Description
I need the solution for three problems 3.1 3.7 and the problem in the paper
Unformatted Attachment Preview
Purchase answer to see full attachment
Explanation & Answer
UFF... I FINALLY MADE IT!
1. The union of two-dimensional intervals 𝐴 = [0,2] × [1,4] and 𝐵 = [1,3] × [0,3]
may be partitioned on finite number of disjoint intervals (not uniquely). For example
𝐴1 = [0,1] × [1,3], 𝐴2 = [0,2] × [3,4], 𝐶 = [1,2] × [1,3], 𝐵1 = [1,3] × [0,1], 𝐵2 = [2,3] × [1,3].
A2 A2
A1 C
B2
A1 C
B2
B1 B1
Here 𝐶 = 𝐴 ∩ 𝐵, 𝐴1 , 𝐴2 ⊂ 𝐴\𝐵, 𝐵1 , 𝐵2 ⊂ 𝐵\𝐴. The given function ℎ = 𝑓 + 𝑔 = 𝑎𝜑𝐴 + 𝑏𝜑𝐵
possesses constant values on 𝐴1 , 𝐴2 , 𝐶, 𝐵1 , 𝐵2 and therefore may be expressed as a sum of characteristic functions of
disjoint intervals (which is by definition a step function).
a. Namely, ℎ = 𝒂𝝋𝑨𝟏 + 𝒂𝝋𝑨𝟐 + (𝒂 + 𝒃)𝝋𝑪 + 𝒃𝝋𝑩𝟏 + 𝒃𝝋𝑩𝟐 .
b. Therefore ∫ ℎ = 𝑎 ∙ 𝑚(𝐴1 ) + 𝑎 ∙ 𝑚(𝐴2 ) + (𝑎 + 𝑏) ∙ 𝑚(𝐶) + 𝑏 ∙ 𝑚(𝐵1 ) + 𝑏 ∙ 𝑚(𝐵2 ) =
= 2𝑎 + 2𝑎 + 2(𝑎 + 𝑏) + 2𝑏 + 2𝑏 = 𝟔𝒂 + 𝟔𝒃.
3.1. Well, the plan is given and we can implement it. Assume for the beginning that 𝑓 ≥ 1.
𝜀
a. 𝑓 is bounded, say by constant 𝑀. For a given 𝜀 > 0 apply the Theorem 3.3 to 𝑀 > 0 and integrable function 𝑓 to
𝜀
find step functions ℎ and 𝑘 such that ℎ ≤ 𝑓 ≤ 𝑘 and ∫(𝑘 − ℎ) < 2(𝑀+1).
𝜀
[it is impossible to “select 𝑘 ≤ 𝑓 ≤ ℎ and ∫(ℎ − 𝑘) < 𝑁 where 𝑘 + ℎ ≤ 𝑁.”
Theorem 3.3 can work only with 𝜀 independent of 𝑘, ℎ.
Instead of this, we need to know that there are ℎ, 𝑘 such that min 𝑓 − 1 ≤ 𝑘 ≤ ℎ ≤ max 𝑓 + 1
I believe we can do this if analyze the proof of the Theorem 3.3.]
We want to apply the Theorem 3.3 to ℎ2 , 𝑘 2 , 𝑓 2 and 𝜀. Because 𝑓 is positive it is obvious that
𝑘 2 ≤ 𝑓 2 ≤ ℎ2 . Further, ∫(ℎ2 − 𝑘 2 ) = ∫(ℎ − 𝑘)(ℎ + 𝑘) ≤ 2(𝑀 + 1) ∫(ℎ − 𝑘) < 𝜀.
We need to prove yet that 𝑘 2 , ℎ2 are step functions. This is almost obvious: on an interval on which 𝑘 is a constant,
𝑘 2 is also a constant.
To finish the proof for any 𝑓, consider 𝑔 = 𝑓 + 𝑀 + 1 ≥ 1. Its square is integrable by the previous and
𝑓 2 = 2𝑓(𝑀 + 1) + (𝑀 + 1)2 − 𝑔2
is integrable as a sum/difference of integrable functions.
1
b. Yes, 𝑓𝑔 = 4 ((𝑓 + 𝑔)2 − (𝑓 − 𝑔)2 ) which is integrable by a.
1
1
3.7. 𝐽𝑛 (𝑥)~𝑥 𝑛 ∫−1(1 − 𝑡 2 )𝑛−2 cos 𝑥𝑡 𝑑𝑡, prove that 𝑥 2 𝐽𝑛′′ (𝑥) + 𝑥𝐽𝑛′ (𝑥) + (𝑥 2 − 𝑛2 )𝐽𝑛 (𝑥) = 0.
𝑢2𝑘
𝑘...
Review
Review
