Context:
Let $\Omega \subset \mathbb{R}$ be open and bounded. Suppose two sequences of functions $f_{n}, g_{n} : \Omega \times [0,T] \to \mathbb{R}$ are such that $f_{n}$ is uniformly bounded in $L^{\infty}(0,T; W^{1,1}(\Omega))$ and $g_{n}$ is uniformly bounded in $L^{\infty}(0,T; H^{1}(\Omega))$.
We can certainly say that there exist $f,g \in L^{\infty}((0,T) \times \Omega))$ such that \begin{aligned}f_{n} \rightharpoonup^{\ast} f \text{ in } L^{\infty}((0,T) \times \Omega)) \\ g_{n} \rightharpoonup^{\ast} g \text{ in } L^{\infty}((0,T) \times \Omega)). \end{aligned}
Question:
I wish to know whether it is true that
$$ f_{n} \partial_{x}g_{n} \longrightarrow f\partial_{x}g$$ in the sense of distributions.
I am not sure if this is true because usually when we try to prove such a convergence we require some time derivative bound on one of the sequences and a space derivative bound on the other, which we don't have.
This is not true. For $T=1$ take $$ f_n(x,t) = \sin(n\pi t), $$ and $$ g_n(x,t) = \sin(n\pi t) \cdot \phi(x), $$ where is not a constant function. Then $f_n$ and $g_n$ converge weak star towards zero, but the product $f_n \partial_x g_n$ not.