Show $f(x) = x^2$ lipschitz on $[0,1]$

Question

I would like to show this result but I am a bit stuck

To show $f(x)$ is lipschitz, show:

$$|x^2 - y^2| \leq L |x-y| \quad \forall x,y \in [0,1]$$

Proceed as usual:

$|x^2 - y^2| = |x-y||x+y|$

But what is $|x+y|?$

Just bound $|x+y|$ on $[0,1]$. Since this is increasing, it is bounded above by 2. — Rellek, Feb 28 '16 at 03:21

score 3 · Accepted Answer · answered Feb 28 '16 at 03:20

3

Hint: $|x+y| \leq |x|+|y|$. Can you continue the last step?

answered Feb 28 '16 at 03:20

DeepSea

Are you suggesting that I should take $x = 1, y= 1$ to get a bound of 2? But wouldn't that make $|x-y|$ part go to zero? – Fraïssé Feb 28 '16 at 03:24
You need to keep $|x-y|$,and bound the $|x+y|$ – DeepSea Feb 28 '16 at 03:28
Can you explain what it means to keep $|x-y|$ but bound $|x+y|$ – Fraïssé Feb 28 '16 at 03:41

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