# Product of weakly convergent sequence and sequence boundedly convergent in measure

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?

## 1 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 $\mathbb L^2$.

• Thank you very much for your answer and for pointing out my (embarrassing) mistake with $A_n$. I had already tried with your way of splitting the integrals but didn't know how to apply dominated convergence... And I still don't get it: with C-S you bound the first integral with an integrable function (because $|\Omega| < \infty$), ok, but then what sequence converging in measure do you take for the DCT? The term $g_n$ multiplies everything (?) I'm (embarrassingly) missing something very obvious, right? – AnCo Mar 16 '14 at 11:46
• You can forget the term $\lVert g_n\rVert_2$ because it can be bounded by a constant independent of $n$. Define $h_n:=(f_n-f)^2|\psi|^2$. By the assumptions, this is an integrable function which converges in measure to $0$, and is dominated by the integrable function $2\sup_n\lVert f_n\rVert_\infty|\psi|^2$. – Davide Giraudo Mar 16 '14 at 12:26
• Ok, I was considering the wrong sequence... Thank you again! – AnCo Mar 16 '14 at 16:19