Yesterday I was looking for problems about improper integrals. In a book by B.P. Demidovich I found this nice problem: Suppose $f$ is a $C^1(R)$ function with bounded derivative (i.e. $|f'(x)|<M$ for all $x\in R$). Prove that if $$ \int_0^\infty |f(x)|\,dx < \infty, $$ then $\displaystyle{\lim_{x\rightarrow \infty} f(x) =0}$.
We can prove this noting that the integral $\int_0^\infty f(x) \, f'(x) \,dx$ is convergent. $\square$
I wonder if we can replace the hypothesis about the absolute convergence of the integral by symply convergence: $$ \lim_{M\rightarrow \infty} \int_0^M f(x)\,dx \mbox{ exists.} $$ I suspect the result is still true but I did not realize how to prove it. I think que can not use now that $\int_0^\infty f(x) \, f'(x) \,dx$ is convergent. I appreciate any suggestions you can give me.
Since $|f'(x)| \leq M$ then by the mean value inequality we have the following estimate \begin{align} |f(x) - f(y)| = |f'(\xi)(x-y)| \leq M|x-y| \end{align} for all $x, y \in \mathbb{R}$. Hence it follows $f$ is uniformly continuous.
Suppose the limit of $f$ as $x \rightarrow \infty$ is not zero. In a more precise manner, there is an $\epsilon>0$ where for all $n \in \mathbb{N}$ there exists $x_n>n$ such that $|f(x_n)|>\epsilon$. Moreover, by uniform continuity of $f$ on $\mathbb{R}$, there exists $\delta>0$ such that for all $|x_n-y|<\delta$ we have that $|f(x_n)-f(y)|<\epsilon/2$ for all $n \in \mathbb{N}$ which means \begin{align} \epsilon<|f(x_n)|<\frac{\epsilon}{2}+|f(y)| \ \ \Rightarrow \ \ |f(y)| >\frac{\epsilon}{2} \end{align}
whenever $|x_n-y|<\delta$ for all $n$. (Heuristically, this means that we could fit infinitely many boxes of volume $\delta* \epsilon/2$ under the curve of $f$.)
Next, consider the sequence $y_n$ defines as follows \begin{align} y_1 = x_1 -\delta, y_2 = x_1 + \delta, y_3 = x_2-\delta, \ldots \end{align} then it follows either \begin{align} \int^{y_{2k}}_0 f(x) \ dx -\int^{y_{2k-1}}_0 f(x)\ dx = \int^{y_{2k}}_{y_{2k-1}} f(x)\ dx> \frac{\epsilon}{2}\delta \ \ \text{ or } \ \ \ \int^{y_{2k}}_{y_{2k-1}} f(x)\ dx< -\frac{\epsilon}{2}\delta \end{align} which means \begin{align} \int^{y_n}_0 f(x)\ dx \end{align} doesn't form a Cauchy sequence, i.e. the limit doesn't exists, a contradiction since the limit does exists.