The question is

Let f be a continuous Lebesgue integrable function on $[0,+∞)$, show $\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = 0$.

My attempt:

Suppose $\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) \neq 0$, then $\exists \epsilon>0 \forall x_0\in[1,+\infty)\exists x>x_0$ st. $|f(x)|>\epsilon$. Thus we can construct a sequence $x_0 < x_1 < x_2 < ...$ st. $f(x_k)>\epsilon$ for all $k=1,2,3...$.

Also by continuity, $\exists \delta_1,\delta_2,\delta_3,...$ st. that $|f(x)|>\frac{\epsilon}{2}$ for each $(x_k-\delta_k,x_k+\delta_k)$.

Thus $\int_1^{ + \infty } f > \sum\limits_k {\varepsilon {\delta _k}} = \varepsilon \sum\limits_k {{\delta _k}} $. However, it is possible for $\sum\limits_k {{\delta _k}} $ to converge, so I am not able to conclude $\int_1^{ + \infty } f = + \infty $.

Thank you!

  • 8
    $\begingroup$ There are continuous positive integrable functions such that $\limsup\limits_{x\to\infty} f(x) = +\infty$. You need further assumptions to deduce $\lim\limits_{x\to\infty} f(x) = 0$. $\endgroup$ – Daniel Fischer Aug 14 '15 at 21:05
  • $\begingroup$ Do you mean that $f$ is Lebesgue integrable? (Not Lebesgue measurable) $\endgroup$ – Michael Burr Aug 14 '15 at 21:08
  • $\begingroup$ Yes. I am sorry. That's my mistake. Thank you! $\endgroup$ – Tony Aug 14 '15 at 21:09
  • 2
    $\begingroup$ Something like uniformly continuous or bounded variation would be helpful in this problem, otherwise, you can use your construction to create a counterexample. $\endgroup$ – Michael Burr Aug 14 '15 at 21:16
  • $\begingroup$ Being integrable does not help much $\endgroup$ – A.Γ. Aug 14 '15 at 21:20

This doesn't really help in the understanding but I love finding explicit examples. The function

$$f(x) = x(\cos^2 x)^{x^5}$$

is Lebesgue integrable on $[0,\infty)$ and $f(2n\pi) =2n\pi, n = 1,2,\dots $

| cite | improve this answer | |
  • 2
    $\begingroup$ Spikes of given bases and heights are just as explicit, if you ask me... $\endgroup$ – Did Aug 15 '15 at 10:28
  • 1
    $\begingroup$ By "explicit" I meant "easy to write down as a formula in terms of elementary functions". $\endgroup$ – zhw. Aug 15 '15 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.