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$f$ 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 ?

  • $\begingroup$ Are you sure that the last limit is for $x \to 0$? I guess it should be $x \to +\infty$. $\endgroup$ – Giuseppe Negro Mar 25 '12 at 10:18
  • $\begingroup$ This is Barbalat's Lemma. A solution can be found here: mathproblems123.wordpress.com/2009/10/01/barbalats-lemma And the last limit should be with $x \to \infty$, because changing the function near $0$ such that it still remains uniformly continuous does not affect the converge of the integral limit. $\endgroup$ – Beni Bogosel Mar 25 '12 at 10:24
  • $\begingroup$ @BeniBogosel: You required $f$ to be positive ($f \colon [0, \infty) \to [0, \infty)$) but this hypothesis is absent here. While your argument pushes through if $f \in L^1([0, \infty))$, I'm afraid it doesn't work if $f$ is allowed to change sign and $\int \lvert f(x)\rvert\, dx=+\infty$. $\endgroup$ – Giuseppe Negro Mar 25 '12 at 10:35
  • $\begingroup$ @Giuseppe Negro: I have edited $\endgroup$ – WLOG Mar 25 '12 at 11:10
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    $\begingroup$ @GiuseppeNegro: I know that the hypothesis is $f$ positive in my proof, but the proof does not make use of that hypothesis anywhere but in the place I say that $f(x_n) \to \ell > 0$, which can be assumed WLOG here (the idea is that we assume that there exists a sequence $x_n \to \infty$ such that $f(x_n)$ does not converge to zero). The proof works just fine if you assume that $f(x_n) \to \ell<0$. $\endgroup$ – Beni Bogosel Mar 25 '12 at 12:04

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 $a,b \to \infty$.

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  • $\begingroup$ I have the same objection I had in comments to the main post: this argument certainly works if $f$ is non-negative, or if it is $L^1$, but I'm afraid it doesn't if $f$ is allowed to change sign and $\int \lvert f \rvert\, dx=+\infty$. Do you agree? $\endgroup$ – Giuseppe Negro Mar 25 '12 at 10:38
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    $\begingroup$ @GiuseppeNegro Why? If $\int_0^\infty f(x) \,dx$ exists as in inproper integral, we need to have $\int_a^b f(x) \, dx \to 0$ for $a, b\to \infty$, which yields that there is an $M$ such that $|\int_a^b f(x)\, dx| < \epsilon\delta$ for all $a,b \ge M$. But now I can choose $x_n$ with $x_n - \delta > M$. I don't see where I could have used $f \in L^1$. $\endgroup$ – martini Mar 25 '12 at 10:44
  • $\begingroup$ Oh yes, yes, you're right. The fact that $\int_a^bf(x)\, dx \to 0$ is certainly true and you argument contradicts it. I had mistakenly seen a contradiction with $\int\lvert f(x)\rvert \, dx <+\infty$. Thank you for clarifying! $\endgroup$ – Giuseppe Negro Mar 25 '12 at 15:09
  • $\begingroup$ @martini Did you mean "Suppose that $$\lim_{x \to \infty} f(x) \neq 0$$" in your first sentence? $\endgroup$ – Rudy the Reindeer May 2 '12 at 20:56
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    $\begingroup$ @MattN. As I wrote in the paratheses: If the limit exists, it has to be 0, for if it is $c > 0$ (wlog) say, then we would have $f(x) > \frac c2$ for $x \ge N$ ($N$ choosen properly) and $\int_0^\infty f(x)\, dx$ would diverge. $\endgroup$ – martini May 2 '12 at 20:58

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