Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am working on this problem from a past examination:

Let $f:[0,\infty)\rightarrow\mathbb R$ be a continuous, non-negative and non-increasing function such that the improper integral $\int_0^\infty (f(x)/\sqrt{x})\ dx$ converges. Show that $\lim_{x\rightarrow\infty}f(x)\sqrt x=0$. Also, prove that $0<\forall \epsilon<1$, $$\lim_{x\rightarrow\infty}\int_{\epsilon x}^x\frac{f(y)}{\sqrt{x-y}}dy = 0.$$

I tried to bound the order of growth of $f$ to show the first assertion, but in vain.

I would be grateful if you could provide a clue (not necessarily a complete proof).

share|cite|improve this question
up vote 2 down vote accepted

Here are some ideas that might help:

If $f(b)\sqrt{b}\ge\varepsilon$, then $f(x)\ge\varepsilon/\sqrt{b}$ for all $x\le b$.

The assumption on the integral implies that $$\lim_{b\to\infty}\int_{b/2}^b\frac{f(x)}{\sqrt{x}}\,dx=0.$$

share|cite|improve this answer
Thank you. Could you please elaborate a little on the second assertion? – Pteromys Aug 7 '13 at 14:23
You mean the second assertion of my answer? The point is that when $g$ is a nonnegative function and $\int_0^\infty g(x)\,dx<\infty$ then $\lim_{a\to\infty}\int_a^\infty g(x)\,dx=0$ (because $\int_a^\infty=\int_0^\infty-\int_0^a$). This elementary fact is often useful and easily overlooked. – Harald Hanche-Olsen Aug 7 '13 at 15:05
I mean the second proposition to prove in the original problem statement. I am sorry for the confusing question. But thank you any way. – Pteromys Aug 8 '13 at 3:20
Oh. Yeah, I was wondering. Just using $f(y)\le f(x)$ in the integral seems to work. Now the constant factor $f(x)$ can be taken out of the integral, and the rest can be evaluated explicitly. In the end, it reduces to the first assertion. – Harald Hanche-Olsen Aug 8 '13 at 9:26

I would try to absorb $\sqrt{x}$ in the integral, into the differential $dx$, by using the relation $d(x^{3/2})=\frac32\sqrt{x}dx$, and then use the fact that the function remains monotone after the change of variable.

share|cite|improve this answer

Based on the accepted answer, I provide a complete solution to the problem.

The first proposition follows from the fact that $$ 0 \le \frac{f(x)\sqrt{x}}{2} = \frac{xf(x)}{2\sqrt{x}}\le\int_{x/2}^x\frac{f(\xi)}{\sqrt \xi}d\xi\longrightarrow 0\ (x \longrightarrow \infty). $$ The inequality on the right follows from the monotonicity of $f$ and thus $\frac{f(\xi)}{\sqrt \xi}$.

Next, $$ 0 \le \int_{\epsilon x}^x\frac{f(y)}{\sqrt{x-y}}dy \le f(x)\int\frac{dy}{x-y} = 2f(x)\left[\sqrt{x - y}\right]^{y=\epsilon x}_{y=x} = 2f(x)\sqrt{(1-\epsilon) x} \le 2f((1-\epsilon) x)\sqrt{(1-\epsilon) x} \rightarrow 0\ (x\rightarrow \infty). $$

share|cite|improve this answer

Your Answer


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.