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I am trying to find out the relation between those spaces. Take $I\subset R$ on the real line. $I$ can be unbounded. Then I have:

We first assume $I$ is bounded. If $u\in C^{0,\alpha}(I)$, for $0<\alpha<1$, then $u$ is uniformly continuous for sure. However, I can not prove $u\in C^{0,\alpha}(I)$ then $u\in AC(I)$, nor $u\in AC(I)$ then $u\in C^{0,\alpha}(I)$. I tried a lot and now I start to think that there are no relations between those two spaces, even if $I$ is bounded.

update: I think I find an example to show that $u\in AC(0,1)\setminus C^{0,\alpha}(0,1)$ by taking $u= x^\beta$ for some $\beta>\alpha$. But I still can not prove the converse, nor find an counterexample.

Moreover, I am wondering that can Holder continuous implies bounded variation? Certainly when $\alpha=1$ we are good. But what about $\alpha<1$, say for $I$ bounded?

Finally, it is clearly that $BV$ can not implies Holder, just take any discontinuous functions for example.

spatially
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1 Answers1

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The Weierstrass function is Hölder continuous, but is not absolutely continuous, and is not of bounded variation either. (Every function of bounded variation is differentiable at almost every point, but the Weierstrass function is nowhere differentiable.)

Conversely, $f(x) = 1/\log x$ is absolutely continuous on $[0,1/2]$ but is not Hölder continuous for any $\alpha\in (0,1)$.

The Lipschitz case, $\alpha=1$, is dramatically different: a Lipschitz function is absolutely continuous, and belongs to $W^{1,p}$ for every $p$.