Given a function $f$ and a constant $M \ge 0$ such that:
$$ |f(x_1) - f(x_2) |\le M {|x_1 - x_2|}^{\alpha} $$
we say that $f$ satisfies a Lipschitz condition of order $\alpha$. We denote by Lip $\alpha$ the set of all such functions. Consider the following problem:
For $\alpha \in (0,1)$, show that $f(x) = x^{\alpha} $ belongs to Lip $\alpha $.
My proof
Suppose $f(x_1) = x_1^\alpha $ and $f(x_2) = x_2^\alpha $ and then $$ |f(x_1) - f(x_2) | = |x_1^\alpha - x_2^\alpha|\le |x_1 - x_2|^\alpha $$ where $x_1 \ne x_2$.
Note that $ |x_1^\alpha - x_2^\alpha|\le |x_1 - x_2|^\alpha\le M|x_1 - x_2|^\alpha $
Hence, $f(x) = x^\alpha $ belongs to Lip $\alpha$.
Is there anything wrong with my proof? If so please let me know!
