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\begin{document}
\begin{corollary}
\label{Corollary1}
Let $f:I\subseteq[0,\infty)\rightarrow\mathbb{R}$ be twice differantiable mapping on
$I^{\circ}$, $a,b\in I$ with $a<b$. If $|f''|^q$ is convex, then
\begin{eqnarray}
\label{C1}
&&\bigg|\int_{a}^bf(x)dx-f(\frac{a+b}{2})\bigg| \\
&& \leq \frac{(b-a)^2}{16(2p+1)^\frac{1}{p}}\bigg[\bigg(\frac{|f''(a)|^q+3|f''(b)|^q}{4}\bigg)^\frac{1}{q}
+\bigg(\frac{|3f''(a)|^q+|f''(b)|^q}{4}\bigg)^\frac{1}{q}\bigg] \nonumber \\
&& \leq \frac{(b-a)^2}{2^{(2+\frac{2}{q})}(2p+1)^\frac{1}{p}}\Big(|f''(a)|+|f''(b)|\Big). \nonumber
\end{eqnarray}
\end{corollary}
%-------------------------------------------------------------------------------
\begin{corollary}
\label{Corollary2}
Let $f:I\subseteq[0,\infty)\rightarrow\mathbb{R}$ be twice differantiable mapping on
$I^{\circ}$, $a,b\in I$ with $a<b$. If $|f''|^q$ is convex,
then
\begin{eqnarray}
\label{C2}
&&\bigg|\int_{a}^bf(x)dx-f(\frac{a+b}{2})\bigg| \\
&&\leq \frac{(b-a)^2}{48} \bigg[\bigg(\frac{3|f''(a)|^q+5|f''(b)|^q}{8}\bigg)^\frac{1}{q}
+\bigg(\frac{|5f''(a)|^q+|3f''(b)|^q}{8}\bigg)^\frac{1}{q}\bigg] \nonumber \\ \nonumber
\end{eqnarray}
\end{corollary}
\end{document}
答案1
要使用您的文档revtex4-1而不是amsart我必须加载amsmath和amsthm包，并重命名您的case环境，因为该名称已经定义，请参阅标有%%%%%%%
\documentclass{revtex4-1}%%%%%%%
\usepackage{amsmath,amsthm}%%%%%%%
\usepackage{amsfonts}
\setcounter{MaxMatrixCols}{10}
\newtheorem{theorem}{Theorem}
\theoremstyle{plain}
\newtheorem{acknowledgement}{Acknowledgement}
\newtheorem{algorithm}{Algorithm}
\newtheorem{axiom}{Axiom}
\newtheorem{mycase}{Case}%%%%%%%