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It is well-known that the function

$$f(x) = \begin{cases} e^{-1/x^2}, \mbox{if } x \ne 0 \\ 0, \mbox{if } x = 0\end{cases}$$

is smooth everywhere, yet not analytic at $x = 0$. In particular, its Taylor series exists there, but it equals $0 + 0x + 0x^2 + 0x^3 + ... = 0$, so while it has radius of convergence $\infty$, it is not equal to $f$ even in a tiny neighborhood of $0$.

There is also a function

$$f(x) = \sum_{n=0}^{\infty} e^{-\sqrt{2^n}} \cos(2^n x)$$

which is smooth everywhere (that is, $C^{\infty}$) yet analytic nowhere. In particular, the Taylor series at every point has radius of convergence $0$. In fact, "most" smooth functions are not analytic.

But this gets me wondering. Could there exist some function which is smooth everywhere, analytic nowhere, yet its Taylor series at any point has nonzero radius of convergence, and so converges to something, but that something is not the function, not even in a tiny neighborhood about the point of expansion? If yes, what is an example of such a function? If no, what is the proof that such a thing is impossible? And also, if no, what sort of restrictions exist on the convergence of the T.s.? At how many/what distribution of points can it converge to something which is not the function? I note that if we multiply together the two functions just given above, we have another smooth-everywhere, analytic-nowhere function, but this time at $0$ we have a convergent Taylor series (the same zero series as before -- just use the generalized Leibniz rule) which doesn't converge to the function in even a tiny neighborhood of $0$.

EDIT (Dec 31, 2013): With some Googling I came across a post to mathoverflow:

http://mathoverflow.net/a/81465

The Taylor series of the Fabius function at any dyadic rational actually has infinite radius of convergence (only finitely many terms are nonzero) but does not represent the function on any interval.

So it seems it is possible to have a function whose Taylor series converges to "the wrong thing" at a dense set of expansion points. But it still doesn't answer the question of whether that is possible for all expansion points on the entire real line.

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What do you mean by "smooth"? That it has $\infty$ derivatives? –  Betty Mock Jan 6 at 2:31
    
@Betty Mock: Yes. –  mike4ty4 Jan 7 at 21:08

1 Answer 1

up vote 13 down vote accepted

No, this is not possible. Dave L. Renfro wrote an excellent historical Essay on nowhere analytic $C^\infty$ functions in two parts (with numerous references). See here: 1 (dated May 9, 2002 6:18 PM), and 2 (dated May 19, 2002 8:29 PM).

As indicated in part 1, in

Zygmunt Zahorski. Sur l'ensemble des points singuliers d'une fonction d'une variable réelle admettant les dérivées de tous les ordres, Fund. Math., 34, (1947), 183–245. MR0025545 (10,23c); and Supplément au mémoire "Sur l'ensemble des points singuliers d'une fonction d'une variable réelle admettant les dérivées de tous les orders", Fund. Math., 36, (1949), 319–320. MR0035329 (11,718a),

Zahorski suggested in 1947 the following classification of points where a function $f$ is $C^\infty$ but not analytic:

  1. A point $a$ is a C-point (for Cauchy) iff the formal Taylor series about $a$ associated to $f$ converges in a neighborhood of $a$, but the resulting analytic function does not coincide with $f$ in any neighborhood of $a$.
  2. The point $a$ is a P-point (for Pringsheim) iff the formal Taylor series of $f$ about $a$ has radius of convergence $0$.

Theorem (Zahorski). Let $C,P$ be sets of real numbers. The following are equivalent:

  1. $C$ and $P$ are the sets of C-points and P-points, respectively, of some $C^\infty$ function $f:\mathbb R\to\mathbb R$.
  2. The following 4 conditions hold:
    • $C$ is a first category $F_\sigma$ set.
    • $P$ is a $G_\delta$ set.
    • $C\cap P=\emptyset$.
    • $C\cup P$ is closed in $\mathbb R$.

As a corollary, note that if $f:\mathbb R\to\mathbb R$ is smooth, and its set of P-points is empty then, since no interval is first category (by the Baire category theorem), in every interval there must be points where $f$ is analytic.

Two other key references you may want to consult (also mentioned in Renfro's essay) are

Gerald Gustave Bilodeau. The origin and early development of nonanalytic infinitely differentiable functions, Arch. Hist. Exact Sci., 27 (2), (1982), 115–135. MR0677684 (84g:26017),

and

Helmut R. Salzmann, and Karl Longin Zeller. Singularitäten unendlich oft differenzierbarer Funktionen, Math. Z., 62 (1), (1955), 354–367. MR0071479 (17,134b).

(The latter contains a simplified proof of Zahorski's result.)

(Coincidentally, last term I had the opportunity to cover some of the results in this area in my analysis class. See also MathOverflow, for a version of this question, and the related question of whether the set of P-points can be $\mathbb R$.)

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