product of weakly convergent sequences

Question

Let $\Omega$ be a bounded subset of $\mathbb{R}^n$. Let $f_\epsilon, g_\epsilon, f, f^*$ and $g$ be real-valued functions. Suppose that $f_\epsilon\left(x\right) \rightharpoonup f(x)$ in $L^p(\Omega)$ and $g_\epsilon(x) \rightharpoonup g(x)$ in $L^q(\Omega)$, where $1/p+1/q=1$. Suppose that $g_\epsilon\geq 0$ and $g\geq 0$, which are not equivalent to 0, and $f_\epsilon g_\epsilon \rightharpoonup f^*(x)g(x)$ in $L^1\left(\Omega\right)$. Can we conclude that $f=f^*$ a.e. when $g\not=0$?

Answer 1

Here is a counterexample. Let $\Omega=(0,2\pi)$ and $f_n(x)=\sin(nx), g_n(x)=1+\sin(nx) \geq 0$. Then $f_n(x)$ converges weakly to $f(x)=0$, and $g_n(x)$ converges weakly to $g(x)=1$. Then $$f_n(x)g_n(x)=\sin(nx)(1+\sin(nx))=\sin(nx)+\sin^2(nx)=\sin(nx)+0.5-0.5\cos(2nx),$$ which converges to $f^*(x)=0.5$ weakly.