Weak convergence exercise.

Question

Let $(f_n)$ be a sequence in $L^2(\mathbb R)$ and let $f\in L^2(\mathbb R)$ and $g\in L^1(\mathbb R)$. Suppose that \begin{eqnarray*} f_n\rightharpoonup f \hbox{ weakly in }L^2(\mathbb R)\,, \\ f_n^2\rightharpoonup g \hbox{ weakly in }L^1(\mathbb R)\,. \end{eqnarray*} Prove that $g\geq f^2$ a.e. in $\mathbb R$.

Thank you for your help.

Answer 1

Your proof should make use of the following facts:

1. The $L^2$-norm (in fact, any norm) is lower semicontinuous w.r.t. weak convergence, i.e. $\Vert h \Vert_2 \leq \liminf_n \Vert h_n \Vert_2$ for $h_n \rightharpoonup h$.
2. The map $h \mapsto h \cdot \chi_A$ is continuous (on $L^1$ and on $L^2$).
3. If $h_n \rightharpoonup h$ and $T$ is linear and continuous, then $T h_n \rightharpoonup Th$.
4. You can characterize the inequality $f^2 \leq g$ by using integrals (how?).

EDIT: Here is a more detailed hint. You have (why?)

$$ \int_A |f|^2 = \Vert f \chi_A \Vert_2^2 \leq \liminf_n \Vert f_n \chi_A \Vert_2^2 = \int |f_n|^2 \chi_A. $$

How does that help you?