Prove that $\lim_{x\to\infty} f(x) = 0$

Question

Let $f:\mathbb{R}\to\mathbb{R}$ which is continuously differentiable ($f\in C^1$). Lets assume $\int_0^\infty f < \infty$ and $f'(x)$ is bounded. Show that $\lim_{x\to\infty}f(x) = 0$.

All I could think of is doing integration by parts:

$$ \int_0^\infty f(x)\,dx = \sum_{n=1}^\infty \int_{n-1}^n 1\cdot f(x)\,dx = \sum_{n=1}^\infty \left( \left.xf(x)\vphantom{\frac11}\right|_{n-1}^n - \int_{n-1}^n xf'(x) \, dx \right)$$

You have to evaluate $xf(x)$ at $n$ and $n-1$ inside the summation sign - thats how you do integration by parts. — user160738, Jan 20 '15 at 15:24
$\sum_{n=1}^\infty \left( xf(x) - \int_0^n xf'(x) dx\right) \le \sum_{n=1}^\infty \left( xf(x) - \int_0^n xM \right)$ --> why is $- \int_0^n xf'(x) dx \leq - \int_0^n xM dx$ ? I would say only $- \int_0^n xf'(x) dx \leq \int_0^n xM dx$ — Olórin, Jan 20 '15 at 15:27
Maybe not the best way to go about it, but your hypotheses imply $f$ is uniformly continuous. You can argue as in this post. — David Mitra, Jan 20 '15 at 15:30
$f$ is uniformly continuous and integrable, this implies it tends to $0$ — Olórin, Jan 20 '15 at 15:32
I'll have a look at the related post, thank you in the meantime. — Elimination, Jan 20 '15 at 15:34
@DavidMitra, I've looked at the accepted answer there. I don't understand why is $f(q)$ must be $>\varepsilon/2$. — AlonAlon, Jan 20 '15 at 15:49
Use the result from this post and you can get the proof immediately. — xpaul, Jan 20 '15 at 16:33
@xpaul, Can you answer my question above regarding this post? — Elimination, Jan 20 '15 at 17:18
Writing $\int_0^\infty f < \infty$ is not useful unless f is nonnegative — zhw., May 16 '16 at 23:08
Let $F(x) = \int_{0}^{x}f(t),dt$ and then use the following question http://math.stackexchange.com/q/396707/72031 — Paramanand Singh, May 17 '16 at 07:23

score 0 · Answer 1 · answered May 03 '17 at 01:46

Define $F(x) = \int^x_0 f(t)dt$ and suppose $\lim_{x \rightarrow \infty} F'(x) \neq 0$. This means there is $\epsilon > 0$ and an increasing sequence $\{x_n\}$ s.t. $x_n\rightarrow \infty$ and $|F'(x_n)| \geq \epsilon$ for all $n$.

Since $f(x) \in C^1$, $F'(x)$ is uniformly continuous. Thus there is $\delta>0$ s.t. for all $n$, if $|x - y| < \delta$, then $|F'(x) - F'(y)| < \epsilon/2$. Note that this means for $x \in [x_n, x_n+\delta]$, by the triangle inequality, $|f'(x)| \geq |f'(x_n)| - |f'(x) - f'(x_n)| > \epsilon/2$.

The assumption $\int^\infty_0 f(t)dt < \infty$ means $\lim_{n\rightarrow\infty} |F(x_n + \delta) - F(x_n)| = 0$. However, $|F(x_n + \delta) - F(x_n)| \geq \left|\int^{x_n+\delta}_{x_n} F'(t)dt\right| \geq \int^{x_n+\delta}_{x_n} (\epsilon/2) dt = \epsilon \delta /2 > 0 $. Hence the contradiction.

Therefore, $\lim_{x \rightarrow \infty} F'(x) = \lim_{x \rightarrow \infty} f(x) = 0$.

Prove that $\lim_{x\to\infty} f(x) = 0$

1 Answers1