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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?

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1 Answer 1

up vote 1 down vote accepted

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

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I edited my question, but I think this is what I needed :) –  MSKfdaswplwq Dec 12 '12 at 15:07
    
What about the case if $y_1 >1, y_2<-1$ –  MSKfdaswplwq Dec 12 '12 at 15:24
    
You edited the question to include $y<-1$? –  Nameless Dec 12 '12 at 15:26
    
yes, but its the same case as $y_1, y_2 > 1$ of course –  MSKfdaswplwq Dec 12 '12 at 15:27

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