# 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 ,...... } ̸∈ ℓ ′ p because it has infinitely many elements. Now, ρ ( 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 ϵ > 0 and ˜ x ∈ C [0 , 1], we can find a δ ( ϵ ) = ϵ such that ∥ x − ˜ x ∥ < δ ⇒ | f ( x ) − f (˜ x ) | < δ = ϵ . Since this is true at every ˜ x ∈ C [0 , 1], f is continuous everywhere . b) YES , it is uniformly continuous , because δ 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 αt } , where α is an arbitrary real number. For α ̸ = β , let us compute the distance between sin αt and sin βt : ∥ sin αt − sin βt ∥ 2 = lim T →∞ 1 T ∫ T − T [ sin αt − sin βt ] 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 ϵ-neighborhood of x o , with say ϵ = 1 2 . Since, by hypothesis, there is only a countable number of such ϵ-neighborhoods, there is at least one such ϵ-neighborhood which contains two distinct elements, sin αt and sin βt , with α ̸ = β . Then, √ 2 = ∥ sin αt − sin βt ∥ ≤ ∥ sin αt − x o ∥ + ∥ x o − sin βt ∥ ≤ 1 2 + 1 2 = 1 which, however, is impossible – leading to a contradiction. 14. x ( t ) = T ( x )( t ) = 1 2 t 3 + α sin πx ( t ); − 2 ≤ t ≤ 2. Note that | T ( x )( t ) | ≤ 1 2 max t | t 3 | + α max x | sin πx ( t ) | ≤ 4 + α. Let S = { x ∈ C [ − 2 , 2] : | x ( t ) | ≤ 4+ α } . 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 ) | = α | sin πx ( t ) − sin πy ( t ) | ≤ 2 α sin πx ( t ) − πy ( t ) 2 · cos πx ( t )+ πy ( t ) 2 ≤ απ | x ( t ) − y ( t ) | cos πx ( t ) + πy ( t ) 2 | {z } ≤ 1 ....
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s2color - ECE 580 Math 587 SPRING 2011 Correspondence 5...

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