Multiresolution analysis, change of integral and summation and convergence a.e. Let $(V_n, \phi)$ be multiresolution analysis with scaling function $\phi$ and $V_n \subset L_2(\mathbb{R})$. I need an explanation why $\phi$ can be written in following way (taken from Bachman's Fourier Analysis, chapter 7 - Wavelets):
$\hat{\phi} (w)= \hat{m}(\frac{w}{2})\hat{\phi}(\frac{w}{2})$, almost everywhere.  $(*)$
Since $\phi \in V_0 \subset V_1$, then it can be written in terms of orthonormal basis vectors of $V_1$ i.e.
\begin{equation} \phi(x)= \sqrt{2} \sum_{k \in \mathbb{Z}} c_k \phi(2x -k) \quad (1) \end{equation}
where $c_n= \langle \phi, \phi_{1,n}\rangle  $ and  $\phi_{1,n}(x)= \sqrt{2}\phi(2x-n)$
Now, we compute the Fourier tranform of $\phi$. Since $\phi \in L_2(\mathbb{R})$,  its Fourier transform   $\hat{\phi}(t)$ is defined as $l.i.m. \int_{-n}^n \phi(x) e^{- i t x } dx$ (the limit in $L_2$ norm). My questions are:

*

*Why  can we change the order of $l.i.m.$ and sum? I found explanation in Bachman's Fourier Analysis that it follows from continuity of Fourier map $F:  L_2(\mathbb{R}) \rightarrow L_2(\mathbb{R})$, $F(f)= \hat{f}$. Can we say that $\phi_n(x)=  \sqrt{2} \sum_{|k|\leq n} c_k \phi(2x -k)$? Then does $\phi_n \rightarrow \phi$ in $L_2$ norm and how can it be proved? This is my attempt: $|| \phi - \phi_n||_2 =  || \sum_{ |k|> n}c_k \sqrt{2} \phi(2t-k) dt||_2$, but I don't know how to formally prove that this converges to $0$ as $n \rightarrow \infty$ (which theorem should be used?).


*The convergence required in (*) is convergence almost everywhere, but in the question (1) we had convergence in norm. How do we obtain a.e. convergence?
Thanks a lot in advance, I would be grateful for help. Any book or link where these questions are answered would be helpful.
 A: Let me try to address your questions one by one. I'm taking the following convention for the Fourier transform of a function (for which the integral converges)
$$\widehat f(\omega)=\int_{\mathbb R}f(x)e^{-i\omega x}dx$$
Convergence of the orthonormal decomposition in a Hilbert space
Because $\{x\mapsto \sqrt{2}\phi(2x - k)\}_{k\in\mathbb Z}$ is an orthonormal basis of $V_1$, you can indeed expand $\phi$ on that basis. Thus, by definition, in $L^2(\mathbb R)$,
$$\phi (x)= \sqrt{2}\sum_{k\in\mathbb Z}c_k \phi(2x - k) \tag{1}$$
Note that this is in the sense of the convergence for the $L^2(\mathbb R)$ norm. In the infinite sum above, the order of summation doesn't matter. In particular, if you define the partial sum $$\phi_n(x)=\sqrt{2}\sum_{|k|\leq n} c_k \phi(2x -k)$$
then in $L^2(\mathbb R)$, we have $$\lim_{n\rightarrow +\infty}\phi_n= \phi$$
Convergence in the Fourier domain
Now let's talk about the Fourier transform. By the Plancherel theorem, it's a multiple of an isometry on $L^2(\mathbb R)$: For all $f\in L^2(\mathbb R)$, $$\|\widehat f\|_{L^2(\mathbb R)}=\sqrt{2\pi}\|f\|_{L^2(\mathbb R)}$$
Consequently, it is continuous, and thus we have $$\lim_{n\rightarrow+\infty}{\widehat \phi}_n =  \widehat\phi\tag{2}$$
Scaling function equation
The Fourier transform of a rescaled function is a rescaled version of the Fourier transform of the original function. The Fourier transform of a shifted function is a the Fourier transform of the original function times a complex exponential. In fact:
$$\widehat{\phi(2\cdot-k)}(\omega) = \frac 1 2 \widehat \phi\left(\frac \omega 2\right)e^{-i\frac{k\omega}2}$$
Thus $(2)$ can be written as
$$\widehat\phi(\omega)=\frac 1 {\sqrt{2}}\sum_{k\in\mathbb Z}c_k \widehat \phi\left(\frac \omega 2\right)e^{-i\frac{k\omega}2} = \widehat m\left(\frac \omega 2\right)\widehat \phi\left(\frac \omega 2\right)\tag{3}$$
with
$$\widehat m(\omega)=\frac 1 {\sqrt{2}}\sum_{k\in\mathbb Z}c_k e^{-i k\omega}\tag{4}$$
Equation $(3)$ is in the sense of the $L^2(\mathbb R)$ norm.
Almost everywhere convergence
For $(3)$, the convergence is in the sense of $L^2(\mathbb R)$, not in terms of pointwise convergence.
Now, notice that because of $(1)$, $$2\sum_{k\in\mathbb Z} |c_k|^2=\|\varphi\|^2_{L^2(\mathbb R)}<+\infty$$
Thus by $(4)$, we have $\hat m\in L^2([0, 2\pi])$. By Carleson's theorem, the convergence in $(4)$ is true almost everywhere, and it's also true in $(3)$.
