Property of function $\varphi(x)=|x|$ on $\mathbb{R}$

Question

Define $\varphi(x)=|x|$ on $[-1,1]$ and extend the definition of $\varphi(x)$ to all real $x$ by requiring that $\varphi(x+2)=\varphi(x).$ How do you prove that for any $s,t$ $$ |\varphi(s)-\varphi(t)|\leqslant |s-t|? $$

I was going to do the following: For any $s,t$ exists $n,m$ such that $s=2n+\theta_s, t=2m+\theta_t$, where $\theta_s, \theta_t\in [-1,+1).$ Then $$|\varphi(s)-\varphi(t)|=|\varphi(s-2n)-\varphi(t-2m)|=|\varphi(\theta_s)-\varphi(\theta_t)|=||\theta_s|-|\theta_t||=$$$$=||s-2n|-|t-2m||=...$$ and I am stuck.

Answer 1

Note that $\phi(x) = \min_{k \in \mathbb{Z}} |x-2k|$.

We have $|x-2k| \le |y-2k| + |x-y|$. Hence $\phi(x) \le |y-2k| + |x-y|$, and since this holds for all $k$ we have $\phi(x) \le \phi(y) + |x-y|$. Repeating this with $x,y$ interchanged gives the desired result.

Answer 2

for any $s,t\in \mathbb R$, $$\phi(s)=\phi(s+2)=|s+2|\leq |s-t|+|t+2|=|s-t|+\phi(t+2)=|s-t|+\phi(t) $$
$$\phi(s)-\phi(t) \leq |s-t|$$

By reversing the role of $s$ and $t$, we will get, $$\phi(t)-\phi(s) \leq |s-t|$$

Which shows that,$$\phi(s)-\phi(t)| \leq |s-t|$$