当第一页空间不足时，我的 latex 文档不会生成第二页 pdf 页面。当我剪切中间部分时，应该出现在第二页上的文本会毫无问题地出现在第一页上。我也没有收到任何错误。
为什么会发生这种情况？
这是我的用于测试的 Tex。
\documentclass[11pt]{article}
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb,amsfonts}
\usepackage{textgreek}
\usepackage{enumitem}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\mP}{\mathbb{P}}
\newcommand{\calI}{\mathcal{I}}
\newcommand{\partialfrac}[1]{\frac{\partial}{\partial #1}}
\newtheorem{lemma}{Lemma}
\newenvironment{problem}[2][Problem]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries #2.}]}{\end{trivlist}}
\begin{document}
We note that,
\begin{equation*}
\begin{align}
G'(s) = (p-1)|s|^{p-2} sgn(s) s + |s|^{p-1} = p |s|^{p-1} \\
w' = (G (v))' = G'(v) v' = p |v|^{p-1} v'
\end{align}
\end{equation*}
Let $v\in C_c^1(\R)$. Following the hints, we show now that $|v(x)|^p \leq p ||v||^{p-1}_{L^p} ||v'||_{L^p}$
Since $v$ has compact support,
\begin{equation}
|v(x)|^p = ||v(x)|^{p-1}v(x)| = ||v(x)|^{p-1}v(x) - \lim\limits_{x \to -\infty}|v(x)|^{p-1}v(x)|= |[w \, ]_{-\infty}^x|
\end{equation}
By the fundamental theorem of integral calculus:
\begin{equation}
|[ w\, ]_{-\infty}^x | = | \int_{-\infty}^x w' | = | \int_{-\infty}^x p|v|^{p-1}v' dt| \leq \int_{-\infty}^x |p|v|^{p-1}v' dt| \leq \int_{\R} |p|v|^{p-1}v'|
\end{equation}
By H\"older inequality, $q = \frac{p}{p-1}$:
\begin{equation}
p \int_{\R} ||v|^{p-1}v'| = \Big(\int_\R (|v|^{p}) \Big)^{\frac{p-1}{p}}\Big(\int_\R |v'|^p\Big)^{\frac{1}{p}} = p ||v||_{L^p}^{p-1}||v'||_{L^p}
\end{equation}
And by Young's inequality:
\begin{equation}
p ||v||_{L^p}^{p-1}||v'||_{L^p} \leq p \Big( (p-1)\frac{||v||_{L^p}^p}{p} + \frac{||v'||_{L^p}^p}{p} \Big) \leq p (||v||_{L^p}^p + ||v'||_{L^p}^p) = p||v||_{W^{1,p}}^p
\end{equation}
So we have,
\begin{equation}
|v(x)| \leq p^{1/p} ||v||_{W^{1,p}}
\end{equation}
\begin{lemma}
For all $p\in \N$, $p^{1/p} \leq e^{1/e}$
\end{lemma}
\begin{proof}
Consider the function $f(x) = x^{1/x} = \exp{ 1/x log (x)}$.
\end{proof}
asdasdasdas
\end{document}
答案1
您有一个align嵌套在equation*环境中的环境，这是被禁止的。您可以使用aligned，但由于您的方程式没有编号，因此不使用 更align*简单equation*。
我借此机会改进了您的代码。如果您有多个方程式，且它们之间有简短的文本，则可以使用单个gather(*)环境和命令内的文本获得更好的垂直间距\intertext。
此外，为了获得正确的间距，请不要在equations 和正文之间添加空行。请注意，对于向量的范数，使用 获得的结果与使用 获得的结果不同|| v ||。\lVert v \rVert我使用命令from定义了一个\norm命令（以及一个\abs 命令）来获取可以轻松调整到内容大小的分隔符。\DeclarePairedDelimitermathtools
因此，无论如何，你的文本最终还是可以放在一页里……
\documentclass[11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{textgreek}
\usepackage[margin=1in]{geometry}
\usepackage{mathtools, amsthm, amssymb}
\DeclarePairedDelimiter{\abs}\lvert\rvert
\DeclarePairedDelimiter{\norm}\lVert\rVert
\DeclareMathOperator{\sgn}{sgn}
\usepackage{enumitem}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\mP}{\mathbb{P}}
\newcommand{\calI}{\mathcal{I}}
\newcommand{\partialfrac}[1]{\frac{\partial}{\partial #1}}
\newtheorem{lemma}{Lemma}
\newenvironment{problem}[2][Problem]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries #2.}]}{\end{trivlist}}
%If you want to title your bold things something different just make another thing exactly like this but replace "problem" with the name of the thing you want, like theorem or lemma or whatever
\begin{document}
\title{Funktionanalysis Blatt 8}
\author{Bernat Sopena Gilboy, Matrikelnummer: - (Gasth\"orerstudent) }
\maketitle
\begin{problem}{1}
\end{problem}
(a) Let $I \subset \R$ be an interval. We want to show that there is a constant $C$ such that
\begin{equation}
\norm{u}_{L^\infty} \leq C \norm{u}_{W^{1,p}} \; \forall u \in W^{1,p}(I), \, p = 1, 2 \dots
\label{ex1}
\end{equation}
Without restriction we can let $I = \R$. Suppose $I \neq \R$ and that we showed (1) for $I = \R$. On one side we have
\begin{gather}
\norm{Eu}_{L^\infty(\R)} \leq C_1 \norm{Eu}_{W^{1,p}(\R)} \leq C_1 C_2(p, I, \R) \norm{Eu}_{W^{1,p}(I)} \\
\intertext{On the other,}
\norm{Eu}_{L^\infty(\R)} \geq \norm{Eu}_{L^\infty(I)} = \norm{u}_{L^\infty(I)}
\end{gather}
So letting $C = C_1 C_2$ (\ref{ex1}) follows.
Henceforth, we assume $I = \R$. Let $G(s)\coloneqq |s|^{p-1}s$ and $w\coloneqq G(v)$. We note that
\begin{align*}
G'(s) & = (p-1)|s|^{p-2} \sgn(s) s + |s|^{p-1} = p |s|^{p-1} \\
w' & = (G (v))' = G'(v) v' = p |v|^{p-1} v'
\end{align*}
Let $v \in C_c^1(\R)$. Following the hints, we show now that $|v(x)|^p \leq p \norm{ v}^{p-1}_{L^p} \norm{v'}_{L^p}^{\vphantom{p}}$.
Since $v$ has compact support,
\begin{gather}
|v(x)|^p = \abs{|v(x)|^{p-1}v(x)}= \abs[\Big]{|v(x)|^{p-1}v(x) - \lim_{x \to -\infty}|v(x)|^{p-1}v(x)}= \abs*{[w \, ]_{-\infty}^x}. \\
\intertext{By the fundamental theorem of integral calculus:}
\abs*{[ w\, ]_{-\infty}^x } = \abs[\Big]{ \int_{-\infty}^x w' } = \abs[\Big]{\int_{-\infty}^x p|v|^{p-1}v' dt} \leq \int_{-\infty}^x \abs*{p|v|^{p-1}v' dt} \leq \int_{\R}\abs*{p|v|^{p-1}v'}. \\
\intertext{By Hölder inequality, $q = \frac{p}{p-1}$:}
p \int_{\R}\abs*{|v|^{p-1}v'} = \Bigl(\int_\R (|v|^{p}) \Bigr)^{\!\frac{p-1}{p}}\Bigl(\int_\R |v'|^p\Bigl)^{\!\frac{1}{p}} = p \norm{v}_{L^p}^{p-1}\norm{v'}_{L^p}, \\
\intertext{and by Young's inequality:}
p \norm{v}_{L^p}^{p-1}\norm{v'}_{L^p} \leq p \Bigl( (p-1)\frac{\norm{v}_{L^p}^p}{p} + \frac{\norm{v'}_{L^p}^p}{p} \Bigr) \leq p (\norm{v}_{L^p}^p + \norm{v'}_{L^p}^p) = p\norm{v}_{W^{1,p}}^p. \\
\intertext{So we have,}
|v(x)| \leq p^{1/p} \norm{v}{W^{1,p}}
\end{gather}
\begin{lemma}
For all $p\in \N$, $p^{1/p} \leq e^{1/e}$.
\end{lemma}
\begin{proof}
Consider the function $f(x) = x^{1/x} = \exp{ 1/x \log (x)}$.
\end{proof}
asdasdasdas
\end{document}
