Convolution inequality with weak Lp

Question

I had the pleasure to stumble upon a "well-known convolution inequality": $$\iint u(x)u(y)f(x-y)dxdy \leq C_1||u||_r^2||f||_{p,\infty}$$ The integrals are over $\mathbb{R}^n \times \mathbb{R}^n$, $r = (2p)/(2p-1)$ and $||\cdot||_{p,\infty}$ is just the weak $L^p$ space. I tried proving it by truncating (in respect to $f = f\mathbb{1}_{|f|>\alpha} + f\mathbb{1}_{|f|\leq \alpha}$) then applying Young on the two sums. I did not succeed in finding the above expression, especially I did not manage to "only" find $u$ in respect to the $L^r$ norm... Any help, ideas or a reference are very much appreciated!

Answer 1

First of all the following weak type Young inequality holds: \begin{align*} &\frac{1}{q}+1=\frac{1}{p}+\frac{1}{r},~~~~1<p,~q,~r<\infty\\ &\|f\ast g\|_{L^{q}}\leq C_{p,q,r}\|g\|_{L^{r,\infty}}\|f\|_{L^{p}}. \end{align*} Plugging in the appropriate constant, one should get \begin{align*} \|u\ast f\|_{L^{2p}}\leq C_{p,r}\|f\|_{L^{p,\infty}}\|u\|_{L^{r}}. \end{align*} We also know that \begin{align*} \|u\cdot(u\ast f)\|_{L^{1}}\leq\|u\|_{r}\|u\ast f\|_{L^{2p}}. \end{align*} Such a version of Young inequality can be found in Loukas Grafakos' Classical Fourier Analysis, 3rd Edition, page 73.