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

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

TRƯỜNG ĐẠI H C TH ỌỦ ĐÔ HÀ NỘI KHOA SƯ PHẠM

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BÀI TI U LU N MÔN: FUNCTIONAL ANALYSIS ỂẬĐỀ TÀI: PROBLEMS ABOUT BIDUAL AND

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

CONCLUSION 35

BIBLIOGRAPHY 36

3 INTRODUCTION 1 Rationale

Although the development of Mathematics has ups and downs at each historical moment, the most brilliant results it achieved were in the twentieth century, which was the development of Mathematical Analysis

With the advent of Mathematical Analysis, especially Functional Analysis, many problems in life, physics and science and technology are solved quickly and accurately The methods and typical results of Functional Analysis have been introduced into all related branches of mathematics On the one hand, that penetration has opened up vast horizons for the field of Functional Analysis, which has the task of synthesizing the results of separate branches of Mathematics to, to some extent, create general and abstract mathematical models Functional analysis is a subject in the program However, because class time is limited, it is difficult to research in depth Through this thesis, I do not dare to have the ambition to learn deeply about Functional Analysis, but only wish to research and understand more deeply about a problem or a space of Functional Analysis That's why I chose the topic "Bidual and Reflexivity" to have the opportunity to learn more deeply about this space with many applications

Function theory and functional analysis 5 Structure of the essay

4 The content of the topic includes 2 chapters: Chapter 1: Theory about Bidual and Reflexivity Chapter 2: Problems about Bidual and Reflexivity

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CHAPTER 1: THEORY ABOUT BIDUAL AND REFLEXIVITY 1.1 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] ≔ 𝑦( ), 𝑥 𝑥 is a linear form which is also continuous since

|𝐽(𝑥)[ ]|𝑦 = 𝑦(𝑥)| ≤ ǁ ǁǁ ǁ| 𝑦 𝑥 for all 𝑦 ∈ 𝐸′.

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} = ǁ𝑥ǁ for 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 ′′ and we shall write 𝐸 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 𝐸 in 𝐸′′ [2, p 52]

Proposition 1.1.3 For every normed space we have: 𝐸

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(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 𝐴 ∈ 𝐿(𝐸 , 𝐹) such that 𝐴 |𝐸= 𝐴 and ǁ𝐴 ǁ = ǁ𝐴ǁ [2, p 52]

Proof [2, pp -52 53]

(a) The closure 𝐸 of 𝐸 in 𝐸′′ is a Banach space since 𝐸′′ is a Banach space Since 𝐸 is dense in 𝐸 we get 𝐸 is a completion of From addition (in 𝐸 𝐸) has a uniquely determined continuous extension to the completion This coincides therefore with the addition induced on 𝐸 by A similar argument can be 𝐸′′ made for scalar multiplication and the norm

(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 ∈ 𝐸 and therefore ǁ𝐴 ǁ ≤ ǁ𝐴ǁ, from which it 𝑥 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]

1.2 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]

7 Remark: [2, p 53]

(a) A Banach space is reflexive if, and only if, for each 𝐸 𝑧 ∈ 𝐸′′ there exists an 𝑥 ∈ 𝐸 such that 𝑧(𝑦) = 𝑦(𝑥) for all 𝑦 ∈ 𝐸′

(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′

Proof [2, p 53]

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:

𝑦(𝜉) = 𝜉(𝜂) for all 𝜉 ∈ 𝐸′′ Since 𝑦|𝐸 = 0 we have

0 = 𝑦(𝑥) = 𝑥(𝜂) = 𝜂(𝑥) for all 𝑥 ∈ 𝐸,

i.e., 𝜂 = 0 Consequently, also 𝑦 = 0 and thus 𝐸𝑜= {0} for the polar of 𝐸 ⊂ 𝐸′′ in ′′′ Since 𝐸 is closed in , we obtain from that 𝐸 𝐸′′ 𝐸 = 𝐸′′.

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

𝐸 ⊂ 𝐸 ⊂ 𝐸′′ (4)⊂ 𝐸(6)⊂ … 𝐸′ ⊂ 𝐸 ⊂ 𝐸′′′ (5)⊂ 𝐸(7)⊂ …

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]

Proof [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: (*) 𝑧 ∘ 𝜌(𝑌) = 𝑌(𝑥) for all 𝑌 ∈ 𝐸′

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

Then, by (*), we have

𝑧(𝑦) = 𝑧 ∘ 𝜌(𝑌) = 𝑌(𝑥) = 𝑦(𝑥) for all 𝑦 ∈ 𝐹′ Thus is reflexive 𝐹

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 Definition 1.2.5 (normed sequence space ): [2, p 54]

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 = (𝛿𝑗, 𝑛)𝑗∈ℕ for each 𝑛 ∈ ℕ

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Clearly every normed sequence space contains the space 𝜑: = {𝑥 ∈ 𝜔 ∶ 𝑥 = (𝑥𝑗)𝑗∈ℕ, 𝑥𝑗= 0 for almost all 𝑗 ∈ ℕ}, of all finite sequences as a subspace

The spaces 𝑙∞, 𝑐0 and are normed sequence spaces Further examples are 𝑐

If 𝑥 ∈ 𝑙𝑝 and 𝑦 ∈ 𝑙𝑞 with 𝑥 ≠ 0, 𝑦 ≠ 0 are given, then we apply what we have shown above to 𝑥′ ∶=𝑥

ǁ𝑥ǁ𝑝 , and 𝑦′∶= 𝑦

ǁ ǁ𝑦𝑞 and obtain the stated inequality after multiplying by ǁ𝑥ǁ𝑝ǁ𝑦ǁ𝑞

Lemma 1.2.7: Let 𝑝, 𝑞 ∈ ]1, ∞[ with 𝑝1+1

𝑞= 1 Then for all 𝑥 ∈ 𝜔 the

𝑗=1 for all 𝑦 ∈ 𝜑 with ǁ𝑦ǁ𝑞≤ 1 Thus the concerned supremum is at most ǁ𝑥ǁ𝑝

On the other hand, if for 𝑥 ∈ 𝜔 , 𝑥 ≠ 0, the supremum in question equals 𝐶 ∈ ℝ+, then we choose 𝜆 ∈ 𝜔, such that |𝜆𝑗| = 1 and 𝜆𝑗𝑥𝑗= |𝑥𝑗|; for all 𝑗 ∈ ℕ Then, for sufficiently large 𝑁 ∈ ℕ the quantity 𝐴 ∶=(∑𝑁 |𝑥𝑗|𝑝

𝑞 for 1 ≤ j ≤ N and 𝑦𝑗= 0 for 𝑗 > 𝑁 Then, by the choice of , we get 𝐴

From this it follows that 𝑥 ∈ 𝑙𝑝 and 𝐶 ≥ ǁ𝑥ǁ𝑝, which implies the result Proposition 1.2.8 For 1 < 𝑝 < ∞, 𝑙𝑝 is a normed sequence space [2, p 56] Proof [2, p 56]

For 𝑞 ∶=𝑝

𝑝−1 we have 𝑝1+1

𝑞= 1 Then, by 1.2.7, we have for arbitrary 𝑥, 𝑦 ∈ 𝑙𝑝 and all 𝑧 ∈ 𝜑 with ǁ𝑧ǁ𝑞≤ 1:

Thus 𝑙𝑝 is a linear subspace of 𝜔 on which ǁ ⋅ ǁ𝑝 is a norm Obviously also 𝑙𝑝 contains the vectors 𝑒𝑛 for all 𝑛 ∈ ℕ

Notation: [2, p 56]

If 𝜆 and are normed sequence spaces, then we write 𝜇 𝜆 = 𝜇′ if: (1) For every 𝑦 ∈ 𝜇 and every 𝑥 ∈ 𝜆 the series ∑∞ 𝑥𝑗𝑦𝑗

𝑗=1 =: 𝑦(𝑥) converges and defines an element 𝑦(⋅) of for which 𝜆′ ǁ𝑦(⋅) ǁ𝜆′= ǁ𝑦ǁ𝜇 holds

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(2) For every 𝜂 ∈ 𝜆′ there exists a 𝑦 ∈ 𝜇 such that 𝑦(⋅) = 𝜂

When we write 𝜇 ∈ 𝜆′ it also signifies that the map 𝑦 ⟼ 𝑦(⋅) is an isometric isomorphism between and 𝜇 𝜆′

Proposition 1.2.9 𝑙′ = 𝑙𝑝 𝑞, for 𝑝, 𝑞 ∈ 1, ∞] [ with 1𝑝+1

𝑗=1 ; as well as the inequality |𝑦(𝑥)| ≤ ǁ𝑥ǁ𝑝ǁ𝑦ǁ𝑞, i.e., ǁ ǁ𝑦𝑙𝑝 ′ ≤ ǁ𝑦ǁ𝑞 follow from Holder's inequality Obviously 𝑦(⋅) = 𝜂, which implies ǁ𝑦(⋅) ǁ𝑙𝑝′= ǁ𝑦ǁ𝑞

𝑐0′ = 𝑙1: For 𝑦 ∈ 𝑙1 and 𝑥 ∈ 𝑐0 we have Thus the series 𝑦(𝑥) ∶= ∑∞ 𝑥𝑗𝑦𝑗

𝑗=1 converges and defines 𝑎 𝑦(⋅) ∈ 𝑐0 for which ǁ𝑦(⋅) ǁ𝑐

0′= ǁ𝑦ǁ𝑙1

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If 𝜂 ∈ 𝑐′0 is given, then we define 𝑦 ∈ 𝜔 by 𝑦𝑗∶= 𝜂(𝑒𝑗), 𝑗 ∈ ℕ

One can then give a proof similar to the one given above to show that 𝜂(𝑥) = ∑∞ 𝑥𝑗𝑦𝑗

𝑗=1 = 𝑦(𝑥) for all 𝑥 ∈ 𝑐0, 𝑦 ∈ 𝑙1, as also, ǁ ǁ𝑦𝑙1= 𝑦(⋅) ǁ ǁ𝑐0′ 𝑙1′= 𝑙∞: The proof is analogous

Corollary 1.2.10 For 1 < 𝑝 < ∞, 𝑙𝑝 is a reflexive Banach space The Banach spaces 𝑐, 𝑙1 𝑎𝑛𝑑 𝑙∞ are not reflexive [2, p 57]

Proof [2, pp 57-58]

By 1.2.9, for 1 ≤ 𝑝 ≤ ∞, the spaces are dual spaces, and therefore, 𝑙𝑝 complete Thus, 𝑐0 and are complete 𝑐

If 𝑝, 𝑞 ∈ ]1, ∞[ with 1𝑝 +1

𝑞 = 1, then, by 1.2.9, 𝑙′ = 𝑙𝑝 𝑞 and 𝑙𝑞′ = 𝑙𝑝 In this sense we also have 𝑙′′= 𝑙𝑝𝑝 That this actually signifies the reflexivity of can be shown by the following: 𝑙𝑝

For 𝑥 ∈ 𝑙𝑝 and 𝑦 ∈ 𝑙′𝑝= 𝑙𝑞 we have

𝑐0 is not reflexive: If by means of 1.2.9 we identify 𝑐0′′ with , then the 𝑙∞ canonical imbedding is precisely the inclusion of 𝐽 𝑐0 in 𝑙∞,, and therefore is not surjective

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Since is a closed subspace of and of , by 1.2.4 and the previous result, 𝑐0 𝑐 𝑙∞ these spaces cannot be reflexive From 1.2.2 and 1.2.9, 𝑙 = 𝑐1 0′ is also not reflexive

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CHAPTER 2: PROBLEMS ABOUT BIDUAL AND REFLEXIVITY 2.1 The James Theorem

Using a consequence of the Hahn-Banach theorem we obtain that there is 𝑥∗∗ ∈ 𝑋∗∗ such that ǁ𝑥∗∗ǁ = 1 and 𝑥∗∗(𝑥∗) = ǁ𝑥∗ǁ

But 𝑋 is reflexive and therefore the canonical embedding 𝐾𝑋: 𝑋 → 𝑋∗∗, 𝐾𝑥(𝑥) = 𝑥 is surjective, and then there is 𝑥 ∈ 𝑋 such that

16 (ii) → (i)

Assume that any linear and continuous functional on X achieves its norm on the closed unit ball of X

Consider the canonical embedding 𝐾𝑋: 𝑋 → 𝑋∗∗, 𝐾𝑥(𝑥) = 𝑥 and 𝑥∗∈ 𝑋∗ and consider the linear functional 𝑥∗: 𝑋 → 𝕂

By the assumption, 𝑥∗∈ 𝑋∗ attains its norm at on the closed unit ball; i.e., there exists 𝑥 ∈ 𝑋 with ∥ 𝑥 ∥= 1 such that 𝑥∗(𝑥) = ǁ𝑥∗ǁ

𝑥∗∈ 𝑋∗, 𝑥∗ ≠ 0 Using a consequence of the Hahn-Banach theorem we obtain that there is 𝑥∗∗ ∈ 𝑋∗∗ such that ǁ𝑥∗∗ǁ = 1 and 𝑥∗∗(𝑥∗) = ǁ𝑥∗ǁ 1 Prove is not reflexive space: 𝒄𝟎

Consider the linear functional 𝑥∗: 𝑐0→ ℝ, 𝑥∗(𝑥) = ∑ 𝑥𝑛

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Suppose 𝑥 achieve its norm on the closed unit ball of , i.e., there is 𝑐0 𝑥 =

Thus, does not achieve its norm on the closed unit ball of 𝑥∗ 𝑐0 Therefore, by James Theorem, 𝑐0 is not reflexive

2 The space 𝒍𝟏 is not reflexive

Consider the linear functional 𝑥∗: 𝑙1→ ℝ, 𝑥∗(𝑥) = ∑ (1 −1

Thus, does not achieve its norm on the closed unit ball of 𝑥∗ 𝑙1 Therefore, by James Theorem, is not reflexive 𝑙1

3 𝑪[𝒂, 𝒃] is not reflexive space:

Consider the linear functional 𝑥∗: 𝐶 𝑎, 𝑏[ ] → ℝ, 𝑥∗(𝑓) = ∫ 𝑓(𝑥) 𝑑𝑥

Suppose achieve its norm on the closed unit ball of 𝑥∗ 𝐶[𝑎, 𝑏], i.e., there is 𝑓 ∈ 𝐶[𝑎, 𝑏] with ǁ𝑓ǁ≤ 1 such that 𝑥∗(𝑓) = ǁ𝑥∗ǁ= b − a, i.e., 𝑓 is continuous,

Thus, does not achieve its norm on the closed unit ball of 𝑥∗ 𝐶[𝑎, 𝑏] Therefore, by James Theorem, 𝐶[𝑎, 𝑏] is not reflexive

iii) Give a counterexample to prove that the assertion from (ii) is no longer true if 𝑋 is not supposed to be reflexive

Solution [3, pp 79-80]

i) We know that if 𝐴 ⊆ 𝑋 is a convex set then 𝐴 𝑤𝑒𝑎𝑘 = 𝐴 ǁ⋅ǁ (the Mazur theorem )

Suppose that 𝑥𝑛→ 𝑥0 weak For any 𝑚 ∈ ℕ, (𝑥𝑛)𝑛≥𝑚 converges weak to 𝑥0 Since 𝑥𝑛∈ 𝐾𝑚 ∀𝑛 ≥ 𝑚, we obtain that 𝑥0 ∈ 𝐾𝑚𝑤𝑒𝑎𝑘 = 𝐾𝑚ǁ⋅ǁ= 𝐾𝑚

⇒ 𝑥 ∈ 𝐾0 𝑚 ∀𝑚 ∈ ℕ i.e., {𝑥0} ⊆ ⋂∞ 𝐾𝑛

Let 𝑦 ∈ ⋂∞𝑛=1 𝐾, and consider 𝑛 𝜀 > 0 and an element 𝑥∗ ∈ 𝑋∗

Since 𝑥𝑛 → 𝑥0 weak there is an 𝑚 ∈ ℕ such that |𝑥∗ (𝑥𝑛 − 𝑥0)| ≤ 𝜀, for any 𝑛 ≥ 𝑚

Let then 𝐴 = {𝑥 ∈ 𝑋 | |𝑥 (𝑥 − 𝑥 )| ≤ 𝜀}∗ We obtain that is a closed

convex set, 𝑥𝑛 ∈ 𝐴 ∀𝑛 ≥ 𝑚, and therefore 𝐾𝑚⊆ 𝐴 Since 𝑦 ∈ 𝐾𝑚 we obtain that 𝑦 ∈ 𝐴,

⇒ |𝑥∗(𝑦 − 𝑥 )| ≤ 𝜀0

Since 𝜀 > 0 was arbitrary, passing to the limit for 𝜀 → 0 we obtain that 𝑥∗(𝑦 − 𝑥 ) = 00 ∀𝑥 ∈ 𝑋∗∗, and therefore 𝑦 = 𝑥0

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ii) We have the general remark: If 𝑋 is a linear topological space and (𝑥𝑛)𝑛∈ℕ⊆ 𝑋, 𝑥 ∈ 𝑋0 are such that for any subsequence (𝑘𝑛)𝑛∈ℕof , we can find a 𝑁 subsequence (𝑝𝑛)𝑛∈ℕ such that 𝑥𝑘𝑝𝑛→ 𝑥0, then 𝑥𝑛→ 𝑥0

Indeed, if we suppose that does not converge towards then there is a 𝑥𝑛 𝑥0 neighborhood 𝑉 of 0 such that ∀𝑛 ∈ ℕ ∃𝑘 ≥ 𝑛 such that 𝑥𝑘− 𝑥 ∉ 𝑉0 Then in a standard way we can construct a subsequence (𝑘𝑛)𝑛∈ℕ of such 𝑁 that for any 𝑛, 𝑥𝑘𝑛− 𝑥𝑜∉ 𝑉

By hypothesis there is a subsequence (𝑝𝑛)𝑛∈ℕ such that 𝑥𝑘

𝑝𝑛 → 𝑥0, hence there is an 𝑚 ∈ ℕ such that 𝑥𝑘𝑝𝑚− 𝑥0∈ 𝑉

This contradicts our choice for the subsequence (𝑘𝑛)𝑛∈ℕ

In order to prove that 𝑥𝑛 → 𝑥0 weak we will use the above remark Consider a subsequence (𝑘𝑛)𝑛∈ℕ of 𝑁

Since the sequence (𝑥𝑘𝑛)𝑛∈ℕ⊆ 𝑋 is bounded and 𝑋 is a reflexive space, using the Eberlein-Smulian theorem we obtain that: weak Now we apply the above remark

iii) Let (𝑒𝑛)𝑛∈ℕ⊆ 𝑙1 be the standard basis

22

Then the set 𝑐𝑜{𝑒𝑛, 𝑒𝑛+1, … } is included in the set of elements from 𝑙1, which have 0 at the first 𝑛 − 1 positions, and therefore 𝐾𝑛 is included in the set of elements from 𝑙1 , which have 0 at the first 𝑛 − 1 positions

Then then ⋂𝑛∈ℕ𝐾𝑛= {0} and but if 𝑥∗: 𝑙

1→ 𝕂, 𝑥∗(𝑥1, 𝑥2, …) = ∑∞ 𝑥𝑛

i.e.,𝑥∗= (1,1, … ∈ 𝑙) 1∗= 𝑙∞, ⇒ 𝑥∗(𝑒𝑛) = 1 ∀𝑛 ∈ ℕ

Therefore (𝑒𝑛)𝑛∈ℕ does not converge weak towards 0 2.4 A hereditary property for reflexive spaces

Problem 2.4.1 [3, p 70]

Let be a reflexive Banach space and 𝑋 𝑌 ⊆ 𝑋 a closed linear subspace Prove that 𝑌 is a reflexive Banach space

Solution [3, p 80]

Let 𝑦∗∗∶ 𝑌∗→ 𝕂 be a linear and continuous functional

For any 𝑥∗: 𝑋 → 𝕂 linear and continuous, we consider its restriction on 𝑌, 𝑥∗|𝑌: 𝑌 → 𝕂 Then 𝑥∗|𝑌 is linear and

ǁ𝑥∗|𝑌 𝑌ǁ∗ = sup

𝑦∈𝑌,ǁ𝑦ǁ≤1|𝑥∗(𝑦)| ≤ sup

𝑥∈𝑋,ǁ𝑥ǁ≤1|𝑥∗(𝑥)| = ǁ𝑥∗ǁ, i.e., 𝑥∗|𝑌∈ 𝑌∗, ǁ𝑥∗|𝑌ǁ𝑌∗ ≤ ǁ𝑥∗ǁ

We define 𝑦∗∗∶ 𝑋∗→ 𝕂 , 𝑦∗∗(𝑥∗) = 𝑦∗∗(𝑥∗|𝑌) ∀𝑥∗∈ 𝑋∗ Then 𝑦∗∗ is well defined and linear We have

|𝑦∗∗(𝑥∗)| = |𝑦∗∗(𝑥∗|𝑌)| ≤ ǁ𝑦∗∗ǁǁ𝑥∗|𝑌ǁ𝑌∗≤ ǁ𝑦∗∗ǁǁ𝑥∗ǁ, i.e., 𝑦∗∗ ∈ 𝑋∗∗ Since is a reflexive space, there is 𝑋 𝑥 ∈ 𝑋 such that 𝑥 = 𝑦, i.e., 𝑦∗∗∗∗(𝑥∗) = 𝑥∗(𝑥) ∀𝑥∗∈ 𝑋∗, and therefore 𝑦∗∗(𝑥∗|𝑌) = 𝑥∗(𝑥)∀𝑥 ∈ 𝑋∗∗