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Is it legitimate to show the Lipshitz continuity of $f(x)=x^2$ as I did below?

$$|x^2-y^2| \leq L|x-y| $$ w.l.o.g. $x>y.$

$$\Rightarrow x^2-y^2 \leq L|x-y|$$

$$\Leftrightarrow \ \dfrac {x^2-y^2}{x-y} \leq L\\ \Leftrightarrow \ \dfrac {(x-y) \cdot (x+y) }{x-y} \leq L\\ \Leftrightarrow \ x+y \leq L.$$

Define $n=x+y$, then we get: $n \leq L$. After Archimedes' principle we are able to find an $n$ which violates the relation between $n$ and $L$.

EDIT: The function isn't Lipschitz continous after my calculation.


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Yes, it's good (modulo some inaccuracies). You are doing proof by way of contradiction, so it's better if you start by saying suppose $f$ is Lipschitz, then ... . Your statements are true for all $x$ and $y$, and as you pointed out there are $x$ and $y$ such that $x+y>L$ (e.g. $x=L+1$ and $y=L$), a contradiction. – Cantor Nov 22 '12 at 20:35
Thanks for the fast response. Am I allowed to unfold the abs of x² - y² like I did by assuming that x > y w.l.o.g or do I have to prove it for both cases (x>y, x<y)? – optional Nov 22 '12 at 20:41
$|x^2-y^2|=|x-y||x+y|$, this is all you need. – Cantor Nov 22 '12 at 20:42
haha, silly me! Thank you very much. – optional Nov 22 '12 at 20:46
To divide though you assume $x\neq y$. – Cantor Nov 22 '12 at 20:51
up vote 6 down vote accepted

The function $f(x)=x^2$ is Lipschitz continuous on compact sets. For example, on $[-L,L]$, we have $$ \begin{align} |x^2-y^2| &=|x+y|\,|x-y|\\ &\le2L|x-y| \end{align} $$ $f$ is not Lipschitz continuous on all of $\mathbb{R}$: $$ |(x+1)^2-x^2|=|2x+1|\,|(x+1)-x| $$ shows that there is no single bound that works over all $\mathbb{R}$.

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