Specific question about Lipschitz condition to be satisfied Given the function 
$$f(x,y) = \begin{cases} 
|y|  & \text{ if $|y|$} \le 1 \\
1 & \text{ if $y$} \gt 1 \text{ or } y\lt -1
\end{cases} $$
Determine whether the function satisfy a uniform Lipschitz condition with respect to $y$ and provide a Lipschitz constant if it exists.
Try:
Lipschitz w.r.t. $y$ means $$|f(x,y_1) - f(x,y_2)| \le L |y_1-y_2|$$ 
I know that $\max{f(x,y)}=1; \min{f(x,y)=0}$, which implies that $|f(x,y_1) - f(x,y_2)| \le 1$ for all $y \in \mathbb{R}$. How can I relate this to $|y_1-y_2|$ ?
$$y_1= y_2+y_1 -y_2 \implies |y_1| \le |y-2|+|y_1-y_2| \implies |y_1| - |y_2| \le |y_1-y_2| $$ 
$$y_2= y_1+y_2 -y_1 \implies |y_2| \le |y_1|+|y_1-y_2| \implies |y_2| - |y_1| \le |y_1-y_2| $$
$$-|y_1 - y_2| \le |y_1|-|y_2| $$ Putting this together gives 
$$ -|y_1 - y_2| \le |y_1|-|y_2|  \le |y_1 -y_2| \implies | |y_1|-|y_2| |  \le |y_1 -y_2| $$ so the function $f(x,y)=|y|)$ is Lipschitz with Lipschitz constant 1. My question now is: How can I show that the whole function (including $y>1$) satisfies Lipschitz condition?

If $\left|y_1\right|,\left|y_2\right|\le 1$,
$$|f(x,y_1) - f(x,y_2)|= |\left|y_1\right|-\left|y_2\right||\le |y_1-y_2|$$
If $\left|y_1\right|,\left|y_2\right|>1$,
$$|f(x,y_1) - f(x,y_2)|= |1-1|=0\le |y_1-y_2|$$
If $\left|y_1\right|\le 1<\left|y_2\right|$,
$$|f(x,y_1) - f(x,y_2)|= |\left|y_1\right|-1|=1-\left|y_1\right|<\left|y_2\right|-\left|y_1\right|\le |y_1-y_2|$$
The proof is pretty much complete
