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\begin{center}
\textbf{\large Multi-Resolution Analysis for the Haar Wavelet}\bigskip\\
\textsc{Helmut Knaust\bigskip\\
Department of Mathematical Sciences\\
University of Texas at El Paso\\
El Paso TX 79968-0514\bigskip\\}
Last edits: October 22, 2019\bigskip\\
\end{center}
\section{The space $L^2([0,1))$ and its scalar product}
We will denote by $\Ltwo$ the vector space of all functions $f:[0,1)\to\real$ satisfying\footnote{More precisely: those (equivalence classes of) measurable functions on $[0,1)$ whose square is Lebesgue-integrable. Since we will be ultimately interested only in the discrete case anyway, you can just think of bounded piecewise-continuous functions with the Riemann integral.}
\begin{equation}
	\int_0^1 |f(x)|^2\,dx <\infty.
\end{equation}
On $\Ltwo$ one can define a \textsc{scalar product} as follows:
\begin{equation}
	<f,g>=\int_0^1 f(x)\cdot g(x)\,dx.
\end{equation}
Similarly to the case of $\real^n$, the scalar product automatically defines a \textsc{norm} on $\Ltwo$ via the definition
\begin{equation}
	\|f\|=\sqrt{<f,f>}=\sqrt{\int_0^1 |f(x)|^2\,dx}
\end{equation}
Finally, we say that a sequence ($f_n)$ of functions in $\Ltwo$ converges to a function $f(x)\in\Ltwo$ \textsc{in the $L^2$-sense}, if 
\begin{equation}
	\lim_{n\to\infty}\|f_n-f\|= 0
\end{equation}
\section{Orthonormal sets}
We say that a set $B$ of elements in $\Ltwo$ is an \textsc{orthonormal set}, if the scalar product of each element in $B$ with itself equals 1, and the scalar product of two different elements in $B$ is equal to 0:
\begin{enumerate}
\item $	<f,f>=1$\quad for all  $f\in B$
\item $<f,g>=0$\quad for all  $f, g\in B$  satisfying $f\neq g$.
\end{enumerate}
An orthonormal set $B$ is automatically linearly independent.\footnote{Indeed assume that 
for some real numbers $a_1,a_2,\ldots, a_n$ and some distinct elements  $g_1, g_2,\ldots, g_n$ in $B$,
\[a_1 g_1(x)+a_2 g_2(x)+\cdots + a_n g_n(x)=0 \mbox{ for all } x. \]
Then for any $k$ with $k\in\{1,\ldots,n\}$
\[	<a_1 g_1+a_2 g_2+\cdots + a_n g_n,g_k>=<0,g_k>=0.
\]
On the other hand, 
\[	<a_1 g_1+a_2 g_2+\cdots + a_n g_n,g_k>=a_1 <g_1,g_k> + a_2 <g_2,g_k> +\cdots + 
a_n <g_n,g_k>=a_k <g_k,g_k>=a_k.
\]
So $a_k=0$ for all $k$.}
Our \textbf{ultimate goal} will be to find a particular orthonormal set $B=\{f_1(x),f_2(x), \ldots\}$ such that we can approximate every function $f(x)\in\Ltwo$ by linear combinations of the elements in $B$; more precisely, given $f(x)\in\Ltwo$,  we will be able to find scalars $(a_k)$ such that the sequence   
\begin{equation}
	\left(\sum_{k=1}^n a_k f_k(x)\right)
\end{equation}
converges to $f(x)$ in the $L^2$-sense.\footnote{We  basically already know one example of such a set: It is known that the set 
\[F=\{\frac{1}{\sqrt{2\pi}},\frac{1}{\sqrt{\pi}}\cos(x),\frac{1}{\sqrt{\pi}} \cos(2x),\ldots,\frac{1}{\sqrt{\pi}} \sin(x),\frac{1}{\sqrt{\pi}}\sin(2x),\ldots\}\]
forms an orthonormal set with which we can approximate all elements in $L^2([-\pi,\pi])$ in this fashion.}
 

\section{The Haar scaling function}
\begin{figure}[h]
\begin{center}
\resizebox{3in}{!} {\includegraphics{nb/001.pdf}}
\end{center}
\caption{The Haar scaling function $\phi(x)$}
\end{figure}
We denote by $\phi(x)$ the following function:
\begin{equation}
	\phi(x)=\left\{\begin{array}{rl}1,&\mbox{ if } x\in [0,1)\\0,&\mbox{ if } x< 0 \mbox{ or } x\geq 1\end{array}\right.
\end{equation}
\begin{figure}[h]
\begin{center}
\resizebox{3in}{!} {\includegraphics{nb/002.pdf}}
\end{center}
\caption{The Haar scaling function $\phi(x)$ restricted to $[0,1)$}
\end{figure}
$\phi:\real\to\real$ is called the \textsc{Haar scaling function}, or the Haar ``father" wavelet. Throughout we will identify $\phi(x)$ with its restriction to $[0,1)$.

Let $V_0$ denote the one-dimensional vector space spanned by $\phi(x)$; this is nothing else but the set of all functions constant on $[0,1)$ (and vanishing elsewhere).

Next we consider the functions $2^{1/2}\phi(2x)$ and $2^{1/2}\phi(2x-1)$.
\begin{figure}[h]
\begin{center}
\resizebox{3in}{!} {\includegraphics{nb/003.pdf}}
\end{center}
\caption{The function $2^{1/2}\phi(2x)$}
\end{figure}
\begin{figure}[h]
\begin{center}
\resizebox{3in}{!} {\includegraphics{nb/004.pdf}}
\end{center}
\caption{The function $2^{1/2}\phi(2x-1)$}
\end{figure}
They span a two-dimensional vector space, denoted by $V_1$, consisting of all functions on $[0,1)$ that are constant both on $[0,\frac{1}{2})$ and on $[\frac{1}{2},1)$. 

Note that $V_0\subset V_1$. Indeed the function $\phi(x)$ is a linear combination of  $2^{1/2}\phi(2x)$ and $2^{1/2}\phi(2x-1)$:
\begin{equation}\phi(x)=\frac{1}{\sqrt{2}}\left(2^{1/2}\phi(2x)\right)+\frac{1}{\sqrt{2}}\left(2^{1/2}\phi(2x-1)\right)
\end{equation}
More generally, an equation of the form

\begin{equation}
\phi(x)=\sum_{k=0}^n a_k \phi(2x-k)
\end{equation} 
is called a \textbf{dilation equation}. Functions $\phi$ that satisfy a dilation equation are rare and hard to find.

Continuing in this fashion, we can define a $2^j$-dimensional vector space $V_j$, spanned by the functions 
\[2^{j/2}\phi(2^j x), 2^{j/2}\phi(2^j x-1),\ldots,2^{j/2}\phi(2^j x-(2^j-1)).\]
The vector space $V_j$ consists of all functions on $[0,1)$ that are constant on intervals of the form $[k2^{-j},(k+1)2^{-j})$ for $k=0,1,2,\ldots 2^j-1$. Figure~\ref{Vthree}\ shows the function $2^{3/2}\phi(2^3 x-5)$ contained in $V_3$.
\begin{figure}[h]
\begin{center}
\resizebox{3in}{!} {\includegraphics{nb/005.pdf}}
\end{center}
\caption{The function $2^{3/2}\phi(2^3x-5)$}\label{Vthree}
\end{figure}
We have $V_0\subset V_1\subset\cdots\subset V_j\subset\cdots$.


You should have wondered by now why the factor $2^{j/2}$ is included. The answer is straightforward: this way the functions form an orthonormal set! 
\begin{ex}
Show that the set $\{2^{j/2}\phi(2^j x), 2^{j/2}\phi(2^j x-1),\ldots,2^{j/2}\phi(2^j x-(2^j-1))\}$ forms an orthonormal set of functions in the vector space $V_j$.
\end{ex}

\section{Using $V_j$ to approximate functions in $\Ltwo$}
A function $f\in V_j$ has the form
\begin{equation}\label{cof}
	f(x)=a_0 2^{j/2}\phi(2^j x)+a_1 2^{j/2}\phi(2^j x-1)+\cdots\ +a_{2^j-1} 2^{j/2}\phi(2^j x-(2^j-1)).
\end{equation}
Since the functions on the right side form an orthonormal set, the coefficients $a_k$ are given by the formula
\begin{equation}\label{coff}
	a_k=<f(x),2^{j/2}\phi(2^j x-k)>=\int_0^1 f(x)\cdot 2^{j/2}\phi(2^j x-k)\,dx
\end{equation}

\begin{ex}
Take the scalar product with $2^{j/2}\phi(2^j x-k)$ on both sides of (\ref{cof}) to verify Formula~(\ref{coff}). 
\end{ex}

\begin{figure}[h]
\begin{center}
\resizebox{5in}{!} {\includegraphics{nb/006.pdf}}
\end{center}
\caption{Approximating a function by an element in $V_4$}\label{app4}
\end{figure}

The same formula for the coefficients can be used to approximate functions in $\Ltwo$ by a function in $V_j$. Let $f(x)$ be a function in $\Ltwo$, and set
\begin{equation}
	f_j(x)=a_0 2^{j/2}\phi(2^j x)+a_1 2^{j/2}\phi(2^j x-1)+\cdots\ +a_{2^j-1} 2^{j/2}\phi(2^j x-(2^j-1)),
\end{equation}
where the coefficients $a_k$ are computed via Formula~(\ref{coff}).

Alfred Haar (1885--1933) showed in 1910 that, if $f(x)$ is continuous, the sequence $(f_j(x))$ converges to $f(x)$ uniformly. If, on the other hand,  $f(x)\in\Ltwo$,  then \[\lim_{j\to\infty}\|f-f_j\|=0.\] 
Figures~\ref{app4} and \ref{app7} show the approximation of a function (dashed line) by an element in $V_4$ and $V_7$, respectively (solid line).
 
\begin{figure}[h]
\begin{center}
\resizebox{5in}{!} {\includegraphics{nb/007.pdf}}
\end{center}
\caption{Approximating a function by an element in $V_7$}\label{app7}
\end{figure} 

While we have found nice orthonormal bases for all the vector spaces $V_j$, we still fall short of our goal: If we take two basis elements from different $V_j$'s, their scalar product will not necessarily equal zero, because the intervals where the basis elements are equal to 1 may overlap.

\section{The Haar wavelet}
Let's see whether we can remedy this deficiency step by step. We want to find a function $\psi(x)$ in $V_1$, such that the linear combinations of $\phi(x)$ and $\psi(x)$ span the vector space $V_1$, and such that the following conditions are satisfied:
\begin{enumerate}
\item $<\phi,\psi>=0$
\item $<\psi,\psi>=1$ 
\end{enumerate}
Since $\psi\in V_1$ we can find scalars $a_1$ and $a_2$ such that
\begin{equation}\label{psi1}
	\psi(x)=a_1 \sqrt{2}\phi(2x)+a_2 \sqrt{2}\phi(2x-1).
\end{equation}
Note also that 
\begin{equation}\label{psi2}
	\phi(x)=\phi(2x)+\phi(2x-1).
\end{equation}
Using (\ref{psi1}) and (\ref{psi2}), the first condition becomes
\begin{equation}\label{psi3}
	<\phi,\psi>=<\phi(2x)+\phi(2x-1),a_1 \sqrt{2}\phi(2x)+a_2 \sqrt{2}\phi(2x-1)>=a_1 /\sqrt{2}+a_2 /\sqrt{2}=0,
\end{equation}
so $a_2=-a_1$. The second condition yields:
\begin{equation}\label{psi4}
	<\psi,\psi>=<a_1 \sqrt{2}\phi(2x)+a_2 \sqrt{2}\phi(2x-1),a_1 \sqrt{2}\phi(2x)+a_2 \sqrt{2}\phi(2x-1)>=a_1^2+a_2^2=1,
\end{equation}
Solving  (\ref{psi3}) and (\ref{psi4}) for $a_1$ and $a_2$, we obtain\footnote{There are actually two solutions; it suffices for us to consider the solution for which $a_1>0$. Why?}
\begin{equation}
\psi(x)	= \phi(2x)-\phi(2x-1)=\left\{\begin{array}{rl}1,&\mbox{ if } x\in [0,\frac{1}{2})\\-1,&\mbox{ if } x\in[\frac{1}{2},1)\\0,&\mbox{ otherwise}\end{array}\right.
\end{equation}
\begin{figure}[h]
\begin{center}
\resizebox{3in}{!} {\includegraphics{nb/008.pdf}}
\end{center}
\caption{The Haar mother wavelet $\psi(x)$}\label{Hmw}
\end{figure}
The function $\psi(x)$ is called the \textsc{Haar ``mother" wavelet}; its graph is depicted in Figure~\ref{Hmw}.

We will denote the vector space spanned by the function $\psi(x)$ as $W_0$. It is customary to write 
\begin{equation}
	V_1=V_0\oplus W_0.
\end{equation}
Here the symbol $\oplus$ is used to indicate  that each element in $V_1$ can be written in a unique way as the sum of an element in $V_0$ and an element in $W_0$ and  that the scalar product of any element in $V_0$ with any element in $W_0$ equals zero.

\begin{ex}
Show that the functions $\sqrt{2}\psi(2x)$ and $\sqrt{2}\psi(2x-1)$ are elements in $V_2$.
\end{ex}
\begin{ex}Show that the set $\{\sqrt{2}\psi(2x),\ \sqrt{2}\psi(2x-1)\}$ forms an orthonormal set.
\end{ex}
\begin{ex}
Show that $<\sqrt{2}\psi(2x), f(x)>=0$ for all functions $f(x)\in V_1$. (The same result holds for $\sqrt{2}\psi(2x-1)$.)
\end{ex}
\noindent These two functions are shown in Figures~\ref{Hmw21} and \ref{Hmw22}, respectively.
\begin{figure}[h]
\begin{center}
\resizebox{3in}{!} {\includegraphics{nb/009.pdf}}
\end{center}
\caption{The function $\sqrt{2}\,\psi(2x)$ in $W_1$}\label{Hmw21}
\end{figure}
\begin{figure}[h]
\begin{center}
\resizebox{3in}{!} {\includegraphics{nb/010.pdf}}
\end{center}
\caption{The function $\sqrt{2}\,\psi(2x-1)$ in $W_1$}\label{Hmw22}
\end{figure}

Let's denote the vector space spanned by $\sqrt{2}\psi(2x)$ and $\sqrt{2}\psi(2x-1)$ as $W_1$. The three exercises above show that
\begin{equation}
	V_2=V_1\oplus W_1=V_0\oplus W_0\oplus W_1, 
\end{equation}
meaning once again that each element in $V_2$ can be written in a unique way as the sum of an element in $V_1$ and an element in $W_1$ and  that the scalar product of any element in $V_1$ with any element in $W_1$ equals zero.

Continuing in this fashion, we can write
\begin{equation}
	V_{j+1}=V_j\oplus W_j, 
\end{equation}
where $W_j$ is the vector space spanned by the functions 
\[2^{j/2}\psi(2^j x), 2^{j/2}\psi(2^j x-1),\ldots,2^{j/2}\psi(2^j x-(2^j-1)).\]
The formula
\[V_j=V_0\oplus W_0 \oplus W_1\oplus \cdots \oplus W_{j-1}\]
is called \textsc{Multi-Resolution Analysis}.
\section{A discrete example}
A function in $V_4$ is determined by its 16 coefficients. Suppose the vector of coefficients is
\begin{equation}\label{v4c}
	(a_0,a_1,a_2,\ldots,a_{15})=(180,167,244,190,159,242,176,192,168,250,175,219,193,232,200,234)
\end{equation}
The corresponding function 
\begin{equation}
f(x)=\sum_{n=0}^{15} a_n\  4 \phi(16x-n)
\end{equation}is shown in Figure~\ref{v4}. 
\begin{figure}[h]
\begin{center}
\resizebox{4.5in}{!} {\includegraphics{nb/011.pdf}}
\end{center}
\caption{A function in $V_4$}\label{v4}
\end{figure}
How can we write this function $f$ as a sum of a function in $V_3$ and a function in $W_3$? Let's start with the component $g$ in $V_3$. By the discussion following the derivation of Formula~(\ref{coff}), the function $g$ is of the form
\begin{equation}
g(x)=\sum_{k=0}^7 b_k\ 2^{3/2}\phi(8x-k),
\end{equation}
with the coefficients $b_k$ computed as $b_k =<f(x),2^{3/2}\phi(8x-k)>$.
Thus we obtain \begin{eqnarray*}
	b_k &=&<f(x),2^{3/2}\phi(8x-k)>\\
	&=&<\sum_{n=0}^{15} a_n\  4 \phi(16x-n), 2^{3/2}\phi(8x-k)>\\
	&=&\sum_{n=0}^{15} a_n\  8\sqrt{2}<\phi(16x-n),\phi(8x-k)>\\
	&=&a_{2k}\  8\sqrt{2}<\phi(16x-2k),\phi(8x-k)>\  
	+\  a_{2k+1}\  8\sqrt{2}<\phi(16x-(2k+1)),\phi(8x-k)>\\\label{calc}
	&=&a_{2k}\  8\sqrt{2} \frac{1}{16}+ a_{2k+1}\  8\sqrt{2} \frac{1}{16}
	\end{eqnarray*}
Here we are using the fact that $<\phi(16x-n),\phi(8x-k)>\neq0$ only if $n=2k$ or $n=2k+1$.
(Draw a picture). More specifically, in these two cases $<\phi(16x-n),\phi(8x-k)>=\frac{1}{16}$. 

Simplifying the last expression we obtain
	\begin{equation}
	b_k=\frac{a_{2k}+a_{2k+1}}{\sqrt{2}}
	\end{equation}
This means that we can compute the vector representing the function in $V_3$ by multiplying the vector in (\ref{v4c}) by the $16\times8$ matrix
\begin{equation}
	\left(
\begin{array}{llllllll}
 \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}}
\end{array}
\right)
\end{equation}
on the right.
In our particular case, the vector (rounded to the nearest integer) representing the function in $V_3$ is given by
\begin{equation}\label{sum}
	(245,307,284,260,296,279,301,307).
\end{equation}
\begin{figure}[h]
\begin{center}
\resizebox{4.5in}{!} {\includegraphics{nb/012.pdf}}
\end{center}
\caption{A function in $V_4$ (black) and its ``orthogonal projection" onto $V_3$ (dashed)}\label{w3}
\end{figure}
Figure~\ref{w3} shows the function in $V_4$ and its ``blurry" counterpart in $V_3$.

\begin{ex}
Let $f=g+h$, with $f\in V_4$, $g\in V_3$, and $h\in W_3$. Show that if
\[f(x)=\sum_{n=0}^{15} a_n\  4 \phi(16x-n)\]
\[\mbox{ and }h(x)=\sum_{k=0}^7 c_k\ 2^{3/2}\psi(8x-k),\]
then
\begin{equation}
	c_k=\frac{a_{2k}-a_{2k+1}}{\sqrt{2}},
\end{equation}
\end{ex}
\noindent You can either use the fact that $h=f-g$, or repeat the computation on p.~\pageref{calc}.

In other words, this time we obtain the vector representing the function in $W_3$ by multiplying the vector in (\ref{v4c}) by the matrix
\begin{equation}
	\left(
\begin{array}{llllllll}
 \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 -\frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & -\frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & -\frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{\sqrt{2}} & 0 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & \frac{1}{\sqrt{2}} \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{\sqrt{2}}
\end{array}
\right)
\end{equation}
In our particular case, the vector (rounded to the nearest integer) representing the function in $W_3$ is given by
\begin{equation}\label{diff}
	(9,38,-59,-11,-58,-31,-28,-24).
\end{equation}
\begin{figure}[h]
\begin{center}
\resizebox{4.5in}{!} {\includegraphics{nb/013.pdf}}
\end{center}
\caption{The ``orthogonal projection" of our function onto $W_3$ (dashed)}\label{ww3}
\end{figure}
Figure~\ref{ww3} shows this as a function in $W_3$.

If we are ``joining" the vectors in (\ref{sum}) and (\ref{diff}), we obtain
\begin{equation}\label{comp}
	(245,307,284,260,296,279,301,307,9,38,-59,-11,-58,-31,-28,-24).
\end{equation}
Since our function in $V_4$ is the sum of its orthogonal projections onto $V_3$ and $W_3$, we will be able to retrieve the vector in (\ref{v4c}) from the vector (\ref{comp}). The cumulative energy plots of both vectors are shown in Figure~\ref{cenergy}, indicating that the vector in (\ref{comp}) has a higher energy concentration than the original vector (\ref{v4c}) and thus may be considered as a compressed version of the vector in (\ref{v4c}).
\begin{figure}[h]
\begin{center}
\resizebox{5in}{!} {\includegraphics{nb/014.pdf}}
\end{center}
\caption{The original vector (\ref{v4c}) is shown dashed, while vector (\ref{comp}) is depicted solid.}\label{cenergy}
\end{figure}
\begin{ex}
This section has shown how to compute vector (\ref{comp}) from vector (\ref{v4c}). Can we reverse the procedure? Suppose our procedure produces as the vector in (\ref{comp})
\[( 200,350,351,130,115,215,122,308,15,35,47,23,-12,-32,67,-23).\]
What does the corresponding original vector (\ref{v4c}) look like? 
\end{ex}
\section{Concluding Remarks}
We have outlined a general procedure: (1) start with a father wavelet $\phi(x)$, (2) use a dilation equation to construct an increasing sequence of vector spaces 
\[V_1\subset V_2\subset \cdots \subset V_j \subset\cdots,\] 
capable of approximating functions in $\Ltwo$, 
(3) construct the corresponding mother wavelet, and (4) ultimately produce a multi-resolution analysis
\[V_j=V_0\oplus W_0 \oplus W_1\oplus \cdots \oplus W_{j-1}\]
Our choice for $\phi(x)$ was the constant 1, with the dilation equation
\[\phi(x)=\phi(2x)+\phi(2x-1).\] As you will see later in the course, in the 1980's other possible candidates emerged. 
\paragraph{Acknowledgment.} This exposition is based on material in  \emph{A First Course in Wavelets with Fourier Analysis} by Albert Boggess \& Francis J. Narcowich.
\end{document}
The figure below shows the (differentiable!) \textsc{D6  wavelet}, named after the Belgian mathematician Ingrid Daubechies.
\begin{figure}[h]
\begin{center}
\resizebox{5in}{!} {\includegraphics{nb/015.pdf}}
\end{center}
\caption{The Daubechies-6 father wavelet}\label{d6}
\end{figure}

