$f:\mathbb{R}\to \mathbb{R} $ uniformly d,ifferentiable function, $\int_{0}^{\infty}f$ converges, and $f'$ is bounded. I need to prove that $\lim f(x)=0$ when $x \to \infty$.

This is what I did so far:

I chose a series $c_n \to \infty$ such that $f(c_n)> \epsilon$, assuming that is$\lim f(x)\neq0$ when $x \to \infty$. then I wrote $f(c_n+t)=f(c_n)+ \int_{c_n}^{c_n+t}f'(x) \geq f(c_n)-tM \geq\epsilon-tM$. I don't really know if this is useful, I can't see the whole picture. I'd love to know if this is useful and to read a possible solutions.



If $\lim_{x\to +\infty}f(x)\neq 0$, then, substituting $f$ by $-f$ is necessary, we can find a sequence $\{x_n\}$ with converges to $+\infty$ and $\delta_0>0$ such that $f(x_n)\geq 2\delta_0$. Put $M:=1+\sup_{x\geq 0}|f'(x)|$. We can extract from $\{x_n\}$ a subsequence, which will be denoted $\{t_n\}$ such that $t_{n+1}-t_n\geq\frac{2\delta_0}M$. We have $|f(t_n)-f(x)|\leq M|t_n-x|$ so if $|t_n-x|\leq \frac{\delta_0}M$ we have $f(x)\geq \delta_0$. Indeed, $|f(t_n)-f(x)|\leq \delta_0$ so $2\delta_0-f(x)\leq f(t_n)-f(x)\leq \delta_0$, and $\delta_0-f(x)\leq 0$.

Hence $$\forall n\in\mathbb N:\int_{t_n-\frac{\delta_0}M}^{t_n+\frac{\delta_0}M}f(t)dt\geq2\frac{\delta_0^2}M,$$ but this contradicts the fact that the integral $\int_0^{+\infty}f(t)dt$ is convergent. Indeed, if it was the case we would be able to find $c$ such that if $x,y\geq c$ then $\left|\int_x^yf(t)dt\right|\leq \frac{\delta_0^2}M$.

In fact, it can be shown by a similar argument that an uniformly continuous function on $[0,+\infty[$ such that the integral $\int_0^{+\infty}f(t)dt$ is convergent has a limit $0$. The arguments are quite the same, since we get $\eta$ such that if $|x-y|\leq \eta$ then $|f(x)-f(y)|\leq \delta_0$.

  • $\begingroup$ Does $t_n$ is a subsequence of $x_n$? $\endgroup$ – Jozef Feb 4 '12 at 11:03
  • $\begingroup$ Yes it's a subsequence of $\{x_n\}$. $\endgroup$ – Davide Giraudo Feb 4 '12 at 11:04
  • $\begingroup$ You chose it in such way that it would be increasing? $\endgroup$ – Jozef Feb 4 '12 at 11:06
  • $\begingroup$ Yes, but in fact I realize it's not necessary. $\endgroup$ – Davide Giraudo Feb 4 '12 at 11:07
  • $\begingroup$ Why does $|t_n-x| \leq \frac{\delta_0}M$ and you're assuming that $f(x) \geq \delta_0 +f(t_n) \geq \delta_0$ why does $f(t_n) \geq 0$? $\endgroup$ – Jozef Feb 4 '12 at 11:11

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