I'm wondering if the weak convergence of a sequence $f_n \rightharpoonup f$ in $L^2(\Omega)$ implies the convergence $ \| f_n \| \rightarrow \| f \| $. I know there's another topic about this question at the link Weak convergence in $L^2$ implies strong convergence? , but I used a different calculation to "demonstrate" the first sentence, which I believe to be false (because it would imply the equivalence of weak and strong convergence in $L^2(\Omega)$), but I can't find where I'm wrong.
This is my argument: consider the functional $\int_\Omega f_m f_n\;:\;f_n\mapsto R$ , with fixed $f_m$. This is a linear bounded functional on $L^2(\Omega)$, then
$(1)\;\; \int_\Omega f_m f_n \rightarrow \int_\Omega f_m f $
and, at the same time, being $f\in L^2(\Omega)$, $\int_\Omega f_m f\;:\;f_m\mapsto R $ is a bounded linear functional on $L^2(\Omega)$, so
$(2)\;\;\int_\Omega f_m f \rightarrow \int_\Omega f^2=\|f\|^2_{L^2(\Omega)} $ .
We are at the point where we can say that $\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\int_\Omega f_m f_n = \|f\|^2_{L^2(\Omega)}$ and it is clear that it's also valid $\lim\limits_{n\to\infty}\lim\limits_{m\to\infty}\int_\Omega f_m f_n = \|f\|^2_{L^2(\Omega)}$ (so we can write $\lim\limits_{n,m\to\infty}\int_\Omega f_m f_n = \|f\|^2_{L^2(\Omega)}$) .
Since I'm not sure that this implies $\lim\limits_{n\to\infty}\int_\Omega f_n f_n = \lim\limits_{n\to\infty}\|f_n\|^2_{L^2(\Omega)} =\|f\|^2_{L^2(\Omega)}$ (it could be valid thinking about $\int_\Omega f_n f_n$ as a subsequence of $\int_\Omega f_m f_n$), I've used the definition of the two limits $(1),\;(2)$ to obtain, for arbitrary $\epsilon$:
$(1')\;\exists\; N_\epsilon$ such that $|\int_\Omega f_m f_n\,-\int_\Omega f_m f|<\epsilon/2\;\;\forall n\geq N_\epsilon$, and
$(2')\;\exists\; M_\epsilon$ such that $|\int_\Omega f_m f\,-\int_\Omega f^2|<\epsilon/2\;\;\forall m\geq M_\epsilon$.
Now we look at the quantity $|\int_\Omega f_m f_n\,-\int_\Omega f^2|\leq|\int_\Omega f_m f_n\,-\int_\Omega f_mf|+|\int_\Omega f_m f\,-\int_\Omega f^2|$ and, because of the two previous consideration,
for all $m,n\geq P_\epsilon := max(N_\epsilon,M_\epsilon)\,:\; |\int_\Omega f_m f_n\,-\int_\Omega f^2|\leq \epsilon$,
and since this is valid for all $ m,n \geq P_\epsilon $ , I can choose $m=n$ . By the arbitrariness of $\epsilon$, this would imply that, for a given $\epsilon>0$ , there exists a natural number $P_\epsilon$ such that $|\int_\Omega f_n f_n\,-\int_\Omega f^2|=|\|f_n\|^2_{L^2(\Omega)}-\|f\|^2_{L^2(\Omega)}|\leq \epsilon$, that is the definition of $lim_{n\to\infty}\|f_n\|^2_{L^2(\Omega)}=\|f\|^2_{L^2(\Omega)}$, and that's all. Anyone can find the mistake?
Best Answer
If you take an orthonormal sequence $f_n$ in $L^2$, such as $f_n(x)=\pi^{-1/2} \cos(nx)$ in $L^2[0,2\pi]$, then https://en.wikipedia.org/wiki/Bessel%27s_inequality implies that $f_n \to 0$ weakly in $L^2$. If you go over your proposed proof line by line with this counterexample in mind, you should be able to locate the error.