Exponents for Hölder functions on metric spaces

Question

Sometimes people talk about Hölder functions on metric spaces without mentioning the allowed range for the exponent. On manifolds, it's traditional to assume an exponent in $[0, 1]$, since all functions with higher exponents are constant. On general metric spaces, however, there are apparently interesting functions with higher exponents.

If I see a result about Hölder functions on metric spaces, with no range of exponents specified, is it safe to assume the result holds for all non-negative exponents? If not, are there any techniques that might help me check what range is allowed? (And what about negative exponents? Silly as they sound, I've learned never to bet against the perversity of mathematics...)

p.s. While writing this question, I came across a related one that could use some attention. If you have something to say about this question, you might be able to contribute to that one as well.

Answer 1

It seems as a general rule of thumb it's safe to assume "$\alpha$-Hölder" means (possibly locally) $\alpha$-Hölder with $\alpha\in]0,1[$; Lipschitz functions are qualitatively much stronger than $\alpha$-Hölder functions with $\alpha<1$ (e.g. in hyperbolic dynamics the stable and unstable distributions are automatically Hölder but not Lipschitz; see Why Do We Care About Hölder Continuity?), and existence of non-constant functions that are $\alpha$-Hölder with $\alpha>1$ indicates that the underlying metric geometry is subtler than Euclidean. For instance here is an excerpt from the appendix by Semmes to Gromov's Metric Structures for Riemannian and Non-Riemannian Spaces (p.423):

When $\alpha = 1$ we have already christened this the Lipschitz condition, but the latter is special and deserves a special name. Traditionally one does not consider $\alpha > 1$ because on many spaces Hölder continuous functions of order $\alpha > 1$ are all constant. This happens on Euclidean spaces, for instance, because Hölder continuity of order $\alpha > 1$ implies that the first derivatives of the function exist and vanish everywhere.

Notice that if we replace $(M,d(x,y))$ with the snowflake $(M,d(x,y)^\alpha)$ then [the $\alpha$-Hölder condition for $d$] becomes simply the Lipschitz condition with respect to the new metric. So all theorems about Lipschitz functions on metric spaces apply to Hölder continuous functions. The reverse fails in a strong way, as we shall see.

Typically when there are plenty of functions which satisfy a condition like [the $\alpha$-Hölder condition for $d$] with $\alpha > 1$ it means that the space that we are working on is already the result of applying the snowflake functor to a metric space. One can formulate precise versions of this statement which are easy to prove.

To detect the range of $\alpha$ it might be useful to be mindful of where the convexity/concavity of $t\mapsto t^\alpha$ is used (if at all).

Regarding non-negative exponents; (for all its worth) I personally never saw the phrase "$0$-Hölder"; it's hard to imagine "$0$-Hölder" meaning anything other than bounded. Functions with negative exponents could be useful as "moduli of total discontinuity"; though it's unclear to me if such things are studied.

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Note that there is a similar situation with $L^p$ spaces; typically $p\in[1,\infty[$, possibly $p\in[1,\infty]$, but at times $p\in]0,\infty]$ (or even $p=0$ too, denoting the space of measurable functions, as in Notation for the set of measurable functions and the related quotient space).