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Unformatted text preview: ECE 580 / Math 587 SPRING 2011 Correspondence # 5 February 17, 2011 Solution Set 2 11. lim i x i = { 1 , 1 2 , 1 4 ,..., 1 2 n ,...... } p because it has infinitely many elements. Now, ( x n ,x m ) = ( n 1 i = m 1 2 ip ) 1 p m,n, m < n 0 as m , n , m < n . A Cauchy sequence . 12. a) For x , x C [0 , 1],  f ( x ) f ( x )  =  max t 1 x ( t ) max t 1 x ( t )  max t 1  x ( t ) x ( t ) = x x . [To see the inequality above, suppose that (without any loss of generality) max t x ( t ) = x ( t max ) max t x ( t ) x ( t ) t [0 , 1] . Then, max t 1  x ( t ) x ( t )   x ( t max ) x ( t max )  = x ( t max ) x ( t max ) x ( t max ) max t 1 x ( t ) =  max t x ( t ) max t x ( t )  which verifies the inequality.] . Given > 0 and x C [0 , 1], we can find a ( ) = such that x x <  f ( x ) f ( x )  < = . Since this is true at every x C [0 , 1], f is continuous everywhere . b) YES , it is uniformly continuous , because does not depend on x . 13. This space is indeed not separable. To prove this, using the given hint, consider a continuum of elements { sin t } , where is an arbitrary real number. For = , let us compute the distance between sin t and sin t : sin t sin t 2 = lim T 1 T T T [ sin t sin t ] 2 dt = 2 Consequently, there exists an uncountable set of elements such that the distance between any two of them is 2. This readily implies that there cannot exist a countable everywhere dense set in this space, as the following argument shows. Say there is such a set, with x o an arbitrarily picked element. Consider an neighborhood of x o , with say = 1 2 . Since, by hypothesis, there is only a countable number of such neighborhoods, there is at least one such neighborhood which contains two distinct elements, sin t and sin t , with = . Then, 2 = sin t sin t sin t x o + x o sin t 1 2 + 1 2 = 1 which, however, is impossible leading to a contradiction. 14. x ( t ) = T ( x )( t ) = 1 2 t 3 + sin x ( t ); 2 t 2. Note that  T ( x )( t )  1 2 max t  t 3  + max x  sin x ( t )  4 + . Let S = { x C [ 2 , 2] :  x ( t )  4+ } . Clearly T : S S , and S is complete (being a closed and bounded subset of another complete metric space, C [ 2 , 2]). Now,  T ( x )( t ) T ( y )( t )  =  sin x ( t ) sin y ( t )  2 sin x ( t ) y ( t ) 2 cos x ( t )+ y ( t ) 2  x ( t ) y ( t )  cos x ( t ) + y ( t ) 2  {z } 1 ....
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