I encountered the following problem: Let $f:\mathbb{R}_+\to \mathbb{R}_+$ be a continuous increasing function. Suppose there is $r_0>0$ (however large you want it to be) such that
- 1 ) $Cr^{\alpha}\ge f(r) \ge \frac{1}{C}r^\alpha$ for some $C>0$ and $\alpha>0$ for $r \in [0,r_0]$.
- 2 ) There are $\lambda_0,C>0$ such that $C\exp(\lambda_0 r)\ge f(r) \ge \frac{1}{C}r^\alpha$ for $r\ge r_0$ with same $\alpha$ as in (1).
For a fixed $\epsilon$, we define $F(r)$ to be \begin{align} F(r) = \int_0^r\exp(\epsilon s)df(s) \int_0^r\exp(-\epsilon s)df(s). \end{align} I am interested in the divergence rate of $F(r)$, do we have something like \begin{align} \lim_{r \to \infty} r^{-n}F(r) = \infty,\quad \forall n \ge 1. \end{align} Where the integration is in Lebesgue-Stieltjes sense since $f$ is of bounded variation on any finite interval.
I first did a naive Cauchy-Schwarz so $\frac{1}{C'} r^\frac{\alpha}{2}\ge F(r) \ge \frac{1}{C'} r^\frac{\alpha}{2}$. Use the integration by parts formula we can get a more workable expression for $F(r)$, namely: \begin{align} F(r) = \left(e^{\epsilon r}f(r) - \epsilon\int_0^re^{\epsilon s}f(s)ds\right)\left(e^{-\epsilon r}f(r) + \epsilon\int_0^re^{-\epsilon s}f(s)ds\right). \end{align} So now we can only worry about the first term. When $f(r) = a\exp(br)$ then a simple calculation shows $F(r)$ goes to infinity exponentially fast. I am not familiar with Gamma functions, but from Wolfram Alpha, when $f(r) = r^n$ for integer $n$'s, it also diverges exponentially fast.
We can also use L'Hospital's rule, but that would involve the quantity $f'(r)$ when $r \to \infty$, for which I don't seem to have control of.
Any help will be predicated, thanks in advance.