Reversed Jensen type integral inequalities for monotone functions (Q2720279)

Let \(f\) be non-decreasing and non-negative in \([0,b]\) and let \(g\) be convex in \([0,\infty)\) with \(g(0) = 0\). Put \(F(x) = \int^x_0 f(s) ds\), \(0\leq x \leq b\). Let \(k\in L^1 [0,b]\) be non-negative in \([0,b]\) and let \(\sigma\) be a positive measure in \([0,b]\). Then, for \(0<x\leq b\), NEWLINE\[NEWLINE \int^x_0 g(f(s)) k(s) d\sigma(s) \leq \sup_{0<c\leq 1} \biggl\{ g(F(x)/cx)\int^x_{(1-c)x} k(s) d\sigma (s)\biggr\}. NEWLINE\]NEWLINE NEWLINENEWLINENEWLINEIn particular: NEWLINENEWLINENEWLINE(i) If, for \(0\leq x \leq b\), \(y>0\) and \(0<c\leq 1\), NEWLINE\[NEWLINE g(y/c) \int^x_{(1-c)x} k(s) d\sigma (s) \leq g(y) \int^x_0 k(s) d\sigma(s), NEWLINE\]NEWLINE then NEWLINE\[NEWLINE \int^x_0 g(f(s)) k(s) d\sigma(s) \leq \biggl(\int^x_0 k(s) d\sigma(s)\bigg) g \biggl(\frac{F(x)}{x}\biggr), \quad 0 < x \leq b. NEWLINE\]NEWLINE NEWLINENEWLINENEWLINE(ii) If, for \(0\leq x \leq b\), \(y>0\) and \(0<c\leq 1\), NEWLINE\[NEWLINE g(y/c) \int^{cx}_0 k(s) ds \leq g(y)\int^x_0 k(s) ds, NEWLINE\]NEWLINE then NEWLINE\[NEWLINE \int^x_0 g(f(s)) k(x-s) ds \leq \biggl(\int^x_0 k(s) ds\biggr) g\biggl(\frac{F(x)}{x}\biggr), \quad 0 < x \leq b. NEWLINE\]NEWLINE These are the main results of the paper.