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Recall the Coifman–Meyer Theorem as stated in Grafakos and Oh - The Kato-Ponce Inequality.

Theorem: Let $m \in L^{\infty}\left(\mathbf{R}^{2 n}\right)$ be smooth away from the origin and satisfy $$ \left|\partial_{\xi}^{\alpha} \partial_{\eta}^{\beta} m\right|(\xi, \eta) \leq C(\alpha, \beta)(|\xi|+|\eta|)^{-|\alpha|-|\beta|} $$ for all $\xi, \eta \in \mathbf{R} \setminus\{0\}$ and $\alpha, \beta \in \mathbf{Z}^{n}$ multi-indices with $|\alpha|,|\beta| \leq 2 n+1$. Then for all $f, g \in \mathcal{S}\left(\mathbf{R}^{n}\right)$, $$ \left\|\int_{\mathbf{R}^{2 n}} m(\xi, \eta) \widehat{f}(\xi) \widehat{g}(\eta) e^{i\langle\xi+\eta,\rangle} d \xi d \eta\right\|_{L^{r}\left(\mathbf{R}^{n}\right)} \leq C(p, q, r, m)\|f\|_{L^{p}\left(\mathbf{R}^{n}\right)}\|g\|_{L^{q}\left(\mathbf{R}^{n}\right)} $$ where $\frac{1}{2}<r<\infty, 1<p, q \leq \infty$ satisfy $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$.

I was wondering if it makes sense to expect the following estimate to hold, $$ \left\|\int_{\mathbf{R}^{2 n}} m(\xi, \eta) \widehat{f}(\xi) \widehat{g}(\eta) e^{i\langle\xi+\eta,\rangle} d \xi d \eta\right\|_{L^{p}\left(\mathbf{R}^{n}\right)} \leq C(p, m)\|fg\|_{L^{p}\left(\mathbf{R}^{n}\right)} $$ for some Schwartz functions $f,g$ and exponent $p\geq 1?$

Edit: Thanks to Terry's remark, the estimate does not necessarily hold for any multiplier. Actually, I had the following estimate which I wanted to show, $$\|D^\alpha(fg)\|_p\leq C(\|gD^\alpha f\|+\|fD^\alpha g\|_p).$$ Here $\alpha\in (0,1)$ and $D^\alpha = (-\Delta)^\alpha.$

Naturally, one would write $D^\alpha(fg)$ as a double integral involving the Fourier transforms of $f$ and $g$. I was hoping that if the modified Coifman-Meyer estimate was true then this estimate would hold after applying Littlewood decomposition.

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No. For instance, if $f,g$ have disjoint supports, the right-hand side vanishes, but there is no reason to expect the left-hand side to vanish; and one can easily construct examples where it does not, for instance by first selecting a pair $f,g$ of bump functions with disjoint support and then localising $m$ to a small region of pairs $(\xi,\eta)$ where $\hat f(\xi)$ and $\hat g(\eta)$ are non-zero and don't vary too much.

More generally, one of the first sanity checks to see if a given estimate $X \leq CY$ is plausible (for some non-negative quantities $X,Y$ and some unspecified constant $C$) is to ask "If $Y$ vanishes, it is then obvious that $X$ must also vanish?" If the answer is "no", it is highly unlikely that the estimate is going to be true. (A "yes" answer, on the other hand, is inconclusive. A good followup question is "If we know $Y$ is very small, does this look like it could force $X$ to be small as well?". Depending on the type of estimate being considered, "bounded" or "finite" can be a more suitable adjective here than "small".)

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  • $\begingroup$ Thank you for your answer, I added some motivation to my question. Perhaps for the multipliers obtained in the proof of the Kato-Ponce inequality one can establish the modified Cofiman-Meyer estimate? $\endgroup$
    – Student
    Commented Dec 31, 2021 at 7:40
  • $\begingroup$ Your proposed revision of the estimate is also false. For instance setting $D^\alpha = \Delta$ it would imply that $fg$ is harmonic whenever $f,g$ are harmonic, which is of course false. (Technically this is not yet a counterexample because non-trivial harmonic functions are not Schwartz, but I am pretty sure that some suitable truncation of this example would still disprove the estimate.) $\endgroup$
    – Terry Tao
    Commented Dec 31, 2021 at 20:12
  • $\begingroup$ Thank you for your comment. I was thinking about the range $\alpha \in (0,1)$. I also thought about the same argument that you mentioned above, though on discussing with some people, I was told about a Liouville theorem for the fractional harmonic functions that states if $u$ is bounded (from above for instance) and $\alpha$-Harmonic then it is a constant. See for instance arxiv.org/pdf/1401.7402.pdf $\endgroup$
    – Student
    Commented Jan 1, 2022 at 9:35
  • $\begingroup$ In that case, if $f$ and $g$ are $\alpha$-harmonic then they are constants, and thus their product is also a constant which implies that $D^\alpha(f g)=0.$ So both sides of the inequality vanish. $\endgroup$
    – Student
    Commented Jan 1, 2022 at 9:37

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