there is one question bothering me for quite a while now.
Let $a_{n},b_{n}\in L^{2}:a_{n}\stackrel{L^{2}}{\rightharpoonup} a\in L^{2} $ weakly $ b_{n}\stackrel{L^{2}}{\rightarrow} b \in L^{2}$ strongly and $a_{n}\cdot b_n\in L_{2}$ weakly and let all the sequences, also the product sequence, be bounded in $L^{2}$, that means w.l.o.g. $a_{n}\cdot b_n\stackrel{L^{2}}{\rightharpoonup}c\in L^{2}$ weakly
I'd like to prove that $a_{n}\cdot b_{n}\stackrel{L^{2}}{\rightharpoonup} a\cdot b$, that means $a\cdot b=c$.
My comments:
It would be enough to show that the nemytskij-operator of $\mathbb{R}\times\mathbb{R}\times\Omega\ni\left(u_{1},u_{2},x\right)\mapsto u_{1}\cdot u_{2}\in\mathbb{R}$ to be weakly closed
EDIT but this is wrong in general. It would STILL BE ENOUGH to prove that this nemytskij-operator is weakly$\times$strongly-closed. (the first function can be considered as the derivative of the first, so by sobolew embeding it converges strongly)
does anyone have an idea or a hint for literature?
Best Answer
In fact $a_nb_n\in L^1(\Omega)$ not $L^2(\Omega)$, and $$ \int_\Omega a_nb_n\,dx \to \int_\Omega ab\,dx. $$ Indeed, $$ a_nb_n-ab=a_n(b_n-b)+(a_n-a)b $$ Then $$ \Big|\int_\Omega a_n(b_n-b)\,dx\,\Big|\le \|a_n\|_{L^2}\|b_n-b\|_{L^2}\le M\|b_n-b\|_{L^2}\to 0, $$ and $$ \int_\Omega (a_n-a)b \to 0, $$
as $a_n-a\rightharpoonup 0$.
EDIT by OP: this argumentation gives $a_{n}\cdot b_{n}\stackrel{L^{1}}{\rightarrow}a\cdot b$.
together with
$a_{n}\cdot b_{n}\stackrel{L^{2}}{\rightarrow}c \Rightarrow a_{n}\cdot b_{n}\stackrel{L^{1}}{\rightarrow}c$
it follows that $a\cdot b = c$