Let $\Omega \in L^{1}(S^{d-1})$ have mean zero. Prove that, if the operator $T_{\Omega}: L^{p} \rightarrow L^{q}$ given by $T_{\Omega}f(x) $:= p.v. $\int_{\mathbb{R}^{d}} \frac{\Omega \left(\frac{y}{|y|} \right)}{|y|^{d}}f(x-y) dy$, $f \in \mathcal{S}(\mathbb{R}^{d})$ is bounded, then $p = q$.
The hint says to use dilations. I set $\tau_{y}(f(x)) = f(x-y)$ and then note that $T_{\Omega}$ is translation invariant. I can show that $\lim_{y \to \infty}||f+ \tau_{y}||_{p} = 2^{p^{-1}}||f||_{p}$. Denoting the norm of $T_{\Omega}$ by $||T_{\Omega}||_{p,q}$. Since $T_{\Omega}$ commutes with translations, we have $||T_{\Omega}f + \tau_{y}T_{\Omega}f||_{q} \leq ||T_{\Omega}||_{p, q}||f + \tau_{y}f||_{p}$ and on taking the limit as $y \to \infty$, we get $||T_{\Omega}f||_{q} \leq 2^{p^{-1} - q^{-1}}||T||_{p,q}||f||_{p} \implies p \leq q$. This does not quite give the equality though.