I want to to prove the classical div-curl lemma of Coifman-Lions-Meyer-Semmes in the following form:
Div-Curl Lemma. Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be a Schwartz function such that $f$ is divergence free ($\nabla\cdot f=0$) and $g:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be a Schwartz function such that $g$ is curl-free ($\nabla\wedge g=0$). Then for any Hormander-Mikhlin multiplier $a(D)$, $$\|a(D)(f\cdot g)\|_{L^{1}}\lesssim_{n,p}\|f\|_{L^{p}}\|g\|_{L^{p'}} \tag{1}$$ where $1<p<\infty$.
For simplicity, let's just consider the case $p=p'=2$. Note that since the Riesz transforms are HM-multipliers, the above inequality implies that $f\cdot g$ belongs to the Hardy space $\mathcal{H}^{1}(\mathbb{R}^{n})$, but let's pretend that I don't know what a Hardy space is.
I read somewhere (don't remember the paper right now) that one can prove the div-curl lemma stated in an equivalent form by dualizing the estimate and using paraproducts; however, no further details were given. I want to try to make this approach work.
By dualizing the inequality, it suffices to show that for any $\|h\|_{L^{\infty}}\leq 1$, $$|\int_{\mathbb{R}^{n}}a(D)(f\cdot g)(x)h(x)dx|\lesssim\|f\|_{L^{2}}\|g\|_{L^{2}} \tag{2}$$
Suppose we can move the $a(D)$ over to $h$. It's not obvious to me why I can do this, as $\widehat{h}$ is a priori a tempered distribution, $a(\xi)$ isn't necessarily $C^{\infty}$, and $L^{\infty}$ doesn't have a nice dense subspace. I know that CZOs map $L^{\infty}\rightarrow BMO$, and I can then rewrite the above estimate as $$|\int_{\mathbb{R}^{n}}((a(D)h)f)\cdot g(x)dx|\lesssim\|f\|_{L^{2}}\|g\|_{L^{2}} \tag{3}$$
By Cauchy-Schwarz, it suffices now to show that $$\|(a(D)h)f\|_{L^{2}}\lesssim\|f\|_{L^{2}} \tag{4}$$
Decompose $(a(D)h)f$ into a sum of low-high, high-low, and high-high Coifman-Meyer paraproducts (see p. 5-6 of Terry Tao's notes 6 here for definitions if unfamiliar with this terminology). $$(a(D)h)f=\pi_{lh}(f,a(D)h)+\pi_{hl}(f,a(D)h)+\pi_{hh}(f,a(D)h)$$
I realize that $f$ is vector-valued, while $a(D)h$ is scalar-valued, but just work coordinate-wise. Using the fact that low-high Coifman-Meyer paraproducts with BMO functions in the second slot are CZOs (see p. 16 Proposition 4.3) and by taking the relevant transposes (see p. 5), so are high-low and high-high paraproducts, we obtain the desired estimate.
This argument can't possibly be right as I never used the hypothesis that $f$ is div-free and $g$ is curl-free. Even though I know that $f\cdot g\in\mathcal{H}^{1}(\mathbb{R}^{n})$ with the strictly weaker hypothesis that $\int f \cdot g=0$, the estimate (1) isn't true under this weaker hypothesis, as $\mathcal{H}^{1}$ is a proper subspace of mean zero $L^{1}$ functions.
Edit 1: I think I found the error in the reasoning above: I mistakenly assumed that if $\pi_{hl}$ is a high-low Coifman-Meyer multiplier, then
$$\|\pi_{hl}(f,g)\|_{L^{2}}\lesssim\|f\|_{L^{2}}\|g\|_{BMO}$$
If $m$ is the symbol of $\pi_{hl}$, and we write $T_{m}(f,g)=\pi_{hl}(f,g)$, then the transpose of $T_{m}$ is the operator $T_{m'}$ satisfying
$$\int_{\mathbb{R}^{n}}T_{m}(f,g)hdx=\int_{\mathbb{R}^{n}}fT_{m'}(g,h)dx$$
with symbol $m'(\xi_{2},\xi_{3})=m(-\xi_{2}-\xi_{3},\xi_{2})$. Observe that $m'$ is a low-high multiplier, but now $g$ is in the first slot.