English

The Hermite-Hadamard inequality on hypercuboid

Classical Analysis and ODEs 2016-04-08 v1

Abstract

Given any a:=(a1,a2,,an){\bf{a}}: = \left( {a_1 ,a_2 , \ldots ,a_n } \right) and b:=(b1,b2,,bn){\bf{b}}: = \left( {b_1 ,b_2 , \ldots ,b_n } \right) in Rn\mathbb{R}^n. The n\textbf{n}-fold convex function defined on [a,b]\left[ {{\bf{a}},{\bf{b}}} \right], a,bRn{\bf{a}},{\bf{b}} \in \mathbb{R}^n with a<b{\bf{a}}<{\bf{b}} is a convex function in each variable separately. In this work we prove an inequality of Hermite-Hadamard type for n\textbf{n}-fold convex functions. Namely, we establish the inequality \begin{align*} f\left( {\frac{{{\bf{a}} + {\bf{b}}}}{2}} \right) \le \frac{1}{{{\bf{b}} - {\bf{a}}}}\int_{\bf{a}}^{\bf{b}} {f\left( {\bf{x}} \right)d{\bf{x}}} \le \frac{1}{{2^n }}\sum\limits_{\bf{c}} {f\left( {\bf{c}} \right)}, \end{align*} where cf(c):=ci{ai,bi}1inf(c1,c2,,cn)\sum\limits_{\bf{c}} {f\left( {\bf{c}} \right)} : = \sum\limits_{\mathop {c_i \in \left\{ {a_i ,b_i } \right\}}\limits_{1 \le i \le n} } {f\left( {c_1, c_2, \ldots ,c_n } \right)}. Some other related result are given.

Keywords

Cite

@article{arxiv.1604.01857,
  title  = {The Hermite-Hadamard inequality on hypercuboid},
  author = {Mohammad W. Alomari},
  journal= {arXiv preprint arXiv:1604.01857},
  year   = {2016}
}

Comments

12 pages

R2 v1 2026-06-22T13:27:06.144Z