$(1)$ If $f$ is a continuous function such that $f > 0, \forall x \in [0,\infty)$ and $\int_0^\infty f $ converges, then $\lim_{x\to \infty}f(x)=0.$

$(2)$ If $f$ is a continuous and monotonic decreasing function such that $f \geq 0, \forall x \in [0,\infty)$ and $\int_0^\infty f $ converges, then $\lim_{x\to \infty}f(x)=0.$

I think $(1)$ is false, but I couldn't find a counter example.

I think $(2)$ is true, tried to prove with Cauchy's criterion for improper integrals and the definition of the limit, but got stuck.

Any help is appreciated.

  • $\begingroup$ (1) give the function a series of humps, of the same height but reducing in width (2) such a function will converge to a limit; what if that limit is nonzero? $\endgroup$ – Lord Shark the Unknown Aug 2 '17 at 5:51

Hint for (1): tall thin triangles.

(2) $\int_0^\infty f(x)\; dx \ge x f(x)$.

  • $\begingroup$ I thought about the triangles function, as suggested here, and I've used it to prove a previous exercise, where $f$ was non-negative. How can I justify the integral converges, if the triangles' bases are at, say $1/2?$ $\endgroup$ – Itay4 Aug 2 '17 at 5:59
  • 1
    $\begingroup$ The integral doesn't converge if the bases of the triangles are elevated above $y = 0$. However, you can simply take that triangles example with bases at $y = 0$ and add any positive function that's integrable on $[0, \infty)$, such as $\frac{1}{(1+x)^2}$. $\endgroup$ – Michael Lee Aug 2 '17 at 6:23
  • 1
    $\begingroup$ That's correct. You can even make the triangle function $f$ such that $g(x) = f(x)+\frac{1}{(1+x)^2}$ is unbounded as $x\to \infty$ (such as by giving the triangles a height of $2^n$ and a base width of $2^{1-2n}$). $\endgroup$ – Michael Lee Aug 2 '17 at 6:36
  • 1
    $\begingroup$ I would tweak it a bit, probably. Consider that a monotonically decreasing function $f$ such that $f > 0$ has a limit $c = \lim_{x\to \infty} f(x)\geq 0$. If $c > 0$, then $f\geq c$, so $$\int_0^a f(x)\,\mathrm{d}x\geq \int_0^a c\,\mathrm{d}x = ca$$ for all $a > 0$. This implies that $$\int_0^{\infty} f(x)\,\mathrm{d}x\geq \lim_{a\to \infty} ca = \infty$$ $\endgroup$ – Michael Lee Aug 2 '17 at 6:43
  • 1
    $\begingroup$ Sure. Consider that $f(x)\geq f(a)$ for all $x\in [0, a]$. Therefore, $$\int_0^a f(x)\,\mathrm{d}x\geq \int_0^a f(a)\,\mathrm{d}x = af(a)$$ Therefore, $$\limsup_{a\to \infty} af(a)\leq \int_0^{\infty} f(x)\,\mathrm{d}x$$ This implies that $\lim_{a\to \infty} f(a) = 0$; otherwise, there is some $\epsilon > 0$ such that for any $a$, there is an $x > a$ such that $f(x) > \epsilon$. Therefore, for any $a$, there is an $x > a$ such that $xf(x) > \epsilon x$, which implies that $$\limsup_{a\to \infty} af(a) = \infty$$ This would contradict the fact that $\limsup_{a\to \infty} af(a)$ is finite. $\endgroup$ – Michael Lee Aug 2 '17 at 7:10

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.