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Unformatted text preview: Ma 450: Mathematics for Multimedia Solution: to Homework Assignment 5 Prof. Wickerhauser Due Friday, April 16th, 2010 1. Draw the graphs of w ( t 3 4) and w ( t 4 3 ) on one set of axes for the Haar function w ( t ) defined in Equation 5.2. Solution: The graphs are shown in Figure 1. 2 2. For each ( a, b ) ∈ Aff , define the linear operator ζ ( a, b ) : L 2 ( R ) → L 2 ( R ) by ζ ( a, b ) f ( t ) def = √ a f ( at + b ) . Is ζ a representation of the group Aff ? If so, is it a faithful representation? Is it unitary? Solution: Show that ζ fails to be a representation because it is not multiplicative: ζ ( a , b ) ζ ( a, b ) f ( t ) = √ a [ ζ ( a, b ) f ]( a t + b ) = √ aa f ( a [ a t + b ] + b ) = √ aa f ( aa t + ab + b ) = ζ ( aa , ab + b ) f ( t ) = ζ (( a, b )( a , b )) f ( t ) . Thus ζ ( a , b ) ζ ( a, b ) = ζ (( a, b )( a , b )) when it should be ζ ( a , b ) ζ ( a, b ) = ζ (( a , b )( a, b )). But Aff is noncommutative: let ( a , b ) = (4 , 0) and ( a, b ) = (1 , 1) to see that ( a , b )( a, b ) = (4 , 4) while ( a, b )( a , b ) = (4 , 1). Choosing f = 1 = 1 [0 , 1) shows that ζ (4 , 4) f ( t ) = 2 × 1 (4 t +4) while ζ (4 , 1) f ( t ) = 2 × 1 (4 t + 1). These functions differ at all . 25 < t < 0, so ζ (4 , 4) 6 = ζ (4 , 1). 2 w ([ t –4]/3) w ( t /3–4) 12 1 t 15 4 7 Figure 1: Graphs of w ( t 3 4) and w ( t 4 3 ) for the Haar function w . 1 3. Let f = f ( a ) = f ( a, b ) be the function on Aff defined by f ( a ) = 1 D ( a ), where 1 D is the indicator function of the region D = { a = ( a, b ) : A < a < A , B < b < B } ⊂ Aff for < A < A and∞ < B < B < ∞ . Evaluate R Aff f ( a ) d a using the normalized left invariant integral on Aff . Solution: Use the integral defined in Equation 5.19: Z Aff f ( a ) d a def = Z ∞ b =∞ Z ∞ a =0 f ( a, b ) dadb a 2 = Z B b = B Z A a = A 1 a 2 dadb = ( B B ) Z A a = A 1 a 2 da = ( B B ) 1 A 1 A . 2 4. Let w = w ( t ) be the Haar mother function and define φ J M,K ( t ) def = M + J X j = M +1 1 2 j w t K 2 j for arbitrary fixed K ∈ R and M, J ∈ Z with J > 0. Prove that lim J →∞ φ J M,K ( t ) = 2 M 1 [ K,K +2 M ) ( t ) def = φ M,K ( t ) , and also that D φ J M,K , u E → h φ M,K , u i as J → ∞ for any function u ∈ L 2 ( R ). Solution: Using φ J as defined in Equation 5.4, evaluate φ J M,K ( t ) = M + J X j = M +1 1 2 j w t K 2 j = J X j =1 1 2 j + M w t K 2 j + M = 1 2 M J X j =1 1 2 j w 1 2 j t K 2 M = 1 2 M φ J ( t K 2 M ) . Thus for all t ∈ R , lim J →∞ φ J M,K ( t ) = 1 2 M lim J →∞ φ J ( t K 2 M ) = 1 2 M 1 ( t K 2 M ) = 2 M 1 [ K,K +2 M ) ( t ) , as claimed. Lemma 5.1 then gives lim J →∞ D φ J M,K , u E = 1 2 M Z ∞∞ 1 ( t K 2 M ) u ( t ) dt = 1 2 M Z K +2 M K u ( t ) dt = h φ M,K , u i ....
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This note was uploaded on 05/16/2010 for the course MATH 450 taught by Professor Wickerhauser during the Spring '10 term at Washington University in St. Louis.
 Spring '10
 Wickerhauser
 Math

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