From sequence of functions to Cantor function. We define a sequence of functions $L_n\colon [0,1]\to [0,1]$ in the following way:
$$L_n(0):=0$$
$$
L_n:=\begin{cases}
\text{linear function with slope}\; \big(\frac{3}{2}\big)^n\;\text{in}\; K_n \\
\text{constant on each component interval of}\; [0,1]\setminus K_n
\end{cases}
$$
where $K_n$ is the union of remaining intervals at each step in the construction of the Cantor set.
For example,
$$
L_1(x):=\begin{cases}
\frac{3}{2}x\quad x\in\big[0,\frac{1}{3}\big]\\
\frac{1}{2}\quad x\in\big(\frac{1}{3},\frac{2}{3}\big)\\
\frac{3}{2}x-\frac{1}{2}\quad x\in\big[\frac{2}{3},1\big]
\end{cases}
$$
in this case $K_1=\big[0,\frac{1}{3}\big]\cup \big[\frac{2}{3},1\big].$
For $n=2$ we have
$$
L_2(x):=\begin{cases}
\frac{9}{4}x\quad x\in\big[0,\frac{1}{9}\big]\\
\frac{1}{4}\quad x\in\big(\frac{1}{9},\frac{2}{9}\big)\\
\frac{9}{4}x-\frac{1}{4}\quad x\in\big[\frac{2}{9},\frac{1}{3}\big]\\
\frac{1}{2}\quad x\in\big(\frac{1}{3},\frac{2}{3}\big)\\
\frac{9}{4}x-1\quad x\in\big[\frac{2}{3},\frac{7}{9}\big]\\
\frac{3}{4}\quad x\in\big(\frac{7}{9},\frac{8}{9}\big)\\
\frac{9}{4}x-\frac{5}{4}\quad x\in\big[\frac{8}{9},1\big]\\
\end{cases}
$$
in this case $K_2=\big[0,\frac{1}{9}\big]\cup\big[\frac{2}{9},\frac{1}{3}\big]\cup\big[\frac{2}{3},\frac{7}{9}\big]\cup\big[\frac{8}{9},1\big]$.

Question 1. We have, for all $n\in\mathbb{N}$ $$L_n(x)=1-L_n(1-x)\quad (x\in[0,1]).$$
How can I prove this?

We observe that, if $m>n$, $L_m(x)=L_n(x)$ for all $x\in[0,1]\setminus K_n$
Now, I would like to show that $\{L_n(x)\}$ is a Cauchy sequence. To do this you need what,first of all, I write:
\begin{split}
\sup_{x\in [0,1]}|L_{n+1}(x)-L_n(x)|&=\sup_{x\in K_n}|L_{n+1}(x)-L_n(x)|\\
&\color{RED}{=}\sup_{x\in [0,1/3^n]} \bigg|\bigg(\frac{3}{2}\bigg)^{n+1}x-\bigg(\frac{3}{2}\bigg)^n x\bigg|\\
&=\bigg(\frac{3}{2}\bigg)^n \bigg(\frac{3}{2}-1\bigg)\frac{1}{3^n}=\frac{1}{2^{n+1}}
\end{split}

Question 2. Could you explain to me the reason for the equality in red?

Thanks!
 A: Question 1
$L_n(x)=\left(\frac{3}2\right)^n\mu\left([0,x]\cap K_n\right)$ with $\mu$ denoting Lebesgue measure. This is because $L_n$ is increasing with constant slope on $K_n$ and is piecewise constant on $[0,1]\setminus K_n$, therefore the value of $L_n(x)$ is proportional to the amount of 'time' that $L_n$ was increasing for before 'time' $x$.
From the left-right symmetry of $K_n$, it follows that $\mu([1-x,1]\cap K_n)=\mu([0,x]\cap K_n)$. Note that $\mu([1-x,1]\cap K_n)=L_n(1)-L_n(1-x)$.
Question 2
The red inequality is not strictly correct.
To prove that it is sufficient to take the supremum over $x \in [0,\frac{1}{3^n}]$, we need to show that $L_{n+1}(\frac{k}{3^n}+x)-L_n(\frac{k}{3^n}+x)=L_{n+1}(x)-L_n(x)$ whenever $x$ and $\frac{k}{3^n}+x$ are both in $[0,1]$. This can be proved using the fact that $L_{n+1}(\frac{k}{3^n})=L_n(\frac{k}{3^n})$ and the fact that $L_{n+1}$ behaves identically within each interval of $K_n$.
In fact, it is sufficient to take the supremum over the interval $[0,\frac{1}{3^{n+1}}]$ because $L_n+1-L_n$has a left-right symmetry over that interval, and because the constant interval in the middle of $L_{n+1}$ will be furthest from $L_n$ at its endpoints. 
$L_{n+1}-L_n$ evaluates to $\left(\frac{3}{2}\right)^{n+1}x-\left(\frac{3}{2}\right)^nx$ on the interval $[0,\frac{1}{3^{n+1}}]$. The supremum is $\frac{1}{3.2^{n+1}}$
Addendum
There is an easier way to prove that $L_n(x)$ is a Cauchy sequence. Let $y_n$ be a sequence of points in $[0,1] \setminus K_n$ which converges to the point $x$. Then $L_n(y_n)$ converges to $L_n(x)$ and $L_{n+1}(y_n)$ converges to $L_{n+1}(x)$. Since $L_n(y_n)=L_{n+1}y_n$, $L_{n+1}(x)$ must converge to $L_n(x)$.
