证明方程

Question

证明中的方程式显示不完整。你能帮我吗？提前谢谢！

我已附加了背面模板和编译模板。

Overleaf 模板：

\usepackage{mathtools,amssymb}
\begin{equation*}
\begin{aligned}
 \max_{1 \leq j \leq n^{2}} \sup_{z \in \left[\frac{j-1}{n^2}, \frac{j}{n^{2}}\right]}  
 \left|\sum^{n}_{t=1} \left[ V_{(t,1)} \left( z \right) -  V_{(t,1)}  \left( \frac{j}{n^{2}} \right) 
 \right] \right| 
 & = \max_{1 \leq j \leq n^{2}} \sup_{z \in \left[\frac{j-1}{n^2}, \frac{j}{n^{2}}\right]} 
 \left|  \sum^{n}_{t=1}  x^{*}_{t} I_{ \left( \left| x^{*}_{t} \right| \leq n^{1/3}h^{\gamma}_{n} 
\right) } \left( K' \left(\frac{z-z_{t}}{h_{n}} \right) - K' \left(\frac{\frac{j}{n^{*}}-z_{t}} {h_{n}} 
   \right)
 \\[1.5ex]  
& + E \left(  K' \left(\frac{\frac{j}{n^{2}}-z_{t}}{h_{n}} \right) \right) - E \left(  K' 
\left(\frac{z-z_{t}}{h_{n}} \right) \right) \right)
\right|
\end{aligned}
\end{equation*}

编译模板：

Answer 1

我建议使用一个align*环境和一个aligned嵌套的内部：

\documentclass{article}
\usepackage{nccmath, mathtools}
\usepackage{amssymb}
\usepackage[svgnames]{xcolor}

\usepackage{showframe}
\renewcommand{\ShowFrameLinethickness}{0.4pt}
\renewcommand{\ShowFrameColor}{\color{Coral}}

\begin{document}

\vspace*{1cm}

    \begin{fleqn}
    \begin{align*}
      \max_{1 \leq j \leq n^{2}} & \sup_{z \in \left[\frac{j-1}{n^2}, \frac{j}{n^{2}}\right]}
     \left|\sum^{n}_{t=1} \biggl[ V_{(t,1)} \left( z \right) - V_{(t,1)} \Bigl( \frac{j}{n^{2}}\Bigr)
     \biggr] \right| = \\
     & \max_{1 \leq j \leq n^{2}} \begin{aligned}[t] \sup_{z \in \left[\frac{j-1}{n^2}, \frac{j}{n^{2}}\right]}
     \Biggl| \sum^{n}_{t=1} x^{*}_{t} I_{(|x^{*}_{t}| \leq n^{1/3}h^{\gamma}_{n})}
     \biggl[ K' \Bigl(\frac{z-z_{t}}{h_{n}} \Bigr) - K' \biggl(\frac{\frac{j}{n^{*}}-z_{t}} {h_{n}}
      & \biggr)
     \\[-1.5ex]
     {} + E \biggl( K' \biggl(\frac{\frac{j}{n^{2}}-z_{t}}{h_{n}} \biggr) \biggr) - E \biggr( K'
    \Bigl(\frac{z-z_{t}}{h_{n}} \Bigr) & \biggr) \biggr] \Biggr|
    \end{aligned}
    \end{align*}
    \end{fleqn}

 \end{document} 

Answer 2

嵌套的对齐multline应该执行以下操作：

\documentclass[a4paper]{article}
\usepackage{mathtools,amssymb}

\begin{document}

\begin{multline*}
\max_{1 \leq j \leq n^{2}}
\sup_{z \in \bigl[\frac{j-1}{n^2}, \frac{j}{n^{2}}\bigr]}  
   \Biggl|
     \sum^{n}_{t=1} \biggl[
       V_{(t,1)} (z) -  V_{(t,1)}  \biggl( \frac{j}{n^{2}} \biggr) 
     \biggr]
   \Biggr| 
\\
= 
\max_{1 \leq j \leq n^{2}} \sup_{z \in \bigl[\frac{j-1}{n^2}, \frac{j}{n^{2}}\bigr]} 
  \begin{aligned}[t]
   & \Biggl|
     \sum^{n}_{t=1}  x^{*}_{t} I_{(|x^{*}_{t}| \leq n^{1/3}h^{\gamma}_{n}) }
     \biggl(
       K' \biggl(\frac{z-z_{t}}{h_{n}} \biggr) - 
       K' \biggl(\frac{\frac{j}{n^{*}}-z_{t}} {h_{n}}\biggr)
     \\
  & + E \biggl( K' \biggl(\frac{\frac{j}{n^{2}}-z_{t}}{h_{n}} \biggr) \biggr) - 
      E \biggl( K' \biggl(\frac{z-z_{t}}{h_{n}} \biggr) \biggr)
    \biggr)
  \Biggr|
\end{aligned}
\end{multline*}

\end{document}

在我看来，手动选择分隔符的大小更好。