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It is quite easy to see that there are plenty of functions $f$ for which

$$ \frac{d^n}{dx^n} f(x) \geq 0 \;\;\;\forall n \in \mathbb{N}, x \in \mathbb{R}_+$$

For starters, it holds for any polynomial with positive or zero coefficients. The stronger condition, with a strict inequality

$$(1) \;\;\;\frac{d^n}{dx^n} f(x) > 0 \;\;\;\forall n \in \mathbb{N}, x \in \mathbb{R}_+$$

holds in less cases. However, sums of exponentials of the form

$$ \;\;\; f(x) = \sum_i A_i e^{B_i x} \;\;\; A_i, B_i > 0$$

satisfy this stronger condition. But there are others

My questions are:

1) Does (1) imply $O(f(x)) \geq O(\exp(x))$,

(in the sense $O(a(x)) >,=,< O(b(x))$ if $\lim_{x\rightarrow\infty} \frac{a(x)}{b(x)} = \infty,\text{constant},0$, respectively)

2) Is there a simple test for (1), and/or, does this condition have an established name?

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Closely related:… – mrf Nov 29 '12 at 23:23
It most certainly does not hold for any polynomial with positive or zero coefficients. Consider $f(x)=x^2$ for a simple counterexample. – Cameron Buie Nov 29 '12 at 23:35
Ah, oops, I meant positive reals – Lucas Nov 29 '12 at 23:40
up vote 1 down vote accepted

Consider the Taylor series expansion of $f(x)$: $$ f(x)=\sum_{i=0}^{\infty}a_ix^{i}. $$ If all coefficients $a_i$ are positive and the series converges for all $x \ge 0$, then $f$ is an entire function and is also completely monotonic (in that all derivatives are positive on the positive real line). This is still true under the slightly weaker condition that all coefficients are non-negative, and infinitely many are positive. The set of functions of this type is clearly closed under multiplication and addition; it's also closed under multiplication by polynomials with non-negative coefficients, and as pointed out in the related MathOverflow question, it's closed under convolution. So functions like $\exp(\exp(x))$ and $\exp(x^2 \exp(x))$ are included. Note that the functions I'm describing form a subset of the ones you're asking about... there may well be completely monotonic functions that aren't also entire.

Your first question asks whether any completely monotonic function grows more slowly than exponentially. The answer is certainly yes, and we can achieve this by letting the $a_i$ decay more quickly than they do in any exponential (where $a_i^{-1} \in O(i!)$). For instance, the function $$ \sum_{i=0}^{\infty}\frac{x^{i}}{(i!)!} $$ grows much more slowly than $e^{kx}$ for any positive $k$.

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Closure under those things seems like an excellent way of describing the set. I have a follow on question, what then is the lowest order possible for f? – Lucas Nov 30 '12 at 17:40

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