Prove that $g\geq f^2$ a.e. in $\mathbb{R}$ (a problem in analysis) Please give some hints not the whole solution.

Problem: Let $(f_n)$ be a sequence in $L^2(\mathbb{R})$ and let $f ∈ L^2(\mathbb{R})$ and $g ∈ L^1(\mathbb{R}).$ Suppose that
  $$ f_n \to f\;\; weakly\; in\; L^2(\mathbb{R}) ,$$
  $$ f^2_n \to g\;\; weakly\; in\; L^1(\mathbb{R}) .$$
  Prove that 
  $g\geq f^2$ a.e. in 
  $\mathbb{R}.$

 A: Edit: I have simplified the following argument, removing the `approximation' component. The essence of the proof is unchanged, and some more detail is added.
It is not hard to see that $g \geq f^2$ a.e. is equivalent to the statement that for any $\phi \in L^\infty(\mathbb R)$, we have that 
$$
(*) \, \, \langle \phi, g \rangle \geq \langle \phi, f^2\rangle \, ,
$$
where $\langle, \rangle$ denotes the dual pairing of $L^1$ and $L^\infty$. To see, at least, that $(*)$ implies $g \geq f^2$ almost surely, set $\phi = \chi_{\{ g < f^2\}}$, so that $\langle \phi, g \rangle < \langle \phi, f^2 \rangle$ holds if $g < f^2$ on a positive measure set, contradicting $(*)$.
Indeed, we'll check that $(*)$ holds directly for nonnegative $\phi \in L^\infty(\mathbb R)$. For signed $\phi$, the same is true, as one can decompose into negative and positive parts.
Assuming $\phi \in L^\infty(\mathbb R)$ is nonnegative, we now check $(*)$. To start, observe that $\sqrt \phi \cdot f_n$ converges weakly to $\sqrt \phi \cdot f$ weakly in $L^2(\mathbb R)$. Moreover, it is a general fact that if $\psi_n \to \psi$ weakly in an $L^2$ space, then $\liminf_n \| \psi_n\|_{L^2} \geq \| \psi \|_{L^2}$  (check this if you don't believe me : P). 
Plugging in,
$$
\liminf_n \int \phi f_n^2 \geq \int \phi f^2 \, .
$$
The RHS is clearly $\langle \phi , f^2 \rangle$. The LHS is
$$
\liminf_n \int \phi f_n^2 = \liminf_n \langle \phi, f_n^2 \rangle = \lim_n \langle \phi, f_n^2 \rangle =  \langle \phi, g \rangle \, ,
$$
since $f_n^2 \to g$ weakly in $L^1$.
A: Since you want hints and not a whole solution, here are some things to try:


*

*Try to rewrite the conclusion (about pointwise inequality) in terms of an inequality involving $\langle f^2,\phi\rangle$ and $\langle g, \phi\rangle$ for arbitrary $\phi\in L^\infty$.

*Use the following inequality, which is a direct application of Cauchy-Schwartz
$$
\int f_n(x) f(x) \phi(x)\,dx \leq \left(\int f_n(x)^2\phi(x)\,dx\right)^{1/2}
\left(\int f(x)^2\phi(x)\,dx\right)^{1/2}.
$$


Hope this helps.
