Bounded variation + injective implies strictly monotonic?

Question

The question is are the following assertions true:

I.(strong version) Consider $a,b\in {\bf R}$, $a<b$.

(1) If $u\in {\rm BV}([a,b])$, and

(2) if $u$ is an injection on $[a,b]$,

then $u$ is strictly monotonic on $[a,b]$.

II.(weak version) If assumptions (1) and (2) are satisfied,

then $u$ is "piecewise strictly monotonic" on $[a,b]$, meaning that there exists at most countably many disjoint sub-intervals $(a_k,b_k)$ such that $u_0$ is strictly monotonic on $(a_k,b_k)$ for every $k\in{\bf N}$ and such that $[a,b]=\cup_{k=1}^{+\infty}(a_k,b_k)\cup N$, where $\lambda(N)=0$.

Remark (A). It is already established that, under assumptions that $u\in {\rm C}([a,b])$ and that $u$ is injective on $[a,b]$, it follows that $u$ is strictly monotonic on $[a,b]$(see

Continuous injective map is strictly monotonic

) But, in our setting, we do not assume continuity.

Remark (B). If we add the assumption that $u([a,b])$ is an interval, then $u$ satisfies the intermediate-value property, and the claim follows even without the assumption that $u\in {\rm BV}([a,b])$, see

Injective functions with intermediate-value property are continuous. Better proof?

Indeed, it follows that $u\in {\rm C}([a,b])$, and so the assertion follows by Remark (A).

Remark (C). Here is somewhat similar(but ultimately different) question which is settled:

Why is a strictly monotonic mapping between intervals continuous?

Remark (D). I observe that, if $u\in {\rm BV}([a,b])$, then there exist two strictly(!) increasing functions $u_1,u_2:[a,b]\rightarrow {\bf R}$ such that $u=u_1-u_2$. Therefore $u_1$ and $u_2$ are injections. I don't know if this helps.

In any case, I was not able to find any similar result about BV-functions in the literature.

Answer 1

I think this is not true: Let $(q_n)_{n=1}^\infty$ be an enumeration of $\mathbb{Q}\cap [a,b]$ with $q_n \not= q_m$ $(n \not= m)$ and define $f:[a,b] \to\mathbb{R}$ the following way.

1. $f(x)=x$ $(x \in [a,b] \setminus \mathbb{Q})$.
2. By induction: For $q_1$ choose $q_{n_1}$ with $n_1 >1$ such that $|q_1- q_{n_1}| \le 1$ and set $f(q_1)=q_{n_1}$,
   For $q_2$ choose $q_{n_2}$ with $n_2 > n_1$ such that $|q_2- q_{n_2}| \le 1/2$ and set $f(q_2)=q_{n_2}$,
   $\dots$
   For $q_{k}$ choose $q_{n_k}$ with $n_k > n_{k-1}$ such that $|q_k- q_{n_{k}}| \le 1/2^k$ and set $f(q_k)=q_{n_k}$
   $\dots$

Now $f$ is injective and $f$ is not monotone on any interval $I\subseteq [a,b]$. Moreover $f$ is of bounded variation: Let $a=x_0<x_1< \dots < x_n=b$. Then $$ \sum_{j=0}^{n-1} |f(x_{j+1})-f(x_j)| \le \sum_{j=0}^{n-1} (|f(x_{j+1})-x_{j+1}| + (x_{j+1}-x_{j}) + |f(x_{j})-x_{j}|) $$ $$ \le \sum_{k=0}^\infty \frac{1}{2^k} + (b-a) + \sum_{k=0}^\infty \frac{1}{2^k} = 4 +(b-a). $$