Does the Leibniz (product) rule hold for the spectral fractional Laplacian?

Question

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

Answer 1

There are precise rules for the composition of pseudo-differential operators and although these fractional derivatives are singular integrals and not always pseudo-differential operators, because of the singularity of the Fourier multiplier at the origin, it is quite likely that the composition formula of pseudo-differential operators can be extended to singular integrals. Let me recall the simplest version, dealing only with two terms: take $A, B$ pseudo-differential operators with respective symbols $a,b$ and respective order $m_a, m_b$. We write $A=\text{Op} a, B=\text{Op} b$. Then $$ AB=\text{Op}\bigl(ab+\frac{1}{i}\partial_\xi a\cdot \partial_x b\bigr)+ R_{m_a+m_b-2}, $$ 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 $$ A (f g)=\text{Op}\bigl(f(x)a(x,\xi) \bigr) g+\frac{1}{i} \text{Op}\bigl((\partial_\xi a)(x,\xi) \cdot f'(x)\bigr) g+ R_{m-2}g. $$ 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 $m-1$. The remainder is of order $m-2$.

Answer 2

Considering the following definition for the fractional Laplacian \begin{eqnarray} \label{pointwisedef} (-\Delta)^{s}u(x) & : = & \mathrm{ \mbox{p. v}} \quad a_{d,s} \int_{\mathbb{R}^d}\frac{(u(x)-u(y))} {|x-y|^{d+2s}}dy \\ \nonumber & : = & a_{d,s}\lim_{\delta\rightarrow 0^{+}}\int_{\{y\in \mathbb{R}^d:\ \lvert x-y\rvert\} \ge \delta}\frac{(u(x)-u(y))} {|x-y|^{d+2s}}dy \end{eqnarray}

The Proposition 1.5 of LOCAL ELLIPTIC REGULARITY FOR THE DIRICHLET FRACTIONAL LAPLACIAN by Umberto Biccari et al., states that if $u, v$ be such that $(-\bigtriangleup)^{s}(u)$ and $(-\bigtriangleup)^{s}(v)$ exist and \begin{eqnarray*} \int_{\mathbb{R}^{d}} \frac{|(u(x)-u(y))(v(x)-v(y))|}{\vert x-y\vert^{d+2s}}dy < \infty. \end{eqnarray*} Then $(-\bigtriangleup)^{s}(uv)$ exist and is given by \begin{eqnarray} (-\bigtriangleup)^{s}(uv)=u(-\bigtriangleup)^{s}(v)+v(-\bigtriangleup)^{s}(u)-I_{s}(u,v) \end{eqnarray} where \begin{eqnarray*} I_{s}(u,v)=a_{s,d}\int_{\mathbb{R}^{d}} \frac{(u(x)-u(y))(v(x)-v(y))}{\vert x-y\vert^{d+2s}}dy. \end{eqnarray*}.

Answer 3

Why should it? In one dimension, the product rule fails for all fractional derivatives $$ \frac{d^s}{dx^s} $$ except $s=1$.

Answer 4

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 $\gamma \neq 1$).