answer In order to show that g n is not equicontinuous we have to find an so

# Answer in order to show that g n is not

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1]? (1) - f n ( y ) | = | 1 - y n | = | 1 - y | · | 1 + y | < 1 2 , then it would have to be true that | 1 - y | < 1 2 | y + 1 | or 1 < 1 2 | 1 + y | + y This is not possible, so there is no way to choose a δ so that | 1 - y | < δ , so g n is not equicontinuos. 6.2.16 Given f n : [0 , 1] R , each f n is bounded (by some M ), and ( f n ) is equicontinuous, show that f m ) contains a uniformly convergence subsequence. (a) Since Q = { r 1 , r 2 , . . . } , and f n is bounded, we can use problem 6.2.14 to find a subsequence f n k = : g n that converges to each rational number in [0 , 1]. (See part (c) of that problem). (b) We are given a set Λ = { r 1 , r 2 , . . . , r m } such that [0 , 1] = [ r Λ V δ ( r ), where | x - y | < δ and | g k ( x ) - g k ( y ) | < 3 for some positive . 4 We want to show that there is an N , so that | g s ( r i ) - g t ( r i ) | < 3 for all s, t N and 4 This is by equicontinuity 28 r i Λ. Each g k is Cauchy since it is uniformly continuous (by Theorem 6.2.5), so we can find an N for any particular r i in Λ. Because it is given that Λ is finite, we may choose the maximum of these N . If Λ was infinite, we would not be be able to choose such an N . (c) We can use the conclusion from part (b) with the fact that g n is equicontinuous to conclude: | g s ( x ) - g t ( x ) | = | g s ( x ) - g s ( r ) + g s ( r ) - g t ( r ) + g t ( r ) - g t ( x ) | | g s ( x ) - g s ( r ) | + | g s ( r ) - g t ( r ) | + | g t ( r ) - g t ( x ) | < 3 3 = Rate of convergence of lefthand Riemann Sums Suppose that f : [0 , 1] R has a continuous second derivative. Prove that Z 1 0 f ( x ) d x - 1 n n - 1 X k =0 f ( k/n ) = f (1) - f (0) 2 n + O 1 n 2 Proof. We can rewrite Z 1 0 f ( x ) d x as a sum of integrals, Z 1 n 0 f ( x ) d x + Z 2 n 1 n f ( x ) d x + · · · + Z 1 n - 1 n f ( x ) d x = n - 1 X k =0 Z k +1 n k n f ( x ) d x This is useful, becuase it allows to write the difference between Z 1 0 f ( x ) d x and the lefthand Riemann sum as a sum of integrals. To do this, consider some arbitrary term 1 n f ( k / n ) from the lefthand sums, and note that Z k +1 k k n f k n d x = f k n Z k +1 k k n d x = 1 n f k n Using this fact, we can write n - 1 X k =0 f ( k/n ) as n - 1 X k =0 Z k +1 n k n f ( k/n ) d x . Then, multiplying the difference between the integral and left hand sums by n , we see n Z 1 0 f ( x ) d x - 1 n n - 1 X k =0 f ( k/n ) = n n - 1 X k =0 Z k +1 n k n f ( x ) d x - n - 1 X k =0 Z k +1 n k n f ( k/n ) d x = n n - 1 X k =0 Z k +1 n k n f ( x ) - f ( k/n ) d x The first order Taylor approximation of some integrand about its lower limit is f ( x ) - f k n = " f k n - f k n # + 29 f 0 k n x - k n + R ( x ) = f 0 k n x - k n + 1 2 f 00 ( c k )( x - k n ) 2 Using this: n n - 1 X k =0 Z k +1 n k n f ( x ) - f ( k/n ) d x = n n - 1 X k =0 Z k +1 n k n f 0 k n x - k n d x + n n - 1 X k =0 Z k +1 n k n 1 2 f 00 ( c k )( x - k n ) 2 d x = n n - 1 X k =0 f 0 k n Z 1 n 0 u d u + n n - 1 X k =0 Z k +1 n k n 1 2 f 00 ( c k )( x - k n ) 2 d x = n n - 1 X k =0 f 0 k n 1 2 n 2 + n n - 1 X k =0 Z k +1 n k n 1 2 f 00 ( c k )( x - k n ) 2 d x = 1 2 n - 1 X k =0 f 0 k n 1 n + n n - 1 X k =0 Z k +1 n k n 1 2 f 00 ( c k )( x - k n ) 2 d x = 1 2 Z 1 0 f 0 ( x ) d x + #### You've reached the end of your free preview.

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• Fall '15
• Topology, yn, Metric space, Xn, lim F
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