Construct a continuous non-monotone function $H\in BV(\mathbb{R})$ satisfying $H'\geq 1$ a.e.
or, prove this is impossible.
My intuition tells me that this is impossible. But I don't know how to start.
If someone could tell me whether this is possible or not, and provide with a hint , that will help.
The following is what I did.
On a closed interval
Let $C(x)$ be the Cantor function on $[0,1]$ and define $h_0:[0,1]\rightarrow\mathbb{R},h_0(x)=x-C(x)$.
$h_0$ is continuous since both $x$ and $C(x)$ are continuous.
$h_0$ is non-monotone since $h_0(0)=h_0(\frac{1}{2})=h(1)=0$ and $h_0(\frac{1}{3})=-\frac{1}{6}$
$h_0'=1-C'=1$ a.e. since $C'=0$ a.e.
Since $x$ is bounded increasing and $-C(x)$ is bounded decreasing, $h_0\in BV([0,1])$.
I was trying to copy, paste and stretch $h_0(x)$ to define $H$ on $\mathbb{R}$, but it seems impossible to have the sum of horizontal invervals diverge while the sum of vertical difference (that measures the total variatiom) to converge without affecting the derivative
On $\mathbb{R}$
$H\in BV\Longrightarrow H=F+G$ where $F$ is bounded increasing and $G$ is bounded decreasing.
$H,F$ and $G$ are differentiable a.e. On the set that they are all differentiable,
$\vert H'\vert=\vert F'+G'\vert\leq\vert F'\vert+\vert G'\vert\leq\vert F'\vert$.
$F$ is bounded increasing $\Longrightarrow F'\in L^1$, which implies $H'\in L^1$ by monotonicity.
But $H'\geq 1$ a.e. implies $\int H'\geq \int 1=\infty\Longrightarrow H'\notin L^1$.
Therefore, this is impossible.
So my new question is, it is indeed impossible to do this on $\mathbb{R}$ right?
