# s5color - ECE 580 / Math 587 SPRING 2011 Correspondence #...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 Hahn-Banach 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 Hahn-Banach 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 ) = −...
View Full Document

## 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.

### Page1 / 6

s5color - ECE 580 / Math 587 SPRING 2011 Correspondence #...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online