For which $a\in\mathbb{R}$ is $f: (0,\infty)\to\mathbb{R}$, $f(x)=x^a$ Lipschitz-continuous?

Question

I want to give all $a\in\mathbb{R}$, for which $f_a: (0,\infty)\to\mathbb{R}$, $f_a(x)=x^a$ is Lipschitz-continuous?

I think, that $f_a$ is only Lipschitz-continuous for $a=0$ and $a=1$, which is trivial.

But what is the best way, to show, that it is not Lipschitz-continuous for every other $a$, using the definition only?

I want to seperate three cases. $a\in (-\infty, 0)$, $a\in (0,1)$ and $a\in (1,\infty)$

For the first case $a\in (-\infty, 0)$ I just observe $(0,1]$. Because, when I can show, that it fails on this interval, $f_a$ can not be Lipschitz.

I have to show, that $\forall L\geq 0\quad\exists x,y\in (0,1]: |f_a(x)-f_a(y)|> L|x-y|$

Without loss of generality, we can take $y=1$ and get $|x^a-1|=x^a-1$, since $x\in (0,1]$.

Also if we can show, that $x^a-1>L$, then $x^a-1> L|x-y|$

This gives us $x^a>L+1\Leftrightarrow x^{-a}<\frac1{L+1}$. Taking the "(-a)th root", leaves us with $x<\sqrt[-a]\frac1{L+1}$.

Now I choose, lets say $x=\sqrt[-a]\frac1{L+2}$ and $y=1$ and conclude the proof.

Would this be sound? Taking the "(-a)th root" feels not that good. Also, can the choice of $x$ depend on $a$ too? I think so, since $a$ is fixed.

Is this correct so far? Thanks in advance.

Answer 1

Your conclusion is right: here's another look:

If we unpack the definition of Lipschitz we'll find that for a function $f$ there must exist $C$ such that $\forall x \neq y$ in the domain we have $$ \Bigg|\frac{f(x)-f(y)}{x-y} \Bigg|\leq C$$ But then we have that the left hand side of this inequality is only a bounded function when we take $f(x)=x^n$, when $x\in 0,1$.

$|\frac{x^n-y^n}{x-y}|\leq C$ but then if we consider $x=y+1$ this becomes

$(y+1)^n-y^n =y^n \bigg( \big(\frac{y+1}{y} \big)^n-1 \bigg)$ and this is unbounded when $n\notin 0,1$.

Answer 2

$f$ is Lipschitz on $\Bbb R^+$ iff $\sup_{(0<x<y)}|g(x,y)|<\infty$ where $g(x,y)=\frac {f(y)-f(x)}{y-x}.$

For $0\ne a\ne 1:$ Consider that when $x>0$ we have $$g(x,2x)=\frac {f(2x)-f(x)}{2x-x}=\frac {(2x)^a-x^a}{x}=(2^a-1)x^{a-1}.$$ Since $a\ne 0\implies 2^a-1\ne 0,$ it suffices to show that

(i). If $a>1$ then $\sup_{(x\geq1)}x^{a-1}=\infty.$

(ii). If $0\ne a<1$ then $\sup_{(0<x\leq 1)}x^{a-1}=\infty.$

To prove (i), if $a>1$ there exist $b,c \in \Bbb Z^+$ with $a-1>b/c.$ Now for any $n\in \Bbb Z^+$ we have $(n^c)^{a-1}\geq(n^c)^{b/c}=n^b\geq n.$ Therefore $$\sup_{x\geq 1} x^{a-1}\geq \sup_{n\in \Bbb Z^+}(n^c)^{a-1}\geq \sup_{n\in \Bbb Z^+}n=\infty$$ because every $r\in \Bbb R$ is less than some $n\in \Bbb Z^+.$

To prove (ii), if $0\ne a<1$ let $a'=2-a.$ Then $a'>1.$ By (i) we have $$\sup_{0<x\leq 1}x^{a-1} =\sup_{x'\geq 1}\;(1/x')^{a-1}=\sup_{x'\geq1}\;(x')^{a'-1}=\infty.$$