08.10.2016 Views

Foundations of Data Science

2dLYwbK

2dLYwbK

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

11.7 Sufficient Conditions for the Wavelets to be Orthogonal<br />

Section 11.6 gave necessary conditions on the b k and c k in the definitions <strong>of</strong> the scale<br />

function and wavelets for certain orthogonality properties. In this section we show that<br />

these conditions are also sufficient for certain orthogonality conditions. One would like a<br />

wavelet system to satisfy certain conditions.<br />

1. Wavelets, ψ j (2 j x − k), at all scales and shifts to be orthogonal to the scale function<br />

φ(x).<br />

2. All wavelets to be orthogonal. That is<br />

∫ ∞<br />

−∞<br />

ψ j (2 j x − k)ψ l (2 l x − m)dx = δ(j − l)δ(k − m)<br />

3. φ(x) and ψ jk , j ≤ l and all k, to span V l , the space spanned by φ(2 l x − k) for all k.<br />

These items are proved in the following lemmas. The first lemma gives sufficient conditions<br />

on the wavelet coefficients b k in the definition<br />

ψ(x) = ∑ k<br />

b k ψ(2x − k)<br />

for the mother wavelet so that the wavelets will be orthogonal to the scale function. That<br />

is, if the wavelet coefficients equal the scale coefficients in reverse order with alternating<br />

negative signs, then the wavelets will be orthogonal to the scale function.<br />

Lemma 11.7 If b k = (−1) k c d−1−k , then ∫ ∞<br />

−∞ φ(x)ψ(2j x − l)dx = 0 for all j and l.<br />

Pro<strong>of</strong>: Assume that b k = (−1) k c d−1−k . We first show that φ(x) and ψ(x − k) are orthogonal<br />

for all values <strong>of</strong> k. Then we modify the pro<strong>of</strong> to show that φ(x) and ψ(2 j x − k) are<br />

orthogonal for all j and k.<br />

Assume b k = (−1) k c d−1−k . Then<br />

∫ ∞<br />

−∞<br />

φ(x)ψ(x − k) =<br />

∫ ∞<br />

∑d−1<br />

∑d−1<br />

c i φ(2x − i) b j φ(2x − 2k − j)dx<br />

−∞ i=0<br />

j=0<br />

∑d−1<br />

∑d−1<br />

∫ ∞<br />

= c i (−1) j c d−1−j φ(2x − i)φ(2x − 2k − j)dx<br />

i=0<br />

j=0<br />

−∞<br />

∑d−1<br />

∑d−1<br />

= (−1) j c i c d−1−j δ(i − 2k − j)<br />

i=0<br />

j=0<br />

∑d−1<br />

= (−1) j c 2k+j c d−1−j<br />

j=0<br />

= c 2k c d−1 − c 2k+1 c d−2 + · · · + c d−2 c 2k−1 − c d−1 c 2k<br />

= 0<br />

367

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!