A Condition for Existence

assuming ∫φtdt≠0 t φ t 0 . For a solution φt φ t to even exist, it is required that the coefficients hn h n add up to 2 2 .

Fourier Form of the Dilation Equation

ℱLHS=ℱRHS ℱ LHS ℱ RHS ℱφt=ℱ2∑nhnφ2t−n ℱ φ t ℱ 2 n h n φ 2 t n Φω=2∑nhnℱφ2t−n Φ ω 2 n h n ℱ φ 2 t n Φω=2∑nhn∫-∞∞φ2t−nⅇ-ⅈωtdt Φ ω 2 n h n t φ 2 t n ω t For τ=2t−n τ 2 t n , Φω=2∑nhn12∫-∞∞φτⅇ-ⅈωτ+n2dτ Φ ω 2 n h n 1 2 τ φ τ ω τ n 2 Φω=12∑nhnⅇ-ⅈω2n∫-∞∞φτⅇ-ⅈω2τdτ Φ ω 1 2 n h n ω 2 n τ φ τ ω 2 τ Φω=12∑nhnⅇ-ⅈω2nΦω2 Φ ω 1 2 n h n ω 2 n Φ ω 2 Φω=12Φω2∑nhnⅇ-ⅈω2n Φ ω 1 2 Φ ω 2 n h n ω 2 n Note that H f ω=DTFThn=∑nhnⅇ-ⅈωn H f ω DTFT h n n h n ω n is the discrete-time Fourier transform (DTFT) of the scaling filter hn h n . We then have the Fourier form of the dilation equation:

When ω=0 ω 0 is put into this equation, we get Φ0=12 H f 0Φ0 Φ 0 1 2 H f 0 Φ 0 or

which is exactly the same equation as ∑nhn=2 n h n 2 that we already got.

From Equation 10, we can write Φω2=12 H f ω4Φω4 Φ ω 2 1 2 H f ω 4 Φ ω 4 or Φω=12 H f ω212 H f ω4Φω4 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 Φω Φ ω ,

From this formula, we can see that the zeros of Φω Φ ω are determined by the zeros of H f ω H f ω .

What is the support of φ(t)?

Given hn h n , what is the support of φt φ t ?

Suppose hn h n is supported on 0≤n≤N−1 0 n N 1 then from the dilation equation φt=2∑n=0N−1hnφ2t−n φ t 2 n 0 N 1 h n φ 2 t n the support of φt φ t is found by matching the support of the left hand side and the right hand side.

Suppose the support of φt φ t is ab a b . Then

The support of the LHS is ab a b by assumption. The support of the RHS is a2b2+N−12 a 2 b 2 N 1 2 . Therefore, matching the endpoints we get a=a2 a a 2 b=b2+N−12 b b 2 N 1 2 which gives a=0 a 0 b=N−1 b N 1 The support of φt φ t is 0N−1 0 N 1 . hn  has finite support  ⇒φt  of finite support   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: limj→∞12j∑lφl2j=∫φtdt j 1 2 j l φ l 2 j t φ t Let us define ∀j,j≥1: S j ≔12j∑lφl2j j j 1 ≔ S j 1 2 j l φ l 2 j Begin with the dilation equation φt=2∑nhnφ2t−n φ t 2 n h n φ 2 t n and substitute l2j l 2 j for tt. φl2j=2∑nhnφl2j−1−n φ l 2 j 2 n h n φ l 2 j 1 n with j≥1 j 1 . Now sum over ll: ∑lφl2j=∑l2∑nhnφl2j−1−n l φ l 2 j l 2 n h n φ l 2 j 1 n ∑lφl2j=2∑nhn∑lφl2j−1−n l φ l 2 j 2 n h n l φ l 2 j 1 n ∑lφl2j=2∑nhn∑lφl2j−1 l φ l 2 j 2 n h n l φ l 2 j 1 ∑lφl2j=2∑lφl2j−1 l φ l 2 j 2 l φ l 2 j 1 Divide both sides by 2j 2 j : 12j∑lφl2j=12j−1∑lφl2j−1 1 2 j l φ l 2 j 1 2 j 1 l φ l 2 j 1 S j = S j − 1 S j S j − 1 For j≥1 j 1 , all the S j S j are the same! limj→∞ S j = S 0 j S j S 0 Therefore

If φt φ t is continuous, then it can also be shown that

for any real t o t o .