Is the function $G(x)=\int_0^x \left| \frac{d}{dt}\int_{t-1}^tf(s)ds \right|dt$ bounded? Let $f:\mathbb{R}\to \mathbb{R}$ be a continously differentiable bounded function. Then the function $F$ defined for each $x>0$ by
$$F(x)=\int_0^x \left( \frac{d}{dt}\int_{t-1}^tf(s)ds \right)dt$$ is bounded since
$$\left| F(x)\right| =\left| \int_{x-1}^xf(s)ds-\int_{-1}^0f(s)ds\right| \leq 2\sup_{t\in \mathbb{R}} \left|f(t) \right|. $$
What about the function $G$ defined for each $x>0$ by
$$G(x)=\int_0^x \left| \frac{d}{dt}\int_{t-1}^tf(s)ds \right|dt$$
is it also bounded ?
 A: Let $f(s)=\cos(\pi s)$ so that $f(s)=-f(s-1)$. Then 
$$
\int_{t-1}^tf(s)ds=\int_{t-1}^t\cos(\pi s)ds=\frac{1}{\pi}(\sin(\pi t)-\sin(\pi(t-1))=\frac{2}{\pi}\sin(\pi t). 
$$
Then 
$$
\frac{d}{dt}\int_{t-1}^tf(s)ds=\frac{d}{dt}\Big[\frac{2}{\pi}\sin(\pi t)\Big]=2\cos(\pi t)
$$
so that 
$$
G(x)=\int_0^x|2\cos(\pi t)|dt\to\infty
$$
A: $$
G(x)=\int_0^x \left| \frac{d}{dt}\int_{t-1}^tf(s)ds \right|dt
= \int_0^x \left| \frac{d}{dt} \left( \int_{0}^tf(s)ds - \int_{0}^{t-1}f(s)ds \right)\right|dt
$$
$$
= \int_0^x \left| {d\over dt}  \int_{0}^tf(s)ds - 
{d\over dt}\int_{0}^{t-1}f(s)ds \right|dt
$$
$$
= \int_0^x \left| f(t) - f(t-1) \right|dt \le
 \int_0^x |f(t)| dt + \int_0^x |f(t-1)| dt   \tag{1}
$$
which is not bounded in the general case. We can choose $f(x)$ such that 
$$
| f(t) - f(t-1) | = | f(t) | + |f(t-1) |,
$$
e.g. $f(x)=\arcsin\sin(\pi x)$, and for this choice of $f(x)$ we have an unbounded $G(x)$ (note that the last $\le$ sign in $(1)$ would actually be equality for such $f$).
On the other hand, we can choose $f(x)=C$, so $|f(t)-f(t-1)|=0$ in $(1)$, and for this choice we have a bounded $G(x)$.
So $G(x)$ may or may not be bounded, depending on the actual choice of $f(x)$.
As shown in the answer to another question, $G(x)$ is bounded by the total variation of $f(x)$ on $[-1,x]$, which is in turn bounded by the total variation of $f(x)$ on $[-1,\infty]$. Also, $G(x)$ may be finite even when the total variation of $f(x)$ is $\infty$; for example, $G(x)=0$ for $f(x)=\cos^2(\pi x)$.
