A Condition for Existence Dilation Equation φ t 2 n h n φ 2 t n t φ t t 2 n h n φ 2 t n t φ t 2 n h n t φ 2 t n t φ t 2 n h n t φ 2 t t φ t 2 n h n t φ 2 t t φ t 2 n h n 1 2 t φ t 1 1 2 n h n assuming t φ t 0 . For a solution φ t to even exist, it is required that the coefficients h n add up to 2 . n h n 2

Fourier Form of the Dilation Equation Dilation Equation φ t 2 n h n φ 2 t n ℱ LHS ℱ RHS ℱ φ t ℱ 2 n h n φ 2 t n Φ ω 2 n h n ℱ φ 2 t n Φ ω 2 n h n t φ 2 t n ω t For τ 2 t n , Φ ω 2 n h n 1 2 τ φ τ ω τ n 2 Φ ω 1 2 n h n ω 2 n τ φ τ ω 2 τ Φ ω 1 2 n h n ω 2 n Φ ω 2 Φ ω 1 2 Φ ω 2 n h n ω 2 n Note that H f ω DTFT h n n h n ω n is the discrete-time Fourier transform (DTFT) of the scaling filter h n . We then have the Fourier form of the dilation equation: Φ ω 1 2 H f ω 2 Φ ω 2 When ω 0 is put into this equation, we get Φ 0 1 2 H f 0 Φ 0 or H f 0 2 which is exactly the same equation as n h n 2 that we already got. From , we can write Φ ω 2 1 2 H f ω 4 Φ ω 4 or Φ ω 1 2 H f ω 2 1 2 H f ω 4 Φ ω 4 If we iterate, we get the infinite-product formula for Φ ω , Φ ω Φ 0 k 1 1 2 H f ω 2 k From this formula, we can see that the zeros of Φ ω are determined by the zeros of H f ω .

What is the support of φ(t)? Given h n , what is the support of φ t ? Suppose h n is supported on 0 n N 1 then from the dilation equation φ t 2 n 0 N 1 h n φ 2 t n the support of φ t is found by matching the support of the left hand side and the right hand side. Suppose the support of φ t is a b . Then φ 2 t has support a 2 b 2 φ 2 t 1 has support a 1 2 b 1 2 ⋮ φ 2 t k has support a k 2 b k 2 ⋮ φ 2 t N 1 has support a N 1 2 b N 1 2 The support of the LHS is a b by assumption. The support of the RHS is a 2 b 2 N 1 2 . Therefore, matching the endpoints we get a a 2 b b 2 N 1 2 which gives a 0 b N 1 The support of φ t is 0 N 1 . h n has finite support φ t of finite support That means that there are finite degrees of freedom in design of finitely supported scaling functions.

Sum-Integral Equality of φ(t) Note: j 1 2 j l φ l 2 j t φ t Let us define j j 1 ≔ S j 1 2 j l φ l 2 j Begin with the dilation equation φ t 2 n h n φ 2 t n and substitute l 2 j for t. φ l 2 j 2 n h n φ l 2 j 1 n with j 1 . Now sum over l: l φ l 2 j l 2 n h n φ l 2 j 1 n l φ l 2 j 2 n h n l φ l 2 j 1 n l φ l 2 j 2 n h n l φ l 2 j 1 l φ l 2 j 2 l φ l 2 j 1 Divide both sides by 2 j : 1 2 j l φ l 2 j 1 2 j 1 l φ l 2 j 1 S j S j − 1 For j 1 , all the S j are the same! j S j S 0 Therefore t φ t k φ k If φ t is continuous, then it can also be shown that t φ t k φ t o k for any real t o .