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arXiv 1411.7859math.CA
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含数值微分公式的Hermite-Hadamard型不等式

Inequalities of the Hermite-Hadamard type involving numerical differentiation formulas

Andrzej Olbryś, Tomasz Szostok

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摘要：

我们发现，以$$f\left(\frac{x+y}{2}\right)\leq\frac{F(y)-F(x)}{y-x}\leq\frac{f(x)+f(y)}{2}$$形式表示的Hermite-Hadamard不等式可以看作是两个求积算子$f\left(\frac{x+y}{2}\right)$ $\frac{f(x)+f(y)}{2}$和一个微分公式$\frac{F(y)-F(x)}{y-x}.$之间的不等式。事实证明，在某些情况下，使用Ohlin引理就足够了，就像最近的一篇论文\cite{Rajba}中所做的那样，但是为了得到更有趣的结果，必须使用一些更通用的工具。为此，我们使用莱文-Stečkin定理，它提供了充要条件，在此条件下，我们考虑的类型的不等式得到满足。

英文摘要：

We observe that the Hermite-Hadamard inequality written in the form $$f\left(\frac{x+y}{2}\right)\leq\frac{F(y)-F(x)}{y-x}\leq\frac{f(x)+f(y)}{2}$$ may be viewed as an inequality between two quadrature operators $f\left(\frac{x+y}{2}\right)$ $\frac{f(x)+f(y)}{2}$ and a differentiation formula $\frac{F(y)-F(x)}{y-x}.$ We extend this inequality, replacing the middle term by more complicated ones. As it turns out in some cases it suffices to use Ohlin lemma as it was done in a recent paper \cite{Rajba} however to get more interesting result some more general tool must be used. To this end we use Levin-Stečkin theorem which provides necessary and sufficient conditions under which inequalities of the type we consider are satisfied.

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