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Determine whether or not the cube root of x is Lipschitz. x^(1/3)


I understand that this is NOT Lipschitz. I am having trouble properly proving this. I know for other problems I have shown functions are not uniformly continuous to prove that they are not Lipschitz, but I can't seem to figure out the right way to go about this problem. Reviewing for a test, Thanks!

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    On which domain?2012-04-24
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    On the domain: All R2012-04-24
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    Hint: Look at the point $0$.2012-04-24
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    No, the cube root of $x$ is $\root3\of x$. The cube root of $({\rm Lipschitz})^3$ is Lipschitz.2012-04-24

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