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Let $ f(x)=x^{\alpha}$ for $\alpha \in \mathbb {R}$. Determine the conditions on $\alpha$ such that f satisfies a local Lipschitz condition on $\mathbb R$. For the local Lipschitz case, I need to find a Lipschitz constant on the interval $[-a,a]$ where $a>0$.

I think that $\alpha\geq 1$ but how do I show this? And I need to find a L.

Please help. Thank you! Klara

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Here are some hints: Note that $|f'(x)|$ is almost a Lipschitz constant in a very small nbh. of $x$ (the mean value theorem is your friend). Also, treat the cases for $\alpha$ separately: $(-\infty,0)$, $0$, $(0,1)$ and $[1,\infty)$. – copper.hat Nov 5 '12 at 5:16
The function $f(x) = \frac{1}{x}$ is not defined at $x=0$, and is certainly not Lipschitz there... – copper.hat Nov 5 '12 at 15:59
@copper so @ x=0 $\alpha \geq 0$, right? – Klara Nov 5 '12 at 16:45
Well, to be defined you need $\alpha\geq 0$, but not all $\alpha \geq 0$ result in locally Lipschitz $f$. Plot $f(x) = \sqrt{x}$, and notice the slope as $x \to 0$. (Actually, I'm not sure how to define $f$ on all of $\mathbb{R}$ for $\alpha \geq 0$, perhaps you meant $f(x) = |x|^\alpha$?). – copper.hat Nov 5 '12 at 16:47
@ copper No the question is right, unless professor wrote it wrong. – Klara Nov 5 '12 at 17:44
up vote 2 down vote accepted

The question is to determine conditions on $\alpha$ so that $f_\alpha(x) = x^{\alpha}$ satisfies a local Lipschitz condition on $\mathbb{R}$.

Note: I suspect that either the function in the problem should have been $f_\alpha(x) = |x|^{\alpha}$, or the interval of the form $[0,a]$, in which case Issue (1) below is moot.

There are two issues here: (1) For what $\alpha$ is $f_\alpha$ well defined on $\mathbb{R}$. (2) For what $\alpha$ is $f_\alpha$ locally Lipschitz.

Issue (1) is most important, but distracts from what I think the purpose of the exercise was. First, if $\alpha<0$, then $f_\alpha$ is not defined at $x=0$, so we have $\alpha \geq 0$.

If $\alpha=0$, we have $f_0(x) = 1$ for all $x \neq 0$. It is not clear what to do with $f_0 (0)$, but common convention is to take $0^0 = 1$, which is what we will do. Since $f_0$ is constant, it is globally Lipschitz with rank $L=0$.

Now consider $\alpha>0$.

Since $f_\alpha$ needs to be defined for $x<0$, we cannot use the equality $x^\alpha = e^{\alpha \ln x}$ to define $f_\alpha$. Since we cannot use $\ln$, we can start by defining $x^\alpha$ for rational $\alpha$ (since we may have $x<0$). If $\alpha = \frac{p}{q}$, with $p,q$ coprime, then define $y=x^{\frac{p}{q}}$ as the real solution to the equation $y^q = x^p$. If $q$ is even, then there are no solutions for $x<0$ (since in this case $p$ is odd, and $x^p <0$), if $q$ is odd, we can define $\sqrt[q]{x} = \text{sgn}(x) \sqrt[q]{|x|} $, in which case $f_{\frac{p}{q}}(x) = \sqrt[q]{x^p}$. Let $A=\{\frac{p}{q} | \frac{p}{q}>0,\ \gcd(p,q)=1,\ 2 \not\mid q \}$. A little bit of work shows that $\overline{A} = [0,\infty)$, and a little more work shows that $\overline{\mathbb{Q} \setminus A} = [0,\infty)$, from which it follows that there is no way to continuously extend the definition of $f_\alpha$ from $\alpha \in A$ to $[0,\infty)$. So, we are stuck with $\alpha \in A$.

Before leaving Issue (1), we note that for $x>0$ and $\alpha \in A$, the above definition of $f_\alpha$ matches the usual definition, ie, we have $f_\alpha(x) = e^{\alpha \ln x}$. Hence, for $x\neq 0$, we have $f_\alpha(x) = \text{sgn}(x) e^{\alpha \ln |x|}$. This simplifies the ensuing calculations, and shows that for $x\neq 0$, $f_\alpha$ is differentiable with $f'(x) = \text{sgn}(x) \alpha e^{(\alpha-1) \ln |x|}$.

Issue (2) is more straightforward. We need some interim results first. Suppose $g:[a,b]\to \mathbb{R}$ is differentiable on $(a,b)$.

First, the mean value theorem shows that $|g(x)-g(y)| \leq \sup_{\xi \in (a,b)} |g'(\xi)|$ for all $x,y \in [a,b]$. Note that the $\sup$ is over the open interval, whereas the bound applies to the closed interval. hence if the derivative is bounded on an open interval, then the function is Lipschitz (with rank $L=\sup_{\xi \in (a,b)} |g'(\xi)|$) on the closed interval.

Second, suppose $K = |g'(\xi)|>0$ for some $\xi \in (a,b)$. Then, since $\lim_{t \to \xi} |\frac{g(t)-g(\xi)}{t-\xi} | = K$, we see that for all $\epsilon>0$, there exists a $t$ near $\xi$ such that $|g(t)-g(\xi)| \geq (K-\epsilon) |t-\xi|$. Hence if $L$ is a local Lipschitz rank for $g$ on an open set containing $\xi$, then we must have $L \geq K$.

Now suppose that $\alpha \in A$ and $\alpha <1$. Then we see that $\lim_{x \downarrow 0} f_\alpha'(x) = \infty$. It follows from the second remark above that $f_\alpha$ cannot be locally Lipschitz on any interval containing $0$.

Now suppose $\alpha \in A$, $\alpha \geq 1$, and that $x \in [0,a]$. Then $L=\sup_{\xi \in (0,a)} |f_\alpha'(\xi)|= \alpha a^{\alpha-1}$, and hence if $x,y \in [0,a]$, we have $|f_\alpha(x)-f_\alpha(y)| \leq L |x-y|$. Since $f_\alpha(-x) = - f_\alpha(x)$, we have the same bound for $x,y \in [-a,0]$. Finally, suppose $x \in [-a,0], y \in [0,a]$. Then we have

\begin{eqnarray} |f_\alpha(x)-f_\alpha(y)| &=& |f_\alpha(x)-f_\alpha(0)+f_\alpha(0)-f_\alpha(y)| \\ &\leq & |f_\alpha(x)-f_\alpha(0)|+|f_\alpha(0)-f_\alpha(y)| \\ & \leq & L|x|+L|y| = L|x-y| \end{eqnarray}

Hence $f_\alpha$ is defined and locally Lipschitz on $[-a,a]$ iff $\alpha = 0$ or $\alpha \in A$ and $\alpha \geq 1$. When $f_\alpha$ is locally Lipschitz, the rank is given by $L= \alpha a^{\alpha-1}$.

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WOW that's all I can say. Thank you so very much! You are right my professor did not even think of negative value issues, when I talked to him tonight. – Klara Nov 7 '12 at 3:15
You are very welcome - I'm glad it was of help! – copper.hat Nov 7 '12 at 3:46

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