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I want to solve following problem

Let $f,g : (-1, 1) \rightarrow \mathbb{R}$ be $C^{\infty}$ functions, and suppose that $f^{(n)}(0)= g^{(n)}(0) $ for $n=0,1,2,\cdots, $. Is there some $\delta>0$ such that $f(x) = g(x)$ for all $x \in [-\delta, \delta]$?

Naively, from taylor series expansion of $C^{\infty}$ functions, I suspect that there is $\delta>0$ but I don't know how to prove this rigorously.

phy_math
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    In other words, you are asking "Is every $C^\infty$ function also real analytic?" As given in the answers, the answer is no. You can find many more such questions if you use that rephrasing. – Brian Moehring Apr 08 '21 at 05:30
  • Another one: https://math.stackexchange.com/q/240026/42969 – Martin R Apr 08 '21 at 06:58

2 Answers2

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No. Take $f(x)=e^{-1/x^2}$ for $x\ne 0$ and $f(0)=0$, $g(x)=0$. Then for any $x>0$ $f(x)\ne g(x)$, but $f^{(n)}(0)=0$ for any $n$.

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Nope.

Here's a counterexample:

$f(x)=0$, the constant function equal to 0 everywhere, and let $g(0)=0$ and for $x\neq0,$

$$g(x)=e^{-\frac1{x^2}}.$$

Then, all derivatives of $g$ exist and equal $0$ at $x=0$, so the taylor series of $f$ and $g$ at $x=0$ are identical. However, clearly, $f(x)=g(x)$ only if $x=0$.

Michael Hartley
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