Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(CMCS 2010 Longlist) Let $f: [0, + \infty) \rightarrow (0, +\infty)$ be a continuous function such that$\lim\limits_{x \rightarrow \infty} \int_{0}^{x}f(t)\,dt=l\in \mathbb{R}$. Prove that $$\lim\limits_{x \rightarrow \infty}\frac{1}{\sqrt {x}}\int_{0}^{x}\sqrt {f (t)}\,dt = 0.$$

share|improve this question

1 Answer 1

up vote 2 down vote accepted

By-parts integration ($u=t,~dv=fdt$) and then L'Hôpital's, in that order, gives

$$\triangle=\frac{1}{x}\int_0^x tf(t)dt=\int_0^xf(t)dt-\frac{1}{x}\int_0^x \left[\int_0^tf(\tau)d\tau\right]dt \xrightarrow{x\to\infty} l-l=0.$$

Equating the above with the result of direct application of L'Hôpital's to $\triangle$ gives

$$\lim_{t\to\infty} tf(t)=0.$$

Now apply L'Hôpital's to the original problem, noting $\sqrt{\cdot}$ is continuous on $[0,\infty)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.