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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2$\begingroup$ 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? $\endgroup$– Gerald EdgarCommented Nov 22, 2011 at 15:51
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5$\begingroup$ 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? $\endgroup$– user16974Commented Nov 22, 2011 at 16:03
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4$\begingroup$ @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... $\endgroup$– fedjaCommented Nov 22, 2011 at 18:03
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6$\begingroup$ 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 $\endgroup$– Dave L RenfroCommented Nov 22, 2011 at 19:39
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4$\begingroup$ 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. $\endgroup$– Bruce BlackadarCommented Nov 23, 2011 at 15:32
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1 Answer
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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 $[s_1,s_0]\subset(1,+\infty)$.
That function is good explicit substitute to Fabius function since it is smooth and nowhere analytic: even better, it is multidimensional and its analytic wave-front-set is all the cotangent space (minus the zero section). To prove this use Gevrey classes.
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.
Bazin.
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$\begingroup$ You can use the Baire category theorem to show that there are no non-analytic functions with a Taylor series with positive radius of convergence everywhere. $\endgroup$ Commented Mar 10, 2012 at 0:32
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$\begingroup$ This is also stated in Dave L Renfro's first link above. See the part where it says "...Therefore, it is not possible to have an example in which every point is a (C)-point." $\endgroup$ Commented Mar 10, 2012 at 0:35
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$\begingroup$ that's just stated for d=1 though $\endgroup$ Commented Mar 10, 2012 at 0:41