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Unformatted text preview: ECE 580 / Math 587 SPRING 2011 Correspondence # 17 April 21, 2011 Solution Set 6 41. (i) We are given that ( x n x o ,x ) 0 (and equivalently that ( x,x n x o ) 0) for all x X , and that x n x . Then, x n x o 2 = x n 2 + ( x n ,x o ) ( x o ,x n x o ) = x n 2 x o 2 ( x n x o ,x o ) ( x o ,x n x o ) (ii) Take x o = 0, without any loss of generality. By weak convergence, for every y X , ( x n ,y ) 0 as n , which means that given y X , we can find N > 0 such that  ( x n ,y )  < for all n > N . Clearly, also, given y 1 ,y 2 ,...,y m X and > 0 there exists N > 0 such that  ( x n ,y i )  < for all n > N , i = 1 ,...,m . Now, given the sequence { x n } , choose a subsequence { x n k } as follows: Choose x n 1 = x 1 . Choose x n 2 such that  ( x n 1 ,x n 2 )  < 1 (such an x n 2 exists from the weak convergence property above). Choose x n 3 such that  ( x n 1 ,x n 3 )  < 1 2 ,  ( x n 2 ,x n 3 )  < 1 2 (again use the weak convergence property above, with y 1 = x n 1 , y 2 = x n 2 , and = 1 2 ). Iteratively pick x n 1 ,x n 2 ,...,x n k , and choose x n k +1 such that  ( x n i ,x n k +1 )  < 1 k , i = 1 , 2 ,...,k Also, since ( x n ,x ) 0 for all x X , ( x n ,x n ) = x n 2 can be uniformly bounded, say by M 2 . Then, y m 2 = 1 m m k =1 x n k 2 ( 1 m ) 2 mM 2 + 2 m i =2 i 1 j =1 ( x n j ,x n i ) ( 1 m ) 2 ( mM 2 + 2( m 1) ) 0 as m Hence, y m converges strongly to 0. 42. (i) We want to show that a linear functional f on a normed space X can be expressed in the form f ( x ) = < x,x > , with x X , if and only if it is weakly continuous. First let f ( x ) = < x,x > , and in the definition of weak continuity given > 0 choose = and x 1 = x . Then,  < x,x 1 >  <  f ( x )  < and hence f is weakly continuous at x = and thereby everywhere (since f is linear). Note that weak continuity is a stronger notion of continuity than regular continuity, in the sense that weak continuity implies continuity in norm, but not vice versa....
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 Spring '08
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