论Opial-Beesack不等式

On Opial-Beesack Inequality

设:f(x)∈AC[o,A),并f(0)＝f(h)＝0.则有integral from n=0 to h(|f(x)f(x)|dx)≤h/4 integral from n=0 to h(|f'(x)|~2dx)这个不等式叫做Opial不等式.许多数学家对它曾进行过研究.在此我们给予有意义的改进:integral from n=0 to h (|ff'|dx)≤1/2(h/2)^(2/Q)(integral from n=0 to h(|f'|~pdx))^((2/p)-(2/Q)){(integral from n=0 to h(|f'|~pdx))~2-1/4(integral from n=0 to h(|f'|~pcos(2πx/h)dx)~2)}((?)/Q)其中I<P≤2,Q＝p/(P—1)．(2)显然比(1)优秀,实际上我们已证得更一般的结果.

:The main result of this paper is:Theorem: Let f(x) be absolutely continuous on [0,h]with f(0) =f(h) =0. Then we haveWhere 1<P≤2,Q=P/(P-1).This is an improvement of Beesack inequality. The P= 2 is an improvement of Opial inequality.In fact,we have proved more generally inequality than (1) in this paper.