# s1color - ECE 580 Math 587 SPRING 2011 Correspondence 3...

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Unformatted text preview: ECE 580 / Math 587 SPRING 2011 Correspondence # 3 February 3, 2011 Solution Set 1 1. Let x ∈ M + N . Then x = m + n with m ∈ M , n ∈ N . Clearly m ∈ M ∪ N , n ∈ M ∪ N and x = m + n ∈ [ M ∪ N ]. Thus M + N ⊂ [ M ∪ N ]. Let x ∈ [ M ∪ N ]. Then x = ∑ k i =1 d i y i with y i ∈ M or y i ∈ N (or both). Thus x = m + n , where m = ∑ y i ∈ M ∪ e N d i y i ∈ M n = ∑ y i ∈ N d i y i ∈ N. Thus [ M ∪ N ] ⊂ M + N , and combined with above we have equality. 2. We must show that (i) K ⊂ co ( S ) (ii) K ⊃ co ( S ) (i) K ⊂ co ( S ) k ∈ K ⇒ k = m ∑ i =1 α i x i , where m ∑ i =1 α i = 1 , α i > ,x i ∈ S. Now x i ∈ S ⇒ x i ∈ co ( S ) . co ( S ) convex ⇒ α 1 x 1 α 1 + α 2 + α 2 x 2 α 1 + α 2 = y 1 ∈ co ( S ) . Similarly, α 1 + α 2 α 1 + α 2 + α 3 y 1 + α 3 x 3 α 1 + α 2 + α 3 = y 2 ∈ co ( S ) . But y 2 = α 1 x 1 α 1 + α 2 + α 3 + α 2 x 2 α 1 + α 2 + α 3 + α 3 x 3 α 1 + α 2 + α 3 . Continuing in this manner, and noting that ∑ n i =1 α i = 1, y n − 1 = α 1 x 1 ∑ n i =1 α i + ··· + α n x n ∑ n i =1 α i = n ∑ i =1 α i x i ∈ co ( S ) . Therefore, K ⊂ co ( S ) . (ii) co ( S ) ⊂ K ∀ x,y ∈ K ∃ α i ,x i ,α ′ i ,x ′ i : x = n ∑ i =1 α i x i and y = n ∑ i =1 α ′ i x ′ i where n ∑ i =1 α i = n ∑ i =1 α ′ i = 1 . Now, for 0 < λ < 1, λy + (1 − λ ) x = λ n ∑ i =1 α ′ i x ′ i + (1 − λ ) n ∑ i =1 α i x i where λ n ∑ i =1 α ′ i + (1 − λ ) n ∑ i =1 α i = 1 . = ⇒ λy +(1 − λ ) x ∈ K . Therefore, K is convex. Obviously S ⊂ K , and hence co ( S ) ⊂ K by definition. 3. By H¨older’s inequality, for x ∈ L p [ − 1 , 1] and y ∈ L q [ − 1 , 1], with (1 /p ) + (1 /q ) = 1, we have ∫ 1 − 1 | x ( t ) y ( t ) | dt ≤ ∥ x ∥ p ∥ y ∥ q with equality if, and only if, | x ( t ) | = ( | y ( t ) | ∥ y ∥ q ) q p ∥ x ∥ p Now, for the problem at hand, take y ( t ) = t 3 and p = 3, which makes q = 3 / 2. Furthermore, ∥ x ∥ p = 2 1 3 . Hence, from H¨older’s inequality, ∫ 1 − 1 | x ( t ) t 3 | dt ≤ 2 1 3 ∥ t 3 ∥ 3 2 = (2) 5 3 (11) − 2 3 , which says that ∫ 1 − 1 | x ( t ) t 3 | dt cannot be larger than (2) 5 3 (11) − 2 3 , and in fact this upper bound can be achieved (from the equality above) by choosing | x ( t ) | = ( | t 3 | (2) − 4 3 (11) 2 3 ) 1 2 2 1 3 = (2) − 1 3 (11) 1 3 | t | 3 2 2 Now, ∫ 1 − 1 t 3 x ( t ) dt ≤ ∫ 1 − 1 | t 3 x ( t ) | dt where we would have equality if x ( t ) is picked so that t 3 x ( t ) is equal to | t 3 x ( t ) | , that is the sign of x ( t ) should be the same as the sign of t 3 , and hence of t . Hence, the unique solution to the maximization problem is x ( t ) = ( 2 ) − 1 3 ( 11 ) 1 3 | t | 3 2 sgn ( t ) 4. Noting the obvious relationship, min ∫ 1 − 1 t 3 x ( t ) dt ≥ min ( − ∫ 1 − 1 | t 3 x ( t ) | dt ) = − max ∫ 1 − 1 | t 3 x ( t ) | dt it readily follows from the solution to Problem 3 above that the following choice of...
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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.

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s1color - ECE 580 Math 587 SPRING 2011 Correspondence 3...

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