Bounded and Continuous but not Holder

Question

This is a follow-up question to this post. Does there exist an example of a bounded and continuous but not necessarily uniformly continuous function defined on all of $\mathbb{R}$, which is not H\"{o}lder continuous?

Answer 1

on $[0,1]$ the function $\sqrt x$ is bounded and continuous but not Lipschitz.

Suppose, if possible, $|\sqrt x -\sqrt y| \leq C|x-y|$. Then $|\frac {x-y} {\sqrt x +\sqrt y}| \leq C|x-y|$ which gives $1 \leq C (\sqrt x +\sqrt y)$ and we get a contradiction by letting $x$ and $y$ tend to $0$.

A bounded continuous function which is not Holder continuous of any order is given by $f(x)=\frac 1 {\ln x}$ for $0<x\leq \frac 1 2$ and $0$ for $x=0$. For more information see https://en.wikipedia.org/wiki/H%C3%B6lder_condition