Skip to main content
8 events
when toggle format what by license comment
Dec 9, 2016 at 4:02 history edited Pedro Lauridsen Ribeiro CC BY-SA 3.0
Small grammar fixes, added technical remark
Dec 7, 2016 at 13:04 comment added Pedro Lauridsen Ribeiro On the other hand, Borel's theorem is not about that - the idea is that its proof consists in truncating the $j$-th term in the would-be Taylor series of $f$ at a point $x_0\in\mathrm{int}\Omega$ by cutoff functions supported in $\Omega$ which equal 1 in a neighborhood of $x_0$ and whose supports shrink very fast with $j$ (depending on how fast the $a_j$'s grow with $j$). Particularly, it's not guaranteed at all that the original Taylor series for f will converge (actually, for this particular $f$, it certainly won't converge to $f$ due to the non-analyticity of the cutoff functions).
Dec 7, 2016 at 13:03 comment added Pedro Lauridsen Ribeiro The argument I originally had in mind for the $L^1$ norms was not quite precise. I've just replaced it with a better one using the Sobolev embedding theorem. As for your second question, the particular kind of blowup provided in Alexander Eremenko's answer is meant to preclude a positive radius of convergence for the Taylor series of $f$ around any point of $\Omega$ by means of the root test, hence to prevent that f is analytic anywhere in $\Omega$.
Dec 7, 2016 at 11:16 history edited Pedro Lauridsen Ribeiro CC BY-SA 3.0
Added explanation
Dec 7, 2016 at 10:50 history edited Pedro Lauridsen Ribeiro CC BY-SA 3.0
Corrected argument for $L^1$ norms
Dec 7, 2016 at 8:05 comment added Amir Sagiv Also, in Alexander Eremenko's answer there is always a subsequanse that blows up. But it seems that from Borel's theorem we can build converging sequences of derivatives. How are the two compatible?
Dec 7, 2016 at 8:02 comment added Amir Sagiv Thanks. The interchange between the $L^{\infty}$ blowups to $L^1$ norms blowups is vie Holder inequality?
Dec 7, 2016 at 1:38 history answered Pedro Lauridsen Ribeiro CC BY-SA 3.0