In this problem we show that pointwise convergence of a sequence Sn( x) does not imply

Chapter 11, Problem 4

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In this problem we show that pointwise convergence of a sequence Sn( x) does not imply mean convergence, and conversely. a. Let Sn( x) = n xenx2/2, 0 x 1. Show that Sn( x) 0 as nfor each x in 0 x 1. Show also that Rn = _ 1 0 _0 Sn( x)_2dx = n 2 _1 en_ and hence Rn as n . Thus pointwise convergence does not imply mean convergence. b. Let Sn( x) = xn for 0 x 1, and let f ( x) = 0 for 0 x 1. Show that Rn = _ 1 0 ( f ( x) Sn( x))2dx = 1 2n + 1 , and hence Sn( x) converges to f ( x) in the mean. Also show that Sn( x) does not converge to f ( x) pointwise throughout 0 x 1. Thus mean convergence does not imply pointwise convergence.