I was wondering if $f:D\rightarrow\mathbb{R}$, where $D$ is bounded, and $f$ is $\alpha-$Holder continuous for all $\alpha\in(0,1)$, does it imply that $f$ is Lipschitz? Thanks for any help.
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No, we can let $$ f(x) =\begin{cases} -x\ln x & x\in (0,1/e]\\ \quad 0 & x=0\end{cases},$$ then $f$ is strictly increasing on $[0,1/e],$ while $f'(x)$ is strictly decreasing on $(0,1/e].$
Claim: For $x\in [0,1/e]$ and $h \in [0,1/e-x],$ $f(x+h)-f(x)\le f(h).$
Proof: Fix such an $x.$ Consider $g(h) = f(x+h)-f(x)- f(h).$ Then $g(0)= 0$ and $g'(h) \le 0$ on $[0,1/e-x].$ Therefore $g(h) \le 0$ on $[0,1/e-x].$
Thus on the interval $[0,1/e]$, we have, for appropriate $h>0,$
$$\frac{f(x+h)- f(x)}{h^\alpha} \le \frac{f(h)}{h^\alpha} = -h^{1-\alpha}\ln h.$$
This is bounded if $\alpha \in (0,1).$ But $f(h)/h = - \ln h.$ Thus $f$ is $\alpha$-Holder continuous for $\alpha \in (0,1)$, but not for $\alpha = 1.$
