WAVELET BASIS IN THE SPACE C ∞ [−1, 1]
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ω n+1 (x) = 2 n+1 (1 − x 2 ) T 1 (x) T 2 (x) T 4 (x) · · · T 2n
ω j 0 (cos kπ 2j
Now the Kilgor-Prestin wavelets are defined as ψ j, k (x) = T 2j
with x j, k = cos (2k+1)π 2j+1
span{ψ j,k , k = 0, 1, · · · , 2 j − 1} = span{T 2j
Π 2j+1
By H we denote the corresponding Hilbert space. Let ε j, n take the value 1 for 1 ≤ n ≤ 2 j − 1 and ε j, 0 = ε j, 2j
2 Xj
Since the decomposition (1) is orthogonal, we get T n = P 2j
j=0 2 Xj
Lemma 4. For any j ∈ N 0 the matrices X j = (T n (x j, k )) n=2 2j+1j
Theorem 1. The system {T 0 , T 1 , (ψ j, k ) ∞, 2 j=0, k=0j
Therefore the functionals {ξ 0 , ξ 1 , (ξ j, k ) ∞, 2 j=0, k=0j
| T n | m ≤ 2 1−j 2 j | T 2j+1
| ψ j, k | p ≤ C| T 2j
| T 2j
is well defined and continuous. If Af = 0, then for any j ∈ N 0 we have P 2j
In the same way, one can easily show that A is surjective. Therefore the operator A is a continuous linear bijection. By the open mapping theorem, A is an isomorphism. Thus the system {T 0 , T 1 , (ψ j, k ) ∞, 2 j=0, k=0j
Remark. Since {T 0 , T 1 , (ψ j, k ) ∞, 2 j=0, k=0j
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