\documentclass[a4paper,12pt]{article}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\mathcal{A}(x) \coloneqq \left\{
\pi(t) \text{ prog. mbl. } \bigg \vert \: \begin{split}
& E(U(X^{\pi,x}(T))^-) < \infty, X^{\pi,x}(T) \geq B \text{ a.s.}, \\
& \int_0^{T} \Vert \pi(t)X^{\pi,x}(t)\Vert^2 dt < \infty a.s.
\end{split} \right\}
\end{equation}
\end{document}
生产
有没有办法调整垂直线,使其横跨两条线,并以 分隔\split
?谢谢。
答案1
您想使用\middle
和aligned
:
\documentclass[a4paper,12pt]{article}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\mathcal{A}(x) \coloneqq
\left\{
\pi(t) \text{ prog. mbl.}
\;\middle\vert\;
\begin{aligned}
& E(U(X^{\pi,x}(T))^-) < \infty, X^{\pi,x}(T) \geq B \text{ a.s.}, \\
& \int_0^{T} \lVert \pi(t)X^{\pi,x}(t)\rVert^2 \,dt < \infty \text{ a.s.}
\end{aligned}
\right\}
\end{equation}
\end{document}
注意\lVert
,而\rVert
不是不合格的\Vert
。关系的正确间距是\;
。
在这个特殊情况下,你可以欺骗 TeX,让它认为方程数字是合适的,参见https://tex.stackexchange.com/a/348745/4427
\documentclass[a4paper,12pt]{article}
\usepackage{mathtools,showframe}
\begin{document}
\begin{equation}
\mathcal{A}(x) \coloneqq
\left\{
\pi(t) \text{ prog. mbl.}
\;\middle\vert\;
\begin{aligned}
& E(U(X^{\pi,x}(T))^-) < \infty, X^{\pi,x}(T) \geq B \text{ a.s.}, \\
& \int_0^{T} \lVert \pi(t)X^{\pi,x}(t)\rVert^2 \,dt < \infty \text{ a.s.}
\end{aligned}
\right\}
\hspace{10000pt minus 1fil}
\end{equation}
\end{document}
我添加showframe
只是为了看看等式是否仍然适合边缘。
答案2
与使用array
环境有关:
\documentclass[a4paper,12pt]{article}
\usepackage{mathtools}
\DeclarePairedDelimiter\norm{\lVert}{\rVert}
\begin{document}
\begin{equation}
\mathcal{A}(x) \coloneqq \left\{
\pi(t) \text{ prog. mbl.} \;
\begin{array}{| l @{}}
E(U(X^{\pi,x}(T))^-) < \infty,\, X^{\pi,x}(T) \geq B \text{ a.s.}, \\
\int\limits_0^{T} \norm*{\pi(t)X^{\pi,x(t)}}^2 dt < \infty \text{ a.s.}.
\end{array} \right\}
\end{equation}
\end{document}