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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).

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$f'$ 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 $\mathbb R^+$.

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  • $\begingroup$ 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

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