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Apr 22, 2019 at 4:06 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Dec 23, 2018 at 3:58 comment added BigM @CPJ thanks for the reference to the original work of Gervey.
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Apr 26, 2018 at 21:24 answer added arriopolis timeline score: 2
Oct 14, 2015 at 20:21 history edited Bazin CC BY-SA 3.0
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Oct 14, 2015 at 14:59 history edited Bazin CC BY-SA 3.0
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Oct 14, 2015 at 14:32 comment added Bazin @CPJ Thanks for your comment. In fact with $M_n=n^{ns}$, we get $(M_n/n^n)^{1/n}=n^{s-1}$ which is increasing for $s\ge 1$. This suggests that the composition algebra property holds for $s\ge 1$ but not for $s<1$.
Oct 13, 2015 at 7:23 comment added Bazin @Piero D'Ancona You mean $s'<s$ since the Gevrey space $G^{(s)}$ with the notation above is increasing with $s$, e.g. analytic is $G^{(1)}$ is included in $G^{(2)}$ which contains compactly supported functions. On the other hand, I believe that the answer to the question is positive and is a matter of writing a precise Faà de Bruno formula.
Oct 11, 2015 at 20:31 comment added Piero D'Ancona If I remember correctly, the answer in general is no. However if you compose a Gevrey function of order s'>s with a Gevrey function of order s you get a Gevrey function of order s.
Oct 11, 2015 at 18:04 history edited Bazin CC BY-SA 3.0
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Oct 9, 2015 at 16:14 comment added CPJ There is a result (Cartan?) on Carleman-Denjoy classes saying that they are stable under composition if the sequence $(M_n/n!)^{1/n}$ is increasing. Letting M_n = (n!)^{s}, this should be true. See this paper. Also Gevrey himself seems to have proved that.
Oct 9, 2015 at 15:34 history edited YCor
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Oct 9, 2015 at 14:06 history edited Bazin CC BY-SA 3.0
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Oct 9, 2015 at 13:59 history asked Bazin CC BY-SA 3.0