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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!

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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.

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  • $\begingroup$ Very thanks for your answer! Then, the value of the modulus of continuity, in this case, is due to "f is decreasing and concave"? $\endgroup$
    – user123043
    Commented Sep 9, 2017 at 6:22
  • $\begingroup$ Since $f$ is decreasing and concave, it is indeed easy to find the simple expression for the modulus of continuity of continuity. $\endgroup$
    – Iosif Pinelis
    Commented Sep 10, 2017 at 1:12

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