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Let $\Omega \subset \mathbb{R}^n$ be an open bounded domain.

Define $$W^1 := W^1(0,T;L^2,H^1) := \{w \in L^2(0,T;H^1(\Omega)) \mid w' \in L^2(0,T;H^{-1}(\Omega))\}$$ where $w'$ means the weak derivative that satisfies $$\int_0^T w(t)\phi'(t) = -\int_0^T w'(t)\phi(t)$$ for all $\phi \in C_c^\infty(0,T).$

Let $u \in W^1$ and $f \in C^1([0,T]\times \Omega).$ Under what conditions on $f$ is it true that $fu \in W^1$? (When $\Omega$ is compact, this follows. But I am not sure when $\Omega$ is open.

I would like to use the product rule for weak derivatives for $f(t)u(t)$.. but am unsure what the weak derivative is in this case (keeping $x \in \Omega$ fixed).

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Really a comment, unfortunately too long for the comment box:

Your terminology is incorrect: the theory of distributions always requires the "domains" to be open. So here your $\Omega$ is an open set, fixed once and for all as the "space domain" of your given distribution $u\in W^1$, and it really makes no sense to consider "$\Omega$ compact" when you speak of $u$. This being said, you still have to decide whether your function $f$ is in $\mathcal{C}^1([0,T]\times\Omega)$ or in the better space $\mathcal{C}^1([0,T]\times\overline{\Omega})$. At this point open/compact makes both sense and a real difference when speaking of $f$ because you assume it is $\mathcal{C}^1$ in the classical sense, so I guess this is what you meant by "$\Omega$ compact".

In the easy case $f\in\mathcal{C}^1([0,T]\times\overline{\Omega})$ the answer to your question is yes as you pointed out, basically because $f\in\mathcal{C}^1([0,T]\times\overline{\Omega})$ is uniformly bounded together with all its derivatives. If you have no boundary regularity in space, the answer is no in general: take for example $u\equiv 1$ in $[0,T]\times \Omega$ (which is indeed a nice element of your $W^1$ space!) and choose any $f=f(x)$ to be constant in time, belonging to $\mathcal{C}^1(\Omega)$ but not to $\mathcal{C}^1(\overline{\Omega})$, and such that $f\notin L^2(\Omega)$ (e.g. blow up at the boundary). Clearly $u,f$ satisfy your hypotheses, but the product $fu$ is not even $L^2$ in space because $f$ is not.

Have you tried googling something like "product rule in Sobolev spaces"?

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  • $\begingroup$ Thank you for your comment. The $f$ I had in mind satisfies $f(x,t) \leq C$ for all $(x,t) \in \Omega \times [0,T]$. And it has the same boundedness condition for $f_t(x,t)$. I guess that suffices? I have tried this googling similar things but the cases considered are usually too restrictive. $\endgroup$
    – soup
    Dec 14, 2013 at 20:27

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