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Iosif Pinelis
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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 deacreasingdecreasing and concave.

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 deacreasing and concave.

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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Iosif Pinelis
  • 127.7k
  • 8
  • 107
  • 229

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 deacreasing and concave.