Holder continuity

Question

Let $A = \{k2^{-n} : 0 \leq k \leq 2^n, n \geq 1\}$ and $f$ be a function defined on $A$ satisfying $$\sup_{n,k} c^n|f((k+1)2^{-n}) - f(k2^{-n})| < \infty$$ for some $c > 1.$ Then there exists unique a continuous function $g : [0,1] \rightarrow \mathbb{R}$ such that $g(t) = f(t) $ for all $t \in A$. Moreover, $g$ is also Holder continuous with exponent $\log_2c.$

I think that this might be a well-known theorem, but I cannot find its name or reference about it. I will be appreciate if anyone can suggest me how to prove it, or provide me the reference about its proof.

Answer 1

Define $M$ to be the supremum of the quantity above.

For $n \in \Bbb N$, write $A_n:=\{ k2^{-n} : 0 \leq k \leq 2^n\}$, so $A = \bigcup_n A_n$.

From now on fix some $n \in \Bbb N$ and some $x,y \in A_n$ with $x<y$.

For $k \leq n$ define $x_k:= \min\{ r \in A_k: r\geq x\}$ and $y_k:= \max\{ r \in A_k: r \leq y\}$. Let $N:= \min\{ k: x_k < y_k\}$. Note that $|x_{k+1}-x_k| \leq 2^{-k}$, similarly $|y_{k+1}-y_k| \leq 2^{-k}$, and also $|x_N-y_N| =2^{-N}$. Also note that $x_n=x$ and $y_n=y$.

Note that $$|f(y)-f(x)| \leq \sum_{k=N}^{n-1} |f(y_{k+1})-f(y_k)|+|f(y_N)-f(x_N)| + \sum_{k=N}^{n-1}|f(x_{k+1})-f(x_k)|$$$$\leq \sum_{k=N}^{n-1} Mc^{-k} +Mc^{-N} + \sum_{k=N}^{n-1} Mc^{-k} \leq 3 M \sum_{k=N}^{\infty} c^{-k} = \frac{3M}{1-c^{-1}} \cdot c^{-N}$$

But by the definition of $N$, it follows that $|y-x| \geq |y_N-x_N| =2^{-N}$, and therefore $c^{-N} = 2^{-N \log_2 c} \leq |y-x|^{\log_2 c}$. Therefore, we showed that for all $x,y \in A$ $$|f(y)-f(x)| \leq \frac{3M}{1-c^{-1}} |y-x|^{\log_2 c}$$ so it follows that we can extend $f$ continuously to a Holder-$\log_2 c$ function. Uniqueness is an easy topology exercise.