Let $f:(a,b)\to\mathbb{R}$. We are given $(k+1)$ continuous functions $a_0,a_1,\ldots,a_k:(a,b)\to\mathbb{R}$ such that for every $c\in(a,b)$ we can write $f(c+t)=\sum_{i=0}^k a_i(c)t^i+o(t^k)$ (for any $t$ in a neighbourhood of $0$). Can we conclude that $f$ is of class $C^k$?

Yes. It's a classical result that goes back to Marcinkiewicz and Zygmund (On the differentiability of functions and summability of trigonometric series, Fund.Math 26 (1936) ). There is a sublety in the form of the remainder: a first and natural characterization of $C^k$ is obtained asking a remainder of the form $t^k \sigma(c,t)$ with $\sigma$ continuous in the pair $(c,t)$ and $\sigma(c,0)=0$ (i.e. the remainder is "$o(t^k)$ locally uniformly wrto $c$"). In this form, the proof is very easy, even for vector valued functions of several variables. But one may state a characterization of $C^k(a,b)$ asking (seemengly) less, that is, for any $c\in(a,b)$ the remainder at $c$ is just $o(t^k)$: it is true, but not a trivial fact, that then the remainder necessarily has the preceding form, so one gets a characterization as well (this is a successive result and I'll add references to it as soon as I recall it; or maybe somebody can do it for me). edit. Given the origin of the question, I'll leave some hints for an elementary proof of: $f\in C^k(a,b)$ if and only $f$ has a polynomial expansion of order $k$ with continuous coefficients $a_i\in C^0(a,b)$ and remainder of the form $$f(c+t)  \sum_{j=0}^k \frac{a_j(c)}{j!}t^j = t^k\sigma(c,t)\, ,$$ with a continuous $\sigma(c,t)$ vanishing identically for $t=0$. One implication is given of course by the Taylor theorem; for the other let $A_0,\dots,A_k$ be antiderivatives of $a_0=f,a_1, \dots a_k$. Fix $a < x < y < b$ and consider the function: $$\phi(t):=A_0(y+t) A_0(x+t) = \int_x^y f(c + t) dc\, .$$ Find two polynomial expansions of order $k$ for the function $\phi$ at zero (you can integrate the polynomial expansion for $f(t+c)$ either wrto $c$ or wrto $t$, this is the idea!). Then use the unicity of polynomial expansions (that is, if a polynomial $P(t)$ of degree not larger than $k$ is $o(t^k)$ for $t\to0$, then it is the zero polynomial), and deduce $A_j(y)A_j(x) = a_{j1}(y)  a_{j1}(x)$. Since $x$ and $y$ are arbitrary, the thesis follows with $f ^{(j)} = a_0^{(j)} = a_j $. 


Here is the answer I posted on https://math.stackexchange.com/questions/876071/multipledifferentiabilityfromtaylorexpansion, before seeing the question here. This question is answered in the affirmative in Abraham, Robbin, Transversal mappings and flows, Ch.1, $\S2$, A criterion for smoothness. They prove this converse to Taylor's theorem for functions between Banach spaces and attribute the onedimensional case to Marcinkiewicz, Zygmund, On the differentiability of functions and summability of trigonometrical series. As I understand after a glimpse at the proof, they prove by induction that $a_j=f^{(j)}$ by proving that $a_j(c+h)−a_j(c)=\int_0^1 a_{j+1}(c+th)h\,dt$. To prove that $f$ is $C^k$ and thus to justify the above, they prove that $a_1$ satisfies the hypothesis of the theorem with $k$ replaced by $k−1$ and then use induction (in the finite dimensional case; a trick using HahnBanach permits to reduce the theorem to that case). The proof of that fact looks elementary but tricky; in particular, they use a polynomial interpolation lemma. 

