# Non-analytic function with convergent Taylor series everywhere

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.

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How about taking an example (like $\exp(-1/x^2)$) with this property at one point only, then doing a series with translates by the rationals and coefficients going to zero fast enough? –  Gerald Edgar Nov 22 '11 at 15:51
I don't understand the whole situation: how can a function with a convergent Taylor series expansion at every point and with positive radius of convergence be non analytic? –  user16974 Nov 22 '11 at 16:03
@Charlie: OK, I'll be giving a graduate course in complex analysis next semester. I'll assign it as homework and let you know the result :). Remind me if I forget. @Bruce: The set $A_{m,n}$ of points $x$ for which $|f^{(k)}(x)|\le mk!n^k$ for all $k$ is closed and contains no interval. Ergo... –  fedja Nov 22 '11 at 18:03
I posted a fairly extensive survey on this topic in May 2002 at mathforum.org/kb/message.jspa?messageID=387148 and mathforum.org/kb/message.jspa?messageID=387149 –  Dave L Renfro Nov 22 '11 at 19:39
After I finally read some definitive history on this subject such as Dave's survey, what strikes me is how consistently many mathematicians (now myself included) have been ignorant of previous work in this area over a rather long time. I don't offhand recall any topic, at least in analysis, where basic examples and results have been rediscovered, reproved, and republished so often, and I would guess that even today many mathematicians would not know about this work. –  Bruce Blackadar Nov 23 '11 at 15:32

We define $$\Psi(x)= \sum_{k\ge 0} 2^{-k}\psi_{\sigma_k}(x-x_k),\quad \psi_{\sigma}(y)=\exp{-{\vert x\vert}^{-\frac{1}{s-1}}},$$ where $(x_k)_{k\ge 0}$ is dense in $\mathbb R^d$ and $(\sigma_k)_{k\ge 0}$ is decreasing and valued in $[s_1,s_0]\subset(1,+\infty)$.
It seems likely that the radius of convergence of the Taylor series is positive on a dense subset of $\mathbb R^d$. Anyhow it is a good candidate.