2
$\begingroup$

Suppose that we have a pseudodifferental operator $A\in \Psi^m(\mathbb{R}^n)$ with symbol $a\in S^m(\mathbb{R}^n\times\mathbb{R}^n)$, and $a$ has an asymptotic expansion $a\sim \sum\limits_{j=0}^\infty a_j.$ Suppose further that each $a_j$ is compactly-supported, with the supports non-increasing in $j$. Is $a\in S^{-\infty}(\mathbb{R}^n\times\mathbb{R}^n)$? I know that compactly-supported symbols are smoothing (but the expansion need not preserve the support structure) and so are symbols where each term in the expansion is zero. I'm not sure if assuming that the supports form a non-increasing chain helps. Here's an attempt which does not use the precise support structure and feels wrong:

Fix $k\in\mathbb{R}.$ Then, we can write $a=\sum\limits_{j=0}^{k-1}a_j+r_k,$ with $r_k\in S^k(\mathbb{R}^n\times\mathbb{R}^n).$ Next,$$|\partial_\xi^\alpha\partial_x^\beta a(x,\xi)|\leq \sum\limits_{j=0}^{k-1} |\partial_\xi^\alpha\partial_x^\beta a_j(x,\xi)|+|\partial_\xi^\alpha\partial_x^\beta r_k(x,\xi)|\leq C_{\alpha,\beta,k}\langle\xi\rangle^{k-|\alpha|},$$ using that $r_k\in S^k(\mathbb{R}^n\times\mathbb{R}^n)$ and that each $a_j$ is compactly-supported. There is an issue in this line of logic, though (I believe). The constant depends on $k$ since it depends on where I truncate the asymptotic expansion.

Is this argument flawed? If so, is the claim true?

$\endgroup$

1 Answer 1

1
$\begingroup$

You have one slight error that may be confusing you, namely that $r_k\in S^{m-k}$ not $S^k$, so the $r_k$ are getting better, not worse.

In particular, you actually don't really need to do any work, it follows just from the fact that $S^l$ closed under summation.

More specifically, to show that $a\in S^{-\infty}$, it suffices to show that $a\in S^{m-N}$ for all $N\in \mathbb{N}$. Then we have by the definition of asymptotic summation that $$a-\sum_{i=0}^{N-1}a_i=r_N\in S^{m-N}.$$ On the other hand $a_i \in S^{-\infty}\subseteq S^{m-N}$ by assumption, so that $\sum_{i=0}^{N-1}a_i\in S^{m-N}$ are well. Thus $a\in S^{m-N}$ which completes the proof.

In particular, you actually don't need to assume anything about the joint support of $a_i$. On the otherhand, if you assume that $a=\sum_{i=1}^{\infty}a_i$ literally, instead of asymptotically, you need your support condition. This follows from the fact that any member of $S^m$ with compact support is in $S^{-\infty}$.

$\endgroup$
4
  • 1
    $\begingroup$ Indeed, I was just above to edit the first mistake! $\endgroup$
    – user900940
    Sep 14, 2021 at 21:50
  • $\begingroup$ Ah, that's quite obvious actually. Do you know, by chance, that if one has the given support properties, can we glean anything about the support of the full symbol? $\endgroup$
    – user900940
    Sep 14, 2021 at 21:52
  • 1
    $\begingroup$ @user900940 You can always add a symbol in $S^{-\infty}$ without affecting the asymptotic expansion, which can be chosen to have arbitrary support, so sadly I don't think you can get any information on the support of $a$. $\endgroup$
    – pax
    Sep 14, 2021 at 21:55
  • $\begingroup$ Of course, thank you! $\endgroup$
    – user900940
    Sep 14, 2021 at 21:58

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.