# Bounding $\int_0^T\int_\Omega v|\nabla u|^2$ given that $v \in L^2(0,T;L^2(\Omega))$ and $u \in L^2(0,T;H^1(\Omega)) \cap L^\infty(0,T;L^2(\Omega))$?

If $v \in L^2(0,T;L^2(\Omega))$ and $$u \in L^2(0,T;H^1(\Omega)) \cap L^\infty(0,T;L^2(\Omega)),\tag{1}$$ is possible to get a bound on the integral $$\int_0^T\int_\Omega v|\nabla u|^2$$ of the form $$\int_0^T\int_\Omega v|\nabla u|^2 \leq C\lVert v \rVert_{L^2(0,T;L^2(\Omega))}\times(\text{some norm of u or \nabla u that is finite (given that u satisfies (1))})$$ (Notice the multiplcaton on the second line above)?

Young's/Holder's inequality are no use here I don't think.

No. Take $\Omega=(0,1)$, one-dimensional. Let $v(t,x)=x^{-1/3}$, i.e., independent of $t$. And $u(t,x)=x^{1/3}$, also independent of $t$. All assumptions are satisfied. But $v(u')^2$ is a multiple of $x^{-1}$, not integrable over $\Omega$. Subsequent integration of $\infty$ over $t$ still gives $\infty$.
The problem is that you are multiplying $|\nabla u|^2$, which is only known to be in $L^1$, by $v$, which is allowed to be unbounded. The product can easily be non-integrable, in every dimension.