Multi-Resolution Analysis
Definition
If a sequence of closed subspaces $L^{2}(\mathbb{R})$ and a function $\phi \in V_{0}$ satisfy the following conditions, then $\left( \left\{ V_{j} \right\}, \phi \right)$ is called a multiresolution analysis.
(a) For each $V_{j}$, $\cdots V_{-1} \subset V_{0} \subset V_{1}\cdots$ holds.
(b) $\overline{\cup_{j\in\mathbb{Z}}V_{j}}=L^{2}(\mathbb{R})$ and $\cap_{j\in\mathbb{Z}}V_{j}=\left\{ 0\right\}$.
(c) $\forall j\in \mathbb{Z}$, $V_{j+1}=D(V_{j})$.
(d) If $\forall k \in \mathbb{Z}$, $f \in V_{0}$, then $T_{k}f \in V_{0}$.
(e) $\left\{ T_{k} \phi\right\}_{k\in \mathbb{Z}}$ is an orthonormal basis of $V_{0}$.
If $(\left\{ V_{j} \right\},\phi)$ is a multiresolution analysis, then $\phi$ is said to generate the multiresolution analysis. $T_{k}$ is a translation, $D$ is a dilation.
Explanation
Condition (b) means that $\cup_{j}V_{j}$ is dense in $L^{2}(\mathbb{R})$, meaning that there exists an approximation $g \in \cup_{j}V_{j}$ by some $f \in L^{2}(\mathbb{R})$. If such $g$ belongs to $V_{J}$, then according to condition (a), $g$ is contained in all $V_{j}$ where $j \ge J$. Moreover, from the definition, the following facts are established.
Theorem
If conditions (c) and (d) of the definition are satisfied, then the following two facts also hold for all $j \in \mathbb{Z}$:
(f) $V_{j}=D^{j}(V_{0})$
(g) $V_{j}=\overline{\text{span}}\left\{ D^{j}T_{k}\phi \right\}$
Proof
(f)
Assuming (c) holds, then for all $j \in \mathbb{N}$
$$ V_{j}=D(V_{j-1})=DD(V_{j-2})=\cdots=D^{j}V(_{0}) $$
Also, for all $j \in \left\{ -1,-2,\cdots \right\}$
$$ V_{j}=D^{-1}(V_{j+1})=D^{-1}D^{-1}(V_{j+2})=\cdots=(D^{-1})^{-j}(V_{0})=D^{j}(V_{0}) $$
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(g)
Since $\left\{ T_{k}\phi \right\}_{k\in \mathbb{Z}}$ is an orthonormal basis of $V_{0}$
$$ V_{0}=\overline{\text{span}}\left\{ T_{k}\phi \right\}_{k\in \mathbb{Z}} $$
is established. Hence, by (f)
$$ V_{j}=D^{j}(V_{0})=D^{j}\left( \overline{\text{span}}\left\{ T_{k}\phi \right\}_{k\in \mathbb{Z}} \right)=\overline{\text{span}}\left\{ D^{j}T_{k}\phi \right\}_{k\in \mathbb{Z}} $$
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