1. Trang chủ
  2. » Luận Văn - Báo Cáo

bài tiểu luận môn functional analysis đề tài problems about bidual and reflexivity

36 0 0
Tài liệu đã được kiểm tra trùng lặp

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Tiêu đề Problems About Bidual And Reflexivity
Tác giả Bùi Phương Hiền
Người hướng dẫn TS. Nguyễn Thị Hồng
Trường học Trường Đại Học Thủ Đô Hà Nội
Chuyên ngành Functional Analysis
Thể loại Essay
Năm xuất bản 2023
Thành phố Hà Nội
Định dạng
Số trang 36
Dung lượng 3,23 MB

Cấu trúc

  • CHAPTER 1: THEORY ABOUT BIDUAL AND REFLEXIVITY (5)
    • 1.1. Bidual space (5)
    • 1.2. Reflexive space (6)
  • CHAPTER 2: PROBLEMS ABOUT BIDUAL AND REFLEXIVITY15 2.1. The James Theorem (15)
    • 2.2. Non reflexive spaces (16)
    • 2.3. The weak convergence in reflexive spaces (20)
    • 2.4. A hereditary property for reflexive spaces (22)
    • 2.5. A Banach space is reflexive if and only if its dual is reflexive (23)
    • 2.6. The annihilator for a set in a normed space (24)
    • 2.7. Tree type of properties for reflexive spaces (27)
    • 2.8. Separable linear subspaces into the dual (30)
    • 2.9. A Cantor type of theorem in reflexive spaces (31)
    • 2.10. Failure for the Cantor type of theorem in non-reflexive spaces (33)

Nội dung

A Banach space is reflexive if and only if its dual is reflexive .... Bidual space: Definition 1.1.1 If

THEORY ABOUT BIDUAL AND REFLEXIVITY

Bidual space

Definition 1.1.1 If 𝐸 is a normed space, then not only its dual space but 𝐸′ especially also the dual space (𝐸 ′ ) ′ of are of interest We call 𝐸′

𝐸 ′′ ≔(𝐸 ′ )′ the bidual (or second dual) space of E [1, p 93]

The following considerations show that can be considered as a subspace 𝐸 of : for 𝐸′′ 𝑥 ∈ 𝐸

𝐽(𝑥): 𝐸 ′ → 𝐾,𝐽( )[𝑦] 𝑥 ≔ 𝑦( ), 𝑥 is a linear form which is also continuous since

Hence 𝐽(𝑥) ∈ 𝐸′′ for all 𝑥 ∈ 𝐸 The map 𝐽: 𝐸 → 𝐸′′ defined thus is also linear From the norm formula, it follows that ǁ𝐽(𝑥)ǁ = sup {|𝑦(𝑥)|: ǁ ǁ𝑦 ≤ 1} = ǁ𝑥ǁfor all 𝑥 ∈ 𝐸 [2, p 52]

Thus, we have proved the following proposition

Proposition 1.1.2 For every normed space E, the map 𝐽:𝐸→𝐸 ′′ ,𝐽(𝑥)∶ 𝑦 ⟼ 𝑦(𝑥), is linear and isometric [2, p 52]

The map 𝐽:𝐸→𝐸 ′′ is calles the canonical imbedding of in its bidual 𝐸 𝐸′′ Moreover, by virtue of the isometry we shall identify with the subspace 𝐽 𝐸 𝐽(𝐸) of 𝐸′′ and we shall write 𝑥(𝑦) instead of 𝐽(𝑥)[𝑦] [2, p 52]

We have 𝐸′′ is a Banach space for every normed space 𝐸 Thereby we can obtain the completion of as the closure of 𝐸 𝐸in 𝐸′′ [2, p 52]

Proposition 1.1.3 For every normed space we have: 𝐸

(a) The complete 𝐸 of is a Banach space in a natural way 𝐸

(b) For every Banach space and for each 𝐹 𝐴 ∈ 𝐿(𝐸,𝐹) there exists a unique

Since E is a Banach space, its closure E in E'' is also a Banach space As E is dense in E'', E is a completion of E Consequently, the addition and scalar multiplication operations on E can be uniquely extended to E These extensions coincide with the operations induced on E by E''.

(b) If 𝐴 ∈ 𝐿(𝐸,𝐹), then, A is uniformly continuous Therefore, 𝐴 has a uniquely determined extension 𝐴 on 𝐸 The linearity of 𝐴 follows easily and the linearity of 𝐴

From ǁ𝐴𝑥ǁ ≤ ǁ𝐴ǁǁ𝑥ǁ for all 𝑥 ∈ 𝐸, we obtain by continuous extension that ǁ𝐴 𝑥ǁ ≤ ǁ𝐴ǁǁ ǁ for all 𝑥 ∈ 𝐸𝑥 and therefore ǁ𝐴 ǁ ≤ ǁ𝐴ǁ, from which it follows that ǁ𝐴 ǁ = ǁ𝐴ǁ, since trivially ǁ𝐴ǁ ≤ ǁ𝐴 ǁ

The statements in 1.1 (b) can also be formulated as: if is a normed 3 𝐸 space, 𝐸 is its completion and is a Banach space, then the restriction map 𝐹

𝑅 : 𝐿(𝐸,𝐹) → 𝐿(𝐸,𝐹), 𝑅(𝐴)∶=𝐴| 𝐸 , is an isometric isomorphism In particular, (𝐸 )′ may be identified with by means of 𝐸′ 𝑅

If the normed space is not complete then 𝐸 𝐸 ⊊ 𝐸′′, since is complete 𝐸′′

A priori it is not clear whether or not 𝐸 = 𝐸′′ when is a Banach space [2, 𝐸 p 53]

Reflexive space

Definition 1.2.1: A Banach space is said to be reflexive if the canonical 𝐸 imbedding 𝐽:𝐸→𝐸 ′′ is surjective, i.e., in case 𝐸=𝐸′′ by the canonical imbedding [2, p 53]

(a) A Banach space is reflexive if, and only if, for each 𝐸 𝑧 ∈ 𝐸′′ there exists an

(b) If 𝐸 is reflexive, then there is an isometric isomorphism between and 𝐸 𝐸" (namely, J) The converse of this statement is not true as shown by James (1951)

We shall give examples of reflexive and non-reflexive spaces in 1.2.10 Before that we shall prove some properties which arise out of the notion of reflexivity

Proposition 1.2.2 A Banach space is reflexive if and only if, its dual 𝐸 space is reflexive [2, p 53] 𝐸′

If 𝐸 is reflexive, then 𝐸=𝐸" and therefore 𝐸 ′ = (𝐸 ′′ )′ = ((𝐸 ′ )′)′ = (𝐸 ′ )′′

If 𝐸′ is reflexive, then for each 𝑦 ∈(𝐸′)′′ = (𝐸 ′′ )′ with 𝑦| 𝐸 = 0 there exists an 𝜂 ∈ 𝐸′ such that:

0 = 𝑦(𝑥) = 𝑥(𝜂) =𝜂(𝑥) for all 𝑥 ∈ 𝐸, i.e., 𝜂= 0 Consequently, also 𝑦 = 0 and thus 𝐸 𝑜 = {0} for the polar of 𝐸 ⊂

𝐸′′ in 𝐸′′′ Since 𝐸 is closed in , we obtain from that 𝐸′′ 𝐸=𝐸′′.

From 1.1.2 we obtain for every Banach space E the following increasing chain of Banach spaces: [2, p 54]

Proposition 1.2.2 says that these chains are either constant or are strictly increasing

Proposition 1.2.3 Let and be isomorphic Banach spaces Then, if is 𝐸 𝐹 𝐸 reflexive, so is [2, p 54] 𝐹

Proposition 1.2.4 If 𝐸 is a reflexive Banach space and is a closed subspace 𝐹 of then and 𝐸 𝐹 𝐸/𝐹 are likewise reflexive [2, p 54]

𝐹 is reflexive: The map, 𝜌: 𝐸 ′ ⟶ 𝐹 ′ , 𝜌(𝑌)∶= 𝑌|𝐹, is linear, continuous and surjective For a given 𝑧 ∈ 𝐹′′ then 𝑧 ∘ 𝜌 is in 𝐸′′

Since is reflexive, there exists an 𝐸 𝑥 ∈ 𝐸 such that:

For every 𝑌 ∈ 𝐹° = 𝑁(𝜌) we have 𝑌 (𝑥) = 0, i.e., 𝑥 ∈ 𝐹°°=𝐹 = 𝐹 Since is surjective, there exists, for each 𝜌 𝑦 ∈ 𝐹′, 𝑎 𝑌 ∈ 𝐸′ with 𝑌| 𝐹 𝜌(𝑌) = 𝑦

𝐸/𝐹 is reflexive: By 1.2.2, is also reflexive when 𝐸 is reflexive By what 𝐸′ has been just shown is also reflexive 𝐹°

We have (𝐸/𝐹)′ ≅ 𝐹° Thus, by 1.2.2 and 1.2.3, 𝐸/𝐹 is reflexive

A normed sequence space is a linear subspace of the space of all 𝜆 𝜔 sequences, which is endowed with a norm ǁ ⋅ ǁ and contains the sequence 𝑒 ∶ 𝑛

Clearly every normed sequence space contains the space

𝜑: = {𝑥 ∈ 𝜔 ∶ 𝑥 = (𝑥 𝑗 ) 𝑗∈ℕ , 𝑥 𝑗 = 0 for almost all 𝑗 ∈ℕ}, of all finite sequences as a subspace

The spaces 𝑙 ∞ , 𝑐 0 and are normed sequence spaces Further examples are 𝑐 the spaces

For 𝑙 1 this is easy to see; to prove that 𝑙 𝑝 , for 1 < p < ∞, is a normed space, we prove next:

𝑙𝑝 and 𝑦 ∈ 𝑙𝑞, then (𝑥𝑗𝑦𝑗)𝑗∈ℕ is in 𝑙 1 , and

For 𝑎,𝑏 > 0 let 𝐴 ∶=𝑝log𝑎,𝐵 ∶= 𝑞log𝑏 Since the exponential function is convex, we have exp (𝐴

If now 𝑥 ∈ 𝑙𝑝 and 𝑦 ∈ 𝑙𝑞 with ǁ𝑥ǁ 𝑝 = 1 = ǁ𝑦ǁ 𝑞 , are given, then we obtain that

If 𝑥 ∈ 𝑙 𝑝 and 𝑦 ∈ 𝑙 𝑞 with 𝑗=1𝑥 ≠ 0,𝑦 ≠ 0 are given, then we apply what we have shown above to 𝑥′∶= 𝑥 ǁ𝑥ǁ 𝑝 , and 𝑦 ′ ∶= 𝑦 ǁ ǁ 𝑦 𝑞 and obtain the stated inequality after multiplying by ǁ𝑥ǁ 𝑝 ǁ𝑦ǁ 𝑞

𝑞= 1 Then for all 𝑥 ∈ 𝜔 the following holds: ǁ ǁ𝑥𝑝= sup{|∑ ∞𝑥𝑗𝑦𝑗

𝑗=1 |:𝑦 ∈ 𝜑, ǁ𝑦ǁ 𝑞 ≤ 1 } (supremum formula), where the equality is meant in [0, ∞ ] [2, p 55]

If 𝑥 ∈ 𝜔 with ǁ𝑥ǁ 𝑝 < ∞ is given, then from Holder's inequality we have:

Thus the concerned supremum is at most ǁ𝑥ǁ 𝑝

On the other hand, if for 𝑥 ∈ 𝜔 , 𝑥≠ 0, the supremum in question equals 𝐶 ∈

ℝ + , then we choose 𝜆 ∈ 𝜔, such that |𝜆 𝑗 | = 1 and 𝜆 𝑗 𝑥𝑗= |𝑥 𝑗 |; for all 𝑗 ∈ℕ Then, for sufficiently large 𝑁 ∈ℕ the quantity 𝐴 ∶=(∑ 𝑁 𝑗=1 |𝑥) 𝑗 − | 𝑝

𝑞 for 1 ≤ j ≤ N and 𝑦 𝑗 = 0 for 𝑗 > 𝑁 Then, by the choice of , we get 𝐴 ǁ ǁ𝑦𝑞= (∑𝐴 𝜆 𝑞 | 𝑗 | 𝑞 |𝑥 𝑗 | 𝑝

𝑞 and the choice of we get for these 𝐴 𝑁 ∈ℕ:

From this it follows that 𝑥 ∈ 𝑙𝑝 and 𝐶 ≥ ǁ𝑥ǁ 𝑝 , which implies the result Proposition 1.2.8 For 1

Ngày đăng: 05/05/2024, 21:57

TÀI LIỆU CÙNG NGƯỜI DÙNG

TÀI LIỆU LIÊN QUAN