Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can someone give me an example of a function which is uniform continuous on its domain, but not Hölder continuous (for any $\alpha$)?

share|improve this question
6  
An example is given in the Wiki page. –  David Mitra May 12 '12 at 17:23

1 Answer 1

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 $x=0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.