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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 $\lim_{x→+∞} f(x) = 0.$

Joe
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kingkongdonutguy
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2 Answers2

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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 $x\to \infty$.

Joel
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    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 $\int_x^\infty f = \int_x^{x + \delta/2} f + \int_{x + \delta/2}^\infty f \implies \int_x^{x + \delta/2} f = \int_x^\infty f - \int_{x + \delta/2}^\infty f = F(x) - F(x + \delta/2)$? – DanZimm Jun 27 '14 at 21:21
  • Right of course. I will make the change. Thanks for catching it. – Joel Jun 28 '14 at 03:35
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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 $f$.

InTransit
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  • 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? – Joel Jun 09 '14 at 18:26
  • For instance, what if $f(x)$ was always negative? It could not possibly hold in that case. – Joel Jun 09 '14 at 18:27
  • 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 $(0,\infty)$. – Joel Jun 09 '14 at 18:58
  • 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! – InTransit Jun 11 '14 at 14:54
  • 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. – Joel Jun 11 '14 at 14:57
  • I think there's an ambiguity as to whether the Lebesgue or the Riemann regime of integration is meant. The OP didn't mention which regime he's operating in. $\sin x /x$ isn't $L^1(0,\infty)$, i.e. isn't Lebesgue integrable but exists as an improper Riemann integral. However, assuming the Riemann integral regime where it's usually called $absolute \ integrability$, then yes, one does require eventually constant sign and the proof is easily adapted by essentially first considering $\int_0^R$, $R<\infty$, doing what I did and then passing to the limit. I will post the details later today. – InTransit Jun 12 '14 at 08:45