How can I make it easier to write LaTeX documents with elaborate mathematics?

Question

I'm a severely visually handicapped student of mathematics. I wonder how I can speed up my latex document creation.

I'm pasting below solution to a problem in a functional analysis book by Kreyszig. This is a question that I yesterday posted on the Maths SE site also. It's been quite stressful on my body to type it all up and remove syntactical errors that I encountered while compiling my work from time to time.

Can anyone please advise me on how I can improve my efficiency and how I can make use of the assistive technology available for the blind and visually impaired during my mathematics studies, and for creating such mathematical content myself?

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
$$
\Vert T \circ S \Vert = 1.
$$

Answer 1

LaTeX can be hard to get started with. I can't comment on assistive technology, but — if a good LaTeX editor with syntax hilighting (or a WYS IWYG editor) will do — then I can give suggestions for how to get better output, and to write it more easily.

Here are a few pointers which might help you to speed up your workflow, or at least get more readable source code and results both. I will give you bits of source code, which are edited parts of the code you gave above.

I haven't edited your entire document, but here is something to get you started. I've made a number of changes to get you a compromise between cleaner (and more versatile) source code, and improved output.

Formatting, formatting, formatting!

I am a strong believer in formatting my source code. I may not be completely consistent, but when writing complicated math, it is nearly essential to structure the math to make the source code easier to read.

You will see my version of your math below, with plenty of indents and newlines. (What I haven't done below, but often do in my own source code, is also put a new line after every sentence.) Structuring your LaTeX source, line-by-line, will make it easier for you to edit your equations. If you are editing the LaTeX source, I highly recommend this or some other style for highly structured mathematics source code.

Define and use macros

You can save time by making often-used bits of code into shorthand. For example, early in your document:

\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}

These will take the place of bits of code which you write more than once, and also will make it easier for you to modify them to maintain a consistent look.

Notice that I've used \renewcommand above for \d. That's because \d is already a macro: the character þ in Icelandic. Assuming you are not using Scandanavian characters, this should be safe. If I hadn't done that, you would get an error which would warn you that you're overriding something. You should be careful what you override, but this is a simple one we can get away with.

Also, I've replaced \colon everywhere that it occurs with : (which to my eyes improves the readability), and replaced \colon = everywhere with \defeq. If you decide to change your assignment for definitions to a different symbol, or if you want to tweak the spacing, you can do it quickly.

Section/paragraph headings and AMSmath commands

You can more easily make a readable document using paragraph headings, and can more quickly align your mathematics and get it looking right using the AMSmath environments (which we make available in the snippet above using \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 $\C$.

Note the usage of \R and \C above, and of \defeq, \norm, and \abs. Also: if you want your equations to be numbered, you can now do so simply by removing the stars from {equation*} and {align*}.

We use the command \; to insert a thin space at one point, to help separate s from the function-application x(s) and distinguish them. (This is the same \; which appears in the definition of \d.) We also use the command \qquad, which adds the horizontal witdth of two letter Ms, rather than manually adding many spaces by \ \ \ (etc.) A \quad is a space of 1em, which is the width of a single letter M, while \qquad is double that. The command \text is more or less the same as \mbox, but will play more nicely with any local changes you make (if you want a whole word in a subscript, for instance).

Make a big deal out of your brackets

Several places, you have nested brackets for operations on functions. You probably want to emphasize those, don't you — and it would be nice if they were easier to read. We may use commands to change the size of the brackets, such as \bigl and \bigr or \Bigl or \Bigr, to make the brackets large enough to stand out. For instance:

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*}

Multiline aligned equations

Finally, to get you off to a good start, it may be helpful to see how you might format multiline equations — and to a limited extent, even apparently separate equations — so that they are nicely aligned. This is precisely what the align/align* environments (and their variants) are for.

I've alluded to the difference between the use of align* and eqnarray* above: align* only wants one tab stop before the = sign (it uses pairs of columns which are right- and left-aligned, with a margin separating each pair of columns from the other pairs). We can use this to align both the equations, and the text legends that accompany them.

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*} 

The command \intertext allows us to interrupt the equation for a bit of prose, while maintaining consistency of alignment before and after the prose.

Note that the legends \text{for all $ ... $} are preceded by two tab stops: this is to make sure they are left-aligned (i.e. they are part of the second of the right-aligned, left-aligned column pair.) I've also removed your comments about applying S or T first: perhaps it is clear from the context what is happening — but in any case it would be difficult to make everything fit on one line, and it is better not to put prose on too many lines of your equations. If you really do want to include those remarks, perhaps you can add more line breaks, and align it with \text{for all $ ... $}.

Your first page

The above (almost) fills the first page. Minus the margins and the page-number, here's how it would look (if you put \end{document} at the end of the code I've given you above and compiled it).

A good LaTeX document will compile without errors: for minimum stress, you should make it your goal to make sure your document compiles cleanly after small edits. A good LaTeX editor will make this easy, and will allow you to quickly check that the result looks about how you would like it to look; as well as giving you syntax highlighting. My personal choice (on a Linux system) is "Kile", but there are others that different users prefer.

Good luck!