This may be a trivial question but I can't see it immediately.
Suppose $\{a_k\}$ is an increasing sequence of positive reals. Does there exist a smooth function $f \in C^{\infty}([0,1])$ such that $\inf \partial^k f \ge a_k$ holds for all $k$?
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4$\begingroup$ Yes, see Borel's lemma. $\endgroup$– abxCommented Apr 19, 2018 at 11:16
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$\begingroup$ the only difference is that here I want the bounds to hold over all the interval where as in Borel's lemma it holds only at one point of the interval. perhaps there is a simple way to extend the result? $\endgroup$– AliCommented Apr 19, 2018 at 11:28
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7$\begingroup$ Over all interval obviously not always. Just estimate $f(1)$ using Taylor formula at $0$ of high order. $\endgroup$– fedjaCommented Apr 19, 2018 at 11:35
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$\begingroup$ Your comment is not the same as what you asked in the question. Your question asks for the $\sup$ to be lower bounded, for that you just need one point. Your comment asks for every point. Did you want $\inf$? $\endgroup$– Willie WongCommented Apr 19, 2018 at 13:43
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1$\begingroup$ @WillieWong $f(1)=\sum_{k=0}^n \frac{f^{(k)}(0)}{k!}+\frac{f^{(n+1)}(\xi)}{(n+1)!}\ge\sum_{k=0}^n \frac{a_k}{k!}$ for every $n$, so if the series $\sum_k\frac{a_k}{k!}$ diverges, you are in trouble. $\endgroup$– fedjaCommented Apr 19, 2018 at 14:45
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