Hölder continuity has never appeared in my formal education and the wikipedia article seems insufficiently general. I want to make sure that this definition of Hölder continuity is correct and standard:

Let $f:X \rightarrow Y$ be a map of metric spaces. Then we say $f$ is Hölder continuous with exponent $r \geq 0$ if there exists $C_r$ such that $d_Y(f(x), f(x'))\leq C_r d_X(x, x')^r$

I am specifically asking this to be sure that there is no reason to require the domain to be all of a Euclidean space, as specified in the Wikipedia article, or even Euclidean at all. (Or is it uninteresting unless you at least have a convex body domain in a Euclidean space or something?) Furthermore, it is my observation that although a lot of functions seem to satisfy Hölder conditions for all exponents in intervals of the form $(0, r)$ or $[0, r)$ that actually in general one does not have that Hölder of any exponent implies Hölder of any other exponent. In particular, a Lipschitz function can be not Hölder-1/2. But all Hölder functions are automatically uniformly continuous. Please let me know if all these are correct. I guess being familiar with Lipschitz continuity for a long time I should not be bothered by this, but it disturbs me that Holder continuity can be violated "just because" the function has bad asymptotic behavior, whereas I think of uniformly continuous as a local adjective. Moreover, many mathematicians have, in my opinion, spoken of Hölder continuity as if it's "harder" to have it for higher exponents. Is there some reason for this, or maybe it's just that I'm misreading their intentions?


First of all, yes your definition is correct and the condition makes sense and is interesting for mappings between any metric spaces, not just Euclidean spaces.

Second of all, just like the Lipschitz condition, the Hölder condition is a global condition; it requires that the inequality holds for all pairs of points, no matter how far away they are. As you say, this is in contrast with uniform continuity, which is a local property. Thus, a function can indeed fail to be Lipschitz or Hölder just because of bad asymptotic behavior.

In analysis, however, it is often good enough to have the condition locally. (e.g. locally Lipschitz means for every compact set $K$ in your space, there is a constant $C_K$ for which the Lipschitz condition holds on $K$ with constant $C_K$, and you can make a similar definition of locally $\alpha$-Hölder.) For example, a locally Lipschitz function on Euclidean space is differentiable a.e., just like a global Lipschitz function, and a locally $\alpha$-Hölder map increases Hausdorff dimension by no more than a factor of $1/\alpha$, just like a global $\alpha$-Hölder map.

If you consider this definition, then if $\beta>\alpha$ being locally $\beta$-Hölder is indeed stronger than being locally $\alpha$-Hölder. (On bounded sets, $x^\beta$ is bounded by a constant factor times $x^\alpha$.)

So in many cases, we do consider Hölder-continuity for higher exponents as "better" regularity. Note that (local) Lipschitz regularity gives better properties than (local) $\alpha$-Hölder regularity ($\alpha<1$), for example differentiability almost everywhere.

Hope that helps.


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