Is this true that, $u \in \bigcap_{ \alpha<1} C^{0, \alpha}$ implies $u \in C^{0,1}$

Question

Is there a holder continuous function with any order less than 1, but not Lipschitz. Or does $u \in \bigcap_{\alpha<1} C^{0, \alpha}$ imply $u \in C^{0,1}$.

Answer 1

No, a log-Lipschitz function, i.e. one that satisfies $$ |u(x) - u(y)| \le C|x-y| |\log|x-y||$$ is $C^{0,\alpha}$ for any $\alpha<1$, but not Lipschitz. See this MSE post. An example of such a function is

$$|x||\log|x||$$