수학 방정식의 수직선

Question

어떻게 하면 이런 수직선을 만들 수 있나요?

Answer 1

노력하다

\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}

방정식을 제공해 주시면 배열을 채울 수 있습니다.

부록: 아래 의견에서 David Carlisle이 언급했듯이 / 패키지 aligned의 환경을 사용하는 것이 디스플레이 스타일 수학 환경을 갖기 때문에 더 나은 선택입니다.amsmathmathtools

\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}

Answer 2

여기서는 새로운 환경을 정의하고 mathtools.

또한 내 의견으로는 부등식 기호가 약간 오른쪽으로 이동하여 규칙을 쓸모 없게 만드는 규칙 없는 버전도 추가합니다.

\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}