target="_blank" rel="nofollow" href="#fb3_img_img_839ae9fb-26bc-5324-bd19-eeb277c6544a.png" alt="StartAbsoluteValue d EndAbsoluteValue"/>. Since is equiregulated, there is such that for every and . Let and . Thus, for and , we have
Hence, .
Let be the supremum of the set . Since , is regulated. Thus, there exists such that for every and . Take and a division , say, , such that (1.1) holds with instead of . Denote . Then, for and , we have
which implies . Thus, we have two possibilities: either or . In the first case, the proof is finished. In the second case, one can use a similar argument as the one we used before in order to find such that , and this contradicts the fact that . Thus, , and we finish the proof of the sufficient condition.
Now, we prove the necessary condition. Given , there exists a division , say, such that the inequality (1.1) is fulfilled, for every and every , with . Then, for every Скачать книгу
target="_blank" rel="nofollow" href="#fb3_img_img_839ae9fb-26bc-5324-bd19-eeb277c6544a.png" alt="StartAbsoluteValue d EndAbsoluteValue"/>. Since 