s2color - ECE 580 / Math 587 SPRING 2011 Correspondence # 5...

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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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s2color - ECE 580 / Math 587 SPRING 2011 Correspondence # 5...

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