I need help understanding this step in the given lemma.

Question

In the proof of Lemma $2.4$, I have shown that $\omega(t)/t$ is a decreasing function but how does it follow that $\omega(2t) \le C \omega(t)$?

Answer 1

$$\omega(t) = 1/\log(e/2t), 0< t \le \frac12$$

$$\omega(2t) = 1/\log(e/4t)$$

Since $\frac{\omega(t)}{t}$ is decreasing.

$$\frac{\omega(2t)}{2t} \le \frac{\omega(t)}{t} $$

We have $$\omega(2t) \le 2 \omega(t)$$

Since $r$ is bounded, say $|r| \le M$, $|r(x)-r(y)| \le 2M$.

We have $Q=\min_{x,y \in \Omega} \omega(|x-y|)>0$,

Hence we have $$|r(x)-r(y)| \le \left( \frac{2M}{Q}\right) \omega(|x-y|)$$