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I want to prove the following.

Let $I:=[a,b)$ be finite interval, $f$ is completely monotone on $I$. Then it can be extended analytically into the complex z-plane ($z=x+iy$), and the function $f(z)$ will be analytic in the circle $|z-a|<b-a$.

Here is what I started. By Taylor's formula for every $a\leq x<b$ we have \begin{equation*} f(x)=\sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x-a)^k+ (-1)^{n+1}\int_{a}^x \frac{(-1)^{n+1}f^{(n+1)}(t)}{n!}(x-t)^{n}dt. \end{equation*} Here \begin{equation*} R_n(x):=\int_{a}^x \frac{(-1)^{n+1}f^{(n+1)}(t)}{n!}(x-t)^{n}dt. \end{equation*} Since $(-1)^{n+1}f^{(n+1)}(t)\geq 0$ and $(-1)^{n+1}f^{(n+1)}$ is monotone decreasing we obtain \begin{equation*} 0\leq R_n(x)\leq \frac{(-1)^{n+1}f^{(n+1)}(a)}{(n+1)!}. \end{equation*}

I don't know how to finish the proof, assuming that it can be finished at all.

Ѕᴀᴀᴅ
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vesszabo
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    Looks like a duplicate of https://math.stackexchange.com/q/1193121/42969 to me. Note that $x \mapsto f(-x)$ has all derivatives nonnegative. – Martin R Jul 17 '24 at 17:13
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    This is called “little Bernshtein theorem” in https://en.wikipedia.org/wiki/Absolutely_and_completely_monotonic_functions_and_sequences – Martin R Jul 17 '24 at 17:23
  • @MartinR Before posting the question I read a lot of paper, books, and everywhere was used $g(x):=f(-x)$, but my idea was to prove the statement without this. But no problem, this trick makes the proof simpler. Thanks for the reference, I couldn't find it because there is no completely-monotone keywords, I tried to use, but SAAD deleted it. That post is nice, but not gives the desired radius of convergence; hopefully I can do with a very little modification of some published proofs. Thanks for your help. – vesszabo Jul 18 '24 at 06:37

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