# Haar wavelets and multiresolution analysis

Jordan Bell
April 3, 2014

## 1 Introduction

Let

 $\psi(x)=\begin{cases}0&x<0,\\ 1&0\leq x<\frac{1}{2},\\ -1&\frac{1}{2}\leq x<1,\\ 0&x\geq 1.\end{cases}$

For $n,k\in\mathbb{Z}$, we define

 $\psi_{n,k}(x)=2^{n/2}\psi(2^{n}x-k),\qquad x\in\mathbb{R}.$

$L^{2}(\mathbb{R})$ is a complex Hilbert space with the inner product

 $\left\langle f,g\right\rangle=\int_{\mathbb{R}}f(x)\overline{g(x)}dx.$

We will prove that $\psi$ satisfies the following definition of an orthonormal wavelet.11 1 Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 303, Definition 6.4.1.

###### Definition 1 (Orthonormal wavelet).

If $\Psi\in L^{2}(\mathbb{R})$, $\Psi_{n,k}(x)=2^{n/2}\Psi(2^{n}x-k)$, and the set $\{\Psi_{n,k}:n,k\in\mathbb{Z}\}$ is an orthonormal basis for $L^{2}(\mathbb{R})$, then $\Psi$ is called an orthonormal wavelet.

###### Lemma 2.

$\{\psi_{n,k}:n,k\in\mathbb{Z}\}$ is an orthonormal set in $L^{2}(\mathbb{R})$.

###### Proof.

If $n,n^{\prime},k,k^{\prime}\in\mathbb{Z}$, then

 $\displaystyle\int_{\mathbb{R}}\psi_{n,k}(x)\overline{\psi_{n^{\prime},k^{% \prime}}(x)}dx$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}}2^{n/2}\psi(2^{n}x-k)2^{n^{\prime}/2}\psi(2^{n^{% \prime}}x-k^{\prime})dx$ $\displaystyle=$ $\displaystyle\int_{\mathbb{R}}2^{(n^{\prime}-n)/2}\psi(x-k)\psi(2^{n^{\prime}-% n}x-k^{\prime})dx$ $\displaystyle=$ $\displaystyle 2^{(n^{\prime}-n)/2}\delta_{k,k^{\prime}}\int_{0}^{1}\psi(x)\psi% (2^{(n^{\prime}-n)/2}x)dx$ $\displaystyle=$ $\displaystyle\delta_{k,k^{\prime}}\cdot\delta_{n,n^{\prime}},$

hence $\{\psi_{n,k}:n,k\in\mathbb{Z}\}$ is an orthonormal set. ∎

Bessel’s inequality states that if $\mathscr{E}$ is an orthonormal set in a Hilbert space $H$, then for any $f\in H$ we have $\sum_{e\in\mathscr{E}}|\left\langle f,e\right\rangle|^{2}\leq\left\|f\right\|_% {2}^{2}$, from which it follows that $\sum_{e\in\mathscr{E}}\left\langle f,e\right\rangle e\in H$. To say that a subset $\mathscr{E}$ of a Hilbert space $H$ is an orthonormal basis is equivalent to saying that $\mathscr{E}$ is an orthonormal set and that

 $\mathrm{id}_{H}=\sum_{e\in\mathscr{E}}e\otimes e$

in the strong operator topology. In other words, for $\mathscr{E}$ to be an orthonormal basis of $H$ means that $\mathscr{E}$ is an orthonormal set and that for every $f\in H$ we have

 $f=\sum_{e\in\mathscr{E}}\left\langle f,e\right\rangle e.$

From Lemma 2 and Bessel’s inequality, we know that for each $f\in L^{2}(\mathbb{R})$,

 $\sum_{n,k\in\mathbb{Z}}|\left\langle f,\psi_{n,k}\right\rangle|^{2}\leq\left\|% f\right\|_{2}^{2},\qquad\sum_{n,k\in\mathbb{Z}}\left\langle f,\psi_{n,k}\right% \rangle\psi_{n,k}\in L^{2}(\mathbb{R}).$

We have not yet proved that $f$ is equal to the series $\sum_{n,k\in\mathbb{Z}}\left\langle f,\psi_{n,k}\right\rangle\psi_{n,k}$, and this will not be accomplished until later in this note.

## 2 Coarser sigma-algebras

For $n,k\in\mathbb{Z}$, let

 $I_{n,k}=\left[\frac{k}{2^{n}},\frac{k+1}{2^{n}}\right),$

and let $\mathscr{F}_{n}$ be the $\sigma$-algebra generated by $\{I_{k,n}:k\in\mathbb{Z}\}$. $\mathbb{R}=\bigcup_{k\in\mathbb{Z}}I_{n,k}$, and if $k\neq k^{\prime}$ then $I_{n,k}\cap I_{n,k^{\prime}}=\emptyset$. If $n then

 $\mathscr{F}_{n}\subset\mathscr{F}_{n^{\prime}}\subset\mathscr{F},$

where $\mathscr{F}$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}$. An element of $L^{2}(\mathbb{R},\mathscr{F}_{n})$ is an element of $L^{2}(\mathbb{R},\mathscr{F})$ that is constant on each set $I_{n,k}$, $k\in\mathbb{Z}$. In other words, an element of $L^{2}(\mathbb{R},\mathscr{F}_{n})$ is a function $f:\mathbb{R}\to\mathbb{C}$ such that if $k\in\mathbb{Z}$ then the image $f(I_{n,k})$ is a single element of $\mathbb{R}$ and such that

 $\left\|f\right\|_{2}^{2}=\int_{\mathbb{R}}|f(x)|^{2}dx=\sum_{k\in\mathbb{Z}}% \int_{I_{n,k}}|f(x)|^{2}dx=\sum_{k\in\mathbb{Z}}\frac{1}{2^{n}}\cdot|f(I_{n,k}% )|^{2}<\infty.$

If $n, then

 $L^{2}(\mathbb{R},\mathscr{F}_{n})\subset L^{2}(\mathbb{R},\mathscr{F}_{n^{% \prime}})\subset L^{2}(\mathbb{R},\mathscr{F}).$

## 3 Integral kernels

We define

 $\phi(x)=\begin{cases}0&x<0,\\ 1&0\leq x<1,\\ 0&x\geq 1.\end{cases}$

For $n\in\mathbb{Z}$ we define

 $K_{n}(x,y)=2^{n}\sum_{k\in\mathbb{Z}}\phi(2^{n}x-k)\phi(2^{n}y-k),\qquad x,y% \in\mathbb{R}.$

We have

 $K_{n}(x,y)\in\{0,2^{n}\}.$

$K_{n}(x,y)=2^{n}$ if and only if there is some $k\in\mathbb{Z}$ such that $2^{n}x-k,2^{n}y-k\in[0,1)$, equivalently there is some $k\in\mathbb{Z}$ with $2^{n}x,2^{n}y\in[k,k+1)$, which is equivalent to there being some $k\in\mathbb{Z}$ such that

 $x,y\in\left[\frac{k}{2^{n}},\frac{k+1}{2^{n}}\right)=I_{n,k}.$

We define

 $P_{n}f(x)=\int_{\mathbb{R}}K_{n}(x,y)f(y)dy.$

If $x\in\mathbb{R}$ then there is a unique $k_{x}\in\mathbb{Z}$ with $x\in I_{n,k_{x}}$, and

 $P_{n}f(x)=2^{n}\int_{I_{n,k_{x}}}f(y)dy.$ (1)

It is straightforward to check that $L^{2}(\mathbb{R},\mathscr{F}_{n})$ is a closed subspace of $L^{2}(\mathbb{R},\mathscr{F})$, and in the following theorem we prove that $P_{n}$ is the orthogonal projection onto $L^{2}(\mathbb{R},\mathscr{F}_{n})$.

###### Lemma 3.

If $n\in\mathbb{Z}$, then $P_{n}$ is the orthogonal projection of $L^{2}(\mathbb{R},\mathscr{F})$ onto $L^{2}(\mathbb{R},\mathscr{F}_{n})$.

###### Proof.

For each $k\in\mathbb{Z}$, the function $P_{n}f$ is constant on the interval $I_{n,k}$, and using (1) and the Cauchy-Schwarz inequality,

 $\displaystyle\left\|P_{n}f\right\|_{2}^{2}$ $\displaystyle=\sum_{k\in\mathbb{Z}}\int_{I_{n,k}}|P_{n}f(x)|^{2}dx$ $\displaystyle=\sum_{k\in\mathbb{Z}}\int_{I_{n,k}}\left|2^{n}\int_{I_{n,k}}f(y)% dy\right|^{2}dx$ $\displaystyle=2^{n}\sum_{k\in\mathbb{Z}}\left|\int_{I_{n,k}}f(y)dy\right|^{2}$ $\displaystyle\leq 2^{n}\sum_{k\in\mathbb{Z}}\left(\int_{I_{n,k}}|f(y)|^{2}dy% \right)\left(\int_{I_{n,k}}dy\right)$ $\displaystyle=\sum_{k\in\mathbb{Z}}\int_{I_{n,k}}|f(y)|^{2}dy$ $\displaystyle=\int_{\mathbb{R}}|f(y)|^{2}dy.$

Therefore, $P_{n}:L^{2}(\mathbb{R},\mathscr{F})\to L^{2}(\mathbb{R},\mathscr{F}_{n})$. Moreover, the left-hand side of the above inequality is equal to $\left\|P_{n}f\right\|_{2}^{2}$ and the right-hand side is equal to $\left\|f\right\|_{2}^{2}$, hence we have $\left\|P_{n}f\right\|_{2}\leq\left\|f\right\|_{2}$, giving $\left\|P_{n}\right\|\leq 1$.

If $f\in L^{2}(\mathbb{R},\mathscr{F}_{n})$, then

 $\displaystyle P_{n}f(x)$ $\displaystyle=\int_{\mathbb{R}}K_{n}(x,y)f(y)dy$ $\displaystyle=2^{n}\int_{I_{n,k_{x}}}f(y)dy$ $\displaystyle=f(I_{n,k_{x}})$ $\displaystyle=f(x),$

hence if $f\in L^{2}(\mathbb{R},\mathscr{F}_{n})$ then $P_{n}f=f$. ∎

For $n\in\mathbb{Z}$, we define

 $L_{n}=K_{n+1}-K_{n},$

and the following lemma gives a different expression for $L_{n}$.22 2 Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 293, §6.3.2.

###### Lemma 4.

If $n\in\mathbb{Z}$, then

 $L_{n}(x,y)=\sum_{k\in\mathbb{Z}}\psi_{n,k}(x)\psi_{n,k}(y),\qquad x,y\in% \mathbb{R}.$
###### Proof.

$\psi(2^{n}x-k)=1$ means that $0\leq 2^{n}x-k<\frac{1}{2}$, which is equivalent to $\frac{k}{2^{n}}\leq x<\frac{k+\frac{1}{2}}{2^{n}}$, which is equivalent to $\frac{2k}{2^{n+1}}\leq x<\frac{2k+1}{2^{n+1}}$, which is equivalent to $x\in I_{n+1,2k}$. $\psi(2^{n}x-k)=-1$ means that $\frac{1}{2}\leq 2^{n}x-k<1$, which is equivalent to $\frac{k+\frac{1}{2}}{2^{n}}\leq x<\frac{k+1}{2^{n}}$, and this is equivalent to $x\in I_{n+1,2k+1}$. $\psi(2^{n}x-k)=0$ if and only if $x\not\in I_{n+1,2k}\cup I_{n+1,2k+1}$. Therefore,

 $\psi_{n,k}(x)\psi_{n,k}(y)=\begin{cases}2^{n}&(x,y)\in I_{n+1,2k}\times I_{n+1% ,2k}\cup I_{n+1,2k+1}\times I_{n+1,2k+1},\\ -2^{n}&(x,y)\in I_{n+1,2k}\times I_{n+1,2k+1}\cup I_{n+1,2k+1}\times I_{n+1,2k% },\\ 0&\mathrm{otherwise.}\end{cases}$

If there is no $k\in\mathbb{Z}$ such that $(x,y)\in I_{n,k}\times I_{n,k}$, then $L_{n}(x,y)=0$. Otherwise, suppose that $k\in\mathbb{Z}$ and that $(x,y)\in I_{n,k}\times I_{n,k}$. We have

 $I_{n,k}=I_{n+1,2k}\cup I_{n+1,2k+1}.$

If $(x,y)\in I_{n+1,2k}\times I_{n+1,2k}$, then

 $L_{n}(x,y)=K_{n+1}(x,y)-K_{n}(x,y)=2^{n+1}-2^{n}=2^{n};$

if $(x,y)\in I_{n+1,2k+1}\times I_{n+1,2k+1}$, then

 $L_{n}(x,y)=K_{n+1}(x,y)-K_{n}(x,y)=2^{n+1}-2^{n}=2^{n};$

if $(x,y)\in I_{n+1,2k}\times I_{n+1,2k+1}$, then

 $L_{n}(x,y)=K_{n+1}(x,y)-K_{n}(x,y)=0-2^{n}=-2^{n};$

and if $(x,y)\in I_{n+1,2k+1}\times I_{n+1,2k}$, then

 $L_{n}(x,y)=K_{n+1}(x,y)-K_{n}(x,y)=0-2^{n}=-2^{n}.$

It follows that

 $L_{n}(x,y)=\sum_{k\in\mathbb{Z}}\psi_{n,k}(x)\psi_{n,k}(y).$

## 4 Continuous functions

Let $C_{0}(\mathbb{R})$ denote those continuous functions $f:\mathbb{R}\to\mathbb{C}$ such that if $\epsilon>0$ then there is some compact subset $K$ of $\mathbb{R}$ such that $x\not\in K$ implies that $|f(x)|<\epsilon$. We say that an element of $C_{0}(\mathbb{R})$ is a continuous function that vanishes at infinity. Let $C_{c}(\mathbb{R})$ denote the set of continuous functions $f:\mathbb{R}\to\mathbb{C}$ such that

 $\mathrm{supp}(f)=\overline{\{x\in\mathbb{R}:f(x)\neq 0\}}$

is a compact set.

In the following lemma, we prove that the larger the intervals over which we average a continuous function vanishing at infinity, the smaller the supremum of the averaged function.33 3 Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 295, Lemma 6.3.2.

###### Lemma 5.

If $f\in C_{0}(\mathbb{R})$, then $\left\|P_{n}f\right\|_{\infty}\to 0$ as $n\to-\infty$.

###### Proof.

If $g\in C_{c}(\mathbb{R})$ and $x\in\mathbb{R}$, then

 $\displaystyle|P_{n}g(x)|$ $\displaystyle=\left|\int_{\mathbb{R}}K_{n}(x,y)g(y)dy\right|$ $\displaystyle=\left|\int_{\mathrm{supp}(g)}K_{n}(x,y)g(y)dy\right|$ $\displaystyle\leq\int_{\mathrm{supp}(g)}K_{n}(x,y)|g(y)|dy$ $\displaystyle\leq\int_{\mathrm{supp}(g)}2^{n}|g(y)|dy$ $\displaystyle\leq 2^{n}\cdot\mu(\mathrm{supp}(g))\cdot\left\|g\right\|_{\infty},$

hence

 $\left\|P_{n}g\right\|_{\infty}\leq 2^{n}\cdot\mu(\mathrm{supp}(g))\cdot\left\|% g\right\|_{\infty}.$ (2)

If $f\in C_{0}(\mathbb{R})$ and $\epsilon>0$ then there is some $g\in C_{c}(\mathbb{R})$ with $\left\|f-g\right\|_{\infty}<\epsilon$. Hence,

 $\left\|P_{n}f\right\|_{\infty}\leq\left\|P_{n}(f-g)\right\|_{\infty}+\left\|P_% {n}g\right\|_{\infty}.$

If $x\in\mathbb{R}$, then

 $|P_{n}(f-g)(x)|=2^{n}\left|\int_{I_{n,k_{x}}}(f-g)(y)dy\right|\leq 2^{n}\int_{% I_{n,k_{x}}}|(f-g)(y)|dy\leq\left\|f-g\right\|_{\infty},$

hence $\left\|P_{n}(f-g)\right\|_{\infty}\leq\left\|f-g\right\|_{\infty}$. Using this and (2) we obtain

 $\left\|P_{n}f\right\|_{\infty}\leq\left\|f-g\right\|_{\infty}+2^{n}\cdot\mu(% \mathrm{supp}(g))\cdot\left\|g\right\|_{\infty}<\epsilon+2^{n}\cdot\mu(\mathrm% {supp}(g))\cdot\left\|g\right\|_{\infty}.$

Hence,

 $\limsup_{n\to-\infty}\left\|P_{n}f\right\|_{\infty}\leq\limsup_{n\to-\infty}% \big{(}\epsilon+2^{n}\cdot\mu(\mathrm{supp}(g))\cdot\left\|g\right\|_{\infty}% \big{)}=\epsilon.$

This is true for every $\epsilon>0$, so

 $\lim_{n\to-\infty}\left\|P_{n}f\right\|_{\infty}=0.$

###### Lemma 6.

If $f\in L^{2}(\mathbb{R})$, then $\left\|P_{n}f\right\|_{2}\to 0$ as $n\to-\infty$.

###### Proof.

If $\epsilon>0$ then there is some $g\in C_{c}(\mathbb{R})$ such that $\left\|f-g\right\|_{2}<\epsilon$. Say $\mathrm{supp}(g)\subseteq[-K,K]$. If $2^{m}>K$, then we have by (1) and because $\mathrm{supp}(g)\subseteq I_{-m,-1}\cup I_{-m,0}$,

 $\displaystyle\left\|P_{-m}g\right\|_{2}^{2}$ $\displaystyle=\int_{\mathbb{R}}\left|2^{-m}\int_{I_{-m,k_{x}}}g(y)dy\right|^{2% }dx$ $\displaystyle=2^{m}\left|2^{-m}\int_{I_{-m,-1}}g(y)dy\right|^{2}+2^{m}\left|2^% {-m}\int_{I_{-m,0}}g(y)dy\right|^{2}$ $\displaystyle=2^{-m}\left|\int_{-K}^{0}g(y)dy\right|^{2}+2^{-m}\left|\int_{0}^% {K}g(y)dy\right|^{2}$ $\displaystyle\leq 2^{-m}\mu([-K,0])\left\|g\right\|_{2}^{2}+2^{-m}\mu([0,K])% \left\|g\right\|_{2}^{2}$ $\displaystyle=2K\cdot 2^{-m}\left\|g\right\|_{2}^{2}.$

Therefore, when $2^{m}>K$ we have $\left\|P_{-m}g\right\|_{2}\leq 2^{-\frac{m}{2}}\sqrt{2K}\left\|g\right\|_{2}$, and so, as the operator norm of $P_{-m}$ on $L^{2}(\mathbb{R})$ is 1,

 $\displaystyle\left\|P_{-m}f\right\|_{2}$ $\displaystyle\leq\left\|P_{-m}(f-g)\right\|_{2}+\left\|P_{-m}g\right\|_{2}$ $\displaystyle\leq\left\|f-g\right\|_{2}+\left\|P_{-m}g\right\|_{2}$ $\displaystyle<\epsilon+2^{-\frac{m}{2}}\sqrt{2K}\left\|g\right\|_{2}.$

Thus, if $\epsilon>0$ then

 $\limsup_{m\to\infty}\left\|P_{-m}f\right\|_{2}\leq\epsilon.$

This is true for all $\epsilon>0$, so we obtain

 $\lim_{m\to\infty}\left\|P_{-m}f\right\|_{2}=0.$

The following lemma shows that if $f\in C_{c}(\mathbb{R})$, then $P_{n}f$ converges to $f$ in the $L^{2}$ norm and in the $L^{\infty}$ norm as $n\to\infty$.44 4 Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 296, Lemma 6.3.3.

###### Lemma 7.

If $f\in C_{c}(\mathbb{R})$, then $P_{n}f\to f$ in the $L^{2}$ norm and in the $L^{\infty}$ norm as $n\to\infty$.

###### Proof.

Suppose that $\mathrm{supp}(f)\subseteq[-2^{M},2^{M}]$ for $M\geq 0$. $f$ is uniformly continuous on the compact set $[-2^{M},2^{M}]$, thus, if $\epsilon>0$ then there is some $\delta>0$ such that $x,y\in[-2^{M},2^{M}]$ and $|x-y|<\delta$ imply that $|f(x)-f(y)|<\frac{\epsilon}{2^{M}}$. Let $2^{-n}\leq\delta$. For each $x\in\mathbb{R}$, there is some $k_{x}\in\mathbb{Z}$ such that $x\in I_{n,k_{x}}$ and we have

 $\displaystyle|P_{n}f(x)-f(x)|$ $\displaystyle=\left|2^{n}\int_{I_{n,k_{x}}}f(y)dy-f(x)\right|$ $\displaystyle=2^{n}\left|\int_{I_{n,k_{x}}}f(y)-f(x)dy\right|$ $\displaystyle\leq 2^{n}\int_{I_{n,k_{x}}}|f(y)-f(x)|dy$ $\displaystyle<2^{n}\int_{I_{n,k_{x}}}\frac{\epsilon}{2^{M}}dy$ $\displaystyle=\frac{\epsilon}{2^{M}}.$

This tells us that if $2^{-n}\leq\delta$ then $\left\|P_{n}f-f\right\|_{\infty}\leq\frac{\epsilon}{2^{M}}$. Therefore, if $\epsilon>0$ then for sufficiently large $n$ we have $\left\|P_{n}f-f\right\|_{\infty}\leq\frac{\epsilon}{2^{M}}$, showing that

 $\lim_{n\to\infty}\left\|P_{n}f-f\right\|_{\infty}=0.$

Furthermore, if $n\geq 0$ then

 $\left\|P_{n}f-f\right\|_{2}^{2}=\int_{\mathbb{R}}|P_{n}f(x)-f(x)|^{2}dx=\int_{% -2^{M}}^{2^{M}}|P_{n}f(x)-f(x)|^{2}dx\leq 2\cdot 2^{M}\cdot\left\|P_{n}f-f% \right\|_{\infty}^{2},$

and because $\left\|P_{n}f-f\right\|_{\infty}\to 0$ as $n\to\infty$ we get $\left\|P_{n}f-f\right\|_{2}\to 0$ as $n\to\infty$. ∎

From Lemma 4, we get

 $\displaystyle(P_{n+1}-P_{n})f(x)$ $\displaystyle=\int_{\mathbb{R}}K_{n+1}(x,y)f(y)dy-\int_{\mathbb{R}}K_{n}(x,y)f% (y)dy$ $\displaystyle=\int_{\mathbb{R}}L_{n}(x,y)f(y)dy$ $\displaystyle=\int_{\mathbb{R}}\sum_{k\in\mathbb{Z}}\psi_{n,k}(x)\psi_{n,k}(y)% f(y)dy$ $\displaystyle=\sum_{k\in\mathbb{Z}}\left\langle f,\psi_{n,k}\right\rangle\psi_% {n,k}(x),$

thus

 $P_{n+1}-P_{n}=\sum_{k\in\mathbb{Z}}\psi_{n,k}\otimes\psi_{n,k}$ (3)

in the strong operator topology. Using (3), we obtain for $n\geq 0$ that

 $\displaystyle P_{n+1}$ $\displaystyle=P_{0}+\sum_{j=0}^{n}P_{j+1}-P_{j}$ $\displaystyle=P_{0}+\sum_{j=0}^{n}\sum_{k\in\mathbb{Z}}\psi_{j,k}\otimes\psi_{% j,k}$

in the strong operator topology. For $n<0$,

 $\displaystyle P_{n}$ $\displaystyle=P_{0}-\sum_{j=-n}^{-1}P_{j+1}-P_{j}$ $\displaystyle=P_{0}-\sum_{j=-n}^{-1}\sum_{k\in\mathbb{Z}}\psi_{j,k}\otimes\psi% _{j,k}$

in the strong operator topology.

We have already shown in Lemma 2 that $\{\psi_{n,k}:n,k\in\mathbb{Z}\}$ is an orthonormal set in $L^{2}(\mathbb{R})$, and we now prove that it is an orthonormal basis for $L^{2}(\mathbb{R})$.

###### Theorem 8.

In the strong operator topology,

 $\mathrm{id}_{L^{2}(\mathbb{R})}=\sum_{n,k\in\mathbb{Z}}\psi_{n,k}\otimes\psi_{% n,k}.$
###### Proof.

Let $f\in L^{2}(\mathbb{R})$ and suppose $\epsilon>0$. By Lemma 6, there is some $M$ such that $m\geq M$ implies that $\left\|P_{-m}f\right\|_{2}<\frac{\epsilon}{2}$. There is some $g\in C_{c}(\mathbb{R})$ satisfying $\left\|f-g\right\|_{2}<\frac{\epsilon}{6}$, and by Lemma 7 there is some $N$ such that $n\geq N$ implies that $\left\|P_{n}g-g\right\|_{2}<\frac{\epsilon}{6}$. Hence, if $n\geq N$ then

 $\displaystyle\left\|P_{n}f-f\right\|_{2}$ $\displaystyle\leq\left\|P_{n}f-P_{n}g\right\|_{2}+\left\|P_{n}g-g\right\|_{2}+% \left\|g-f\right\|_{2}$ $\displaystyle\leq 2\left\|f-g\right\|_{2}+\left\|P_{n}g-g\right\|_{2}$ $\displaystyle<\frac{2\epsilon}{6}+\frac{\epsilon}{6}$ $\displaystyle=\frac{\epsilon}{2}.$

Therefore, if $m\geq M$ and $n\geq N$, then

 $\left\|(P_{n}-P_{-m}-\mathrm{id}_{L^{2}(\mathbb{R}})f\right\|_{2}\leq\left\|P_% {n}f-f\right\|_{2}+\left\|P_{-m}f\right\|_{2}<\frac{\epsilon}{2}+\frac{% \epsilon}{2}=\epsilon.$

For $m,n>0$, we have

 $\displaystyle P_{n+1}-P_{-m}$ $\displaystyle=\sum_{j=0}^{n}\sum_{k\in\mathbb{Z}}\psi_{j,k}\otimes\psi_{j,k}+% \sum_{j=-m}^{-1}\sum_{k\in\mathbb{Z}}\psi_{j,k}\otimes\psi_{j,k}$ $\displaystyle=\sum_{j=-m}^{n}\sum_{k\in\mathbb{Z}}\psi_{j,k}\otimes\psi_{j,k}$

in the strong operator topology. ∎

## 5 Other function spaces

Let $C_{b}(\mathbb{R})$ denote those continuous functions $\mathbb{R}\to\mathbb{C}$ that are bounded. We have

 $C_{c}(\mathbb{R})\subset C_{0}(\mathbb{R})\subset C_{b}(\mathbb{R})\subset C(% \mathbb{R}).$
###### Lemma 9.

If $n\in\mathbb{Z}$ and $f\in C_{b}(\mathbb{R})$, then $\left\|P_{n}f\right\|_{\infty}\leq\left\|f\right\|_{\infty}$.

###### Proof.

If $x\in\mathbb{R}$, then there is a unique $k_{x}\in\mathbb{Z}$ with $x\in I_{n,k_{x}}$, and

 $|P_{n}f(x)|=\left|2^{n}\int_{I_{n,k_{x}}}f(y)dy\right|\leq 2^{n}\int_{I_{n,k_{% x}}}|f(y)|dy\leq\left\|f\right\|_{\infty}.$

###### Theorem 10.

If $f\in C_{0}(\mathbb{R})$, then the series $\sum_{n,k\in\mathbb{Z}}\left\langle f,\psi_{n,k}\right\rangle\psi_{n,k}$ converges to $f$ uniformly on $\mathbb{R}$.

###### Proof.

If $\epsilon>0$ then there is some $g\in C_{c}(\mathbb{R})$ with $\left\|f-g\right\|_{\infty}<\frac{\epsilon}{6}$. By Lemma 5, there is some $M$ such that $m\geq M$ implies that $\left\|P_{-m}g\right\|_{\infty}<\frac{\epsilon}{3}$, hence

 $\displaystyle\left\|P_{-m}f\right\|_{\infty}$ $\displaystyle\leq\left\|P_{-m}f-P_{-m}g\right\|_{\infty}+\left\|P_{-m}g\right% \|_{\infty}$ $\displaystyle\leq\left\|f-g\right\|_{\infty}+\left\|P_{-m}g\right\|_{\infty}$ $\displaystyle<\frac{\epsilon}{6}+\frac{\epsilon}{3}$ $\displaystyle=\frac{\epsilon}{2}.$

By Lemma 7, there is some $N$ such that $n\geq N$ implies that $\left\|P_{n}g-g\right\|_{\infty}<\frac{\epsilon}{6}$, hence

 $\displaystyle\left\|P_{n}f-f\right\|_{\infty}$ $\displaystyle\leq\left\|P_{n}f-P_{n}g\right\|_{\infty}+\left\|P_{n}g-g\right\|% _{\infty}+\left\|g-f\right\|_{\infty}$ $\displaystyle\leq 2\left\|f-g\right\|_{\infty}+\left\|P_{n}g-g\right\|_{\infty}$ $\displaystyle<\frac{\epsilon}{2}.$

Therefore, if $n\geq N$ and $m\geq M$, then

 $\left\|P_{n}f-P_{-m}f-f\right\|_{\infty}\leq\left\|P_{n}f-f\right\|_{\infty}+% \left\|P_{-m}f\right\|_{\infty}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$

The following theorem states that $P_{n}$ is an operator on $L^{p}(\mathbb{R})$ with operator norm $\leq 1$.55 5 Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 297, Lemma 6.3.9. In particular, it asserts that if $f\in L^{p}(\mathbb{R})$ then the averaged function $P_{n}f$ is also an element of $L^{p}(\mathbb{R})$.

###### Theorem 11.

If $1\leq p<\infty$, $n\in\mathbb{Z}$, and $f\in L^{p}(\mathbb{R})$, then $\left\|P_{n}f\right\|_{p}\leq\left\|f\right\|_{p}$.

###### Proof.

Let $\frac{1}{p}+\frac{1}{q}=1$, so $q=\frac{p}{p-1}$. (If $p=1$ then $q=\infty$.) If $x\in\mathbb{R}$, then there is a unique $k_{x}\in\mathbb{Z}$ with $x\in I_{n,k_{x}}$, and using Hölder’s inequality we get

 $\displaystyle|P_{n}f(x)|$ $\displaystyle=\left|2^{n}\int_{I_{n,k_{x}}}f(y)dy\right|$ $\displaystyle\leq 2^{n}\left(\int_{I_{n,k_{x}}}|f(y)|^{p}dy\right)^{1/p}(\mu(I% _{n,k_{x}}))^{1/q}$ $\displaystyle=2^{n}\left(\int_{I_{n,k_{x}}}|f(y)|^{p}dy\right)^{1/p}2^{-n/q}.$

Therefore, if $k\in\mathbb{Z}$ then

 $\displaystyle\int_{I_{n,k}}|P_{n}f(x)|^{p}dx$ $\displaystyle\leq\int_{I_{n,k}}2^{np}2^{-np/q}\int_{I_{n,k_{x}}}|f(y)|^{p}dydx$ $\displaystyle=\int_{I_{n,k}}2^{np}2^{-np/q}\int_{I_{n,k}}|f(y)|^{p}dydx$ $\displaystyle=2^{-n}2^{np}2^{-np/q}\int_{I_{n,k}}|f(y)|^{p}dy$ $\displaystyle=\int_{I_{n,k}}|f(y)|^{p}dy.$

We obtain

 $\displaystyle\left\|P_{n}f\right\|_{p}^{p}$ $\displaystyle=\sum_{k\in\mathbb{Z}}\int_{I_{n,k}}|P_{n}f(x)|^{p}dx$ $\displaystyle\leq\sum_{k\in\mathbb{Z}}\int_{I_{n,k}}|f(y)|^{p}dy$ $\displaystyle=\int_{\mathbb{R}}|f(y)|^{p}dy$ $\displaystyle=\left\|f\right\|_{p}^{p},$

giving $\left\|P_{n}f\right\|_{p}\leq\left\|f\right\|_{p}$. ∎

## 6 Multiresolution analysis

For $a\in\mathbb{R}$, we define $m_{a}:\mathbb{R}\to\mathbb{R}$ by $m_{a}(x)=ax$, and we define $\tau_{a}:\mathbb{R}\to\mathbb{R}$ by $\tau_{a}(x)=x-a$.

###### Definition 12 (Multiresolution analysis).

A multiresolution analysis of $L^{2}(\mathbb{R})$ is a set $\{V_{n}:n\in\mathbb{Z}\}$ of closed subspaces of the Hilbert space $L^{2}(\mathbb{R})$ and a function $\Phi\in L^{2}(\mathbb{R})$ satisfying

1. 1.

If $n\in\mathbb{Z}$, then $f\in V_{n}$ if and only if $f\circ m_{2}\in V_{n+1}$.

2. 2.

$V_{n}\subseteq V_{n+1}$.

3. 3.

$\overline{\bigcup_{n\in\mathbb{Z}}V_{n}}=L^{2}(\mathbb{R})$.

4. 4.

$\bigcap_{n\in\mathbb{Z}}V_{n}=\{0\}$.

5. 5.

$\{\Phi\circ\tau_{k}:k\in\mathbb{Z}\}$ is an orthonormal basis for $V_{0}$.

It is straightforward to prove the following theorem using what we have established so far.

###### Theorem 13.

The closed subspaces $\{L^{2}(\mathbb{R},\mathscr{F}_{n}):n\in\mathbb{Z}\}$ of $L^{2}(\mathbb{R})$ and the function $\phi=\chi_{[0,1)}$ is a multiresolution analysis of $L^{2}(\mathbb{R})$.

The following lemma shows that if $P_{n}$ is the projection onto $V_{n}$, where $V_{n}$ is a closed subspace of a multiresolution analysis of $L^{2}(\mathbb{R})$, then $P_{n}\to 0$ in the strong operator topology as $n\to-\infty$.66 6 Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 313, Lemma 6.4.28.

###### Lemma 14.

If $\{V_{n}:n\in\mathbb{Z}\}$ and $\Phi\in L^{2}(\mathbb{R})$ is a multiresolution analysis of $L^{2}(\mathbb{R})$, $P_{n}:L^{2}(\mathbb{R})\to V_{n}$ is the orthogonal projection onto $V_{n}$, and $f\in L^{2}(\mathbb{R})$, then

 $\lim_{n\to-\infty}P_{n}f=0.$
###### Proof.

Define $\Phi_{n,k}(x)=2^{n/2}\Phi(2^{n}x-k)$. The set $\{\Phi_{0,k}:k\in\mathbb{Z}\}$ is an orthonormal basis for $V_{0}$, and one checks that the set $\{\Phi_{n,k}:k\in\mathbb{Z}\}$ is an orthonormal basis for $V_{n}$. Therefore

 $P_{n}=\sum_{k\in\mathbb{Z}}\Phi_{n,k}\otimes\Phi_{n,k}$

in the strong operator topology.

For $R>0$, let $f_{R}=f\chi_{[-R,R]}$. If $2^{n}R<\frac{1}{2}$, then, using the Cauchy-Schwarz inequality,

 $\displaystyle\left\|P_{n}f_{R}\right\|_{2}^{2}$ $\displaystyle=\sum_{k\in\mathbb{Z}}|\left\langle P_{n}f_{R},\Phi_{n,k}\right% \rangle|^{2}$ $\displaystyle=\sum_{k\in\mathbb{Z}}|\left\langle f_{R},\Phi_{n,k}\right\rangle% |^{2}$ $\displaystyle=\sum_{k\in\mathbb{Z}}|\left\langle f_{R},\chi_{[-R,R]}\Phi_{n,k}% \right\rangle|^{2}$ $\displaystyle\leq\sum_{k\in\mathbb{Z}}\left(\int_{-R}^{R}|f_{R}(x)|^{2}dx% \right)\left(\int_{-R}^{R}|\Phi_{n,k}(x)|^{2}dx\right)$ $\displaystyle=\left\|f_{R}\right\|_{2}^{2}\sum_{k\in\mathbb{Z}}\int_{-R}^{R}|% \Phi_{n,k}(x)|^{2}dx$ $\displaystyle=\left\|f_{R}\right\|_{2}^{2}\sum_{k\in\mathbb{Z}}2^{n}\int_{-R}^% {R}|\Phi(2^{n}x-k)|^{2}dx$ $\displaystyle=\left\|f_{R}\right\|_{2}^{2}\sum_{k\in\mathbb{Z}}\int_{-2^{n}R-k% }^{2^{n}R-k}|\Phi(x)|^{2}dx$ $\displaystyle=\left\|f_{R}\right\|_{2}^{2}\int_{U_{n}}|\Phi(x)|^{2}dx,$

where

 $U_{n}=\bigcup_{k\in\mathbb{Z}}(-k-2^{n}R,-k+2^{n}R);$

the intervals are disjoint because $2^{n}R<\frac{1}{2}$. Define $F_{n}(x)=|\Phi(x)|^{2}\chi_{U_{n}}(x)$. For all $x\in\mathbb{R}$ we have $|F_{n}(x)|\leq|\Phi(x)|^{2}$, and if $x\in\mathbb{R}$ then

 $\lim_{n\to-\infty}F_{n}(x)\to|\Phi(x)|^{2}\chi_{\mathbb{Z}}(x),$

where $\mathbb{Z}=\bigcap_{n\in\mathbb{Z}}U_{n}$. Thus by the dominated convergence theorem we get

 $\lim_{n\to-\infty}\int_{\mathbb{R}}F_{n}(x)dx=\int_{\mathbb{R}}|\Phi(x)|^{2}% \chi_{\mathbb{Z}}(x)dx=0,$

because $\mu(\mathbb{Z})=0$. Therefore,

 $\lim_{n\to-\infty}\left\|P_{n}f_{R}\right\|_{2}=0.$

If $\epsilon>0$ then there is some $R$ such that $\left\|f-f_{R}\right\|_{2}<\epsilon$. We have, because $P_{n}$ is an orthogonal projection,

 $\displaystyle\limsup_{n\to-\infty}\left\|P_{n}f\right\|_{2}$ $\displaystyle\leq\limsup_{n\to-\infty}\left\|P_{n}f-P_{n}f_{R}\right\|_{2}+% \limsup_{n\to-\infty}\left\|P_{n}f_{R}\right\|_{2}$ $\displaystyle=\limsup_{n\to-\infty}\left\|P_{n}f-P_{n}f_{R}\right\|_{2}$ $\displaystyle\leq\limsup_{n\to-\infty}\left\|f-f_{R}\right\|_{2}$ $\displaystyle<\epsilon.$

This is true for all $\epsilon>0$, so we obtain

 $\lim_{n\to-\infty}\left\|P_{n}f\right\|_{2}=0.$

If $S_{\alpha},\alpha\in I$, are subsets of a Hilbert space $H$, we denote by $\bigvee_{\alpha\in I}S_{\alpha}$ the closure of the span of $\bigcup_{\alpha\in I}S_{\alpha}$. If $S$ is a subset of $H$, let $S^{\perp}$ be the set of all $x\in H$ such that $y\in S$ implies that $\left\langle x,y\right\rangle=0$. If $S_{n},n\in\mathbb{Z}$, are mutually orthogonal closed subspaces of a Hilbert space, we write

 $\bigoplus_{n\in\mathbb{Z}}S_{n}=\bigvee_{n\in\mathbb{Z}}S_{n}.$

The following theorem shows a consequence of Definition 12.

###### Theorem 15.

If $\{V_{n}:n\in\mathbb{Z}\}$ are the closed subspaces of a multiresolution analysis of $L^{2}(\mathbb{R})$ and $W_{n}=V_{n+1}\cap V_{n}^{\perp}$, then

 $L^{2}(\mathbb{R})=\bigoplus_{n\in\mathbb{Z}}W_{n}.$
###### Proof.

Because $W_{n}=V_{n+1}\cap V_{n}^{\perp}$ is the intersection of two closed subspaces, it is itself a closed subspace. Suppose that $n, $f\in W_{n},g\in W_{n^{\prime}}$. $n+1\leq n^{\prime}$, and hence $V_{n+1}\subseteq V_{n^{\prime}}$. Therefore

 $W_{n^{\prime}}=V_{n^{\prime}+1}\cap V_{n^{\prime}}^{\perp}\subset V_{n^{\prime% }}^{\perp}\subseteq V_{n+1}^{\perp}.$

But $f\in W_{n}\subset V_{n+1}$ and $g\in W_{n^{\prime}}\subset V_{n+1}^{\perp}$, so $\left\langle f,g\right\rangle=0$. Therefore $W_{n}\perp W_{n^{\prime}}$.

If $f\in V_{n}$ and $f\neq 0$, then there is a minimal $N$ such that $f\in V_{N}$; this minimal $N$ exists because $V_{n}\subseteq V_{n+1}$ and $\bigcap_{n\in\mathbb{Z}}V_{n}=\{0\}$. We have

 $V_{N}=V_{N-1}\oplus W_{N-1},$

hence $f=f_{N-1}+g_{N-1}$, with $f_{N-1}\in V_{N-1}$ and $g_{N-1}\in W_{N-1}$. Likewise,

 $V_{N-1}=V_{N-2}\oplus W_{N-2},$

hence $f_{N-1}=f_{N-2}+g_{N-2}$, with $f_{N-2}\in V_{N-2}$ and $g_{N-2}\in W_{N-2}$. In this way, for any $M\geq 0$ we obtain

 $f=f_{N-M}+\sum_{m=1}^{M}g_{N-m},$

where $f_{N-M}\in V_{N-M}$ and $g_{N-m}\in W_{N-m}$. Check that $f_{N-M}$ is the orthogonal projection of $f$ onto $V_{N-M}$. It thus follows from Lemma 14 that $f_{N-M}\to 0$ as $M\to\infty$. Thus, for any $\epsilon>0$ there is some $M$ with $\left\|f_{N-M}\right\|_{2}<\epsilon$ and $f\in f_{N-M}+\bigoplus_{m=1}^{M}W_{N-m}$. Therefore, if $f\in\bigcup_{n\in\mathbb{Z}}V_{n}$ then there is some $g\in\bigoplus_{n\in\mathbb{Z}}W_{n}$ satisfying $\left\|f-g\right\|_{2}<\infty$. Thus

 $\overline{\bigcup_{n\in\mathbb{Z}}V_{n}}\subseteq\bigoplus_{n\in\mathbb{Z}}W_{% n},$

and so

 $L^{2}(\mathbb{R})=\bigoplus_{n\in\mathbb{Z}}W_{n}.$

## 7 The unit interval

$L^{2}([0,1))$ is a Hilbert space with the inner product

 $\left\langle f,g\right\rangle=\int_{0}^{1}f(x)\overline{g(x)}dx.$

If $n\geq 0$, then $I_{n,0}=\left[0,\frac{1}{2^{n}}\right)$ and $I_{n,2^{n}-1}=\left[1-\frac{1}{2^{n}},1\right)$, and we have

 $[0,1)=\bigcup_{k=0}^{2^{n}-1}I_{n,k}.$

Let $n\geq 0$, let $\mathscr{G}_{n}$ be the $\sigma$-algebra generated by $\{I_{n,k}:0\leq k\leq 2^{n}-1\}$, and let $\mathscr{G}$ be the $\sigma$-algebra of Lebesgue measurable subsets of $[0,1)$. If $n, then

 $\mathscr{G}_{n}\subset\mathscr{G}_{n^{\prime}}\subset\mathscr{G}.$

An element of $L^{2}([0,1),\mathscr{G}_{n})$ is an element of $L^{2}([0,1),\mathscr{G})$ that is constant on each set $I_{n,k},0\leq k\leq 2^{n}-1$. Equivalently, an element of $L^{2}([0,1),\mathscr{G}_{n})$ is a function $f:[0,1)\to\mathbb{C}$ that is constant on each set $I_{n,k},0\leq k\leq 2^{n}-1$; because $[0,1)$ is a union of finitely many $I_{n,k}$, any such function will be an element of $L^{2}([0,1),\mathscr{G})$. It is apparent that

 $L^{2}([0,1),\mathscr{G}_{n})\subset L^{2}([0,1),\mathscr{G}_{n^{\prime}})% \subset L^{2}([0,1),\mathscr{G}).$

We check that $L^{2}([0,1),\mathscr{G}_{n})$ is a complex vector space of dimension $2^{n}$.

$I_{n,k}=I_{n+1,2k}\cup I_{n+1,2k+1}$. If $x\in I_{n+1,2k}$, then $\frac{2k}{2^{n+1}}\leq x<\frac{2k+1}{2^{n+1}}$, so $\frac{k}{2^{n}}\leq x<\frac{k}{2^{n}}+\frac{1}{2^{n+1}}$, hence $0\leq 2^{n}x-k<\frac{1}{2}$. If $x\in I_{n+1,2k+1}$, then $\frac{2k+1}{2^{n+1}}\leq x<\frac{2k+2}{2^{n+1}}$, hence $\frac{k}{2^{n}}+\frac{1}{2^{n+1}}\leq x<\frac{k+1}{2^{n}}$, and so $\frac{1}{2}\leq 2^{n}x-k<1$. Thus, if $x\in I_{n+1,2k}$ then

 $\psi_{n,k}(x)=2^{n/2}\psi(2^{n}x-k)=2^{n/2}$

and if $x\in I_{n+1,2k+1}$ then

 $\psi_{n,k}(x)=2^{n/2}\psi(2^{n}x-k)=-2^{n/2}.$

Otherwise $x\not\in I_{n,k}$, for which $\psi_{n,k}(x)=0$. It follows that $\psi_{n,k}\in L^{2}([0,1),\mathscr{G}_{n+1})$.

###### Theorem 16.

If

 $\mathscr{B}_{0}=\{\chi_{[0,1)}\}$

and, for $n\geq 0$,

 $\mathscr{B}_{n+1}=\{\psi_{n,k}:0\leq k\leq 2^{n}-1\},$

then

 $\bigcup_{n=0}^{N}\mathscr{B}_{n}$

is an orthonormal basis of $L^{2}([0,1),\mathscr{G}_{N})$.

###### Proof.

It follows from Lemma 2 that $\bigcup_{n=1}^{N}\mathscr{B}_{n}$ is orthonormal in $L^{2}([0,1))$, as it is a subset of an orthonormal set. If $0\leq n\leq N$ then $\mathscr{B}_{n}\subset L^{2}([0,1),\mathscr{G}_{N})$, hence $\bigcup_{n=1}^{N}\mathscr{B}_{n}$ is orthonormal in $L^{2}([0,1),\mathscr{G}_{N})$. If $0 and $0\leq k\leq 2^{n-1}-1$, then $\psi_{n-1,k}\in\mathscr{B}_{n}$ and

 $\displaystyle\left\langle\psi_{n-1,k},\chi_{[0,1)}\right\rangle$ $\displaystyle=\int_{0}^{1}\psi_{n-1,k}(x)\overline{\chi_{[0,1)}(x)}dx$ $\displaystyle=\int_{0}^{1}\psi_{n-1,k}(x)dx$ $\displaystyle=\int_{I_{n,2k}}\psi_{n-1,k}(x)dx+\int_{I_{n,2k+1}}\psi_{n-1,k}(x% )dx$ $\displaystyle=\int_{I_{n,2k}}2^{(n-1)/2}dx+\int_{I_{n,2k+1}}-2^{(n-1)/2}dx$ $\displaystyle=0.$

Therefore, $\bigcup_{n=0}^{N}\mathscr{B}_{n}$ is orthonormal in $L^{2}([0,1),\mathscr{G}_{N})$.

$|\mathscr{B}_{0}|=1$, and if $n\geq 1$ then $|\mathscr{B}_{n}|=2^{n-1}$. Therefore the number of elements of $\bigcup_{n=0}^{N}\mathscr{B}_{n}$ is

 $1+\sum_{n=1}^{N}2^{n-1}=1+\sum_{n=0}^{N-1}2^{n}=2^{N}.$

As $\dim L^{2}([0,1),\mathscr{G}_{N})=2^{N}$, the orthonormal set $\bigcup_{n=0}^{N}\mathscr{B}_{n}$ is an orthonormal basis for $L^{2}([0,1),\mathscr{G}_{N})$. ∎

By Theorem 16, if $N\geq 0$ then $\bigcup_{n=0}^{N}\mathscr{B}_{n}$ is an orthonormal set in $L^{2}([0,1))$. Hence

 $\mathscr{B}=\bigcup_{n=0}^{\infty}\mathscr{B}_{n}$

is an orthonormal set in $L^{2}([0,1))$: if $f,g\in\mathscr{B}$ then there is some $N$ with $f,g\in\bigcup_{n=0}^{N}\mathscr{B}_{n}$, which is an orthonormal set. The following theorem shows that $\mathscr{B}$ is an orthonormal basis for the Hilbert space $L^{2}([0,1))$.77 7 John K. Hunter and Bruno Nachtergaele, Applied Analysis, p. 177, Lemma 7.13.

###### Theorem 17.

$\mathscr{B}$ is an orthonormal basis for $L^{2}([0,1))$.

###### Proof.

If $f\in L^{2}([0,1))$ and $\epsilon>0$ then there is some $g\in C([0,1])$ with $\left\|f-g\right\|_{2}<\frac{\epsilon}{2}$. $g$ is uniformly continuous on the compact set $[0,1]$, so there is some $\delta>0$ such that $|x-y|<\delta$ implies that $|g(x)-g(y)|<\frac{\epsilon}{2}$. Let $2^{-n}\leq\delta$, and define $h:[0,1)\to\mathbb{C}$ by

 $h(x)=\sum_{k=0}^{2^{n}-1}g\left(\frac{k}{2^{n}}\right)\chi_{I_{n,k}}(x).$

If $x\in[0,1)$ then there is a unique $k_{x},0\leq k_{x}\leq 2^{n}-1$, with $x\in I_{n,k_{x}}$, and for this $k_{x}$ we have $\left|x-\frac{k_{x}}{2^{n}}\right|<2^{-n}\leq\delta$, and hence

 $|g(x)-h(x)|=\left|g(x)-g\left(\frac{k_{x}}{2^{n}}\right)\right|<\frac{\epsilon% }{2}.$

Therefore $\left\|g-h\right\|_{\infty}\leq\frac{\epsilon}{2}$.

We have $h\in L^{2}([0,1),\mathscr{G}_{n})$, and

 $\left\|f-h\right\|_{2}\leq\left\|f-g\right\|_{2}+\left\|g-h\right\|_{2}<\frac{% \epsilon}{2}+\left\|g-h\right\|_{\infty}\leq\epsilon.$

We have shown that if $f\in L^{2}([0,1))$ and $\epsilon>0$ then there is some $n$ and some $h\in L^{2}([0,1),\mathscr{G}_{n})$ with $\left\|f-h\right\|_{2}<\epsilon$. This tells us that $\bigcup_{n=0}^{\infty}L^{2}([0,1),\mathscr{G}_{n})$ is a dense subset of $L^{2}([0,1))$. Since $\mathscr{B}$ is orthonormal and $\mathrm{span}\,\mathscr{B}=\bigcup_{n=0}^{\infty}L^{2}([0,1),\mathscr{G}_{n})$, $\mathscr{B}$ is an orthonormal basis for $L^{2}([0,1))$. ∎

## 8 References

Useful references on wavelets and multiresolution analysis are Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets; P. Wojtaszczyk, A Mathematical Introduction to Wavelets; Yves Meyer, Wavelets and Operators; Eugenio Hernández and Guido Weiss, A First Course on Wavelets.