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

ℱ⁢LHS=ℱ⁢RHS ℱ LHS ℱ RHS ℱ⁢φ⁢t=ℱ⁢2⁢∑nh⁢n⁢φ⁢2⁢t−n ℱ φ t ℱ 2 n h n φ 2 t n Φ⁢ω=2⁢∑nh⁢n⁢ℱ⁢φ⁢2⁢t−n Φ ω 2 n h n ℱ φ 2 t n Φ⁢ω=2⁢∑nh⁢n⁢∫−∞∞φ⁢2⁢t−n⁢e−(i⁢ω⁢t)dt Φ ω 2 n h n t φ 2 t n ω t For τ=2⁢t−n τ 2 t n , Φ⁢ω=2⁢∑nh⁢n⁢12⁢∫−∞∞φ⁢τ⁢e−(i⁢ω⁢τ+n2)dτ Φ ω 2 n h n 1 2 τ φ τ ω τ n 2 Φ⁢ω=12⁢∑nh⁢n⁢e−(i⁢ω2⁢n)⁢∫−∞∞φ⁢τ⁢e−(i⁢ω2⁢τ)dτ Φ ω 1 2 n h n ω 2 n τ φ τ ω 2 τ Φ⁢ω=12⁢∑nh⁢n⁢e−(i⁢ω2⁢n)⁢Φ⁢ω2 Φ ω 1 2 n h n ω 2 n Φ ω 2 Φ⁢ω=12⁢Φ⁢ω2⁢∑nh⁢n⁢e−(i⁢ω2⁢n) Φ ω 1 2 Φ ω 2 n h n ω 2 n Note that H f ⁢ω=DTFT⁢h⁢n=∑nh⁢n⁢e−(i⁢ω⁢n) H f ω DTFT h n n h n ω n is the discrete-time Fourier transform (DTFT) of the scaling filter h⁢n h n . We then have the Fourier form of the dilation equation:

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

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

From Equation 10, we can write Φ⁢ω2=12⁢ H f ⁢ω4⁢Φ⁢ω4 Φ ω 2 1 2 H f ω 4 Φ ω 4 or Φ⁢ω=12⁢ H f ⁢ω2⁢12⁢ H f ⁢ω4⁢Φ⁢ω4 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 ,

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

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

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

Suppose the support of φ⁢t φ t is a b a b . Then

The support of the LHS is a b a b by assumption. The support of the RHS is a2 b2+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 φ⁢t φ t is 0 N−1 0 N 1 . h⁢n⁢  has finite support  ⇒φ⁢t⁢  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.

Note: limit  j→ ∞ 12j⁢∑lφ⁢l2j=∫φ⁢tdt j 1 2 j l φ l 2 j t φ t Let us define ∀j,j≥1: S j ≔12j⁢∑lφ⁢l2j j j 1 ≔ S j 1 2 j l φ l 2 j Begin with the dilation equation φ⁢t=2⁢∑nh⁢n⁢φ⁢2⁢t−n φ t 2 n h n φ 2 t n and substitute l2j l 2 j for tt. φ⁢l2j=2⁢∑nh⁢n⁢φ⁢l2j−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: ∑lφ⁢l2j=∑l2⁢∑nh⁢n⁢φ⁢l2j−1−n l φ l 2 j l 2 n h n φ l 2 j 1 n ∑lφ⁢l2j=2⁢∑nh⁢n⁢∑lφ⁢l2j−1−n l φ l 2 j 2 n h n l φ l 2 j 1 n ∑lφ⁢l2j=2⁢∑nh⁢n⁢∑lφ⁢l2j−1 l φ l 2 j 2 n h n l φ l 2 j 1 ∑lφ⁢l2j=2⁢∑lφ⁢l2j−1 l φ l 2 j 2 l φ l 2 j 1 Divide both sides by 2j 2 j : 12j⁢∑lφ⁢l2j=12j−1⁢∑lφ⁢l2j−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! limit  j→ ∞ S j = S 0 j S j S 0 Therefore

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

for any real t o t o .