Non-analytic function with convergent Taylor series everywhere - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:35:18Z http://mathoverflow.net/feeds/question/81613 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81613/non-analytic-function-with-convergent-taylor-series-everywhere Non-analytic function with convergent Taylor series everywhere Bruce Blackadar 2011-11-22T14:49:44Z 2012-12-30T15:06:14Z <p>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.</p> http://mathoverflow.net/questions/81613/non-analytic-function-with-convergent-taylor-series-everywhere/90773#90773 Answer by Bazin for Non-analytic function with convergent Taylor series everywhere Bazin 2012-03-09T22:30:40Z 2012-03-09T22:30:40Z <p>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 $[s_1,s_0]\subset(1,+\infty)$.</p> <p>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.</p> <p>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.</p> <p>Bazin.</p>