TITLE: Show that if $|x-y| \leq 2^{-n}\implies |f(x) - f(y)|\leq 2^{-n}$ then $|f(x)-f(y)| \leq|x-y|$
QUESTION [2 upvotes]: This question is actually part of a proof of Kolmogorov’s Continuity theorem that I found in a book. Now, during the proof, the author claims the following. Consider $D\in [0,T]$ as an open interval of the dyadic number (i.e $x\in D$ if $x=k2^{-m}$).
The author then proved that for every $n>N$, for $x,y\in D$ and $|x-y|\leq 2^{-n}$, then $|f(x) -f(y)|\leq 2^{-n}$. He then claims that this implies also that $|x-y|\geq |f(x)-f(y)|$.
I don’t understand how one proves this last assertion.

REPLY [3 votes]: (The idea is to split the interval $[x, y]$ into smaller intervals of length $\le 2^{-n}$, so that the given estimate can be applied to each subinterval. Making the subintervals much smaller leads to a worse estimate, therefore we choose the partition such that all but one subinterval have the exact length $2^{-n}$.)
For $x < y$ and $n \ge N$ define a partition
$$
x=x_1 < x_2 < \cdots < x_m = y
$$
such that $x_j-x_{j-1} = 2^{-n}$ for $1 \le j \le m-1$, and $0 < x_m-x_{m-1} \le 2^{-n}$. Then
$$
 |f(x) - f(y)| \le \sum_{j=1}^m |f(x_j) - f(x_{j-1})| \le m 2^{-n} 
$$
and
$$
 |x-y| \ge \sum_{j=1}^{m-1} (x_j-x_{j-1}) = (m-1)2^{-n} \, ,
$$
so that
$$
  |f(x) - f(y)| \le |x-y| + 2^{-n} \, .
$$
This holds for all integers $n \ge N$. Taking the limit $n \to \infty$ gives the desired estimate.
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