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?

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    $\begingroup$ mathworld.wolfram.com/PoissonSumFormula.html $\endgroup$ – user82588 Mar 11 at 21:44
  • $\begingroup$ I used it by defining $f(y) = y^m\phi(x-y)$ and almost found the result. However, instead of $M_\phi^l$, I got $i^k e^{i l x} \frac{d^k\hat{\phi}}{d\xi^k} (-l)$, which (I think) I cannot turn into $M_\phi^l$... $\endgroup$ – AspiringMathematician Mar 13 at 13:28

The identity in the OP does not hold for any $m$, but only for $m< N$ where $N$ is the number of vanishing moments of the wavelet function. To complete the Poisson-summation derivation, one needs the socalled Strang-Fix condition, which says that $\frac{d^k\hat{\phi}}{d\xi^k}(-l)=0$ for integer $l$ unequal to 0 and $k< N$.

The identity says that integer shifts of the scaling function can reproduce polynomials of order $N$. For a proof, see theorems 4.26 and 4.27 in this book. (The identity is equivalent to the recursion relation 4.51.)

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    $\begingroup$ I couldn't have a look at it during the weekend, but this does look like what I was needing. I'll check the theorems and award the bounty later. $\endgroup$ – AspiringMathematician Mar 16 at 13:59

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