Proving this statement for improper integral

Question

The statement says that if $f(x)$ is differentiable, $\ |f´(x)|$ is bounded for all $x$, and $$\int_0^{\infty} |f(x)| \,dx < +\infty,$$then $$\lim_{x\to \infty} f(x) = 0.$$ I´m not sure how to use the hypothesis here, any hint or help?

Answer 1

It is sufficient to require that $|f(x)-f(y)| \le L |x-y|$ for some constant $L$. (This is implied by the hypotheses.)

Note that $\lim_{x \to \infty} \int_x^\infty |f(x)|dx = 0$.

Suppose $c=|f(x_0)| >0$. Then $\int_{x_0}^\infty |f(x)| dx \ge {c^2 \over 2L}$.

In particular, $\int_{x_0}^\infty |f(x)| dx \ge {(\sup_{x \ge x_0} |f(x)|)^2 \over 2L}$.

Combining these results shows that $\limsup_x |f(x)| = 0$, or equivalently, $f(x) \to 0$.