$f: x \mapsto ax + b$ on $\mathbb{R}$ ($a,b\in\mathbb{R}$)
$f: x \mapsto x^2$ on $(0, 1)$
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carmichael561
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Saoirse
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$x \mapsto ax + b$ is Lipschitz continuous with constant $a$, and equality holds.
$x \mapsto x^2$ is Lipschitz continuous. If $0 < x < y < 1$, then $y^2 - x^2 = (y-x)(y+x)$ can be controlled below what multiple of $y-x$?
Henricus V.
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And in fact $x\mapsto x^2$ is Lipschitz on $[0,1]$. – carmichael561 Apr 08 '16 at 04:41