TITLE: Is a function with nowhere vanishing derivatives analytic?
QUESTION [48 upvotes]: My question is the following: Let $f\in C^\infty(a,b)$, such that $f^{(n)}(x)\ne 0$, for every $n\in\mathbb N$, and every $x\in (a,b)$. Does that imply that $f$ is real analytic?
EDIT. According to a theorem of Serge N. Bernstein (Sur les fonctions absolument monotones, Acta Mathematica, 52 (1928) pp. 1–66) if $f\in C^\infty(a,b)$ and $f^{(n)}(x)\ge 0$, for all $x\in(a,b)$, then $f$ extends analytically in a ball centered at $(a,0)\in\mathbb C$ and radius $b-a$!

REPLY [55 votes]: If $f$ is $C^{\infty}$ every derivative is continuous, so the hypothesis on $f$ implies that each derivative $f^{(n)}$ has constant sign.  Such functions were studied by S. Bernstein and called regularly monotonic.  In particular he proved in 1926 that a regularly monotonic function is real analytic.
This 1971 AMM article by R.P. Boas provides a proof, more history, and further results along these lines.  See also this 1975 PAMS article of J. McHugh.