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Is the function $x\cdot|x| \equiv \text{sgn}(x)\cdot x^2$ Locally-Lipschitz?

I believe is indeed locally Lipschitz since is continuous ans its slope is bounded on every compact-supported section.... but since $x\cdot|x| \equiv x\sqrt{x^2}$ and $\sqrt{x}$ is not locally Lipschitz, I am not $100\%$ sure... hope you can corroborate it.

daw
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Joako
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    It's continuously differentiable, so it is indeed locally Lipschitz. – Richard Jensen Jun 15 '22 at 08:15
  • @RichardJensen but it isn't differentiable at $x=0$, right? – Joako Nov 26 '22 at 04:06
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    It is differentiable at $x=0$. For $x<0$, the function is $-x^2$, and for $x>0$, the function is $x^2$. The derivatives of those functions are $-2x$ and $2x$ respectively, so you can see they agree at $0$. – Richard Jensen Nov 26 '22 at 13:31

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Consider $[-a,a]$. Take $C=2a$. Then, $|sgn (x)x^{2}-sgn (y)y^{2}| \leq C|x-y|$ is clear if $x$ and $y$ have the same sign. Suppose $x$ is positive and $y$ is negative. Then $x^{2}+y^{2} \leq ax+a|y|=a(x-y)=a|x-y|\leq C|x-y|$. A similar argument holds if $y$ is positive and $x$ is negative.

geetha290krm
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