HW6.pdf - Introductory Real Analysis Math 327 Winter 2018...

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Introductory Real Analysis University of Washington Math 327, Winter 2018 c 2018, Dr. F. Dos Reis Homework 7 Due at the beginning of the class on Friday March 9th. Exercise 1. 1. f n ( x ) = 0 , x = 0 x, 0 < x 6 1 n x 2 , 1 n < x 2. h n ( x ) = x - 1 n 2 for x in [0 , 1]. 3. k n ( x ) = x - 1 n 2 for x in R . Find the limits of the sequences of functions. Prove that f n and h n are uniformly convergent. Key: Pointwise convergence of f n : If x = 0, then for any n N , f n (0) = 0 and lim n →∞ f n (0) = 0 = f (0). If x > 0, then as long as n < 1 x , f n ( x ) = x but if n > 1 x , then f n ( x ) = x 2 . Therefore lim n →∞ f n ( x ) = x 2 = f ( x ). Remark for x = 0 f (0) = 0 = 0 2 f n is pointwise convergent to f ( x ) = x 2 for any x [0 , ). Uniform convergence of f n : f n ( x ) - f ( x ) = 0 if x = 0 x - x 2 if 0 < x 6 1 n 0 if x > 1 n Pointwise convergence of h n : For any x in [0 , 1] lim n →∞ x - 1 n = 1 therefore lim n →∞ h n ( x ) = x 2 . The sequence of functions h n is pointwise convergent to the function x 2 . Uniform convergence of h n : | h n ( x ) - h ( x ) | = x 2 - 2 x n + 1 n 2 - x 2 = - 2 x n + 1 n 2 6 2 x n + 1 n 2 If x [0 , 1], 2 x n 6 2 n and 1 n 2 6 1 n . > 0 . N = 3 , such that n > N, x [0 , 1] , | h n ( x ) h ( x ) | 6 3 n < Therefore h n is uniformly convergent to h ( x ) = x 2 on [0 , 1].
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  • Fall '08
  • NAGY,KRISZTINA
  • Calculus, n1, Dr. F. Dos Reis, lim ⁡x

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