Characterization of locality in Fourier multiplier

Question

Obviously Laplacian $-\Delta$ is a local operator, and it can be written as a Fourier multiplier $|\xi|^2$. Similarly fractional Laplacian $(-\Delta)^s$ is with Fourier multiplier $|\xi|^{2s}$ and is nonlocal, and same goes with Hilbert transform with Fourier multiplier $\textrm{sign}(\xi)$ etc.

However, for a given Linear operator $L$ defined via Fourier multiplier, say $f(\xi)$, how can one tell that $L$ is local/nonlocal, from the shape of $f$? What is the characteristic of local/nonlocality we can infer from $f$?

Thank you in advance.

Answer 1

Let $a$ be the Fourier transform of $f(\xi)$, in general $a$ will be a distribution. Then the action of $f(D)$ on a function is a convolution with $a$, that is $f(D)u=a*u$. Now, $a$ is compactly supported iff $f$ is an entire function with exponential growth; in this case if you apply $f(D)$ to a function with compact support, you get another function with a larger compact support (this is the case for instance of the fundamental solution of the wave equation). This is not yet a local operator. The only case when you get a local operator is when $a$ is supported at 0, and this means it is a finite combination of derivatives of the Dirac delta. And hence, the original function $f$ is a polynomial.

Answer 2

The only local pseudo-differential operators are the differential operators and this entails that the only local Fourier multipliers are polynomials. It is a classical result due to J. Peetre, Math. Scand. 8 (1960), 116–120, [MR0124611].