答案1
尝试
\documentclass{article}
\begin{document}
\[
\left| \begin{array}{l}
\alpha \\
\gamma \\
\delta \\
%\displaystyle% for display style of equations had to be added in each row
+ \frac{1}{T-1}\left[\;\int\limits_{\{h<|u_n|\}} |f(x)|^m \right]^{\frac{1}{m}} \dots
\end{array}\right.
\]
\end{document}
如果你能提供你的方程式,我可以用它们填充数组。
附录:
正如 David Carlisle 在下面的评论中提到的那样,使用/包aligned
中的环境是更好的选择,因为有了它你就有了 displaystyle 数学环境:amsmath
mathtools
\documentclass{article}
\usepackage{amsmath}
\begin{document}
\[
\left| \begin{aligned}
& \alpha \\
& \beta \\
& \gamma \\
& \delta \\
& + \frac{1}{T-1}\left[\;\int\limits_{\{h<|u_n|\}} |f(x)|^m \right]^{\frac{1}{m}} \dots
\end{aligned}\right.
\]
\end{document}
答案2
这里我定义了一个新的环境,并在排版(你图片里的括号太大了)和编码上做了一些改进,借助mathtools
。
我还添加了一个没有该规则的版本,其中不等式符号稍微向右移动了一点,在我看来,这使得该规则变得毫无用处。
\documentclass{article}
\usepackage{amsmath,mathtools}
\usepackage{newtxtext,newtxmath}
\DeclarePairedDelimiter{\abs}{\lvert}{\rvert}
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
\newcommand{\intl}{\int\limits}
\newenvironment{ruledaligned}
{\left|\aligned}
{\endaligned\right.}
\begin{document}
\begin{equation*}
\begin{ruledaligned}
& (2\lambda\alpha-\gamma)
\intl_\Omega \abs{DG_h(u_n)}^2 e^{2\lambda\abs{DG_h(u_n)}}
+\mu\intl_{\{h<\abs{u_n}\}} \abs{u_n}(e^{2\lambda\abs{DG_h(u_n)}}-1)
\\
&\le
T\intl_{\{h<\abs{u_n}\}} \abs{f(x)}(e^{2\lambda\abs{DG_h(u_n)}}-1)^2
+\frac{1}{T-1}\intl_{\{h<\abs{u_n}\}} \abs{f(x)}
\\
&\le
T\Biggl(\,\intl_{\{h<\abs{u_n}\}} \abs{f(x)}^m\Biggr)^{\!\frac{1}{m}}
\Biggl(\,\intl_{\{h<\abs{u_n}\}} (e^{\lambda\abs{DG_h(u_n)}}-1)^{2m'}\Biggr)^{\!\frac{1}{m'}}
+\frac{1}{T-1}\intl_{\{h<\abs{u_n}\}} \abs{f(x)}
\\
&\le
T\Biggl(\,\intl_{\{h<\abs{u_n}\}} \abs{f(x)}^m\Biggr)^{\!\frac{1}{m}}
\norm[\big]{e^{\lambda\abs{DG_h(u_n)}}-1}_{L^{2^*}(\Omega)}^{2\theta}
\Biggl(\,\intl_{\{h<\abs{u_n}\}} (e^{\lambda\abs{DG_h(u_n)}}-1)^2\Biggr)^{\!1-\theta}
\\
&\qquad+
\frac{1}{T-1}\Biggl(\,\intl_{\{h<\abs{u_n}\}} \abs{f(x)}^m\Biggr)^{\!\frac{1}{m}}
\abs[\big]{\{h<\abs{u_n}\}}^{1-\frac{1}{m}}
\end{ruledaligned}
\end{equation*}
\begin{equation*}
\begin{aligned}
& (2\lambda\alpha-\gamma)
\intl_\Omega \abs{DG_h(u_n)}^2 e^{2\lambda\abs{DG_h(u_n)}}
+\mu\intl_{\{h<\abs{u_n}\}} \abs{u_n}(e^{2\lambda\abs{DG_h(u_n)}}-1)
\\
&\quad\le
T\intl_{\{h<\abs{u_n}\}} \abs{f(x)}(e^{2\lambda\abs{DG_h(u_n)}}-1)^2
+\frac{1}{T-1}\intl_{\{h<\abs{u_n}\}} \abs{f(x)}
\\
&\quad\le
T\Biggl(\,\intl_{\{h<\abs{u_n}\}} \abs{f(x)}^m\Biggr)^{\!\frac{1}{m}}
\Biggl(\,\intl_{\{h<\abs{u_n}\}} (e^{\lambda\abs{DG_h(u_n)}}-1)^{2m'}\Biggr)^{\!\frac{1}{m'}}
+\frac{1}{T-1}\intl_{\{h<\abs{u_n}\}} \abs{f(x)}
\\
&\quad\le
T\Biggl(\,\intl_{\{h<\abs{u_n}\}} \abs{f(x)}^m\Biggr)^{\!\frac{1}{m}}
\norm[\big]{e^{\lambda\abs{DG_h(u_n)}}-1}_{L^{2^*}(\Omega)}^{2\theta}
\Biggl(\,\intl_{\{h<\abs{u_n}\}} (e^{\lambda\abs{DG_h(u_n)}}-1)^2\Biggr)^{\!1-\theta}
\\
&\quad\qquad+
\frac{1}{T-1}\Biggl(\,\intl_{\{h<\abs{u_n}\}} \abs{f(x)}^m\Biggr)^{\!\frac{1}{m}}
\abs[\big]{\{h<\abs{u_n}\}}^{1-\frac{1}{m}}
\end{aligned}
\end{equation*}
\end{document}