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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 c overTilde element-of upper B. Thus, we have two possibilities: either c overTilde equals b or c overTilde less-than b. 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 e element-of left-parenthesis c overTilde comma b right-bracket such that e element-of upper B, and this contradicts the fact that c overTilde equals sup upper B. Thus, c overTilde equals b, and we finish the proof of the sufficient condition.

      left-parenthesis left double arrow right-parenthesis Now, we prove the necessary condition. Given epsilon greater-than 0, there exists a division d prime element-of upper D Subscript left-bracket a comma b right-bracket, say, d prime colon a equals t 0 less-than t 1 less-than midline-horizontal-ellipsis less-than t Subscript StartAbsoluteValue d Sub Superscript prime Subscript EndAbsoluteValue Baseline equals b such that the inequality (1.1) is fulfilled, for every f element-of script í’œ and every left-bracket t comma t Superscript prime Baseline right-bracket subset-of left-parenthesis t Subscript j minus 1 Baseline comma t Subscript j Baseline right-parenthesis, with j equals 1 comma 2 comma ellipsis comma StartAbsoluteValue d prime EndAbsoluteValue. Then, for every j equals 1 comma 2 comma ellipsis comma StartAbsoluteValue d prime EndAbsoluteValue, take tau Subscript j Baseline element-of left-parenthesis t Subscript j minus 1 Baseline comma t Subscript j Baseline right-parenthesis and delta greater-than 0 such that left-parenthesis tau Subscript j Baseline minus delta comma tau Subscript j Baseline plus delta right-parenthesis subset-of left-parenthesis t Subscript j minus 1 Baseline comma t Subscript j Baseline right-parenthesis. Thus, (1.1) is satisfied, for all t comma t prime element-of left-parenthesis tau Subscript j Baseline minus delta comma tau Subscript j Baseline plus delta right-parenthesis. In particular, if either t equals tau Subscript j and t prime element-of left-parenthesis tau Subscript j Baseline minus delta comma tau Subscript j Baseline right-bracket, or t equals tau Subscript j and t prime element-of left-bracket tau Subscript j Baseline comma tau Subscript j Baseline plus delta right-parenthesis, then the inequality (1.1) holds. Thus, script í’œ is equiregulated.

      The next result describes an interesting property of equiregulated sets of upper G left-parenthesis left-bracket a comma b right-bracket comma upper X right-parenthesis. 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

parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis t Superscript minus Baseline right-parenthesis parallel-to less-than-or-slanted-equals gamma Subscript t Baseline comma t element-of left-parenthesis a comma b right-bracket comma and parallel-to f left-parenthesis t Superscript plus Baseline right-parenthesis minus f left-parenthesis t right-parenthesis parallel-to less-than-or-slanted-equals gamma Subscript t Baseline comma t element-of left-bracket a comma b right-parenthesis period

       Then, there is a constant such that, for every ,

parallel-to f left-parenthesis t right-parenthesis minus f left-parenthesis a right-parenthesis parallel-to less-than-or-slanted-equals upper K comma t element-of left-bracket a comma b right-bracket period

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