# Given a continuous function which it's improper integral converges, prove/disprove $\lim_{x\to \infty}f(x)=0$

Prove/Disprove:

$(1)$ 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.

• (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? – Lord Shark the Unknown Aug 2 '17 at 5:51

(2) $\int_0^\infty f(x)\; dx \ge x f(x)$.
• 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 $1/2?$ – Itay4 Aug 2 '17 at 5:59
• 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 $\frac{1}{(1+x)^2}$. – Michael Lee Aug 2 '17 at 6:23
• 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 $2^{1-2n}$). – Michael Lee Aug 2 '17 at 6:36
• 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 $$\int_0^{\infty} f(x)\,\mathrm{d}x\geq \lim_{a\to \infty} ca = \infty$$ – Michael Lee Aug 2 '17 at 6:43
• 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. – Michael Lee Aug 2 '17 at 7:10