Modulus of Continuity of a $x^{\alpha}$

Question

How do I show that modulus of continuity, $\omega(f,\delta)$ where $f(x):=x^{\alpha}$, is $\leq c\delta^{\alpha}$ where c is some constant independent of $\delta$. ($0 <\alpha < 1$)

Answer 1

Since domain of $f$ is positive integers we have $$ \omega(f,\delta) =\sup_{|x-y|\leq 2\delta,\;\;x,y\in D(f)}|f(x)-f(y)| =\sup_{|x-y|\leq 2\delta,\;\;x,y\geq 0}|f(x)-f(y)| $$ Since $f$ is strictly increasing, then $$ \begin{align} \omega(f,\delta) &=\sup_{|x-y|\leq 2\delta,\;\;x,y\geq 0}|f(x)-f(y)|\\ &=\sup_{|x-y|= 2\delta,\;\;x,y\geq 0}|f(x)-f(y)|\\ &=\sup_{x\geq 0}|f(x+2\delta)-f(x)|\\ &=\sup_{x\geq 0}(f(x+2\delta)-f(x))\\ \end{align} $$ Using the fact that $\alpha-1<0$ it is routine task to check that $f(x+2\delta)-f(x)$ is strictly decreasing on $[0,+\infty)$. Hence $$ \omega(f,\delta) =\sup_{x\geq 0}(f(x+2\delta)-f(x)) =f(0+2\delta)-f(0)=(2\delta)^{\alpha} $$