I want to prove that
$$\left| \left| \frac{1}{|x|^a} \ast f \right| \right|_q \lesssim ||f||_p$$
with $1 < p < q < \infty$ and $a= n \left(1 + \frac{1}{q}- \frac{1}{p} \right)$ using a version of the Young inequality in Lorenz spaces, that is:
$$||f \ast g||_{p_1 , q_1} \lesssim ||f||_{p_2 , q_2} ||g ||_{p_3 , q_3}$$
where $|| \cdot||_{p,q}$ is the norm of the Lorenz space $L^{p,q}(\mathbb{R}^n)$ and I have the relations:
$$ \frac{1}{p_1}+ 1= \frac{1}{p_2} + \frac{1}{p_3} \quad \frac{1}{q_1}=\frac{1}{q_2}+ \frac{1}{q_3}$$
So the problem is: if I choose $p_2= n/a$ and $q_2 = \infty$ so that I have $||\frac{1}{|x|^a}||_{\frac{n}{a} , \infty}$ bounded, I either have to consider $|| |x|^{-a} \ast f ||_{q,p}$ and $||f||_p $, or $|| |x|^{-a} \ast f ||_q$ and $||f||_{p,q}$. I can't take $|| |x|^{-a} \ast f ||_q$ and $||f||_p$ together.
Is there some obvious norm-inequality in Lorenz spaces that I am missing?
P.S. The inequality is due to O'Neil (he invented it in the sixties I think), I don't know the reference though
Edit: Lorenz spaces are defined with the following norm:
$$||f||_{p,q}= p^{\frac{1}{q}} \left|\left| \lambda \mu\{|f(x)|>\lambda\}^{\frac{1}{p}}\right|\right|_{L^q (\mathbb{R}_+, \frac{d\lambda}{\lambda}}$$
with $1 \leq p < \infty$ and $1 \leq q \leq \infty$. If $q=\infty$ they are the weak $L^p$ spaces. There is anothere characterization of the Lorenz norm though, which is more useful in order to prove some inequalities. It is based on the so called "decreasing rearrangement function". See here if interrested https://en.wikipedia.org/wiki/Lorentz_space#Definition.
The refined Young inequality that you need is the following; $$\lVert f\ast g\rVert_r\le C \lVert f\rVert_p\lVert g\rVert_{q, \infty}, $$ where the numerology is the same as in the standard Young, that is $$1+\frac1r = \frac1p+\frac1q.$$ I took this from Bahouri-Chemin-Danchin, "Fourier analysis and nonlinear PDEs" https://www.springer.com/gp/book/9783642168291, Theorem 1.5.