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 $C^k$?
"Converse" of Taylor's theorem
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calculus
real-analysis
functions
derivatives
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1If 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. – 2012-02-09
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0This 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](http://books.google.ch/books?id=i4vw2STJl2QC&lpg=PP1&hl=de&pg=PP1#v=onepage&q&f=false)*. 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](http://www.zentralblatt-math.org/zmath/en/search/?q=an:0518.46018&format=complete)). – 2012-02-12
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0Answered on Mathoverflow: http://mathoverflow.net/questions/88501/converse-of-taylors-theorem – 2012-02-15