Show that if $\int_0^x f(y)dy \sim Ax^\alpha$ then $f(x)\sim \alpha Ax^{\alpha -1}$

Let $f$ be a real, continuous function defined on $[0,\infty)$ such that $xf(x)$ is increasing for all sufficiently large values of $x$. Show that if

$$\int_0^x f(y)\,dy \sim Ax^\alpha \quad \left(\,x\to \infty\right)$$

for some positive constants $A$ and $\alpha$, then

$$f(x)\sim \alpha Ax^{\alpha -1} \quad \left(\,x\to \infty\right).$$

Clearly, I have to use differentiation somewhere, but I don't know how to manipulate $\lim_{x\to \infty}\frac{\int_0^x f(y)\,dy}{Ax^\alpha}=1$ to get the desired result.

From suggestions given, I know that by L'hospital's rule, it's enough to show that the limit $\lim \frac{f(x)}{\alpha Ax^{\alpha -1}}$ exists, and this limit equals $ \lim \frac{xf(x)}{\alpha Ax^{\alpha}}$. From here, I'll need to use the given assumption that $xf(x)$ is eventually increasing. But how can I show the existence of the limit based on these facts?

I would greatly appreciate any solutions, hints or suggestions.

Solution 1:

Ooh, a tauberian theorem with a simple elementary proof, cool.

Assume $x$ is always so large below that $xf(x)$ is increasing. Fix $\delta>0$. It follows that $$\int_x^{(1+\delta)x}f(y)\,dy\sim A((1+\delta)^\alpha-1)x^\alpha.$$In particular, if $x$ is large enough we have $$\int_x^{(1+\delta)x}f(y)\,dy\le(1+\delta)A((1+\delta)^\alpha-1)x^\alpha.$$But $yf(y)$ increasing shows that $$\int_x^{(1+\delta)x}f(y)\,dy\ge\int_x^{(1+\delta)x}\frac{xf(x)}{y}\,dy=xf(x)\log(1+\delta),$$and combining this with the previous inequality shows that if $x$ is large enough then$$\frac{f(x)}{x^{\alpha-1}} \le\frac{(1+\delta)A((1+\delta)^\alpha-1)}{\log(1+\delta)}.$$Letting $\delta\to0$ now shows that $$\limsup\frac{f(x)}{x^{\alpha-1}}\le\alpha A.$$The inequality $\liminf f(x)/x^{\alpha-1}\ge\alpha A$ is proved similarly, starting with $$\int_{(1-\delta)x}^xf(y)\,dy\sim A(1-(1-\delta)^\alpha)x^\alpha.$$