Is this an equivalent definition of Hölder continuity?

Question

I am wondering if this is an equivalent definition of Hölder continuity for a real function $f:\mathbb{R}\rightarrow \mathbb{R}$:

Assume that there exists $\Delta,\alpha,K>0$ such that if $|x-y|<\Delta$ then

$$|f(x)-f(y)|\le K|x-y|^\alpha.$$

So in order to show that function is Hölder continuous we need to show that there exists $K_2,\alpha_2$ such that

$$|f(x)-f(y)|\le K_2|x-y|^{\alpha_2}.$$

I think it holds if we also require the domain of the function to be compact. Do you agree with this? But if the domain is not compact it may fail to hold?

If the domain is compact this is trivial, with $\alpha_2=\alpha$. — David C. Ullrich, May 17 '21 at 22:41
@DavidC.Ullrich Could you also please tell me what $K_2$ would be in that case? — user394334, May 17 '21 at 22:44

score 1 · Answer 1 · answered May 17 '21 at 23:37

I don't have reputation for comment, I will write something more and post as answer.

The notions you describe are different: the first property is called locally Hölder, similar definition exists also for locally Lipschitz function. For example $f(x)=x^2$ is locally Hölder but not globally. As usual when you have a compact set you can have a uniform estimate for the $K$ and get global Hölder property (very much as Heine Cantor theorem). See for example In $\mathbb{R}^n$, locally lipschitz on compact set implies lipschitz

If you don't have compactness you may fail global Hölder for any exponent $\alpha$, take for example the exponential function.

Is this an equivalent definition of Hölder continuity?

1 Answers1