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MTH2140/MTH3140 Real Analysis – Assignment 2 (2023)

发布时间:2023-06-10

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MTH2140/MTH3140 Real Analysis – Assignment 2 (2023)

Due date:  12pm on Thursday 25 May

1. For each of the following limit, find its value or explain why it does not exist. In either case, justify your response (using theorems and results covered in the unit).

(a) limx 0

(b) limx 0  .

(c)  [MTH3140]   limx 0 arctan(exp( )).

2. We consider the sequence of functions fn  : [0, ∞) → R defined by fn (北) = arctan(n北) for all n ∈ N and 北 ∈ [0, ∞), (a) Prove that (fn )nN  converges pointwise on [0, ∞) towards a function f that you will determine.

(b) Let a > 0. Prove that fn  → f uniformly on [a,∞) as n → ∞ .

(c)  [MTH3140]  Does fn  → f uniformly on [0, 1]?

3. The purpose of this exercise is to analyse the following theorem.

Theorem 1  If f : [0, 1] → R is continuous, then:

For all ε > 0 there  exists δε  > 0 such that, for all x,y ∈ [0, 1],

if |x y| ≤ δε , then |f(x) − f(y)| ≤ ε .               (1)

(a) What is the difference between the statement (1) and the statement“for all x ∈ [0, 1], f is continuous at x”?

We now turn to the proof of Theorem 1, which is done by contradiction. We therefore assume that there exists ε > 0 such that, for all δ > 0, we can find x,y ∈ [0, 1] such that |x y| ≤ δ but |f(x) − f(y)| ≥ ε . Applying this statement to δ = 1/n for all n ∈ N, we construct sequences (xn )nN  and (yn )nN  in [0, 1] such that |xn − yn | ≤ 1/n and |f(xn ) − f(yn )| ≥ ε for all n N.

(b) Explain why we can find subsequences (xnk )kN  of (xn )nN  and (ynk )kN  of (yn )nN   (with the same indices nk  for both sequences!) which both converge to the same a ∈ [0, 1].

(c) Prove that f(xnk ) → f(a) and f(ynk ) → f(a) as k → ∞, and conclude the proof of Theorem 1.

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 )nN   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.

We now turn to the proof of Theorem 2. Let ε > 0.

(b)  [MTH3140.  Students in MTH2140 will assume it to be true.]  Prove that f is non-decreasing.

(c) Using Theorem 1, show that there exists points 0 = x0  < x1  < ··· < xℓ  = 1 (the number ℓ + 1 = ℓε + 1 of these points depend on ε) such that, for all i = 0, . . . ,ℓ − 1, f(xi ) ≤ f(xi+1) ≤ f(xi ) + ε .

(d) Prove that there exists N  N (depending on ε) such that, if n N then |fn (xi ) − f(xi )| ≤ ε for all i = 0, . . . ,ℓ .

(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 xi  ≤ x xi+1 . Show that f(xi ) − e fn (x) ≤ f(xi ) + 2e.

(f) Deduce that f(x) − 2e fn (x) ≤ f(x) + 2e, and thus that Theorem 2 holds.