# Extending of analytic function from $[0,\infty)$ on $\mathbb{R}$

Assume that $g: [0, \infty) \rightarrow \mathbb{R}$ is $\mathbb{R}$-analytic on $[0, \infty)$ (i.e. for every point of $a \in [0,\infty)$ there exist $R_a>0$ and a real sequence $(a_n)_{n=0}^\infty$ (depending on $a$) such that the series $\sum_{n=0}^\infty a_n (x-a)^n$ is convergent for all $x \in(a-R_a, a+R_a)$ and $\sum_{n=0}^\infty a_n (x-a)^n=f(x)$ for all $x \in(a-R_a, a+R_a)$. Does $g$ can be extended to analytic function on the whole $\mathbb{R}$ ?

P.S. Please without complex analysis, if possible.

Thanks.

-
Wouldn't a function like $g(x)=e^{-1/(x+1)^2}, g(-1)=0$ give the usual headache with all its derivatives vanishing at $x=-1$? – Jyrki Lahtonen Nov 21 '11 at 12:55

Since the Taylor series converges in a neighbourhood of 0, the function can be extended to $(-\varepsilon,\infty)$ for a small $\varepsilon>0$, but one cannot go beyond that; simply take $$f(x):=\frac{1}{x+\varepsilon}$$ as a counterexample; it has a pole at $-\varepsilon$.