5
$\begingroup$

Possible Duplicate:
$f$ uniformly continuous and $\int_a^\infty f(x)\,dx$ converges imply $\lim_{x \to \infty} f(x) = 0$

This is an exercise from Berkeley preliminary exams, Fall 1983

Let $ f : [0, \infty ) \rightarrow \mathbb{R} \ $ be a uniformly continuous function with the property that

$ \lim_{b \to \infty}\int_{0}^{b} f(x)dx \ $

exists (as a finite limit). Show that

$\lim_{x \to \infty}f(x) = 0 $

Obviously if the limit exists, it must be $0 \ $; so the problem is to prove that the limit exists. Any hint ?

  • 0
    @BeniBogosel: Yes I got it, see comments in martini's answer. Thank you for this comment too.2012-03-25

1 Answers 1

9

Suppose that $\lim_{x\to \infty} f(x)$ doesn't exist. Then there is an $\epsilon > 0$ and a sequence $x_n \to \infty$ such that $|f(x_n)| > \epsilon$ for all $n$ (because the limit, if existing, has to be 0). By uniform continuity there is a $\delta > 0$ such that $|f(x) - f(y)| < \frac{\epsilon}2$ if $|x-y| < \delta$. It follows that $|f(x)| > \frac{\epsilon}2$ if $|x-x_n| < \delta$ for some $n$. But now $|\int_{x_n-\delta}^{x_n+\delta} f(x)\, dx| > \epsilon\delta$ for all $n$ contradicting $\int_a^b f(x)\,dx \to 0$ for $a,b \to \infty$.

  • 0
    @martini Oh my, of course! No further questions : )2012-05-03