Given $u \in L^2(0,T;H^1(\Omega))\cap L^\infty(0,T;L^2(\Omega))$ and $v\in H^1([0,T]\times \Omega) \cap L^2(0,T;H^1(\Omega))$, can one conclude $u \cdot\nabla v \in L^2([0,T]\times \Omega)$?
Some proof i read seems to use this but i think this is wrong? The best i can see is $u \cdot \nabla v \in L^1([0,T] \times \Omega )$, since $u, \nabla v \in L^2.$
Also one could show $u \in L^4,$ but again this is not enough.
Thank you for any hints.
The assumptions on $v$ imply $\nabla v\in L^2(0,T;L^2(\Omega))$. But $u$ is at best in $L^4(0,T;L^4(\Omega))$ with $\Omega\subset \mathbb R^2$. Hence, it holds $u\cdot \nabla v \in L^{4/3}(0,T;L^{4/3}(\Omega))$ at best, $u\cdot \nabla v$ is not in $L^2(0,T;L^2(\Omega))$ in general.