我正在尝试编译一些我使用 Latex（WinEdit 9.1 + MikeTex 2.9.5845 + Latex：AMS 文章）写的笔记。
虽然它已经编译好了（当我在附件菜单上选择编译时），但是有一个坏框我仍然无法删除。
这是我编译后控制台中对该坏框的引用。
Underfull \vbox (badness 2573) has occurred while \output is active [5]
[6] [7] [8] [9] (ProjectionTheorem.aux) )
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Output written on ProjectionTheorem.pdf (9 pages, 173212 bytes).
SyncTeX written on ProjectionTheorem.synctex.
Transcript written on ProjectionTheorem.log.
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现在我的问题是，如何移除这个仅存的坏框？这个坏框的来源是什么？而且，生成的 PDF 有什么不对劲的地方？
这是我的代码。
[ 代码开始 ]
% ----------------------------------------------------------------
% AMS-LaTeX Paper ************************************************
% **** -----------------------------------------------------------
\documentclass{amsart}
\usepackage{graphicx}
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\vfuzz2pt % Don't report over-full v-boxes if over-edge is small
\hfuzz2pt % Don't report over-full h-boxes if over-edge is small
% THEOREMS -------------------------------------------------------
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\theoremstyle{definition}
\newtheorem{defn}[thm]{Definition}
\theoremstyle{remark}
\newtheorem{rem}[thm]{Remark}
\numberwithin{equation}{section}
% MATH -----------------------------------------------------------
\newcommand{\norm}[1]{\left\Vert#1\right\Vert}
\newcommand{\abs}[1]{\left\vert#1\right\vert}
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\Real}{\mathbb R}
\newcommand{\eps}{\varepsilon}
\newcommand{\To}{\longrightarrow}
\newcommand{\BX}{\mathbf{B}(X)}
\newcommand{\A}{\mathcal{A}}
% ----------------------------------------------------------------
\begin{document}
\today
\title[Projection Theorem]{Projection Theorem}%
\author{Saaqib Mahmood}%
% ----------------------------------------------------------------
\maketitle
% ----------------------------------------------------------------
\section*{Preliminaries}
We begin with some preliminary definitions and results.
\subsection*{Definition}
Let $X$ be an inner product space, and let $M$ be a non-empty subset of $X$. Then the set $M^\perp$ is
defined as follows:
$$M^\perp \ \colon= \ \{ \ x \in X \ \colon \ \langle
x, v \rangle = 0 \ \}.$$
\subsection*{Lemma}
Let $X$ be an inner product space, and let $M$ be a
non-empty subset of $X$. Then $M^\perp$ is a (vector)
subspace of $X$.
\subsection*{Proof}
Let $\theta_X$ denote the zero vector in $X$. Then
since $\langle \theta_X, v \rangle = 0$ for all $v \in
X$, therefore $\langle \theta_X, v \rangle = 0$ for all
$v \in M$ in particular and hence $\theta_X \in
M^\perp$; so $M^\perp$ is non-empty.
Supose that $x, y \in M^\perp$ and $\alpha, \beta$ are
scalars. Then we have
$$\langle x, v \rangle \ = \ \langle y, v \rangle \ =
\ 0 \ \mbox{ for all } \ v \in M.$$
So, for all $v \in M$, we have
$$\langle \alpha x + \beta y, \ v \rangle = \alpha
\langle x, v \rangle + \beta \langle y, v \rangle =
0,$$
showing that $\alpha x + \beta y \in M^\perp$.
Hence $M^\perp$ is a subspace of $X$.
\subsection*{Lemma}
Let $X$ be a vector space, and let $Y$ and $Z$ be
subspaces of $X$. Then the set $Y+Z$ defined by
$$Y+Z \ \colon= \ \{ \ y+z \ \colon \ y \in Y, \ z \in
Z \ \}$$
is also a subspace of $X$.
\subsection*{Proof}
Let $\theta_X$ denote the zero vector in $X$. Since
both $Y$ and $Z$ are subspaces, we have $\theta_X \in
Y$ and $\theta_X \in Z$. Moreover,
$$\theta_X = \theta_X + \theta_X.$$
So $\theta_X \in Y + Z$.
Suppose that $v_1, v_2 \in Y+Z$ and $\alpha_1,
\alpha_2$ are scalars. Then, by definition of $Y+Z$,
$$v_1 = y_1 + z_1 \ \mbox{ for some } \ y_1 \in Y \
\mbox{ and for some } \ z_1 \in Z,$$
and
$$v_2 = y_2 + z_2 \ \mbox{ for some } \ y_2 \in Y \
\mbox{ and for some } \ z_2 \in Z.$$
Then since $Y$ and $Z$ are subspaces, we must have
$$\alpha_1 y_1 + \alpha_2 y_2 \in Y \ \ \ \mbox{ and }
\ \ \ \alpha_1 z_1 + \alpha_2 z_2 \in Z.$$
Now
$$\alpha_1 v_1 + \alpha_2 v_2 = \alpha_1 ( y_1 + z_1) +
\alpha_2 (y_2 + z_2) = (\alpha_1 y_1 + \alpha_2 y_2) +
(\alpha_1 z_1 + \alpha_2 z_2),$$
and therefore $\alpha_1 v_1 + \alpha_2 v_2 \in Y+Z$,
showing that $Y+Z$ is a subspace of $X$ as well.
The above result can be generalised to any finite
number of subspaces.
\subsection*{Definition}
Let $X$ be a vector space, and let $Y$ and $Z$ be
subspaces of $X$. Then $X$ is said to be the direct sum
of $Y$ and $Z$ if each element $x \in X$ can be written
uniquely as a sum $y+z$ of some element $y \in Y$ and
some element $z \in Z$.
That is, $X$ is the direct sum of $Y$ and $Z$ if, for
each element $x \in X$ there exist a unique element $y
\in Y$ and a unique element $z \in Z$ such that $x = y
+ z$.
If $X$ is the direct sum of $Y$ and $Z$, then we write
$$X = Y \oplus Z.$$
For example, let $X \colon= \mathbb{R}^3$, and let $Y$
and $Z$ be defined by
$$Y \ \colon= \ \{ \ (\xi_1, \xi_2, \xi_3) \in
\mathbb{R}^3 \ \colon \ \xi_2 = \xi_3 = 0 \ \}$$
and
$$Z \ \colon= \ \{ \ (\xi_1, \xi_2, \xi_3) \in
\mathbb{R}^3 \ \colon \ \xi_1 = 0 \ \}.$$
That is, $Y$ is the $x$-axis and $Z$ is the $yz$-plane.
Then $X = Y \oplus Z$.
\subsection*{Lemma}
Let $X$ be a vector space, and let $Y$ and $Z$ be
subspaces of $X$. Then $X = Y \oplus Z$ if and only if
$X = Y + Z$ and $Y \cap Z = \{ \theta_X \}$.
\subsection*{Proof}
Suppose that $X = Y \oplus Z$. Then each $x \in X$ can
be written as $$x = y + z \ \ \ \mbox{ for a unique } \
y \in Y \ \mbox{ and a unique } \ z \in Z.$$
So $X = Y+Z$.
Since $Y$ and $Z$ are subspaces of $X$, they both
contain the zero vector $\theta_X$; so $\theta_X \in Y
\cap Z$.
Now suppose that $v \in Y \cap Z$. Then $v \in Y$ and
$v \in Z$. But after all $v$ is an element of $X$. So
we have
$$v = v + \theta_X, \ \mbox{ where } \ v \in Y \ \mbox{
and } \ \theta_X \in Z;$$
similarly,
$$v = \theta_X + v, \ \mbox{ where } \ \theta_X \in Y \
\mbox{ and } \ v \in Z.$$
Since the representation of $v$ as the sum of an
element of $Y$ and an element of $Z$ must be unique, we
must have $v = \theta_X$.
But $v \in Y \cap Z$ was arbitrary. Hence $Y \cap Z =
\{ \theta_X \}$.
Conversely, suppose that $X = Y + Z$ and $Y \cap Z = \{
\theta_X \}$. Now suppose that an element $x \in X$ has
representations
$$x = y_1 + z_1, \ \mbox{ where } \ y_1 \in Y \ \mbox{
and } \ z_1 \in Z,$$
and
$$x = y_2 + z_2, \ \mbox{ where } \ y_2 \in Y \ \mbox{
and } \ z_2 \in Z.$$
Then
$$y_1 + z_1 = y_2 + z_2;$$
so
$$y_1 - y_2 = z_2 - z_1.$$
Since $y_1, y_2 \in Y$ and $Y$ is a vector subspace,
the element $y_1 - y_2 \in Y$ also.
And, since $z_1, z_2 \in Z$ and $Z$ is a subspace, the
element $z_2 - z_1 \in Z$ also.
So from the last equality we can conclude that
$$y_1 - y_2 = z_2 - z_1 \in Y \cap Z.$$
But $Y \cap Z = \{ \theta_X \}$ by our hypothesis.
Hence $y_1 = y_2$ and $z_1 = z_2$ and the
representation of any element $x \in X$ as a sum of an element $y \in Y$ and
an element $z \in Z$ is unique.
\subsection*{Definition}
Let $X$ be a vector space over the field $\mathbb{R}$
of real numbers or the field $\mathbb{C}$ of complex
numbers. Then given two elements $x, y \in X$,
the \emph{segment} joining $x$ and $y$ is defined to be
the following subset of $X$:
$$\{ \ (1-\alpha)x + \alpha y \ \colon \ \alpha \in
\mathbb{R}, \ 0 \leq \alpha \leq 1 \ \}.$$
In fact, the above set is the segment from $x$ to $y$.Note that for $\alpha = 0$ we have the point $x$ and
for $\alpha = 1$ we have the point $y$.
The special linear combination $\mu x + \nu y$, where
$\mu, \nu$ are non-negative scalars such that $\mu+ \nu
= 1$, is called a \emph{convex combination} of $x$ and
$y$.
Thus, $(1-\alpha)x + \alpha y$ is a convex combination
of $x$ and $y$.
\subsection*{Definition}
Let $X$ be a vector space over $\mathbb{R}$ or $\mathbb{C}$, and let $M$ be a non-empty subset of $X$. Then
$M$ is said to be \emph{convex} if, for all $x, y \in
M$ and for all scalars $\alpha \in [0,1]$, the element
$(1-\alpha)x + \alpha y \in M$.
If $M$ is a vector subspace of $X$, then $M$ is convex.
However, not every convex subset $M$ of $X$ is a
subspace of $X$. For example, let $X \colon=
\mathbb{R}^2$, and let $M$ be defined by
$$M \ \colon= \ \{ \ (\xi_1, \xi_2 ) \in \mathbb{R}^2 \
\colon \ 0 \leq \xi_1 \leq 1, \ 0 \leq \xi_2 \leq 1 \
\}.$$
Geometrically, $M$ is the interior and boundary of the
unit square with the origin $(0,0)$ as one of the
vertices and sides parallel to the coordinate axes.
The set $M$ is not a subspace of $\mathbb{R}^2$: the
point $x \colon= (1, 0) \in M$, but $2x \not\in M$, so
that $M$ is not closed under scalar multiplication.
However, $M$ is convex: This we show now. Suppose that
$x \colon= (\xi_1, \xi_2), \ y \colon= (\eta_1,
\eta_2) \in M$ and suppose that $\alpha \in [0,1]$.
Then, for $i = 1, 2$, we have
$$0 \leq \xi_i \leq 1 \ \mbox{ and } \ 0 \leq \eta_i
\leq 1.$$
Now
\begin{align*}
(1-\alpha) x + \alpha y \
&= \ (1- \alpha) \left( \xi_1, \xi_2 \right) + \alpha \left( \eta_1, \eta_2 \right) \\
&= \ \left( \ (1-\alpha) \xi_1 + \alpha \eta_1, \ (1- \alpha) \xi_2 + \alpha \eta_2 \ \right).
\end{align*}
Since $0 \leq \alpha \leq 1$, therefore, we have
$-1 \leq -\alpha \leq 0$ and hence $0 \leq 1- \alpha
\leq 1$.
In particular, $\alpha \geq 0$ and $1-\alpha \geq 0$.
Moreover, for $i = 1, 2$, we have
$$0 \leq \xi_i \leq 1, \ \ \mbox{ and } \ \ \ 0 \leq
\eta_i \leq 1. $$
Upon multiplying the first one of the last two
inequalities by $1- \alpha$ and the second one by
$\alpha$ and then adding together the resulting
inequalities, we get
$$(1-\alpha) \cdot 0 + \alpha \cdot 0 \ \leq \
(1-\alpha) \xi_i + \alpha \eta_i \ \leq \ (1- \alpha)
\cdot 1 + \alpha \cdot 1,$$
which simplifies to
$$0 \ \leq \ (1-\alpha) \xi_i + \alpha \eta_i \ \leq \
1 \ \ \ \mbox{ for } \ i = 1, 2.$$
Thus, by the definition of the set $M$, we can conclude
that the element $(1-\alpha)x + \alpha y \in M$. Hence
the set $M$ is convex.
\subsection*{Definition}
Let $M \neq \emptyset$ be a subset of a metric space
$(X, d)$. Let $x \in X$. Then the distance $D(x, M)$
between $x$ and $M$ is defined to be
$$D(x, M) \ \colon= \ \inf \{ \ d(x, v) \ \colon \ v
\in M \ \}.$$
If $X$ is a normed space, then this definition becomes
$$D(x, M) \ \colon= \ \inf \{ \ \Vert x-v \Vert_X \
\colon \ v \in M \ \}.$$
Since an inner product space $(X, \langle \cdot
\rangle)$ is also a normed space, with the norm defined
by
$$\Vert x \Vert_X \colon= \sqrt{ \langle x, x \rangle}
\ \ \ \mbox{ for all } \ x \in X, $$
the above definition takes the following form in an inner product space:
$$D(x, M) \ \colon= \ \inf \{ \ \sqrt{\ \langle \
x-v,\ x-v\ \rangle} \ \ \colon \ v \in M \ \}.$$
\section*{The Main Result}
Here's our main result. It is also called the \emph{projection theorem}.
\subsection*{Theorem}
Let $X$ be an inner product space, let $x \in X$, and
let $M$ be a non-empty subset of $X$ such that $M$ is
complete in the metric induced by the inner product, that is, $M$ is complete in the metric which is
the restriction to $M \times M$ of the metric
$d$ on $X$ given by
$$d(u,v) \colon= \sqrt{ \ \langle u-v, \ u-v \rangle \
} \ \ \ \mbox{ for all } u, v \in X.$$
Then we have the following:
(a) If $M$ is also a convex set, then there exists a unique element $y \in M$ such that
$$D(x,M) = \Vert x - y \Vert;$$
that is,
$$D(x,M) = \sqrt{ \ \langle x-y,\ x-y \rangle \ }.$$
(b) If $M$ is a vector subspace of $X$, then (a) still
holds and the unique element $y$ whose existence is
asserted in (a) also satisfies
$ (x-y ) \perp M$, that is,
$$\langle x-y, \ v \rangle = 0 \ \ \ \mbox{ for all } \
v \in M.$$
And,
(c) if $X$ is a Hilbert space (i.e. if $X$ is complete
in the metric induced by the inner product), if $M$ is
a vector subspace of $X$, and if $M$ is closed with
respect to the metric induced by the inner product on
$X$, then there are unique elements $y \in M$ and $z
\in M^\perp$ such that $x = y + z$, that is, then
$$X = M \oplus M^\perp.$$
\subsection*{Proof}
Let $X$ be an inner product space, let $x \in X$, and
let $M$ be a non-empty subset of $X$ such that $M$ is
complete in the metric induced by the inner product.
\subsubsection*{(a)}
Suppose that $M$ is also convex.
Since, by definition,
$$
D(x,M) = \inf \{ \ \Vert x-v \Vert
\ \colon \ v \in M \ \},
$$
we have the following:
$$D(x,M) \leq \Vert x-v \Vert \ \ \ \mbox{ for all } \
v \in M,$$
and for every $\delta > 0$, there exists some element
$y_\delta \in M$ such that
$$D(x,M) \leq \Vert x - y_\delta \Vert < D(x, M ) +
\delta.$$
In particular, for each $n \in \mathbb{N}$, there is
some $y_n \in M$ such that
$$D(x,M) \leq \Vert x - y_n \Vert < D(x,M) +
\frac{1}{n}.$$
Thus we have a sequence $(y_n)$ in $M$ such that
$$\lim_{n \to \infty} \Vert x - y_n \Vert = D(x,M). \ \
\ \ \mbox{ [ using the sandwiching theorem ]}$$
So,
$$\lim_{n \to \infty} \Vert x - y_n \Vert^2 = \left(
D(x,M) \right)^2.$$
So, given $\epsilon> 0$, there exists a natural number
$N$ such that
$$ \left\vert \Vert x-y_n \Vert^2 - \left( D(x,M)
\right)^2 \right\vert < \frac{\epsilon^2}{4} \ \ \
\mbox{ for all } \ n \in \mathbb{N} \ \mbox{ such that
} \ n > N.$$
Now we show that the sequence $(y_n)$ is a Cauchy
sequence.
Note that, for all $m, n \in \mathbb{N}$,
since $y_m, y_n \in M$ and $M$ is convex, and since
$$\frac{1}{2} ( y_m + y_n ) = \frac{1}{2} y_m +
\frac{1}{2} y_n = (1-\frac{1}{2}) y_m + \frac{1}{2}
y_n, $$
therefor we can conclude also that
$$
\frac{1}{2} (y_m + y_n ) \in M.
$$
We will be using this fact in the remainder of this
proof.
For each $m, n \in \mathbb{N}$ such that $m > N$ and
$n > N$, we have
\begin{align*}
\Vert y_m - y_n \Vert^2
&= \Vert y_m - x + x - y_n
\Vert^2 \\
&= \Vert - (x-y_m )\ + \ (x-y_n) \Vert^2 \\
&= \Vert (x-y_n) - (x- y_m) \Vert^2 \\
&= 2 \Vert x-y_n \Vert^2 + 2 \Vert x-y_m \Vert^2 -
\Vert (x-y_n) \ + \ (x-y_m) \Vert^2 \\
& \ \ \ \mbox{ [the parallelogram identity ] } \\
&= 2 \Vert x-y_n \Vert^2 + 2 \Vert x-y_m \Vert^2 -
\Vert 2x - ( y_n + y_m) \Vert^2 \\
&= 2 \Vert x-y_n \Vert^2 + 2 \Vert x-y_m \Vert^2 -
\left\Vert 2 \left( \ x \ - \ \frac{1}{2}( y_n + y_m) \
\right) \right\Vert^2 \\
&= 2 \Vert x-y_n \Vert^2 + 2 \Vert x-y_m \Vert^2 - 2^2
\left\Vert x - \frac{1}{2}( y_n + y_m) \right\Vert^2 \\
&= 2 \Vert x-y_n \Vert^2 + 2 \Vert x-y_m \Vert^2 - 4
\left\Vert x - \frac{1}{2}( y_n + y_m) \right\Vert^2 \\
&\leq 2 \Vert x-y_n \Vert^2 + 2 \Vert x-y_m \Vert^2 - 4
\left( D(x,M) \right)^2 \\
& \ \ \ \mbox{ [Since $M$ is convex,
$\frac{1}{2}(y_m + y_n) \in M$ also.] } \\
&= 2 \left( \Vert x - y_n \Vert^2 - \left( D(x,M)
\right)^2 \right) \ + \ 2 \left( \Vert x - y_m \Vert^2
- \left( D(x,M) \right)^2 \right) \\
&\leq 2 \left\vert \ \Vert x - y_n \Vert^2 - \left(
D(x,M) \right)^2 \ \right\vert \ + \ 2 \left\vert \
\Vert x - y_m \Vert^2 - \left( D(x,M) \right)^2 \
\right\vert \\
&< 2 \frac{\epsilon^2}{4} + 2 \frac{\epsilon^2}{4} \\
&= \epsilon^2.
\end{align*}
This we have shown that, for every given $\epsilon > 0$, there exists a natural number $N$ such that
$$
\Vert y_m - y_n \Vert < \epsilon \ \mbox{ for all } \ m, n \in \mathbb{N} \ \mbox{ such that } \ m > N \ \mbox{ and } \ n > N.
$$
So the sequence $(y_n)$ is a Cauchy sequence in $M$, and since
$M$ is complete, this sequence converges to some
element $y$ of $M$; that is, there exists an element $y \in M$ such that
$$\lim_{n \to \infty} \Vert y_n - y \Vert = 0.$$
Now we show that $D(x,M) = \Vert x - y \Vert$.
Since $y \in M$, we have
$$D(x,M) \leq \Vert x-y \Vert.$$
Now we show that $D(x,M) \geq \Vert x -y \Vert$ also.
For all $n \in \mathbb{N}$, we have
\begin{align*}
\Vert x - y \Vert &= \Vert x-y_n + y_n - y \Vert \\
&\leq \Vert x - y_n \Vert + \Vert y_n - y \Vert.
\end{align*}
Now since $\lim_{n \to \infty} \Vert x-y_n \Vert =
D(x,M)$ and since $\lim_{n\to \infty} \Vert y_n - y
\Vert = 0$, we have
$$\lim_{n \to \infty} \left( \Vert x - y_n \Vert +
\Vert y_n - y \Vert \right) = D(x,M) + 0 = D(x,M).$$
And, as we have shown above, since
$$
\Vert x - y \Vert \leq \Vert x - y_n \Vert + \Vert y_n - y \Vert \ \mbox{ for all } \ n \in \mathbb{N},
$$
therefore upon taking the limit as $n \to \infty$, we get
$$\lim_{n \to \infty} \Vert x - y \Vert \ \leq \ \lim_{n \to \infty} \left( \Vert x - y_n \Vert + \Vert y_n - y \Vert \right) $$
or
$$\Vert x-y \Vert \leq D(x,M).$$
Hence $$D(x,M) = \Vert x-y \Vert.$$
Now we show that this point $y \in M$ is unique.
Suppose that there is some point $y^\prime \in M$
such that
$$D(x,M) = \Vert x-y^\prime \Vert$$
also.
Then we have
\begin{align*}
\Vert y - y^\prime \Vert^2
&= \Vert y -x + x - y^\prime
\Vert^2 \\
&= \Vert -(x-y) + (x-y^\prime) \Vert^2 \\
&= \Vert (x-y^\prime) - (x-y) \Vert^2 \\
&= 2 \Vert x-y^\prime \Vert^2 + 2 \Vert x-y \Vert^2 -
\Vert (x-y^\prime) + (x-y) \Vert^2 \\
&= 2 \left( D(x,M) \right)^2 + 2 \left( D(x, M)
\right)^2 - \Vert 2x - (y^\prime + y) \Vert^2 \\
&= 4 \left( D(x,M) \right)^2 - \left\Vert 2 \left( x -
(y^\prime + y) \right) \right\Vert^2 \\
&= 4 \left( D(x,M) \right)^2 - 2^2 \left\Vert x -
\frac{1}{2} (y^\prime + y) \right\Vert^2 \\
&= 4 \left( D(x,M) \right)^2 - 4 \left\Vert x -
\frac{1}{2} (y^\prime + y) \right\Vert^2 \\
&\leq 4 \left( D(x,M) \right)^2 - 4 \left( D(x,M)
\right)^2 \\
& \ \ \ \mbox{ [since $y, y^\prime \in M$ and $M$ is convex, $\frac{1}{2}(y+y^\prime) \in M$ also.]}\\
&= 0.
\end{align*}
Thus,
$$\Vert y-y^\prime \Vert \leq 0.$$
But
$$\Vert y-y^\prime \Vert \geq 0 \ \ \mbox{ by the
property N1 of the norm }.$$
So $\Vert y-y^\prime \Vert = 0$ and hence $y =
y^\prime$.
\subsubsection*{(b)}
Suppose that $M$ is a vector subspace of $X$. Then $M$
is convex, and therefore using (a) above, we can
conclude that there exists a unique $y \in M$ such that
$$D(x,M) = \Vert x-y \Vert.$$
Let $z \colon= x-y$. We now show that this $z$ is
orthogonal to all $v \in M$, that is, we show that $\langle z, v
\rangle = 0$ for all $v \in M$.
Suppose that there is some element $v_0 \in M$ such
that $\langle z, v_0 \rangle \neq 0$.
This element $v_0$ cannot be the zero vector $\theta_X$
in $X$ because
$$\langle z, \theta_X \rangle = \langle z, 0z \rangle =
\overline{0} \langle z,z \rangle = 0 \Vert z \Vert^2 =
0.$$
Since $v_0 \neq \theta_X$, using the property IP4 of an
inner product, we can conclude that $$\Vert v_0 \Vert^2
= \langle v_0, v_0 \rangle > 0.$$
Now for any scalar $\alpha$, we have
\begin{align*}
\Vert z- \alpha v_0 \Vert^2 &= \langle z-\alpha v_0, \
z-\alpha v_0 \rangle \\
&= \langle z, z-\alpha v_0 \rangle - \langle \alpha
v_0, z-\alpha v_0 \rangle \\
&= \langle z, z \rangle - \overline{\alpha} \langle z,
v_0 \rangle - \alpha \left( \langle v_0, z \rangle -
\overline{\alpha} \langle v_0, v_0 \rangle \right) \\
&= \Vert z \Vert^2 - \overline{\alpha} \langle z, v_0
\rangle - \alpha \left( \overline{\langle z, v_0
\rangle } - \overline{\alpha} \Vert v_0 \Vert^2 \right)
\end{align*}
Now let's take
$$\alpha \colon= \frac{\langle z, v_0 \rangle}{\langle
v_0, v_0 \rangle} = \frac{\langle z, v_0 \rangle}{\Vert
v_0 \Vert^2} .$$
Then
\begin{align*}
\Vert z - \alpha v_0 \Vert^2
&= \Vert z \Vert^2 - \overline{\frac{\langle z, v_0 \rangle}{\Vert v_0
\Vert^2}} \langle z, v_0 \rangle - \alpha \cdot 0 \\
&= \Vert z \Vert^2 - \frac{ \left\vert \langle z, v_0
\rangle \right\vert^2 }{\Vert v_0 \Vert^2} \\
&< \Vert z \Vert^2 \\
& \ \ \ \mbox{ [Since the absolute value
of a non-zero complex number is positive.]} \\
&= \left(D(x,M) \right)^2.
\end{align*}
So
$$
\Vert z-\alpha v_0 \Vert < D(x,M).$$
But
$$z- \alpha v_0 = (x-y) - \alpha v_0 = x - (y+ \alpha
v_0 ).$$
And since $y, v_0 \in M$, since $\alpha$ is a scalar,
and since $M$ is a vector subspace of $X$, therefore
we must have $y + \alpha v_0 \in M$ also and hence
$$\Vert x - (y + \alpha v_0 ) \Vert \geq D(x,M);$$
that is,
$$\Vert z- \alpha v_0 \Vert \geq D(x,M).$$
Thus we have a contradiction.
Therefore our supposition that there exists some $v_0
\in M$ such that $\langle z, v_0 \rangle \neq 0$ is
wrong and hence $\langle z, v \rangle = 0$ for all $v
\in M$.
Thus we have shown that if $M$ is a complete subspace
of an inner product space $X$, then every $x \in X$ can
be written as $x = y + z$, where $y \in M$ and $z
\colon= x-y \in M^\perp$ and $y \in M$ is unique. So $z
\in M^\perp$ is also unique.
Another way to see the uniqueness of the $y \in M$ and the $z \in M^\perp$ is
as follows:
Suppose that, for some $x \in X$, there exist $y_1, y_2
\in M$ and $z_1, z_2 \in M^\perp$ such that
$$x = y_1 + z_1 \ \ \ \mbox{ and } \ \ \ x = y_2 +
z_2.$$
Then
$$y_1 + z_1 = y_2 + z_2,$$
and so
$$y_1 - y_2 = z_2 - z_1.$$
Now since both $M$ and $M^\perp$ are vector subspaces
of $X$, therefore $y_1 - y_2 \in M$ for all $y_1, y_2
\in M$ and $z_2 - z_1 \in M^\perp$ for all $z_1, z_2
\in M^\perp$.
Then since $y_1 - y_2 = z_2 - z_1$, we can conclude
that $y_1 - y_2 = z_2 - z_1 \in M \cap M^\perp$, which implies that
$$\langle y_1 - y_2, y_1 - y_2 \rangle = 0,$$
which in turn implies that $y_1 - y_2 = \theta_X$, the
zero vector in $X$; therefore $y_1 = y_2$ and hence
$z_1 = z_2$ also.
Hence the representation of any given $x \in X$ as a
sum of an element $y \in M$ and an element $z \in
M^\perp$ is unique whenever $X$ is any inner product
space and $M$ is a subspace of $X$ such that $M$ is
complete in the metric induced by the inner product.
Therefore, we can write
$$X = M \oplus M^\perp.$$
\subsubsection*{(c)}
Now suppose that $X$ is a Hilbert space (i.e. an inner product space that is complete with respect to the metric induced by the inner product) and suppose that $M$ is a (vector) subspace of $X$ such that $M$ is closed (in the metric induced by the inner product on $X$). Then $M$ is
complete, by Theorem 1.4-7 in Kreyszig. Therefore, as we have already shown in (b) above,
$$X = M \oplus M^\perp.$$
\section*{Consequences}
Here are some of the applications of the theorem in the previous section.
\subsection*{Definition}
Let $X$ be a Hilbert space, and let $M$ be a closed subspace of $X$. Then $X = M \oplus M^\perp$. Now we can define the maps $P_M \colon X \to M$ and $P_{M^\perp} \colon X \to M^\perp$ as follows: Let $x \in X$. Then there are unique elements $y \in M$ and $z \in M^\perp$ such that $x = y+z$. So let's define
$$P_M (x) \colon= y \ \ \ \mbox{ and } \ \ \ P_{M^\perp} (x) \colon= z.$$
The subspaces $M$ and $M^\perp$ are then called the \emph{orthogonal complements} of each other, and the maps $P_M$ and $P_{M^\perp}$ are called the \emph{projection maps}, or the \emph{projections}, of $X$ onto $M$ and $M^\perp$, respectively.
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\end{document}
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[ 代码结束 ]
答案1
垂直盒子未充满意味着垂直空白处没有足够的可拉伸性来使垂直盒子填充到其指定的长度。
当它在输出例程中出现时，它要么是主页主体，要么是页眉或页脚，很可能主页主体只是一页。
最可能的原因是该页面仅包含文本，并且文档的页面尺寸不正确，\textheight-\topskip不是 的倍数\baselineskip，因此在只有文本的页面上不可能完全填满一页。
如果你添加
\showboxdepth=1
\showboxbreadth=1000
在文档中你可以看到下面这个框的内容：
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这表明显示数学上方和下方的空白空间没有足够的灵活性，段落之间的空间也完全没有灵活性来精确填充页面。基本上，如果您希望在一页上显示那么多内容，这可能是意料之中的，您可以忽略这一点，或者您可以增加 abovedisplayskip 中的可拉伸性，但如果只有一页，我可能不会这么做。
请注意，您永远不应该在数学显示前留下空白行，也永远不要$$在乳胶中使用，总是[......\]
日期应按定义设置\maketitle，但如果确实需要手动设置，请使用\begin{center}..\end{center}