Following problem in a mock exam:
Prove or falsify: (1) If $f_j \rightharpoonup f$, $g_j \rightarrow g$ in $L^4(\mathbb{R})$ then $f_j g_j \rightharpoonup fg$ in $L^2(\mathbb{R})$.
(2) If $f_j \rightharpoonup f$, $g_j \rightharpoonup g$ in $L^4(\mathbb{R})$ then $f_j g_j \rightharpoonup fg$ in $L^2(\mathbb{R})$.
My try:
We want to show that $\int_\mathbb{R} f_j g_j \psi dx \rightarrow \int_\mathbb{R} f g \psi dx$ for all $\psi \in L^2(\mathbb{R})$, so lets start:
$\vert \int_\mathbb{R} f_j g_j \psi dx - \int_\mathbb{R} f g \psi dx \vert \leq \vert \int_\mathbb{R} (f_j g_j - fg_j) \psi dx\vert + \vert \int_\mathbb{R} (f g_j - fg) \psi dx\vert$
The first summand equals $\int (f_j - f) g_j \psi dx$ and tends to zero because the assumption and $g_j \psi \in L^2$. The second summand equals $\int (g_j - g)f \psi dx$ and with $f\psi \in L^2$ it tends also to zero because $\int (g_j - g)f \psi dx \leq \vert\vert g_j - g\vert\vert_{L^4} \vert\vert f\psi\vert\vert_{L^{4/3}} \rightarrow 0$.
So that would be a proof for (1) but (2) would fail because this last estimate is no more true. Does anyone have a counterexample? Everything true in my steps above? :) Thanks for comments!
It is not too hard to see for
$$ f_j (x) = e^{2\pi i j x} 1_{(0,1)}(x) $$ and $$ g_j (x) = e^{-2\pi i j x} 1_{(0,1)}(x) $$ that $f_j, g_j \rightharpoonup 0$. But $f_j g_j = 1_{(0,1)} $ does not converge weakly to $0$.
Also, in your proof, the integral $$\int(f_j - f) g_j \psi $$ needs some further thought, since $g_j \psi $ depends on $j $.