Strong convergence of a product of weakly converging functions

Question

Let $(f_j)\in L^4(\Omega)$ be a sequence that weakly converges to $f$ in $L^4(\Omega)$. Let also $(g_j)\in L^4(\Omega)$ be a sequence that weakly converges to $g$ in $L^4(\Omega)$. Let us assume that the product $(f_j g_j)\in L^2(\Omega)$ strongly converges to 1 in $L^2(\Omega)$. Can we conclude that $fg=1$ a.e.? I do thank you for your attention, All the best.

Answer 1

No: Just take $\Omega=(0,1)$, $f_j=g_j=sign(\sin(j \pi x))$. Then $f_jg_j=1$, $f_j,g_j\rightharpoonup 0$.

Answer 2

I assume that $g_j\to g$ strongly in $L^4(\Omega)$, but only weak $L^2$-convergence of $f_jg_j\to 1$. For $\Omega$ with finite measure $L^4(\Omega)\subset L^2(\Omega)\subset L^{4/3}(\Omega)$. Enough to show $\int_\Omega(fg)\varphi dx = \int_\Omega 1\varphi dx$ for every $\varphi\in L^4(\Omega)$. By Cauchy-Schwarz the integral is welldefined since $f,g\in L^4(\Omega).$ We can show that $\int_\Omega 1\varphi dx=\lim_{j\to\infty}\int_\Omega (f_jg_j)\varphi dx=\int_\Omega (fg)\varphi dx$. The first equality follows by weak $L^2$-convergence of $f_jg_j$ to $1$. For the second equality consider $$ \int_\Omega (f_jg_j \pm f_jg-fg)\varphi dx= \int_\Omega (g_j -g)f_j\varphi dx+\int_\Omega (f_j-f)g\varphi dx. $$ The second integral goes to $0$ by weak $L^4$-convergence as $\varphi g\in L^{4/3}(\Omega)$ (by Cauchy-Schwarz $\int_\Omega |\varphi g|^{4/3}dx\le \|\varphi^{4/3}\|_{L^2}\|g^{4/3}\|_{L^{2}}<\infty$). For the first integral we compute \begin{align} \int_\Omega |(g_j -g)f_j\varphi| dx&\le\|(g_j -g) \|_{L^4} \|f_j\varphi\|_{L^{4/3}}\\ &= \|(g_j -g) \|_{L^4} \left( \int_\Omega |f_j|^{4/3}|\varphi|^{4/3}dx\right)^{3/4}\\ &\le \|(g_j -g) \|_{L^4} \left( \|f_j^{4/3}\|_{L^3}\|\varphi^{4/3}\|_{L^{3/2}}\right)^{3/4}\\ &\le \|(g_j -g) \|_{L^4} \left(K\|\varphi^{4/3}\|_{L^{3/2}}\right)^{3/4}, \end{align} which goes to $0$ by strong $L^4$-convergence of $g_j\to g$. We used Hölder's inequality for the first two inequalities.In the last equality we use $K:=\sup_j\|f_j\|_{L^4}<\infty$, which holds as a consequence of weak $L^4$ convergence of the $f_j$'s.