6
$\begingroup$

Can someone give me an example of a function $f: I\to \mathbb R$ which is uniform continuous but not Hölder continuous (for any $\alpha$)? Here $I$ is an interval. If $I$ is closed and bounded then we are essentially assuming that $f$ is merely continuous.

By saying that $f$ is not Hölder continuous for any $\alpha$, I mean for all $\alpha >0$,

$$\sup_{x,y\in I, x\neq y} \frac{|f(x) - f(y)|}{|x-y|^\alpha} = \infty.$$

That is, I need to find a function $f$ so that for all $\alpha$ and $M>0$, there are $x, y\in I$ so that

$$ \frac{|f(x) - f(y)|}{|x-y|^\alpha} \ge M.$$

I know that $f(x) = x^\alpha$ are $\alpha$-Hölder continuous. Thus, in some sense, I am looking for $f$ which are worst than the functions $x^\alpha$ for all $\alpha >0$.

  • 7
    An example is given in the [Wiki page](http://en.wikipedia.org/wiki/H%C3%B6lder_condition).2012-05-12

1 Answers 1