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:

1. $\int_0^\infty f(x) dx=0$
2. $f(x)$ 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.

Answer 1

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 $f(x_n)=n, f(y_n)=-n.$

Answer 2

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.