I want to show that if $f:\mathbb R\longrightarrow \mathbb R$ is $\alpha -$Holder continuous for $\alpha >1$, then $f$ is constant.

This is my proof:

Let $\alpha =1+\varepsilon$. Then, there is a $C$ s.t. $$|f(x)-f(y)|\leq C|x-y||x-y|^\varepsilon\implies \left|\frac{f(x)-f(y)}{x-y}\right|\leq C|x-y|^\varepsilon.$$

I want to says that it implies that $$|f'(y)|=\lim_{x\to y}\left|\frac{f(x)-f(y)}{x-y}\right|=0,$$ but since $f$ is not supposed differentiable, I'm not sure if I can.

  • $\begingroup$ If $x\to y$ then $x-y\to0$, so this is a problem. $\endgroup$ Jul 7, 2016 at 11:59

3 Answers 3


I learned this beautiful proof from my advisor, Piotr Hajlasz (https://mathoverflow.net/users/121665/piotr-hajlasz) that avoids differentiability:

For fixed $x<y$, for a natural number $N$, apply the Hölder assumption to the points $$ x_o=x, x_1=x_0+\frac{y-x}{N}, x_2=x+2\frac{y-x}{N}, \cdots, x_N=x+N\frac{y-x}{N}=y, $$ to see that $$ |f(y)-f(x)|\leq \sum|f(x_{i+1})-f(x_i)| \leq \sum|x_{i+1}-x_i|^\alpha = N \times(\frac{y-x}{N})^\alpha = \frac{(y-x)^{\alpha}}{N^{\alpha-1}}. $$ The left-hand side is independent of $N$, so, by letting $N \rightarrow \infty$, we get $$ |f(y)-f(x)|=0. $$

  • 2
    $\begingroup$ My first thought was "is it beautiful?" My second thought was "yes, damn, it is." Thanks for sharing. $\endgroup$ Sep 22, 2020 at 19:29

Given $x \in \Bbb R$, we have for any $y \neq x$,

$$\left| \frac{f(y) - f(x)}{y-x} \right| \le C |x - y|^{\epsilon}$$

Taking $y \to x$, we get that:

$$\lim_{y \to x} \left| \frac{f(y)-f(x)}{y-x} \right| = 0 $$

Which gives:

$$\lim_{y \to x} \frac{f(y)-f(x)}{y-x} = 0$$

So, $f$ is differentiable at $x$ and $f'(x) = 0$. Therefore $f$ is constant.


The idea is that the derivative is $0$ over an interval.

$f'(x_0)=0\space \forall x_0 \in A \subset \mathbb R$

$|f'(x_0)|=\lim_{h\to 0}\frac{|f(x_0+h)-f(x_0)|}{|h|}\le\lim_{h\to 0}\frac{C|h|^{\alpha}}{|h|}=0$ if $\alpha \gt 1$ .


Not the answer you're looking for? Browse other questions tagged .