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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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    [For lazy people like me...](http://www.ams.org/tran/1989-315-01/S0002-9947-1989-1008470-5/S0002-9947-1989-1008470-5.pdf)2010-12-09