Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be a real function, and let $0\leq n\leq+\infty$. We make the following assumption:

For every $a \in\mathbb{R}$ and for $k=n$ (resp., in the case $n=+\infty$: for any $k\geq 0$), there exist real numbers $c_0(a),\ldots,c_k(a)$ such that

$$f(x) = c_0(a) + c_1(a)\,(x-a) + \frac{1}{2}c_2(a)\,(x-a)^2 + \cdots + \frac{1}{k!}c_k(a)\,(x-a)^k + o((x-a)^k)$$

where, as usual, $o((x-a)^k)$ means $(x-a)^k\,\varepsilon_{a,k}(x)$ for some function $\varepsilon_{a,k}$ tending to $0$ when $x \to a$.

In other words, we assume that $f$ has a power expansion of order $k$ with $o$ error term at (every) $a$. Note that no assumption is made on uniformity of the $o$ error term when $a$ varies (e.g., we do not assume that $\varepsilon_{a,k}(x)$ is bounded by some function of $x-a$): we only assume that for every $a$ there exists an expansion of order $k$ as above, nothing more.

Naturally, the $c_i(a)$ are uniquely determined, we have $c_0 = f$ (that is, $c_0(a) = f(a)$ for every $a$) and $f$ is continuous; and moreover, as soon as $n\geq 1$, clearly, $f$ is differentiable with derivative $f' = c_1$.

We cannot deduce that $f$ is twice differentiable, or even $C^1$, from the above hypothesis alone, no matter how large $n$ is. The simple example of $f(x) = x^{n+1} \sin(x^{-n})$ provides a counterexample (it is $o(x^n)$ at $0$ and analytic everywhere else, so it has a power expansion of order $n$ everywhere, yet it is easily seen that it is not even $C^1$ at $0$); a slightly more complicated counterexample works for $n=\infty$.

Now here is my question. Let us make the following additional assumption (which is not satisfied for the above counterexample):

For each $0\leq k\leq n$, the function $c_k$ (that is, $a\mapsto c_k(a)$) is continuous.

(In particular, if $n\geq 1$, it is now clear that $f$ is $C^1$.)

Can I conclude from both assumptions that $f$ is $C^n$? (Or, if not, can I conclude something non-trivial?)

share|improve this question
1  
This question received and answer on Mathoverflow: mathoverflow.net/questions/88501/converse-of-taylors-theorem –  Siminore Aug 20 at 15:50

2 Answers 2

up vote 4 down vote accepted

This question is answered in the affirmative in Abraham, Robbin, Transversal mappings and flows, Ch.1, $\S$2, 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 $c_k = f^{(k)}$ by proving that $c_k(a+h) - c_k(a) = \int_0^1 c_{k+1}(a+th)h \, dt$. To prove that $f$ is $C^n$ and thus to justify the above, they prove that $c_1$ satisfies the hypothesis of the theorem with $n$ replaced by $n-1$ 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.

share|improve this answer

To repeat what I wrote on MO, 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 (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.

(Surprisingly, Mukhopadhyay wrote in 2012 a whole book on Higher-Order Derivatives and does not seem to mention these theorems anywhere!)

share|improve this answer
1  
FYI, some names you can google (along with "Peano derivative") are: J. Marshall Ash, Ernesto Corominas (1946 paper), Michael Jon Evans, Miklos Laczkovich, Cheng-Ming Lee, Richard John O'Malley, Clifford Edward Weil, and Ralph Eugene Svetic. I've given their full names for identification purposes, but of course their papers will appear under various versions (e.g. C. E. Weil, Cliff Weil, Cliff E. Weil, Clifford E. Weil, etc.). –  Dave L. Renfro Sep 8 at 20:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.