Let $(f_n)_n \in L^2(\mathbb{R})$ such that $f_n$ converges weakly to $f$ in $L^2(\mathbb{R})$ and $f_n^2 \rightarrow g$ in $L^1(\mathbb{R})$. Prove that $f^2 \le g$ almost everywhere in $\mathbb{R}$.
In order to prove the thesis, I have to show that $\int_{\mathbb{R}} f^2 \psi dx \le \int_{\mathbb{R}} g \psi dx$, for a suitable $\psi$.
My first idea was to apply Mazur's lemma to $(f_n)_n$ in order to obtain a strongly convergent sequence in $L^2$. However, I cannot conclude the strong convergence of the aforementioned sequence squared to $g$ in $L^1$.
I think that I should use the fact that $L^2$ is an Hilbert space and thus it is uniformly convex and reflexive. I know that there exists a theorem which states that if I have a weakly convergent sequence on an Hilbert space and I manage to prove that I have also convergence in norm, then the sequence is strongly convergent.
However, I can't prove the convergence in norm (if the sequence converges in that sense) and hence I am not able to conclude.
Any hint would be greatly appreciated.
Best Answer
Let $\phi\in C_c(\mathbb R)$ with $\phi\ge0$. The functional $f\mapsto \int_{\mathbb R} f^2 \phi\ dx $ is convex and continuous on $L^2$, hence weakly sequentially lower semi-continuous. This proves $$ \int_{\mathbb R} f^2 \phi\ dx \le \liminf_{n\to\infty} \int_{\mathbb R} f_n^2\phi\ dx. $$ Since $\liminf_{n\to\infty} \int_{\mathbb R} f_n^2\phi \ dx=\int_{\mathbb R} g\phi \ dx$, we have $$ \int_{\mathbb R} (g-f^2) \phi \ dx\ge 0. $$ Since $\phi\ge0$ was arbitrary, $g\ge f^2$ a.e. follows.
The proof also works if $f_n^2\rightharpoonup g$ in $L^1$ only.