2

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.

azimut
  • 22,696
user29999
  • 5,211

2 Answers2

4

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.

Daniel Lorenz
  • 3
  • 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 00:56
  • @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 $\alpha<1$. –  Jun 16 '12 at 01:11
  • Thanks, that does clarify some things. – Dejan Govc Jun 16 '12 at 01:21
  • 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? –  Jun 16 '12 at 16:04
3

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 $C<0$!

HorizonsMaths
  • 16,526