bounding an integral by the function's derivative Say $f\in C^1$ and has compact support in [-1, 1]. Furthermore,
$$\int_{-\infty}^\infty \lvert f'(x)\rvert dx \le 1$$
I want to show that
$$\int_{-\infty}^\infty \lvert f(x+y)-f(x) \rvert dx \le |y|$$
First I tried using the mean value theorem, then I tried the fundamental theorem of calculus. Either I made a mistake somewhere, or neither is the right approach. Seems like this should be simple, so just a hint is enough.
 A: Actually it should be $\int_{-\infty}^{\infty}|f(x+y)-f(x)|dx \le 2+ |y|$ $\quad$ If $f'$ exists at every point and not necessarily continuous.
Proof:
If $f'$ exists at every point
Then according to Lemma 6.14  from 'The Integrals of Lebesgue, Denjoy , Perron , and Henstock (Graduate Studies in Mathematics Volume 4 )' by Russell A. Gordon:$\quad$$|f(x+y)-f(x)| \le \int_{x}^{x+|y|} |f'(t)|dt$
$|f(x+y)-f(x)| \le \int_{x}^{x+|y|} |f'(t)|dt\le\int_{-\infty}^{\infty} |f'(t)|dt\le1$
$B=$ $[-1 -y, 1-y]$ $\bigcup$ $[-1 , 1]$
$|f(x+y)-f(x)|$ $\le 1$$\quad$for $x \in B$ 
$|f(x+y)-f(x)|$ $=0$ $\quad$for $x \notin B$ 
$\mu^*(B) < 2+|y|$
we conclude $\int_{-\infty}^{\infty}|f(x+y)-f(x)|dx \le 2+ |y|$
A: I decided the summarize comments, which basically contained an answer. First assume that $y\ge 0$. The proof where $y<0$ is very similar. Since $f'$ is continuous, we can use the fundamental theorem of calculus
$$f(x+y) - f(x) = \int_x^{x+y}f'(s)ds$$
Thus
$$\begin{align}
\int_{-\infty}^\infty |f(x+y)-f(x)|dx &= \int_{-\infty}^\infty \left| \int_x^{x+y}f'(s)ds \right| dx\\
&\le \int_{-\infty}^\infty \int_x^{x+y}\lvert f'(s)\rvert ds dx\\
&= \int_{-\infty}^\infty \int_{s-y}^s \lvert f'(s)|dx ds\\
&= y\int_{-\infty}^\infty \lvert f'(s)| ds\\
&\le y
\end{align}
$$
When $y<0$, we use the same argument, but get $\ge y$ at the end. Thus we have the result
$$\int_{-\infty}^\infty |f(x+y)-f(x)|dx \le |y|$$
