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Show that none of the functions $f(x)=\sqrt[n]{x}$, for $n\in \Bbb N$, $n\ge2$ are Lipschitz continuous.

So i did it this way: $|\sqrt[n]{x}-\sqrt[n]{y}|=|x-y||\frac{1}{x^{\frac{n-1}{n}}+x^{\frac{n-2}{n}}y^{\frac{1}{n}}+...+x^{\frac{1}{n}}y^{\frac{n-2}{n}}+y^{\frac{n-1}{n}}}| \le (?)L|x-y|$ and argued that for very small x and y (close to 0) the denominator is very small, therefore the whole expressions is big, and the whole fraction cannot be limited by a real number (L), so these functions are not Lipschitz continuous.

Am I corrected? Is there any nicer, more formal way to show it?

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    There's no information about x, so i assumed $x \in \Bbb R$.2012-12-25

2 Answers 2

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That looks fine to me. A similar way is to note that being Lipschitz would mean there's an $L$ such that $\bigg|{\sqrt[n]{y} - \sqrt[n]{x} \over y - x} \bigg| \leq L$ Letting $y = 2x$ means that for all $x$ one must have $(\sqrt[n]{2} - 1)x^{{1 \over n} - 1} \leq L$ Letting $x \rightarrow 0$ gives a contradiction

If you're including $x = 0$ in your domain (you didn't say) then you could take $x = 0$ in the above and let $y \rightarrow 0$ to get a similar contradiction.

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Your argument shows that if $x,y$ are bounded away from zero, then the function is indeed Lipschitz (i.e. it is Lipschitz on any interval $(\delta,\infty)$ with $\delta>0$).

So the problem can only arise at $0$. There, with $y=0$, the Lipschitz condition amounts to $ \sqrt[n]x\leq x, $ which doesn't hold for $x\in[0,1)$ and $n>1$.