The question: The function $f(x) = x^\alpha $ belongs to the Lipschitz class of order $\alpha$ gives an example of $\alpha$-lipschitz function for $\alpha \in ]0,1[$. This is $x^{\alpha}$.
I guess this does not hold when $\alpha > 1$. What is an easy argument to show this? Also, what are examples of $\alpha$-Lipstchitz functions when $\alpha > 1$?
Recall $\alpha$-Lipschitz means there is a $K$ such that $\|f(x)-f(y) \| \le K \| x-y \|^{\alpha}$.
The correct name is $\alpha$-Hölder, not Lipschitz. Lipschitz refers to the case $\alpha=1$. Anyway, for $\alpha>1$, you can show that $\alpha$-Hölder functions defined on "good sets" (see below) are constant.
Assume that $E$ is a convex set. Obviously, $\mathbb{R}$, $\mathbb{R}^n$ or open balls in these spaces all are convex sets.
$\forall y \in E$, our assumption implies $$\forall x\in E: \,\,\,\|f(x)-f(y) \| \le K \| x-y \|^{\alpha}$$
From the properties of real numbers we know that when $\alpha>1$,
$$\forall r\in\mathbb{R}:0<r<1 \implies r^{\alpha}<r$$
Now given $\epsilon>0$, take $\delta=\frac{1}{2}\min\{1,\frac{2\epsilon}{K}\}$. This ensures that $\delta < 1$ holds. Therefore,
$$\| x-y \|< \delta \implies \| x-y \|^{\alpha}<\delta^{\alpha}<\delta<\frac{\epsilon}{K}$$
$$\|f(x)-f(y) \| \le K \| x-y \|^{\alpha} < K \frac{\epsilon}K = \epsilon$$
Since the above inequality is true for all $\epsilon >0$, we get $\|f(x)-f(y) \|=0$ when $\|x-y\|< \delta$.
Hence $f(x)=f(y)$ for all $x\in E \cap B_{\delta}(y)$ for all $y\in E$, i.e. $f$ is locally constant.
If $d(x,y) \geq \delta$, since $d(x,y)$ is a finite number and $E$ is convex, you can choose $n$ points $\{z_0=x,z_1,\cdots,z_n=y\} \in E$ to connect $x$ to $y$ such that $$d(z_i,z_{i+1})<\delta \text{ for } i=0,1,\cdots,n-1$$
Then $\forall i\in\{1,\cdots,n-1\}: \,f(x)=f(z_i)=f(y)$, which means that $f(x)=f(y)$ for all $x,y \in E$. Hence, $f$ is constant on $E$.
Generalization to connected sets
Using the theorem in topology that locally constant functions on connected spaces are constant, we can assume $E$ to be connected only. Our argument will be simplified then. We won't need intermediate points anymore. Instead, we argue that we have shown $f$ to be locally constant at each point $y \in E$. Since $E$ is connected, it must be constant on all of $E$.
An example of a bad set
To see an example of a bad set $E$, take $E = (-\infty,-1] \cup [+1,+\infty) \subset \mathbb{R}$ and let $f: E \to \{\pm 1\}$ be $$f(x)=\begin{cases} +1 & \text{if}\, x\geq +1 \\ -1 & \text{if}\, x\leq -1\end{cases}$$
Then the assumptions of being $\alpha$-Hölder are satisfied for $K=1$, but $f$ is not constant obviously. So, more generally, one can say that if $E$ has several connected components, then $f$ is constant on each connected component.
