# If a function is log-lipschitz, then it is $\alpha$-Hölder for all 0<$\alpha$<1 and isn't Lipschitz continuous

Consider $C^{0,\alpha}$ the class of $\alpha$-Hölder continuous functions. Let $\Omega \subset \mathbb{R}$ be a bounded subset and let $u: \Omega \rightarrow \mathbb{R}$ a continuous function. Prove that if there is $C\in \mathbb{R}$ such that $$|u(x)-u(y)| \leq C \left|x-y\right| \ln {\frac{1}{\left|x-y\right|}}$$ For $|x-y|<\frac{1}{2}$, then $u\in C^{0,\alpha}$, i.e., $\forall \alpha \in (0,1)\exists K_{\alpha}\in \mathbb{R}$ such that $$|u(x)-u(y)| \leq K_{\alpha} \left|x-y\right|^{\alpha}$$ My doubt:

First: How I prove that $K$ depends on $\alpha$?

Second: I think that we can use the subbaditive property of $ln$ function. But I can't do this

Third: How I prove that it isn't Lipschitz continuous?

Forth: why is the supremum of the distance $1/2$? Can I improve the distance, i.e. $\exists d>1/2$ such that the inequality is true for $1/2<|x-y|<d$? Or, can I suppose that this inequality are well-defined at $|x-y|<1$?

This definition I found in this link, page 80.

• Do you need to prove $|u(x)-u(y)| \leq K_{\alpha} \left|x-y\right|^{\alpha}$ only when $|x-y| < 1/2$? What are you talking about in 4-th question? What is $d$?.. Commented May 20, 2018 at 16:42
• I saw the statement that $|x-y|<1/2$ in a paper, but I don't no why this is necessary. I think that I can improve to a greater distance. Commented May 20, 2018 at 21:06
• Take $u(x) = const, x \in (0,1)$, then property $|u(x)-u(y)| \leq C \left|x-y\right| \ln {\frac{1}{\left|x-y\right|}}$ is clearly satisfied even for those $x,y$ for which $|x - y| > 1/2$. So it's definitely not a necessary condition. Commented May 20, 2018 at 22:53
• If you saw it in the paper, what don't you provide a reference to this paper in the question? :) Commented May 20, 2018 at 22:54
• this link, page 80 Commented May 21, 2018 at 5:12

First: How I prove that K depends on $\alpha$?

I will show that if $u$ is Log-Lipschitz, then it's $\alpha$-Hölder for $\alpha \in (0,1)$. Actually, this is pretty straightforward. Let's assume that $$|u(x)-u(y)| \leq C \left|x-y\right| \ln {\frac{1}{\left|x-y\right|}} \le K_\alpha |x - y|^\alpha$$ for some $\alpha \in (0,1)$. Then we must have $$|x-y|^{1-\alpha}\ln {\frac{1}{\left|x-y\right|}} \le \frac{K_\alpha}{C}.$$ Consider a function $f(t) = t^{1-\alpha}\ln (1/t) = - t^{1-\alpha}\ln t$ for $t < 1/2$. Restriction $t < 1/2$ comes from the definition you provided (see link in the question). This function behaves nicely in zero, since $$\lim_{t \to +0} -t^{1-\alpha} \ln t = 0$$ and so it's bounded on the interval $(0,1/2)$. That's why there exist a constant $M_\alpha$ such that $$t^{1-\alpha} \ln (1/t) < M_\alpha$$ for all $t\in (0,1/2)$. Now we see that one can take $K_\alpha = C\cdot M_\alpha$ and $u$ is $\alpha$-Hölder continuous.

Third: How I prove that it isn't Lipschitz continuous?

It can be Lipschitz continuous. For example, take a constant function. Though, I think, you can think of an example of the function that is log-Lipschitz but not Lipschitz.

Forth: why is the supremum of the distance 1/2?

Because it is a definition introduced by authors of the article. I suppose they are interested only in the local behavior of functions. Since such conditions like Lipschitz or Hölder measure local smoothness of the function, they consider only "close" points.

• Thanks @Virtuoz , I find this same answer but You confirm with elegant solution. Thanks, again Commented May 21, 2018 at 15:22