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\begin{document}
\title{Elementary Wavelet Analysis and an Application to the Study of $H^{1}\left(
\Bbb{R}^{n}\right) $}
\author{A. G. Marasingha}
\department{Department of Mathematics}
\thisdegree{MSci}
\university{University of London}
\degreemonth{June}
\degreeyear{2000}
\date{16 June 2000}
\chairmanname{.}
\chairmantitle{\qquad }
\super{T J Lyons}
\supertitle{Wallis Professor of Mathematics}
\maketitle
\begin{abstract}
A discussion of orthonormal wavelet bases and wavelet series, utilising the
notion of multiresolution approximations. The regular wavelets of compact
support are seen to form an unconditional basis of the space $H^{1}\left(
\Bbb{R}^{n}\right) $ of Stein and Weiss.
\end{abstract}
\tableofcontents
\chapter{Introduction}
In this article, I will present an overview of elementary wavelet analysis
and conclude by using wavelet methods to characterize the real Hardy space $%
H^{1}(\Bbb{R}^{n})$.
What are wavelets? At the definition level, they are simply a basis (usually
orthonormal) of $L^{2}(\Bbb{R})$ with certain properties which express their
regularity, oscillatory nature (hence `wave') and localization in space
(hence `-let'). Wavelet analysis can be viewed as a monumental improvement
on Fourier analysis.
The primary goal of this article is to construct and discuss the properties
of orthonormal wavelet bases and wavelet series. A\ fundamental tool in this
process is the multiresolution approximation, which, as the name suggests,
is way of describing a given $L^{2}\left( \Bbb{R}^{n}\right) $ function at
different resolutions. In fact the concept was developed in connection with
image processing, in which the interpretation is direct.
We will see that any multiresolution approximation gives rise to a wavelet
basis of $L^{2}\left( \Bbb{R}^{n}\right) $ via a `scaling function' $\phi $.
In particular, we will demonstrate how to find multiresolution
approximations which lead to wavelets (and scaling functions) of compact
support.
Finally, it will be demonstrated that the wavelets of compact support are
just what we need to form an unconditional basis of the space $H^{1}\left(
\Bbb{R}^{n}\right) $ of Stein and Weiss.
\chapter{Fourier analysis}
Though wavelet analysis is in many ways superior to Fourier analysis,
Fourier methods are indispensable in the development of the wavelet theory.
In this section, I will state the results which will be needed in later
sections. Though I will only look at Fourier analysis on the real line,
analogous results hold in $\Bbb{R}^{n}$, and I will use such results later
with neither statement nor proof.
\section{Fourier Integrals}
\begin{definition}
If $f\in L^{1}\left( \Bbb{R}\right) $ then its Fourier transform is, by
definition:
\[
\hat{f}(\xi ):=\int_{-\infty }^{\infty }e^{-i\xi x}f(x)dx
\]
\end{definition}
\begin{theorem}
If $f\in L^{1}\left( \Bbb{R}\right) $ then
\begin{enumerate}
\item $\hat{f}\in L^{\infty }\left( \Bbb{R}\right) $ and $\Vert \hat{f}%
\Vert _{\infty }\leq \left\| f\right\| _{1}$ ;
\item $\hat{f}$ is uniformly continuous on $\Bbb{R}$ ;
\item if $f^{\prime }$ exists and $f^{\prime }\in L^{1}\left( \Bbb{R}%
\right) $, then $\widehat{f^{\prime }}(\xi )=i\xi \hat{f}(\xi )$; and
\item $\hat{f}(\xi )\rightarrow 0$, as $\xi \rightarrow \pm \infty $.
\end{enumerate}
\end{theorem}
I will frequently have occasion to call upon the following simple but useful
formula:
\begin{remark}
If $f\in L^{1}\left( \Bbb{R}\right) $ and $g(x):=\frac{x-b}{a}$, for some
constants $a$ and $b$, then $f\circ g$ $\in L^{1}(\Bbb{R)}$ and
\[
\widehat{f\circ g}(\xi )=ae^{-ib\xi }\hat{f}(a\xi )
\]
\end{remark}
\begin{definition}
If $f\in L^{1}\left( \Bbb{R}\right) $ and its Fourier transform $\hat{f}$ is
also in $L^{1}\left( \Bbb{R}\right) $, then the inverse Fourier transform is
defined by:
\[
(\mathcal{F}^{-1}\hat{f})(x):=\frac{1}{2\pi }\int_{-\infty }^{\infty
}e^{i\xi x}\hat{f}(\xi )d\xi
\]
\end{definition}
\begin{theorem}[Fourier Inversion]
If $f\in L^{1}\left( \Bbb{R}\right) $ and $\hat{f}\in L^{1}\left( \Bbb{R}%
\right) $ then $f(x)=(\mathcal{F}^{-1}\hat{f})(x)$ at every point where $f$
is continuous.
\end{theorem}
The proof of the theorems mentioned above can be found in \cite{M3P6}.
\subsection{Fourier transforms on $L^{2}(\Bbb{R)}$}
Importantly, the Fourier transform can be extended to $L^{2}\left( \Bbb{R}%
\right) $. One approach I will sketch is to firstly note that if $f$ is in
the Schwartz space $\mathcal{S}\left( \Bbb{R}\right) $, then also its
Fourier transform is in $\mathcal{S}\left( \Bbb{R}\right) $. The inversion
formula applies and we have that ($\hat{f},\hat{g})=2\pi \left( f,g\right) $
for $f,g\in \mathcal{S}\left( \Bbb{R}\right) $. Then thinking of $\mathcal{S}%
\left( \Bbb{R}\right) $ as a subspace of $L^{2}\left( \Bbb{R}\right) $, we
have that $\left( 2\pi \right) ^{-1/2}\mathcal{F}$ is an isometry. As $%
\mathcal{S}\left( \Bbb{R}\right) $ is dense in $L^{2}\left( \Bbb{R}\right) $%
, we can extend $\mathcal{F}$ to $L^{2}\left( \Bbb{R}\right) $. The
important results we will need in subsequent chapters are the following:
\begin{theorem}[Parseval]
If $f,g\in L^{2}\left( \Bbb{R}\right) $, then $\left( f,g\right) =\frac{1}{%
2\pi }(\hat{f},\hat{g})$
\end{theorem}
\begin{theorem}
The mapping $\mathcal{F}:L^{2}\left( \Bbb{R}\right) \rightarrow L^{2}\left(
\Bbb{R}\right) $ is a bijection
\end{theorem}
\section{Fourier Series}
Let $f\in L^{2}[0,2\pi ]$. Its Fourier series is $\sum_{k\in \Bbb{Z}%
}c_{k}e^{ikx}$, where $c_{k}=\left( f,e^{ikx}\right) $. It is a standard
result that the sequence $\left\{ \left( 2\pi \right) ^{-1/2}e^{ikx}\right\}
$ forms an orthonormal basis of $L^{2}[0,2\pi ]$, hence the Fourier series
of $f$ converes to $f$ in $L^{2}[0,2\pi ]$. We also have a Parseval's
identity for Fourier series.
For $N\in \Bbb{Z}_{\geq 0}$, define the $N^{th}$ partial sum of the Fourier
series $S_{N}f$ by
\[
S_{N}f\left( x\right) =\sum_{k=-N}^{N}c_{k}e^{ikx}
\]
and define the $N^{th}$ Ces\`{a}ro sum $\sigma _{N}f=\left( N+1\right)
^{-1}\sum_{0}^{N}S_{i}f\left( x\right) $, and the Fej\'{e}r kernel
\[
K_{N}\left( x\right) =\frac{1}{N+1}\frac{\sin ^{2}\left( \frac{N+1}{2}%
x\right) }{2\sin ^{2}\left( \frac{x}{2}\right) }
\]
it can easily be shown that $\sigma _{N}f$ is the convolution of $f$ with $%
K_{N}$. (see Chui\cite{chui} for details).
Let $\Bbb{T}$ denote the unit circle in the complex plane, then we have the
following very deep result, due to Carleson\cite{carleson}.
\begin{theorem}[Carleson]
If $f\in L^{2}\left( \Bbb{T}\right) $, then the partial sums $S_{N}f$
converge to $f$ almost everywhere
\end{theorem}
\chapter{Multiresolution Approximations and Scaling Functions}
Fundamental to wavelet theory is the notion of a multiresolution
approximation of $L^{2}\left( \Bbb{R}^{n}\right) $, a concept crystallised
by S. Mallat \cite{mallat}.
\begin{definition}
A multiresolution approximation of $L^{2}\left( \Bbb{R}^{n}\right) $ is an
increasing sequence $V_{j}$, $j\in \Bbb{Z}$ of closed subspaces of $%
L^{2}\left( \Bbb{R}^{n}\right) $ such that:
\begin{enumerate}
\item $\bigcap\limits_{-\infty }^{\infty }V_{j}=\left\{ 0\right\} $;
\item \label{denseness}$\bigcup\limits_{-\infty }^{\infty }V_{j}$ is dense
in $L^{2}\left( \Bbb{R}^{n}\right) $;
\item \label{double frequency}$f(x)\in V_{j}\Longleftrightarrow f(2x)\in
V_{j+1}$,\qquad for any $f\in L^{2}\left( \Bbb{R}^{n}\right) $ and any $j\in
\Bbb{Z}$;
\item $f(x)\in V_{0}\Longleftrightarrow f(x-k)\in V_{0}$,\qquad for any $%
f\in L^{2}\left( \Bbb{R}^{n}\right) $ and any $j\in \Bbb{Z}$; and
\item there exists a function $g(x)$, known as the scaling function, such
that $\{g(x-k):k\in \Bbb{Z}^{n}\}$ is a Riesz basis of $L^{2}\left( \Bbb{R}%
^{n}\right) $.
\end{enumerate}
\end{definition}
Because of property (\ref{denseness}), any function in $L^{2}\left( \Bbb{R}%
^{n}\right) $ can be approximated arbitrarily well by its projection onto
some space $V_{j}$. Moreover, as we step from a space $V_{j}$ to its
successor $V_{j+1}$, we introduce functions with double the `frequency', as
guaranteed by property (\ref{double frequency}). Roughly speaking, this
means that a function $f$ can be approximated at resolution $2^{-j}$ by an
element of $V_{j}.$
The significance of multiresolution approximations in the wavelet theory is
that
\begin{definition}
\bigskip \label{Riesz basis}If $H$ is a Hilbert space, then a sequence $%
\left\{ e_{i}\right\} _{i\in \Bbb{N}}$ is said to be a Riesz basis of $H$ if
its span is dense in $H$ and if there exist constants $00$%
\[
\left| \hat{T}\left( \xi \right) \right| \leq C_{N}\left( 1+\left| \xi
\right| \right) ^{N}\left\{
\begin{array}{ll}
e^{b\func{Im}\xi } & \func{Im}\xi \geq 0 \\
e^{a\func{Im}\xi } & \func{Im}\xi \leq 0
\end{array}
\right.
\]
\end{theorem}
\begin{lemma}
Let $m_{0}$ be a trigonometric polynomial, say there exists $T\in \Bbb{N}$
and scalars $\alpha _{k}$ such that
\[
m_{0}=\sum_{\left| k\right| \leq T}\alpha _{k}e^{ik\xi }
\]
and that $m_{0}\left( 0\right) =1$. Define $\hat{\phi}\left( \xi \right)
=\prod_{j=1}^{\infty }m_{0}\left( 2^{-j}\xi \right) $, then $\phi $ is
compactly supported
\end{lemma}
\begin{proof}
We can use the argument I presented at the beginning of this section to show
that the partial products converge locally uniformly to $\hat{\phi}$, since
each of the functions $m_{0}\left( 2^{-j}\xi \right) $ is entire, we have
that $\hat{\phi}$ is (see \cite{M3P4}). Again using the previous arguments,
we have
\[
\left| \hat{\phi}\left( \xi \right) \right| \leq e^{C\left| \xi \right| }
\]
One can derive other inequalities to get both of the polynomial bounds as
required (see Daubechies\cite{daubechies}, for example), and we have that $%
\phi $ is compactly supported.
\end{proof}
Finally we have:
\begin{lemma}
If $m_{0}\left( \xi \right) $ is a trigonometric polynomial and $\left|
m_{0}\left( \xi \right) \right| ^{2}+\left| m_{0}\left( \xi +\pi \right)
\right| ^{2}=1$ and $m_{0}\left( 0\right) =1$ and $m_{0}\left( \xi \right)
\neq 0$ on $[-\frac{\pi }{2},\frac{\pi }{2}]$, then $\left\{ \phi (x-k):k\in
\Bbb{Z}\right\} $ is an orthonormal sequence.
\end{lemma}
\begin{proof}
Put $\alpha \left( \xi \right) =\sum_{-\infty }^{\infty }\left| \hat{\phi}%
\left( \xi +2k\pi \right) \right| ^{2}$, then working lemma \ref{lemma about
sums of fourier transforms of scaling function} backwards, we only have to
show that $\alpha \equiv 1$. This is a consequence of the identity $\alpha
\left( 2\xi \right) =\alpha \left( \xi \right) g\left( \xi \right) +\alpha
\left( \xi +\pi \right) g\left( \xi +\pi \right) $. See Meyer\cite{meyer}
for details.
\end{proof}
Why is $\psi $ compactly supported? Well $\hat{\psi}\left( 2\xi \right)
=e^{-i\xi }\overline{m_{0}\left( \xi +\pi \right) }\hat{\phi}\left( \xi
\right) $ and a Paley-Wiener argument gives compact support. The tricky
parts which are left to show are that the $V_{j}$ do indeed form a
multiresolution approximation of $L^{2}\left( \Bbb{R}^{n}\right) $, and that
the mutliresolution approximation is $r$-regular. Again, see either Meyer
\cite{meyer} or Daubechies\cite{daubechies} for the proofs.
\section{Multi-dimensional wavelets\label{multidimensional wavelets}}
There are two ways to proceed in the construction of multidimensional
wavelets. One approach is to start with a multiresolution approximation of $%
L^{2}\left( \Bbb{R}^{n}\right) $, and then to generalise the construction of
the previous section. The problem with this method is that it won't in
general lead to compactly supported wavelets, even if the scaling function $%
\phi $ is compactly supported. Consequently, I will use the alternative,
`tensor product' approach in which we construct a higher dimensional
multiresolution approximation as the algebraic tensor product of
one-dimensional \ multiresolution approximations
Specifically, suppose that $\mathcal{V}_{j}$ is a multiresolution
approximation of $L^{2}\left( \Bbb{R}\right) $. We define $V_{j}\subset
L^{2}\left( \Bbb{R}^{n}\right) $ by $V_{j}=\overline{\bigotimes_{i=1}^{n}%
\mathcal{V}_{j}}$. Then by definition, we have that
\[
\left\{ \phi (x_{1}-k_{1})\ldots \phi (x_{2}-k_{n}):(k_{1},\ldots ,k_{n})\in
\Bbb{Z}^{n}\right\}
\]
is an orthonormal basis of $V_{0}$. It is easily seen that $V_{j}$ is a
multiresolution approximation of $L^{2}\left( \Bbb{R}^{n}\right) $. We let $%
W_{j}$ denote the orthogonal complement of $V_{j}$ in $V_{j+1}$, and
analogously to the one-dimensional case, we will seek a collection of basic
wavelets $\psi ^{\varepsilon }$ (a single basic wavelet isn't sufficient in
the multidimensional case) which generate an orthonormal basis of $W_{0}$.
Now let $\mathcal{W}_{j}$ denote the orthogonal complement of $\mathcal{V}%
_{j}$ in $\mathcal{V}_{j+1}$. We have
\begin{eqnarray*}
V_{1} &=&\overline{\bigotimes_{i=1}^{n}\mathcal{V}_{1}} \\
&=&\overline{\bigotimes_{i=1}^{n}\left( \mathcal{V}_{0}\oplus \mathcal{W}%
_{0}\right) } \\
&=&\overline{\bigotimes\limits_{i=1}^{n}\mathcal{V}_{0}}\oplus
\bigoplus\limits_{\varepsilon \in E}W^{\varepsilon } \\
&=&V_{0}\oplus \bigoplus\limits_{\varepsilon \in E}W^{\varepsilon }
\end{eqnarray*}
where we define $E=\left\{ \left( \varepsilon _{1},\ldots ,\varepsilon
_{n}\right) :\varepsilon _{j}\in \left\{ 0,1\right\} \right\} \backslash
\left\{ \left( 0,\ldots ,0\right) \right\} $ and
\[
W^{\varepsilon }=\overline{\bigotimes\limits_{i=1}^{n}A_{i}}
\]
with $A_{i}=V_{0}$ if $i=0$ and $A_{i}=W_{0}$ if $i=1$. Consequently, we
have that $W_{0}=\bigoplus\limits_{\varepsilon \in E}W^{\varepsilon }$, and
each $W^{\varepsilon }$ has an orthonormal basis which we get from
multiplying bases of $\mathcal{V}_{0}$ and $\mathcal{W}_{0}$, that is, if we
define $\psi ^{0}=\phi $ and $\psi ^{1}=\psi $, then $W^{\varepsilon }$ has
basis
\[
\left\{ \psi ^{\varepsilon }\left( x-k\right) :k\in \Bbb{Z}^{n}\right\}
:=\left\{ \prod\limits_{i=1}^{n}\psi ^{\varepsilon _{i}}\left(
x_{i}-k_{i}\right) :\left( k_{1},\ldots ,k_{n}\right) \in \Bbb{Z}%
^{n}\right\}
\]
and so we define the collection $\left\{ \psi ^{\varepsilon }:\varepsilon
\in E\right\} $ to be the set of basic wavelets. Correspondingly, if we
define $\mathcal{Q}$ to be the set of dyadic cubes as in \ref{dyadic cubes},
then the wavelets are given by
\[
\psi _{Q}^{\varepsilon }\left( x\right) =2^{nj/2}\psi ^{\varepsilon
_{1}}\left( 2^{j}x_{1}-k_{1}\right) \ldots \psi ^{\varepsilon _{n}}\left(
2^{j}x_{n}-k_{n}\right)
\]
we see that these wavelets are compactly supported (assuming that $\phi $
and $\psi $ are compactly supported). Indeed suppose that the supports of $%
\phi $ and $\psi $ are contained in the interval $\left[ \frac{1}{2}-\frac{m%
}{2},\frac{1}{2}+\frac{m}{2}\right] $ for some $m\geq 1$. Now each function $%
\psi ^{\varepsilon _{i}}\left( 2^{j}x_{i}-k_{i}\right) $ is formed from $%
\psi ^{\varepsilon _{i}}$ by stretching by $2^{-j}$ and then moving it over
to $k_{i}2^{-j}$, so is supported in an interval centered around $%
k_{i}2^{-j} $, of diameter $m$, and hence the product $\psi
_{Q}^{\varepsilon }$ is supported in the cube $mQ$. One can verify that the
wavelets constructed in this manner satisfy the properties which one expects
from wavelets (i.e. that they form a basis of $L^{2}\left( \Bbb{R}%
^{n}\right) $, are regular, localised (which we've just shown) and the $%
k^{th}$ moments vanish for $0\leq k\leq r$).
\chapter{Unconditional Bases and Khinchin's Inequality}
\begin{definition}
Let $\left\{ x_{j}\right\} _{j\in \Bbb{N}}$ be a sequence in a normed linear
space.
\begin{enumerate}
\item \label{unconuncon}The sequence is said to be unconditionally summable
if for any permutation $\sigma $ of $\Bbb{N}$, $\sum x_{\sigma (j)}$
converges.
\item \label{unconsign}The sequence is said to be sign summable if for any
sequence $\left\{ \varepsilon _{j}\right\} _{j\in \Bbb{N}}$ of $1$s and $-1$%
s, $\sum \varepsilon _{j}x_{j}$ converges.
\item \label{unconfish}The sequence is said to be fishy summable if there
exists a constant $C$ such that for any finite subset $F\subset \Bbb{N}$,
and for any sequence $\varepsilon (\lambda ),\lambda \in F$ of $1$s and $-1$%
s, $\left\| \sum_{\lambda \in F}\varepsilon (\lambda )x_{\lambda }\right\|
\leq C$.
\end{enumerate}
\end{definition}
\begin{theorem}
\label{equivalence sums}Suppose $\left\{ x_{k}\right\} $ is a sequence in a
Banach space, then the following are equivalent
\begin{enumerate}
\item $\left\{ x_{k}\right\} $ is unconditionally summable;
\item $\left\{ x_{k}\right\} $ is sign summable;
\item $\left\{ x_{k}\right\} $ is fishy summable.
\end{enumerate}
\end{theorem}
The reader is referred to \cite{diestel}\ for the proof.
\begin{definition}
Let $\left\{ e_{k}\right\} $ be a sequence of vectors in a Banach space $X$.
The sequence is said to be an unconditional basis of $X$ if
\begin{itemize}
\item For any vector $x\in X$, there exists a unique sequence of scalars $%
\left\{ \alpha _{k}\right\} $ such that $x=\sum \alpha _{k}e_{k}$ (with
convergence of the sum occurring in the Banach space norm);
\item The series above is unconditionally convergent.
\end{itemize}
\end{definition}
\section{Khinchin's Inequality}
\begin{definition}
For $n\in \Bbb{N}$, we define the Rademacher function, $r_{n}:[0,1]%
\longmapsto \Bbb{R}$ \ by
\[
r_{n}(t)=sign\left( \sin \left( 2^{n}\pi t\right) \right)
\]
and by definition $r_{0}\equiv 1$. Alternatively, one has
\[
r_{n}(t)=\sum_{2^{n}\leq l<2^{n+1}}q_{l}(t)
\]
where $q_{l}\left( t\right) =2^{-n/2}h_{l}(t)$ and $h_{l}$ is the $l^{th}$
Haar function.
\end{definition}
\begin{lemma}
The collection of Rademacher functions are a set of independent random
variables on $\left( [0,1],\mathcal{B},\lambda \right) $, where $\mathcal{B}$
are the Borel sets of $[0,1]$, and $\lambda $ is the restriction of the
Lebesgue measure to $\mathcal{B}$.
\end{lemma}
\begin{proof}
For each $j\in \Bbb{N}$, define $l_{j}=2^{-j}$, and let $A_{jk}$ be the
interval $((2k-1)l_{j},2kl_{j}]$. Define $B_{j}=\coprod\limits_{1\leq k\leq
2^{j-1}}A_{jk}$.
For any real number $x$, and any non-negative integer $n$, the event
\[
\left( r_{n}\leq x\right) =\left\{
\begin{array}{ll}
\left[ 0,1\right] & x\geq 1 \\
B_{n} & -12^{k}\right\} $, where $k\in \Bbb{Z}$.These will be used to
determine the support of $\alpha $ (where by definition, the support of $%
\alpha $ is the set of $Q\in \mathcal{Q}$ for which $\alpha \left( Q\right)
\neq 0$), and to get the bases for the atoms in the atomic decomposition.
But the $E_{k}$ are themselves fairly arbitrary sets, so we need some way of
cubifying them. Fix some constant $\beta \in \left( 0,\gamma \right) $, then
pick $k\in \Bbb{Z}$. We define the collection $\mathcal{C}_{k}$ to be the
set of dyadic cubes which are nearly contained in $E_{k}$, that is,
\[
\mathcal{C}_{k}=\left\{ Q\in \mathcal{Q}:\left| Q\cap E_{k}\right| \geq
\beta \left| Q\right| \right\}
\]
and we then define the magnification $E_{k}^{\ast }$ of $E_{k}$ by $%
E_{k}^{\ast }=\bigcup_{Q\in \mathcal{C}_{k}}Q$.
Now, to get maximality, observe that the elements of $\mathcal{C}_{k}$ have
their volume bounded from above (suppose $Q\in \mathcal{C}_{k}$ then $\left|
Q\right| \leq \beta ^{-1}\left| Q\cap E_{k}\right| \leq \beta ^{-1}\left|
E_{k}\right| $) and that if any distinct pair of dyadic cubes have non-empty
intersection, then one must be contained in the other and the volume of the
larger one is at least twice that of the smaller one. From the Hausdorff
maximal principle, each $Q\in \mathcal{C}_{k}$ is contained in some maximal
chain, and the above comments show that this chain has a maximal element,
say $Q(k,l)\in \mathcal{C}_{k}$.
The $Q\left( k,l\right) $ are disjoint, as $l$ varies (say $Q\left(
k,l_{1}\right) $ and $Q\left( k,l_{2}\right) $ have non-empty intersection;
without loss of generality, assume $Q\left( k,l_{1}\right) \subset $ $%
Q\left( k,l_{2}\right) $, then by maximality of $Q\left( k,l_{1}\right) $, $%
Q\left( k,l_{1}\right) =$ $Q\left( k,l_{2}\right) $). It is also clear that
each $Q\in \mathcal{C}_{k}$ is contained in some $Q\left( k,l\right) \in
\mathcal{C}_{k}$. That is, the sets $Q\left( k,l\right) $, as $l$ varies,
form a partition of $\mathcal{C}_{k}$.
Denote the support of $\alpha $ by $\mathcal{C}$. We'll show that $\mathcal{C%
}\subset \bigcup_{k}\mathcal{C}_{k}$. For let $Q^{\prime }\in \mathcal{Q}$
such that $\alpha \left( Q^{\prime }\right) \neq 0$, there exists $k\in \Bbb{%
Z}$ small enough such that $\left| \alpha \left( Q^{\prime }\right) \right|
\left| Q^{\prime }\right| ^{-1/2}>2^{k}$. Suppose $R^{\prime }\subset
Q^{\prime }$ with $\left| R^{\prime }\right| \geq \gamma \left| Q^{\prime
}\right| $ and that $x\in R^{\prime }$, then
\begin{eqnarray*}
\sigma (x) &=&\left( \sum_{Q\in \mathcal{Q}}\left| \alpha \left( Q\right)
\right| ^{2}\left| Q\right| ^{-1}\chi _{R}\left( x\right) \right) ^{1/2} \\
&\geq &\left( \left| \alpha \left( Q^{\prime }\right) \right| ^{2}\left|
Q^{\prime }\right| ^{-1}\right) ^{1/2} \\
&>&2^{k}
\end{eqnarray*}
so $E_{k}\supset R^{\prime }$, from which we get $Q^{\prime }\cap
E_{k}\supset R^{\prime }$, then $\left| Q^{\prime }\cap E_{k}\right| \geq
\left| R^{\prime }\right| \geq \gamma \left| Q^{\prime }\right| >\beta
\left| Q\right| $, consequently $Q^{\prime }\in \mathcal{C}_{k}$.
Define $\Delta _{k}=\mathcal{C}_{k}\backslash \mathcal{C}_{k+1}$, (so that $%
\Delta _{k}$ is approximately is the set of $Q$ for which $Q$ is nearly $%
\left\{ x:2^{k}<\sigma (x)\leq 2^{k+1}\right\} $). Define $\Delta
(k,l)=\left\{ Q\in \Delta _{k}:Q\subset Q\left( k,l\right) \right\} $. The $%
\Delta \left( k,l\right) $ will be the bases of the atoms in our atomic
decomposition. Fix $k\in \Bbb{Z}$, then the set of $\Delta \left( k,l\right)
$ as $l$ varies forms a partition of $\Delta _{k}$. (Reason: certainly if $%
Q\in \Delta \left( k,l\right) $, then by definition $Q\in \Delta _{k}$, and
if $Q\in \Delta \left( k,l\right) \cap \Delta (k,l^{\prime })$, then $Q\in
Q\left( k,l\right) $ and $Q\in Q\left( k,l^{\prime }\right) $, but, as we've
shown, this implies $Q\left( k,l\right) =Q\left( k,l^{\prime }\right) $).
Suppose, more generally, that $Q\in \Delta (k,l)\cap \Delta (k^{\prime
},l^{\prime })$. Now if $k\left| R\right| (1-\frac{\beta }{\gamma }%
)=\left| R\right| \gamma ^{-1}\left( \gamma -\beta \right) >\left| Q\right|
\left( \gamma -\beta \right) $. Using this in the above displayed equation
gives the desired result.
\begin{lemma}
\label{long-time lemma}Let $\sigma \left( x\right) $ be a non-negative
measurable function on $\Bbb{R}^{n}$. For $k\in \Bbb{Z}$, define $%
E_{k}=\left\{ x:\sigma \left( x\right) >2^{k}\right\} $, then
\[
\sum_{-\infty }^{\infty }2^{k}\left| E_{k}\right| \leq 2\int_{\Bbb{R}%
^{n}}\sigma \left( x\right) dx
\]
\end{lemma}
\begin{proof}
I'll prove the result for any bounded simple function $\sigma $, then the
general result holds by approximating a measurable function by a simple
function.
Write $\sigma =\sum_{i=1}^{N}a_{i}\chi _{i}$, where the $\chi _{i}$ are
characteristic functions of measurable sets $A_{i}$ write $l_{j}=\left|
A_{j}\right| $, we may assume that the $a_{i}$ are increasing. Put $%
b_{i}=\lceil \log _{2}\left( a_{i}\right) \rceil $. Then
\begin{eqnarray*}
\sum_{-\infty }^{\infty }2^{k}\left| E_{k}\right| =\sum_{k=-\infty
}^{b_{1}-1}2^{k}\left( \sum_{j=1}^{N}l_{j}\right) +2^{b_{1}}\left(
\sum_{a_{j}>2^{b_{1}}}l_{j}\right) +2^{b_{1}+1}\left(
\sum_{a_{j}>2^{b_{1}+1}}l_{j}\right) +\ldots \\
=l_{1}\left( \sum_{k=-\infty }^{b_{1}-1}2^{k}\right) +l_{2}\left(
\sum_{k0$ such that
for any $\varepsilon \in E$, there exists a cube $A^{\varepsilon }\subset
\lbrack 0,1)^{n}$ such that $\left| A^{\varepsilon }\right| \geq \gamma $
and $\left| \psi ^{\varepsilon }(x)\right| \geq c$ whenever $x\in
A^{\varepsilon }$ (where, if necessary, we have replaced each $\psi
^{\varepsilon }$ by an appropriate translated version). Now we define the
cubes $R\left( \lambda \right) $ by $R\left( \lambda \right) =\left\{
x:2^{j}x-k\in A^{\varepsilon }\right\} $ (where, as always, $\lambda
=k2^{-j}+\varepsilon 2^{-j-1}$).So certainly $\left| R\left( \lambda \right)
\right| \geq \gamma \left| Q\left( \lambda \right) \right| $, and for each $%
x\in R(\lambda )$, $\left| \psi _{\lambda }\left( x\right) \right| =\left|
2^{nj/2}\psi ^{\varepsilon }\left( 2^{j}x-k\right) \right| \geq
c2^{nj/2}=c\left( 2^{-nj}\right) ^{-1/2}=c\left| Q\right| ^{-1/2}$. That is,
on $R\left( \lambda \right) $, $\left| Q\right| ^{-1}\leq c^{-2}\left| \psi
_{\lambda }(x)\right| ^{2}$.So we have
\begin{eqnarray*}
\left( \sum_{\lambda \in \Lambda }\left| \alpha \left( \lambda \right)
\right| ^{2}\left| Q\left( \lambda \right) \right| ^{-1}\chi _{R\left(
\lambda \right) }\left( x\right) \right) ^{1/2} &\leq &\left( \sum_{\lambda
\in \Lambda }\left| \alpha \left( \lambda \right) \right| ^{2}c^{-2}\left|
\psi _{\lambda }(x)\right| ^{2}\chi _{R\left( \lambda \right) }\left(
x\right) \right) ^{1/2} \\
&\leq &c^{-1}\left( \sum_{\lambda \in \Lambda }\left| \alpha \left( \lambda
\right) \right| ^{2}\left| \psi _{\lambda }(x)\right| ^{2}\right) ^{1/2}
\end{eqnarray*}
by hypotheses, the right-hand side is in $L^{1}\left( \Bbb{R}^{n}\right) $,
hence also the left-hand side is.
(\ref{finalR})$\Rightarrow $(\ref{finalatomd}): Think of the function $%
\alpha :\Lambda \rightarrow \Bbb{C}$ as a sequence. Then it is determined by
the $2^{n}-1$ sequences $\alpha ^{\varepsilon }:\mathcal{Q}\rightarrow \Bbb{C%
}$. Then
\[
\left( \sum_{Q\in \mathcal{Q}}\left| \alpha ^{\varepsilon }\left( Q\right)
\right| ^{2}\left| Q\right| ^{-1}\chi _{R}\left( x\right) \right) ^{1/2}\leq
\left( \sum_{\lambda \in \Lambda }\left| \alpha \left( \lambda \right)
\right| ^{2}\left| Q\left( \lambda \right) \right| ^{-1}\chi _{R\left(
\lambda \right) }\left( x\right) \right) ^{1/2}
\]
the right-hand side is in $L^{1}\left( \Bbb{R}^{n}\right) $, so the
left-hand side is. We apply theorem \ref{atomicsequence} and see that the
\emph{sequence }$\alpha ^{\varepsilon }$ has a decomposition by atoms of $%
T_{1}$, say $\alpha ^{\varepsilon }\left( Q\right) =\sum_{i=0}^{\infty
}\lambda _{i}^{\varepsilon }\alpha _{i}^{\varepsilon }\left( Q\right) $,
where $\sum_{i=0}^{\infty }\lambda _{i}<\infty $ and the $\alpha
_{i}^{\varepsilon }$ are atoms of $T_{1}$. Of course, we want to show that $%
f(x)=\sum_{\lambda \in \Lambda }\alpha \left( \lambda \right) \psi _{\lambda
}\left( x\right) $ has a decomposition by atoms of $H^{1}\left( \Bbb{R}%
^{n}\right) $. Indeed, we will show that the functions $a_{i}^{\varepsilon
}\left( x\right) =\sum_{Q\in \mathcal{Q}}\alpha _{i}^{\varepsilon }\left(
Q\right) \psi _{Q}^{\varepsilon }\left( x\right) $ are atoms of $H^{1}\left(
\Bbb{R}^{n}\right) $ (well, they're nearly atoms, we need to tweak them by a
uniform constant), then we'd have
\begin{eqnarray*}
f(x) &=&\sum_{\lambda \in \Lambda }\alpha \left( \lambda \right) \psi
_{\lambda }\left( x\right) \\
&=&\sum_{\varepsilon \in E}\sum_{Q\in \mathcal{Q}}\alpha ^{\varepsilon
}\left( Q\right) \psi _{Q}^{\varepsilon }(x) \\
&=&\sum_{\varepsilon \in E}\sum_{Q\in \mathcal{Q}}\sum_{i=0}^{\infty
}\lambda _{i}^{\varepsilon }\alpha _{i}^{\varepsilon }\left( Q\right) \psi
_{Q}^{\varepsilon }(x) \\
&=&\sum_{\varepsilon \in E}\sum_{i=0}^{\infty }\lambda _{i}^{\varepsilon
}a_{i}^{\varepsilon }(x)
\end{eqnarray*}
which is then an atomic decomposition of $f$ (well, a finite sum of atomic
decompositions, which amounts to the same thing).
For the first time, we use the fact that our basic wavelets $\psi
^{\varepsilon }$ are compactly supported. As in section \ref
{multidimensional wavelets}, we know that there exists a constant $m\geq 1$
such that for each $Q\in \mathcal{Q}$, and each $\varepsilon \in E$, the
support of $\psi _{Q}^{\varepsilon }$ is contained in $mQ$. Define $%
Q_{i}^{\varepsilon }$ to be the base of the atom $\alpha _{i}^{\varepsilon
}\left( Q\right) $. Then $a_{i}^{\varepsilon }\left( x\right) =\sum_{Q\in
\mathcal{Q}}\alpha _{i}^{\varepsilon }\left( Q\right) \psi _{Q}^{\varepsilon
}\left( x\right) $ has support which is contained in $mQ_{i}^{\varepsilon }$
also$\left\| a_{i}^{\varepsilon }\right\| _{2}=\left( \sum_{Q\in \mathcal{Q}%
}\left| \alpha _{i}^{\varepsilon }\left( Q\right) \right| ^{2}\right) ^{1/2}$
(by orthonormality of the wavelets!) $\leq \left| Q_{i}^{\varepsilon
}\right| ^{-1/2}$ (by definition of an atom in $T_{1}$). Equally, $\left|
mQ_{i}^{\varepsilon }\right| ^{-1/2}=m^{-n/2}\left| Q_{i}^{\varepsilon
}\right| ^{-1/2}\geq \left\| m^{-n/2}a_{i}^{\varepsilon }\right\| _{2}$.
\begin{remark}
This doesn't quite prove atomicity in $H^{1}\left( \Bbb{R}^{n}\right) $, as
we've used dyadic cubes instead of balls; however each cube $Q$ is contained
in a unique minimal ball $B$, and the ratio $\zeta _{n}=\left| B\right|
/\left| Q\right| $ is independent of the choice of cube. If a function $%
\frak{a}$ $\in L^{2}\left( \Bbb{R}^{n}\right) $ satisfies the conditions
that for some cube $Q$, the support of $\frak{a}$ is contained in $Q$ and $%
\left\| \frak{a}\right\| _{2}\leq \left| Q\right| ^{-1/2}$ and $\int \frak{a}%
dx=0$. Then we also have that the support of $\frak{a}$ is contained in $B$,
and $\left| B\right| ^{-1/2}=\zeta _{n}^{-1/2}\left| Q\right| ^{-1/2}\geq
\left\| \xi _{n}^{-1/2}\frak{a}\right\| _{2}$
\end{remark}
Because $a_{i}^{\varepsilon }$ has compact support and is in $L^{2}\left(
\Bbb{R}^{n}\right) $, it is in $L^{1}\left( \Bbb{R}^{n}\right) $, and the
Lebesgue dominated convergence theorem applies,
\[
\int a_{i}^{\varepsilon }\left( x\right) dx=\sum_{Q\in \mathcal{Q}}\alpha
_{i}^{\varepsilon }\left( Q\right) \int \psi _{Q}^{\varepsilon }\left(
x\right) dx=0
\]
Hence by the remarks, we know that $m^{-n/2}\zeta ^{-1/2}a_{i}^{\varepsilon
} $ is an atom of $H^{1}\left( \Bbb{R}^{n}\right) $.
(\ref{finalR})$\Rightarrow $(\ref{finalS}): As in the proof of the previous
part of the theorem, we know that each of the sequences $\alpha
^{\varepsilon }$ belongs to the tent space $T_{1}$. Consequently, the same
is true for the sequence $\alpha $, and the result follows.
(\ref{finalatomd})$\Rightarrow $(\ref{finalother}): We want to show that if $%
f$ has an atomic decomposition, then
\[
Tf\left( x\right) :=\left( \sum_{\lambda \in \Lambda }\left| \alpha \left(
\lambda \right) \right| ^{2}\left| \psi _{\lambda }\left( x\right) \right|
^{2}\right) ^{1/2}
\]
is Lebesgue integrable. First, let's show the result holds if $f$ is an atom
of $H^{1}\left( \Bbb{R}^{n}\right) $, then the result will follow for any $f$
with an atomic decomposition.
An initial observation is that by Pythagoras, $\left\| f\right\|
_{2}^{2}=\sum_{\lambda \in \Lambda }\left| \alpha \left( \lambda \right)
\right| ^{2}$, and
\[
\left\| Tf\right\| _{2}^{2}=\int \left( \sum_{\lambda }\left| \alpha \left(
\lambda \right) \right| ^{2}\left| \psi _{\lambda }\left( x\right) \right|
^{2}\right) dx=\sum_{\lambda }\left[ \left| \alpha \left( \lambda \right)
\right| ^{2}\int \left| \psi _{\lambda }\left( x\right) \right| ^{2}dx\right]
=\left\| f\right\| _{2}^{2}
\]
Suppose $f$ has support contained in the ball $B=\left\{ x:\left|
x-x_{0}\right| \leq r\right\} $. Choose $C>1$. We make $\Bbb{R}^{n}$ into an
onion centered about $x_{0}$ by defining the core of the onion to be the
ball $\tilde{B}=CB$, the rest of $\Bbb{R}^{n}$ is then decomposed into
layers $R_{k}=\left\{ x:2^{k}Cr<\left| x-x_{0}\right| \leq 2^{k+1}Cr\right\}
$, for $k\in \Bbb{Z}_{\geq 0}$. So we split up the integral of $Tf$ over $%
\Bbb{R}^{n}$ as sum of the integrals on the parts of the onion. Using the
above observation, and the fact that $f$ is an atom, we estimate the
integral on the core as follows:
\[
\int_{\tilde{B}}1.Tf(x)dx\leq \left| \tilde{B}\right| ^{1/2}\left\|
Tf\right\| _{2}\leq C^{n/2}\left| B\right| ^{1/2}\left\| f\right\| _{2}\leq
C^{n/2}
\]
Now to estimate the integral on each of the layers $R_{k}$, we will show
that for many $\lambda \in \Lambda $, $\left| \alpha \left( \lambda \right)
\right| ^{2}\left| \psi _{\lambda }\left( x\right) \right| ^{2}=0$, given
that $x\in R_{k}$. Indeed, as we've mentioned before, there exists a
constant $m$, such that the support of $\psi _{\lambda }$ is contained in $%
mQ\left( \lambda \right) $ and by hypothesis, the support of $f$ is
contained in $B$; we have $\alpha \left( \lambda \right) =\left( f,\psi
_{\lambda }\right) $, so that $\alpha \left( \lambda \right) $ is zero if
the intersection of $mQ\left( \lambda \right) $ with $B$ is empty, and if
the intersection of $mQ\left( \lambda \right) $ with $R_{k}$ is emtpy, then $%
\left| \psi _{\lambda }\left( x\right) \right| ^{2}=0$, for $x\in R_{k}$. So
to evaluate $Tf$ on $R_{k}$, we merely need to sum over the $\lambda \in
\Lambda $ such that $mQ\left( \lambda \right) \cap B\cap R_{k}\neq \emptyset
$. If $\lambda $ satisfies this property, then $Q\left( \lambda \right) $
must be fairly big. Precisely, if we choose $C$ sufficiently large, then
there exists a positive constant $c$ such that $2^{-j}\geq cr2^{k}$.
It can then be shown that $\left| \left( f,\psi _{\lambda }\right) \right|
\leq C2^{nj/2}2^{j}r$, and hence that for $x\in R_{k}$, $Tf\left( x\right)
\leq C^{\prime }2^{-k\left( n+1\right) }r^{-n}$.
Then
\[
\int_{R_{k}}Tf\left( x\right) dx\leq \left| R_{k}\right| C^{\prime
}2^{-k\left( n+1\right) r^{-n}}
\]
But the volume of $\left| R_{k}\right| $ is the difference between the
volume of a ball of radius $2^{k+1}Cr$ and that of a ball of radius $2^{k}Cr$%
, so there exists a constant $D$ (depending on the dimension $n$) such that
\begin{eqnarray*}
\left| R_{k}\right| &=&D\left( 2^{k+1}Cr\right) ^{n}-D\left( 2^{k}Cr\right)
^{n} \\
&=&DC^{n}r^{n}2^{kn}\left( 2^{n}-1\right)
\end{eqnarray*}
and we have
\begin{eqnarray*}
\int_{R_{k}}Tf\left( x\right) dx &\leq &DC^{n}r^{n}2^{kn}\left(
2^{n}-1\right) C^{\prime }2^{-k\left( n+1\right) }r^{-n} \\
&\leq &DC^{n}C^{\prime }\left( 2^{n}-1\right) 2^{-k} \\
&=&D^{\prime }2^{-k}
\end{eqnarray*}
where $D^{\prime }=DC^{n}C^{\prime }\left( 2^{n}-1\right) $. Then $%
\sum_{k=0}^{\infty }\int_{R_{k}}Tf\left( x\right) dx\leq 2D^{\prime }<\infty
$. So indeed $Tf$ is integrable if $f$ is an atom of $H^{1}\left( \Bbb{R}%
^{n}\right) $.
(\ref{finalR})$\Rightarrow $(\ref{finaluncon}): By what has gone before, we
have proved the inequality
\[
\left\| \sum_{\lambda \in \Lambda }\alpha \left( \lambda \right) \psi
_{\lambda }\left( x\right) \right\| _{1}\leq C\left\| \left( \sum_{\lambda
\in \Lambda }\left| \alpha \left( \lambda \right) \right| ^{2}\left| Q\left(
\lambda \right) \right| ^{-1}\chi _{R\left( \lambda \right) }\left( x\right)
\right) ^{1/2}\right\| _{1}
\]
Replace $\alpha \left( \lambda \right) $ with $\varepsilon \left( \lambda
\right) \alpha \left( \lambda \right) $, the right-hand remains the same,
and the left hand becomes the quantity we wished to show the finiteness of.
\chapter{Notes and Comments}
Though the wavelet concept is relatively new, a tremendous amount of
literature already exists on the subject, perhaps due to its
interdisciplinary nature. Many applications of the wavelet theory have been
discovered, ranging from image compression methods, mathematical models of
human hearing, through to fast integral transform techniques.
From a pure mathematical point of view, one of the most appealing aspects of
wavelets is that they are effective in most spaces of functions and
distributions. I've shown that they form an unconditional basis of $%
H^{1}\left( \Bbb{R}^{n}\right) $, but the same is true for the spaces $%
L^{p}\left( \Bbb{R}^{n}\right) $ with $1