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Let $f:\mathbb{R}\to\mathbb{R}$, $f\in C^1(\mathbb{R})$.

Suppose $\int_0^\infty f$ converges and $f'$ is bounded.

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

My attempt:

W.l.o.g. assume that $\lim_{x\to\infty}f(x)=L>0$.

from some point $x_0\in [0,\infty)$ we get that $f$ is positive then we can write:

$\int_0^\infty f$ = $\int_0^{x_0} f$ + $\int_{x_0}^\infty f$

$f'$ is bounded, thus $f$ is uniformly continuous in $[0,x_0]$ and the integral $\int_0^{x_0} f$ is equal to $I$ for some $I$.

$\lim_{x\to\infty}\frac{f(x)}{1}=L>0$. And we get that $\int_{x_0}^\infty f$ and $\int_{x_0}^\infty 1$ converge and diverge together.

$\int_{x_0}^\infty 1$ diverges, so $\int_{x_0}^\infty f$ diverges, too. contradiction.

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    WLOG assume that the limit exists, you say. What if the limit does not exist? – 2017-02-03
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    Why do you not lose generality in assuming that $\lim_{x\to\infty} f(x)$ exists? – 2017-02-03
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    Oh yes my mistake, what to do if the limit do not exist? – 2017-02-03
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    In fact, that's where you need to use the hypothesis on $f'$: $f$ is uniformly continuous on $[0,x_0]$ regardless of even being differentiable, thanks to Heine-Cantor theorem. – 2017-02-03
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    Can you help? @G.Sassatelli – 2017-02-03
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    Right now nothing too valuable comes to mind. – 2017-02-03
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    and Do you have some advice? @GiuseppeNegro – 2017-02-03

1 Answers 1

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Since $I=\lim\limits_{x\to\infty}\int_0^x f(t)\,dt$ exists finite, you have $\limsup\limits_{x\to\infty} f(x)\ge0$ and $\liminf\limits_{x\to\infty} f(x)\le 0$.

Assume as a contradiction that $\limsup\limits_{x\to\infty} f(x)>0$. Since $\liminf\limits_{x\to\infty} f(x)\le 0$, you have that $f'(x)$ is negative in at least one point (actually, it must be frequently negative as $x\to\infty$). Therefore, $$A=\inf_{x> 0} f'(x)<0$$

Again, since $\liminf\limits_{x\to\infty} f(x)\le 0<\limsup\limits_{x\to\infty} f(x)$ and $f$ is continuous, there is a positive real number $M$ such that $f(x)=M$ frequently often as $x\to\infty$. Id est, there is a sequence $x_n\to\infty$ such that $f(x_n)=M$.

By Lagrange theorem, $$\begin{align}&f(x)\ge A(x-x_n)+M&\text{if }x>x_n\\\end{align}$$

A fortiori, this holds for $x_n< x

Hence, $$\int_{x_n}^{x_n-\frac MA}f(t)\,dt\ge \left[\frac A2t^2+(M-Ax_n)t\right]^{x_n-\frac MA}_{x_n}=-\frac{M^2}{2A}$$

But $x_n-\frac MA\stackrel{n\to\infty}\longrightarrow \infty$ and $$\liminf\limits_{n\to\infty}\int_0^{x_n-\frac MA}f(t)\,dt=\liminf\limits_{n\to\infty}\left(\int_0^{x_n}f(t)\,dt+\int_{x_n}^{x_n-\frac MA}f(t)\,dt\right)=\\=\lim\limits_{n\to\infty}\int_0^{x_n}f(t)\,dt+\liminf\limits_{n\to\infty}\int_{x_n}^{x_n-\frac MA}f(t)\,dt\ge I-\frac{M^2}{2A}>I$$

Which contradicts $\lim\limits_{x\to\infty} \int_0^x f(t)\,dt=I$. Therefore it must hold $\limsup\limits_{x\to \infty}f(x)\le0$ and, hence, $=0$.

Now, by using the previous result on $g=-f$ (which satisfies the hypothesis of the theorem), we obtain that $$0\ge\limsup\limits_{x\to\infty} g(x)=-\liminf\limits_{x\to\infty} f(x)$$

So $\liminf\limits_{x\to\infty}f(x)\ge 0$ as well. Hence $\liminf\limits_{x\to\infty} f(x)=0$.

This proves that $\lim\limits_{x\to\infty} f(x)=0$.

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    Good solution. Thank you! – 2017-02-03
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    @GiuseppeNegro The underlying idea is that the bounded derivative implies that there must frequently be "triangular cusps" all similar one to another. Basically, you have one way to go and it works. – 2017-02-03