weak convergence of product of weakly and strongly convergent $L^{2}$ sequences in $L^{2}$ 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?
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 $a\cdot b = c$
There is a disproof of this statement in general. Take $ L^2([-1,1],\mathbb{C})$, and $a_n(x)= \exp(2\pi i n x)$ and $b_n (x) = \exp(-2\pi i n x)$. In this case, both $a_n \rightharpoonup a = 0$ and $b_n\rightharpoonup b=0$, but $a_n (x)\cdot b_n(x) =1$ for all $x$. Clearly $c_n=c =1$. Then 
$$\lim\limits_{n \to \infty} \int_{-1}^1 a_n(x)\cdot b_n(x)\,dx=2$$
but 
$$\int_{-1}^1 a(x)\cdot b(x)\,dx =0$$