We know that $f \in \mathcal{S({\mathbb{R^n}})}$ implies that $f$ has rapidly decreasing derivatives. But does this, by some comparison argument, give us that $f$ is analytic (i.e. equal to its power series expansion)? Why or why not?


1 Answer 1


A Schwartz class function $f$ can be equal to $0$ on a whole interval around $0$, thus its Taylor series at $0$ would be the zero function too, yet $f$ could be non-zero outside of said interval, thus that $f$ can't be analytic (at $0$ at least). You should be able to construct such a $f$ using an adequate bump function I'm sure.

  • 2
    $\begingroup$ Any nonzero smooth and compactly supported function is also a counterexample for similar reasons. $\endgroup$
    – Alex Ortiz
    Feb 4 at 16:24
  • $\begingroup$ In your example, $f$ is certainly analytic in a neighborhood of the origin (wherein it is identically 0). I guess I should've specified this in my original question, but is it possible for such an $f$ to exist that is not analytic anywhere? $\endgroup$
    – algebroo
    Feb 4 at 20:13
  • $\begingroup$ Hmm it's probably worthy of being posted as another question. I'm not experienced enough to be able to provide an answer to that, though there is this question which might help figuring that out? $\endgroup$
    – Bruno B
    Feb 4 at 20:18
  • $\begingroup$ @algebroo Someone found a nowhere analytic Schwartz class function here, in case you're still interested. $\endgroup$
    – Bruno B
    Feb 19 at 11:06
  • $\begingroup$ @BrunoB Thanks for asking the question! I had a really busy week so I appreciate it. $\endgroup$
    – algebroo
    Feb 19 at 17:44

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