If $\int_0^\infty f(x) dx=0$ , $f(x)$ continuous for $x\in[0,\infty)$, can it be proved that $\lim_{x\to\infty}f(x)=0$?

Question

As stated the condition is:

$\int_0^\infty f(x) dx=0$
$f(x)$ continuous on $x\in[0,\infty)$

What I would like to prove is $\lim_{x\to\infty}f(x)=0$

It is easy to prove that if $\lim_{x\to\infty}f(x)$ exists, or $\lim_{x\to\infty}f(x)=\infty$.

But I would like to ask if only with the condition of continuity, which is $\lim_{x\to x_0}f(x)=f(x_0)$, the statement is true or not.

Consider $f(x)$ on $[0,\infty)$ defined by $f(x) = \sin x, x\in [0,2\pi],$ $f(x) = \sin^3 x, x\in [2\pi,4\pi],$ $f(x) = \sin^5 x, x\in [4\pi,6\pi], \dots.$ — zhw., Aug 26 '18 at 01:37
Consider the integral $\int_0^\infty \sin(x^2)-\cos(x^2) dx$. There exist both complex and real analytic proofs that this integral comes out to be zero. I believe this function is continuous, but you can verify that it does not take on a limiting value. — user196574, Aug 26 '18 at 01:49
Thanks all of you, I think this is on the right track. I'll do some more readings. — SHORE SHEN, Aug 26 '18 at 02:14
@zhw If you could show me the counter example that is unbounded... Thank you! — SHORE SHEN, Aug 26 '18 at 10:37

W. mu · Answer 1 · 2018-08-26T02:06:15.210

1

The condition $\int fdx=0$ is equivalent to $\int fdx<+\infty$.

$f(x)=\begin{cases}0,\quad n\leq x<n+1-1/n^2,\\\sin(n^2\pi(x-(n+1-1/n^2))),\quad n+1-1/n^2\leq x<n+1.\end{cases}$ $n\in \mathbb{Z}.$

$\int fdx$ exists, but $\lim_{x\rightarrow+\infty}f(x)$ doesn't exist.

edited Aug 26 '18 at 02:06

answered Aug 26 '18 at 01:42

W. mu

923

Mate, I think this function is discontinues at the point of $x=n+1-\frac{1}{n^2}$ ?? – SHORE SHEN Aug 26 '18 at 01:57
@SHORESHEN No, it continuous there with value $0$. – amsmath Aug 26 '18 at 02:33

score 1 · Accepted Answer · answered Aug 26 '18 at 16:37

As requested, here is a counterexample that is unbounded.

For $n=1,2,\dots $ define the function $f_n$ on the interval $[n,n+1/n^2]$ to be an up-triangle of height $n$ over the first half of the interval, a down-triangle of height $n$ on the second half of the interval. Each of these triangles has area $(1/2)n(1/(2n^2) = 1/(4n).$ Define $f_n=0$ everywhere else. Then

$$\int_0^\infty f_n = 1/(4n) - 1/(4n) = 0.$$

Now define $f=\sum_{n=1}^{\infty}f_n.$ Then $f$ is continuous on $[0,\infty),$ and $\int_0^\infty f = 0$ (that needs verification that I'll leave to reader for now). But there are sequences $x_n,y_n\to \infty$ along which $f(x_n)=n, f(y_n)=-n.$

If $\int_0^\infty f(x) dx=0$ , $f(x)$ continuous for $x\in[0,\infty)$, can it be proved that $\lim_{x\to\infty}f(x)=0$?

2 Answers2