Fixed $a>0$, define $f(x):=(1-x)^{a}$ for each $x\in [0,1]$. Recalling that the modulus of continuity of $f$ of order $\varepsilon>0$ is given by
$\omega(f;\varepsilon):=\sup\{|f(x)-f(y)|: |x-y|\leq \varepsilon\}$,
How can we find an upper bound for $\omega(f;\varepsilon)$? I have seen a proof in the post
Modulus of Continuity of a $x^{\alpha}$
for the funcion $x^{\alpha}$, but I do not see clear the proof. Can you help me, please?
Very thanks in advance for your comments.
