Let $H$ be the completion (in the $L^2$ norm) of the space of divergence-free smooth functions with compact support in some bounded set $\Omega \subset \mathbb{R}^3$, and $V = H \cap (H^1_0)^3$. I am following a proof of the following lemma:
If $u \in L^\infty(0, T; H) \cap L^2(0, T; V)$ then $$(u \cdot \nabla)u \in L^{4/3}(0,T; L^{6/5}).$$
I have understand everything aside from somehow bounding the $L^6$ norm by $L^2$. More specifically, let $u \in L^\infty(0, T; H) \cap L^2(0, T; V)$, then for fixed time I would like to prove that $$\|\nabla u\|_{L^2} \|u\|_{L^2}^{1/2}\|u\|_{L^6}^{1/2} \leq C \|u\|_{H^1}^{3/2}\|u\|_{L^2}^{1/2}.$$ Using the fact that $$\|u\|_{H^1} = \|u\|_{L^2} + \|\nabla u\|_{L^2}$$ it follows that $$\|\nabla u\|_{L^2} \|u\|_{L^2}^{1/2}\|u\|_{L^6}^{1/2} \leq \|u\|_{H^1} \|u\|_{H^1}^{1/2} \|u\|_{L^6}^{1/2} = \|u\|_{H^1}^{3/2} \|u\|_{L^6}^{1/2}.$$ However this is where I am stuck, how would I then bound the $L^6$ norm by some multiple of the $L^2$ norm, that is $$\|u\|_{L^6} \leq C \|u\|_{L^2}?$$
I am thinking that this might follow from the assumption that $\Omega$ is bounded, because then $L^q \subset L^p$ for all $1 \leq p \leq q \leq \infty$. So applying this to $q = 6$, if $\|u\|_{L^6} < \infty$, then that $\|u\|_{L^2} < \infty$ as well. Hence choose $C$ sufficiently large so that $\|u\|_{L^6} \leq C\|u\|_{L^2}$. But since $u$ is arbitrary I feel uneasy about this, as $C$ depends on $u$.
If $\Omega$ is nice enough, we can use Sobolev's inequality $$ \Vert u \Vert_{L^6(\Omega)} \leq C \Vert u \Vert_{W^{1,2}(\Omega)}$$ for some constant $C>0$, since $$ \frac{1}{6} = \frac{1}{2} - \frac{1}{3}.$$ I guess this bound should suffice here.
EDIT: The inequality was falsely stated as $\Vert u \Vert_{L^6(\Omega)} \leq C \Vert \nabla u \Vert_{L^2(\Omega)}$. As @peek-a-boo pointed out, constant functions are a counterexample to this one.