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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 $C^k$?

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Have you tried doing this for $k = 0$ and $k = 1$? –  Deane Yang Feb 15 '12 at 9:09
This question is really more suitable for math.stackexchange.com than here. –  Deane Yang Feb 15 '12 at 9:10
I know, but there nobody could answer. Btw yes, if $k\ge 1$, it is easy to check that $f$ is of class $C^1$. –  Mizar Feb 15 '12 at 10:49
Yes, I spoke too soon. Nice answer by Pietro. –  Deane Yang Feb 15 '12 at 11:28
Also, I seem to recall that is is important that these are functions on a whole interval. Doing this at one point is not enough to show the function is $k$ times differentiable at that point. –  Gerald Edgar Feb 15 '12 at 13:14

2 Answers 2

up vote 10 down vote accepted

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 $.

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meta.mathoverflow.net/a/779/763 –  Yemon Choi Sep 5 '13 at 16:40

Here is the answer I posted on https://math.stackexchange.com/questions/876071/multiple-differentiability-from-taylor-expansion, 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 one-dimensional 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 Hahn-Banach 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.

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Note that the proof for functions between Banach spaces follows easily from the case of functions between finite dimensional spaces, as shown in Abraham-Robbin. However, the finite dimensional case has a two lines proof (compare with the proof in A-R, several pages of hard analysis via partitions of unity!). Mollify, use the uniqueness of polynomial expansion, apply the theorem of limit under the sign of derivative. –  Pietro Majer Aug 20 at 18:17

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