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Does the function $F(x,y) = xy^1/3$ satisfy the Lipschitz condition on the rectangle $ {(x,y) : |x| \le h, |y| \le k} $ where $h < 0$ and $k < 0$?

I have tried using the mean value theorem to show this:

|F(x, u) − F(x, v)| = |$F_y (x, w)$ (u − v)| ≤ K|u − v|

However I found that, $F_y (x, w) > A$ where A is a real number. as when $y \to 0$, $F_y (x, w) \to \infty$. Does this mean it does not satisfy the Lipschitz condition?

Many thanks.

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  • $\begingroup$ I imagine that you mean $h > 0$ and $k>0$ and not negative. $\endgroup$ – mathcounterexamples.net Sep 30 '15 at 7:45
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No it doesn't as for $x=\frac{h}{2}$ you have $$f(x,y)=f(\frac{h}{2},y)=\frac{h}{2}y^{\frac{1}{3}}$$ and $$\lim\limits_{y \to 0} \frac{\frac{h}{2}y^{\frac{1}{3}}}{y} = +\infty$$

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