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Let $f:[0,\infty)\to\mathbb{R}$ be integrable in everywhere.

  1. Suppose $\int\limits_0^{\infty}|f(t)|dt$ converges. Show that there exists a sequence $x_n$ such that $x_n\to\infty$ while $f(x_n)\to 0$.

  2. Show that if we require $f$ to be continuous, then the above holds even if $\int\limits_0^{\infty}f(t)dt$ converges (not necessarily absolutely).

  3. Is the statement in (1) true if we drop the requirement for continuity?

For (1), suppose every sequence such that $x_n\to\infty$, $f(x_n)$ does not tend to $0$. Then necessarily there exists $L>0$ such that from some point $x_0$, for all $x>x_0$, $|f(x)| \geq L$. But then $$ |f(x)| > L \Rightarrow \int\limits_{x_0}^\infty |f(t)|dt \geq \int\limits_{x_0}^{\infty}Ldx = \infty $$ a contradiction for the convergence of the improper integral.

For (2), I am not really sure. The only conclusion I got is that if $f$ is continuous and there is no such sequence $x_n$, then necessarily $f$ intersects the $x$ axis only finitely many times.

Edit: For (3) it is proved here.

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Joshhh
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  • For (3) it is actually proven in your linked answer that the improper integral converges. – Dark Jun 22 '16 at 17:46
  • It is proven only for a certain sequence of the integrals, it needs to be true for every sequence. – Joshhh Jun 22 '16 at 17:51
  • "Since $\epsilon$ is arbitrary, $\displaystyle;\lim\limits_{y\to\infty}\int_0^y f(x)dx;$ exists and equal to $\displaystyle;\Delta = \sum_{k=1}^\infty F_k;$" The convergence is proven. – Dark Jun 22 '16 at 17:58
  • @Joshhh: Re-read the part after "For any $y > ...$" – RRL Jun 22 '16 at 18:04
  • For $(2)$: consider how many zeros $f$ has. If it has an infinite number of zeros then you have your sequence. If it has a finite number of zeros then there is an $x_0$ such that $f(x) \geq 0$ (or $\leq$) for all $x>x_0$ and you can use (1) to conclude. – Winther Jun 22 '16 at 18:05
  • @Dark I probably missed that part, thank you. – Joshhh Jun 22 '16 at 18:06

1 Answers1

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For (2) by the same reasoning by contradiction, there exist $L>0$ and $x_0$ such that for all $x>x_0$, $|f(x)| \geq L$. Because $f$ is continuous, either $f(x) \geq L$ for all $x>x_0$ or $f(x) \leq -L$ for all $x>x_0$.

Then the conclusion is similar than (1) (distinguish the two cases).

For (3) I think the counter-example you provide actually works.

Dark
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  • But, in the counter example it is proven only for a specific sequence of integrals. As I said, the example with $\int\limits_0^{2\pi n}\sin{x}dx$ provides the same information as in the other answer, yet it is not convergent. – Joshhh Jun 22 '16 at 17:49