Prove that, given $b \in (0,1)$, there exists $C_{b} \in \mathbb{R}$ such that, for all $x,y \in [0,1]$, $|x^b-y^b|\leq C_b|x-y|^b$

Question

I'm trying to show this inequality, but I'm running into trouble since exponents don't seem to behave in the usual way when all the numbers involved are so small.

I realize that it is sufficient to show that $\frac{|x^b-y^b|}{|x-y|^b}$ is bounded, but I don't see an easy way out here either.

Answer 1

Assume that $x\geq y\geq 0$, then \begin{align*} x^{b}-y^{b}&=\int_{0}^{1}\dfrac{d}{dt}(tx+(1-t)y)^{b}dt\\ &=\int_{0}^{1}b(tx+(1-t)y)^{b-1}(x-y)dt\\ &=(x-y)\int_{0}^{1}b(t(x-y)+y)^{b-1}dt\\ &\leq(x-y)\int_{0}^{1}b(t(x-y))^{b-1}dt\\ &=(x-y)^{b}\int_{0}^{1}bt^{b-1}dt\\ &=(x-y)^{b}. \end{align*}