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:
- 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 $a$.
- 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:
- $C$ and $P$ are the sets of C-points and P-points, respectively, of some $C^\infty$
function $f:\mathbb R\to\mathbb R$.
- 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 $\mathbb R$.)
Hardly anything you wrote is true.
First of all, an analytic function is not the same thing as a function which has derivatives of all orders. It is stronger than that: it is a function $f$ such that, for each $x_0$ in its domain, the Taylor series of $f$ converges to $f$ in some interval $(x_0-\varepsilon,x_0+\varepsilon)$.
Besides, the function $f\colon\mathbb{R}\longrightarrow\mathbb R$ defined by $f(x)=\frac1{1+x^2}$ is analytic, but the radius of convergence of its Taylor series at $0$ is $1$, not $+\infty$. And, of course, this shows that your second claim is also false.
Best Answer
No. A function $f(x)$ is analytic at a point $x_0$ if its Taylor series exists at $x_0$ and converges to $f(x)$ in a neighborhood of $x_0$. The Taylor series may exist but not converge, or it may converge but not to $f(x)$.
The basic idea is that $f(x)$ decreases so quickly as $x \to 0$ that all of its derivatives exist and are equal to $0$ at $x = 0$; more formally we have $\lim_{x \to 0} \frac{f(x)}{x^m} = 0$, which can be established e.g. by taking logarithms. So the Taylor series at $x = 0$ has all coefficients zero, which means it doesn't converge to $f(x)$ in any neighborhood of $0$. On the other hand, for $x > 0$, $f(x)$ is a composition of smooth functions and hence smooth.
The previous function is an example; the Taylor series at $x = 0$ converges to zero in any neighborhood but the function doesn't. Generally, the way we prove that a Taylor series converges (when it does) is to use Taylor's theorem with remainder, which bounds the error of the approximation given by taking the first $n$ terms of the Taylor series. But it can happen that this bound doesn't go to $0$ as $n \to \infty$; if that's the case then Taylor's theorem doesn't tell us that the Taylor series converges to the function.
Not necessarily. For example, the function $f(x) = x^k |x|$ is differentiable at $x = 0$ exactly $k$ times, but not $k+1$ times.
Yes; more precisely it means that at every point the Taylor series exists and converges to $f(x)$ in some neighborhood.
As an additional comment on question 2, if a function $f(x)$ is analytic at $x = 0$ then it must decay at worst like $x^m$ as $x \to 0$, where $m$ is the first index such that $f^{(m)}(0) \neq 0$. So analytic functions can't decay faster than polynomially at any point.