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Let $f:(a,b)\to\mathbb R$. We know that for every $c\in(a,b)$ we can write $f(t)=\sum_{i=0}^k a_i(c)(t-c)^i+o\left((t-c)^k\right)$ and $\forall i$ $a_i(c)$ is continuous (with respect to $c$). Can we conclude that $f$ is of class $C^k$?

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If we assume $f$ continuous then we should have $a_0(c)=f(c)$ so $\frac{f(t)-f(c)}{t-c}=a_1(c)+\sum_{i=2}^ka_i(c)(t-c)^{i-1}+o((t-c)^{k-1})$ and we get that $f$ is differentiable. –  Davide Giraudo Feb 9 '12 at 21:51
    
This converse of Taylor's theorem is stated as Theorem 2.1.6 (without proof) in Krantz and Parks' book A Primer on Real Analytic Functions. They write that a detailed proof of this can be found in S. G. Krantz, Lipschitz spaces, smoothness of functions, and approximation theory, Expo. Math. 1 (1983), 193-260 (Entry on Zentralblatt). –  Sam Feb 12 '12 at 18:42
    
Answered on Mathoverflow: mathoverflow.net/questions/88501/converse-of-taylors-theorem –  Bill Cook Feb 15 '12 at 14:07

1 Answer 1

EDIT: What is written below doesn't actually answer the original question because the resulting functions $a_i(c)$ are not continuous. I will leave it posted for now because it does suffice to illustrate that the claim can fail without continuity of the $a_i$.


This "converse" is false, as I learned from Fedor Petrov's contribution to MathOverflow's list of common false beliefs in math (perhaps one of the best questions ever asked on that site). In an embarrassed email to Prof. Petrov I admitted to holding this belief, so he provided me with the counterexample \[ f(x) = \begin{cases} \sin (e^{1/x^4}) e^{-1/x^2} & x\neq 0, \\ 0 & x=0. \end{cases} \] This function is $C^\infty$ everywhere except at $x=0$. It goes to zero very quickly as $x\to 0$, but within that $e^{-1/x^2}$ envelope it wiggles faster than it goes to zero. The result is that $f(x) = o(x^n)$ for all $n$ as $x\to 0$, but a quick computation shows that $f'(x)$ is not continuous at zero. So $f$ is differentiable, but not even $C^1$.

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Did you miss the part about the $a_i(c)$ being continuous? –  Robert Israel Feb 10 '12 at 1:54
    
@RobertIsrael: Oops, I did. Thanks! I will edit to make it clear that this does not answer the question, but leave the counterexample to illustrate why continuity is needed. –  Noah Stein Feb 10 '12 at 2:12
    
I don't see why the $a_i(c)$ have to be continuous. Actually, if we consider the function $f(x)$ from Noah, say on the interval $(-1,1)$, then we have the following: for any $c\ne0$, the function is analytic on any interval around $c$ that doesn't contain 0, and so can be written as the question asks (i.e. by it's degree $k$ Taylor polynomial); for $c=0$, we have $a_i(0)=0$ for all $i$, and since $f$ is $o(t^k)$, the condition is also satisfied; but, as Noah said, $f$ is not even $C^1$. –  Martin Argerami Feb 10 '12 at 4:37
    
@Martin: The $a_i$ being continuous is one of the premises (though this is perhaps not immediately clear because of the unorthodox wording of the question). Since the intended counterexample doesn't satisfy this premise, it's not a counterexample. –  joriki Feb 12 '12 at 14:30
    
@joriki: I totally missed that in the statement. –  Martin Argerami Feb 12 '12 at 16:41

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