Let $f\colon \mathbb{R}^2\to \mathbb{R}^2$ be smooth divergence free and decaying at infinity. I want to show that
$$ X=\{g\in L^2(\mathbb{R}^2): f\cdot \nabla g\in L^2(\mathbb{R}^2)\}$$
is dense in $L^2(\mathbb{R}^2)$. My idea is first to show that this set is dense in $W^{1,2}$. So let $h\in W^{1,2}(\mathbb{R}^2)$. We know that $C_c^{\infty}(\mathbb{R}^2)$ is dense in $W^{1,2}(\mathbb{R}^2)$ so we can find a sequence $(h_n)_{n\in \mathbb{N}}\subset C_c^{\infty}(\mathbb{R}^2)$ so that $h_n\to h$ and $\nabla h_n\to \nabla h$ in $L^2(\mathbb{R}^2)$. If $f$ is smooth and decaying at infinity do we have
$$ f\cdot \nabla h_n\to f\cdot \nabla h$$
in $L^2$? Indeed
$$\int_{\mathbb{R}^2} |f|^2|\nabla h_n-\nabla h|^2 dx \leq \|f\|_{L^{\infty}(B_R)}^2\|\nabla(h_n-h)\|_{L^2(B_R(0))}+\int_{B_R(0)^C}|f|^2|\nabla h_n-\nabla h|^2 dx.$$
Due to the decay of $f$ we know that there exists some $R>0$ so that $|f|\leq 1$ for all $|x|>R$. Hence the latter integral is now
$$\int_{B_R(0)^C} |f|^2|\nabla h_n-\nabla h|^2dx \leq \int_{B_R(0)^C}|\nabla h_n-\nabla h|^2dx.$$
This now proves that $X$ is dense in $W^{1,2}$. Then as $W^{1,2}$ is dense in $L^2$ this would conclude the result.
$\textbf{Questions}:$
- Is this correct?
- Is there a faster way to show density of such a set?