By Borel's theorem, for any sequence of real numbers $a_n,$ there is a $C^{\infty}$function $f:\mathbb{R}\to\mathbb{R}$ whose Taylor series at 0 is $\sum a_nx^n.$ In particular, there are $C^{\infty}$functions whose Taylor series at a point has 0 radius of convergence. Motivated by this I have the following question. Is there a $C^{\infty}$function $f:\mathbb{R}\to\mathbb{R}$ whose Taylor series has 0 radius of convergence at every point in $\mathbb{R}?$ I realize that this might sound like a homework problem, but, well, it's not.

3$\begingroup$ There is a book called Counterexamples in Analysis, I don't immediately remember the authors. I think it is available as a Dover reprint. $\endgroup$ – Will Jagy Oct 25 '10 at 2:41

1$\begingroup$ The authors are Gelbaum and Olmsted (Gelbaum had the office next to mine at SUNY Buffalo for some years), but I don't think this is in the book. Closest thing seems to be a function whose Maclaurin series converges at only one point. $\endgroup$ – Gerry Myerson Oct 25 '10 at 3:44
S.S. Kim and K.H. Kwon gave an explicit example of a monotone smooth but nowhere analytic function (link), which is an antiderivative of the function $$\psi(x)=\sum\limits_{k=1}^{\infty} \frac{1}{k!}\phi(2^k(x[x])),$$ where $$\phi(x) = \begin{cases} \exp{\left(\frac{1}{x^2}\frac{1}{(x1)^2}\right)},\qquad & 0 < x < 1,\ \\\ \\\ 0, & \mbox{otherwise.} \end{cases}$$
In fact, the set of smooth but nowhere analytic functions on $\mathbb R$ is of the second category in $C^{\infty}(\mathbb R)$ (just like the set of all continuous but nowhere differentiable functions is of the second category in $C(\mathbb R)$). See a one page note by R. Darst "Most infinitely differentiable functions are nowhere analytic".
Edit. Kim and Kwon mention in their paper that the first concrete example of smooth
but nowhere analytic function dates back to A Pringsheim ("Zur Theorie der Taylor'schen Reihe und der analytischen Functionen mit beschränktem Existenzbereich." Math. Ann. 42 (1893), no. 2, 153–184.)

1$\begingroup$ As defined, $\psi$ is a finite sum locally at nonintegers. Don't you want to make $\phi$ 1periodic (and as defined on [0,1])? Or else, write $\phi(2^k x  [2^k x])$? $\endgroup$ – Benoit Jubin Nov 21 '11 at 3:05
The Fabius function (Wikipedia and MO; a nonprobabilistic description can be found here) is everywhere $C^\infty$ and nowhere analytic.

4$\begingroup$ Nice answer. I did notice that the wikipedia article on the subject is quite minimal. Perhaps some good (and erudite) samaritan would like to flesh it out a bit? $\endgroup$ – Pete L. Clark Oct 25 '10 at 3:36
Neither of the responses exactly answers the question asked. The question asks whether the Taylor series of a smooth function at every point can have radius of convergence 0. This is more restrictive than not being analytic anywhere. The responses just treat this weaker question. 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.
It should be noted that the proof in the article of Kim and Kwon is incorrect, and although it is highly likely the function they consider is not analytic anywhere, I don't see how to prove it. (Their "proof" assumes that the sum of a tail of the series is analytic, which essentially contradicts the conclusion they are trying to prove!)
In fact, I do not know an example of what the questioner is asking, although the example of Kim and Kwon is a candidate. There is an example in Big Rudin (Chapter 19, problem 13) of a complexvalued smooth function on R whose Taylor series has radius of convergence zero at every point, but the argument there does not seem to be adaptable to getting a realvalued example; in particular, it is not at all clear that the real and imaginary parts of this example have the desired property. The question asked seems like a very interesting one, perhaps quite difficult and possibly open.

$\begingroup$ It's not necessarily wrong to make an assumption that contradicts your conclusion. Every proof by contradiction does this. If the assumption follows from the negation of the conclusion, then it is valid to use. That seems to me to be the case here. $\endgroup$ – Toby Bartels Feb 9 '17 at 16:13
OK, Bruce, once you decided to revive it, let's construct an example addressing the question as asked from scratch.
Let $f_k(x)=(k!)^{3k+2}\cos ((k!)^3 x)$. Note that $f_k$ and its first $k1$ derivatives are less than $(k!)^{1}$ while at every point $f_k^{(k)}(x)^2+f_k^{(k+1)}(x)^2\ge (k!)^4$. Now just write $$ f=\sum_{k\in K} f_k $$ where $K$ is some sparse set of integers. We clearly have convergence in $C^\infty$ no matter how we choose $K$. Now we just want to make sure that the $k$th and $k+1$st derivatives of $f_k$ one of which is huge are not killed by other terms. The subsequent terms are never a problem if $K$ is $2$separated. To get rid of the previous terms, we can construct $K$ inductively. If we already have a first few members $k_1,\dots, k_p$, we can notice that the sum of these first $p$ terms is entire and periodic (the head, not the tail!), so its $k$th and $k+1$st derivatives on $\mathbb R$ are uniformly bounded by $C(k+1)!$ with $C$ depending only on the beginning segment $k_1,\dots, k_p$. But this is not enough to kill (or even to interfere noticeably with) the uniform lower bound of order $(k!)^2$ we have for the maximum of the $k$th and $k+1$st derivatives of $f_k$ if the next $k$ is large enough.
If you want a strictly monotone function, just add a big constant multiple of $x$.

$\begingroup$ Nice example! I assume this is original, but if not, can you give a reference? If $I$ is an interval in $\mathbb R$, does the set of functions with the proerty that all Taylor series have radius of convergence zero form a residual set in $C^\infty(I)$ in its usual topology? $\endgroup$ – Bruce Blackadar Nov 21 '11 at 16:29

$\begingroup$ Sure. The set $A_{m,n}$ of functions such that there exists a point $x$ for which $f^{(k)}(x)\le nk!m^k$ is closed in $C^\infty(I)$ for fixed $m,n$ (assuming that "an interval" means "a compact interval"; otherwise take some countable compact exhaustion $I_p$, fix $p$ as well, and add $x\in I_p$ to the condition). The argument above allows one to find a bad function in any open set (since open sets are controlled by just finitely many derivatives). So, the union of $A_{m,n}$ is of first category. $\endgroup$ – fedja Nov 22 '11 at 0:06
Good. I thought something like this might work.
One more question before I get off this subject, which might be harder (or easier!): what about the other extreme? Is there a smooth function on an interval in $\mathbb R$, not analytic on any subinterval, whose Taylor series at every point has positive radius of convergence? The Fabius function might be an example, but this is questionable since the $n$'th derivative has maximum $2^{\sigma(n)}$, where $\sigma(n)=\frac{n(n+1)}{2}$, which is not quite good enough using a crude estimate for radius of convergence if there are points where many derivatives are close to the maximum.

1$\begingroup$ Bruce, you should get an account (I see your two posts have different reputations, so are not linked). Also, you should not ask new questions as "answers" to someone else's very old question, but just post a new question with a link back here. $\endgroup$ – Ryan Reich Nov 22 '11 at 6:12

$\begingroup$ I suggest you use the "ask question" button near the top of the page, and delete this answer. $\endgroup$ – S. Carnahan♦ Nov 22 '11 at 7:05

$\begingroup$ Sorry, I'll do this. I'm new to this site. $\endgroup$ – Bruce Blackadar Nov 22 '11 at 14:15