Consider the Cantor function $f:[0,1]\to [0,1]$ and an element of the Cantor set $x\in C$, we want to prove that $$f(x+3r)-f(x-3r)\leq 4 (f(x+r)-f(x-r))$$ for all $r>0$ such that the formula makes sense, that is, both $x+3r$ and $x-3r$ are in $[0,1]$.
We claim ( and it is verifiable? ) that $$\forall n\geq 3\ , \forall \frac{1}{3^n}< r< \frac{1}{3^{n-1}} :\ f(x+r)-f(x-r)\in [\frac{1}{2^n}, \frac{1}{2^{n-1}}]$$
In other words, if we move from $x$ by a quantity in the order of $\frac{1}{3^n}$ then necessarily $f(x)$ move in the order of $\frac{1}{2^n}$ .
This immediately shows the former inequality, but is it even true? Is there a reason to reject that claim?
I am not sure how to prove it. I probably can prove it using the ternary representation of $r$, adding that to the ternary representation of $x$ and 'push' the way through... But I want a formal proof.
$\bullet \bullet \bullet$ The digits representation attempt is below : $\bullet \bullet \bullet$
$r=\sum_{k=1}^\infty \frac{b_k}{3^k}, \ b_k\in \{0,1, 2\}$ where $b_1=b_2=...=b_{n-1}= 0$ and $b_n\geq 1$ and if it is the digit $1$ then for some $k\geq n+1: b_{k}\neq 0$
$x=\sum_{k=1}^\infty \frac{x_k}{3^k}, \ x_k\in \{0, 2\}$ and $f(x)= \sum_{k=1}^\infty \frac{x'_k}{2^k}$ where $x'_k$ is from $x_k$ by replacing $2$'s with $1$'s.
write $x+r, x-r, x+3r, x-3r$ in ternary, then write the ternary expression of the greatest number in $C$ which is a lower bound for each one of them. Compare $f(y)$ of that expression, with $f(x)$, and find that $f(x+r)-f(x-r) \geq \frac{1}{2^{n}}$
The same goes for the other inequality.