Let $f:\mathbb{R}^+ \to \mathbb{R}^+$ be a continuous function which is integrable on $\mathbb{R}^+$. Ie. $\int_{\mathbb{R}^+} f \in \mathbb{R}$. Moreover suppose that $f'$ is absolutely integrable, i.e.

$$\int_{\mathbb{R}^+} |f'|\in \mathbb{R}.$$

Then do we have $$\lim_{x \to \infty} f(x) = 0?$$

This should be false, yet I am unable to find a counterexample. I've tried to create the usual counterexample of a function composed of triangles such that the sum of the area of these triangles is a convergent serie but it doesn't work ($\mid f' \mid$ is not integrable).

(Note: all the integrals are taken in Riemann sense not Lebesgue).


$f'$ absolutely integrable means $f'$ integrable, so $$x\mapsto \int_0^x f'(t){\rm d}t=f(x)-f(0)$$ has a finite limit $\ell$ for $x\to+\infty$.

If this limit were to be different from $0$, it should be easy tu prove that $f$ can't be integrable on $\mathbb R^+$.

  • $\begingroup$ Is this true? Take the Cantor singular function $f$ and extend it to a function $g$ on $\mathbb R^+$ by reflecting it in the line $y=1/2$ and translating it $1$ unit to the right; in general, if $g$ is defined on $[0,n]$, extend it to $[n,n+1]$ by reflecting $gf_{\big|[n-1,n]}$ the line $y=n+1/2$ and translating by $1$ unit. Then, $g'=0$ $\endgroup$ – Matematleta Nov 11 '18 at 15:27
  • $\begingroup$ @Matematleta : this does not contradict what I said, does it ? $\endgroup$ – Nicolas FRANCOIS Nov 12 '18 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.