Question: Let $\Omega \subset \mathbb{R}^d$ be open and bounded, $f, f_n \in
L^2 (\Omega)$ and $f_n \rightarrow f$ boundedly in measure
(meaning that $f_n \rightarrow f$ in measure and $sup\ ||f_n||_\infty < \infty $.
Let $g, g_n \in L^2 (\Omega)$ too and $g_n \rightharpoonup g$ in $L^2 (\Omega)$
(weak convergence). Then $f_n g_n, fg \in L^2 (\Omega)$ and
$$ f_n g_n \rightharpoonup fg \text{ in } L^2 (\Omega) . $$
Attempt at solution: The first statements are trivial because all $f_n$ are essentially bounded. Therefore these are elements of the dual $L^2$ and we may indeed attempt to
show weak convergence. To this end we take $\psi \in L^2 (\Omega)$ arbitrarily
and try to estimate
$$ \int_{\Omega} g_n f_n \psi – gf \psi d x = \underbrace{\int_{\Omega}
(g_n f_n \psi – gf_n \psi) d x}_{= : A_n} + \underbrace{\int_{\Omega}
(gf_n \psi – gf \psi) d x}_{= : B_n} . $$
On the one hand
$$ A_n \leqslant M \int_{\Omega} (g_n \psi – g \psi) d x \rightarrow 0
\text{ because } g_n \rightharpoonup g \text{ in } L^2 (\Omega) . $$
On the other, using Hölder's inequality (again,
$f_n \psi, f \psi \in L^2$ thanks to the uniform boundedness of $f_n$):
$$ B_n \leqslant \int_{\Omega} G (f_n \psi – f \psi) d x \leqslant \| g
\|_{L^2} \left( \int_{\Omega} | f_n \psi – f \psi |^2 d x \right)^{1
/ 2} . $$
But I don't know how to estimate the last integral.
I tried splitting the domain of integration and using the convergence in measure, but to no avail. I've also tried skipping the use of Hölder, and using that a subsequence $f_{n_k}$ converges a.e. to $f$, but then I get the bound for this subsequence…
Ideas? Solutions?
Best Answer
There is a problem in the bound of $A_n$; there should be absolute values in the integral hence we cannot use directly weak convergence assumption.
However, we may write $$\tag{1}\int_{\Omega}(g_nf_n\psi-gf\psi)dx=\int_\Omega g_n(f_n-f)\psi dx+\int_{\Omega}(g_nf\psi-gf\psi)dx.$$ Since $f\phi\in\mathbb L^2$, using weak convergence definition with this function we obtain that $$\tag{2}\lim_{n\to+\infty}\int_{\Omega}(g_nf\psi-gf\psi)dx=0.$$ Notice that using Cauchy-Schwarz inequality, $$\left|\int_\Omega g_n(f_n-f)\psi dx\right|^2\leqslant \lVert g_n\rVert_2^2 \int_{\Omega}|f_n-f|^2|\psi|^2dx.$$ Since $(g_n)_{n\geqslant 1}$ is weakly convergent, it's in particular bounded in $\mathbb L^2$. By the version of dominated convergence theorem with convergence in measure, we obtain $$\tag{3}\lim_{n\to +\infty}\int_\Omega g_n(f_n-f)\psi dx=0.$$ Combining (1), (2) and (3), we get $f_ng_n\to fg$ weakly in $\mathbb L^2$.