Does the Leibniz (product) rule hold in some sense for the spectral fractional Laplacian (at least in 1 dimension)?

$\begingroup$ Is there a variant of Leibnitz rule for the usual Laplacian that would do the job for you? If not, then you can forget about the fractional Laplacian. (Your question is kind of vague; you may try to be more specific here, as well as in your other questions). $\endgroup$ – Mateusz Kwaśnicki Jun 29 at 16:48

$\begingroup$ @MateuszKwaśnicki I'm thinking of this one: $$\Delta(uv)=u\Delta v+2\nabla u\cdot\nabla v+v\Delta u $$ $\endgroup$ – user139845 Jun 29 at 17:15

$\begingroup$ In the spirit of my answer below, if one looks at what I think is the simplest natural example which one can use as a test, then one sees that the above holds because $(m+n)^2=m^2+2 m n +n^2$—there might be some difficulty for the case of non integral powers of the Laplacian $\endgroup$ – user131781 Jun 29 at 17:43

$\begingroup$ @Lao: Then "carré du champ" is perhaps the right keyword for you. $\endgroup$ – Mateusz Kwaśnicki Jun 29 at 20:37

$\begingroup$ @MateuszKwaśnicki Do you have any references in mind? $\endgroup$ – user139845 Jun 29 at 22:52
There are precise rules for the composition of pseudodifferential operators and although these fractional derivatives are singular integrals and not always pseudodifferential operators, because of the singularity of the Fourier multiplier at the origin, it is quite likely that the composition formula of pseudodifferential operators can be extended to singular integrals. Let me recall the simplest version, dealing only with two terms: take $A, B$ pseudodifferential operators with respective symbols $a,b$ and respective order $m_a, m_b$. We write $A=\text{Op} a, B=\text{Op} b$. Then $$ AB=\text{Op}\bigl(ab+\frac{1}{i}\partial_\xi a\cdot \partial_x b\bigr)+ R_{m_a+m_b2}, $$ where $R_j$ is a pseudo with order $j$. A simple consequence is that the principal symbol of $AB$ is the product $ab$ and that the principal symbol of the commutator $[A,B]$ is $$ \text{$(i)\times $ the Poisson bracket}\quad\frac{1}{i} \\{a, b \\}=\frac{1}{i}\bigl(\partial_\xi a\cdot \partial_x b\partial_x a\cdot \partial_\xi b \bigr). $$ As a result, taking two functions of $x$, $f, g$ and $A$ a pseudodifferential operator of order $m$, we have $$ A (f g)=\text{Op}\bigl(f(x)a(x,\xi) \bigr) g+\frac{1}{i} \text{Op}\bigl((\partial_\xi a)(x,\xi) \cdot f'(x)\bigr) g+ R_{m2}g. $$ In the first term of the rhs, if $A$ is a fractional derivative of order $m$, you get $fAg$ whereas the second term is $ f'Bg, $ where the order of $B$ is $m1$. The remainder is of order $m2$.
Why should it? In one dimension, the product rule fails for all fractional derivatives $$ \frac{d^s}{dx^s} $$ except $s=1$.

$\begingroup$ Where can I find a reference for this statement? Also, is there any "similar" formula that holds instead? $\endgroup$ – user139845 Jun 29 at 16:23
It is usually a good idea to try out such a conjecture in the simplest possible case. In order to make your question precise you require some boundary conditions on the Laplacian. I tried the case of the differential operator $i D$ with periodic conditions. The resulting space has as a basis the trigonometric functions $(e^{i n s})$. If one modestly tries out the Leibniz rule for such functions (never mind finite or infinite linear combinations thereof), then one finds that it only works for the classical case (has to do with the fact that the beginner‘s binomial theorem $(m+n)^\gamma = m^\gamma + n^\gamma$ is usually false if $\gamma \neq 1$).