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

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  • 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)

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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

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