Question

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 $C^k$?

Answer 1

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 $A_j(y)-A_j(x) = a_{j-1}(y) - a_{j-1}(x)$. Since $x$ and $y$ are arbitrary, the thesis follows with $f ^{(j)} = a_0^{(j)} = a_j $.