hw1solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online