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?
2026-04-03 06:05:05.1775196305
Are Schwartz class functions analytic?
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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.