Example of smooth function that is nowhere analytic without Fourier series

Question

Can someone provide me an example of a real smooth nowhere analytic function that isn't constructed from Fourier series? Every example I find depends on it.

Thanks.

Answer 1

Let $f(x) = e^{-1/x}$ for $x>0$, $f(x)=0$ otherwise. This fails be to be analytic at zero, though it is analytic everywhere else.

Letting $q_n$ enumerate the rationals, then $g(x)= \sum_n 2^{-n}f(x-q_n)$ fails to be analytic at all rationals, so $g$ is nowhere analytic.

Edit: It is simple enough to prove that $g$ is smooth. For a reference that $g$ is nowhere analytic, see Is this function nowhere analytic?.