Does there exist an infinitely differentiable function $f:U\to\mathbb{R}$, where $U$ is open subset of $\mathbb{R}$, such that

  1. the Taylor series of $f$ at $x=x_0\in U$ has radius of convergence $R>0$
  2. $f$ equals its Taylor series only on the subinterval $(x_0-r,x_0+r)$, where $\color{red}{0<}r<R$

The customary examples of smooth real functions that fail to be analytic, e.g. $e^{-1/x}$ or $e^{-1/x^2}$ at $x=0$, have $R=\infty$ but $r=0$. The substance of the question is whether we can find a less extreme example for which analyticity at $x=x_0$ gives out only at a nonzero radius smaller than the radius of convergence of the Taylor series.

Note: I don't really know complex analysis, but I know that the easiest path to whatever the truth is here is probably through the complex domain.

  • 2
    $\begingroup$ There is a $C^\infty$ version of Urysohn lemma. I think it satisfies your restriction. $\endgroup$ – Cave Johnson Aug 16 '16 at 1:25
  • 1
    $\begingroup$ @CaveJohnson: yes, you're right: those functions are good examples. You should add that as an answer. $\endgroup$ – symplectomorphic Aug 16 '16 at 2:05

$\newcommand{\Reals}{\mathbf{R}}$Yes: Fix $r > 0$. The function $f:\Reals \to \Reals$ defined by $$ f(x) = \begin{cases} 0 & |x| \leq r, \\ e^{-1/(|x| - r)^{2}} & |x| > r, \end{cases} $$ has Taylor series equal to $0$ (radius $\infty$), but agrees with its Taylor series only on $(-r, r)$.

If you want finite radius instead, add your favorite analytic function with radius $R > r$, e.g., $$ g(x) = \frac{1}{x^{2} + R^{2}}. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.