Are Schwartz class functions analytic?

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?

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.
• 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? Feb 4 at 20:13