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I've got to prove $$f(x,y):=y^{2/3}$$ doesn't satisfy Lipschitz condition in $$G:=\{(x,y): 0\leq x\leq 1,-1\leq y \leq 2\}$$. But I have problems with denying the definition (not sure if that's the right translation to english) for 2 variables. Sadly this means I have to use this definition: Let $$G\in \mathbb{R}^{n+1}$$ an arbitrary set $$f(x,y)$$ a real function in $$y(y_0,...,n-1)$$ (this subindex might be wrong now I see it).

$$F:G\rightarrow\mathbb{R}$$ satisfies Lipschitz condition if exists a real R positive or equal t o 0 that $$|f(x,y)-f(x,y^*)|\leq R|y-y^*|$$ only when y and y* are in G.

José Osorio
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  • Pick a Lipschitz constant $K$ and prove that $K$ is not a Lipschitz constant for $f$ by contradiction (and perhaps an explicit construction). – Michael Burr Mar 28 '16 at 17:28

2 Answers2

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Intuitively: $f$ is not Lipschitz because its derivative blows up near $0$.

In detail: since $f$ doesn't depend on $x$, it suffices to consider the map $$g(y) = y^{2/3}, \,\,\,\,\,\, y \in [-1,2].$$ We see that $g$ is continuously differentiable on $[-1,0)$ and $(0,2]$ with $$g'(y) = \frac{1}{y^{1/3}}, \,\,\,\, y \neq 0.$$ We see $$\lvert g'(y) \rvert \to \infty \,\,\,\, \text{ as } \,\,\,\, y \to 0.$$ In particular, for $K > 0$, there is $\delta > 0$ such that $0 < \lvert y \rvert < \delta$ gives $\lvert g'(y) \rvert > K.$ Take $0 < y_1 < y_2 < \delta$. Then by the mean value theorem, there is $c \in (y_1, y_2)$ with $$\lvert g(y_1) - g(y_2) \rvert = \lvert g'(c) \rvert \lvert y_1 - y_2 \rvert > K\lvert g(y_1) - g(y_2) \rvert. $$ But this shows that $g$ is not Lipschitz with constant $K$. Since $K >0$ was arbitrary, this shows that $g$ does not have a Lipschitz constant; i.e., $g$ is not Lipschitz.

User8128
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  • Apreciated, I just edited, I am supposed to use the 2 variable definition :/ – José Osorio Mar 28 '16 at 17:59
  • The proof is the exact same, just use $(x,y_1)$ and $(x,y_2)$ for any fixed $x \in [ 0,1]$ and these points will prove that $f$ is not Lipschitz. – User8128 Mar 28 '16 at 19:23
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Hint: $\displaystyle \lim_{c \to 0^{+}} f_y(c) = +\infty$

DeepSea
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