If $f_n \to f$ in $H^{-1}$ and $u_n \rightharpoonup^* u$ in $L^\infty$ with $u_n \in H^1_0$, is there any space in which $f_nu_n \rightharpoonup fu$?

Question

Let $\Omega$ be a bounded and smooth domain.

If $f_n \to f$ in $H^{-1}(\Omega)$ (strongly) and $u_n \rightharpoonup^* u$ in $L^\infty(\Omega)$ (weak-star), is there any space in which $f_nu_n \rightharpoonup fu$ if we know that $u_n \in H^1_0(\Omega)$ for each $n$?

We do not know if the limit $u \in H^1_0$, so the convergence doesn't hold in $H^{-1}(\Omega)$, but I wonder if there any other larger/weaker spaces in which it does hold?

If it helps, $0 \leq u_n \leq 1$ a.e.

Answer 1

This seems not to be the case.

Let $\Omega=(0,1)$, $f_n(x)=u_n(x) = \sin(2\pi n x)$. Then $f_n$ and $u_n$ converge weakly in all $L^p$, $p<\infty$, to zero. Also $u_n\rightharpoonup^* 0$ in $L^\infty$. By compact embedding, $f_n\to 0$ in $H^{-1}(\Omega)$. However, $f_nu_n \rightharpoonup \frac12 \ne 0$ in all $L^p$, $p<\infty$.