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.

Here is another solution. The main idea is to obtain an integral estimate, and use an approximation to the identity to reduce it to a pointwise estimate.
Let $E$ be the set of points $x$ such that $x$ is a Lebesgue point of both $f$ and $g$. Since both $f$ and $g$ are locally integrable, $\mathbb{R}-E$ has measure zero.
Now choose $x\in E$ and consider a cube $C_x(r)$ of side length $r$ centered at $x$. Then
$$\phi_{\epsilon}=\frac{1}{|C_{x}(\epsilon)|}\chi_{C_{x}(\epsilon)}$$
lies in $L^2(\mathbb{R})\cap L^{\infty}(\mathbb{R})$, so
$$(f_n,\phi_{\epsilon})\to(f,\phi_{\epsilon})\quad\text{and}\quad(f_n^2,\phi_{\epsilon})\to(g,\phi_{\epsilon}).$$
By Cauchy-Schwarz inequality applied to $f_n\sqrt{\phi_{\epsilon}}\cdot\sqrt{\phi_{\epsilon}}$,
$$(f_n,\phi_{\epsilon})^2=\left( \int f_n\phi_{\epsilon}\right)^2\leq\int f_n^2\phi_{\epsilon}=(f_n^2,\phi_{\epsilon}).$$
Thus taking $n\to\infty$ we have
$$(f,\phi_{\epsilon})^2\leq(g,\phi_{\epsilon}).$$
Now taking $\epsilon\to 0$ gives $f^2(x)\leq g(x)$, completing the proof.
P.S. I saw a same question on AoPS. I hope that my answer is not a duplicate.
