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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? –  Rasmus Apr 24 '12 at 0:33
    
On the domain: All R –  Mark Apr 24 '12 at 0:34
    
Hint: Look at the point $0$. –  Rasmus Apr 24 '12 at 0:34
    
No, the cube root of $x$ is $\root3\of x$. The cube root of $({\rm Lipschitz})^3$ is Lipschitz. –  Gerry Myerson Apr 24 '12 at 0:39

4 Answers 4

Let $f:\mathbb{R}\to\mathbb{R},f(x)=x^{1/3}$. Assume $L\geq 0$ is given. Let us show that $f$ is not $L$-Lipschitz. Pick a so small $x>0$ that $f'(x)=(1/3)x^{-2/3}>L$. Then pick any $y$ with $0<y<x$. By the mean value theorem $f(x)-f(y)=f'(z)(x-y)$ for some $z$ with $y\leq z\leq x$. Note that $|f'(z)|=f'(z)\geq f'(x)$ since $z\leq x$. It follows that $|f(x)-f(y)|=|f'(z)||x-y|\geq f'(x)|x-y|>L|x-y|$. Hence $f$ is not $L$-Lipschitz. Because $L\geq 0$ was arbitrary, $f$ is not Lipschitz.

Note that the crucial property is that the derivative of $f$ is not bounded near $0$.

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Since $f(x) = x^{1/3}$ has an infinite derivative at $x=0$, we don't expect to have bounded difference quotients on any interval containing $x=0.$ In fact, we can be led to this expectation by using only precalculus ideas. Since the graph of $y = x^3$ straightens out horizontally at the origin, the graph of the inverse function (of $y = x^3$), namely that graph of $y = x^{1/3},$ straightens out vertically at the origin.

Thus, we're led to consider difference quotients based at the origin:

$$\frac{f(x) - f(0)}{x-0} \; = \; \frac{x^{1/3}}{x} \; = \; \frac{1}{x^{2/3}}$$

Since these difference quotients are not bounded on any interval containing $x=0$, for example the interval $[0,1],$ the function $f(x)$ is not Lipschitz continuous on any interval containing the origin.

More generally, $f(x)$ is not Lipschitz on any $E \subseteq {\mathbb R}$ such that $0$ is a limit point of $E.$ For this last result, simply note that the values of

$$\frac{f(r) - f(s)}{r-s} \; = \; \frac{1}{r^{2/3} + (rs)^{1/3} + s^{2/3}}$$

will be unbounded if both $r$ and $s$ can be chosen arbitrarily close to $0.$

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Let me give you a hint that I hope helps. Intuitively a Lipschitz function is one that is BASICALLY subadditive--there exists constants $a,b$ such that $f(x)\leqslant ax+b$. Now, does being subadditive react well to the taking of cube roots?

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Well we haven't exactly covered what you are talking about. But can I say something like: the slope is unbounded at x = 0 , therefore f(x) is not lipschitz? –  Mark Apr 24 '12 at 1:11

Too see it without using derivatives, observe that

[ x^{1/3}-y^{1/3} = (x-y)/(x^{2/3} + (xy)^{1/3} + y^{2/3} ) ]

Now as x or y goes to zero, you can make the denominator as large as you wish (as cube root is a continuous function at zero). Thus, no single L will suffice.

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