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# hw1solutions - Math 318 HW#1 Solutions 1 Exercise 6.2.3...

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Math 318 HW #1 Solutions 1. Exercise 6.2.3. Consider the sequence of functions h n ( x ) = x 1 + x n over the domain [0 , ). (a) Find the pointwise limit of ( h n ) on [0 , ). Answer. Let h be the function h ( x ) = x if 0 x < 1 1 / 2 if x = 1 0 if x > 1 . Then the claim is that h n h pointwise. To see this, let ± > 0. There are three cases to consider: 0 x < 1 : If x = 0, then h n ( x ) = 0 for all n . Otherwise, let N > ln ± ln x , which is equivalent to assuming that x N +1 < ± . Then, for any n N , we have that | h n ( x ) - h ( x ) | = ± ± ± ± x 1 + x n - x ± ± ± ± = ± ± ± ± - x n +1 1 + x n ± ± ± ± = x n +1 1 + x n < x n +1 x N +1 < ±. x = 1 : Clearly, h n (1) = 1 / 2 for all n , so | h n (1) - h (1) | = 0 < ± for any n . x > 1 : Let N > ln(1 ) ln x +1, which is equivalent to having 1 x N - 1 < ± . Then for any n N , | h n ( x ) - h ( x ) | = ± ± ± ± x 1 + x n - 0 ± ± ± ± = x 1 + x n < x x n = 1 x n - 1 1 x N - 1 < ±. In all three cases, there exists N N such that n N implies that | h n ( x ) - h ( x ) | < ±. Since the choice of ± > 0 was arbitrary, we conclude that h n h pointwise. (b) Explain how we know that the convergence cannot be uniform on [0 , ). Answer. Each h n is continuous, whereas the function h is clearly not continuous at x = 1. If the convergence were uniform, then the function h would have to be continuous by Theorem 6.2.6. Since that isn’t the case, we know the convergence cannot be uniform. (c) Choose a smaller set over which the convergence is uniform and supply an argument to show that this is indeed the case. Answer. Pick r > 1. Then I claim that h n h uniformly on [ r, ). To see this, ﬁrst note that h ( x ) = 0 on [ r, ). Now, let ± > 0 and choose N > ln(1 ) ln r + 1. Then for all n N we know that | h n ( x ) - 0 | = x 1 + x n < 1 x n - 1 1 r n - 1 1 r N - 1 < ±. Since the choice of ± > 0 was arbitrary, we conclude that h n 0 uniformly on [ r, ). 1

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2. Exercise 6.2.6. Using the Cauchy Criterion for convergent sequences of real numbers (Theorem 2.6.4), supply a proof for Theorem 6.2.5. (First, deﬁne a candidate for f ( x ), and then argue that f n f uniformly.) Proof. ( ) Suppose ( f n ) converges uniformly on A to some function f . Let ± > 0. Then, by deﬁnition of uniform convergence, there exists N N such that n N implies that | f n ( x ) - f ( x ) | < ±/ 2 for any x A . In particular, if n,m N , then | f n ( x ) - f m ( x ) | = | f n ( x ) - f ( x ) + f ( x ) - f m ( x ) | ≤ | f n ( x ) - f ( x ) | + | f ( x ) - f m ( x ) | by the triangle inequality, but then the fact that n,m N implies that the right hand side is less than ±/ 2 + ±/ 2 = ± . In other words, n,m N implies that | f n ( x ) - f m ( x ) | < ± for any x A , which was the desired conclusion. ( ) Suppose, on the other hand, that for every
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hw1solutions - Math 318 HW#1 Solutions 1 Exercise 6.2.3...

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