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Unformatted text preview: ECE 580 / Math 587 SPRING 2011 Correspondence # 15 April 12, 2011 Solution Set 5 34. i) ∥ f ∥ = sup ∥ x ∥≤ 1  f ( x )  = sup x  x ( t ) ≤ 1  x (1 / 4) + 3 x (1 / 2) + 2 ∫ 1 tx ( t ) dt  = 1 + 3 + 2 ∫ 1 tdt = 5 which is attained by choosing x ( t ) ≡ 1. ii) Let v = v 1 + v 2 + v 3 , where x (1 / 4) = ∫ 1 x ( t ) dv 1 ( t ) 3 x (1 / 2) = ∫ 1 x ( t ) dv 2 ( t ) 2 ∫ 1 tx ( t ) dt = ∫ 1 x ( t ) dv 3 ( t ) . Clearly, v 1 ( t ) = { 0 0 ≤ t < 1 / 4 1 1 / 4 ≤ t ≤ 1 v 2 ( t ) = { 0 0 ≤ t < 1 / 2 3 1 / 2 ≤ t ≤ 1 dv 3 ( t ) dt = 2 t, v 3 (0) = 0 ⇒ v 3 ( t ) = t 2 . Hence, v ( t ) = t 2 , ≤ t < 1 / 4 1 + t 2 , 1 / 4 ≤ t < 1 / 2 4 + t 2 , 1 / 2 ≤ t < 1 . Since v , given above, is a monotonically nondecreasing function, its total variation is v (1) − v (0) = 5, which is clearly equal to ∥ f ∥ in part (i). 35. Let S := { s 1 = g 1 ( x ) ,s 2 = g 2 ( x ) ,...,s n = g n ( x ) : x ∈ X } which is a subset of IR n . It is in fact a subspace, because θ ∈ S , and y,z ∈ S, α,β ∈ IR ⇒ αy + βz ∈ S . Define a linear functional, F , on S by F ( s ) = f ( x ), which is clearly linear (because f and g i ’s are, and g i ( x ) = 0 ∀ i ⇒ f ( x ) = 0), and is bounded (because its domain is finite dimensional). By the HahnBanach Theorem (extension form), F can be extended from S to IR n . Since the dual of IR n is itself, this extension, say y ∗ , can be uniquely identified with a vector λ = ( λ 1 ,...,λ n ) ′ , that is < y,y ∗ > = y ∗ ( y ) = n ∑ i =1 λ i y i Also, since y ∗ agrees with F on S , < s,y ∗ > = n ∑ i =1 λ i s i = n ∑ i =1 λ i g i ( x ) = F ( s ) = f ( x ) and this establishes the desired result. 36. i) First note that ∥ x ∥ = max { 1 , 1 , 2 3 , 1 2 , ···} = 1. Furthermore, since c ∗ = ℓ 1 , f sought is in ℓ 1 . It should be aligned with x , as the given constraint can be written as f ( x ) = ∥ f ∥∥ x ∥ . A solution exists by Corollary 2 to the HahnBanach theorem, and by the alignment condition f = { a i } ∞ i =1 will have nonzero elements in the sequence only for i = 1 and 2. In view of this, we rewrite the constraint as: ∥ f ∥ = ∞ ∑ i =1  a i  =  a 1  +  a 2  = 1 ( ∗ ) Also, f ( x ) = −...
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This note was uploaded on 10/11/2011 for the course ECE 580 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
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