# Integer translates of a scaling function

I think this is asked as a standard exercise in books about wavelets (e.g. exercise 7.2 in Mallat's book), but I couldn't find a proof. Let $\phi$ be a scaling function (see definition below). I would like to learn why $$\sum_{k\in\mathbb Z} \phi(x-k) = 1$$
almost everywhere.

Definition. A sequence of subspaces $\{V_j: j\in \mathbb{Z}\}$ of $L^2(\mathbb R)$ is called a multiresolution analysis if it satisfies the following:

• $V_j \subset V_{j+1}$
• $\bigcap_{j}V_j = \{0\}$
• $\overline{\bigcup_jV_j} = L^2(\mathbb R)$
• $f(x)\in V_j$ if and only if $f(2x) \in V_{j+1}$
• There exists a function $\phi \in V_0$ such that $\{\phi(x-k)\}_{k\in\mathbb Z}$ is an orthogonal basis for $V_0$

The function $\phi$ here is called as a scaling function.

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I think you will find the proof for this in Mallat 1989, 'Multiresolution approximations and wavelet orthonormal bases of L^2'. Theorem 1 (in particular Equations (23), (36)) is what you are after. It is not trivial, longer to prove than I immediately thought. Perhaps there is a very fast proof but I can't think of it now.

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@Glen: But $1\notin L^2$. If it were, a formal argument would be $1 = \sum c_k \phi(x-k)$, and $c_k = \int 1\cdot \phi(x-k) =$const. Some tricks are required to make it rigorous. This is not homework, you may give an outline if you have an idea. –  AgCl Dec 7 '10 at 16:17
@AgCl: What was in my mind when I wrote that was possibly using a sequence of scaled characteristic functions on longer and longer intervals, something like f_l = 2^-l\chi_{[-l,l]}. But now I think about it some more, this doesn't seem to work. I think I should just delete my answer until I think of a better idea, yes? –  Glen Wheeler Dec 8 '10 at 10:01
@AgCl: OK. I just looked it up in a paper, Mallat 1989, 'Multiresolution approximations and wavelet orthonormal bases of L^2'. Theorem 1 (in particular Equations (23), (36)) is, I think, what you are after. It is far from immediate, longer to prove than I immediately thought. Perhaps there is a very fast proof but I can't think of it now. I could reproduce the proof there if you wish, although it would probably be better to just read the paper. –  Glen Wheeler Dec 8 '10 at 10:56
@Glen: Thanks for the reference, that seems to be what I was looking for! Could you edit your answer with a reference to this paper, so that I can accept it? –  AgCl Dec 8 '10 at 23:31
For lazy people like me... –  Guess who it is. Dec 9 '10 at 10:04