Quasi-analytic classes (Denjoy-Carleman)

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Let f∈C∞([a,b]),f≢0f \in C^{\infty }(\left[a, b\right]), f \not\equiv 0. Prove that if the function ff vanishes on some nonempty interval, then

sup⁡n,x∣f(n)(x)∣=+∞ \sup _{n, x}\left|f^{(n)}(x)\right|=+\infty

Does there exist a point xx such that sup⁡n∣f(n)(x)∣=+∞\sup_{n}\left|f^{(n)}(x)\right|=+\infty?

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Prove that C{n!}=C{nn}C\left\{ n!\right\} = C\left\{ n^{n}\right\}.

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Prove that there are quasi-analytic classes which are larger than C{n!}C\left\{ n!\right\}.

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Put λn=Mn−1/Mn\lambda_{n} = M_{n-1} / M_{n}, as in the proof of Theorem 19.11. Pick g0∈Cc(R1)g_{0} \in C_{c}\left(\mathbb {R}^{1}\right), and define

gn(x)=(2λn)−1∫−λnλngn−1(x−t)dt(n=1,2,3,…) g_{n}(x) = \left(2 \lambda _{n}\right)^{-1} \int _{-\lambda _{n}}^{\lambda _{n}} g_{n-1}(x-t) d t \quad (n = 1,2,3, \ldots )

Prove directly (without using Fourier transforms or holomorphic functions) that g=lim⁡gng = \lim g_{n} is a function which demonstrates that (e) implies (a) in Theorem 19.11. (You may choose any g0g_{0} that is convenient).

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Suppose C{Mn}C\left\{ M_{n}\right\} is quasi-analytic, f∈C{Mn}f \in C\left\{ M_{n}\right\}, and f(x)=0f(x) = 0 for infinitely many x∈[0,1]x \in [0,1]. What follows?

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References

[МГЛ+04] Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. Избранные задачи по вещественному анализу. 2-е изд., Невский Диалект; БХВ-Петербург, 2004.
[Rud87] Rudin W. Real and Complex Analysis. 3rd ed., McGraw-Hill, 1987.