I'm trying to compile some notes I've written using Latex (WinEdit 9.1 + MikeTex 2.9.5845 + Latex: AMS Article).
Although it has compiled (when I select Compile on the Accessories menu), there is one bad box that I've still not been able to remove.
Here's the reference to that bad box in the console after I compile.
Underfull \vbox (badness 2573) has occurred while \output is active [5]
[6] [7] [8] [9] (ProjectionTheorem.aux) )
(see the transcript file for additional information)<C:/Program Files/MiKTeX 2.
9/fonts/type1/public/amsfonts/cm/cmbx10.pfb><C:/Program Files/MiKTeX 2.9/fonts/
type1/public/amsfonts/cm/cmcsc10.pfb><C:/Program Files/MiKTeX 2.9/fonts/type1/p
ublic/amsfonts/cm/cmex10.pfb><C:/Program Files/MiKTeX 2.9/fonts/type1/public/am
sfonts/cm/cmmi10.pfb><C:/Program Files/MiKTeX 2.9/fonts/type1/public/amsfonts/c
m/cmmi7.pfb><C:/Program Files/MiKTeX 2.9/fonts/type1/public/amsfonts/cm/cmr10.p
fb><C:/Program Files/MiKTeX 2.9/fonts/type1/public/amsfonts/cm/cmr7.pfb><C:/Pro
gram Files/MiKTeX 2.9/fonts/type1/public/amsfonts/cm/cmr8.pfb><C:/Program Files
/MiKTeX 2.9/fonts/type1/public/amsfonts/cm/cmsy10.pfb><C:/Program Files/MiKTeX
2.9/fonts/type1/public/amsfonts/cm/cmsy5.pfb><C:/Program Files/MiKTeX 2.9/fonts
/type1/public/amsfonts/cm/cmsy7.pfb><C:/Program Files/MiKTeX 2.9/fonts/type1/pu
blic/amsfonts/cm/cmti10.pfb><C:/Program Files/MiKTeX 2.9/fonts/type1/public/ams
fonts/symbols/msbm10.pfb>
Output written on ProjectionTheorem.pdf (9 pages, 173212 bytes).
SyncTeX written on ProjectionTheorem.synctex.
Transcript written on ProjectionTheorem.log.
_____________________________________________________________________
PDFTeXify Compilation Report (Pages: 9)
Errors: 0 Warnings: 0 Bad Boxes: 1
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Now my question is, how to remove this only remaining bad box? What is the source of this bad box? And, what is not going to be right about the PDF that is getting generated?
Here's my code.
[ Beginning of Code ]
% ----------------------------------------------------------------
% AMS-LaTeX Paper ************************************************
% **** -----------------------------------------------------------
\documentclass{amsart}
\usepackage{graphicx}
% ----------------------------------------------------------------
\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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Best Answer
an underfull vbox means that there is not enough stretchability in vertical white space to make a vertical box be padded to its specified length.
As it occurs in the output routine it is either a main page body or a head or foot, most likely the main page body as just one page.
The most likely cause is that the page just consists of text and the page dimensions for the document are incorrect, with
\textheight-\topskipnot being a multiple of\baselineskipso that on a page with just text it is impossible to completely fill a page.If you add
to the document you see the contents of the underful box as:
which shows that there is not enough flexibility in the white space above and below display math, and no flexibility at all in space between paragraphs to exactly fill a page. Basically if you want that many displays on a page this might be expected and you could just ignore this, or you could increase the stretchability in abovedisplayskip, but if it is just one page I probably wouldn't bother.
Note you should never leave a blank line before a math display and never use
$$in latex, always[..\].The date should be set by the
\maketitledefinition but if you do need to centre by hand use\begin{center}..\end{center}