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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.

Greetings.

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    Okay I will try to improve till my next question. I appreciate your criticism, thanks @Cantor.2012-11-22

1 Answers 1

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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 $\mathbb{R}$.