On an index set function (Q5929900)

Let \(I\) be a non-empty subset of the positive integers, an index set, \(\underline a = (a_1, \ldots)\), \(\underline b, \underline w\) be positive sequences, and let\( f( \underline a) = (f(a_1),\ldots)\) and define NEWLINE\[NEWLINEF_I( \underline a, \underline b) = W_I \Bigl(f\bigl( A_I( \underline b)\bigr)-A_I\bigl(f( \underline b)\bigr)-f\bigl( A_I( \underline a)\bigr)-A_I\bigl(f( \underline a)\bigr)\Bigr),NEWLINE\]NEWLINE where \(W_I = \sum_{i\in I} w_i\), and \(A_I( \underline a)= {1\over W_I}\sum_{i\in I}w_ia_i\); when \(I = (1,\ldots ,n)\) we write just \(F_n( \underline a, \underline b)\), etc. Levinson's inequality is: if \(f\) is 3-convex and if \(a_i+b_i = 2c, 1\leq i\leq n\), \(\max \underline a \leq \max \underline b\) then \(F_n( \underline a, \underline b)\leq 0\). The proof makes intricate use of the properties of 3-convex functions. NEWLINENEWLINENEWLINEIn the present paper the authors generalize this by showing that \(F_I( \underline a, \underline b)\) is sub-additive on the class of index sets. Further, the proof seems to be the ultimate simplification possible for this result, being almost a 1-line proof.