1873_solutions

F on 0 but that f 2 fails to converge uniformly to f

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Unformatted text preview: 0, 2 for which the sequence sin a n x diverges. Given any number x, if the sequence sin a n x converges then the sequence f n x must converge to the number 0 and we know from part a that there are numbers x for which f n x does not converge to 0. 3. Suppose that for each positive integer n we have n j0 fn x  x j 1x1 if if x  1 1 2 . a. Prove that if 1 1x converges boundedly to the function f on the interval 0, 1 . fx  for 1  x 1 then the sequence f n This statement is obvious. b. Explain why each function f n is integrable on the interval 0, 1 and why Þ0 fn  n 1 j0 1j j1 and deduce that Ý 1j  log 2. j1 j0 Each function f n is bounded and is continuous at all but one number in the interval 0, 1 . Therefore each function f n is integrable on 0, 1 . We see at once that, for each n, Þ0 fn  Þ0 1 1 n n x j dx  j0 j0 1j . j1 From the bounded convergence theorem we deduce that Ý j0 1j nÝ lim j1  Þ0 1 n j0 1j nÝ lim j1 Þ0 fn 1 1 dx  log 2. 1x c. Given any number x satisfying 1  x  1, prove that the series f n converges boundedly to f on the closed interval running between 0 and x and deduce that the equation Ý j0 holds whenever 1  x 1 j x j1  log 1  x j1 1. 370 This part of the exercise is simpler than part b because the comparison test guarantees that the series 1 j t j1 j1 converges absolutely and uniformly in t on the closed interval running between 0 and x. We repeat the argument used in part b but with integration from 0 to x. 4. Suppose that for each positive integer n we have n j0 fn x  x2 j x1 if 0 if x  1 1 2 . Repeat the steps of the preceding for this sequence of functions and deduce that Ý j0 1j  . 4 2j  1 The sequence f n converges boundedly to the function f defined by the equation 1 fx  1  x2 for 0 x 1. Applying the bounded convergence theorem to this sequence of functions yields the desired result. 5. Prove that if 1 n fn x  if 0 x n if x  n 0 whenever n is a positive integer then the sequence f n converges uniformly on the interval 0, Ý to the constant function 0 even though lim nÝ Þ0 Ý Þ0 fn Ý 0. There really isn’t much to do in this exercise. All the assertions should be obvious, or nearly so. The point of the exercise is to point out that uniform convergence isn’t enough to guarantee interchangability of limits and integrals on an unbounded domain interval. 6. Suppose that   0 and that 1 fn x  x n n 1 x 1 n if x if x  0 1 n n or x  n whenever n is a positive integer. a. Prove that if x is any positive number then lim f nÝn x  e xx 1 and that for each n we have |f n x | e xx 1. Solution: A simple application of L’Hôpital’s rule can be used to show that if x  0 then x n  e x. n x n we have x n e x. 1n lim 1 nÝ We now want to show that whenever 0 Suppose that x is a positive number. We define ft  1 371 x t t whenever t x. The fact that f t e x whenever t increasing. Now whenever t x we have x will follow if we can show that the function f is x t f t...
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