Can anybody give me an example of function which is holder continuous but is not bounded variation function?
Thanks
I think that the function $\displaystyle x\sin\frac1x$ will satisfy both conditions. It is not of bounded variation, because it varies by an amount of approximately 1/k in each interval of the form $\displaystyle \Bigl[\frac1{(2k+\frac12)\pi},\frac1{(2k-\frac12)\pi}\Bigr]$. And it ought to satisfy a Hölder continuity condition with exponent 1/2 (or maybe a bit less).
I haven't checked carefully that the above example works. For an alternative construction of an example which definitely does work, look at q.2 in this pdf file.
I have a problem with this example, I don't really understand what the function in the pdf is supposed to be like. But more importantly I have this:
Since $\displaystyle W^{1,p}(0,1)=C^{0,\alpha}(0,1)$ as sets (obviously we pick just one representative in the class of $\displaystyle u\in W^{1,p}(0,1)$) for $\displaystyle p> 1, \alpha = 1-\frac{1}{p}$ and because $\displaystyle W^{1,p}(0,1)\subset W^{1,1}(0,1)=AC(0,1)\subset BV(0,1)$ for all $\displaystyle p\geq 1$ (the only thing we need here is the interval to be bounded) so, since we might as well have picked $\displaystyle [0,1]$ instead, we would get that every Hölder continous function over a bounded interval is of bounded variation, which would contradict your examples.
I'm not sure I haven't made a mistake, but not really understanding the pdf example I can't say for sure where I made a mistake (or where he did).