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Assume that $f:(a,b) \rightarrow \mathbf{R}$, $x_0 \in (a,b)$ and $f(x)=\sum_{n=0}^\infty a_n (x-x_0)^n$ for $x \in (a,b)$ and radius of convergence $R$ of this power series is infinite. (Then $f$ is smooth, $a_n=\frac{f^{(n)}(x_0)}{n!}$ for $n=0, 1, 2, \ldots$ and $\frac{1}{R}=\lim_{n \rightarrow \infty} \sqrt[n] { \frac{ |f^{(n)} (x_0)|} {n!} }=0$). Is it true that

$ \lim_{n \rightarrow \infty} \sqrt[n] { \frac{\sup_{x \in K} |f^{(n)} (x)|} {n!} }=0$ for every compact $K \subset (a,b)$ ?

Thanks.

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up vote 2 down vote accepted

It is true for any compact $K\subset \mathbb C\;$. Let $C_R=\{|z|=R\}\;$ be a circle and $M_R=\sup_{C_R}|f(z)|\ $. Take $R$ large enough, so $K$ is inside $C_R$ and $\mathrm{dist}(C_R,K)\ge R/2\ $. Differentiating the Cauchy integral formula we have $$ f^{(n)}(x)=\frac{n!}{2\pi i}\int_{C_R}\frac{f(z)}{(z-x)^{n+1}}\,dz. $$ It leads to the estimate $$\sup_{x \in K} |f^{(n)} (x)|\le \frac{2\;n!M_R}{ (R/2)^{n}},$$ so $$\lim_{n \rightarrow \infty} \sqrt[n] { \frac{\sup_{x \in K} |f^{(n)} (x)|} {n!} }\;\le 2/R.$$ Since $R$ can be taken arbitrary large the limit is zero.

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