Let $f:[0,2)\to \mathbb{R}$ such that $f$ restricted to $(1,2)$ is Lipschitz. Prove that $f$ is uniformly continuous?

Question

I am trying to solve the following problem:

Let $f:[0,2)\to \mathbb{R}$ continuous such that $f$ restricted to $(1,2)$ is Lipschitz. Prove that $f$ is uniformly continuous.

I know that as $f:(1,2)\to \mathbb{R}$ is Lipschitz, it is uniformly continuous. But why does $f:(1,2)\to \mathbb{R}$ being Lipschitz forces $f:[0,2)\to \mathbb{R}$ to be uniformly continuous?

Answer 1

All you need are these three facts:

1. a continuous function on a compact set is uniformly continuous;
2. all Lipschitz functions are uniformly continuous;
3. if a function is uniformly continuous on $A$ and $B$, and continuous on $A\cup B$, then it is uniformly continuous on $A\cup B$.

Now take $A=[0,\frac32]$ and $B=(1,2)$.

Answer 2

In general I dont think that this statement is true. One could for example take the following map: $$f:[0,2)\to \mathbb{R}, x\mapsto \begin{cases} 1 & x\in \mathbb{Q}\cap [0,1] \\ 0 & x\in (\mathbb{R}\setminus\mathbb{Q}) \cap [0,1] \\ x & x\in (1,2) \end{cases}$$ then this map is Lipschitz-continous on $[1,2)$ however not (unif.) continous on $[0,2)$.