Fixed $0<a<1$, define $f(x):=(1-x)^{a}$ for every $x\in [0,1]$. Recalling that the modulus of continuity of $f$ of order $\varepsilon$ is given by
$\omega(f,\varepsilon):=\sup\{|f(x)-f(y)|:|x-y|\leq \varepsilon\}$, How can I find an upper bound for this function $f(x)$?


Thanks in advance for your comments!



Modulus of continuity an exponential type function
The exact upper bound on the decreasing function $f$ is $\sup\_{x\in[0,1]} f(x)=f(0)=1$. The exact value of $\omega(f,\varepsilon)$ for $\varepsilon>0$ is $f(1-1\wedge\varepsilon)-f(1)=1\wedge\varepsilon^a$, because $f$ is decreasing and concave. 


