I am reading an article on wavelet connection coefficients (G. Beylkin, "On the representation of operators in bases of compactly supported wavelets", 1992 (MSN)) and I came across Equation (3.31): \begin{equation} \sum_{l=-\infty}^\infty l^m\phi(x-l) = x^m + \sum_{l=1}^m (-1)^l \begin{pmatrix} m\\l \end{pmatrix} M_l^\phi x^{m-l} \end{equation} where $\phi(x)$ is the scaling function and \begin{equation} M_l^\phi = \int_{-\infty}^\infty x^l\phi(x)\,dx \end{equation} is the $l$-th momentum of $\phi$.

The author claims that the equation is well-known if $M_l^\phi = 0$ for $l=0,\dotsc,m$, and the general case follows from taking Fourier transforms. However, I could not find it, and trying to prove it myself is not working.

I recognize that both sides are kinds of convolutions, but when taking the Fourier transform the expressions (apparently) lead nowhere. Is there some trick I need to be aware of, or is it simply lack of practice/knowledge?

I do not know if this information should help, but the ultimate goal is to prove that \begin{equation} \sum_{l=-\infty}^\infty lr_l = -1 \end{equation} where \begin{equation} r_l = \int_{-\infty}^\infty \phi(x-l)\phi'(x)\,dx. \end{equation}

EDIT: Using the Poisson summation as @Nemo suggested in a comment, I was able to find that \begin{equation} \sum_{l=-\infty}^\infty l^m\phi(x-l) = \sum_{k=0}^m \sum_{l=-\infty}^\infty (-1)^k \begin{pmatrix} m\\k \end{pmatrix} e^{-ilx} i^k \frac{d^k\hat{\phi}}{d\xi^k}(-l) x^{m-k}. \end{equation}

Now, I know that $i^k\frac{d^k\hat{\phi}}{d\xi^k}(0) = M_l^\phi$ but I'm still stuck with the terms $i^k e^{-ilx} \frac{d^k\hat{\phi}}{d\xi^k}(-l)$ for $l\ne0$. Is there any identity I am not aware of?