Haar wavelets and multiresolution analysis

1 Introduction

Let

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

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

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

$$⟨f,g⟩={\int }_{ℝ}f(x)\overline{g(x)}𝑑x.$$

We will prove that $\psi$ satisfies the following definition of an orthonormal wavelet.

Definition 1 (Orthonormal wavelet).

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

Lemma 2.

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

Proof.

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

${\displaystyle {\int }_{ℝ}}{\psi }_{n,k}(x)\overline{{\psi }_{n^{\prime },k^{\prime }}(x)}𝑑x$ $=$ ${\displaystyle {\int }_{ℝ}}{2}^{n/2}\psi (2^{n}x-k)2^{n^{\prime }/2}\psi (2^{n^{\prime }}x-k^{\prime })𝑑x$

$=$ ${\displaystyle {\int }_{ℝ}}{2}^{(n^{\prime }-n)/2}\psi (x-k)\psi (2^{n^{\prime }-n}x-k^{\prime })𝑑x$

$=$ $2^{(n^{\prime }-n)/2}{\delta }_{k,k^{\prime }}{\displaystyle {\int }_0^1}\psi (x)\psi (2^{(n^{\prime }-n)/2}x)𝑑x$

$=$ ${\delta }_{k,k^{\prime }}\cdot {\delta }_{n,n^{\prime }},$

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

Bessel’s inequality states that if $ℰ$ is an orthonormal set in a Hilbert space $H$, then for any $f\in H$ we have ${\sum }_{e\in ℰ}{|⟨f,e⟩|}^2\le {\parallel f\parallel }_2^2$, from which it follows that ${\sum }_{e\in ℰ}⟨f,e⟩e\in H$. To say that a subset $ℰ$ of a Hilbert space $H$ is an orthonormal basis is equivalent to saying that $ℰ$ is an orthonormal set and that

$${\mathrm{id}}_H=\sum _{e\in ℰ}e\otimes e$$

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

$$f=\sum _{e\in ℰ}⟨f,e⟩e.$$

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

$$\sum _{n,k\in \mathbb{Z}}{|⟨f,{\psi }_{n,k}⟩|}^2\le {\parallel f\parallel }_2^2,\sum _{n,k\in \mathbb{Z}}⟨f,{\psi }_{n,k}⟩{\psi }_{n,k}\in L^2(ℝ).$$

We have not yet proved that $f$ is equal to the series ${\sum }_{n,k\in \mathbb{Z}}⟨f,{\psi }_{n,k}⟩{\psi }_{n,k}$, and this will not be accomplished until later in this note.

2 Coarser sigma-algebras

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

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

and let ${ℱ}_n$ be the $\sigma$-algebra generated by $\{I_{k,n}:k\in \mathbb{Z}\}$. $ℝ={\bigcup }_{k\in \mathbb{Z}}{I}_{n,k}$, and if $k\ne k^{\prime }$ then $I_{n,k}\cap I_{n,k^{\prime }}=\mathrm{\varnothing }$. If then

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

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

If , then

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

3 Integral kernels

We define

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

$$K_n(x,y)=2^n\sum _{k\in \mathbb{Z}}\varphi (2^{n}x-k)\varphi (2^{n}y-k),x,y\in ℝ.$$

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

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

We define

$$P_{n}f(x)={\int }_{ℝ}{K}_n(x,y)f(y)𝑑y.$$

If $x\in ℝ$ 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 $L^2(ℝ,{ℱ}_n)$ is a closed subspace of $L^2(ℝ,ℱ)$, and in the following theorem we prove that $P_n$ is the orthogonal projection onto $L^2(ℝ,{ℱ}_n)$.

Lemma 3.

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

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,

${\parallel P_{n}f\parallel }_2^2$ $={\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{I_{n,k}}}{|P_{n}f(x)|}^2𝑑x$

$={\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{I_{n,k}}}{\left|2^n{\displaystyle {\int }_{I_{n,k}}}f(y)𝑑y\right|}^2𝑑x$

$=2^n{\displaystyle \sum _{k\in \mathbb{Z}}}{\left|{\displaystyle {\int }_{I_{n,k}}}f(y)𝑑y\right|}^2$

$\le 2^n{\displaystyle \sum _{k\in \mathbb{Z}}}\left({\displaystyle {\int }_{I_{n,k}}}{|f(y)|}^2𝑑y\right)\left({\displaystyle {\int }_{I_{n,k}}}𝑑y\right)$

$={\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{I_{n,k}}}{|f(y)|}^2𝑑y$

$={\displaystyle {\int }_{ℝ}}{|f(y)|}^2𝑑y.$

Therefore, $P_n:L^2(ℝ,ℱ)\to L^2(ℝ,{ℱ}_n)$. Moreover, the left-hand side of the above inequality is equal to ${\parallel P_{n}f\parallel }_2^2$ and the right-hand side is equal to ${\parallel f\parallel }_2^2$, hence we have ${\parallel P_{n}f\parallel }_2\le {\parallel f\parallel }_2$, giving $\parallel P_n\parallel \le 1$.

If $f\in L^2(ℝ,{ℱ}_n)$, then

$P_{n}f(x)$ $={\displaystyle {\int }_{ℝ}}{K}_n(x,y)f(y)𝑑y$

$=2^n{\displaystyle {\int }_{I_{n,k_x}}}f(y)𝑑y$

$=f(I_{n,k_x})$

$=f(x),$

hence if $f\in L^2(ℝ,{ℱ}_n)$ 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 $L_n$.

Lemma 4.

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

$$L_n(x,y)=\sum _{k\in \mathbb{Z}}{\psi }_{n,k}(x){\psi }_{n,k}(y),x,y\in ℝ.$$

Proof.

$\psi (2^{n}x-k)=1$ means that , which is equivalent to , which is equivalent to , which is equivalent to $x\in I_{n+1,2k}$. $\psi (2^{n}x-k)=-1$ means that , which is equivalent to , and this is equivalent to $x\in I_{n+1,2k+1}$. $\psi (2^{n}x-k)=0$ if and only if $x\notin I_{n+1,2k}\cup I_{n+1,2k+1}$. Therefore,

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 $C_0(ℝ)$ denote those continuous functions $f:ℝ\to ℂ$ such that if $ϵ>0$ then there is some compact subset $K$ of $ℝ$ such that $x\notin K$ implies that . We say that an element of $C_0(ℝ)$ is a continuous function that vanishes at infinity. Let $C_c(ℝ)$ denote the set of continuous functions $f:ℝ\to ℂ$ such that

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

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.

Lemma 5.

If $f\in C_0(ℝ)$, then ${\parallel P_{n}f\parallel }_{\mathrm{\infty }}\to 0$ as $n\to -\mathrm{\infty }$.

Proof.

If $g\in C_c(ℝ)$ and $x\in ℝ$, then

$|P_{n}g(x)|$ $=\left|{\displaystyle {\int }_{ℝ}}{K}_n(x,y)g(y)𝑑y\right|$

$=\left|{\displaystyle {\int }_{\mathrm{supp}(g)}}{K}_n(x,y)g(y)𝑑y\right|$

$\le {\displaystyle {\int }_{\mathrm{supp}(g)}}{K}_n(x,y)|g(y)|𝑑y$

$\le {\displaystyle {\int }_{\mathrm{supp}(g)}}{2}^n|g(y)|𝑑y$

$\le 2^n\cdot \mu (\mathrm{supp}(g))\cdot {\parallel g\parallel }_{\mathrm{\infty }},$

hence

$${\parallel P_{n}g\parallel }_{\mathrm{\infty }}\le 2^n\cdot \mu (\mathrm{supp}(g))\cdot {\parallel g\parallel }_{\mathrm{\infty }}.$$ (2)

If $f\in C_0(ℝ)$ and $ϵ>0$ then there is some $g\in C_c(ℝ)$ with . Hence,

$${\parallel P_{n}f\parallel }_{\mathrm{\infty }}\le {\parallel P_n(f-g)\parallel }_{\mathrm{\infty }}+{\parallel P_{n}g\parallel }_{\mathrm{\infty }}.$$

If $x\in ℝ$, then

$$|P_n(f-g)(x)|=2^n\left|{\int }_{I_{n,k_x}}(f-g)(y)𝑑y\right|\le 2^n{\int }_{I_{n,k_x}}|(f-g)(y)|𝑑y\le {\parallel f-g\parallel }_{\mathrm{\infty }},$$

hence ${\parallel P_n(f-g)\parallel }_{\mathrm{\infty }}\le {\parallel f-g\parallel }_{\mathrm{\infty }}$. Using this and (2) we obtain

Hence,

$$\underset{n\to -\mathrm{\infty }}{lim\; sup}{\parallel P_{n}f\parallel }_{\mathrm{\infty }}\le \underset{n\to -\mathrm{\infty }}{lim\; sup}\left(ϵ+2^n\cdot \mu (\mathrm{supp}(g))\cdot {\parallel g\parallel }_{\mathrm{\infty }}\right)=ϵ.$$

This is true for every $ϵ>0$, so

$$\underset{n\to -\mathrm{\infty }}{lim}{\parallel P_{n}f\parallel }_{\mathrm{\infty }}=0.$$

∎

Lemma 6.

If $f\in L^2(ℝ)$, then ${\parallel P_{n}f\parallel }_2\to 0$ as $n\to -\mathrm{\infty }$.

Proof.

If $ϵ>0$ then there is some $g\in C_c(ℝ)$ such that . Say $\mathrm{supp}(g)\subseteq [-K,K]$. If $2^m>K$, then we have by (1) and because $\mathrm{supp}(g)\subseteq I_{-m,-1}\cup I_{-m,0}$,

${\parallel P_{-m}g\parallel }_2^2$ $={\displaystyle {\int }_{ℝ}}{\left|2^{-m}{\displaystyle {\int }_{I_{-m,k_x}}}g(y)𝑑y\right|}^2𝑑x$

$=2^m{\left|2^{-m}{\displaystyle {\int }_{I_{-m,-1}}}g(y)𝑑y\right|}^2+2^m{\left|2^{-m}{\displaystyle {\int }_{I_{-m,0}}}g(y)𝑑y\right|}^2$

$=2^{-m}{\left|{\displaystyle {\int }_{-K}^0}g(y)𝑑y\right|}^2+2^{-m}{\left|{\displaystyle {\int }_0^K}g(y)𝑑y\right|}^2$

$\le 2^{-m}\mu ([-K,0]){\parallel g\parallel }_2^2+2^{-m}\mu ([0,K]){\parallel g\parallel }_2^2$

$=2K\cdot 2^{-m}{\parallel g\parallel }_2^2.$

Therefore, when $2^m>K$ we have ${\parallel P_{-m}g\parallel }_2\le 2^{-\frac{m}{2}}\sqrt{2K}{\parallel g\parallel }_2$, and so, as the operator norm of $P_{-m}$ on $L^2(ℝ)$ is 1,

${\parallel P_{-m}f\parallel }_2$ $\le {\parallel P_{-m}(f-g)\parallel }_2+{\parallel P_{-m}g\parallel }_2$

$\le {\parallel f-g\parallel }_2+{\parallel P_{-m}g\parallel }_2$

Thus, if $ϵ>0$ then

$$\underset{m\to \mathrm{\infty }}{lim\; sup}{\parallel P_{-m}f\parallel }_2\le ϵ.$$

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

$$\underset{m\to \mathrm{\infty }}{lim}{\parallel P_{-m}f\parallel }_2=0.$$

∎

The following lemma shows that if $f\in C_c(ℝ)$, then $P_{n}f$ converges to $f$ in the $L^2$ norm and in the $L^{\mathrm{\infty }}$ norm as $n\to \mathrm{\infty }$.

Lemma 7.

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

Proof.

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

$|P_{n}f(x)-f(x)|$ $=\left|2^n{\displaystyle {\int }_{I_{n,k_x}}}f(y)𝑑y-f(x)\right|$

$=2^n\left|{\displaystyle {\int }_{I_{n,k_x}}}f(y)-f(x)dy\right|$

$\le 2^n{\displaystyle {\int }_{I_{n,k_x}}}|f(y)-f(x)|𝑑y$

$={\displaystyle \frac{ϵ}{2^M}}.$

This tells us that if $2^{-n}\le \delta$ then ${\parallel P_{n}f-f\parallel }_{\mathrm{\infty }}\le \frac{ϵ}{2^M}$. Therefore, if $ϵ>0$ then for sufficiently large $n$ we have ${\parallel P_{n}f-f\parallel }_{\mathrm{\infty }}\le \frac{ϵ}{2^M}$, showing that

$$\underset{n\to \mathrm{\infty }}{lim}{\parallel P_{n}f-f\parallel }_{\mathrm{\infty }}=0.$$

Furthermore, if $n\ge 0$ then

$${\parallel P_{n}f-f\parallel }_2^2={\int }_{ℝ}{|P_{n}f(x)-f(x)|}^2𝑑x={\int }_{-2^M}^{2^M}{|P_{n}f(x)-f(x)|}^2𝑑x\le 2\cdot 2^M\cdot {\parallel P_{n}f-f\parallel }_{\mathrm{\infty }}^2,$$

and because ${\parallel P_{n}f-f\parallel }_{\mathrm{\infty }}\to 0$ as $n\to \mathrm{\infty }$ we get ${\parallel P_{n}f-f\parallel }_2\to 0$ as $n\to \mathrm{\infty }$. ∎

From Lemma 4, we get

$(P_{n+1}-P_n)f(x)$ $={\displaystyle {\int }_{ℝ}}{K}_{n+1}(x,y)f(y)𝑑y-{\displaystyle {\int }_{ℝ}}{K}_n(x,y)f(y)𝑑y$

$={\displaystyle {\int }_{ℝ}}{L}_n(x,y)f(y)𝑑y$

$={\displaystyle {\int }_{ℝ}}{\displaystyle \sum _{k\in \mathbb{Z}}}{\psi }_{n,k}(x){\psi }_{n,k}(y)f(y)dy$

$={\displaystyle \sum _{k\in \mathbb{Z}}}⟨f,{\psi }_{n,k}⟩{\psi }_{n,k}(x),$

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 $n\ge 0$ that

$P_{n+1}$ $=P_0+{\displaystyle \sum _{j=0}^n}{P}_{j+1}-P_j$

$=P_0+{\displaystyle \sum _{j=0}^n}{\displaystyle \sum _{k\in \mathbb{Z}}}{\psi }_{j,k}\otimes {\psi }_{j,k}$

in the strong operator topology. For ,

$P_n$ $=P_0-{\displaystyle \sum _{j=-n}^{-1}}{P}_{j+1}-P_j$

$=P_0-{\displaystyle \sum _{j=-n}^{-1}}{\displaystyle \sum _{k\in \mathbb{Z}}}{\psi }_{j,k}\otimes {\psi }_{j,k}$

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 $L^2(ℝ)$, and we now prove that it is an orthonormal basis for $L^2(ℝ)$.

Theorem 8.

In the strong operator topology,

$${\mathrm{id}}_{L^2(ℝ)}=\sum _{n,k\in \mathbb{Z}}{\psi }_{n,k}\otimes {\psi }_{n,k}.$$

Proof.

Let $f\in L^2(ℝ)$ and suppose $ϵ>0$. By Lemma 6, there is some $M$ such that $m\ge M$ implies that . There is some $g\in C_c(ℝ)$ satisfying , and by Lemma 7 there is some $N$ such that $n\ge N$ implies that . Hence, if $n\ge N$ then

${\parallel P_{n}f-f\parallel }_2$ $\le {\parallel P_{n}f-P_{n}g\parallel }_2+{\parallel P_{n}g-g\parallel }_2+{\parallel g-f\parallel }_2$

$\le 2{\parallel f-g\parallel }_2+{\parallel P_{n}g-g\parallel }_2$

$={\displaystyle \frac{ϵ}{2}}.$

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

For $m,n>0$, we have

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

$={\displaystyle \sum _{j=-m}^n}{\displaystyle \sum _{k\in \mathbb{Z}}}{\psi }_{j,k}\otimes {\psi }_{j,k}$

in the strong operator topology. ∎

5 Other function spaces

Let $C_b(ℝ)$ denote those continuous functions $ℝ\to ℂ$ that are bounded. We have

$$C_c(ℝ)\subset C_0(ℝ)\subset C_b(ℝ)\subset C(ℝ).$$

Lemma 9.

If $n\in \mathbb{Z}$ and $f\in C_b(ℝ)$, then ${\parallel P_{n}f\parallel }_{\mathrm{\infty }}\le {\parallel f\parallel }_{\mathrm{\infty }}$.

Proof.

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

$$|P_{n}f(x)|=\left|2^n{\int }_{I_{n,k_x}}f(y)𝑑y\right|\le 2^n{\int }_{I_{n,k_x}}|f(y)|𝑑y\le {\parallel f\parallel }_{\mathrm{\infty }}.$$

∎

Theorem 10.

If $f\in C_0(ℝ)$, then the series ${\sum }_{n,k\in \mathbb{Z}}⟨f,{\psi }_{n,k}⟩{\psi }_{n,k}$ converges to $f$ uniformly on $ℝ$.

Proof.

If $ϵ>0$ then there is some $g\in C_c(ℝ)$ with . By Lemma 5, there is some $M$ such that $m\ge M$ implies that , hence

${\parallel P_{-m}f\parallel }_{\mathrm{\infty }}$ $\le {\parallel P_{-m}f-P_{-m}g\parallel }_{\mathrm{\infty }}+{\parallel P_{-m}g\parallel }_{\mathrm{\infty }}$

$\le {\parallel f-g\parallel }_{\mathrm{\infty }}+{\parallel P_{-m}g\parallel }_{\mathrm{\infty }}$

$={\displaystyle \frac{ϵ}{2}}.$

By Lemma 7, there is some $N$ such that $n\ge N$ implies that , hence

${\parallel P_{n}f-f\parallel }_{\mathrm{\infty }}$ $\le {\parallel P_{n}f-P_{n}g\parallel }_{\mathrm{\infty }}+{\parallel P_{n}g-g\parallel }_{\mathrm{\infty }}+{\parallel g-f\parallel }_{\mathrm{\infty }}$

$\le 2{\parallel f-g\parallel }_{\mathrm{\infty }}+{\parallel P_{n}g-g\parallel }_{\mathrm{\infty }}$

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

∎

The following theorem states that $P_n$ is an operator on $L^p(ℝ)$ with operator norm $\le 1$. In particular, it asserts that if $f\in L^p(ℝ)$ then the averaged function $P_{n}f$ is also an element of $L^p(ℝ)$.

Theorem 11.

If , $n\in \mathbb{Z}$, and $f\in L^p(ℝ)$, then ${\parallel P_{n}f\parallel }_p\le {\parallel f\parallel }_p$.

Proof.

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

$|P_{n}f(x)|$ $=\left|2^n{\displaystyle {\int }_{I_{n,k_x}}}f(y)𝑑y\right|$

$\le 2^n{\left({\displaystyle {\int }_{I_{n,k_x}}}{|f(y)|}^p𝑑y\right)}^{1/p}{(\mu (I_{n,k_x}))}^{1/q}$

$=2^n{\left({\displaystyle {\int }_{I_{n,k_x}}}{|f(y)|}^p𝑑y\right)}^{1/p}{2}^{-n/q}.$

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

${\displaystyle {\int }_{I_{n,k}}}{|P_{n}f(x)|}^p𝑑x$ $\le {\displaystyle {\int }_{I_{n,k}}}{2}^{np}{2}^{-np/q}{\displaystyle {\int }_{I_{n,k_x}}}{|f(y)|}^p𝑑y𝑑x$

$={\displaystyle {\int }_{I_{n,k}}}{2}^{np}{2}^{-np/q}{\displaystyle {\int }_{I_{n,k}}}{|f(y)|}^p𝑑y𝑑x$

$=2^{-n}{2}^{np}{2}^{-np/q}{\displaystyle {\int }_{I_{n,k}}}{|f(y)|}^p𝑑y$

$={\displaystyle {\int }_{I_{n,k}}}{|f(y)|}^p𝑑y.$

We obtain

${\parallel P_{n}f\parallel }_p^p$ $={\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{I_{n,k}}}{|P_{n}f(x)|}^p𝑑x$

$\le {\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{I_{n,k}}}{|f(y)|}^p𝑑y$

$={\displaystyle {\int }_{ℝ}}{|f(y)|}^p𝑑y$

$={\parallel f\parallel }_p^p,$

giving ${\parallel P_{n}f\parallel }_p\le {\parallel f\parallel }_p$. ∎

6 Multiresolution analysis

For $a\in ℝ$, we define $m_a:ℝ\to ℝ$ by $m_a(x)=ax$, and we define ${\tau }_a:ℝ\to ℝ$ by ${\tau }_a(x)=x-a$.

Definition 12 (Multiresolution analysis).

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

1. If $n\in \mathbb{Z}$, then $f\in V_n$ if and only if $f\circ m_2\in V_{n+1}$.

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

3. $\overline{{\bigcup }_{n\in \mathbb{Z}}{V}_n}=L^2(ℝ)$.

4. ${\bigcap }_{n\in \mathbb{Z}}{V}_n=\{0\}$.

5. $\{\mathrm{\Phi }\circ {\tau }_k:k\in \mathbb{Z}\}$ 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 $\{L^2(ℝ,{ℱ}_n):n\in \mathbb{Z}\}$ of $L^2(ℝ)$ and the function $\varphi ={\chi }_{[0,1)}$ is a multiresolution analysis of $L^2(ℝ)$.

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 $L^2(ℝ)$, then $P_n\to 0$ in the strong operator topology as $n\to -\mathrm{\infty }$.

Lemma 14.

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

$$\underset{n\to -\mathrm{\infty }}{lim}{P}_{n}f=0.$$

Proof.

Define ${\mathrm{\Phi }}_{n,k}(x)=2^{n/2}\mathrm{\Phi }(2^{n}x-k)$. The set $\{{\mathrm{\Phi }}_{0,k}:k\in \mathbb{Z}\}$ is an orthonormal basis for $V_0$, and one checks that the set $\{{\mathrm{\Phi }}_{n,k}:k\in \mathbb{Z}\}$ is an orthonormal basis for $V_n$. Therefore

$$P_n=\sum _{k\in \mathbb{Z}}{\mathrm{\Phi }}_{n,k}\otimes {\mathrm{\Phi }}_{n,k}$$

in the strong operator topology.

For $R>0$, let $f_R=f{\chi }_{[-R,R]}$. If , then, using the Cauchy-Schwarz inequality,

${\parallel P_n{f}_R\parallel }_2^2$ $={\displaystyle \sum _{k\in \mathbb{Z}}}{|⟨P_n{f}_R,{\mathrm{\Phi }}_{n,k}⟩|}^2$

$={\displaystyle \sum _{k\in \mathbb{Z}}}{|⟨f_R,{\mathrm{\Phi }}_{n,k}⟩|}^2$

$={\displaystyle \sum _{k\in \mathbb{Z}}}{|⟨f_R,{\chi }_{[-R,R]}{\mathrm{\Phi }}_{n,k}⟩|}^2$

$\le {\displaystyle \sum _{k\in \mathbb{Z}}}\left({\displaystyle {\int }_{-R}^R}{|f_R(x)|}^2𝑑x\right)\left({\displaystyle {\int }_{-R}^R}{|{\mathrm{\Phi }}_{n,k}(x)|}^2𝑑x\right)$

$={\parallel f_R\parallel }_2^2{\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{-R}^R}{|{\mathrm{\Phi }}_{n,k}(x)|}^2𝑑x$

$={\parallel f_R\parallel }_2^2{\displaystyle \sum _{k\in \mathbb{Z}}}{2}^n{\displaystyle {\int }_{-R}^R}{|\mathrm{\Phi }(2^{n}x-k)|}^2𝑑x$

$={\parallel f_R\parallel }_2^2{\displaystyle \sum _{k\in \mathbb{Z}}}{\displaystyle {\int }_{-2^{n}R-k}^{2^{n}R-k}}{|\mathrm{\Phi }(x)|}^2𝑑x$

$={\parallel f_R\parallel }_2^2{\displaystyle {\int }_{U_n}}{|\mathrm{\Phi }(x)|}^2𝑑x,$

where

$$U_n=\bigcup _{k\in \mathbb{Z}}(-k-2^{n}R,-k+2^{n}R);$$

the intervals are disjoint because . Define $F_n(x)={|\mathrm{\Phi }(x)|}^2{\chi }_{U_n}(x)$. For all $x\in ℝ$ we have $|F_n(x)|\le {|\mathrm{\Phi }(x)|}^2$, and if $x\in ℝ$ then

$$\underset{n\to -\mathrm{\infty }}{lim}{F}_n(x)\to {|\mathrm{\Phi }(x)|}^2{\chi }_{\mathbb{Z}}(x),$$

where $\mathbb{Z}={\bigcap }_{n\in \mathbb{Z}}{U}_n$. Thus by the dominated convergence theorem we get

$$\underset{n\to -\mathrm{\infty }}{lim}{\int }_{ℝ}{F}_n(x)𝑑x={\int }_{ℝ}{|\mathrm{\Phi }(x)|}^2{\chi }_{\mathbb{Z}}(x)𝑑x=0,$$

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

$$\underset{n\to -\mathrm{\infty }}{lim}{\parallel P_n{f}_R\parallel }_2=0.$$

If $ϵ>0$ then there is some $R$ such that . We have, because $P_n$ is an orthogonal projection,

$\underset{n\to -\mathrm{\infty }}{lim\; sup}{\parallel P_{n}f\parallel }_2$ $\le \underset{n\to -\mathrm{\infty }}{lim\; sup}{\parallel P_{n}f-P_n{f}_R\parallel }_2+\underset{n\to -\mathrm{\infty }}{lim\; sup}{\parallel P_n{f}_R\parallel }_2$

$=\underset{n\to -\mathrm{\infty }}{lim\; sup}{\parallel P_{n}f-P_n{f}_R\parallel }_2$

$\le \underset{n\to -\mathrm{\infty }}{lim\; sup}{\parallel f-f_R\parallel }_2$

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

$$\underset{n\to -\mathrm{\infty }}{lim}{\parallel P_{n}f\parallel }_2=0.$$

∎

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 $S^{⟂}$ be the set of all $x\in H$ such that $y\in S$ implies that $⟨x,y⟩=0$. If $S_n,n\in \mathbb{Z}$, are mutually orthogonal closed subspaces of a Hilbert space, we write

$$\underset{n\in \mathbb{Z}}{\oplus }{S}_n=\underset{n\in \mathbb{Z}}{\bigvee }{S}_n.$$

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 $L^2(ℝ)$ and $W_n=V_{n+1}\cap V_n^{⟂}$, then

$$L^2(ℝ)=\underset{n\in \mathbb{Z}}{\oplus }{W}_n.$$

Proof.

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

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

But $f\in W_n\subset V_{n+1}$ and $g\in W_{n^{\prime }}\subset V_{n+1}^{⟂}$, so $⟨f,g⟩=0$. Therefore $W_n⟂W_{n^{\prime }}$.

If $f\in V_n$ and $f\ne 0$, 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 $M\ge 0$ 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 $M\to \mathrm{\infty }$. Thus, for any $ϵ>0$ there is some $M$ with and $f\in f_{N-M}+{\oplus }_{m=1}^M{W}_{N-m}$. Therefore, if $f\in {\bigcup }_{n\in \mathbb{Z}}{V}_n$ then there is some $g\in {\oplus }_{n\in \mathbb{Z}}{W}_n$ satisfying . Thus

$$\overline{\bigcup _{n\in \mathbb{Z}}{V}_n}\subseteq \underset{n\in \mathbb{Z}}{\oplus }{W}_n,$$

and so

$$L^2(ℝ)=\underset{n\in \mathbb{Z}}{\oplus }{W}_n.$$

∎

7 The unit interval

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

$$⟨f,g⟩={\int }_0^{1}f(x)\overline{g(x)}𝑑x.$$

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

$$[0,1)=\bigcup _{k=0}^{2^n-1}{I}_{n,k}.$$

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

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

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

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

We check that $L^2([0,1),{𝒢}_n)$ 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 , so , hence . If $x\in I_{n+1,2k+1}$, then , hence , and so . 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 $x\notin I_{n,k}$, for which ${\psi }_{n,k}(x)=0$. It follows that ${\psi }_{n,k}\in L^2([0,1),{𝒢}_{n+1})$.

Theorem 16.

If

$${ℬ}_0=\{{\chi }_{[0,1)}\}$$

and, for $n\ge 0$,

$${ℬ}_{n+1}=\{{\psi }_{n,k}:0\le k\le 2^n-1\},$$

then

$$\bigcup _{n=0}^N{ℬ}_n$$

is an orthonormal basis of $L^2([0,1),{𝒢}_N)$.

Proof.

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

$⟨{\psi }_{n-1,k},{\chi }_{[0,1)}⟩$ $={\displaystyle {\int }_0^1}{\psi }_{n-1,k}(x)\overline{{\chi }_{[0,1)}(x)}𝑑x$

$={\displaystyle {\int }_0^1}{\psi }_{n-1,k}(x)𝑑x$

$={\displaystyle {\int }_{I_{n,2k}}}{\psi }_{n-1,k}(x)𝑑x+{\displaystyle {\int }_{I_{n,2k+1}}}{\psi }_{n-1,k}(x)𝑑x$

$={\displaystyle {\int }_{I_{n,2k}}}{2}^{(n-1)/2}𝑑x+{\displaystyle {\int }_{I_{n,2k+1}}}-2^{(n-1)/2}dx$

$=0.$

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

$|{ℬ}_0|=1$, and if $n\ge 1$ then $|{ℬ}_n|=2^{n-1}$. Therefore the number of elements of ${\bigcup }_{n=0}^N{ℬ}_n$ is

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

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

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

$$ℬ=\bigcup _{n=0}^{\mathrm{\infty }}{ℬ}_n$$

is an orthonormal set in $L^2([0,1))$: if $f,g\in ℬ$ then there is some $N$ with $f,g\in {\bigcup }_{n=0}^N{ℬ}_n$, which is an orthonormal set. The following theorem shows that $ℬ$ is an orthonormal basis for the Hilbert space $L^2([0,1))$.

Theorem 17.

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

Proof.

If $f\in L^2([0,1))$ and $ϵ>0$ then there is some $g\in C([0,1])$ with . $g$ is uniformly continuous on the compact set $[0,1]$, so there is some $\delta >0$ such that implies that . Let $2^{-n}\le \delta$, and define $h:[0,1)\to ℂ$ 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 $k_x,0\le k_x\le 2^n-1$, with $x\in I_{n,k_x}$, and for this $k_x$ we have , and hence

Therefore ${\parallel g-h\parallel }_{\mathrm{\infty }}\le \frac{ϵ}{2}$.

We have $h\in L^2([0,1),{𝒢}_n)$, and

We have shown that if $f\in L^2([0,1))$ and $ϵ>0$ then there is some $n$ and some $h\in L^2([0,1),{𝒢}_n)$ with . This tells us that ${\bigcup }_{n=0}^{\mathrm{\infty }}{L}^2([0,1),{𝒢}_n)$ is a dense subset of $L^2([0,1))$. Since $ℬ$ is orthonormal and $\mathrm{span}ℬ={\bigcup }_{n=0}^{\mathrm{\infty }}{L}^2([0,1),{𝒢}_n)$, $ℬ$ 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.