Possible Duplicate:
“Converse” of Taylor's theorem
I need some hints for this problem.
Let $I\subseteq \mathbb{R}$ an open nonempty interval, $f:I\to \mathbb{R}$ and $N\in \mathbb{N}$.
Then $f\in C^N(I)$ iff there exist $a_0,\ldots ,a_N \in C^0(I)$ such that: $\tag{1} f(x)=\sum_{n=0}^N a_n(x_0)\ (x-x_0)^n + \text{o}((x-x_0)^N)$ for all $x_0\in I$.
Now, part $\Rightarrow$ is an immediate consequence of the Taylor's theorem.
The converse $\Leftarrow$, on the other hand, seems to be harder. Actually, if (1) holds, then $f(x_0)=a_0(x_0)$ for all $x_0\in I$, hence $f\in C^0(I)$. Moreover: $\begin{split} \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0} &= \lim_{x\to x_0} \frac{\sum_{n=0}^N a_n(x_0)\ (x-x_0)^n + \text{o}((x-x_0)^N)-a_0 (x_0)}{x-x_0} \\ &= \lim_{x\to x_0} \frac{\sum_{n=1}^N a_n(x_0)\ (x-x_0)^n + \text{o}((x-x_0)^N)}{x-x_0} \\ &= a_1(x_0)\end{split}$ thus $f$ is differentiable in $I$ and $f^\prime (x_0)=a_1(x_0)$ for all $x_0\in I$; therefore $f\in C^1(I)$. So, when $N\leq 1$ there are no problems in the proof.
Problems arise when one wants to prove higher order differentiability in $I$. I thougth a possible way to win could consist in considering something like the limit: $\lim_{x\to x_0} \frac{f(x)-\sum_{n=0}^k a_n(x_0)\ (x-x_0)^n}{(x-x_0)^k}\; ,$ for it exists and equals $a_k(x_0)$... But I didn't succeed in the proof.
Any hints?