prove that $g\geq f^2$ The problem is this:
Let $(f_n)$ a sequence in $L^2(\mathbb R)$ and let $f\in L^2(\mathbb R)$ and $g\in L^1(\mathbb R)$. Suppose that $$f_n \rightharpoonup f\;\text{ weakly in } L^2(\mathbb R)$$ and $$f_n^2 \rightharpoonup g\;\text{ weakly in } L^1(\mathbb R).$$ Show that $$f^2\leq g$$ almost everywhere on $\mathbb R$.
I admit I'm having problems with this since It's quite a long time I don't deal with these kind of problems. Even an hint is welcomed. Thank you very much.

Well it is a property of the weak convergence that every weak convergent sequence is bounded and
$$||f||\leq \lim\inf ||f_{n}||$$
then for every $\Omega \in \mathbb{R}^n$ measurable with finite measure we have
$$\left(\int_\Omega f^2\right)^{\frac{1}{2}}\leq \lim\inf\left(\int_\Omega f_n^2\right)^{\frac{1}{2}}$$
That implies 
$$\int_\Omega f^2\leq \lim\inf\int_\Omega f_n^2\tag{1}$$
Note that the function $h$ constant equal to 1 in $\Omega$ and 0 case contrary is in $L^{\infty}{(\mathbb{R})}$ which is the dual of $L^{1}{(\mathbb{R})}$ then 
$$\int_{\Omega}f_n^2\rightarrow\int_\Omega g\tag{2}$$
Jointing (1) and (2) we get 
$$\int_{\Omega}f^2\leq\int_\Omega g$$
Since $\Omega$ is arbitrary we have $f^2\leq g$ almost everywhere.$\blacksquare$
