Orthonormal basis for $W_j$ Given $\{V_j: j \in \mathbb{Z}\}$ is multi-resolution analysis in $L^2(\mathbb{R})$. Suppose $W_0$ is an orthogonal complement of $V_0$ relative to $V_1$.
So that we can write, $V_1 = V_0 \bigoplus W_0$.
For every $j \in \mathbb{Z}$, define $$W_j = \{f(2^j x): f \in W_0\}.$$
How can I show:
if $\{ \psi(x-k): k \in \mathbb{Z}\}$ is an orthonormal basis for $W_0$, then for every $j \in \mathbb{Z}$, the set $$\{2^{j/2}\psi(2^jx-k): k \in \mathbb{Z}\}$$ is an orthonormal basis for $W_j$?
Can I use induction method? Or are there any methods that can be used effectively? Thanks in advance.
 A: First we show that $g\in W_0$ if and only if $g(2^j\cdot)\in W_j$ for $j\in \mathbb{Z}$. Since
$$V_1=V_0\oplus W_0$$
then $g\in W_0$ if and only if $g\in V_1$ and
$$\int_{\mathbb{R}} g(x)\overline{f}(x)dx=0,\,\forall f\in V_0$$
Also since
$$V_{j+1}=V_{j}\oplus W_{j}$$
then $\tilde{g}\in W_{j}$ if and only if  $\tilde{g}\in V_{j+1}$ and
$$\int_{\mathbb{R}} \tilde{g}(x)\overline{f}dx=0,\, \forall f\in V_j.$$
Now observe that $g\in V_0$ if and only if $g(2^j\cdot)\in V_j$ and
$$\int_{\mathbb{R}} g(x)\overline{f}(x)dx=0\Leftrightarrow \int_{\mathbb{R}} g(2x)\overline{f}(2x)dx=0.$$
So we proved
$$g\in W_0 \text{ if and only if } g(2^j\cdot)\in W_j\text{ for } j\in \mathbb{Z}.$$
Since $\{\varphi(\cdot-k):k\in \mathbb{Z}\}$ is an ortonormal basis for $W_0$ then
$$g(x)=\sum_{k\in \mathbb{Z}} \alpha_k \varphi(x-k)$$
where the sum is finite. So
$$g(2^jx)=\sum_{k\in \mathbb{Z}} 2^{-j/2}\alpha_k 2^{j/2}\varphi(2^jx-k).$$
So we proved they generate $W_j$. With a change of variables you prove they have norm one and also
$$\int_{\mathbb{R}} \varphi(x-k)\overline{\varphi}(x-m)dx=0\Leftrightarrow \int_{\mathbb{R}} \varphi(2^jx-k)\overline{\varphi}(2^jx-m)dx=0.$$
Therefore, $\{2^{j/2}\varphi(2^jx-k):k\in \mathbb{Z}\}$ is an orthonormal basis for $W_j$.
