我是一名视力严重受损的数学学生。我想知道如何才能加快我的 latex 文档创建速度。
我在下面粘贴了 Kreyszig 的一本功能分析书中一道题的答案。这也是我昨天在 Maths SE 网站上发布的一道题。我得把所有答案都打出来，还要时不时地删除我在编写作业时遇到的语法错误，这让我感到压力很大。
谁能给我一些建议，如何提高我的效率，以及如何在学习数学的过程中利用盲人和视障人士可用的辅助技术，以及我自己创建这样的数学内容？
Prob. 2.7-10
On $C[0,1]$ define $S$ and $T$ by
$$
y(s) = \ s \int_0^1 \ x(t) \ \mathrm{d} t \ \ \ \ \ \ \ \ \ y(s) \ = \ s x(s),
$$
respectively. Do $S$ and $T$ commute? Find $\Vert S \Vert$, $\Vert T \Vert$, $\Vert ST \Vert$ and $\Vert TS \Vert$.
Solution
Recall that $C[0,1]$ denotes the normed space of all (real- or complex-valued) functions defined and continuous on the closed unit interval $[0,1]$ of the real line $\mathbb{R}$, with the norm defined by
$$
\Vert x \Vert_{C[0,1]} \colon= \max_{t\in[0,1]} \vert x(t) \vert \ \ \ \mbox{ for all } \ x \in C[0,1].
$$
Let $K$ denote either $\mathbb{R}$ or $\mathbb{C}$.
The operator $S \colon C[0,1] \to C[0,1]$ is defined as follows: for each $x \in C[0,1]$, let the map $S(x)
\colon [0,1] \to K$ be defined by
$$
\left( S(x) \right) (s) \ \colon= \ s \int_0^1 x(t) \, \mathrm{d} t \ \ \
\mbox{ for all } \ s \in [0,1].
$$
And, the operator $T \colon C[0,1] \to C[0,1]$ is defined as follows: for each $x \in C[0,1]$, let the map
$T(x) \colon [0,1] \to K$ be defined by
$$
\left( T(x) \right) (s) \ \colon= \ s \cdot x(s) \ \ \
\mbox{ for all } \ s \in [0,1].
$$
By $ST$ and $TS$, Kreyszig intends the composite maps $S \circ T$ and $T \circ S$, respectively. Thus $S$ and $T$ commute if and only if the maps $S \circ T$ and $T \circ S$ are equal.
For any $x \in C[0,1]$, the map $\left( S \circ T \right) (x) \colon [0,1] \to K$ is computed as follows:
\begin{eqnarray*}
\left( \left( S \circ T \right) (x) \right) (s)
&=& \left( \ S \left( T (x) \right) \ \right) (s) \\
&=& S(sx(s)) \ \ \ \mbox{ [We apply $T$ first, followed by $S$. ] } \\
&=& s \int_0^1 tx(t) \ \mathrm{d} t \ \ \
\mbox{ for all } \ s \in [0,1],
\end{eqnarray*}
while the map $\left( T \circ S \right) (x) \colon [0,1] \to K$ is computed as follows:
\begin{eqnarray*}
\left( \left( T \circ S \right) (x) \right) (s)
&=& \left( \ T \left( S (x) \right) \ \right) (s) \\
&=& T\left(s \int_0^1 \ x(t) \ \mathrm{d} t \ \right) \ \ \ \mbox{ [We first apply $S$ and then $T$.]} \\
&=& s^2 \int_0^1 \ x(t) \ \mathrm{d} t \ \ \
\mbox{ for all } \ s \in [0,1].
\end{eqnarray*}
Thus, $T \circ S \neq S \circ T$ because for the point $x_0 \in C[0,1]$, where $x_0 \colon [0,1] \to K$ is defined as $x_0(t) \colon= 1$ for all $t \in [0,1]$, we see that,
$$
\left( \left( T \circ S \right) (x_0) \right) (s) = s^2 \int_0^1 \ 1 \ \mathrm{d} t
= s^2 \ \ \ \mbox{ for all } \ s \in [0,1],
$$
whereas
$$
\left( \left( S \circ T \right) (x_0) \right) (s) = s \int_0^1 \ t \ \mathrm{d} t
= \frac{s}{2} \ \ \ \mbox{ for all } \ s \in [0,1];
$$
Now $s^2 = s/2$ if and only if $s = 1/2$ or $s=0$;
thus if $s \neq 1/2$ and $s \neq 0$, then
$$
\left( \left( S \circ T \right) (x_0) \right) (s) \ \not= \ \left( \left( T \circ S \right) (x_0) \right) (s). $$
Hence $S$ and $T$ do not commute.
Before we compute the norms, we mention a couple of results from integration theory.
\subsubsection*{Result 1}
Let $f$ be a real or complex-valued function defined and continuous on a closed interval $[a,b]$. Then
the integral $\int_a^b \ f(t) \ \mathrm{d} t $ exists.
Refer to Theorem 6.8 in \textit{Principles of Mathematical Analysis} by Walter Rudin, 3rd edition. We have already used this result.
Now we are also going to use the following two results:
\subsubsection*{Result 2}
If $f$ and $g$ are two real-valued functions integrable on $[a,b]$ such that $f(t) \leq g(t)$ for all $t \in [a,b]$, then we also have $\int_a^b \ f(t) \ \mathrm{d} t \ \leq \ \int_a^b \ g(t) \ \mathrm{d} t \ $.
Refer to Theorem 6.12(b) in \textit{Principles of Mathematical Analysis} by Walter Rudin, 3rd edition.
\subsubsection*{Result 3}
If $f$ is a real or complex-valued function defined on a closed interval $[a,b]$ such that $\int_a^b \ f(t) \ \mathrm{d} t $ exists, then the integral $\int_a^b \ \vert f(t) \vert \ \mathrm{d} t $ exists too and we also have the inequality
$$
\left\vert \ \int_a^b \ f(t) \ \mathrm{d} t \ \right\vert
\ \leq \ \int_a^b \ \vert f(t) \vert \ \mathrm{d} t.
$$
Refer to Theorem 6.13(b) in \textit{Principles of Mathematical Analysis} by Walter Rudin, 3rd edition.
To compute the norms, we see that, for any $x \in C[0,1]$,
\begin{eqnarray*}
\Vert S(x) \Vert_{C[0,1]}
&=& \max_{s \in [0,1]} \left\vert \left( S (x) \right) (s) \right\vert \\
&=& \max_{s\in [0,1]} \left\vert s \int_0^1 \ x(t) \ \mathrm{d} t \right\vert \\
&=& \max_{s \in [0,1]} \left( \vert s \vert \ \cdot \ \left\vert \int_0^1 \ x(t) \ \mathrm{d} t \right\vert \right) \\
&=& \left( \max_{s\in[0,1]} \vert s \vert \right) \ \cdot \ \left\vert \int_0^1 \ x(t) \ \mathrm{d} t \right\vert \\
&=& 1 \ \cdot \ \left\vert \int_0^1 \ x(t) \ \mathrm{d} t \right\vert \\
&\leq& \int_0^1 \ \vert x(t) \vert \ \mathrm{d} t \\
&\leq & \int_0^1 \ \max_{\tau\in[0,1]} \vert x(\tau) \vert \ \mathrm{d} t \\
&=& \int_0^1 \ \Vert x \Vert_{C[0,1]} \ \mathrm{d} t \\
&=& \Vert x \Vert_{C[0,1]} \ \int_0^1 \ \mathrm{d} t \\
&=& \Vert x \Vert_{C[0,1]}.
\end{eqnarray*}
So, the operator $S$ is bounded, by Definition 2.7-1 in Kreyszig.
Now, from the last relation (upon division on both sides by $\Vert x \Vert_{C[0,1]}$), we conclude that, for all non-zero $x \in C[0,1]$,
$$
\frac{\Vert S(x) \Vert_{C[0,1]}}{\Vert x \Vert_{C[0,1]}} \leq 1;
$$
thus the real number $1$ is an upper bound of the following subset of $\mathbb{R}$:
$$
\left\{ \ \frac{\Vert S(x) \Vert_{C[0,1]}}{\Vert x \Vert_{C[0,1]}} \ \colon \ x \in C[0, 1] \ x \neq \theta_{C[0,1]} \ \right\}.
$$
where $ \theta_{C[0,1]} $ denotes the "zero vector" in $C[0,1]$ (i.e. the zero function defined on $[0,1]$ ).
and since $\Vert S \Vert$ is by definition the supremum (or least upper bound) of this set, therefore
\begin{equation}
\Vert S \Vert \leq 1. \ \label{2.7-10-1}
\end{equation}
But for $x_0 \in C[0,1]$ such that $x_0(t) \colon= 1$ for all $t \in [0,1]$, we see that
$$
\Vert x_0 \Vert_{C[0,1]} = \max_{t\in[0,1]} \vert x_0(t) \vert = 1.
$$
Moreover the map $S(x_0) \colon [0,1] \to K$ is given by
$$
\left( S(x_0) \right) (s) \ = \ s \ \int_0^1 \ x_0(t) \ \mathrm{d} t \ = \
s \ \int_0^1 \ 1 \ \mathrm{d} t \ = \ s \ \ \ \mbox{ for all } \ s \in [0,1].
$$
That is, the map $S(x_0) \colon [0,1] \to K$ is given by
$$
\left( S(x_0) \right) (s) \ = \ s \ \ \ \mbox{ for all } \ s \in [0,1].
$$
So,
$$
\left\Vert S(x_0) \right\Vert_{C[0,1]} = \max_{s\in[0,1]} \left\vert \left( S(x_0) \right) (s) \right\vert = \max_{s\in[0,1]} \vert s \vert = 1.
$$
But by definition of the norm $\Vert S \Vert$, we also have
\begin{equation}
\Vert S \Vert \geq \frac{\left\Vert S(x_0) \right\Vert_{C[0,1]}}{\Vert x_0 \Vert_{C[0,1]}} = \frac{1}{1}
= 1.
\label{2.7-10-2}
\end{equation}
Hence from (\ref{2.7-10-1}) and (\ref{2.7-10-2}), we conclude that
$$
\Vert S \Vert = 1.
$$
Now, for any $x \in C[0,1]$, we have
\begin{eqnarray*}
\Vert T(x) \Vert_{C[0,1]}
&=& \max_{s \in [0,1]} \left\vert \left( T(x) \right) (s) \right\vert \\
&=& \max_{s\in[0,1]} \vert s x(s) \vert \\
&=& \max_{s\in[0,1]} \left(\ \vert s \vert \ \vert x(s) \vert \ \right) \\
&\leq& \max_{s\in[0,1]} \vert x(s) \vert \\
&=& \Vert x \Vert_{C[0,1]},
\end{eqnarray*}
which shows that $T$ is bounded, and upon taking the supremum over all $x \in C[0,1]$ such that
$\Vert x \Vert_{C[0,1]} = 1$, we obtain
\begin{equation}
\Vert T \Vert \leq 1.
\label{2.7-10-3}
\end{equation}
But for the point $x_0 \in C[0, 1]$, where $x_0 \colon [0,1] \to K$ is defined as \newline
$x_0(t) \colon= 1$ for all $t \in [0,1]$, we see as before that
$$
\Vert x_0 \Vert_{C[0,1]} = 1.
$$
And the map $T(x_0) \colon [0,1] \to K$ is given by
$$
\left( T(x_0) \right) (s) \ = \ s x_0(s) = s \ \ \ \mbox{ for all } \ s \in [0,1].
$$
So
$$
\Vert T(x_0) \Vert_{C[0,1]} \ = \ \max_{s\in [0,1]} \left\vert \ \left( T(x_0) \right) (s) \ \right\vert
\ = \ \max_{s\in [0,1]} \vert s \vert = 1,
$$
and therefore by definition
\begin{equation}
\Vert T \Vert \ \geq \ \frac{\Vert T(x_0) \Vert_{C[0,1]} }{ \Vert x_0 \Vert_{C[0,1]} } = \frac{1}{1} = 1.
\label{2.7-10-4}
\end{equation}
Hence (\ref{2.7-10-3}) and (\ref{2.7-10-4}) together yield
$$
\Vert T \Vert = 1.
$$
Now comes the turn of finding the norms of the composite operators. Since both $S$ and $T$ are bounded linear operators [Verify the linearity of $S$ and $T$ for yourself.], we can conclude that both $S \circ T$ and $T \circ S$ are linear and bounded operators, by Problem 2.6-6 and Problem 2.7-7 above.
We now show directly the boundedness of these composite operators and find their norms. We have found earlier in this solution that, for each $x \in C[0,1]$, the maps $\left( S \circ T \right) (x) \colon [0,1] \to K$ and $\left( T \circ S \right) (x) \colon [0,1] \to K$ are given, respectively by
\begin{equation}
\left( \ \left( S \circ T \right) (x) \ \right) (s) \ = \ s \ \int_0^1 \ t x(t) \ \mathrm{d} t
\ \ \ \mbox{ for all } \ s \in [0, 1],
\label{2.7-10-5}
\end{equation}
and
\begin{equation}
\left( \ \left( T \circ S \right) (x) \ \right) (s) \ = \ s^2 \ \int_0^1 \ x(t) \ \mathrm{d} t
\ \ \ \mbox{ for all } \ s \in [0, 1].
\label{2.7-10-6}
\end{equation}
So
\begin{eqnarray*}
\left\Vert \left( \ \left( S \circ T \right) (x) \ \right) \right\Vert_{C[0,1]}
&=& \max_{s \in [0,1]} \left\vert \ \left( \ \left( S \circ T \right) (x) \ \right) (s) \ \right\vert \\
&=& \max_{s \in [0,1]} \left\vert \ s \ \int_0^1 \ t x(t) \ \mathrm{d} t
\ \right\vert \\
&=& \max_{s \in [0,1]} \left( \ \vert s \vert \ \left\vert \int_0^1 \ t x(t) \ \mathrm{d} t
\ \right\vert \ \right) \\
&=& \left( \max_{s \in [0,1]} \vert s \vert \right) \ \left\vert \ \int_0^1 \ t x(t) \ \mathrm{d} t
\ \right\vert \\
&=& \left\vert \ \int_0^1 \ t x(t) \ \mathrm{d} t
\ \right\vert \\
&\leq & \int_0^1 \ \left\vert \ t x(t) \ \right\vert \ \mathrm{d} t \\
&=& \int_0^1 \ \left( \ \vert t \vert \ \vert x(t) \vert \ \right) \ \mathrm{d} t \\
&\leq& \int_0^1 \ \left( \ \vert t \vert \ \left( \max_{\tau \in [0,1]} \vert x(\tau) \vert \right) \ \right) \ \mathrm{d} t \\
&=& \int_0^1 \ \left( \ \vert t \vert \ \Vert x \Vert_{C[0,1]} \ \right) \ \mathrm{d} t \\
&=& \Vert x \Vert_{C[0,1]} \ \int_0^1 \ \vert t \vert \ \mathrm{d} t \\
&=& \Vert x \Vert_{C[0,1]} \ \int_0^1 \ t \ \mathrm{d} t \\
&=& \frac{1}{2} \Vert x \Vert_{C[0,1]} \ \ \ \mbox{ for all } \ x \in C[0,1],
\end{eqnarray*}
which shows that the operator $S \circ T$ is bounded and upon taking the supremum over all $x \in C[0,1]$ of norm one, we obtain
\begin{equation}
\Vert S \circ T \Vert \leq \frac{1}{2}.
\label{2.7-10-7}
\end{equation}
Now for the point $x_0 \in C[0,1]$ such that $x_0(t) \colon= 1$ for all $t \in [0,1]$, we have seen earlier on in this solution that the map $\left( S \circ T \right) (x_0) \colon [0,1] \to K$ is given by
$$
\left( \ \left( S \circ T \right) (x_0) \ \right) (s) = \frac{s}{2} \ \ \
\mbox{ for all } \ s \in [0,1].
$$
So
\begin{eqnarray*}
\left\Vert \ \left( \ S \circ T \ \right) (x_0) \ \right\Vert_{C[0,1]}
&=& \max_{s \in [0,1]} \left\vert \ \left( \ \left( \ S \circ T \ \right) (x_0) \ \right) (s) \ \right\vert \\
&=& \max_{s \in [0,1]} \left\vert \ \frac{s}{2} \ \right\vert \\
&=& \max_{s \in [0,1]} \frac{s}{2} \\
&=& \frac{1}{2},
\end{eqnarray*}
and $\Vert x_0 \Vert_{C[0,1]} = 1$. So we have
\begin{equation}
\Vert S \circ T \Vert \ \geq \ \frac{\left\Vert \ \left( \ S \circ T \ \right) (x_0) \ \right\Vert_{C[0,1]} }{\Vert x_0 \Vert_{C[0,1]} } \ = \frac{ 1/2 }{1} = \frac{1}{2}.
\label{2.7-10-8}
\end{equation}
Therefore (\ref{2.7-10-7}) and (\ref{2.7-10-8}) together yield
$$ \Vert S \circ T \Vert = \frac{1}{2}. $$
Now using (\ref{2.7-10-6}), we see that
\begin{eqnarray*}
\left\Vert \ \left( T \circ S \right) (x) \ \right\Vert_{C[0,1]}
&=& \max_{s \in [0,1]} \left\vert \ \left( \ \left( T \circ S \right) (x) \ \right) (s) \ \right\vert \\
&=& \max_{s \in [0,1]} \left\vert \ s^2 \ \int_0^1 \ x(t) \ \mathrm{d} t \ \right\vert \\
&=& \max_{s \in [0,1]} \left( \ \vert s^2 \vert \ \left\vert \ \int_0^1 \ x(t) \ \mathrm{d} t \ \right\vert \ \right) \\
&=& \left( \ \max_{s \in [0,1]} \vert s^2 \vert \ \right) \ \left\vert \ \int_0^1 \ x(t) \ \mathrm{d} t \ \right\vert \\
&=& \left\vert \ \int_0^1 \ x(t) \ \mathrm{d} t \ \right\vert \\
&\leq & \int_0^1 \ \vert x(t) \vert \ \mathrm{d} t \\
&\leq& \int_0^1 \ \max_{\tau \in [0,1]} \vert x(\tau) \vert \ \mathrm{d} t \\
&=& \int_0^1 \ \Vert x \Vert_{C[0,1]} \ \mathrm{d} t \\
&=& \Vert x \Vert_{C[0,1]} \ \int_0^1 \ \mathrm{d} t \\
&=& \Vert x \Vert_{C[0,1]} \ \ \ \mbox{ for all } \ x \in C[0,1],
\end{eqnarray*}
showing that $T \circ S$ is bounded and it also follows that
\begin{equation}\label{2.7-10-9}
\Vert T \circ S \Vert \leq 1.
\end{equation}
Now for the point $x_0 \in C[0,1]$, where $x_0(t) \colon = 1$ for all $t \in [0,1]$, we have shown earlier that the map $\left( \ T \circ S \ \right) (x_0) \colon [0,1] \to K$ is given by
$$
\left( \ \left( \ T \circ S \ \right) (x_0) \ \right) (s) \ = \ s^2 \ \ \
\mbox{ for all } \ s \in [0,1].
$$
So
\begin{eqnarray*}
\left\Vert \ \left( \ T \circ S \ \right) (x_0) \ \right\Vert_{C[0,1]}
&=& \max_{s \in [0,1]} \left\vert \ \left( \ \left( \ T \circ S \ \right) (x_0) \ \right) (s) \ \ \right\vert \\
&=& \max_{s \in [0,1]} \vert s^2 \vert \\
&=& 1 \\
&=& \Vert x_0 \Vert_{C[0,1]}.
\end{eqnarray*}
Therefore
$$
\Vert T \circ S \Vert \ \geq \ \frac{ \left\Vert \ \left( \ T \circ S \ \right) (x_0) \ \right\Vert_{C[0,1]} }{ \Vert x_0 \Vert_{C[0,1]} } \ = \ \frac{1}{1} = 1,
$$
which along with (\ref{2.7-10-9}) implies that
$$
\Vert T \circ S \Vert = 1.
$$
答案1
LaTeX 入门可能比较难。我无法对辅助技术发表评论，但如果有一款带语法高亮功能的优秀 LaTeX 编辑器（或 WYS IWYG 编辑器）可以满足需求，那么我可以就如何获得更好的输出以及如何更轻松地编写提供建议。
以下几点可能会帮助您加快工作流程，或者至少获得更易读的源代码和结果。我将为您提供一些源代码，这些源代码是对您上面提供的代码进行编辑的部分。
我还没有编辑你的整个文档，但这里有一些内容可以让你入门。我做了一些更改，以便在更干净（和更通用）的源代码和改进的输出之间取得平衡。
格式化，格式化，格式化！
我坚信格式化源代码。我可能并不完全一致，但在编写复杂的数学运算时，对数学运算进行结构化几乎是必不可少的，以使源代码更易于阅读。
您将在下面看到我的数学版本，其中有大量缩进和换行符。（下面我没有做，但在我自己的源代码中经常做的是，在每个句子后也放一个新行。）逐行构建您的 LaTeX 源代码将使您更容易编辑您的方程式。如果您正在编辑 LaTeX 源代码，我强烈建议您使用这种或其他高度结构化的数学源代码样式。
定义并使用宏
您可以将常用的代码片段简化为简写，从而节省时间。例如，在文档开头：
\documentclass{article}
\usepackage{amsmath,amssymb}
\renewcommand\d{\;\mathrm d} % differentials
\newcommand\R{\mathbb R} % real numbers
\newcommand\C{\mathbb C} % complex numbers
\newcommand\norm[1]{\lVert #1 \rVert} % norms
\newcommand\abs[1]{\lvert #1 \rvert} % absolute values
\newcommand\defeq{\;\mathrel{:=}\;} % definition-style assignment
\begin{document}
它们将代替您多次编写的代码片段，并且还使您更容易修改它们以保持一致的外观。
\renewcommand请注意，我在上面使用了\d。这是因为\d已经是一个宏：冰岛语中的 þ 字符。假设您不使用斯堪的纳维亚字符，这应该是安全的。如果我没有这样做，您将收到一个错误，警告您正在覆盖某些内容。您应该小心覆盖的内容，但这是一个我们可以避免的简单问题。
此外，我已将\colon出现的所有位置替换为:（在我看来，这提高了可读性），并将\colon =所有位置替换为\defeq。如果您决定将定义的分配更改为其他符号，或者想要调整间距，您可以快速完成。
章节/段落标题和 AMSmath 命令
您可以更轻松地使用段落标题制作可读的文档，并且可以更快地对齐您的数学并使其使用 AMSmath 环境（我们在上面的代码片段中使用）看起来正确\usepackage{amsmath}。
\paragraph{Prob. 2.7-10}
On $C[0,1]$ define $S$ and $T$ by
\begin{align*} % similar to eqnarray* but:
y(s) &= s \int_0^1 x(t) \d{t} % notice only one tab stop before =
& % this tab stop jumps one column right
y(s) &= s \; x(s), % and again only one tabe stop before =
\end{align*}
respectively. Do $S$ and $T$ commute?
Find $\norm{S}$, $\norm{T}$, $\norm{ST}$ and $\norm{TS}$. % Norms, huzzah!
\paragraph{Solution}
Recall that $C[0,1]$ denotes the normed space of all (real-
or complex-valued) functions defined and continuous on the
closed unit interval $[0,1]$ of the real line $\R$, with the
norm defined by
\begin{equation*}
\norm{x}_{C[0,1]}
\defeq
\max_{t\in[0,1]} \abs{x(t)}
\qquad
\text{for all $x \in C[0,1]$}.
\end{equation*}
Let $K$ denote either $\R$ or $\C$.
\R注意上面的和 的用法\C，以及\defeq、\norm和的用法\abs。另外：如果你想要对方程式进行编号，现在只需从{equation*}和中删除星号即可{align*}。
我们使用命令\;在某一点插入一个细小的空格，以帮助s与函数应用程序分离并区分它们。（这与 的定义中出现的x(s)相同。）我们还使用命令，它添加两个字母的水平宽度，而不是手动添加许多空格（等等）。A是 1em 的空格，即单个字母的宽度，而是其两倍。该命令或多或少与 相同，但会更适合你所做的任何本地更改（例如，如果你想要下标中的整个单词）。\;\d\qquadM\ \ \\quadM\qquad\text\mbox
让你的括号变得很重要
有几处地方，函数操作时会使用嵌套括号。你可能想强调这些，不是吗——如果它们更容易阅读就好了。我们可以使用命令来更改括号的大小，例如\bigland\bigr或\Biglor \Bigr，以使括号足够大以突出。例如：
The operator $S : C[0,1] \to C[0,1]$ is defined as follows: for each
$x \in C[0,1]$, let the map $S(x) : [0,1] \to K$ be defined by
\begin{equation*}
\Bigl[ S(x) \Bigr] (s)
\defeq
s \int_0^1 x(t) \d{t}
\qquad
\text{for all $s \in [0,1]$}.
\end{equation*}
And, the operator $T : C[0,1] \to C[0,1]$ is defined as follows: for
each $x \in C[0,1]$, let the map $T(x) : [0,1] \to K$ be defined by
\begin{equation*}
\Bigl[ T(x) \Bigr] (s)
\defeq
s \cdot x(s)
\qquad
\text{for all $s \in [0,1]$}.
\end{equation*}
多行对齐方程
最后，为了让你有一个好的开始，了解如何格式化多行方程可能会有所帮助——在有限的范围内，甚至显然是分开的方程式 — 以便它们很好地对齐。这正是align/align*环境（及其变体）的用途。
align*我在上面提到了使用和的区别eqnarray*：align*只需要在符号前有一个制表位=（它使用右对齐和左对齐的列对，每对列与其他列对之间有一个边距）。我们可以使用它来对齐方程式以及伴随方程式的文本图例。
By $ST$ and $TS$, Kreyszig intends the composite maps $S \circ T$
and $T \circ S$, respectively. Thus $S$ and $T$ commute if and only
if the maps $S \circ T$ and $T \circ S$ are equal. For any $x \in C[0,1]$,
the map $\left( S \circ T \right) (x) : [0,1] \to K$ is computed as follows:
\begin{align*}
\Bigl[ \left( S \circ T \right) (x) \Bigr] (s)
=
\Bigl[ S \left( T (x) \right) \Bigr] (s)
&=
S(s \; x(s))
\\&=
s \int_0^1 t \; x(t) \d{t}
&&
\text{ for all $s \in [0,1]$},
\intertext{%
while the map $\left( T \circ S \right) (x) : [0,1] \to K$ is
computed as follows:
}
\Bigl[ \left( T \circ S \right) (x) \Bigr) (s)
=
\Bigl[ T \left( S (x) \right) \Bigr](s)
&=
T \left(s \int_0^1 x(t) \d{t} \right)
\\&=
s^2 \int_0^1 x(t) \d{t}
&&
\text{for all $s \in [0,1]$.}
\end{align*}
该命令\intertext允许我们打断一段散文的等式，同时保持散文前后的对齐一致性。
请注意，图例\text{for all $ ... $}前面有两个制表位：这是为了确保它们左对齐（IE它们是右对齐、左对齐列对中的第二个列的一部分。）我还删除了您关于首先应用S或的评论T：也许从上下文中可以清楚地看出发生了什么——但无论如何，很难将所有内容放在一行上，最好不要在方程式的太多行上写散文。如果您确实想包含这些注释，也许您可以添加更多换行符，并将其与对齐 \text{for all $ ... $}。
您的第一页
上面的内容（几乎）填满了第一页。减去边距和页码，它看起来是这样的（如果你把它放在\end{document}我上面提供的代码的末尾并编译它）。
好的 LaTeX 文档在编译时不会出现错误：为了尽量减少压力，您应该确保文档在经过少量编辑后也能顺利编译。好的 LaTeX 编辑器将使这变得简单，并允许您快速检查结果是否符合您的期望；同时还为您提供语法高亮显示。我个人的选择（在 Linux 系统上）是“Kile”，但不同用户更喜欢其他编辑器。
祝你好运！