1) Prove that there exists NEN (depending on e) such that, if n > N then |fn (xi) - f(x₁)| ≤ e for all i = 0,..., l.
1) Prove that there exists NEN (depending on e) such that, if n > N then |fn (xi) - f(x₁)| ≤ e for all i = 0,..., l.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Need help with part (d). Please explain each step and neatly type up. Thank you :)
![4. The purpose of this exercise is to analyse the following theorem.
Theorem 2 For all n N, let fn [0,1] → R be a continuous non-decreasing function. We assume that (fn)neN converges
pointwise to some f, and that f : [0, 1] → R is also continuous. Then fn →f uniformly on [0, 1].
(a) Find a counter-example to the theorem if we just remove the assumption that the limit f is continuous.
Solution:
We now turn to the proof of Theorem Let & > 0.
(b) [MTH3140. Students in MTH2140 will assume it to be true.] Prove that f is non-decreasing.
Solution:
(c) Using Theorem 1 show that there exists points 0 = 20 <₁ < ... < xe = 1 (the number l + 1 = lɛ +1 of these points depend
on e) such that, for all i = 0,..., l 1, f(xi) ≤ f(xi+1) ≤ f(xi) + ε.
Solution: 2
(d) Prove that there exists NEN (depending on e) such that, if n ≥ N then fn(xi) - f(x₁)| ≤ e for all i = 0,..., l.
Solution:
(e) [MTH3140. Students in MTH2140 will assume it to be true.] Let n ≥ N and x € [0,1]. Take i = 0,..., l-1 such
that x¡ ≤ x ≤ Fi+1. Show that f(xi) - € ≤ fn(x) ≤ f(xi) + 2€.
Solution:
(f) Deduce that f(x) − 2ɛ ≤ fn(x) ≤ f(x) + 2€, and thus that Theorem 2 holds.
Solution:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3cb672f7-47ed-4ee3-be4e-71db737c6150%2F7f00a5f3-c865-4e40-8bf4-c3af67a3ed7a%2Fjon9m3t_processed.png&w=3840&q=75)
Transcribed Image Text:4. The purpose of this exercise is to analyse the following theorem.
Theorem 2 For all n N, let fn [0,1] → R be a continuous non-decreasing function. We assume that (fn)neN converges
pointwise to some f, and that f : [0, 1] → R is also continuous. Then fn →f uniformly on [0, 1].
(a) Find a counter-example to the theorem if we just remove the assumption that the limit f is continuous.
Solution:
We now turn to the proof of Theorem Let & > 0.
(b) [MTH3140. Students in MTH2140 will assume it to be true.] Prove that f is non-decreasing.
Solution:
(c) Using Theorem 1 show that there exists points 0 = 20 <₁ < ... < xe = 1 (the number l + 1 = lɛ +1 of these points depend
on e) such that, for all i = 0,..., l 1, f(xi) ≤ f(xi+1) ≤ f(xi) + ε.
Solution: 2
(d) Prove that there exists NEN (depending on e) such that, if n ≥ N then fn(xi) - f(x₁)| ≤ e for all i = 0,..., l.
Solution:
(e) [MTH3140. Students in MTH2140 will assume it to be true.] Let n ≥ N and x € [0,1]. Take i = 0,..., l-1 such
that x¡ ≤ x ≤ Fi+1. Show that f(xi) - € ≤ fn(x) ≤ f(xi) + 2€.
Solution:
(f) Deduce that f(x) − 2ɛ ≤ fn(x) ≤ f(x) + 2€, and thus that Theorem 2 holds.
Solution:
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