A function is Log-Lipschitz if there exists a constant $C$ such that \begin{equation} |u(x) - u(y)| \le C|x-y| \log|x-y| \end{equation} Is a Log-Lipschitz function $C^{0,\alpha}$ for any $\alpha \in (0,1) $(Hölder continuous)? If you need, assume hypothesis. Thank you.
Does Log-Lipschitz regularity imply Hölder continuity?
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1Dejan, No, because we can assume a C<0 such that satisfy this condition, yet. – 2018-05-18
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Yes, it is -- assuming that you think and act locally. Think in terms of moduli of continuity $\omega$, i.e., functions such that $|u(x)-u(y)|\le \omega(|x-y|)$. The function $\delta\log (1/\delta)$ is smaller than $\delta^{\alpha}$ ($\alpha<1$) near $0$.
Edit: For $|x-y| < 1$ it is clear that $\log|x-y| < 0$. Commonly this is fixed by adding the modulus: $|\log |x-y||$. In the notation of moduli of continuity the issue is resolved by writing $\log(1/\delta) = - \log(\delta)$, where $\delta$ is small.
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0@Marcos As other people already pointed out, logarithm is negative for small values of its argument. Do you see why this is a problem for the inequality that you stated? – 2012-06-16
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I'm not answering the question, just pointing out that that $\log|x-y|$ should be replaced by $|\log|x-y||$, otherwise the function $u$ is just a constant unless of course $C<0$!