# A “comparison theorem” for real analytic functions ? 2

Let $$B(x_{0},R)$$ designate the open ball of center $$x_{0}$$ and radius $$R>0$$ in $$\mathbb{R}^{k}$$ ($$k\geq1$$). Let $$f$$ and $$g$$ be two real analytic functions on a neighborhood of the closure of $$B(x_{0},R)$$. Suppose we have $$0\leq f(x)\leq g(x)$$ on the whole of $$B(x_{0},R)$$. Now, if the radius of convergence of the Taylor's series of $$g$$ is $$R$$, can we conclude that the radius of convergence for the Taylor's series of $$f$$ is also $$R$$?

Let $$n=1$$, $$x_0=0$$, $$R=1$$ and $$f(x)=\frac{1}{1+4\,x^2},\quad g(x)=\frac{1}{1+x^2}.$$ Both $$f$$ and $$g$$ are real analytic on $$\Bbb R$$ and $$f(x)\le g(x)$$ for all $$x$$. The radius of convergence of $$g$$ around $$0$$ is $$1$$, but the radius of convergence of $$f$$ around $$0$$ is $$1/2$$.