Leibniz rule for square root of Laplacian

Question

Let $(M,g)$ be a compact Riemannian manifold (e.g. $M=S^3$ the 3-sphere) and let $\Delta$ be the metric Laplacian on $M$. Then $\Delta$ has an $L^2(M)$ basis of eigenfunctions $\pi_m$, $$ \Delta \pi_m = - \lambda^2_m \pi_m. $$ Define the square root of the Laplacian, $(-\Delta)^{1/2}$, by $$ (-\Delta)^{1/2} \pi_m = \lambda_m \pi_m. $$ Then I understand that $(-\Delta)^{1/2}$ is a pseudodifferential operator of order 1 on $M$. My question is, does $(-\Delta)^{1/2}$ satisfy the Leibniz rule $$ (-\Delta)^{1/2}(fg) = f(-\Delta)^{1/2} g + g (-\Delta)^{1/2} f, $$ perhaps up to a smoothing operator of some kind? Additionally, intuitively it feels like $(-\Delta)^{1/2}$ should essentially be the gradient (i.e. exterior derivative $d$ on functions), but of course it is a scalar operator unlike the exterior derivative. Is there a sense in which $$ (-\Delta)^{1/2} f = d f + \text{stuff} ?$$ Any references would be appreciated.

Answer 1

For the Leibniz rule, it is not true that \begin{equation} (-\Delta)^{\frac12}(fg) = f (-\Delta)^{\frac12}g + g (-\Delta)^{\frac12}f \end{equation} even in the simplest possible case, i.e. in $\mathbb{R}$. I think the easiest way to see it is to consider $f,g$ sufficiently "good" functions, i.e. Schwarz functions in this case, then applying the Fourier transform, we have \begin{align} \widehat{(-\Delta)^{\frac12}(fg)}(\xi) - \widehat{ f (-\Delta)^{\frac12}g}(\xi) - \widehat{ g (-\Delta)^{\frac12}f}(\xi) & = |\xi| \int_\mathbb{R} \hat{f}(\omega)\hat{g}(\xi-\omega)d\omega \\ & - \int_\mathbb{R} \hat{f}(\omega)\hat{g}(\xi-\omega)|\xi-\omega|d\omega \\ &- \int_\mathbb{R} \hat{f}(\omega)|\omega|\hat{g}(\xi-\omega)d\omega \\ & = \int_\mathbb{R} \hat{f}(\omega)\hat{g}(\xi-\omega)[|\xi|-|\omega|-|\xi-\omega|]d\omega. \end{align} Then I let someone else verify that you in any reasonable sense this is not a smoothing operator. For example one cannot control the $L^1(\mathbb{R})$ norm of the above expression in terms of the $L^2(\mathbb{R})$ norms of $f,g$.

What is reasonable to expect and in fact it is true in $\mathbb{R}^n$ and it could be true also in manifolds (where you have good boundedness properties of the Riesz transforms) is that you can control some kind of norm of $(-\Delta)^{\frac12}(fg)$ in terms of norms of $f, g, (-\Delta)^\frac12 f, (-\Delta)^\frac12 g$. See for example, Chapter 7.6, of the book of Grafakos, Modern Fourier Analysis, (3rd Edition)