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Consider the class of functions from $\mathbb R$ to $\mathbb R$, such that the function is positive everywhere and its $n$th derivative is positive everywhere for all $n$.

The only examples I can construct are the functions $ae^{bx}+c$ for $a,b,c>0$.

Are these functions the only examples?

If not, for which nonlinear functions $g$ does $e^{g(x)}$ have this property?

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    $\begingroup$ what happens for exp(exp(x))? $\endgroup$ – Yemon Choi Jan 26 '12 at 17:51
  • $\begingroup$ Very interesting question. Just out of curiosity: what is the signicance of function itself being positive?(hopefully people still care about this question after 4 years) $\endgroup$ – BigM Oct 14 '16 at 5:07
  • $\begingroup$ @BigM If you remove the condition that the function is positive, then its derivative is a function satisfying my stated condition, so you get only a slight variation of Bernstein's theorem. $\endgroup$ – Will Sawin Oct 14 '16 at 7:55
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See completely monotonic in the literature. Function $f(x)$ is completely monotonic if and only if $f(-x)$ is the sort of function you're looking for.

S.N. Bernstein (1928). "Sur les fonctions absolument monotones". Acta Mathematica 52: 1–66. doi:10.1007/BF02592679.

http://mathworld.wolfram.com/CompletelyMonotonicFunction.html

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    $\begingroup$ You have to be careful, though, because in the literature completely monotonic functions are assumed to be defined only on the half-line, not on the whole line as in the question. $\endgroup$ – Mark Meckes Jan 26 '12 at 18:49
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    $\begingroup$ Yet we can reduce to the notion of completely monotonic functions on the positive half-line because $f:\mathbb{R}\to\mathbb{R}$ has all derivatives non-negative iff $f(c-x):[0,\infty)\to\mathbb{R}$ is completely monotonic for all $c\in\mathbb{R}$; this e.g. allows to exploit Bernstein's representation in the OP's case. $\endgroup$ – Pietro Majer Jan 26 '12 at 19:28
  • $\begingroup$ So we would say that for all $c$ there is a finite measure $\mu_c$ such that for $t\leq c$, $f(t)=\int_0^\infty e^{(t-c)x} d\mu$? We would then make sure that this implies that there is a measure $\mu$ such that for all $t$, $f(t)=\int_0^\infty e^{tx} d\mu$. $\endgroup$ – Will Sawin Jan 26 '12 at 19:45
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Well, there are certainly more. If you look at the chain rule then you see that the $n$-th derivative is a linear combinations of products of derivatives of the two functions you compose with positive coefficients. Thus if you have two functions with your property, then their composition will again have only positive derivatives. So you can go on...

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If $f(x)$ is a function with positive derivatives, then the function $f(-x)$ is completely monotonic. The completely monotonic functions are classified by Bochner's theorem; see Nimza's question On the generalisation of Bernstein's theorem on monotone functions.

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