# A uniform continuous function which is not Hölder continuous

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 $\alpha >0$.

• An example is given in the Wiki page. May 12 '12 at 17:23

As mentioned in the wiki entry, the function defined by $f(x)=\cases{{1\over \ln x},&$0<x\le 1/2$\cr \strut0, &$x=0$}$ is an example of a uniformly continuous function that is not Hölder continuous for any $\alpha>0$.
$f$ is continuous on $[0,1/2]$; and thus, since $[0,1/2]$ is compact, uniformly continuous on $[0,1/2]$.
But, $f$ is not Hölder continuous for any $\alpha>0$ at $x=0$. If it were, then there would exist positive $C$ and $\alpha$ such that $\bigl|0- {1\over \ln x}\bigr|\le C|x|^\alpha$ for all $0< x\le1/2$. But then, we would have $C|x|^\alpha |\ln x|\ge 1$, which can't happen as $\lim\limits_{x\rightarrow0^+}C|x|^\alpha |\ln x|=0$.
According to the Wiki definition, $f$ is Hölder continuous for $\alpha=0$. That is, it is bounded. But one may extend $f$ to an unbounded, uniformly continuous function on $\Bbb R^+\cup\{0\}$ which is still not Hölder continuous at $x=0$.