Prove that if $f$ is an element of Lip $\alpha$, then $f$ is uniformly continuous

Question

I am required to prove that if $f$ is an element of Lip $\alpha$, then $f$ is uniformly continuous. Where Lip means a Lipschitz function. I also have to prove that if $f$ is an element of Lip $\alpha$ and $\alpha>1$ then $f$ is constant where $\alpha$ is actually a greek symbol I dont know how to type

i) Prove that if $f \in$ Lip $\alpha$, then $f$ is uniformly continuous.

ii) Prove that if $f \in $ Lip $\alpha$ and $\alpha > 1 $, then $f$ is constant.

Answer 1

For the second part, if $f \in Lip(\alpha > 1)$, then

$$| f(x) - f(y)| \leq | x - y|^\alpha \ \ \Longrightarrow \ \ \frac{| f(x) - f(y)|}{|x-y|} \leq |x - y|^{\alpha - 1}$$

where $\alpha - 1 > 0$

Now what can you deduce from that?