# Interesting stipulation about completely monotone functions

This question relates to a question I asked here. I thought of a well thought out generalization which appears to follow in the situations I've encountered it. I tried to generalize the answer provided in the first question, but I think it requires a type of finesse that I haven't thought of yet.

Recall that if $f(x)$ is completely monotone then

$$(-1)^n\frac{d^n}{dx^n} f(x) > 0$$

Let $0 < q<1$; if $f(x)$ is completely monotone and satisfies the following equations

$$f(1) = q$$ $$qf(x) = f(x+1)$$

then $f(x) = q^x$. This was the base of the question I asked before. The result can be restated: if $\theta(x)$ is a $1$ periodic function, and $q^{x+\theta(x)}$ is completely monotone, then $\theta$ is a constant function.

Under this rephrasing we can pose a similar question in another manner. What if $f$ is a series of exponentials?

Take the sequence $\{a_n\}_{n=1}^\infty$ where $a_n \in \mathbb{R}$, let $f$ be defined by the series $$f(x) = \sum_{n=1}^\infty a_nq^{nx}$$

Assume $f$ is completely monotone.

Let $\theta$ be $1$ periodic, if $f(x+\theta(x))$ is completely monotone must $\theta$ be constant?

An important note is that necessarily these functions are analytic because they are completely monotone by a theorem of Bernstein. I'm also assuming $f:\mathbb{R}^+ \to \mathbb{R}^+$.

I'm half expecting a negative of this result. I think there must be some kind of exception. Nonetheless it would help identify why the functions I'm dealing with seem to satisfy this.

• @GeraldEdgar I was being short, I'll add that in. I was taking $\theta$ to be defined as it was earlier. – user78249 Feb 17 '17 at 18:17