The HardyLittlewoodSobolev Inequality says that $$\text{for $p,q,r\in (1,+\infty)$ such that }\quad 1\frac1p+1\frac1q=1\frac1r,\tag {$\sharp$} $$ $$ \exists C, \forall u\in L^p(\mathbb R^n),\quad \Vert{u\ast\vert\cdot\vert^{n/q}}\Vert_{L^r(\mathbb R^n)}\le C \Vert u\Vert_{L^p(\mathbb R^n)}. $$ Setting $v_q(x)=\vert x\vert^{n/q}$, we see that $v_q\in L^q_w(\mathbb R^n)$, which is also the Lorentz space $L^{q,\infty}(\mathbb R^n)$ (the latter space is a Banach space when $q\in (1,+\infty)$). Is there a generalization of Young's inequality such as $$ \exists C, \forall u\in L^p(\mathbb R^n), \forall v\in L^{q,\infty}(\mathbb R^n),\quad \Vert{u\ast v}\Vert_{L^r(\mathbb R^n)}\le C \Vert u\Vert_{L^p(\mathbb R^n)}\Vert v\Vert_{L^{q,\infty}(\mathbb R^n)}, $$ with $p,q,r$ satisfying $(\sharp)$?

$\begingroup$ Do you have a typo in the first line of your $\sharp$ statement? $\endgroup$ – John Bentin Sep 23 '14 at 20:18

$\begingroup$ @John Bentin : No, this is the same condition as the one for Young's inequality $L^p\ast L^q\subset L^r$ under $(\sharp)$. $\endgroup$ – Bazin Sep 24 '14 at 10:23
See Corollary 2.15 in the paper
V.F. Kovalenko, M.A. Perelmuter, Yu.A. Semenov, Schrödinger Operators with $L^{l/2}_w (R^l)$Potentials, J. Math. Phys., Vol. 22, No. 5, 1981, pp. 10331044

$\begingroup$ A textbook reference is Theorem 1.4.24 in Grafakos: "Classical Fourier Analysis". One way to prove Young's inequality is using the RieszThorin interpolation theorem and similarly one can use Marcinkiewicz' interpolation theorem to prove the weak Young inequality. Observe that you can allow the endpoints in (#) for Young's inequality but not for the weak Young inequality. $\endgroup$ – gsa Sep 24 '14 at 13:57
The weak type Young inequality $$ \Vert{u\ast v}\Vert_{L^r(\mathbb R^n)}\le C_{p,q,r} \Vert u\Vert_{L^p(\mathbb R^n)}\Vert v\Vert_{L^{q,\infty}(\mathbb R^n)}, \quad 1<p,q,r<\infty, \quad \frac1p+\frac1q = 1 + \frac1r $$ is in fact a direct consequence of the HardyLittlewoodSobolev inequality. By the Riesz rearrangement inequality (a special case of the BrascampLiebLuttinger inequlaity) we have $$ \iint f(x) g(xy) h(y) dxdy \leq \iint f^*(x) g^*(xy) h^*(y) dxdy, $$ where $f^*$ is the symmetric nonincreasing rearrangement of a function $f$. Now, $$ f^*(x) \leq C_n x^{n/q}\f\_{q,\infty} $$ and the claim follows by duality.
This argument has the advantage that it gives the optimal constant provided that the optimal constant in the HLS inequality is known (thus at least in the case $p=r'$), as opposed to interpolation arguments.
Inequalities like this are discussed in this paper of Nakanishi: http://epubs.siam.org/doi/pdf/10.1137/S0036141000369083
(See section two.)
As others have mentioned already, one can get estimates like this from the standard estimates plus interpolation.