac{1}{|x|}*f\in L^\infty$ since the dual of $L^{3,\infty}$ is $L^{3/2,1}$. On the other hand, since $f\in L^r$ for any $r\in(1,3/2)$, by the Hardy$\unicode{x2013}$Littlewood$\unicode{x2013}$Sobolev's inequality $$ \|\frac{1}{|x|}*f\|_{L^p} \leq C \|f\|_{L^r} $$ with $\frac{1}{p} = \frac{1}{r} - \frac{2}{3}$, and $r>1$ is equivalent to $p>3$.

---

Then as you say, a longer but more elementary proof can be obtained by cutting the functions. You can write $\frac{1}{|x|}$ as a sum in $L^2+L^\infty$, and more explicitly $$ \frac{1}{|x|} = \frac{1}{|x|} \,\mathbf{1}_{|x|<1} + \frac{1}{|x|} \,\mathbf{1}_{|x|\geq 1} = G_1+G_2 \in (L^1\cap L^{3-\epsilon}) + (L^\infty\cap L^{3+\epsilon}) $$ Then again you have to prove separately that $f\in L^\infty$, and then that $f\in L^p$.

• $G_1*f\in L^\infty$ by Young's inequality (actually Hölder's inequality in this case) because the Hölder conjugate of $q$ verifies $q'<3/2$, so $G_1\in L^{q'}$ and $f\in L^{q}$.
• $G_2*f\in L^\infty$ because $G_2\in L^\infty$ and $f\in L^1$.

So $\frac{1}{|x|}*f \in L^\infty$.

• $G_1*f\in L^p$ because $G_1\in L^1$ and $f\in L^p$.
• $G_2*f\in L^p$ because $G_2\in L^{3+\epsilon}$ and $f\in L^1$.

So $\frac{1}{|x|}*f \in L^p$ with $p>3$.