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Proof:

Let $a,b\in \mathbb{R}$ with $a<b$. $f$ is Lipschitz continuous since $\forall x,y\in [a,b]:$

$$|f(x)-f(y)|=|x^2-y^2|=|x+y|\cdot |x-y| \overset{(*)}\leq (|x|+|y|)\cdot |x-y| \leq 2\max \{|x|,|y|\}\cdot |x-y|$$ Finally $2\max \{|x|,|y|\}\cdot |x-y|\leq 2\max \{|a|,|b|\}\cdot |x-y|$.

I understand nearly everything, except why $\max \{|a|,|b|\}$ is being multiplied by $2$. What's the reasoning behind that?

Kind regards

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    $\begingroup$ since $x,y\in[a,b]$, then $x+y\le2b$ $\endgroup$ – WrongMath Jul 16 at 19:50
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Since $x,y\in [a,b],$ it follows that $|x|,|y|\leq \max\{|a|,|b|\}.$ Hence, $$|x|+|y|\leq 2\max\{|x|,|y|\}\leq 2\max\{|a|,|b|\}.$$

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Without loss of generality, $0\le a<b.$ By the mean value theorem, there is a $c\in (x,y)\subset [a,b]$ such that $|f(x)-f(y)|=2c|x-y|<2b|x-y|.$

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