# Power functions and $\lim_{x \to +\infty}{\frac{\int_{0}^{x}{f(t)\mathrm{d}t}}{x f(x)}}=\frac{1}{1 + \beta}$

integrationlimitsreal-analysis

Let $$f : \mathbb{R}_+ \rightarrow \mathbb{R}_+$$ an increasing continuous function such that $$f(0) = 0$$ and $$\beta \geq 0$$. When we have for all $$x > 0$$, $$\displaystyle\frac{\int_{0}^{x}{f(t)\mathrm{d}t}}{x f(x)} = \frac{1}{1 + \beta}$$, it is straightforward that $$f$$ is a power function, i.e. there is $$\lambda > 0$$ such that for any $$x \geq 0$$, $$f(x) = \lambda x^{\beta}$$.

My question is:

Does $$\lim\limits_{x \rightarrow +\infty}{\displaystyle\frac{\int_{0}^{x}{f(t)\mathrm{d}t}}{x f(x)}} = \displaystyle\frac{1}{1 + \beta}$$ imply that there is some $$\lambda \geq 0$$ such that $$\lim\limits_{x \rightarrow +\infty}{\displaystyle\frac{f(x)}{x^{\beta}}} = \lambda$$ ?

In general, $$f$$ will be “almost” a power function. More precisely, it will be regularly varying with index $$\beta$$, i.e., of the form $$f(x)=x^\beta\cdot\ell(x)$$ where $$\ell$$ is a slowly varying function, such as $$\ell(x):=\log(1+x)$$. This is in essence the content of Karamata's theorem (see Theorems 1.5.11 and 1.6.1 of Bingham, Goldie, and Teugel's book “Regular variation”): Karamata's theorem states that a positive, locally bounded function $$f:[0,\infty)\to[0,\infty)$$ is a regularly varying function with index $$\beta\ge0$$ if and only if $$\frac{x^{\sigma+1}f(x)}{\int_0^xt^\sigma\,f(t)\,\mathrm dt}\xrightarrow[x\to\infty]{}\sigma+\beta+1\tag{C_\sigma}$$ holds for some $$\sigma>-(\beta+1)$$. In this case, $$(C_\sigma)$$ holds for all $$\sigma\ge-(\beta+1)$$.
The representation theorem (Theorem 1.3.1 in the book) says that we can write any slowly varying function $$\ell$$ in the form $$\ell(x)=c(x)\exp\left(\int_0^x\varepsilon(u)\frac{\mathrm du}u\right)$$ for some $$c$$ measurable with $$c(x)\to c\in(0,\infty)$$ and $$\varepsilon(x)\to0$$ as $$x\to\infty$$.