I have the following problem. I need to estimate the following quantity:
$$|\nabla|^{-1}A u \nabla u$$
I know that $A\in L^6(B_R)$, $u \in L^{14/5}\cap L^2(B_R)$ and $\nabla u \in L^2(B_R)$ where $B_R$ is a ball of radius $R$ in $\mathbb{R}^3$. Clearly, $\frac{1}{6}+\frac{5}{14} +\frac{1}{2} >1$, so this is not in any $L^p(B_R)$ with $p\geq 1$. However I don't have to obtain an estimate in $L^p$. I would also be satisfied with an estimate in $H^{s}(B_R)$ or $W^{s,p}(B_R)$ with $s<0$, i.e. a negative order Sobolev space (I'm using the Aubin-Lions lemma but that's not the point here). So I thought about estimating, for $-1<s<-1/14$,
$$ \|Au\nabla u\|_{H^s} \leq \|Au\|_{H^{-1/14}}^{\theta} \|\nabla u\|_{H^{-1}}^{1-\theta} \\ \leq \|Au\|_{L^{21/11}}^{\theta} \|u\|_{L^2}^{1-\theta} \\ \leq \|A\|_{L^6}^{\theta} \|u\|_{L^{14/5}}^{\theta} \|u\|_{L^2}^{1-\theta} .$$
where in the second step I would use Sobolev's inequality with $\frac{1}{2}+\frac{1}{14}\cdot\frac{1}{3} = \frac{11}{21}$ and in the third step Hölder with $\frac{1}{6}+ \frac{5}{14} = \frac{11}{21}$. But I was wondering whether the first step is even remotely allowed. I know we can't really take products of distributions but the functions I'm dealing with here really are functions in $L^p$ spaces? I know this is all pretty hand-wavey and this is why I'm asking to clarify this.