alt="parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis t Superscript prime Baseline right-parenthesis parallel-to less-than-or-slanted-equals parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis c overTilde Superscript minus Baseline right-parenthesis parallel-to plus parallel-to f left-parenthesis t Superscript prime Baseline right-parenthesis minus f left-parenthesis c overTilde Superscript minus Baseline right-parenthesis parallel-to less-than-or-slanted-equals epsilon comma"/>
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 , take and such that . Thus, (1.1) is satisfied, for all . In particular, if either and , or and , then the inequality (1.1) holds. Thus, is equiregulated.
The next result describes an interesting property of equiregulated sets of . Such result can be found in [97, Proposition 3.8].
Theorem 1.12: Assume that a set is equiregulated and, for any , there is a number such that
Then, there is a constant such that, for every ,
Proof. Take as the set of all numbers fulfilling the condition that there exists a positive number for which we have , for every and every Since is an equiregulated set, there exists such that for every and every . From this fact and the hypotheses, we can infer that for every
