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Recall, given a metric space $X$, a function $f:X \rightarrow \mathbb{R}$ has (uniform) modulus of continuity $w:[0,\infty) \rightarrow [0,\infty]$ if $|f(x) - f(y)| < w(|x-y|)$ for all $x,y \in X$.

In particular, $f$ is $K-$Lipschitz $\implies$ $w(\delta) = K\delta$ works (and in particular, we can choose the 'tightest' $K$ can be chosen). And $f$ is uniformly continuous on $X$ $\implies w(\delta) = \epsilon_{\delta}$, where $\epsilon_{\delta} = \inf\{ \epsilon: |f(x)-f(y)| \leq \epsilon \text{ whenever } |x-y| \leq \delta\}$.

Now, for (uniform) continuity, we also require that $\lim_{\delta \rightarrow 0} w(\delta)=0$. Also, given a uniformly continuous function $f$, we can think about associating it with the best case modulus of continuity, $w_f(\delta) := \inf \{w(\delta): w \text{ is a modulus of cty. for } f \}$.

Now, consider the class of uniformly continuous $f$ with `worse-than-linear' or sublinear $w_f$, such that $\lim_{\delta \rightarrow 0} \frac{w_f(\delta)}{\delta} = \infty $, i.e., as points get closer, the distance between them decreases at a worse-than-linear rate.

Similarly, let $f$ have superlinear modulus of continuity if $\lim_{\delta \rightarrow 0} \frac{w_f(\delta)}{\delta} = 0 $.

Question: Like how the notion of Lipschitz functions = functions with linear modulus of continuity, of which functions with bounded derivatives form a subclass, I'm looking for any sort of alternate characterization (sufficient and/or necessary) of the class of superlinear/sublinear classes.

P.S. Hopefully I'm using sublinear and superlinear in the ways that match usual mathematical intuition!

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    $\begingroup$ I would guess that a superlinear modulus of continuity is extremely rare, probably not much more than constant functions. For the sublinear notion, there is of course the class of Hoelder (or little Hoelder) functions and the like. $\endgroup$
    – Hannes
    Commented Aug 6 at 11:51
  • $\begingroup$ For many interesting metric spaces (e.g. length spaces), superlinear implies that the function is constant. (The sufficient condition is: there exists $M \geq 1$ such that given $x, y \in X$ and $\delta > 0$, there exists a finite collection of points $x = z_0, z_1, \ldots, z_N = y$ such that $d(z_i, z_{i+1}) < \delta$ and $\sum d(z_i, z_{i+1}) < M d(x,y)$.) There's even an urban legend based on this fact. $\endgroup$ Commented Aug 6 at 16:09
  • $\begingroup$ @WillieWong that’s hilarious, thank you for sharing that. Short of the amazing results, that’s basically what this post is. $\endgroup$
    – algebroo
    Commented Aug 6 at 16:36

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