1
$\begingroup$

By the Schwartz Kernel Theorem, we know that the kernel of an integral operator belongs to distribution space $\mathscr{S}'(\mathbb{R}^n)$. Moreover, we know that the kernel $K$ is $C^{\infty}$ off the diagonal in $\mathbb{R}^n \times \mathbb{R}^n$.

Now my questions are:

  1. Does the Schwartz kernel of a pseudo-differential operator of arbitrary order belong to the Schwartz space?

  2. Can we have the Schwartz Kernel Theorem be true with the kernel belonging to $\mathscr{S}(\mathbb{R}^n)$ as opposed to $\mathscr{S}'(\mathbb{R}^n)$?

$\endgroup$
3
$\begingroup$

Your preliminary statement needs clarification: a linear continuous operator $T$ from $\mathscr S(\mathbb R^n)$ into $\mathscr S'(\mathbb R^n)$ has a kernel $k$ in $\mathscr S'(\mathbb R^n\times\mathbb R^n )$ so that for $\phi, \psi\in \mathscr S(\mathbb R^n)$, $$ \langle T\psi,\phi\rangle_{\mathscr S'(\mathbb R^n),\mathscr S(\mathbb R^n)} =\langle k(x,y),\phi(x) \psi (y)\rangle_{\mathscr S'(\mathbb R^{2n}),\mathscr S'(\mathbb R^{2n})}. $$ So some operators may have a very singular kernel everywhere and not only on the diagonal. A simple example of an operator with a singular kernel off diagonal is the translation of vector $h$: $$ (T\phi)(x)=\phi(x-h), $$ whose kernel is $k(x,y)=\delta_0(x-h-y)$, singular at $x=h+y$.

Your statement (1) is incorrect: the identity has the kernel $\delta_0(x-y)$ and this answers negatively (2).

What is true about pseudodifferential operators is that their kernels are indeed smooth off diagonal but more and more singular at the diagonal when you increase the order of the operator. The simplest example is to consider, say in one dimension, the operator $(d/dx)^N$, a differential operator of order $N$: its kernel is $$ \delta_0^{(N)}(x-y). $$ Operators with kernels in $\mathscr S(\mathbb R^{2n})$ are some particular type of regularizing operators.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ $\delta_0$ is not smooth. Any operator operaor with smooth kernel is smoothing. The identity is not. $\endgroup$ – Liviu Nicolaescu Sep 20 '15 at 23:54
  • $\begingroup$ There was a typo (now erased) with an unwanted ' in the last sentence. $\endgroup$ – Bazin Sep 21 '15 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.