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?