# Construct a continuous non-monotone function

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?

• Do you know of any continuous functions that, for example, have derivative zero a.e. but are not constant? – Andrew D. Hwang Mar 27 '15 at 12:20
• "Is this impossible on $\mathbf{R}$?": There's an obvious continuous extension of the Cantor function to $\mathbf{R}$ whose derivative is $0$ a.e. :) (Separately, it's probably a good idea to post your own construction as an answer and mark it as such to resolve this question.) – Andrew D. Hwang Mar 27 '15 at 14:23
• @user86418 but then the function will fail to be in BV – Justin Zhou Mar 27 '15 at 16:09
• Ah, yes, I was tacitly thinking "locally of bounded variation". (If $H' \geq 1$ a.e. on $\mathbf{R}$, then the total variation of $H$ on $[a, b]$ is at least $b - a$, so $H$ certainly isn't of bounded variation on $\mathbf{R}$.) – Andrew D. Hwang Mar 27 '15 at 17:04

Consider the Cantor function as defined for instance here: http://en.wikipedia.org/wiki/Cantor_function This function is continuous, has derivative 0 a.e., and takes all values between 0 and 1. With this and the identity $x \mapsto x$ it is rather easy to construct an example.