Real Analysis Chapter 6 Solutions Jonathan Conder
نویسنده
چکیده
3. Since Lp and Lr are subspaces of CX , their intersection is a vector space. It is clear that ‖ · ‖ is a norm (this follows directly from the fact that ‖ · ‖p and ‖ · ‖r are norms). Let 〈fn〉n=1 be a Cauchy sequence in Lp ∩ Lr. Since ‖fm − fn‖p ≤ ‖fm − fn‖ and ‖fm − fn‖r ≤ ‖fm − fn‖ for all m,n ∈ N, it is clear that 〈fn〉n=1 is a Cauchy sequence in both Lp and Lr. Let gp ∈ Lp and gr ∈ Lr be the respective limits of this sequence. Given ε ∈ (0,∞), there exists N ∈ N such that ‖fn − gp‖p < ε(p+1)/p for all n ∈ N with n ≥ N. If n ∈ N and n ≥ N
منابع مشابه
Real Analysis Chapter 3 Solutions Jonathan Conder
2. Let E be measurable, and suppose that E is ν-null but |ν|(E) 6= 0. Then ν+(E) − ν−(E) = ν(E) = 0 and ν+(E) + ν−(E) = |ν|(E) > 0, so ν+(E) = ν−(E) > 0. Since ν+ ⊥ ν−, there exist disjoint measurable sets A and B covering X such that A is ν+-null and B is ν−-null. In particular ν+(E ∩ B) = ν+(E ∩ B) + ν+(E ∩ A) = ν+(E) > 0 but ν−(E∩B) ≤ ν−(B) = 0. This implies that ν(E∩B) > 0, which is a contr...
متن کاملReal Analysis Chapter 8 Solutions Jonathan Conder
3. (a) Note that η(0)(t) = 1 · e−1/t for all t ∈ (0,∞), where 1 is a polynomial of degree 0. Given k ∈ N, suppose that η(k−1)(t) = Pk−1(1/t)e −1/t for all t ∈ (0,∞), where Pk−1(x) is some polynomial of degree 2(k − 1). By the product rule and the chain rule, η(k)(t) = P ′ k−1(1/t)(−t)e + Pk−1(1/t)et = Pk(1/t)e for all t ∈ (0,∞), where Pk(x) := x(Pk−1(x) − P ′ k−1(x)) is a polynomial of degree 2...
متن کاملReal Analysis Chapter 2 Solutions
1. Suppose f is measurable. Then f−1({−∞}) ∈ M and f−1({∞}) ∈ M, because {−∞} and {∞} are Borel sets. If B ⊆ R is Borel then f−1(B) ∈M, and hence f−1(B) ∩ Y ∈M (since R is also Borel). Thus f is measurable on Y. Conversely, suppose that f−1({−∞}) ∈ M, f−1({∞}) ∈ M and f is measurable on Y. Let B ⊆ R be Borel. Then f−1(B) ∩ Y ∈ M, and f−1(B) = (f−1(B) ∩ Y ) ∪ (f−1(B) \ Y ). Clearly f−1(B) \ Y = ...
متن کاملReal Analysis Chapter 1 Solutions
3. (a) Let M be an infinite σ-algebra of subsets of some set X. There exists a countably infinite subcollection C ⊆M, and we may choose C to be closed under taking complements (adding in missing complements if necessary). For each x ∈ X, define Dx := ∩{C ∈ C | x ∈ C}, so that Dx ∈ M. Let x, y ∈ X and suppose y ∈ Dx. Then y ∈ C for all C ∈ C with x ∈ C. Moreover, if y ∈ C for some C ∈ C, then y ...
متن کاملReal Analysis Chapter 4 Solutions
Since X is infinite it contains two distinct points x and y. Suppose there exist disjoint open sets A and B (in the cofinite topology) such that x ∈ A and y ∈ B. Then A ⊆ Bc, which is finite so A is also finite. This is a contradiction because Ac is finite but X = A ∪Ac is infinite. Hence, the cofinite topology on X is not T2. Suppose that X is countable, and let x ∈ X. Choose a surjection q : ...
متن کامل