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I have recently been studying $\alpha$-Holder continuous paths for $\alpha \in (0,1]$, where when we $\alpha = 1$, we have Lipschitz continuity.

My definition of a smooth path is one that has infinitely many continuous derivatives.

My question is: can a path that is $\alpha$-Holder continuous for any $\alpha \in (0,1]$ be smooth? Intuitively I thought the answer was no. My thought process was that the sample paths of Brownian motion have $\alpha$ holder continuity for $\alpha = 1/2$. However, these are also famously non-differentiable - and i guess that this holds for $\alpha < \frac{1}{2}$.

I also know that not all smooth functions are Lipschitz continuous. So is there a relation in general between $\alpha$ and smoothness?

Thanks

Jamal
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  • On this wikipedia page it lists continuously differentiable functions as subset of $\alpha$-holder continuous functions. Clearly, smooth functions are subsets of continuously differentiable functions. Caveat here is that these conditions only hold for compact (i.e. closed & bounded) domain. But per definition a path has a closed interval as domain right? – student91 Mar 03 '23 at 14:26
  • @student91 So then it would be possible to have a continuously differentiable function that is holder continuous with $\alpha <1/2$? Yes a path has a closed interval as domain, usually $[0,T]$. – Jamal Mar 03 '23 at 15:12
  • Yes, the constant function $f(x)=0$ is the most boring example and there's also more interesting examples. – student91 Mar 03 '23 at 15:18
  • @student91 sorry by $\alpha < 1/2$ I mean a function that is not holder continuous for $\alpha \geq 1/2$, since obviously if a function is alpha holder continuous for $\alpha = a$, then it is alpha holder continuous for all $\alpha \leq a$. – Jamal Mar 03 '23 at 15:29
  • Then the answer by Nick should address that question. If it does, don't forget to accept it – student91 Mar 03 '23 at 15:30

1 Answers1

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In general, no. If a path is Lipschitz, the in fact it is $\alpha-$Hölder continuous for all $\alpha\in(0,1)$. But Lipschitz paths are not guaranteed to be smooth, just consider as a simple example the graph of $f(x)=|x|$ around the origin.

Nick
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  • Thank you, so there appears to be no real relation between $\alpha$ - holder continuity and smoothness? A function that is Lipschitz may or may not be smooth while a function that is $\alpha$-holder continuous ONLY for $\alpha <= a$ may or may not be smooth. – Jamal Mar 03 '23 at 15:30
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    @Jamal: In fact, a Lipschitz function $f:{\mathbb R} \rightarrow {\mathbb R}$ -- even if it is also strictly increasing -- can fail to be differentiable at every point of a set that is dense in ${\mathbb R}.$ Moreover, "dense in $\mathbb R$" can be significantly strengthened (e.g. the points of non-differentiability can have cardinality continuum in every interval, and much more than even this). See this answer for more details. – Dave L. Renfro Mar 03 '23 at 15:41
  • @Jamal: You might also keep in mind that to say there is no relation would mean that the concepts are logically independent of each other, which is not the case in this situation. But there is a relation: smoothness implies $\alpha$-Hölder continuity for all $\alpha$. The point is that the converse implication is false, as one shows using counterexamples. – Lee Mosher Mar 03 '23 at 16:23