$f, f'\in L^{1}(\mathbb R) \implies \lim_{x\to \infty} f(x)=0 ?$ Suppose $f\in L^{1}(\mathbb R) $ and $f'\in L^{1}(\mathbb R).$
My Question is: 

Can we show, $\lim_{x\to \infty} f(x)=0$ ?

Thanks,

$f'\in L^1(\Bbb R)$, then $\int_{-\infty}^0|f'(t)|dt$ and $\int_0^{+\infty}|f'(t)|dt$ converges, in particular $\int_{-\infty}^0f'(t)dt$ and $\int_0^{+\infty}f'(t)dt$ converges.


*

*The convergence of the integral $\int_0^{+\infty}f'(t)dt$ implies that the function $x\to \int_0^{x}f'(t)dt$ admit a finite limit $l$ (as $x\to +\infty$). But $\int_0^{x}f'(t)dt=f(x)-f(0)$, this implies that $f$ admit a finite limit $M$ (as $x\to +\infty$), if $M\neq 0$ then $|f(x)|\sim |M|$ (in $+\infty$) this is impossible because a constant function $x\to |M|$ is not integrable on $[0,+\infty[$ but $f$ is integrable on this interval . It follow that $M=0$, that is $f\to 0$ as $x\to +\infty$.

*By the same argument (by working in $]-\infty,0]$) we can show that $f(x)\to 0$ as $x\to -\infty$.
