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Let $f$ be a $\alpha$-Hölder function in $\mathbb{R}^n$. Question : does it have distributional derivatives in a $L^p$ space ? (modulo a suitable relationship between $\alpha$ and $n$).

I know the other way around : if we look at elements of Sobolev spaces, then we get a Hölder regularity depending on the dimension $n$ and exponent $p$. Is there a converse there ? When I tried googling the question, I only found an article with an example of nowhere differentiability in the classical sense.

Rick
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1 Answers1

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If you want an easy example, take the Cantor function, which is Hölder for exponents small enough.

If you want a more elaborate example, take this or this function, which are not Lipschitz, but are Hölder for every exponent less than $1$ for careful choices of the parameters (they are is actually quasi-Lipschitz). However they are not BV, so their distributional derivative cannot be in any Sobolev space.

wisefool
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  • thanks for your answer, I wasn't clear on the properties of BV functions. what if I add the property that $n=2$ and $f$ is subharmonic (and let's say $1/2$ hölder) ? could such a function have a distributional derivative not in $L_{\mathrm{loc}}^1$ ? – Rick Oct 05 '14 at 00:39
  • You could just forget BV and look at the first example: the Cantor function is, I think, one of the most famous examples of function for which there is no function which you could call its derivative. – wisefool Oct 05 '14 at 08:28