If $u_n \to u$ in $L^2(\Omega)$ and $v_n \rightharpoonup^* v$ in $L^\infty(\Omega)$, does $u_nv_n \rightharpoonup uv$ in $L^2(\Omega)$?

Question

Take $\Omega$ to be a bounded smooth domain.

If $u_n \to u$ in $L^2(\Omega)$ and $v_n \rightharpoonup^* v$ in $L^\infty(\Omega)$, does $u_nv_n \rightharpoonup uv$ in $L^2(\Omega)$?

I can show this for a subsequence $n_j$. Does it also hold for the full sequence?

Answer 1

Take $w\in L^2(\Omega)$. Then $$ \int_\Omega(u_nv_n - uv)w = \int_\Omega u(v_n - v)w + \int_\Omega v_n(u_n - u)w . $$ The first integral vanishes because $uw\in L^1(\Omega)$ and $v_n-v \rightharpoonup^* 0$ in $L^\infty(\Omega)=L^1(\Omega)^*$. The second one vanishes since $(v_n)$ is bounded in $L^\infty(\Omega)$, hence $(v_nw)$ is bounded in $L^2(\Omega)$, and $u_n\to u $ in $L^2(\Omega)$.

Answer 2

Yes, it is true. Since $v_n\to v$ weakly $L^\infty$, you have that $\langle u_n v_n,x\rangle=\langle u_n,v_n x\rangle$ (I'm assuming you are working with real valued functions) and $v_n x\to v x$ weakly $L^2$, and therefore conclusion follows from the fact that a strong convergence and a weak one allow to have pass to limit in the scalar product.