Question:
Determine the values of $\alpha$ for which $f$ is Lipschitz :
(1) Let $f(x)=x^\alpha$, $\forall x\in(0,1)$ and $\alpha\in \mathbb{R}$.
(2) Let $f(x)=(1-x^2)^\alpha$, $\forall x\in(0,1)$ and $\alpha\in \mathbb{R}$.
I know Lipschitz condition means $|f(x)-f(y)|\leq L|x-y|$ for some constant L.
I tried to apply Mean Value Theorem : $|f(x)-f(y)|= |f'(z)||x-y|$ where $z$ lies between $x$ and $y$, but I'm not able to show that $|f'(z)|$ is bounded $\forall z\in (0,1)$
Is there any other way to solve this question. I need help in solving this question.
Best Answer
Hint: Use the MVT: $$f(x)-f(y)=\alpha\xi^{\alpha-1}(x-y)$$ and for your second function: $$f(x)-f(y)=-2\xi\alpha(1-\xi^2)^{\alpha-1}(x-y)$$ and $\xi \in (0,1)$ If $$\alpha-1<0$$ then $\xi^{\alpha-1}$ is not bounded. If $\alpha-1\geq 0$ then we get $|\xi^{\alpha-1}|\le 1$