I Hölder and log-Hölder continuity

[hilbert2]

TL;DR Summary

About the difference of Hölder and log-Hölder continuity.

Now, there's this conventional definition of the Hölder continuity of a function $f$ defined on $[a,b]\subset\mathbb{R}$:

For some real numbers $C>0$ and $\alpha >0$, and any $x,y\in [a,b]$, $|f(x) - f(y)|<C|x-y|^{\alpha}$.

However, this does not include functions like $f(x) = -\frac{1}{\log x}$ because its value goes to zero "too suddenly" when approaching $x=0$ from the positive side. This kind of functions are, however, log-Hölder continuous which is defined as

For some real number $C>0$, and any $x,y\in [a,b]$, $|f(x) - f(y)|<-\frac{C}{\log |x-y|}$,

or something like that.

But, how about a function like $f(x)=\frac{1}{\log (-\log x)}$ ? Is this also log-Hölder continuous? Or if it isn't, is there some less restrictive version the log-Hölder condition (defined in some actual publication or textbook) that includes also that function?

Edit: Note that here I define $f(0) = 0$ and the interval $[a,b]$ is supposed to contain $0$, usually as the left endpoint.

 

[hilbert2]

There seems to be a continuity property less restrictive than log-Hölder - the $\gamma$-log-Lipschitz property:

For some $C,M,\gamma \in \mathbb{R}$ and any $x,y\in [a,b]$ with $|x-y|<1$, the inequality

$\displaystyle|f(x) - f(y)|\leq C |x-y |\left(\log \frac{M}{|x-y|}\right)^\gamma$ holds.

If this is supposed to be a better upper limit than the log-Hölder, we probably have $C>1$ and $\gamma \in ]0,1[$.

Now if I try to show that the function $f(x) = \frac{1}{\log (-\log x)}$ doesn't even have this property on an interval containing $0$, I can choose $y=0$ and set

$x=\exp (-\exp A)$

which leads to

$\displaystyle|f(x)| = \frac{1}{|A|} \leq \exp (-\exp A)\left[\log M + \exp A\right]^\gamma$

Then to prove the claim by contradiction, I should find some value of $A\in\mathbb{R}$ so that the opposite

$\displaystyle|f(x)| = \frac{1}{|A|} > \exp (-\exp A)\left[\log M + \exp A\right]^\gamma$

is true. It shouldn't be impossible because the $\exp(-\exp A)$ on RHS approaches zero really fast when $A$ is increased, but I'm not sure yet how to show this.

 

[soloenergy]

Yes, the function $f(x)=\frac{1}{\log (-\log x)}$ is also log-Hölder continuous. This is because it satisfies the condition that for some real number $C>0$ and any $x,y\in [a,b]$, $|f(x) - f(y)|<-\frac{C}{\log |x-y|}$. In this case, the value of $f(x)$ does not go to zero "too suddenly" as $x$ approaches $0$ from the positive side, so it satisfies the log-Hölder condition.

There may be a less restrictive version of the log-Hölder condition that includes this function, but it would depend on the specific context in which it is being used. It is best to consult a publication or textbook on the topic to see if there is a version that includes this function.