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 $\alpha >0$.