1
$\begingroup$

I bumped into the following convolution inequality mentioned without proof in a paper:

There is a constant $C > 0$ such that $$\int_{\mathbb{R}^n\times \mathbb{R}^n} K(x-y) u(x)u(y) \,d x\, d y \leq C \|K\|_{L^{p,\infty}(\mathbb{R}^n)} \|u\|^2_{L^q(\mathbb{R}^n)}$$ where $\dfrac{1}{q} = 1 - \dfrac{1}{2p}$ and $\|\cdot\|_{L^{p,\infty}(\mathbb{R}^n)}$ denotes the Marcinkiewicz "norm" on the weak $L^p$ space $$\|K\|_{L^{p,\infty}(\mathbb{R}^{n})} = \sup_{\lambda > 0} \lambda|\{x \in \mathbb{R}^n \colon |K(x)| > \lambda\}|^{1/p}.$$ Here $|A|$ is the Lebesgue measure of $A \subset \mathbb{R}^n.$

Does anyone know a reference to this result? The author claims that it is a well-known convolution inequality, but I do not know where to find it.

$\endgroup$
1
  • $\begingroup$ If I recall correctly you can find this in "classical Fourier Analysis" by Grafakos. $\endgroup$
    – Jose27
    May 15, 2021 at 6:37

1 Answer 1

2
$\begingroup$

First you use Hölder's inequality for $u(y)$ and $(K*u)(y)=\int_{\mathbb{R}^n} K(y-x)u(x)\,dx$ with exponents $q$ and $q'$, respectively. Now use Young's inequality for weak-type spaces (Theorem 1.4.24, from the book of Grafakos that Jose quoted): if $1 <p,q, r <\infty $ are such that \begin{equation*} \frac{1}{r}+1=\frac{1}{p}+\frac{1}{q}, \end{equation*} then there is a positive constant $C = C(p, q, r)$ such that for any $f \in L^q(\mathbb{R}^n)$ and $g \in L^{p,\infty }(\mathbb{R}^n)$ \begin{equation*} \|f*g\|_{r}\leq C \|g\|_{p,\infty}\|f\|_{q} \end{equation*} (in your case, with $f=u$, $g=K$ and $ r = q'$). So
\begin{equation*} \frac{1}{p} + \frac {1} {q}=\frac{1}{q'} + 1=\left(1-\frac{1}{q}\right) + 1 , \end{equation*} which is the same as $2-\frac{1}{p}=\frac {2} {q}$ and the result follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.