我正在使用我所在机构提供的模板撰写论文（https://gradschool.utexas.edu/academics/theses-and-dissertations/digital-submission-requirement/latex-document-preparation)。当我使用内联数学时，垂直间距似乎会出现一些奇怪的现象；我之所以这么说，是因为当我将一些内联方程式“对齐”时，这种情况往往会得到改善。奇怪的是，它似乎使间距“更小”而不是更大（这很有道理）。话虽如此，我无法找出其中的韵律或原因，它可能与段落的长度有关，或者 latex 不想在多个页面上“打破”某些环境。我附上了一个例子。有人对如何改善这种情况有什么建议吗？整个内容应该是双倍行距，但你可以看到最后一段被挤压了……
主文件：
\documentclass[12pt]{report} % The documentclass must be ``report''.
\usepackage{utdiss2} % Dissertation package style file.
\usepackage{amsmath,amsthm,amsfonts,amscd}
%
\usepackage{eucal} % Euler fonts
\usepackage{verbatim} % Allows quoting source with commands.
\usepackage{makeidx} % Package to make an index.
\usepackage{epsfig} % Allows inclusion of eps %
\usepackage{url} % Allows good typesetting of web URLs.
\usepackage{IEEEtrantools}
\topmargin 0.125in
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{ax}{Axiom}
\begin{document}
\chapter{Introduction}
we must demonstrate that the KL divergences $D_{\mathrm{KL}}(\left.\mathbf{q}_{{t}}||\mathbf{q}\right|\boldsymbol{\delta}_{{t}}),D_{\mathrm{KL}}(\mathbf{q}_{{t}}||\mathbf{q})$ tend to $0$ as ${t} \rightarrow \infty$. This is the subject of the following lemma.
\begin{lemma}
We have that both \begin{align}
\lim_{{t}\rightarrow\infty} D_{\mathrm{KL}}(\left.\mathbf{q}_{{t}}||\mathbf{q}\right|\boldsymbol{\delta}_{{t}}) =0 \text{ and }\lim_{{t}\rightarrow\infty} D_{\mathrm{KL}}(\mathbf{q}_{{t}}||\mathbf{q}) =0.
\end{align}
\end{lemma}
\begin{proof}
It can be shown via Jensen's inequality that if $\mathbf{a},\mathbf{b}$ are random variables that are absolutely continuous with respect to Lebesgue measure such that $\mathbf{a}$ is absolutely continuous with respect to $\mathbf{b}$, then $D_{\mathrm{KL}}(Q_{{\Delta}}(\mathbf{a})||Q_{{\Delta}}(\mathbf{b}))\le D_{\mathrm{KL}}(\mathbf{a}||\mathbf{b})$. Thus,
we have $D_{\mathrm{KL}}(\mathbf{q}_{{t}}||\mathbf{q}) \le D_{\mathrm{KL}}(C\mathbf{e}_{{t}}+\boldsymbol{\delta}_{{t}}||C\mathbf{e}+\boldsymbol{\delta})$. Since $\boldsymbol{\delta}_{{t}}$ and $\boldsymbol{\delta}$ are identically distributed, $\mathbf{e}_{{t}}\perp\boldsymbol{\delta}_{{t}}$, and $\mathbf{e}\perp\boldsymbol{\delta}$, the data processing inequality (DPI) for KL divergences (cf. [44, Theorem 2.15]) gives $ D_{\mathrm{KL}}(C\mathbf{e}_{{t}}+\boldsymbol{\delta}_{{t}}||C\mathbf{e}+\boldsymbol{\delta})\le D_{\mathrm{KL}}(\mathbf{e}_{{t}}||\mathbf{e})$.
The proof that $D_{\mathrm{KL}}(\left.\mathbf{q}_{{t}}||\mathbf{q}\right|\boldsymbol{\delta}_{{t}})\le D_{\mathrm{KL}}(\mathbf{e}_{{t}}||\mathbf{e})$ is analogous. To begin, recognize that for each $\delta \in [-\frac{{\Delta}}{2},\frac{{\Delta}}{2}]^n$, $D_{\mathrm{KL}}(\left.\mathbf{q}_{{t}}||\mathbf{q}\right|\boldsymbol{\delta}_{{t}}=\delta)\le D_{\mathrm{KL}}(\left.C\mathbf{e}_{{t}}+\delta|| C\mathbf{e}+\delta \right|\boldsymbol{\delta}_{{t}}=\delta)$ where both $\mathbf{e}_{{t}}\perp \boldsymbol{\delta}_{{t}}$ and $\mathbf{e}\perp \boldsymbol{\delta}_{{t}}$. Applying the DPI for every realization $\delta$ and using the fact that, by independence, $\mathbb{P}_{\mathbf{e}_{{t}}|\boldsymbol{\delta}_{{t}}} = \mathbb{P}_{\mathbf{e}_{{t}}}$ and likewise $\mathbb{P}_{\mathbf{e}|\boldsymbol{\delta}} = \mathbb{P}_{\mathbf{e}}$ completes the argument. Thus, we can prove the lemma by demonstrating that $\lim_{{t}\rightarrow\infty} D_{\mathrm{KL}}(\mathbf{e}_{{t}}||\mathbf{e})=0$.
Let $\{\boldsymbol{\nu}_{{t}}\}$ denote an IID sequence of random variables uniformly distributed on $[-{\Delta}/2,{\Delta}/2]^{{m}}$, let $\{\boldsymbol{\omega}_{{t}}\}$ be IID with $\boldsymbol{\omega}_{{t}}\in\mathcal{N}(0_{{m}},W)$, and let $\boldsymbol{\lambda}\sim\mathcal{N}(0_{{m}},X_{0})$. Assume $\{\boldsymbol{\omega}_{{t}}\}$, $\{\boldsymbol{\nu}_{{t}}\}$, and $\boldsymbol{\lambda}$ are mutually independent. Let ``$\overset{D}{=}$" denote ``equality in distribution", e.g., we write $\mathbf{a}\overset{D}{=}\mathbf{b}$ to imply $\mathbf{a}$ and $\mathbf{b}$ are identically distributed. From (22), we have $\mathbf{e}_{{t}} =R\mathbf{e}_{{t}-1} -L\mathbf{v}_{{t}-1}+\mathbf{w}_{{t}-1}$. Via Prop. 4 Claim (iii) and the factorization of system variables in (44), it can be verified that $\mathbf{w}_{{t}}\perp \mathbf{e}^{{t}}, \mathbf{v}^t,\mathbf{w}^{{t}-1}$ and $\mathbf{v}_{{t}}\perp \mathbf{e}^{{t}}, \mathbf{v}^{{t}-1},\mathbf{w}^{{t}}$.
Thus, by this recursive definition of $\{\mathbf{e}_{{t}}\}$, $ \mathbf{e}_{{t}} \overset{D}{=} R^{{t}}\boldsymbol{\lambda}+\sum_{{i}=0}^{{t}-1}R^{{i}}(\boldsymbol{\omega}_{{i}}-L\boldsymbol{\nu}_{{i}}). $
Likewise, by definition of $\mathbf{e}$, we have that both $\mathbf{e} \overset{D}{=} \lim_{{t}\rightarrow\infty}R^{{t}}\boldsymbol{\lambda}+\sum_{{i}=0}^{{t}-1}R^{{i}}(\boldsymbol{\omega}_{{i}}-L\boldsymbol{\nu}_{{i}})$ and $\mathbf{e} \overset{D}{=}\lim_{{t}\rightarrow\infty}\sum_{{i}=0}^{{t}-1}R^{{i}}(\boldsymbol{\omega}_{{i}}-L\boldsymbol{\nu}_{{i}})$, which follows since Lemma 5's convergence in total variation implies weak convergence.
We have
\begin{IEEEeqnarray}{rCl}
D_{\mathrm{KL}}(\mathbf{e}_{{t}}|| \mathbf{e} ) &=& D_{\mathrm{KL}}(R^{{t}}\boldsymbol{\lambda}+\mathbf{g}_{\le {t}} +\mathbf{u}_{\le {t}}||\mathbf{g}_{\le {t}} +\mathbf{u}_{\le {t}}+\mathbf{s}_{>{t}} )\nonumber \\ &\le& D_{\mathrm{KL}}(R^{{t}}\boldsymbol{\lambda}+\mathbf{g}_{\le {t}}||\mathbf{g}_{\le {t}} +\mathbf{s}_{>{t}} )\label{eq:usedidiv} \\ &\le& D_{\mathrm{KL}}(\left.R^{{t}}\boldsymbol{\lambda}+\mathbf{g}_{\le {t}} || \mathbf{g}_{\le {t}}+\mathbf{s}_{>{t}} \right| \mathbf{s}_{>{t}} ),\label{eq:condincdiv}
\end{IEEEeqnarray} where (\ref{eq:usedidiv}) follows from the data processing inequality for KL divergence and (\ref{eq:condincdiv}) follows since conditioning increases KL divergence (see [44, Theorem 2.14 (e)]).
\end{proof}
\end{document}
标头：
% Dimension Setting %
%%%%%%%%%%%%%%%%%%%%%
% Top margin is now set in the disstemplate.tex file.
% 29 Nov 2001 cwm
\oddsidemargin 0.625in \evensidemargin 0.625in
\textwidth 5.5in
\headheight .21in
\headsep .29in % \headheight + \headsep = 0.5in
\textheight 8in % \textheight includes \footskip
\footskip .5in
\skip\footins 22.8pt plus 4pt minus 2pt
\parindent .5in
\parskip 5 pt plus 1.5pt minus .5pt
\itemsep 0 pt plus .5pt
\pagenumbering{roman}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Macros to Modify Some Default Settings of `report.sty'. %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%
% Macros to make:
% the text height of the first page of every chapter
% = \textheight - \footskip
%
\newlength{\regular@textheight}
\setlength{\regular@textheight}{\textheight}
%
\addtolength{\regular@textheight}{-\footskip}
% Added -\footskip to move bottom
% page numbers on non-first-page-of-chapter
% pages up so they meet the Graduate School
% spec. 24 Sep 2001 cwm
\newlength{\short@textheight}
\setlength{\short@textheight}{\textheight}
\addtolength{\short@textheight}{-\footskip}
\def\short@page{%
\setlength{\textheight}{\short@textheight}
\global \pagestyle{plain} % Added to put page numbers
% at bottom of short pages.
% 24 Sep 2001 cwm
\global \@colht\textheight
\global \@colroom\textheight
\global \vsize\textheight}
\def\regular@page{%
\setlength{\textheight}{\regular@textheight}
\global \pagestyle{plain} % Added to put page numbers
% at bottom of regular pages.
% 24 Sep 2001 cwm
\global \@colht\textheight
\global \@colroom\textheight
\global \vsize\textheight}
\newcounter{regular@short}
\let\old@opcol\@opcol
\def\@opcol{%
\ifnum\value{regular@short}=0 \regular@page\old@opcol
\else\short@page\old@opcol\fi}
% Regarding the space after the zero in the value setting above:
%
% "For best results, ALWAYS PUT A BLANK SPACE AFTER A NUMERIC CONSTANT;
% this blank space tells TeX that the constant is complete, and
% such a space will never "get through" to the output."
% " ... a missing space ... might cause TeX to expand macros when you
% don't want any expansion, and such anomalies can cause subtle and
% confusing errors"
% - the TeXbook, p. 208. cwm
%
% Macros to set spacing of text and quotations.
% ^^^^ ^^^ ^^^^^^^^^^
\def\single@space{%
\vskip\baselineskip
\baselineskip=15.5pt plus .5pt minus .2pt
\vskip-\baselineskip}
\def\endsingle@space{\par}
\def\oneandonehalf@space{%
\vskip\baselineskip
\baselineskip=20.5pt plus .5pt minus .2pt
\vskip-\baselineskip}
\def\endoneandonehalf@space{\par}
\def\double@space{%
\vskip\baselineskip
\baselineskip=23.5pt plus .5pt minus .2pt
\vskip-\baselineskip}
\def\enddouble@space{\par}
%
% Regular text spacing
%
\def\doublespacing{%
\def\default@spacing{\baselineskip=23.5pt plus .5pt minus .2pt}}
\def\oneandonehalfspacing{%
\def\default@spacing{\baselineskip=20.5pt plus .5pt minus .2pt}}
\def\singlespacing{%
\def\default@spacing{\baselineskip=15.5pt plus .5pt minus .2pt}}
%
% Stolen from `report.sty'. Modified to change the size of left skip
% for \chapter title which is longer than one line.
%
\newlength{\toc@chap@indent}
\settowidth{\toc@chap@indent}{\bf\@chapapp\space\thechapter.\hspace*{1em}}
\def\l@chapter#1#2{\addpenalty{-\@highpenalty}
\vskip 1.0em plus 1pt
\@tempdima=\toc@chap@indent
\begingroup
\parindent \z@
\rightskip \@pnumwidth
\parfillskip -\@pnumwidth
\bf \leavevmode
\advance\leftskip\@tempdima \hskip -\leftskip #1\nobreak\hfil
\nobreak\hbox to\@pnumwidth{\hss #2}\par
\penalty\@highpenalty \endgroup}
\def\@makechapterhead#1{ \vspace*{0pt} {\parindent 0pt \centering
\ifnum \c@secnumdepth >\m@ne \Large\bf \@chapapp{} \thechapter
\par \vskip 15pt \fi \Large \bf #1\par \nobreak \vskip 35pt}}
\def\@makeschapterhead#1{\vspace*{0pt} {\parindent 0pt \centering
\Large \bf #1\par \nobreak \vskip 30pt}}
\def\chapter{%
\clearpage
\short@page \setcounter{regular@short}{0} %<<--
\default@spacing
%
% Comment out one of the two lines below to determine indention of
% first lines of chapters. 28 Sep 2001 cwm
%
\global\@topnum\z@ \@afterindenttrue % Indent 1st line of chapter.
% \global\@topnum\z@ \@afterindentfalse % Do NOT indent 1st line
% of chapter.
\secdef\@chapter\@schapter}
\def\shortchapter{%
\clearpage
\short@page \setcounter{regular@short}{1} %<<--
\default@spacing
%
% Comment out one of the two lines below to determine indention of
% first lines of short chapters. 28 Sep 2001 cwm
%
\global\@topnum\z@ \@afterindenttrue % Indent 1st line of short
% chapter.
% \global\@topnum\z@ \@afterindentfalse % Do NOT indent 1st line
% of short chapter.
\secdef\@chapter\@schapter}
\newcount\chap@or@app\chap@or@app=1
% Redefinition of \@part for using by \part (M.A.L.)
%
\newcount\with@parts\with@parts=0 % <- this would allow to restart
% page number with Chapter 1 (MAL)
\def\@part[#1]#2{%
\short@page % <- Added 4/21/97 (MAL)
\ifnum \c@secnumdepth >-2 \relax
\refstepcounter{part}% % Added \partname~ below (M.A.L.)
\addcontentsline{toc}{part}{\partname~\thepart\hspace{1em}#1}%
\else
\addcontentsline{toc}{part}{#1}%
\fi
\markboth{}{}
{\centering
\interlinepenalty \@M
\reset@font
\ifnum \c@secnumdepth >-2 \relax
\huge\bfseries \partname~\thepart
\par
\vskip 20\p@
\fi
\Huge \bfseries #2\par}%
\ifnum\value{part}=1 \pagenumbering{arabic}% % if Part I
\with@parts=1 \fi% % <- to avoid restarting page number with chapter 1
\@endpart}
%
%%%% End redefinition of \@part
\newcounter{no@chapters} % (7/16/97 M.A.L.)
\setcounter{no@chapters}{0} %
\newcommand\nochapters{% %
\setcounter{no@chapters}{1} % (7/16/97 M.A.L.)
\fpage \setcounter{regular@short}{0} %
%
% Comment out one of the two lines below to determine indention of
% first lines of text with no chapters.
% 28 Sep 2001 cwm
%
\global\@topnum\z@ \@afterindenttrue % Indent 1st lines.
% \global\@topnum\z@ \@afterindentfalse % Do NOT indent 1st lines.
%
\pagenumbering{arabic} % Added for using with
\markboth{}{}\pagestyle{myheadings} % no chapters
\addtocontents{toc}{\protect\addvspace{10pt}} % (4/22/97, MAL)
\default@spacing % (6/30/97 M.A.L.)
%
} %
\def\@chapter[#1]#2{\ifnum \c@secnumdepth >\m@ne
\refstepcounter{chapter}
\typeout{\@chapapp\space\thechapter.}
\addcontentsline{toc}{chapter}{\@chapapp\space\thechapter.\hspace*{1em}#1}\else
\addcontentsline{toc}{chapter}{#1}\fi
% \chaptermark{#1}
\addtocontents{lof}{\protect\addvspace{10pt}}
\addtocontents{lot}{\protect\addvspace{10pt}}
\@makechapterhead{#2} \@afterheading
\ifnum\chap@or@app=1 \ifnum\value{chapter}=1 % if Chapter 1
\markboth{}{}\pagestyle{myheadings}
\ifnum\with@parts=0 \pagenumbering{arabic}\fi% Restart page number only
% if there are no parts (MAL)
\fi\fi
\thispagestyle{plain}}
\def\@schapter#1{\thispagestyle{plain}
\@makeschapterhead{#1} \@afterheading}
\doublespacing
编辑：添加了一个示例来说明问题（尽管在多个页面上）。另请参阅https://www.overleaf.com/read/ddgbkzmnqsjn。我已经修剪了标题以消除多余的命令等。
答案1
它与内联数学无关，从示例中删除不相关的代码会产生更合理的示例：
\documentclass[12pt]{report} % The documentclass must be ``report''.
%
\usepackage{IEEEtrantools}
\makeatletter
\def\doublespacing{%
\def\default@spacing{\baselineskip=23.5pt plus .5pt minus .2pt}}
\doublespacing\default@spacing
\topmargin 0.125in
\begin{document}
\doublespacing\default@spacing
\chapter{Introduction}
It can be shown via Jensen's inequality that if It can be shown via Jensen's inequality that if
It can be shown via Jensen's inequality that if It can be shown via Jensen's inequality that if
It can be shown via Jensen's inequality that if It can be shown via Jensen's inequality that if
It can be shown via Jensen's inequality that if It can be shown via Jensen's inequality that if
It can be shown via Jensen's inequality that if It can be shown via Jensen's inequality that if
It can be shown via Jensen's inequality that if It can be shown via Jensen's inequality that if
denote an IID sequence of random variables uniformly distributed on
denote an IID sequence of random variables uniformly distributed on
denote an IID sequence of random variables uniformly distributed on
denote an IID sequence of random variables uniformly distributed on
denote an IID sequence of random variables uniformly distributed on
denote an IID sequence of random variables uniformly distributed on
denote an IID sequence of random variables uniformly distributed on
denote an IID sequence of random variables uniformly distributed on
denote an IID sequence of random variables uniformly distributed on
\begin{IEEEeqnarray}{rCl}
a&=&b
\end{IEEEeqnarray}
\end{document}
问题在于utdss2定义\doublespace更改 baslineskip 的方式，这意味着任何 latex size 命令（例如）\normalsize都将丢失设置。您可以改用包setspace，或者为了快速修复，在 eqnarray 之前放置一个空行。