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I think the following question is probably fairly easy but can't think of an easy way of proving it. Some help would be awesome. This question comes from an old qual. Thanks.

Let $f$ be an integrable function on $]0, +\infty[.$

(a) Assume that $f$ is uniformly continuous. Prove that $\lim_{x→+∞} f(x) = 0.$

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  • $\begingroup$ See this. $\endgroup$ Jun 9, 2014 at 16:03

2 Answers 2

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Since $f$ is integrable on $(0,\infty)$ we have that the function $F(x) = \int_{x}^\infty f(t) dt$ is well defined. Moreover, $\lim_{x\to\infty} F(x) = 0$.

Now let $\epsilon > 0$ and by uniform continuity take $\delta > 0$ to be such that $|f(x) - f(y)| < \epsilon$ when $|x-y| < \delta$.

Now consider $F(x) - F(x + \delta/2) = \int_{x}^{x+\delta/2} f(t) dt = \delta/2 f(\xi_x)$ for some $\xi_x \in [x, x+\delta/2]$ by the mean value theorem for integrals. In particular $|f(\xi_x) - f(x)| < \epsilon$ for all $x$.

Now $F(x + \delta/2) - F(x) \to 0$ as $x \to \infty$, since $F(x) \to 0$. This means $f(\xi_x) \to 0$ as $x \to \infty$. Which means that $f(x)$ is inside $B_\epsilon(0)$ for sufficiently large $x$. Since $\epsilon$ was arbitrary, this means $f(x) \to 0$ as $x\to \infty$.

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    $\begingroup$ This is probably just a small technical detail but wouldn't $F(x+\delta/2) - F(x) = \int_{x + \delta/2}^x f(t) dt$? Since $\int_x^\infty f = \int_x^{x + \delta/2} f + \int_{x + \delta/2}^\infty f \implies \int_x^{x + \delta/2} f = \int_x^\infty f - \int_{x + \delta/2}^\infty f = F(x) - F(x + \delta/2)$? $\endgroup$
    – DanZimm
    Jun 27, 2014 at 21:21
  • $\begingroup$ Right of course. I will make the change. Thanks for catching it. $\endgroup$
    – Joel
    Jun 28, 2014 at 3:35
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This solution doesn't require the mean value theorem. Suppose the conclusion is false. Then there exists $\epsilon>0$ and an infinite sequence $x_n\in(0,\infty)$ such that $x_n\rightarrow \infty$ and $|f(x_n)|>\epsilon$. We may assume that the $x_n$'s are spaced at least one unit apart. Since $f$ is continuous, $|f(x)|>0$ on some nbd $U_n$ of $x_n$. The $U_n$'s can be assumed to be disjoint. From uniform continuity, $|f(x)-f(t)|<\epsilon/2$ whenever $|x-t|<\delta$. In particular, $|f(x)-f(x_n)|<\epsilon/2$ whenever $|x-x_n|<\delta$ independently of $n$. Take $\delta$ small enough so that $I_n:= (x_n-\delta, x_n+\delta)\subset U_n$. So, $|f(x)|>\epsilon/2$ on $I_n$. Thus,

$\int_0^\infty |f(x)|\,dx \ge \sum_{n=1}^\infty\int\limits_{I_n}|f(x)|\,dx \ge \sum_{n=1}^\infty\epsilon\delta$ which contradicts the integrability of $f$.

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  • $\begingroup$ There are certainly ways around the mean value theorem. It is a valuable tool to learn to use, especially in an introductory real analysis class. @InTransit, I have doubts about the inequality at the end, could you justify that? $\endgroup$
    – Joel
    Jun 9, 2014 at 18:26
  • $\begingroup$ For instance, what if $f(x)$ was always negative? It could not possibly hold in that case. $\endgroup$
    – Joel
    Jun 9, 2014 at 18:27
  • $\begingroup$ Consider: $$f(x) = \frac{\sin(x)}{x}$$Here is an example where the integral can converge, but the integral over all of the positive parts diverges. Moreover, the function is uniformly continuous on $(0,\infty)$. $\endgroup$
    – Joel
    Jun 9, 2014 at 18:58
  • $\begingroup$ Obviously I'd missed putting in the modulus sign under the integral esp. since the second term in the line has the required absolute value sign. I've corrected the omission. And yes, no overkill in using the MVT! $\endgroup$
    – InTransit
    Jun 11, 2014 at 14:54
  • $\begingroup$ this still doesn't quite do it. What you have proved is that $f \not\in L^1(\mathbb{R}_+)$ which is not the same as integrability. Your proof would work for functions where $f$ was strictly non-negative. However, there are functions that are integrable, but not $L^1$. For instance $f(x) = \sin(x)/x$ above. Here cancelation by negative terms is essential. $\endgroup$
    – Joel
    Jun 11, 2014 at 14:57

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