I'm reading the First Course in Wavelets with Fourier Analysis by Boggess & Narcowich. In chapter 5 section 1.2, it defines $$\newcommand{\scal}[1]{\langle{#1}\rangle}p_k = \sqrt2 \scal{ \phi, \phi_{1,k} } = 2 \int \phi(x) \overline{\phi(2x-k)}\, dx$$ for a given scaling function $\phi$ so that $$\phi(x) = \sum_k p_k \phi(2x-k),$$ where the limit is presumably taken in $L^2$. And it states that $\sum_k p_k = 2$ in Theorem 5.9. I believe there is a missing assumption here. We do not know if $\sum_k p_k$ converges in the first place. The question now is, just exactly what would be the right assumption to make this work.
My best guess is to require the convergence of $\phi(x) = \sum_k p_k \phi(2x-k)$ in $L^1$. Is that sufficient? Any better suggestions? And what about the resulting convergence of $\sum_k p_k$? It seems to me that we only have conditional convergence here, but without absolute convergence, I am not sure if the proof of the subsequent statements, $\sum_k p_{2k}=1$ and $\sum_k p_{2k+1}=1$, would still work.
EDIT The integral stated at the bottom of the proof must exist because $\phi$ would not be Fourier transformable otherwise since it's Fourier Transform at frequency 0 is $$\int_{-\infty}^{\infty}\phi(t)\cdot e^{-2\pi n \cdot 0}dt = \int_{-\infty}^{\infty}\phi(t)dt$$
In the copy of the edition I have the proof goes as follows:
"Theorem 5.9 Suppose $\{V_j;j\in Z\}$ is a multiresolution analysis with scaling function $\phi$. Then the following equalities hold:
$\displaystyle\sum_{k\in Z} p_{k-2l} {\bar{p_k} } = 2\delta_{l0}$
$\displaystyle \sum_{k\in Z} |p_k|^2 = 2$
$\displaystyle \sum_{k\in Z} p_k = 2$
$\displaystyle \sum_{k\in Z} p_{2k} = 1$ and $\displaystyle \sum_{k\in Z} p_{2k+1} = 1$
Proof: The first equation follows from the two-scale relation (Theorem 5.6) and the fact that $\{\phi(x-k), k\in Z\}$ is orthonormal. We leave the details to Exercise 4a. By setting $l=0$ in the first equation we get the second equation. The proof of the third equation uses Theorem 5.6 as follows:
$$\int_{-\infty}^{\infty}\phi(x)dx = \sum_{k\in Z} p_k \int_{-\infty}^{\infty} \phi(2x-k)dx$$
By letting $t = 2x-k$ and $dx = dt/2$ we obtain:
$$\int_{-\infty}^{\infty}\phi(x)dx = 1/2\sum_{k\in Z} p_k \int_{-\infty}^{\infty} \phi(t)dt$$
Now $\int\phi(t)dt$ cannot be zero (see Exercise 6) for otherwise we could never approximate functions $f\in L^2(R)$ with $\int f(t)dt \neq 0$ by functions in $V_j$. Therefore the factor $\int \phi(t)dt$ on the right can be cancelled with $\int \phi(x)dx$ on the left (cancelled as divided away, my remark). The third equation now follows."
The proof of 4 is a little lengthier but also seems sound.