Properties of the sequence

Question

Given a sequence $\{a_i\}_{i\in \mathbb{Z}},$ consider the sequence defined by $b_i:=F(a_{i-1},a_{i},a_{i+1}),$ where $F:\mathbb{R}^3 \rightarrow \mathbb{R}$ is increasing in each of the variable, and $F(a,a,a)=a.$ Suppose total variation of the sequence $\{a_i\}_{i\in \mathbb{Z}}$ is finite i.e. $$\sup\limits_{k\in \mathbb{N}} \sum_{i=-k}^k |a_i-a_{i-1}| < \infty$$ then how to prove the following

$$\sup\limits_{k\in \mathbb{N}} \sum_{i=-k}^k |b_i-b_{i-1}| \leq \sup\limits_{k\in \mathbb{N}} \sum_{i=-k}^k |a_i-a_{i-1}| $$

I have an intuitive idea of the proof but some how unable to give a rigorous proof.

The idea is local extremas of $b_i$ are bounded below and above by some $a_m$ and $a_n$ because of the monotonicity..How to give a rigorous proof?

I could not succeed with mathematical induction either.. :(

Answer 1

What you are claiming is not true... due to the following counter example

Define $F(x,y,z)=max(x,z),$ clearly $F$ is increasing in each of the variable.

define the sequence by \begin{equation}a_i= { \left\{ \begin{array}{ccl} 1 & \, \mbox{if}\,i=0, \\ 0 & \, \mbox{otherwise}\,, \end{array}\right.} \end{equation} Then $b_i$ turns out to be

\begin{equation}b_i= { \left\{ \begin{array}{ccl} 1 & \, \mbox{if}\,i=-1,1, \\ 0 & \, \mbox{otherwise}\,, \end{array}\right.} \end{equation} and we have, $$\sup\limits_{k\in \mathbb{N}} \sum_{i=-k}^k |a_i-a_{i-1}|=2$$

$$\sup\limits_{k\in \mathbb{N}} \sum_{i=-k}^k |b_i-b_{i-1}| =4$$