Let $f\in L^1(\mathbb{R})$. Find $\lim_{n\rightarrow\infty} \int_\mathbb{R}\vert f(x+n)-f(x)\vert dx$.
I think the limit is $2\Vert f\Vert_{L^1}$, as this result holds for simple functions. If we first consider $f$ is a simple function, after some large $n$, support of $f(x+n)$ and $f(x)$ are going to be disjoint which will give the result I guess.
But I could not generalize my idea by density, because I could get only one side inequality by density,
Maybe my guess is wrong, Thanks in advance for any help/hint suggestions.
Hint: You see how to do this when $f$ is compactly supported. Now approximate arbitrary $f$ by $f\chi_{[-N,N]}$, where $N$ is large enough to make the integral of $|f-f\chi_{[-N,N]}|$ less than $\epsilon$.
Solution: Note
$$\left| \int |f_N(x+n) -f_N(x)| - \int |f(x+n)-f(x)| \right| \le \int ||f_N(x+n) -f_N(x)| - |f(x+n)-f(x)||.$$
Now apply the reverse triangle inequality $||a|-|b||<|a-b|$.
$$\le \int |f_N(x+n) -f_N(x) -f(x+n)+f(x)| \le \int |f_N(x+n) - f(x+n)| + \int |f(x)-f_N(x)|.$$
We chose $N$ so that both terms on the right hand side are less than $\epsilon$. So the problem reduces to the compactly supported case, which you know how to do.