# “Converse” of Taylor's theorem

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
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
right, for instance $f(t):=t^{k+1}\chi_{\mathbb{Q}}(t)$ is $o(t^k)$ for $t\to 0$ but it is only differentiable at $0$. – Pietro Majer Feb 15 '12 at 13:59

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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The Marcinkiweicz & Zygmund paper is available from EUDML, but I can't find the appropriate theorem in it (as I mentioned in my answer I could find it in a later paper by Oliver, however): all conclusions seem to be "almost everywhere" (e.g., theorem 1). Did I miss something, or were you perhaps thinking of a different paper? – Gro-Tsen Aug 30 '14 at 23:02
According to Abraham & Robbin's Transversal mapping and flows the result (with the local uniform assumption on the remainder) is due to Marcinkiweicz & Zygmund; but I could have copied the reference wrongly. My fault. The above proof is just home-made. – Pietro Majer Aug 31 '14 at 15:38

I was unable to get my hands on the book by Abraham & Robbin mentioned in the answer by Benoit Jubin, nor locate the exact statement in the paper by Marcinkiewicz & Zygmund mentioned in the answer by Pietro Majer (every theorem in this paper, as well as a later related solo paper by Marcinkiewicz, "Sur les séries de Fourier", Fund. Math. 27 (1937) 38–69, seems to have "almost everywhere" in the conclusion).

However, the keyword is "de la Vallée-Poussin derivative" or "Peano derivative" to designate the $k$-th coefficient of an asymptotic expansion of $f$ around a point: i.e., if $f(a+h) = f(a) + f_{1}(a)\,h + \cdots + \frac{1}{n!} f_{n}(a)\,h^n + o(h^n)$ (with no assumption of uniformity w.r.t. $a$ on the error term), we say that $f_{k}(a)$ is the $k$-th dlVP/Peano derivative of $f$ at $a$. Given this, is fairly easy to use Google to find a theorem in the literature that answers the question and much more:

H. William Oliver, "The Exact Peano Derivative", Trans. Amer. Math. Soc. 76 (1954) 444–456: theorem 3:

If $f_n(x)$ exists and is bounded above or below throughout $[a,b]$, then $f_n(x) = f^{(n)}(x)$, the ordinary $n$-th derivative, at every point $x \in [a,b]$.

The same paper proves various other interesting properties of these dlVP/Peano derivatives: they are Baire class 1, have the Darboux property (=intermediate values property), satisfy a mean value theorem (stated in the paper) and have the Denjoy property (=the inverse image of an open interval is either empty or has positive measure).

The proofs in Oliver's paper involves rather tedious case distinctions. A different, and possibly simpler, proof of the above-quoted statement is given in: Verblunsky, "On the Peano Derivatives", Proc. London Math. Soc. 22 (1971) 313–324 (see theorem 1(ii)).

Analogous properties for approximate Peano derivatives (same definition as above except that the asymptotic expansion is only assumed on a set of $h$ having density $1$ at $0$) are proved in Babcock, "On Properties of the Approximate Peano Derivatives", Trans. Amer. Math. Soc. 212 (1975) 279–294.

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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 '14 at 18:17

Another answer: You are prescribing a Whitney jet, and the Whitney extension theorem gives the answer:

• Whitney, Hassler: Analytic extensions of differentiable functions defined in closed sets, Trans. AMS 36 (1934), 63--89.

Also, another reference building on the paper of Marcinkiewicz and Zygmund, is

• G. Glaeser: Etude de quelques Algebres Tayloriennes, J. Anal. Math. 11 (1958), 1-118.
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