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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.

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$\log|x-y|$ is negative for $|x-y|<1$. Doesn't that cause a problem? –  Dejan Govc Jun 15 '12 at 23:56
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up vote 2 down vote accepted

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 $0$.

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By "locally", do you by any chance mean something like: $u:D\to\Bbb R^n$ is log-Lipschitz if for each $x\in D$ there exist a constant $C\in\Bbb R$ and an open neighborhood $U$ of $x$, such that for all $y\in U$ the inequality $|u(x) - u(y)| \le C|x-y| \log|x-y|$ holds? –  Dejan Govc Jun 16 '12 at 0:56
    
@DejanGovc Not really: I just meant that Holder continuity is usually considered for not-too-large distances, say $|x-y|<1$. Otherwise we'd have to say that $f(x)=x$ is not Holder continuous on $\mathbb R$ with exponents $\alpha<1$. –  user31373 Jun 16 '12 at 1:11
    
Thanks, that does clarify some things. –  Dejan Govc Jun 16 '12 at 1:21
    
Thank you but, insted $\delta \log(1/\delta) $ should not be $\delta \log(\delta) $? –  user29999 Jun 16 '12 at 14:10
    
@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? –  user31373 Jun 16 '12 at 16:04
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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 $C<0$!

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