Cantor ternary function

Question

For $x \in [0,1]$, there is a ternary expansion $$ x = \sum_{n=1}^{\infty} \frac{x_n}{3^n} \;, x_n \in \{0,1,2\} . $$ Let $N_x=\infty$ if none of the $x_n$ are $1$, otherwise let $N_x$ be the smallest value $n$ s.t. $x_n=1$. Define the Cantor ternary function $f:[0,1] \longrightarrow \mathbb{R}$ by $$ f(x)=\frac{1}{2^{N_x}} + \sum_{n=1}^{N_x-1} \frac{x_n}{2^{n+1}} \;. $$ Show that $f$ is an increasing function.

My attempt: Let $$ x=\sum_{n=1}^{\infty}\frac{x_n}{3^n} < y=\sum_{n=1}^{\infty}\frac{y_n}{3^n} \; , $$ let $N=\min\{n \in \mathbb{N} \mid x_n \neq y_n \}$.

Step $1$: We claim $x_N < y_N$. $$ \begin{aligned} 0 > x-y &= \sum_{n=N}^{\infty} \frac{x_n-y_n}{3^n} \\ &\geq \frac{x_N - y_N}{3^N} - \sum_{n=N+1}^{\infty} \frac{2}{3^n} \\ &= \frac{x_N - y_N}{3^N} - \frac{1}{3^N} \; , \end{aligned} $$ so $x_N - y_N < 1$, since $x_N, y_N \in \{0,1,2\}$, and $x_N \neq y_N$, then $x_N - y_N < 0$.

Step $2$: We claim $$ \sum_{n=1}^{\infty} \frac{x_n-y_n}{2^{n+1}} \leq 0 \; . $$ $$ \begin{aligned} \sum_{n=1}^{\infty} \frac{x_n-y_n}{2^{n+1}} &= \sum_{n=N}^{\infty} \frac{x_n-y_n}{2^{n+1}} \\ &\leq \frac{-1}{2^{N+1}} + \sum_{n=N+1}^{\infty} \frac{2}{2^{n+1}} \\ &= 0 \; . \end{aligned} $$ Step $3$: We claim $f(x) \leq f(y)$. Then I got stuck here.

Answer 1

Things become more clear if we write:$$f(x)=\sum_{n=1}^{N_x-1}2^{-n}\left(\frac12x_n\right)+\sum_{n=N_x+1}^{\infty}2^{-n}$$It indicates that it will be handsome to write $f(x)$ as a binary number and it is already clear at first sight that writing $x$ as a ternary number is handsome as well.

Doing so let us have a look at the following examples where $f$ is applied at ternary numbers at LHS and its value is written as a binary number on RHS.

• $0.20220\dots\mapsto0.10110\dots$
• $0.20221\dots\mapsto0.10110111\dots=0.10111000\dots$
• $0.20222\dots\mapsto0.10111\dots$

Increasing order on LHS evidently induces increasing order on RHS.

I hope this is enough for you. If not then at least it is inspiring.