# Bounded variation + injective implies strictly monotonic?

The question is are the following assertions true:

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

(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.

• The obvious counterexample to the "strong" version: $$f(x)=\cases{1+x\,,&x\in[0,1]\,,\\x\,,&x>1\,.}$$ Commented Aug 5 at 11:56
• @KurtG. Isn't $f(1)=f(2)$ in your example? Commented Aug 5 at 12:04
• @geetha290krm oops. So let's fix that $$f(x)=\cases{x+1\,,&x\in[0,1]\cup[2,+\infty)\,,\\x-1\,,&x\in(1,2)\,.}$$ Commented Aug 5 at 12:13

## 1 Answer

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 $$a=x_0. Then $$\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).$$

• I thank you for the counterexample(s). Now, I am wondering if we can, with a slight reformulation, still get some kind of strict monotonicity. More precisely, my next question is: is it true that, if $u\in {\rm BV}([a,b])$, if $u$ is injective on $[a,b]$, then there exists $u_0:[a,b]\rightarrow {\bf R}$ such that $u_0$ is piecewise strictly monotonic on $[a,b]$, and such that $u(s)=u_0(s)$ (a.e. $s\in [a,b]$)? Note that examples from of Gerd and Kurt G. does not provide a counterexample for this reformulated version of the question. Commented Aug 5 at 13:11
• Maybe you should open a new question for this.
– Gerd
Commented Aug 5 at 13:16
• here is the new question with the reformulated statement: math.stackexchange.com/questions/4954802/… Commented Aug 5 at 14:22